Variational calculus on time scales
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MATHEMATICS RESEARCH DEVELOPMENTS
VARIATIONAL CALCULUS ON TIME SCALES
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MATHEMATICS RESEARCH DEVELOPMENTS
VARIATIONAL CALCULUS ON TIME SCALES
SVETLIN G. GEORGIEV
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Published by Nova Science Publishers, Inc. † New York
Contents Preface
ix
1 Elements of the Time Scale Calculus 1.1. Forward and Backward Jump Operators, Graininess Function . . . . . . . . . . . . . 1.2. Differentiation . . . . . . . . . . . . . . . . 1.3. Integration . . . . . . . . . . . . . . . . . . 1.4. The Exponential Function . . . . . . . . . . 1.5. Hyperbolic and Trigonometric Functions . . 1.6. The Multidimensional Time Scale Calculus 1.7. Line Integrals . . . . . . . . . . . . . . . . 1.8. Green’s Formula . . . . . . . . . . . . . . 1.9. Advanced Practical Problems . . . . . . . .
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1 5 12 19 29 30 66 72 75
2 Dynamic Systems on Time Scales 79 2.1. Structure of Dynamic Systems on Time Scales . . . . . . . . . . 79 2.2. Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . 115 2.3. Advanced Practical Problems . . . . . . . . . . . . . . . . . . . 127 3 Functionals 3.1. Definition for Functionals . . . . . . . . . . . 3.2. Self-Adjoint Second Order Matrix Equations 3.3. Jacobi’s Condition . . . . . . . . . . . . . . 3.4. Sturmian Theory . . . . . . . . . . . . . . .
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131 131 133 144 153
vi
Contents
4 Linear Hamiltonian Dynamic Systems 4.1. Linear Symplectic Dynamic Systems 4.2. Hamiltonian Systems . . . . . . . . 4.3. Conjoined Bases . . . . . . . . . . 4.4. Riccati Equations . . . . . . . . . . 4.5. Picone’s Identity . . . . . . . . . . 4.6. “Big” Linear Hamiltonian Systems . 4.7. Positivity of Quadratic Functionals .
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157 157 163 166 177 182 198 211
5 The First Variation 5.1. The Dubois-Reymond Lemma 5.2. The Variational Problem . . . 5.3. The Euler-Lagrange Equation . 5.4. Legendre’s Condition . . . . . 5.5. Jacobi’s Condition . . . . . . 5.6. Advanced Practical Problems .
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219 219 225 233 241 245 248
6 Higher Order Calculus of Variations 6.1. Statement of the Variational Problem . . . . . . . . . . . . . . . 6.2. Euler’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3. Advanced Practical Problems . . . . . . . . . . . . . . . . . . .
251 251 254 260
7 Double Integral Calculus of Variations 7.1. Statement of the Variational Problem 7.2. First and Second Variation . . . . . 7.3. Euler’s Condition . . . . . . . . . . 7.4. Advanced Practical Problems . . . .
261 262 263 268 272
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Noether’s Second Theorem 8.1. Invariance under Transformations . . . . . . . . . . . . . 8.2. Noether’s Second Theorem without Transformations of Time . . . . . . . . . . . . . . . . . . 8.3. Noether’s Second Theorem with Transformations of Time 8.4. Noether’s Second Theorem-Double Delta Integral Case . . . . . . . . . . . . . . . . . . . . . . . .
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275 . . . 275 . . . 279 . . . 281 . . . 285
Contents
vii
References
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Author Contact Information
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Index
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Preface This book encompasses recent developments of variational calculus on time scales. It is intended for use in the field of variational calculus and dynamic calculus on time scales. It is also suitable for graduate courses in the above fields. The book contains eight chapters. The chapters in the book are pedagogically organized. This book is specially designed for those who wish to understand variational calculus on time scales without having extensive mathematical background. The basic definitions of forward and backward jump operators are due to Hilger. In Chapter 1 are given examples of jump operators on some time scales. The graininess function, which is the distance from a point to the closed point on the right, is introduced in this chapter. They are given the definitions for delta derivative and delta integral and they are deducted some of their properties. They are introduced the exponential function and the trigonometric and hyperbolic functions. In this chapter is given an exposition of the multidimensional dynamic calculus on time scales. They are introduced line integrals on time scales and Green’s formula. The basic results in this chapter can be found in [4] and [5]. Chapter 2 introduces dynamic systems on time scales. It is considered the case of constant coefficients. Chapter 3 deals with functionals and self-adjoint second order matrix equations. It is formulated and proved Jacobi’s condition. It is introduced Sturmian theory. Chapter 4 is concerned with linear Hamiltonian dynamic systems. They are deducted some of the basic properties of the symplectic dynamic systems and Hamiltonian dynamic systems. They are introduced Riccati equations and it is proved Picconi’s identity. They are given some criterions for positive definiteness of quadratic functionals. Chapter 5 is devoted on the first and second variation. It is formulated and proved
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Svetlin G. Georgiev
an analogue of Dubois-Reymond lemma on time scales. They are deducted the Euler-Lagrange equation, Legendre’s condition and Jacobi’s condition. Chapter 6 deals with higer-order calculus of variations on some classes of time scales. The double integral calculus of variations is considered in Chapter 7. Chapter 8 deals with Noether’s second theorem without and with transformations of time. It is considered the double delta integral case of Noether’s second theorem. The aim of this book is to present a clear and well-organized treatment of the concept behind the development of mathematics and solution techniques. The text material of this book is presented in highly readable, mathematically solid format. Many practical problems are illustrated displaying a wide variety of solution techniques. The author welcomes any suggestions for the improvement of the text. Svetlin G. Georgiev Paris, France February 14, 2018
Chapter 1
Elements of the Time Scale Calculus This chapter is devoted to a brief exposition of the time scale calculus. A detailed discussion of the time scale calculus is beyond the scope of this book, for this reason the author confine to outlining a minimal set of properties needed in the further proceeding. The presentation in this chapter follows the books [4] and [5].
1.1. Forward and Backward Jump Operators, Graininess Function Definition 1.1.1. A time scale is an arbitrary nonempty closed subset of the real numbers. We will denote a time scale by the symbol T. We suppose that a time scale T has the topology that inherits from the real numbers with the standard topology. Example 1.1.2. [1, 2],
R,
N are time scales.
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Svetlin G. Georgiev
Definition 1.1.3. For t ∈ T we define the forward jump operator σ : T → T as follows σ(t) = inf{s ∈ T : s > t}. We note that σ(t) ≥ t for any t ∈ T. Definition 1.1.4. For t ∈ T we define the backward jump operator ρ : T → T by ρ(t) = sup{s ∈ T : s < t}. We note that ρ(t) ≤ t for any t ∈ T. Definition 1.1.5. We set infØ = supT,
supØ = infT.
Definition 1.1.6. For t ∈ T we have the following cases. 1. If σ(t) > t, then we say that t is right-scattered. 2. If t < supT and σ(t) = t, then we say that t is right-dense. 3. If ρ(t) < t, then we say that t is left-scattered. 4. If t > infT and ρ(t) = t, then we say that t is left-dense. 5. If t is left-scattered and right-scattered at the same time, then we say that t is isolated. 6. If t is left-dense and right-dense at the same time, then we say that t is dense. √ √ Example 1.1.7. Let T = { 2n + 1 : n ∈ N}. If t = 2n + 1 for some n ∈ N, t2 − 1 then n = and 2 √ √ √ √ σ(t) = inf{l ∈ N : 2l + 1 > 2n + 1} = 2n + 3 = t 2 + 2 for n ∈ N, ρ(t) = sup{l ∈ N :
√
2l + 1
2n+1 , l ∈ N = 2n+2 = 2t.
Hence,
µ(t) = σ(t) − t = 2t − t = t or µ 2n+1 = 2n+1 , n ∈ N. n√ o 3 Exercise 1.1.13. Let T = n + 2 : n ∈ N0 . Find µ(t), t ∈ T. Answer. µ
√ √ √ 3 n + 2 = 3 n + 3 − 3 n + 2, n ∈ N0 .
Definition 1.1.14. If f : T → R is a function, then we define the function f σ : T → R by f σ (t) = f (σ(t)) for any t ∈ T. i.e., f σ = f ◦ σ. Example 1.1.15. Let T = t = 2n+2 : n ∈ N , f (t) = t 2 + t − 1. Then n o l+2 l+2 n+2 σ(t) = inf 2 : 2 > 2 , l ∈ N = 2n+3 = 2t.
Hence,
f σ (t) = f (σ(t)) = (σ(t))2 + σ(t) − 1 = (2t)2 + 2t − 1 = 4t 2 + 2t − 1, t ∈ T. o n √ 3 Exercise 1.1.16. Let T = t = n + 2 : n ∈ N , f (t) = 1 − t 3 , t ∈ T. Find f (σ(t)), t ∈ T. Answer. −t 3 . Definition 1.1.17. We define the set T\(ρ(supT), supT] κ T = T otherwise.
if
supT < ∞
5
Elements of the Time Scale Calculus 1 : n ∈ N ∪ {0}. Then supT = 1 and Example 1.1.18. Let T = n 1 1 1 ρ(1) = sup , 0 : , 0 < 1, l ∈ N = . l l 2 Therefore
i 1 T = T\ , 1 = : n ∈ N, n ≥ 2 ∪ {0}. 2 n κ
1
Definition 1.1.19. We assume that a ≤ b. We define the interval [a, b] in T by [a, b] = {t ∈ T : a ≤ t ≤ b}. Open intervals, half-open intervals and so on, are defined accordingly. Example 1.1.20. Let [a, b] be an interval in T and b be a left-dense point. Then sup[a, b] = b and since b is a left-dense point, we have that ρ(b) = b. Hence, [a, b]κ = [a, b]\(b, b] = [a, b]\Ø = [a, b]. 1 Exercise 1.1.21. Let T = : n ∈ N ∪ {0}. Find Tκ . 2n + 1 1 Answer. : n ∈ N, n ≥ 2 ∪ {0}. 2n + 1
1.2. Differentiation Definition 1.2.1. Assume that f : T → R is a function and let t ∈ Tκ . We define f ∆ (t) to be the number, provided it exists, as follows: for any ε > 0 there is a neighbourhood U of t, U = (t − δ,t + δ) ∩ T for some δ > 0, such that | f (σ(t)) − f (s) − f ∆ (t)(σ(t) − s)| ≤ ε|σ(t) − s| for
all
s ∈ U,
s 6= σ(t).
We say f ∆ (t) the delta or Hilger derivative of f at t. We say that f is delta or Hilger differentiable, shortly differentiable, in T κ if f ∆ (t) exists for all t ∈ Tκ . The function f ∆ : T → R is said to be delta derivative or Hilger derivative, shortly derivative, of f in T κ .
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Remark 1.2.2. If T = R, then the delta derivative coincides with the classical derivative. Theorem 1.2.3. The delta derivative is well defined. Remark 1.2.4. Let us assume that supT < ∞ and f ∆ (t) is defined at a point t ∈ T\Tκ with the same definition as given in Definition 1.2.1. Then the unique point t ∈ T\Tκ is supT. Hence, for any ε > 0 there is a neighbourhood U = (t − δ,t + δ) ∩ (T\Tκ ), for some δ > 0, such that f (σ(t)) = f (s) = f (σ(supT)) = f (supT),
s ∈ U,
s 6= σ(t).
Therefore for any α ∈ R and s ∈ U we have | f (σ(t)) − f (s) − α(σ(t) − s)| = | f (supT) − f (supT) − α(sup T − sup T)| ≤ ε|σ(t) − s|,
i.e., any α ∈ R is the delta derivative of f at the point t ∈ T\Tκ .
Example 1.2.5. Let f (t) = α ∈ R. We will prove that f ∆ (t) = 0 for any t ∈ Tκ . Really, for t ∈ Tκ and for any ε > 0 there is a δ > 0 such that s ∈ (t − δ,t + δ) ∩ T, s 6= σ(t), implies | f (σ(t)) − f (s) − 0(σ(t) − s)| = |α − α| ≤ ε|σ(t) − s|.
Exercise 1.2.6. Let f (t) = t ∈ Tκ , t > 0.
√ 1 t, t ∈ T,t > 0. Prove that f ∆ (t) = √ p for t + σ(t)
Theorem 1.2.7. Assume f : T → R is a function and let t ∈ Tκ . Then we have the following. 1. If f is differentiable at t, then f is continuous at t. 2. If f is continuous at t and t is right-scattered, then f is differentiable at t with f (σ(t)) − f (t) f ∆ (t) = . µ(t)
Elements of the Time Scale Calculus
7
3. If t is right-dense, then f is differentiable iff the limit lim
s−→t
f (t) − f (s) t −s
exists as a finite number. In this case f ∆ (t) = lim
s−→t
f (t) − f (s) . t −s
4. If f is differentiable at t, then f (σ(t)) = f (t) + µ(t) f ∆(t). 1 : n ∈ N0 ∪ {0}, f (t) = σ(t), t ∈ T. We will 2n + 1 1 find f ∆ (t), t ∈ Tκ . We have supT = 1, ρ(1) = , 3 [ 1 :n∈N {0}. Tκ = 2n + 1
Example 1.2.8. Let T =
1 1 −t ,n= , n ≥ 1, we have 2n + 1 2t 1 1 1 1 σ(t) = inf ,0 : ,0 > , l ∈ N0 = 2l + 1 2l + 1 2n + 1 2n − 1
For t ∈ Tκ , t =
= i.e., any point t = f ∆ (t) =
1 2 1−t 2t
−1
=
t > t, 1 − 2t
1 , n ≥ 1, is right-scattered. At these points 2n + 1
f (σ(t)) − f (t) σ(σ(t)) − σ(t) (σ(t))2 = =2 σ(t) − t σ(t) − t (1 − 2σ(t))(σ(t) − t)
= 2
1−
2 t 1−2t t 2t 1−2t 1−2t
−t
=2
t2 (1−2t)2 1−4t 2t 2 1−2t 1−2t
=2
t2 1 = . 2 2t (1 − 4t) 1 − 4t
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Svetlin G. Georgiev
Let now t = 0. Then
1 1 σ(0) = inf ,0 : , 0 > 0, l ∈ N0 2l + 1 2l + 1
= 0.
Consequently t = 0 is right-dense. Also, σ(h) − σ(0) = lim h−→0 h h−→0 lim
Therefore σ0 (0) = 1.
h 1−2h
h
−0
1 = 1. h−→0 1 − 2h
= lim
√ Exercise 1.2.9. Let T = { 5 n + 1 : n ∈ N0 }, f (t) = t + t 3 , t ∈ T. Find f ∆ (t), t ∈ Tκ . q p 5 Answer. 1 + 5 (t 5 + 1)2 + t t 5 + 1 + t 2 , t ∈ T. Theorem 1.2.10. Assume f , g : T → R are differentiable at t ∈ Tκ . Then 1. the sum f + g : T → R is differentiable at t with ( f + g)∆ (t) = f ∆ (t) + g∆ (t). 2. for any constant α, α f : T → R is differentiable at t with (α f )∆ (t) = α f ∆ (t). 3. the product f g : T → R is differentiable at t with ( f g)∆ (t) = f ∆ (t)g(t) + f (σ(t))g∆(t) = f (t)g∆(t) + f ∆ (t)g(σ(t)). 4. if f (t) f (σ(t)) 6= 0, we have that
1 : T → R is differentiable at t and f
∆ 1 f ∆ (t) (t) = − . f f (t) f (σ(t)) 5. if g(t)g(σ(t)) 6= 0, we have that
f : T → R is differentiable at t with g
∆ f f ∆ (t)g(t) − f (t)g∆(t) (t) = . g g(t)g(σ(t))
Elements of the Time Scale Calculus
9
Example 1.2.11. Let f , g, h : T → R be differentiable at t ∈ Tκ . Then ( f gh)∆(t) = (( f g)h)∆(t) = ( f g)∆(t)h(t) + ( f g)(σ(t))h∆(t) = ( f ∆ (t)g(t) + f (σ(t))g∆(t))h(t) + f (σ(t))g(σ(t))h∆(t) =
f ∆ (t)g(t)h(t) + f σ(t)g∆(t)h(t) + f σ(t)gσ(t)h∆(t). 2
κ
Definition 1.2.12. Let f : T → R and t ∈ (T κ ) = Tκ . We define the second derivative of f at t, provided it exists, by 2
f∆ = f∆
∆
2
: Tκ → R. n
n
Similarly we define higher-order derivatives f ∆ : Tκ → T. (n)
Theorem 1.2.13. (Leibniz’s Formula) Let Sk be the set consisting of all possible strings of length n, containing exactly k times σ and n − k times ∆. If fΛ
exists for
then ( f g)
∆n
n
=
all
∑ ∑
k=0
(n)
Λ∈Sk
(n)
Λ ∈ Sk ,
k
f Λ g∆ .
Example 1.2.14. Let µ is differentiable at t ∈ Tκ and t is right-scattered. Then σ f (σ(σ(t))) − f (σ(t)) f (σ(t)) − f (t) ∆σ ∆ σ = f (t) = f (t) = σ(t) − t σ(σ(t)) − σ(t) =
f (σ(σ(t))) − f (σ(t)) 1 σ(σ(t))−σ(t) σ(t) − t σ(t)−t
= ( f σ )∆ (t)
1 σ∆(t)
= f σ∆ (t)
1 , 1 + µ∆ (t)
i.e., f σ∆ (t) = (1 + µ∆ (t)) f ∆σ(t).
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Svetlin G. Georgiev
Also, f σσ∆ (t) = ( f σ )σ∆ (t) = (1 + µ∆ (t)) f σ∆ f σ∆σ (t) = ( f σ ) =
∆σ
σ ∆ (t) = ( f σ ) (t)
(1 + µ∆ (t))( f ∆)σ (t)
σ
σ
(t) = (1 + µ∆ (t)) f σ∆σ(t),
= 1 + µ∆σ (t) f ∆σσ (t).
Theorem 1.2.15. (Chain Rule) Assume g : R → R is continuous, g : T → R is delta differentiable on Tκ , and f : R → R is continuously differentiable. Then there exists c ∈ [t, σ(t)] with ( f ◦ g)∆ (t) = f 0 (g(c))g∆(t). Example 1.2.16. Let T = Z, f (t) = t 3 + 1, g(t) = t 2 . We have that g : R → R is continuous, g : T → R is delta differentiable on Tκ , f : R → R is continuously differentiable, σ(t) = t + 1. Then g∆ (t) = σ(t) + t, ( f ◦ g)∆ (1) = f 0 (g(c))g∆(1) = 3(g(c))2(σ(1) + 1) = 9c4 .
(1.2.1)
Here c ∈ [1, σ(1)] = [1, 2]. Also, f ◦ g(t) = f (g(t)) = (g(t))3 + 1 = t 6 + 1,
( f ◦ g)∆ (t) = (σ(t))5 + t(σ(t))4 + t 2 (σ(t))3 + t 3 (σ(t))2 + t 4 σ(t) + t 5 ,
( f ◦ g)∆ (1) = (σ(1))5 + (σ(1))4 + (σ(1))3 + (σ(1))2 + σ(1) + 1 = 63. Hence and (1.2.1), we get 63 = 9c4
or c4 = 7,
or c =
√ 4
7 ∈ [1, 2].
Exercise 1.2.17. Let T = Z, f (t) = t 2 + 2t + 1, g(t) = t 2 − 3t. Find a constant c ∈ [1, σ(1)] such that ( f ◦ g)∆ (1) = f 0 (g(c))g∆(1).
Elements of the Time Scale Calculus
11
Answer. ∀c ∈ [1, 2]. Theorem 1.2.18. (Chain Rule) Assume v : T → R is strictly increasing and T˜ = ˜ v(T ) is a time scale. Let w : T˜ → R. If v∆ (t) and w∆ (v(t)) exist for t ∈ Tκ , then ˜
(w ◦ v)∆ = (w∆ ◦ v)v∆ . Example 1.2.19. Let T = 22n : n ∈ N0 , v(t) = t 2 , w(t) = t 2 + 1. Then v : T → R is strictly increasing, T˜ = v(T ) = 24n : n ∈ N0 is a time scale. For t ∈ T, t = 22n , n ∈ N0 , we have n o σ(t) = inf 22l : 22l > 22n , l ∈ N0 = 22n+2 = 4t, v∆ (t) = σ(t) + t = 5t.
˜ t = 24n , n ∈ N0 , we have For t ∈ T, n o ˜ = inf 24l : 24l > 24n , l ∈ N0 = 24n+4 = 16t. σ(t)
Also, for t ∈ T, we have
(w ◦ v)(t) = w(v(t)) = (v(t))2 + 1 = t 4 + 1,
(w ◦ v)∆ (t) = (σ(t))3 + t(σ(t))2 + t 2 σ(t) + t 3 = 64t 3 + 16t 3 + 4t 3 + t 3 = 85t 3 , ˜ ˜ w∆ ◦ v(t) = σ(v(t)) + v(t) = 16v(t) + v(t) = 17v(t) = 17t 2 , ˜ w∆ ◦ v(t) v∆ (t) = 17t 2 (5t) = 85t 3.
Consequently
˜
(w ◦ v)∆ (t) = (w∆ ◦ v(t))v∆(t), t ∈ Tκ . Exercise 1.2.20. Let T = 23n : n ∈ N0 , v(t) = t 2 , w(t) = t. Prove ˜
(w ◦ v)∆ (t) = (w∆ ◦ v(t))v∆(t),
t ∈ Tκ .
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Svetlin G. Georgiev
Theorem 1.2.21. (Derivative of the Inverse) Assume v : T → R is strictly in˜ := v(T) is a time scale. Then creasing and T ˜
(v−1 )∆ ◦ v(t) =
1 v∆ (t)
for any t ∈ Tκ such that v∆ (t) 6= 0. Example 1.2.22. Let T = N, v(t) = t 2 + 1. Then σ(t) = t + 1, v : T → R is strictly increasing and v∆ (t) = σ(t) + t = 2t + 1, Hence, v−1
∆˜
◦ v(t) =
1 v∆ (t)
=
t ∈ T.
1 , 2t + 1
t ∈ T.
Exercise 1.2.23. Let T = {n + 5 : n ∈ N0 }, v(t) = t 2 + t. Find v−1 Answer.
1 . 2t + 2
∆˜
◦ v(t).
1.3. Integration Definition 1.3.1. A function f : T → R is called regulated provided its rightsided limits exist(finite) at all right-dense points in T and its left-sided limits exist(finite) at all left-dense points in T. Example 1.3.2. Let T = N and f (t) =
t2 , t −1
g(t) =
t , t +1
t ∈ T.
We note that all points of T are right-scattered. The points t ∈ T, t 6= 1, are left-scattered. The point t = 1 is left-dense. Also, lim f (t) is not finite and t−→1−
lim g(t) exists and it is finite. Therefore the function f is not regulated and
t−→1−
the function g is regulated.
Elements of the Time Scale Calculus
13
Exercise 1.3.3. Let T = R and 11 for t = 1 f (t) = 1 for t ∈ R\{1}. t−1
Determine if f is regulated. Answer. No.
Definition 1.3.4. A continuous function f : T → R is called pre-differentiable with region of differentiation D, provided 1. D ⊂ Tκ , 2. Tκ \D is countable and contains no right-scattered elements of T, 3. f is differentiable at each t ∈ D. Example 1.3.5. Let T = Pa,b =
∞ [
[k(a + b), k(a + b) + a] for a > b > 0,
k=0
f : T → R be defined by S∞ 0 i f t ∈ k=0 [k(a + b), k(a + b) + b] f (t) = t − (a + b)k − b i f t ∈ [(a + b)k + b, (a + b)k + a].
Then f is pre-differentiable with D\
∞ [
{(a + b)k + b}.
k=0
Exercise 1.3.6. Let T = R and 0 if f (t) = 1 t+3
t = −3 if
t ∈ R\{−3}.
Check if f : T → R is pre-differentiable and if it is, find the region of differentiation.
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Svetlin G. Georgiev
Definition 1.3.7. A function f : T → R is called rd-continuous if it is continuous at right-dense points in T and its left-sided limits exist(finite) at left-dense points in T. The set of rd-continuous functions f : T → R will be denoted by Crd (T). The set of functions f : T → R that are differentiable and whose derivative is 1 rd-continuous is denoted by Crd (T). If f is continuous on T and differentiable κ κ on T except at t1 , . . .,tm ∈ T such that f ∆ is regulated on Tκ , we abbreviate this piecewise rd-continuous differentiability by writing f ∈ C p (T). Some results concerning rd-continuous and regulated functions are contained in the following theorem. Since its statements follow directly from the definitions, we leave its proof to the reader. Theorem 1.3.8. Assume f : T → R. 1. If f is continuous, then f is rd-continuous. 2. If f is rd-continuous, then f is regulated. 3. The jump operator σ is rd-continuous. 4. If f is regulated or rd-continuous, then so is f σ . 5. Assume f is continuous. If g : T → R is regulated or rd-continuous, then f ◦ g has that property. Theorem 1.3.9. Every regulated function on a compact interval is bounded. Theorem 1.3.10. (Induction Principle) Let t0 ∈ T and assume that {S(t) : t ∈ [t0 , ∞)} is a family of statements satisfying (i) S(t0 ) is true, (ii) If t ∈ [t0 , ∞) is right-scattered and S(t) is true, then S(σ(t)) is true, (iii) If t ∈ [t0 , ∞) is right-dense and S(t) is true, then there is a neighbourhood U of t such that S(s) is true for all s ∈ U ∩ (t, ∞),
15
Elements of the Time Scale Calculus (iv) If t ∈ (t0 , ∞) is left-dense and S(s) is true for s ∈ [t0,t), then S(t) is true. Then S(t) is true for all t ∈ [t0, ∞).
Theorem 1.3.11. (Dual Version of Induction Principle) Let t0 ∈ T and assume that {S(t) : t ∈ (−∞,t0 ]} is a family of statements satisfying (i) S(t0 ) is true, (ii) If t ∈ (−∞,t0 ] is left-scattered and S(t) is true, then S(ρ(t)) is true, (iii) If t ∈ (−∞,t0 ] is left-dense and S(t) is true, then there is neighbourhood U of t such that S(s) is true for all s ∈ U ∩ (−∞,t), (iv) If t ∈ (−∞,t0 ) is right-dense and S(s) is true for s ∈ (t,t0), then S(t) is true. Then S(t) is true for all t ∈ (−∞,t0 ]. Theorem 1.3.12. Let f and g be real-valued functions defined on T, both predifferentiable with D. Then | f ∆ (t)| ≤ |g∆ (t)| for
all
t ∈D
implies | f (s) − f (r)| ≤ g(s) − g(r) for all
r, s, ∈ T,
r ≤ s.
(1.3.1)
Theorem 1.3.13. Suppose f : T → R is pre-differentiable with D. If U is a compact interval with endpoints r, s ∈ T, then ( ) | f (s) − f (r)| ≤
sup | f ∆ (t)| |s − r|.
t∈U k ∩D
Theorem 1.3.14. Let f is pre-differentiable with D. If f ∆ (t) = 0 for all t ∈ D, then f is a constant function.
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Svetlin G. Georgiev
Theorem 1.3.15. Let f and g are pre-differentiable with D and f ∆ (t) = g∆ (t) for all t ∈ D. Then g(t) = f (t) +C
for
t ∈ T,
all
where C is a constant. Theorem 1.3.16. Suppose f n : T → R is pre-differentiable with D for each n ∈ N. Assume that for each t ∈ Tκ there exists a compact interval U(t) such that the sequence { f n∆ }n∈N converges uniformly on U(t) ∩ D.
(i) If { f n }n∈N converges at some t0 ∈ U(t) for some t ∈ Tκ , then it converges uniformly on U(t).
(ii) If { f n }n∈N converges at some t0 ∈ T, then it converges uniformly on U(t) for all t ∈ Tκ . (iii) The limit mapping f = lim fn is pre-differentiable with D and we have n−→∞
f ∆ (t) = lim fn∆ (t) for
all
n−→∞
t ∈ D.
Theorem 1.3.17. Let t0 ∈ T, x0 ∈ R, f : Tκ → R be given regulated map. Then there exists exactly one pre-differentiable function F satisfying F ∆ (t) = f (t) for
all
t ∈ D,
F(t0 ) = x0 .
Definition 1.3.18. Assume f : T → R is a regulated function. Any function F by Theorem 1.3.17 is called a pre-antiderivative of f . We define the indefinite integral of a regulated function f by Z
f (t)∆t = F(t) + c,
where c is an arbitrary constant and F is a pre-antiderivative of f . We define the Cauchy integral by Z s τ
f (t)∆t = F(s) − F(τ) for
all
r, s ∈ T.
A function F : T → R is called an antiderivative of f : T → R provided F ∆ (t) = f (t) holds for
all
t ∈ Tκ .
Elements of the Time Scale Calculus
17
Example 1.3.19. Let T = Z. Then σ(t) = t + 1, f (t) = 3t 2 + 5t + 2, g(t) = t 3 + t 2 , t ∈ T. Since
t ∈ T. Let also,
g∆ (t) = = = = we conclude that
Z
(σ(t))2 + tσ(t) + t 2 + σ(t) + t (t + 1)2 + t(t + 1) + t 2 + t + 1 + t t 2 + 2t + 1 + t 2 + t + t 2 + 2t + 1 3t 2 + 5t + 2,
(3t 2 + 5t + 2)∆t = t 3 + t 2 + c.
Exercise 1.3.20. Let T = N30 . Prove that Z
√ √ 3 (2t + 3 t 2 + 3 3 t + 2)∆t = t 2 + t + c.
Theorem 1.3.21. Every rd-continuous function f : T → R has an antiderivative. In particular, if t0 ∈ T, then F defined by F(t) =
Z t t0
f (τ)∆τ for t ∈ T,
is an antiderivative of f . Theorem 1.3.22. If f ∈ Crd (T) and t ∈ Tκ , then Z σ(t)
f (τ)∆τ = µ(t) f (t).
t
Theorem 1.3.23. Let [a, b] ⊂ T. If f ∆ ≥ 0 on [a, b], then f is nondecreasing on [a, b]. If f ∆ ≤ 0 on [a, b], then f is nonincreasing on [a, b]. Theorem 1.3.24. If a, b, c ∈ T, α ∈ R and f , g ∈ Crd (T), then (i)
Z b
( f (t) + g(t))∆t =
a
(ii)
Z b a
Z b
f (t)∆t +
a
(α f )(t)∆t = α
Z b a
f (t)∆t,
Z b a
g(t)∆t,
18
Svetlin G. Georgiev Z b
(iii)
a
Z b
(iv)
f (t)∆t = − f (t)∆t =
a
Z a
Z c a
f (t)∆t,
b
f (t)∆t +
Z b
f (t)∆t,
c
(v)
Z b
f (σ(t))g (t)∆t = ( f g)(b) − ( f g)(a) −
(vi)
Z b
Z b
a
a
(vii)
Z a
∆
f (t)g∆(t)∆t = ( f g)(b) − ( f g)(a) −
Z b
f ∆ (t)g(t)∆t,
a
f ∆ (t)g(σ(t))∆t,
a
f (t)∆t = 0,
a
(viii) if | f (t)| ≤ g(t), t ∈ [a, b], then Z b Z b ≤ g(t)∆t, f (t)∆t a
a
(ix) if f (t) ≥ 0 for all a ≤ t < b, then
Z b a
f (t)∆t ≥ 0.
Exercise 1.3.25. Let a, b ∈ T and f ∈ Crd (T). (i) If T = R, prove Z b a
f (t)∆t =
Z b
f (t)dt,
a
where the integral on the right is the usual Riemann integral from calculus. (ii) If [a, b] consists only isolated points, then ∑t∈[a,b) µ(t) f (t) i f a < b Z b f (t)∆t = 0 if a = b a − ∑t∈[b,a) µ(t) f (t) i f a > b.
Elements of the Time Scale Calculus
19
Definition 1.3.26. (Improper Integral) If a ∈ T, supT = ∞, and f is rdcontinuous on [a, ∞), then we define improper integral by Z ∞ a
f (t)∆t := lim
Z b
b−→∞ a
f (t)∆t
provided this limit exists, and we say that the improper integral converges in this case. If this limit does not exist, then we say that the improper integral diverges. Example 1.3.27. Let T = qN0 , q > 1. Then σ(t) = qt. Since all points are isolated, then Z ∞ 1 1 q−1 ∆t = ∑ 2 µ(t) = ∑ = ∞. 2 t t t 1 t∈qN0 t∈qN0
1.4. The Exponential Function Definition 1.4.1. Let h > 0. 1. The Hilger complex numbers we define as follows 1 Ch = {z ∈ C : z 6= − }. h 2. The Hilger real axis we define as follows 1 Rh = {z ∈ C : z > − }. h 3. The Hilger alternative axis we define as follows 1 Ah = {z ∈ C : z < − }. h 4. The Hilger imaginary circle we define as follows 1 1 Ih = z ∈ C : z + = . h h
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Svetlin G. Georgiev
For h = 0 we set C0 = C,
R0 = R,
A0 = Ø,
I0 = iR.
Definition 1.4.2. Let h > 0 and z ∈ Ch . We define the Hilger real part of z by Reh (z) =
|zh + 1| − 1 h
and the Hilger imaginary part by Imh (z) =
Arg(zh + 1) , h
where Arg(z) denotes the principal argument of z, i.e., −π < Arg(z) ≤ π. We note that 1 − < Reh (z) < ∞ and h
−
π π < Imh (z) < . h h
In particular, Reh (z) ∈ Rh .
π π Definition 1.4.3. Let − < w ≤ . We define the Hilger purely imaginary numh h ber by ◦ eiwh − 1 . iw= h ◦
Theorem 1.4.4. Let z ∈ Ch . Then i Imh (z) ∈ Ih . Theorem 1.4.5. We have ◦
lim [Reh (z)+ i Imh (z)] = Re(z) + iIm(z).
h→0
π π Theorem 1.4.6. Let − < w ≤ . Then h h ◦
| i w|2 =
4 wh sin2 . 2 h 2
Elements of the Time Scale Calculus
21
Definition 1.4.7. The ”circle plus” addition ⊕ on Ch is defined by z ⊕ w = z + w + zwh. Theorem 1.4.8. (Ch , ⊕) is an Abelian group. Example 1.4.9. Let z ∈ Ch and w ∈ C. We will simplify the expression A = z⊕
w . 1 + hz
We have w zw + h 1 + hz a + hz (1 + hz)w = z+ 1 + hz = z + w.
A = z+
Theorem 1.4.10. For z ∈ Ch we have ◦
z = Reh (z)⊕ i Imh (z). Definition 1.4.11. Let n ∈ N and z ∈ Ch . We define ”circle dot” multiplication by n z = z ⊕ z ⊕ z ⊕ · · · ⊕ z. Theorem 1.4.12. Let n ∈ N and z ∈ Ch . Then n z =
(zh + 1)n − 1 . h
(1.4.1)
Definition 1.4.13. Let z ∈ Ch . The additive inverse of z under the operation ⊕ is defined as follows −z z = . 1 + zh Theorem 1.4.14. Let z ∈ Ch . Then ( z) = z.
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Svetlin G. Georgiev
Definition 1.4.15. Let z, w ∈ Ch . We define ”circle minus” substraction as follows z w = z ⊕ ( w). For z, w ∈ Ch we have z w = z ⊕ ( w)
= z + ( w) + z( w)h w zwh − = z− 1 + wh 1 + wh z + zwh − w − zwh = 1 + wh z−w = , 1 + wh
i.e., z w =
z−w . 1 + wh
(1.4.2)
Theorem 1.4.16. Let z ∈ Ch . Then z = z iff z ∈ Ih . Theorem 1.4.17. Let −
π π < w ≤ . Then h h ◦
◦
( i w) = i w. Definition 1.4.18. Let z ∈ Ch . The generalized square of z is defined as follows z = −z( z).
We have z = −z
−z z2 = . 1 + zh 1 + zh
Theorem 1.4.19. For z ∈ Ch we have ( z) = z .
Elements of the Time Scale Calculus
23
Theorem 1.4.20. For z ∈ Ch we have 1 + zh =
z2 . z
Theorem 1.4.21. For z ∈ Ch we have z + ( z) = z h. Theorem 1.4.22. For z ∈ Ch we have z ⊕ z = z + z2 .
Theorem 1.4.23. Let −
π π < w ≤ . Then h h
4 −( i ) = 2 sin2 h ◦
wh . 2
Exercise 1.4.24. Let z ∈ Ch . Prove that z ∈ R iff z ∈ Rh ∪ Ah ∪ Ih . For h > 0, we define the strip π π Zh := {z ∈ C : − < Im(z) ≤ }. h h For h = 0, we set Z0 := C. Definition 1.4.25. For h > 0, we define the cylindrical transformation ξh : Ch → Zh by 1 ξh (z) := Log(1 + zh), h where Log is the principal logarithm function. For h = 0, we define ξ0 (z) = z for all z ∈ C. We note that ξ−1 h (z) = for z ∈ Zh .
ezh − 1 h
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Svetlin G. Georgiev
Definition 1.4.26. We say that a function f : T → R is regressive provided 1 + µ(t)p(t) 6= 0
f or
all
t ∈ Tκ
holds. The set of all regressive and rd-continuous functions f : T → R will be defined by R (T) or R . In R we define ”circle plus” addition as follows ( f ⊕ g)(t) = f (t) + g(t) + µ(t) f (t)g(t). Exercise 1.4.27. Prove that (R , ⊕) is an Abelian group. Definition 1.4.28. The group (R , ⊕) will be called the regressive group. Definition 1.4.29. For f ∈ R we define ( f )(t) = −
f (t) 1 + µ(t) f (t)
f or
all
t ∈ Tκ .
Exercise 1.4.30. Let f ∈ R . Prove that ( f )(t) ∈ R for all t ∈ Tκ . Definition 1.4.31. We define ”circle minus” substraction on R as follows ( f g)(t) = ( f ⊕ ( g))(t)
f or
all
For f , g ∈ R , we have f g = = = =
f ⊕ ( g) g f⊕ − 1 + µg g µfg f− − 1 + µg 1 + µg f −g . 1 + µg
Theorem 1.4.32. Let f , g ∈ R . Then
t ∈ Tκ .
Elements of the Time Scale Calculus
25
1. f f = 0, 2. ( f ) = f , 3. f g ∈ R , 4. ( f g) = g f , 5. ( f ⊕ g) = ( f ) ⊕ ( g), 6. f ⊕
g = f + g. 1+µf
Definition 1.4.33. If f ∈ R , then we define the generalized exponential function by Rt e f (t, s) = e s ξµ(τ) ( f (τ))∆τ f or s,t ∈ T. In fact, using the definition for the cylindrical transformation we have Rt 1 s µ(τ) Log(1+µ(τ) f (τ))∆τ
e f (t, s) = e
s,t ∈ T.
f or
Theorem 1.4.34. (Semigroup Property) If f ∈ R , then e f (t, r)e f (r, s) = e f (t, s)
f or
all
t, r, s ∈ R .
Exercise 1.4.35. Let f ∈ R . Prove that e0 (t, s) = 1 and e f (t,t) = 1. Theorem 1.4.36. Let f ∈ R and fix t0 ∈ T. Then e∆f (t,t0) = f (t)e f (t,t0). Corollary 1.4.37. Let f ∈ R and fix t0 ∈ T. Then e f (t,t0) is a solution to the Cauchy problem y∆ (t) = f (t)y(t), y(t0 ) = 1. (1.4.3) Corollary 1.4.38. Let f ∈ R and fix t0 ∈ T. Then e f (t,t0) is the unique solution of the problem (1.4.3).
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Svetlin G. Georgiev
Theorem 1.4.39. Let f ∈ R . Then e f (σ(t), s) = (1 + µ(t) f (t))e f (t, s). Theorem 1.4.40. Let f ∈ R . Then 1 = e f (s,t). e f (s,t)
e f (t, s) =
Theorem 1.4.41. Let f , g ∈ R . Then e f (t, s)eg(t, s) = e f ⊕g (t, s). Theorem 1.4.42. Let f , g ∈ R . Then e f (t, s) = e f g (t, s). eg (t, s) Theorem 1.4.43. Let f ∈ R . Then 1 e f (t, r). 1 + µ(s) f (s)
e f (t, σ(s))e f (s, r) = Theorem 1.4.44. Let f , g ∈ R . Then e∆f g (t,t0) =
( f (t) − g(t))e f (t,t0) . eg (σ(t),t0)
Theorem 1.4.45. Let f ∈ R and a, b, c ∈ T. Then (e f (c,t))∆ = − f (t)(e f (c,t))σ = − f (t)e f (c, σ(t)) and
Z b a
f (t)e f (c, σ(t))∆t = e f (c, a) − e f (c, b).
2 5 Exercise 1.4.46. Assume 1 + µ(t) = 6 0, 1 + µ(t) 6= 0 for all t ∈ T ∩ (0, ∞). t t Let also, t0 ∈ T ∩ (0, ∞). Evaluate the integral I=
Z t e 5 (s,t ) 0 s
t0
seσ2 (s,t0) s
∆s.
Elements of the Time Scale Calculus
27
Solution. We have
5 2 − s s
e 5 (s,t ) 0 s
eσ (s,t 2 s
0)
=
3 e 5s (s,t0) s eσ2 (s,t0) s
= e∆5 2 (s,t0) s
s
∆
(s,t0)
= e∆
(s,t0).
= e
3 s 1+µ(s) 2s 3 s+2µ(s)
Hence, 1 t ∆ e 3 (s,t0)∆s 3 t0 s+2µ(s) s=t 1 e 3 (s,t0) s+2µ(s) 3 s=t0 1 1 e 3 (t,t0) − . t+2µ(t) 3 3 Z
I = = =
Exercise 1.4.47. Let α ∈ R. Let also, the exponentials e α2 t
− (α−1) σ(t)
2
(t,t0) and e αt (t,t0)
exist for all t ∈ T ∩ (0, ∞). Prove that 1.
e α2 t
− (α−1) σ(t)
2
(t,t0)
e αt (t,t0) 2.
= e α−1 (t,t0), σ(t)
e α−1 (t,t0) σ(t)
e αt (t,t0)
= e−
1 σ(t)
(t,t0) =
t0 t
for all t,t0 ∈ T ∩ (0, ∞). Let α : T → R be regulated and 1 + α(t)µ(t) 6= 0 for all t ∈ T. Let also, t0 ,t ∈ T, t0 < t.
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Svetlin G. Georgiev
1. T = hZ, h > 0. Every point in T is isolated and µ(t) = h for every t ∈ T. Then Rt
1
eα (t,t0) = e t0 µ(τ)
Log(1+α(τ)µ(τ))∆τ 1
= e∑s∈[t0 ,t) µ(s)
Log(1+α(s)µ(s))µ(s)
= e∑s∈[t0 ,t) Log(1+hα(s)) = ∏ (1 + hα(s)). s∈[t0 ,t)
If α is a constant, then
∏
eα (t,t0) =
(1 + hα)
s∈[t0 ,t)
= (1 + hα)t−t0 . 2. T = qN0 , q > 1. Every point of T is isolated and µ(t) = (q − 1)t for all t ∈ T. Then Rt
1
eα (t,t0) = e t0 µ(τ)
Log(1+α(τ)µ(τ))∆τ 1
= e∑s∈[t0 ,t) µ(s)
Log(1+α(s)µ(s))µ(s)
= e∑s∈[t0 ,t) Log(1+α(s)µ(s)) = e∑s∈[t0 ,t) Log(1+(q−1)sα(s)) =
∏ s∈[t0 ,t)
(1 + (q − 1)sα(s)).
Exercise 1.4.48. Let T = qN0 ,
0 < q < 1. Prove that 1−q eα (t,t0) = ∏ 1 + α(s)s . q s∈[t ,t) 0
3. T = Nk0 ,
k ∈ N. Every point of T is isolated and µ(t) =
√ k k t +1 .
Elements of the Time Scale Calculus
29
Then Rt
eα (t,t0) = e
1 t0 µ(τ) Log(1+α(τ)µ(τ))∆τ 1
= e∑s∈[t0 ,t) µ(s)
Log(1+α(s)µ(s))µ(s)
= e∑s∈[t0 ,t) Log(1+α(s)µ(s)) =
∏
(1 + α(s)µ(s))
s∈[t0 ,t)
=
∏ s∈[t0 ,t)
√ k 1 + k s + 1 − s α(s) .
1.5. Hyperbolic and Trigonometric Functions Definition 1.5.1. (Hyperbolic Functions) If f ∈ Crd and −µ f 2 ∈ R , then we define the hyperbolic functions cosh f and sinh f by cosh f =
e f + e− f 2
and
sinh f =
e f − e− f . 2
Theorem 1.5.2. Let f ∈ Crd . If −µ f 2 ∈ R , then we have 1. cosh∆f = f sinh f , 2. sinh∆f = f cosh f , 3. cosh2f − sinh2f = e−µ f 2 . Definition 1.5.3. (Trigonometric Functions) If f ∈ Crd and µ f 2 ∈ R , then we define the trigonometric functions cos f and sin f by cos f =
ei f + e−i f 2
and
sin f =
ei f − e−i f . 2i
Theorem 1.5.4. Let f ∈ Crd and −µ f 2 ∈ R . Prove that 1. cos∆f = − f sin f .
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Svetlin G. Georgiev 2. sin∆f = f cos f . 3. cos2f + sin2f = eµ f 2 .
Exercise 1.5.5. Let f ∈ Crd and µ f 2 ∈ R . Show Euler’s formula ei f = cos f +i sin f .
1.6. The Multidimensional Time Scale Calculus Let n ∈ N be fixed. For each i ∈ {1, 2, . . ., n}, we denote by Ti a time scale. Definition 1.6.1. The set Λn = T1 × T2 × · · · × Tn = {t = (t1 ,t2 , . . .,tn ) : ti ∈ Ti , i = 1, 2, . . ., n} is called an n-dimensional time scale. Example 1.6.2. (R, Z, N) is a 3-dimensional time scale. Remark 1.6.3. We equip Λn with the metric n
d(t, s) =
∑ |ti − si |2
i=1
! 21
fort, s ∈ Λn .
The set Λn with this metric is a complete metric space. Therefore, we have for Λn the fundamental concepts such as open balls, neighbourhoods of points, open sets, compact sets, and so on. Also, we have for functions f : Λn → R the concepts of limits, continuity, and properties of continuous functions on general metric spaces. Definition 1.6.4. Let σi , i ∈ {1, 2, . . ., n}, be the forward jump operator in Ti . The operator σ : Λn → Rn defined by σ(t) = (σ1 (t), σ2 (t), . . ., σn (t)), is said to be the forward jump operator in Λn .
t ∈ Λn ,
31
Elements of the Time Scale Calculus Example 1.6.5. Let n = 4 and Λ4 = R × Z × Z × N20 . Then T1 = R,
T2 = Z,
T3 = Z,
T4 = N20 .
We have that σ1 (t1 ) = t1 ,
t1 ∈ R,
and σ3 (t3 ) = t3 + 1,
t3 ∈ Z,
Hence,
σ2 (t2 ) = t2 + 1,
t2 ∈ Z,
√ 2 σ4 (t4 ) = 1 + t4 ,
t4 ∈ N20 .
√ 2 σ(t) = σ(t1 ,t2 ,t3,t4 ) = t1 ,t2 + 1,t3 + 1, 1 + t4 , (t1 ,t2 ,t3 ,t4) ∈ T1 × T2 × T3 × T4 . Exercise 1.6.6. Let n = 3 and Λ3 = Z × N × 4N . Find σ(t). Definition 1.6.7. Let ρi , i ∈ {1, 2, . . ., n}, be the backward jump operator in Ti . The operator ρ : Λn → Rn defined by ρ(t) = (ρ1 (t1 ), ρ2 (t2 ), . . ., ρn (tn )),
t = (t1,t2 , . . .,tn ) ∈ Λn ,
is said to be the backward jump operator in Λn . Example 1.6.8. Let n = 4 and Λ4 = R × Z × 3N × N30 . Here, T1 = R,
T2 = Z,
T4 = N30 .
T3 = 3N ,
Then ρ1 (t1 ) = t1 , ρ3 (t3 ) =
ρ4 (t4 ) =
t1 ∈ T1 , t3 , 3
ρ2 (t2) = t2 − 1,
t3 ∈ T3 \ {3},
3 √ 3 t4 − 1 ,
t2 ∈ T2 ,
ρ(3) = 3,
t4 ∈ T4 \ {0},
ρ4 (0) = 0.
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Hence, ρ(t) = (ρ1 (t1 ), ρ2 (t2 ), ρ3(t3 ), ρ4 (t4 ))
=
√ t1 ,t2 − 1, t33 , ( 3 t4 − 1)3 ift1 ∈ T1 ,
t2 ∈ T2 ,
t3 ∈ T3 \ {3},
t4 ∈ T4 \ {0},
√ (t1 ,t2 − 1, 3, ( 3 t4 − 1)3 ) ift1 ∈ T1 ,
t2 ∈ T2 ,
t1 ,t2 − 1, t33 , 0 ift1 ∈ T1 , t2 ∈ T2 , (t1 ,t2 − 1, 3, 0) ift1 ∈ T1 , t2 ∈ T2 ,
t3 = 3,
t4 ∈ T4 \ {0},
t3 ∈ T3 \ {3},
t3 = 3,
t4 = 0,
t4 = 0.
Exercise 1.6.9. Let n = 2 and Λ2 = Z × Z. Find ρ(t), t ∈ Λ2 . Definition 1.6.10. For x = (x1 , x2 , . . ., xn ) ∈ Rn and y = (y1 , y2 , . . ., yn ) ∈ Rn , we write x≥y whenever xi ≥ yi for alli = 1, 2, . . ., n. In a similar way, we understand x > y and x < y, and x ≤ y. Definition 1.6.11. Let t ∈ Λn , t = (t1 ,t2 , . . .,tn ). 1. If σ(t) > t, then t is called strictly right-scattered. 2. If σ(t) ≥ t and there are j, l ∈ {1, 2, . . ., n} such that σ j (t j ) > t j and σl (tl ) = tl , then t is called right-scattered. 3. If t < supΛn and σ(t) = t, then t is called right-dense.
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4. If ρ(t) < t, then t is called strictly left-scattered. 5. If ρ(t) ≤ t and there are l, j ∈ {1, 2, . . ., n} such that ρ j (t j ) < t j and ρl (tl ) = tl , then t is called left-scattered. 6. If t > infΛn and t = ρ(t), then t is called left-dense. 7. If t is strictly right-scattered and strictly left-scattered, then t is said to be strictly isolated. 8. If t is right-dense and left-dense, then t is said to be dense. 9. If t is right-scattered and left-scattered, then t is said to be isolated. Example 1.6.12. Let Λ3 = Z × Z × Z. Then, for t = (t1 ,t2 ,t3 ) ∈ Λ3 , we have σ(t) = (σ1 (t1), σ2 (t2 ), σ3 (t3 )) = (t1 + 1,t2 + 1,t3 + 1) > (t1 ,t2 ,t3 ), i.e., all points of Λ3 are strictly right-scattered. Also, ρ(t) = (ρ1 (t1), ρ2 (t2 ), ρ3 (t3 )) = (t1 − 1,t2 − 1,t3 − 1) < (t1 ,t2 ,t3 ), i.e., all points of Λ3 are strictly left-scattered. Consequently, all points of Λ3 are strictly isolated. Exercise 1.6.13. Classify each point t ∈ Λ2 = N30 × R as strictly right-scattered, right-scattered, right-dense, strictly left-scattered, left-scattered, left-dense, strictly isolated, dense, and isolated, respectively. Definition 1.6.14. The graininess function µ : Λn → [0, ∞)n is defined by µ(t) = (µ1 (t1 ), µ2 (t2 ), . . ., µn (tn )),
t = (t1 ,t2 , . . .,tn ) ∈ Λn .
Example 1.6.15. Let Λ3 = 3Z × R × N40 . Then T1 = 3Z, σ1 (t1 ) = t1 + 3,
T2 = R,
t1 ∈ T1 ,
T3 = N40 ,
σ2 (t2) = t2 ,
√ 4 σ3 (t3 ) = 1 + 4 t3 ,
t3 ∈ T3 .
t2 ∈ T2 ,
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Hence, µ1 (t1 ) = σ1 (t1 ) − t1 = t1 + 3 − t1 = 3, µ2 (t2 ) = σ2 (t2 ) − t2 = t2 − t2 = 0,
t1 ∈ T1 ,
t2 ∈ T2 ,
µ3 (t3 ) = σ3 (t3 ) − t3 √ 4 1 + 4 t3 − t3
=
q q √ 4 3 = t3 + 4 t3 + 6 4 t32 + 4 4 t3 + 1 − t3 q √ √ = 6 t3 + 4 4 t3 + 4 4 t33 + 1,
t3 ∈ T3 ,
µ(t) = (µ1 (t1 ), µ2 (t2), µ3 (t3 )) q √ √ 4 3 4 = 3, 0, 6 t3 + 4 t3 + 4 t3 + 1 ,
t = (t1 ,t2 ,t3 ) ∈ Λ3 .
Exercise 1.6.16. Let Λ3 = 2Z × 3Z × 4Z. Find µ(t). Definition 1.6.17. Let f : Λ → R. We introduce the following notations. f σ (t) =
f (σ1 (t1 ), σ2 (t2 ), . . ., σn (tn )),
fiσi (t) =
f (t1 , . . .,ti−1 , σi (ti ),ti+1, . . .,tn) ,
σi σi ...σil
fi1 i12 ...i2l
(t) =
f (. . ., σi1 (ti1 ), . . ., σi2 (ti2 ), . . ., σil (til ), . . .),
where 1 ≤ i1 < i2 < . . . < il ≤ n, im ∈ N, m ∈ {1, 2, . . ., l}, l ∈ N.
Elements of the Time Scale Calculus Example 1.6.18. Let Λ2 = 2Z × 2N . Here, T1 = 2Z, σ1 (t1) = t1 + 2,
t1 ∈ T1 ,
T2 = 2N , σ2 (t2 ) = 2t2 ,
Let f (t1 ,t2 ) = t12 + t2 ,
(t1,t2 ) ∈ Λ2 .
Hence, f σ (t) =
f (σ1 (t1 ), σ2 (t2 ))
= (σ1 (t1 ))2 + σ2 (t2 ) = (t1 + 2)2 + 2t2 = t12 + 4t1 + 2t2 + 4, f1σ1 (t) =
f (σ1 (t1 ),t2)
= (σ1 (t1 ))2 + t2 = (t1 + 2)2 + t2 = t12 + 4t1 + t2 + 4, f2σ2 (t) =
f (t1 , σ2 (t2))
= t12 + σ2 (t2 ) = t12 + 2t2 ,
t ∈ Λ2 .
t2 ∈ T2 .
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Exercise 1.6.19. Let Λ3 = N × N × 3Z and f : Λ3 → R be defined as f (t) = t1t2t3 ,
t = (t1,t2 ,t3 ) ∈ Λ3 .
Find 1. f σ (t), 2. f 1σ1 (t), 3. f 2σ2 (t), 4. f 3σ3 (t), σ1 σ2 5. f 12 (t), σ1 σ3 (t), 6. f 13 σ2 σ3 7. f 23 (t),
8. g(t) = f σ (t) + f 1σ1 (t). Definition 1.6.20. Let f : Λn → R. We introduce the following notations. f ρ (t) = ρ
fi i (t) = ρi ρi ...ρil
fi1 i12 ...i2l
(t) =
f (ρ1 (t1 ), ρ2 (t2), . . ., ρn (tn )), f (t1 , . . .,ti−1 , ρi (ti),ti+1 , . . .,tn ) , f (. . ., ρi1 (ti1 ), . . ., ρi2 (ti2 ), . . ., ρil (til ), . . .) ,
where 1 ≤ i1 < i2 < . . . < il ≤ n, im ∈ N, m ∈ {1, 2, . . ., n}. Definition 1.6.21. We set Λκn = Tκ1 × Tκ2 × . . . × Tκn , Λκi i n = T1 × . . . × Ti−1 × Tκi × Ti+1 × . . . × Tn , κi κi ...κil n
Λi1 i12 ...i2 l
i = 1, 2, . . ., n,
= . . . × Tκi1 × . . . × Tκi2 × . . . × Tκil × . . . ,
where 1 ≤ i1 < i2 < . . . < il ≤ n, im ∈ N, m = 1, 2, . . ., l.
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Remark 1.6.22. If (i1 , i2 , . . ., il ) = (1, 2, . . ., n), then 2 ...κl n Λκi11iκ2 ...i = Λκn . l
Definition 1.6.23. Assume that f : Λn → R is a function and let t ∈ Λκi i n . We define ∂ f (t1 ,t2 , . . .,tn) ∂ f (t) ∂f = = (t) = ft∆i i (t) ∆i t i ∆i t i ∆i t i to be the number, provided it exists, with the property that for any εi > 0, there exists a neighbourhood Ui = (ti − δi ,ti + δi ) ∩ Ti , for some δi > 0, such that f (t1 , . . .,ti−1 , σi (ti ),ti+1, . . .,tn ) − f (t1 , . . .,ti−1 , si,ti+1 , . . .,tn)
− ft∆i i (t)(σi (ti) − si ) ≤ εi |σi (ti) − si |for allsi ∈ Ui . (1.6.1)
We call ft∆i i (t) the partial delta derivative (or partial Hilger derivative) of f with respect to ti at t. We say that f is partial delta differentiable (or partial Hilger differentiable) with respect to ti in Λκi i n if ft∆i i (t) exists for all t ∈ Λκi i n . The function ft∆i i : Λκi i n → R is said to be the partial delta derivative (or partial Hilger derivative) with respect to ti of f in Λκi i n . Theorem 1.6.24. The partial delta derivative is well defined. Example 1.6.25. Let f (t) = t1t2t3 , t = (t1 ,t2 ,t3 ) ∈ Λ3 . We will prove that ft∆1 1 (t) = t2t3 . Indeed, for every ε1 > 0, there exists a δ1 > 0 such that for every s1 ∈ (t1 −
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Svetlin G. Georgiev
δ1 ,t1 + δ1 ), s1 ∈ T1 , we have | f (σ1 (t1),t2 ,t3 ) − f (s1 ,t2 ,t3 ) − t2 t3 (σ1 (t1 ) − s1 )| = |σ1 (t1 )t2t3 − s1 t2 t3 − t2 t3 (σ1 (t1 ) − s1 )| = 0 ≤ ε1 |σ1 (t1 ) − s1 |. Remark 1.6.26. For t ∈ Λn and si ∈ Ti , we write tsi = (t1 , . . .,ti−1 , si ,ti+1 , . . .,tn ),
i ∈ {1, 2, . . ., n}.
Then we can rewrite (1.6.1) in the form | f iσi (t) − f (tsi ) − ft∆i i (t)(σi(ti ) − si )| ≤ εi |σi (ti) − si |.
(1.6.2)
Theorem 1.6.27. Let f : Λn → R be a function and t ∈ Λκi i n . If f is delta differentiable with respect to ti at t, then lim f (tsi ) = f (t).
(1.6.3)
si →ti
Theorem 1.6.28. Let f : Λn → R, t ∈ Λκi i n , and lim f (tsi ) = f (t).
(1.6.4)
si →ti
If ti < σi (ti ), then f is delta differentiable with respect to ti at t and ft∆i i (t) =
fiσi (t) − f (t) . µi (ti )
(1.6.5)
Example 1.6.29. Let Λ2 = N × 2N and define f : Λ2 → R by f (t) = t12t2 − 2t1 ,
t = (t1 ,t2 ) ∈ Λ2 .
Then σ1 (t1) = t1 + 1,
t1 ∈ T1 ,
σ2 (t2 ) = 2t2 ,
t 2 ∈ 2N .
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39
Hence, t1 < σ1 (t1 ),
t1 ∈ N,
t2 < σ2 (t2 ),
t 2 ∈ 2N .
Therefore, ft∆1 1 (t) = =
f1σ1 (t) − f (t) σ1 (t1 ) − t1 (σ1 (t1 ))2t2 − 2σ1 (t1 ) − t12 t2 + 2t1 t1 + 1 − t1
= (t1 + 1)2t2 − 2(t1 + 1) − t12 t2 + 2t1 = t12 t2 + 2t1 t2 + t2 − 2t1 − 2 − t12 t2 + 2t1 = 2t1t2 + t2 − 2, ft∆2 2 (t) =
(t1 ,t2 ) ∈ Λκ1 1 2 ,
f2σ2 (t) − f (t) σ2 (t2 ) − t2
=
t12 σ2 (t2 ) − 2t1 − (t12 t2 − 2t1 ) 2t2 − t2
=
2t12t2 − 2t1 − t12 t2 + 2t1 t2
= t12 ,
(t1 ,t2) ∈ Λκ2 2 2 .
Theorem 1.6.30. Let t ∈ Λκi i n and ti = σi (ti). Then f is partial delta differentiable with respect to ti at t if and only if the limit lim
si →ti
f (t) − f (tsi ) ti − si
exists as a finite number. In this case, ft∆i i (t) = lim
si →ti
f (t) − f (tsi ) . ti − si
(1.6.6)
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Example 1.6.31. Let 2
Λ = where
N ! 1 1 ∪ , 1 × N, 2 2
1 2 2 , 1 is the real number interval, and define f : Λ → R by f (t) = t12 t2 ,
We will find ft∆1 1
Hence, σ1
1 2
1 2 ,t2
t = (t1 ,t2 ) ∈ Λ2 .
, t2 ∈ N. We note that N 1 , T1 = 2
T2 = N.
= 12 . Thus,
lim
s1 → 21
f
1 2 ,t2 1 2
− f (s1 ,t2 )
− s1
= = =
lim
s1 → 12
1 2 4 t2 − s1 t2 1 2 − s1 1 2
lim
s1 → 12
lim
s1 → 12
− s1
1 2
1 2
+ s1 t2
− s1 1 + s1 t2 2
= t2 . Consequently, ft∆1 1
1 ,t2 = t2 . 2
Exercise 1.6.32. Let Λ2 = N × 4Z and define f : Λ2 → R by f (t) = t13 + 3t12 + t1 t2 , Find ft∆1 1 (1,t2).
t = (t1 ,t2 ) ∈ Λ2 .
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Elements of the Time Scale Calculus
Theorem 1.6.33. Let t ∈ Λκi i n . Suppose f : Λn → R is a function that is partial delta differentiable with respect to ti at t. If α ∈ R, then α f is partial delta differentiable with respect to ti at t and (α f )t∆i i (t) = α ft∆i i (t).
Theorem 1.6.34. Let t ∈ Λκi i n . Assume f , g : Λn → R are partial delta differentiable with respect to ti at t. Then f + g is partial delta differentiable with respect to ti at t and ( f + g)t∆i i (t) = ft∆i i (t) + gt∆i i (t). Theorem 1.6.35. Let t ∈ Λκi i n . Assume f , g : Λn → R are partial delta differentiable with respect to ti at t. Then f g is partial delta differentiable with respect to ti at t and ( f g)t∆i i (t) = ft∆i i (t)g(t) + f iσi (t)gt∆i i (t) = f (t)gt∆i i (t) + ft∆i i (t)gσi i (t). Example 1.6.36. Let Λ2 = N × N20 and define h : Λ2 → R by h(t) = (t12 + 2t1 )(t13 + t2 ),
t ∈ Λ2 .
Here, T1 = N and T2 = N20 . Then σ1 (t1 ) = t1 + 1,
t1 ∈ T1 ,
√ σ2 (t2 ) = (1 + t2 )2 ,
We will find ht∆11 (t), t ∈ Λκ1 1 2 . Let f (t) = t12 + 2t1 ,
g(t) = t13 + t2 ,
t ∈ Λ2 .
t2 ∈ T2 .
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Svetlin G. Georgiev
Hence, h(t) = f (t)g(t), t ∈ Λ2 . Then, for t ∈ Λκ1 1 2 , we have ht∆11 (t) =
ft∆1 1 (t)g(t) + f 1σ1 (t)gt∆11 (t)
= (σ1 (t1) + t1 + 2)g(t) + ((σ1 (t1 ))2 + 2σ1 (t1 ))((σ1(t1 ))2 + t1 σ1 (t1 ) + t12 ) = (t1 + 1 + t1 + 2)(t13 + t2 ) + ((t1 + 1)2 + 2t1 + 2)((t1 + 1)2 + t1 (t1 + 1) + t12 ) = (2t1 + 3)(t13 + t2 ) + (t12 + 4t1 + 3)(3t12 + 3t1 + 1) = 2t14 + 2t1 t2 + 3t13 + 3t2 + 3t14 + 3t13 + t12 + 12t13 + 12t12 + 4t1 + 9t12 + 9t1 + 3 = 5t14 + 18t13 + 22t12 + 13t1 + 2t1 t2 + 3t2 + 3. Theorem 1.6.37. Let f , g : Λn → R be partial delta differentiable with respect to ti at t ∈ Λκi i n . Assume gσi i (t)g(t) 6= 0. Then gf is partial delta differentiable with respect to ti at t and ∆i ft∆i (t)g(t) − f (t)gt∆i i(t) f (t) = i . g ti gσi i (t)g(t) Exercise 1.6.38. Let Λ2 = N × N0 and define h : Λ2 → R by h(t) =
t1 + t2 , t1 − t2 + 1
t ∈ Λ2 .
Find ht∆11 (t) for t ∈ Λκ1 1 2 , (t1 − t2 + 2)(t1 − t2 + 1) 6= 0, and ht∆22 (t) for t ∈ Λκ2 2 2 , (t1 − t2 )(t1 − t2 + 1) 6= 0. Definition 1.6.39. For a function f : Λn → R, we shall talk about the second∆∆ order partial delta derivative with respect to ti and t j , i, j ∈ {1, 2, . . ., n}, ftitij j , κκ n provided ft∆i i is partial delta differentiable with respect to t j on Λi ji j = κn Λκi i n j j with partial delta derivative ∆ j ∆∆ κκ n ftitij j = ft∆i i : Λi ji j → R. tj
43
Elements of the Time Scale Calculus For i = j, we will write ∆2
ft∆itii ∆i = ft 2 i . i
Similarly, we define higher-order partial delta derivatives ∆ ∆ ...∆l
ftitij ...tj l
κ κ ...κl n
i j : Λi j...l
→ R.
For t ∈ Λn , we define σ2 (t) = σ(σ(t)) = (σ1 (σ1 (t1 )), σ2 (σ2 (t2)), . . ., σn (σn (tn))) and ρ2 (t) = ρ(ρ(t)) = (ρ1 (ρ1 (t1 )), ρ2 (ρ2 (t2 )), . . ., ρn (ρn (tn ))), and σm (t) and ρm (t) for m ∈ N are defined accordingly. Finally, we put ρ0 (t) = σ0 (t) = t,
∆0
fti i = f ,
κ0 n
Λi i = Λn .
Example 1.6.40. Let Λ3 = N × N0 × Z. Here T1 = N,
T2 = N0 ,
T3 = Z
and σ1 (t1) = t1 + 1,
t1 ∈ T1 ,
σ2 (t2 ) = t2 + 1,
σ3 (t3 ) = t3 + 1,
t2 ∈ T2 ,
t3 ∈ T3 .
Hence, σ2 (t) = (σ1 (σ1 (t1 )), σ2 (σ2 (t2)), σ3 (σ3 (t3 ))) = (σ1 (t1 ) + 1, σ2 (t2 ) + 1, σ3 (t3 ) + 1) = (t1 + 1 + 1,t2 + 1 + 1,t3 + 1 + 1) = (t1 + 2,t2 + 2,t3 + 2).
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Exercise 1.6.41. Let Λ2 = 3N × Z. Find σ2 (t) and ρ2 (t), t ∈ Λ2 . (m)
Theorem 1.6.42 (Leibniz’s Formula). Let Sik be the set consisting of all possible strings of length m, containing exactly k times σi and m − k times ∆i . If ∆k
(m)
ftαm−k exists for any α ∈ Sik and gt ki exists for any k ∈ {0, 1, . . ., m}. Then i
i
∆m i tim
( f g) holds for any m ∈ N.
m
=
∑ ∑
k=0
(m)
α∈Sik
∆k
ftαm−k gt ki i
(1.6.7)
i
Remark 1.6.43. We note that (1.6.7) holds for m = 0 with the convention that
∑ ftα = f .
α∈0/
k i
Definition 1.6.44. We put Λnκ = T1κ × T2κ × . . . × Tnκ , Λniκi = T1 × . . . × Ti−1 × Tiκ × Ti+1 × . . . × Tn , Λni1 i2 ...il κi
κ ...κil 1 i2
i = 1, 2, . . ., n,
= . . . × Ti1 κ × . . . × Ti2 κ × . . . × Til κ × . . .,
where 1 ≤ i1 < i2 < . . . < il ≤ n, im ∈ N, m = 1, 2, . . ., l. Remark 1.6.45. If (i1 , i2 , . . ., il ) = (1, 2, . . ., n), then Λni1 i2 ...il κ1 κ2 ...κl = Λnκ . Definition 1.6.46. Assume that f : Λn → R is a function and let t ∈ Λiκi n . We define ∂f ∂ f (t1 ,t2 , . . .,tn) ∂ f (t) = = (t) = ft∇i i (t) ∇iti ∇iti ∇i ti to be the number, provided it exists, with the property such that for any εi > 0, there exists a neighbourhood Ui = (ti − δi ,ti + δi ) ∩ Ti
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45
for some δi > 0 such that f (t1 , . . .,ti−1 , ρi (ti),ti+1 , . . .,tn ) − f (t1 , . . .,ti−1 , si ,ti+1 , . . .,tn )
− ft∇i i (t)(ρi(ti ) − si ) ≤ εi |ρi (ti) − si | (1.6.8)
for all si ∈ Ui. We call ft∇i i (t) the partial nabla derivative of f with respect to ti at t. We say that f is partial nabla differentiable with respect to ti in Λniκi if ft∇i i (t) exists for all t ∈ Λniκi . The function ft∇i i : Λniκi → R is said to be the partial nabla derivative with respect to ti of f in Λniκi . Theorem 1.6.47. The partial nabla derivative is well defined. Theorem 1.6.48. Let f : Λn → R be a function and t ∈ Λniκi . If f is nabla differentiable with respect to ti at t, then lim f (tsi ) = f (t).
si →ti
Theorem 1.6.49. Let f : Λn → R, t ∈ Λniκi , and lim f (tsi ) = f (t).
si →ti
If ρi (ti ) < ti , then f is nabla differentiable with respect to ti at t and ρ
ft∇i i (t) =
fi i (t) − f (t) . ρi (ti) − ti
Theorem 1.6.50. Let t ∈ Λniκi and ti = ρi (ti). Then f is partial nabla differentiable with respect to ti at t if and only if the limit lim
si →ti
f (t) − f (tsi ) ti − si
exists as a finite number. In this case, ft∇i i (t) = lim
si →ti
f (t) − f (tsi ) . ti − si
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Svetlin G. Georgiev
Theorem 1.6.51. Let t ∈ Λniκi . Suppose f : Λn → R is a function that is partial nabla differentiable with respect to ti at t. If α ∈ R, then α f is partial nabla differentiable with respect to ti at t and (α f )t∇i i (t) = α ft∇i i (t). Theorem 1.6.52. Let t ∈ Λniκi . Suppose f , g : Λn → R are partial nabla differentiable with respect to ti at t. Then f g is partial nabla differentiable with respect to ti at t and ρ
ρ
( f g)t∇i i (t) = ft∇i i (t)g(t) + f i i (t)gt∇i i (t) = f (t)gt∇i i (t) + ft∇i i (t)gi i (t). Theorem 1.6.53. Let f , g : Λn → R be partial nabla differentiable with respect ρ to ti at t ∈ Λniκi . Assume gi i (t)g(t) 6= 0. Then gf is partial nabla differentiable with respect to ti at t and ∇i ft∇i (t)g(t) − f (t)gt∇i i(t) f (t) = i . ρ g ti gi i (t)g(t) (m)
Theorem 1.6.54 (Leibniz’s Formula). Let Qik be the set consisting of all possible strings of length m, containing exactly k times ρi and m − k times ∇i . If β
∇k
(m)
ft m−k exists for any β ∈ Qik and gt k i exists for any k ∈ {0, 1, . . ., m}. Then i
i
∇m i tim
( f g) holds for any m ∈ N.
m
=
∑ ∑
k=0
(m)
β∈Qik
β
∇k
ft m−k gt k i i
i
We can define higher-order nabla derivatives and also mixed derivatives obtained by combining both delta and nabla differentiations such as, for instance, ∇∇ ∆ ∇∆ ftit ji j or ftit ji tl j l . Example 1.6.55. Let Λ2 = N × Z, f (t) = t12t2 + t1t22 + t22 , t ∈ Λ2 . Here, T1 = N,
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47
T2 = Z, and σ1 (t1 ) = t1 + 1,
t1 ∈ T1 ,
σ2 (t2 ) = t2 + 1,
t2 ∈ T2 ,
ρ1 (t1 ) =
t1 − 1 1
ρ2 (t2 ) = t2 − 1,
ift1 ∈ T1 ,
t1 ≥ 2,
ift1 = 1, t2 ∈ T2 .
Hence, ft∆1 1 (t) = (σ1 (t1 ) + t1 )t2 + t22 = (t1 + 1 + t1 )t2 + t22 = 2t1t2 + t22 + t2 , ft∆1t12∇2 (t) =
ft∆1 1
∇2 t2
t ∈ Λκ1 1 2 ,
(t)
= 2t1 + ρ2 (t2 ) + t2 + 1 = 2t1 + t2 − 1 + t2 + 1 = 2t1 + 2t2 ,
12 t ∈ Λκ12κ . 2
12 . Exercise 1.6.56. Let Λ2 = 3N × Z, f (t) = t12t22 , t ∈ Λ2 . Find ft∆1 t12∇2 (t), t ∈ Λκ12κ 2
Answer. 4t1 (2t2 − 1). Now we will introduce the conception for multiple integration on time scales. Suppose ai < bi are points in Ti and [ai , bi ) is the half-closed bounded
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Svetlin G. Georgiev
interval in Ti , i ∈ {1, . . ., n}. Let us introduce a “rectangle” in Λn = T1 × T2 × . . . × Tn by R = [a1 , b1 ) × [a2 , b2 ) × . . . [an , bn ) = {(t1 ,t2 , . . .,tn) : ti ∈ [ai , bi ), i = 1, 2, . . ., n}. Let ai = ti0 < ti1 < . . . < tiki = bi . Definition 1.6.57. We call the collection of intervals n o ji −1 ji Pi = [ti ,ti ) : ji = 1, . . ., ki , i = 1, 2, . . ., n,
a ∆i -partition of [ai , bi) and denote the set of all ∆i -partitions of [ai, bi ) by Pi ([ai, bi)). Definition 1.6.58. Let j −1
R j1 j2 ... jn = [t11
We call the collection
j
j −1
,t11 ) × [t22
1 ≤ ji ≤ ki ,
j
j −1
,t22 ) × . . . × [tnn
j
,tnn ) (1.6.9)
i = 1, 2, . . ., n.
P = R j1 j2 ... jn : 1 ≤ ji ≤ ki , i = 1, 2, . . ., n
(1.6.10)
a ∆-partition of R, generated by the ∆i -partitions Pi of [ai , bi), and we write P = P1 × P2 × . . . × Pn .
The set of all ∆-partitions of R is denoted by P (R). Moreover, for a bounded function f : R → R, we set M = sup{ f (t1,t2 , . . .,tn ) : (t1 ,t2 , . . .,tn ) ∈ R}, m = inf{ f (t1 ,t2 , . . .,tn ) : (t1,t2 , . . .,tn ) ∈ R}, M j1 j2 ... jn
= sup{ f (t1,t2 , . . .,tn ) : (t1 ,t2 , . . .,tn ) ∈ R j1 j2 ... jn },
m j1 j2 ... jn
= inf{ f (t1 ,t2 , . . .,tn ) : (t1,t2 , . . .,tn ) ∈ R j1 j2 ... jn }.
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Elements of the Time Scale Calculus
Definition 1.6.59. The upper Darboux ∆-sum U( f , P) and the lower Darboux ∆-sum L( f , P) with respect to P are defined by k1
U( f , P) =
k2
kn
∑ ∑ . . . ∑ M j j ... j (t1j
1
1 2
j1 =1 j2 =1
jn =1
k1
kn
n
− t1j1 −1 )(t2j2 − t2j2 −1 ) . . .(tnjn − tnjn −1 )
and L( f , P) =
k2
∑ ∑
...
j1 =1 j2 =1
j −1
j
∑ mj
1 j2 ... jn
jn =1
(t11 − t11
j
j −1
)(t22 − t22
) . . .(tnjn − tnjn −1 ).
Remark 1.6.60. We note that k1
U( f , P) ≤ M
k2
∑ ∑
kn
...
j1 =1 j2 =1
j1
∑ (t1
jn =1
j −1
− t11
j
j −1
)(t22 − t22
) . . .(tnjn − tnjn −1 )
≤ M(b1 − a1 )(b2 − a2 ) . . .(bn − an ) and k1
L( f , P) ≥ m
kn
k2
∑ ∑
j1 =1 j2 =1
...
j1
∑ (t1
jn =1
j −1
− t11
j
j −1
)(t22 − t22
) . . .(tnjn − tnjn −1 )
≥ m(b1 − a1 )(b2 − a2 ) . . .(bn − an ), i.e., m(b1 − a1 )(b2 − a2 ) . . .(bn − an ) ≤L( f , P) ≤U( f , P)
(1.6.11)
≤M(b1 − a1 )(b2 − a2 ) . . .(bn − an ). Definition 1.6.61. The upper Darboux ∆-integral U( f ) of f over R and the lower Darboux ∆-integral L( f ) of f over R are defined by U( f ) = inf{U( f , P) : P ∈ P (R)}andL( f ) = sup{L( f , P) : P ∈ P (R)}.
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From (1.6.11), it follows that U( f ) and L( f ) are finite real numbers. Definition 1.6.62. We say that f is ∆-integrable over R provided L( f ) = U( f ). In this case, we write Z
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆ntn
for this common value. We call this integral the Darboux ∆-integral. Remark 1.6.63. For a given rectangle V = [c1 , d1 ) × [c2 , d2 ) × . . . × [cn , dn ) ⊂ Λn , the “area” of V , i.e., (d1 − c1 )(d2 − c2 ) . . .(dn − cn ), is denoted by m(V ). Definition 1.6.64. Let P, Q ∈ P (R) and P = P1 × P2 × . . . × Pn ,
Q = Q1 × Q2 × . . . × Qn ,
where Pi , Qi ∈ P ([ai, bi )). We say that Q is a refinement of P provided Qi is a refinement of Pi for all i ∈ {1, 2, . . ., n}. Theorem 1.6.65. Let f be a bounded function on R. If P and Q are ∆-partitions of R and Q is a refinement of P, then L( f , P) ≤ L( f , Q) ≤ U( f , Q) ≤ U( f , P), i.e., refining of a partition increases the lower sum and decreases the upper sum. Definition 1.6.66. Suppose P = P1 × P2 × . . . × Pn andQ = Q1 × Q2 × . . . × Qn , where Pi , Qi ∈ P ([ai, bi )), i ∈ {1, 2, . . ., n}, are two ∆-partitions of R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ). If Pi is generated by a set {ti0 ,ti1 , . . .,tiki }, whereai = ti0 < ti1 < . . . < tiki = bi ,
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Elements of the Time Scale Calculus and Qi is generated by a set p
p
{τ0i , τ2i , . . ., τi i }, whereai = τ0i < τ1i < . . . < τi i = bi , then, by P + Q = (P1 + Q1 ) × (P2 + Q2 ) × . . . × (Pn + Qn ),
we denote the ∆-partition of R generated by
p
Pi + Qi = {ti0 ,ti1 , . . .,tiki } ∪ {τ0i , τ1i , . . ., τi i },
i = 1, 2, . . ., n.
Remark 1.6.67. Obviously, P + Q is a refinement of both P and Q. Theorem 1.6.68. If f is a bounded function on R and if P and Q are any two ∆-partitions of R, then L( f , P) ≤ U( f , Q), i.e., every lower sum is less than or equal to every upper sum. Theorem 1.6.69. If f is a bounded function on R, then L( f ) ≤ U( f ). Theorem 1.6.70. If L( f , P) = U( f , P) for some P ∈ P (R), then the function f is ∆-integrable over R and Z
R
f (t1 ,t2 , . . .,tn)∆1 t1 ∆2t2 . . .∆n tn = L( f , P) = U( f , P).
Theorem 1.6.71. A bounded function f on R is ∆-integrable if and only if for each ε > 0, there exists a P ∈ P (R) such that U( f , P) − L( f , P) < ε.
(1.6.12)
Remark 1.6.72. Let T be a time scale with forward jump operator σ. We note that for every δ > 0, there exists at least one partition P1 ∈ P ([a, b]) generated by a set {t0 ,t1,t2 , . . .,tn } ⊂ [a, b], wherea = t0 < t1 < . . . < tn = b, such that for each i ∈ {1, 2, . . ., n} either ti − ti−1 < δ or ti − ti−1 > δandσ(ti−1 ) = ti .
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Svetlin G. Georgiev
Definition 1.6.73. We denote by Pδ ([a, b)) the set of all P1 ∈ P ([a, b)) that possess the property indicated in Remark 1.6.72. Further, by Pδ (R), we denote the set of all P ∈ P (R) such that P = P1 × P2 × . . . × Pn , wherePi ∈ Pδ ([ai, bi )),
i = 1, 2, . . ., n.
Theorem 1.6.74. Let P0 ∈ P (R) be given by P0 = P10 × P20 × . . . × Pn0 in which Pi0 ∈ P ([ai, bi)), i ∈ {1, 2, . . ., n}, is generated by a set A0i = {ti00 ,ti01 , . . .,ti0n } ⊂ [ai , bi ], whereai = ti00 < ti01 < . . . < ti0n = bi . i
i
Then, for each P ∈ Pδ (R), we have L( f , P0 + P) − L( f , P) ≤ (M − m)Dn−1 (n1 + n2 + . . . + nn − n)δ and U( f , P) −U( f , P+ P0 ) ≤ (M − m)Dn−1 (n1 + n2 + . . . + nn − n)δ, where D = maxi∈{1,2,...,n} {bi − ai }, and M and m are defined as above. Theorem 1.6.75. A bounded function f on R is ∆-integrable if and only if for each ε > 0, there exists a δ > 0 such that P ∈ Pδ (R)impliesU( f , P) − L( f , P) < ε.
(1.6.13)
Theorem 1.6.76. For every bounded function f on R, the Darboux ∆-sums L( f , P) and U( f , P) evaluated for P ∈ Pδ (R) have limits as δ → 0, uniformly with respect to P, and lim L( f , P) = L( f )and lim U( f , P) = U( f ).
δ→0
δ→0
Definition 1.6.77. Let f be a bounded function on R and P ∈ P (R). In each “rectangle” R j1 j2 ... jn , 1 ≤ ji ≤ ki , i = 1, 2, . . ., n, choose a point ξ j1 j2 ... jn and form the sum n
S=∑
ki
∑
i=1 ji =1
j
j −1
f (ξ j1 j2 ... jn )(t1i − t1i
) . . .(tnjn − tnjn −1 ).
(1.6.14)
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Elements of the Time Scale Calculus
We call S a Riemann ∆-sum of f corresponding to P ∈ P (R). We say that f is Riemann ∆-integrable over R if there exists a number I such that, for each ε > 0, there exists a δ > 0 such that |S − I| < ε
for every Riemann ∆-sum S of f corresponding to any P ∈ Pδ (R), independent of the choice of the point ξ j1 j2 ... jn ∈ R j1 j2 ... jn for 1 ≤ ji ≤ ki , i = 1, 2, . . ., n. The number I is called the Riemann ∆-integral of f over R. We write I = lim S. δ→0
Theorem 1.6.78. The Riemann ∆-integral is well defined. Remark 1.6.79. Note that in the Riemann definition of the integral, we need not assume the boundedness of f in advance. However, it follows that the Riemann integrability of a function f over R implies its boundedness on R. Theorem 1.6.80. A bounded function on R is Riemann ∆-integrable if and only if it is Darboux ∆-integrable, in which case the values of the integrals are equal. Remark 1.6.81. In the definition of Z
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆ntn
with R = [a1 , b1 ) × . . . × [an , bn ), we assumed that ai < bi , i ∈ {1, . . ., n}. We extend the definition to the case ai = bi for some i ∈ {1, 2, . . ., n} by setting Z
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆ntn = 0
(1.6.15)
if ai = bi for some i ∈ {1, . . ., n}. Theorem 1.6.82. Let a = (a1 , . . ., an ) ∈ Λn andb = (b1 , . . ., bn ) ∈ Λn with ai ≤ bi for all i ∈ {1, . . ., n}. Every constant function f (t1 ,t2, . . .,tn ) = Afor(t1 ,t2, . . .,tn ) ∈ R = [a1 , b1 ) × . . . × [an , bn ) is ∆-integrable over R and Z
n R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆n tn = A ∏(bi − ai ). i=1
(1.6.16)
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Svetlin G. Georgiev
Theorem 1.6.83. Let t 0 = (t10 , . . .,tn0 ) ∈ Λn . Every function f : Λn → R is ∆integrable over R = R(t 0) = [t10, σ1 (t10 )) × . . . × [tn0 , σn (tn0 )), and Z
n
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn = ∏ µi (ti0 ) f (t10 , . . .,tn0).
(1.6.17)
i=1
Theorem 1.6.84. Let a = (a1 , . . ., an ) ∈ Λn andb = (b1 , . . ., bn ) ∈ Λn with ai ≤ bi for all i ∈ {1, 2, . . ., n}. If Ti = R for every i ∈ {1, 2, . . ., n}, then every bounded function f on R = [a1 , b1 )×[a2 , b2 )×. . .×[an , bn ) is ∆-integrable if and only if f is Riemann integrable on R in the classical sense, and in this case Z
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆ntn =
Z
R
f (t1 ,t2 , . . .,tn )dt1 dt2 . . .dtn ,
where the integral on the right-hand side is the ordinary Riemann integral. Theorem 1.6.85. Let a = (a1 , . . ., an ) ∈ Λn andb = (b1 , . . ., bn ) ∈ Λn with ai ≤ bi for all i ∈ {1, 2, . . ., n}. If Ti = Z for all i ∈ {1, 2, . . ., n}, then every function defined on R = [a1 , b1 ) × [a2, b2 ) × . . .× [an , bn ) is ∆-integrable over R, and Z
R
=
f (t1 ,t2 , . . .,tn)∆1 t1 ∆2t2 . . .∆n tn
0
if ai = bi for some i ∈ {1, 2, . . ., n}
b1 −1 b2 −1 −1 f (r1 , r2 , . . ., rn ) otherwise. ∑r1 =a1 ∑r2 =a2 . . . ∑brnn=a n
(1.6.18)
55
Elements of the Time Scale Calculus Example 1.6.86. Let T1 = T2 = Z. We consider I=
Z 4Z 8 0
1
t2 (2t1 + 1)∆1t1 ∆2 t2 .
Here, f (t1 ,t2 ) = t2 (2t1 + 1), σ1 (t1) = t1 + 1,
t1 ∈ T1 ,
(t1 ,t2 ) ∈ T1 × T2 , σ2 (t2 ) = t2 + 1,
t2 ∈ T2 .
We note that, for g(t1,t2 ) = t2t12 , we have gt∆11 (t1 ,t2 ) = t2 (σ1 (t1 ) + t1 ) = t2 (t1 + 1 + t1 ) = t2 (2t1 + 1). Therefore, Z 8 1
t2 (2t1 + 1)∆1 t1 = g(8,t2) − g(1,t2 ) = 64t2 − t2 = 63t2 .
Hence, I=
Z 4 0
63t2∆2 t2 = 63
Z 4 0
t 2 ∆2 t 2 .
Since, for h(t2 ) = t22 , we have 1 ∆2 (h (t2 ) − 1) = 2
1 (σ2 (t2 ) + t2 − 1) 2
=
1 (t2 + 1 + t2 − 1) 2
= t2 ,
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Svetlin G. Georgiev
we get I = 63 = =
63 2
Z 4 1
2
0
Z 4 0
(h∆2 (t2 ) − 1)∆2t2
h∆2 (t2 )∆2t2 −
63 2
Z 4 0
∆2 t 2
63 (h(4) − h(0)) − 126 2
= 504 − 126 = 378. Exercise 1.6.87. Let T1 = 2N and T2 = 4Z. Compute the integral −2
Z 8Z 8 t2 0
2
t1
sin
t1 3t1 sin ∆1t1 ∆2t2 . 2 2
Answer. −32 sin3 sin5. Note that Λn is a complete metric space with the metric d defined by s n
d(t, s) =
∑ (ti − si )2fort = (t1,t2, . . .,tn),
i=1
s = (s1 , s2 , . . ., sn ) ∈ Λn ,
and also with the equivalent metric d(t, s) =
max i∈{1,2,...,n}
{|ti − si |} .
Definition 1.6.88. A function f : Λn → R is said to be continuous at t ∈ Λn if for every ε > 0, there exists a δ > 0 such that | f (t) − f (s)| < ε for all points s ∈ Λn satisfying d(t, s) < δ. Remark 1.6.89. If t is an isolated point of Λn , then every function f : Λn → R is continuous at t. In particular, if Ti = Z for all i ∈ {1, 2, . . ., n}, then every function f : Λn → R is continuous at each point of Λn .
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Theorem 1.6.90. Every continuous function on K = [a1 , b1 ] × [a2 , b2 ] × . . . × [an , bn ] is ∆-integrable over R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ). Definition 1.6.91. We say that a function φ : [α, β] → R satisfies the Lipschitz condition if there exists a constant B > 0, so-called Lipschitz constant, such that |φ(u) − φ(v)| ≤ B|u − v|for allu, v ∈ [α, β]. Example 1.6.92. Let φ : [0, 1] → R be defined by φ(x) = x2 + 1, x ∈ [0, 1]. Then, for x, y ∈ [0, 1], we have |φ(x) − φ(y)| = |x2 + 1 − y2 − 1| = |x2 − y2 | = |x − y||x + y| ≤ |x − y|(|x| + |y|) ≤ 2|x − y|, i.e., φ satisfies the Lipschitz condition with Lipschitz constant B = 2. Theorem 1.6.93. Let f be bounded and ∆-integrable over R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ) and let M and m be its supremum and infimum over R, respectively. If φ : [m, M] → R is a function satisfying the Lipschitz condition, then the composite function h = φ ◦ f is ∆-integrable over R. Theorem 1.6.94. Let f be a bounded function that is ∆-integrable over R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ). If a0i , b0i ∈ [ai , bi ] with a0i < b0i for all i ∈ {1, 2, . . ., n}, then f is ∆-integrable over R0 = [a01 , b01 ) × [a02 , b02 ) × . . . × [a0n , b0n ).
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Theorem 1.6.95. Let f be a bounded function that is ∆-integrable on R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ). If α ∈ R, then α f is ∆-integrable on R and Z
R
α f (t1,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn = α
Z
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆n tn . (1.6.19)
Theorem 1.6.96. If f and g are bounded functions that are ∆-integrable over R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ), then f + g is ∆-integrable over R and Z
R
( f + g)(t1 ,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn
=
Z
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn +
Z
R
g(t1,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn . (1.6.20)
Theorem 1.6.97. If f and g are bounded functions that are ∆-integrable over R with f (t1 ,t2, . . .,tn ) ≤ g(t1 ,t2, . . .,tn )for all(t1 ,t2, . . .,tn ) ∈ R, then Z
R
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆ntn ≤
Z
R
g(t1,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆ntn .
Theorem 1.6.98. If f is a bounded function that is ∆-integrable over R, then so is | f |, and Z Z f (t1 ,t2, . . .,tn )∆1t1 ∆2 t2 . . .∆ntn ≤ | f (t1,t2 , . . .,tn)|∆1t1 ∆2 t2 . . .∆ntn . R
R
Theorem 1.6.99. If f is a bounded function that is ∆-integrable over R, then so is f 2 .
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59
Theorem 1.6.100. If f and g are ∆-integrable over R, then so is f g. Theorem 1.6.101. Let the rectangle R be the union of two disjoint rectangles R1 and R2 . If f is a bounded function that is ∆-integrable on each of R1 and R2 , then f is ∆-integrable over R and Z
R
f (t1 ,t2 , . . .,tn)∆1 t1 ∆2t2 . . .∆n tn =
Z
R1
+
Z
f (t1 ,t2, . . .,tn )∆1t1 ∆2 t2 . . .∆ntn
R2
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆n tn .
Now we will extend the definition for the multiple Riemann ∆-integral over more general sets in Λn , called Jordan ∆-measurable sets. Definition 1.6.102. Let E ⊂ Λn . A point t = (t1 ,t2 , . . .,tn ) ∈ Λn is called a boundary point of E if every open ball B(t, r) = {x ∈ Λn : d(t, x) < r} of radius r with center t contains at least one point of E \ {t} and at least one point of Λn \ E. The set of all boundary points of E is called the boundary of E and is denoted by ∂E. Definition 1.6.103. Let E ⊂ Λn . A point t = (t1 ,t2 , . . .,tn) ∈ Λn is called a ∆boundary point of E if every rectangle of the form V = [t1,t10 ) × [t2 ,t20 ) × . . . × [tn ,tn0 ) ⊂ Λn with ti0 ∈ Ti , t10 > ti, i = 1, 2, . . ., n, contains at least one point of E and at least one point of Λn \ E. The set of all ∆-boundary points of E is called the ∆boundary and is denoted by ∂∆ E. For i ∈ {1, 2, . . ., n}, we set T0i = Ti \ {max Ti }. If Ti does not have a maximum, then T0i = Ti , i ∈ {1, 2, . . ., n}. Evidently, for every point ti ∈ T0i , there exists an interval [αi, βi ) ⊂ Ti that contains the point ti , i ∈ {1, 2, . . ., n}. Definition 1.6.104. A point t 0 = (t10 ,t20 , . . .,tn0 ) ∈ Λn is called ∆-dense if every rectangle of the form V = [t10 ,t1 ) × [t20 ,t2 ) × . . . × [tn0 ,tn ) ⊂ Λn with ti ∈ Ti , ti > ti0 , i ∈ {1, 2, . . ., n}, contains at least one point distinct from t 0 . Otherwise, the point t 0 is called ∆-scattered.
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Remark 1.6.105. In fact, every ∆-dense point is a dense point in Λn and every ∆-scattered point is a scattered point in Λn . Let Λn0 = T01 × T02 × . . . × T0n . Definition 1.6.106. Let E ⊂ Λn0 be a bounded set and let ∂∆ E be its boundary. Assume R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn )
is a rectangle in Λn such that E ∪ ∂∆ E ⊂ R. Suppose P (R) denotes the set of all ∆-partitions of R of the type (1.6.9) and (1.6.10). For every P ∈ P (R), denote with J∗ (E, P) the sum of the areas of those subrectangles of P which are entirely contained in E, and let J ∗ (E, P) be the sum of the areas of the subrectangles of P each of which contains at least one point of E ∪ ∂∆ E. The numbers J∗ (E) = sup{J∗ (E, P) : P ∈ P (R)}andJ ∗ (E) = inf{J ∗ (E, P) : P ∈ P (R)} are called the inner and outer Jordan ∆-measure of E, respectively. The set E is said to be Jordan ∆-measurable if J∗ (E) = J ∗ (E), in which case this common value is called the Jordan ∆-measure of E, denoted by J(E). We note that 0 ≤ J∗ (E) ≤ J ∗ (E)
for every bounded set E ⊂ Λn0 . If E has Jordan ∆-measure zero, then J∗ (E) = J ∗ (E) = 0. Therefore, we have the following statement. Proposition 1.6.107. A bounded set E ⊂ Λn0 has Jordan ∆-measure zero if and only if for every ε > 0, the set E can be covered by a finite collection of rectangles of type j
j
j
j
V j = [α1 , β1 ) × [α2 , β2 ) × . . . × [αnj , βnj ) ⊂ Λn ,
j = 1, 2, . . ., m,
the sum of whose areas is less than ε: E⊂
m [
j=1
m
V j and ∑ m(V j ) < ε. j=1
If E is a set of Jordan ∆-measure zero, then so is any set E˜ ⊂ E.
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Proposition 1.6.108. The union of a finite number of bounded subsets E1 , E2 , . . ., Ek ⊂ Λn0 each of which has Jordan ∆-measure zero is in turn a set of Jordan ∆-measure zero. The empty set is a Jordan ∆-measurable set, and its Jordan ∆-measure is zero. Theorem 1.6.109. For each point t 0 = (t10 ,t20 , . . .,tn0 ) ∈ Λn0 , the single point set {t 0 } is Jordan ∆-measurable, and its Jordan ∆-measure is given by J({t 0 }) = (σ1 (t10 ) − t10 )(σ2 (t20 ) − t20 ) . . .(σn (tn0 ) − tn0 ).
(1.6.21)
Theorem 1.6.110. If E ⊂ Λn0 is a bounded set with ∆-boundary ∂∆ E, then J ∗ (∂∆ E) = J ∗ (E) − J∗ (E).
(1.6.22)
Moreover, E is Jordan ∆-measurable iff its ∆-boundary ∂∆ E has Jordan ∆measure zero. Proposition 1.6.111. Let E1 , E2 ∈ Λn . Then we have the following relations. 1. ∂∆ (E1 ∪ E2 ) ⊂ ∂∆ E1 ∪ ∂∆ E2 , 2. ∂∆ (E1 ∩ E2 ) ⊂ ∂∆ E1 ∪ ∂∆ E2 , 3. ∂∆ (E1 \ E2 ) ⊂ ∂∆ E1 ∪ ∂∆ E2 . Theorem 1.6.112. The union of a finite number of Jordan ∆-measurable sets is a Jordan ∆-measurable set. Theorem 1.6.113. The intersection of a finite number of Jordan ∆-measurable sets is a Jordan ∆-measurable set. Theorem 1.6.114. The difference of two Jordan ∆-measurable sets is a Jordan ∆-measurable set. Definition 1.6.115. Let f be defined and bounded on a bounded Jordan ∆measurable set E ⊂ Λn0 . Let R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ) ⊂ Λn
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be a rectangle containing E and put K = [a1 , b1 ] × [a2 , b2 ] × . . . × [an , bn ]. Define f on K by F(t) =
f (t) ift ∈ E 0
(1.6.23)
ift ∈ K \ E.
(here t = (t1 ,t2 , . . .,tn )). In this case, f is said to be Riemann ∆-integrable over E if F is Riemann ∆-integrable over R. We write Z
E
f (t1 ,t2, . . .,tn )∆1t1 ∆2 t2 . . .∆ntn =
Z
R
F(t1 ,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn .
Definition 1.6.116. Assume that f is defined and bounded on a bounded Jordan ∆-measurable set E ⊂ Λn0 . Let R = [a1 , b1 ) × [a2, b2 ) × . . . × [an , bn ) ⊂ Λn be a rectangle such that E ⊂ R. Let P ∈ P (R) be given by (1.6.9) and (1.6.10). Some of the rectangles of P will be entirely within E, some will be outside of E, and some will be partly within and partly outside E. Let P0 = {R1 , R2 , . . ., Rk } be the collection of subrectangles in P that lie completely within the set E. The collection P0 is called the inner ∆-partition of the set E, determined by the partition P of the rectangle R. We choose an arbitrary point ξl = (ξl1 , ξl2 , . . ., ξln ) in the subrectangle Rl of P0 for l ∈ {1, 2, . . ., k}. Let m(Rl ) denote the area of Rl . We set k
S = ∑ f (ξl )m(Rl ). l=1
We call S a Riemann ∆-sum of f corresponding to the partition P ∈ P (R). We say that f is Riemann ∆-integrable over E ⊂ Λn0 if there exists a number I such that for each ε > 0, there exists a δ > 0 such that |S − I| < ε for every Riemann ∆-sum S of f corresponding to any inner ∆-partition P0 = {R1 , R2 , . . ., Rk } of E, determined by a partition P ∈ Pδ (R), independent of the way in which ξl ∈ Rl for 1 ≤ l ≤ k is chosen. The number I is called the Riemann multiple ∆-integral of f over E. We write I = limδ→0 S. Proposition 1.6.117. Let E ⊂ Λn0 be a bounded set with ∆-boundary ∂∆ E. If R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ) ⊂ Λn
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δ→0
δ→0
for any P ∈ Pδ (R). Proposition 1.6.118. Let Γ ⊂ Λn0 be a set of Jordan ∆-measure zero. Moreover, let R = [a1 , b1 ) × [a2 , b2 ) × . . . × [an , bn ) be a rectangle in Λn that contains Γ. Then, for each ε > 0, there exists δ > 0 such that for every partition P ∈ Pδ (R), the sum of areas of subrectangles of P which have a common point with Γ is less than ε. Theorem 1.6.119. If E ⊂ Λn0 is a bounded and Jordan ∆-measurable set and f is a bounded function on E, then Definition 1.6.115 and Definition 1.6.116 of the Riemann ∆-integrability of f over E are equivalent to each other. Theorem 1.6.120. If E ⊂ Λn0 is a bounded and Jordan ∆-measurable set, then the integral Z E
1∆1t1 ∆2t2 . . .∆n tn
exists. Moreover, we have J(E) =
Z
E
1∆1t1 ∆2 t2 . . .∆n tn .
(1.6.24)
Theorem 1.6.121. Let E1 , E2 ⊂ Λn0 be bounded Jordan ∆-measurable sets such that J(E1 ∩ E2 ) = 0, and let E = E1 ∪ E2 . If f : E → R is a bounded function which is ∆-integrable over each of E1 and E2 , then f is ∆-integrable over E and Z
E
f (t1,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆ntn =
Z
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆ntn
E1
+
Z
E2
f (t1 ,t2 , . . .,tn)∆1 t1 ∆2t2 . . .∆n tn .
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Exercise 1.6.122. Let f and g be ∆-integrable over E, and let α, β ∈ R. Prove that α f + βg is also ∆-integrable over E and Z
E
(α f + βg)(t1 ,t2 , . . .,tn)∆1t1 ∆2t2 . . .∆n tn Z
=α
E
f (t1 ,t2, . . .,tn )∆1t1 ∆2 t2 . . .∆ntn + β
Z
E
g(t1,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn .
Exercise 1.6.123. Let f and g are ∆-integrable over E. Prove that so is their product f g. Exercise 1.6.124. Let f be ∆-integrable over E. Prove that so is | f | with Z Z f (t1 ,t2, . . .,tn )∆1t1 ∆2 t2 . . .∆ntn ≤ | f (t1 ,t2, . . .,tn )|∆1t1 ∆2t2 . . .∆n tn . E
E
Exercise 1.6.125. Let f and g be ∆-integrable over E and
f (t1 ,t2 , . . .,tn ) ≤ g(t1 ,t2 , . . .,tn )for all(t1 ,t2 , . . .,tn ) ∈ E. Prove that Z
E
f (t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆ntn ≤
Z
E
g(t1 ,t2 , . . .,tn )∆1t1 ∆2 t2 . . .∆n tn .
Exercise 1.6.126. Let f and g be ∆-integrable over E and let g be nonnegative (or nonpositive) on E. Prove that there exists a real number Λ ∈ [m, M] such that Z
E
f (t1 ,t2, . . .,tn )g(t1,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn
=Λ
Z
E
g(t1,t2 , . . .,tn )∆1t1 ∆2t2 . . .∆n tn ,
where and
m = inf{ f (t1,t2 , . . .,tn ) : (t1 ,t2 , . . .,tn ) ∈ E} M = sup{ f (t1 ,t2 , . . .,tn ) : (t1,t2 , . . .,tn ) ∈ E}.
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Elements of the Time Scale Calculus Theorem 1.6.127. Let [ai , bi ] ⊂ T0i , i = 1, 2, . . ., n − 1, and let φ : [a1 , b1 ] × [a2 , b2 ] × . . . × [an−1 , bn−1] → T0n and ψ : [a1 , b1 ] × [a2 , b2 ] × . . . × [an−1 , bn−1 ] → T0n be continuous functions such that φ(t) < ψ(t) for all t = (t1 ,t2, . . .,tn−1) ∈ [a1 , b1 ] × [a2 , b2 ] × . . . × [an−1 , bn−1]. Let E be the bounded set in Λn given by n E = (t1 ,t2 , . . .,tn ) ∈ Λn : ai ≤ ti < bi , i ∈ {1, 2, . . ., n − 1}, o φ(t1,t2 , . . .,tn−1) ≤ tn < ψ(t1 ,t2 , . . .,tn−1) .
In this case, E is Jordan ∆-measurable, and if f : E → R is ∆-integrable over E and if the single integral Z ψ(t1 ,t2 ,...,tn−1 ) φ(t1 ,t2 ,...,tn−1 )
f (t1 ,t2 , . . .,tn−1,tn )∆ntn
exists for each (t1,t2 , . . .,tn−1) ∈ [a1 , b1 ) × [a2, b2 ) × . . . × [an−1, bn−1 ), then the iterated integral Z b1 Z b2 a1
...
a2
Z bn−1 an−1
∆1t1 ∆2t2 . . .∆n−1tn−1
Z ψ(t1 ,t2 ,...,tn−1 ) φ(t1 ,t2 ,...,tn−1 )
f (t1 ,t2 , . . .,tn )∆ntn
exists. Moreover, we have Z
f (t1 ,t2, . . .,tn )∆1t1 ∆2 t2 . . .∆ntn
=
Z b1 Z b2
E
a1
a2
...
Z bn−1 an−1
∆1 t1 ∆2t2 . . .∆n−1tn−1
Z ψ(t1 ,t2 ,...,tn−1 ) φ(t1 ,t2 ,...,tn−1 )
f (t1 ,t2 , . . .,tn )∆ntn .
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1.7. Line Integrals Let T be a time scale with the forward jump operator σ and the delta operator ∆. Let a, b ∈ T with a < b. Assume that φi : [a, b] → R is continuous, i ∈ {1, . . ., m}. Definition 1.7.1. The system of functions xi = φi (t),
t ∈ [a, b] ⊂ T,
i ∈ {1, . . ., m}
(1.7.1)
is said to define a time scale continuous curve Γ. The points A(x1 , x2 , . . ., xm ) with the coordinates x1 , x2 , . . ., xm given by (1.7.1) are called the points of the curve. The set of all points of the curve will be referred to as the curve. The points A0 (φ1 (a), . . ., φm(a))andA1 (φ1 (b), . . ., φm(b)) are called the initial point and the final point of the curve, respectively. A0 and A1 are called the end points of the curve. Example 1.7.2. The system x1 = t 2 + t,
x2 = t − 2,
x3 = t,
t ∈ [0, 2],
defines a continuous curve. Here, φ1 (t) = t 2 + t,
φ2 (t) = t − 2,
φ3 (t) = t.
Thus, φ1 (0) = 0, φ1 (2) = 6,
φ2 (0) = −2, φ2 (2) = 0,
φ3 (0) = 0, φ3 (2) = 2.
Moreover, A0 (0, −2, 0) is the initial point and A1 (6, 0, 2) is the final point. Definition 1.7.3. If the initial and final points coincide, then the curve is said to be closed.
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Example 1.7.4. Consider the curve x1 = t 2 − 3t,
x2 = t 3 − 4t 2 + 5t,
t ∈ [1, 2].
Here, φ1 (t) = t 2 − 3t,
φ2 (t) = t 3 − 4t 2 + 5t.
Thus, φ1 (1) = −2,
φ2 (1) = 2,
φ1 (2) = −2,
φ2 (2) = 2.
Hence, A1 (−2, 2) is the initial point and A2 (−2, 2) is the final point. The considered curve is closed. Definition 1.7.5. The time scale parameter t is called the parameter of the curve. Definition 1.7.6. The equations (1.7.1) are called the parametric equations of the curve Γ. Definition 1.7.7. We say that a curve Γ is an oriented curve in the sense that a point (x01 , x02 , . . ., x0m ) = (φ1 (t 0), φ2 (t 0), . . .φm (t 0 )) is regarded as distinct from a point (x001 , x002 , . . ., x00m ) = (φ1 (t 00 ), φ2(t 00 ), . . ., φm(t 00 )) if t 0 6= t 00 and as preceding (x001 , x002 , . . ., x00m ) if t 0 < t 00 . The oriented curve Γ is then said to be “traversed in the direction of increasing t.” Definition 1.7.8. Two curves Γ1 and Γ2 with equations xi = φi(t),
i ∈ {1, . . ., m},
t ∈ T1 ,
xi = ψi (t),
i ∈ {1, . . ., m},
t ∈ T2 ,
and respectively, are regarded as identical if the equations of one curve can be transformed into the equations of the other curve by means of a continuous strictly
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increasing change of the parameter, i.e., if there is a continuous increasing function τ = λ(t), t ∈ [a, b], with the range [α, β], such that ψi (λ(t)) = φi (t),
t ∈ [a, b].
We then say that the two curves have the same direction. We say that the two curves have opposite direction if the function λ is decreasing. In this case, the initial point of Γ1 is the final point of Γ2 , and vice versa. The curve differing by Γ only by the direction in which it is traversed is denoted by −Γ. Example 1.7.9. The curves x3 = t 2 − 1,
x1 = t,
x2 = t − 1,
x1 = t 2 ,
x2 = t 2 − 1,
t ∈ [0, 1],
and are identical.
x3 = t 4 − 1,
t ∈ [0, 1],
Definition 1.7.10. If the same point (x1 , . . ., xm ) corresponds to more than one parameter value in the half-open interval [a, b), then we say that (x1 , . . ., xm ) is a multiple point of the curve (1.7.1). Definition 1.7.11. A curve with no multiple points is called a simple curve or Jordan curve. Example 1.7.12. The curve x1 = t,
x2 = 4 − t,
x3 = 5 + t,
t ∈ [1, 4],
is a Jordan curve. Let Γ be a time scale continuous curve with equations (1.7.1). Consider a partition of the interval [a, b], P = {t0 ,t1 , . . .,tn } ⊂ [a, b], wherea = t0 < t1 < . . . < tn = b.
(1.7.2)
For a given partition P, we denote by A0 , A1 , . . ., An the corresponding points of the curve Γ, i.e., A j (φ1 (t j ), φ2 (t j ), . . ., φm (t j )), j ∈ {0, 1, . . ., n}. We set s n
l(Γ, P) =
∑ d(A j−1, A j ) = j=1
n
m
j=1
k=1
∑ ∑ (φk (t j ) − φk (t j−1))2 ,
where d(A j−1 , A j ) denotes the distance from the point A j−1 to A j .
(1.7.3)
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Definition 1.7.13. The curve Γ is said to be rectifiable if l(Γ) = sup{l(Γ, P) : Pis a partition of[a, b]} < ∞. In this case, the nonnegative number l = l(Γ) is called the length of the curve Γ. If the supremum does not exist, then the curve is said to be nonrectifiable. Proposition 1.7.14. If P and Q are partitions of [a, b] and Q is a refinement of P, then l(Γ, P) ≤ l(Γ, Q). Theorem 1.7.15. If the curve Γ is rectifiable, then its length does not depend on the parameterization of this curve. Theorem 1.7.16. If a rectifiable curve Γ is split by means of a finite number of points A0 , A1 , . . ., An into a finite number of curves Γi and if the points Ai correspond to the values ti of the parameter t and a = t0 < t1 < . . . < tn = b, then each curve Γi is rectifiable and n
l(Γ) = ∑ l(Γi ). i=1
Theorem 1.7.17. Assume the functions φi , i ∈ {1, . . ., m}, are continuous on [a, b] and ∆-differentiable on [a, b). If their ∆-derivatives φ∆i are bounded and ∆-integrable on [a, b), then the curve Γ, defined by the parametric equations (1.7.1), is rectifiable, and its length l(Γ) can be evaluated by the formula s Z b m 2 l(Γ) = ∑ φ∆k (t) ∆t. a
k=1
Let Γ be a curve defined by the equations (1.7.1). Put
A = (φ1 (a), . . ., φm(a))andB = (φ1 (b), . . ., φm (b)). Suppose that the function f (x1 , . . ., xm ) is defined and continuous on the curve Γ. Let P = {t0,t1 , . . .,tn } ⊂ [a, b],
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where a = t0 < t1 < . . . < tn = b, be a partition of the interval [a, b], and put Ai = (φ1 (ti ), . . ., φm(ti )),
i ∈ {0, 1, . . ., n}.
Take any τk ∈ [tk−1,tk ), k ∈ {1, . . ., n}. Denote by lk the length of the piece of the curve Γ between its points Ak−1 and Ak . Then, using Theorem 1.7.17, we have s Z tk m 2 lk = ∑ φ∆l (t) ∆t. tk−1
We introduce
l=1
n
S1 =
∑ f (φ1(τk ), . . ., φm (τk ))lk. k=1
Definition 1.7.18. We say that a complex number I1 is the line delta integral of the first kind of the function f along the curve Γ if for each ε > 0, there exists δ > 0 such that |S1 − I1 | < ε for every integral sum S1 of f corresponding to a partition P ∈ Pδ ([a, b]) independent of the way in which τk ∈ [tk−1,tk ) for k ∈ {1, 2, . . ., n} is chosen. We denote the number I1 symbolically by Z
Γ
f (x1 , . . ., xm )∆tor
Z
AB
f (x1 , . . ., xm )∆l,
(1.7.4)
where A and B are the initial and final points of the curve Γ, respectively. Theorem 1.7.19. Suppose that the curve Γ is given by the parametric equations (1.7.1), where φi , i ∈ {1, . . ., m}, are continuous on [a, b] and ∆-differentiable on [a, b). If φ∆i , i ∈ {1, . . ., m}, are bounded and ∆-integrable over [a, b) and if the function f is continuous on Γ, then the line integral (1.7.4) exists and can be computed by the formula s Z Z b m 2 f (x1 , . . ., xm )∆l = f (φ1 (t), . . ., φm(t)) ∑ φ∆k (t) ∆t. Γ
a
k=1
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Theorem 1.7.20 (Linearity). Let α, β ∈ R. If the functions f and g are ∆integrable along the curve Γ, then α f + βg is also ∆-integrable along the curve Γ and Z
Γ
(α f + βg)(x1 , . . ., xm)∆l = α
Z
Γ
f (x1 , . . ., xm )∆]l + β
Z
Γ
g(x1 , . . ., xm )∆l.
Theorem 1.7.21 (Additivity). If the curve AB consists of two parts AC and CB and if the function f is ∆-integrable along the curve AB, then it is ∆-integrable along each of the curves AC and CB and Z
AB
f (x1 , . . ., xm )∆l =
Z
AC
f (x1 , . . ., xm )∆l +
Z
CB
f (x1 , . . ., xm )∆l.
Theorem 1.7.22 (Existence of Modulus of Integral). If f is ∆-integrable along the curve Γ, then so is | f | and Z Z f (x1 , . . ., xm)∆l ≤ | f (x1 , . . ., xm )|∆l. (1.7.5) Γ
Γ
Assume that gi , i ∈ {1, . . ., m}, are defined and continuous on the curve Γ. We introduce the integral sums m
S2i =
∑ gi(φ1(τk ), . . ., φm(τk ))(φi(tk) − φi (tk−1)),
i = 1, . . ., m.
k=1
Definition 1.7.23. We say that a complex number Ii , i ∈ {1, . . ., m}, is the line delta integral of the second kind of the function gi along the curve Γ if for each ε > 0, there exists δ > 0 such that |S2i − Ii | < ε for every integral sum S2i of gi corresponding to a partition P ∈ Pδ ([a, b]) independent of the way in which τk ∈ [tk−1,tk) for k ∈ {1, . . ., n} is chosen. We denote the number Ii symbolically by Z
Γ
gi (x1 , . . ., xm )∆i xi or
Z
AB
gi (x1 , . . ., xm )∆ixi .
(1.7.6)
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The sum
Z
Γ
g1 (x1 , . . ., xm )∆1 x1 + · · · +
Z
Γ
gm (x1 , . . ., xm )∆m xm
is called a general line delta integral of the second kind, and it is denoted by Z
Γ
g1 (x1 , . . ., xm )∆1 x1 + · · · + gm (x1 , . . ., xm)∆m xm .
Theorem 1.7.24. Suppose that the curve Γ is given by the parametric equations (1.7.1), where φi are continuous on [a, b] and ∆-integrable on [a, b). If φ∆i are bounded and ∆-integrable on [a, b) and if the functions gi are continuous on Γ, then the line integrals (1.7.6) exist and can be computed by the formula Z
Γ
gi (x1 , . . ., xm )∆i xi =
Z b a
gi (φ1 (t), . . ., φm(t))φ∆i (t)∆t.
Theorem 1.7.25 (Linearity). Let α, β ∈ R. If f i and gi , i ∈ {1, . . ., m}, are line ∆-integrable of the second kind along the curve Γ, then α f i + βgi is line ∆-integrable along the curve Γ and Z
Γ
(α f i + βgi )(x1 , . . ., xm )∆ixi
=α
Z
Γ
fi (x1 , . . ., xm )∆i xi + β
Z
Γ
gi (x1 , . . ., xm )∆ixi .
Theorem 1.7.26 (Additivity). If the curve AB consists of two curves AC and CB and if the functions gi , i ∈ {1, . . ., m}, are line ∆-integrable of the second kind along the curve AB, then they are line ∆-integrable of the second kind along each of the curves AC and CB and Z
AB
gi (x1 , . . ., xm )∆i xi =
Z
AC
gi (x1 , . . ., xm )∆i xi +
Z
CB
gi (x1 , . . ., xm )∆i xi .
1.8. Green’s Formula Let T1 and T2 be two time scales. Since T1 and T2 are closed subsets of R, the set T1 × T2 is a complete metric space with metric D defined by q d((x1 , x2 ), (x01, x02 )) = (x1 − x01 )2 + (x2 − x02 )2 .
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Definition 1.8.1. If M is a metric space, then any continuous mapping h : [a, b] → M is called a continuous curve in M . Definition 1.8.2. Let [a, b] ⊂ T1 with a, b ∈ T1 and y0 ∈ T2 . The set {(x, y0 ) : x ∈ [a, b]} is called a horizontal line segment in T1 × T2 and denoted by AB, where A = (a, y0 ) and B = (b, y0). We take x0 ∈ T1 and [c, d] ⊂ T2 and define a vertical line segment in T1 × T2 by {(x0 , y) : y ∈ [c, d]}
and denote it by CD, where C = (x0 , c) and D = (x0 , d).
Definition 1.8.3. A finite sequence A1 B1 , A2 B2 , . . ., An Bn , each of whose term Ak Bk , k ∈ {1, . . ., n}, is a horizontal or vertical line segment in T1 × T2 , is said to form a polygonal path (or broken line) in T1 × T2 with terminal points A1 and Bn if B1 = A2 , B2 = A3 , . . ., Bn−1 = An . Definition 1.8.4. A set of points of T1 × T2 is said to be connected if any two of its points are terminal points of a polygonal path of points contained in the set. Definition 1.8.5. A component of a set Ω ⊂ T1 × T2 is a nonempty maximal connected subset of Ω. Definition 1.8.6. A nonempty open connected Set of points of T1 × T2 is called a domain. Definition 1.8.7. A closed domain is a subset of T1 × T2 which is the closure of a domain in T1 × T2 . For the rectangle R in T1 × T2 defined by R = [a1 , b1 ) × [a2 , b2 ), we set L1 = {(x, a2 ) : x ∈ [a1 , b1 ]},
L2 = {(b1 , y) : y ∈ [a2 , b2 ]},
L3 = {(x, b2 ) : x ∈ [a1 , b1 ]},
L4 = {(a1 , y) : y ∈ [a2 , b2 ]}.
Each of L j , j ∈ {1, 2, 3, 4}, is an oriented line segment.
(1.8.1)
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Svetlin G. Georgiev
Definition 1.8.8. The closed curve Γ = L1 ∪ L2 ∪ (−L3 ) ∪ (−L4 )
(1.8.2)
is called the positively oriented fence of R, i.e., the rectangle Γ remains on the “left” side along the fence curve Γ. Definition 1.8.9. We say that the set E ⊂ T1 × T2 is a set of type ω if it is a connected set in T1 × T2 which is the union of a finite number of rectangles of the form (1.8.1) that are pairwise disjoint and adjoining to each other. Let E ⊂ T1 × T2 be a set of type ω so that E=
m [
Rk ,
k=1
where Rk = [ak1 , bk1 ) × [ak2 , bk2 ) ⊂ T1 × T2 ,
k ∈ {1, . . ., m},
and R1 , R2 , . . ., Rm are pairwise disjoint and adjoining to each other. Let Γk be the positively oriented fence of the rectangle Rk . We set X=
m [
Γk .
k=1
Let X0 consist of a finite number of line segments each of which serves as a common part of fences of two adjoining rectangles belonging to {R1 , R2 , . . ., Rm}. Definition 1.8.10. The set Γ = X \ X0 forms a positively oriented closed “polygonal curve”, which is called the positively oriented fence of the set E. Theorem 1.8.11 (Green’s Formula). Let E ⊂ T1 × T2 be a set of type ω and let Γ be its positively oriented fence. If the functions M and N are continuous and have continuous partial delta derivatives My∆22 and Nx∆11 on E ∪ Γ, then Z Z
E
Nx∆11 (x1 , x2 ) − Mx∆22 (x1 , x2 )
∆1 x1 ∆2 x2 =
Z
Γ
M(x1 , x2 )∆1 x1 +N(x1 , x2 )∆2 x2 .
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Elements of the Time Scale Calculus
1.9. Advanced Practical Problems √ 4 Problem 1.9.1. Classify each points t ∈ T = { 7n : n ∈ N0 } as left-dense, leftscattered, right-dense or right-scattered. √ 4 Answer. Each points t = 7n, n ∈ N, are isolated, t = 0 is right-scattered. n√ o 4 Problem 1.9.2. Let T = n + 7 : n ∈ N0 . Find µ(t), t ∈ T. √ √ √ Answer. µ 4 n + 7 = 4 n + 8 − 4 n + 7. n o √ 4 Problem 1.9.3. Let T = t = 2n + 1 : n ∈ N , f (t) = 1 + 2t 4 , t ∈ T. Find f (σ(t)), t ∈ T. Answer. 5 + 2t 4 .
2 Problem 1.9.4. Let T = : n ∈ N ∪ {0}. Find Tκ . 4n + 3 2 Answer. : n ∈ N, n ≥ 2 ∪ {0}. 4n + 3
Problem 1.9.5. Let f (t) = t + t 2 + t 3 , t ∈ T. Prove that f ∆ (t) = 1 + t + t 2 + (1 + t)σ(t) + (σ(t))2 ,
t ∈ Tκ .
Problem 1.9.6. Let T = {n3 : n ∈ N0 }, f (t) = t 2 + 2t, t ∈ T. Find f ∆ (t), t ∈ Tκ . √ 3 Answer. 2 + t + 3 t + 1 . Problem 1.9.7. Let T = {n + 2 : n ∈ N0 }, f (t) = t 2 + 2, g(t) = t 2 . Find a constant c ∈ [2, σ(2)] such that
Answer. c =
r
( f ◦ g)∆ (2) = f 0 (g(c))g∆(2). 13 . 2
Problem 1.9.8. Let T = 24n+2 : n ∈ N0 , v(t) = t 3 , w(t) = t 2 + t. Prove ˜
(w ◦ v)∆ (t) = (w∆ ◦ v(t))v∆(t),
t ∈ Tκ .
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Svetlin G. Georgiev
Problem 1.9.9. Let ∆˜ v−1 ◦ v(t).
Answer.
T = {n + 9 : n ∈ N0 },
v(t) = t 2 + 7t + 8.
Find
1 . 2t + 8
Problem 1.9.10. Let T = R and 1 for t = 2 f (t) = 10 for t ∈ R\{2}. t−2
Determine if f is regulated. Answer. No.
Problem 1.9.11. Let T = R and 0 if f (t) = 1 t−5
t =5 if
R\{5}.
Check if f : T → R is pre-differentiable and if it is, find the region of differentiation. Answer. No. Problem 1.9.12. Let T = 3N . Prove that −
Z
1 sint sin2t∆t = cost + c. t
Problem 1.9.13. Let p : T → R be rd-continuous and 1 + µ(t)p(t) 6= 0 for all t ∈ T. Let also, φ(t) be a nontrivial solution to the equation φ∆ − p(t)φ = 0. Prove that
1 is a solution to the equation φ(t) ψ∆ + p(t)ψσ = 0.
Elements of the Time Scale Calculus
77
Problem 1.9.14. Let T1 = Z and T2 = 4Z. Compute Z 4Z 8 −4 4
t2 log
t1 + 1 ∆1 t 1 ∆2 t 2 . t1
Problem 1.9.15. Let φ : [2, 8] → R, φ(x) = x3 . Prove that φ satisfies the Lipschitz condition. Find a Lipschitz constant. Problem 1.9.16. Let
φ(x) =
1 x3 −8
forx ∈ [0, 2),
−1
forx = 2.
Check if φ satisfies the Lipschitz condition. Problem 1.9.17. Let T1 = 2N0 and T2 = Z. Prove that Z 3Z 8 −1 1
(t13 + 3t12t2 + t1 t22 + t23 )∆1t1 ∆2 t2 =
Z 3
∆2 t 2
=
Z 8
∆1 t 1
−1
1
Z 8
(t13 + 3t12 t2 + t1 t22 + t23 )∆1t1
Z 3
(t13 + 3t12 t2 + t1 t22 + t23 )∆2t2 .
1
−1
Problem 1.9.18. Let T1 = T2 = Z, Find J(E) and
R
E = [1, 2] × [2, 4].
E (2t1 − t2 )∆1t1 ∆2 t2 .
Problem 1.9.19. Let T1 = T2 = 2N0 . Compute the integral Z 3 Z 2t2 1
t2
(t12 + t1 t2 + 3t22 )∆1 t1 ∆2t2 .
Problem 1.9.20. Let T1 = T2 = 4N0 ∪ {0}. Compute the integral Z 4 Z t1 0
0
sinh f (t1 , 1) − 5t2 cosh f 2 (t1 , 1) ∆2t2 ∆1 t1 , where f (t1 ) = t1 .
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Svetlin G. Georgiev
Problem 1.9.21. Let T1 = 2N0 ∪ {0}, T2 = 3N0 ∪ {0}. Compute the integral Z 1 Z 3t1 0
0
(t1 + 2t22 )∆2t2 ∆1 t1 .
Problem 1.9.22. Let T1 = N0 and T2 = 2N0 ∪ {0}. Compute the integral Z 2 Z t1 0
0
cosh f (t1 , 1))∆2t2 ∆1 t1 , where f (t1 ) = t1 .
Problem 1.9.23. Let T1 = Z and T2 = 3N0 ∪ {0}. Compute the integral Z 1 Z t1 −1 0
e f (t1 , 1)∆2t2 ∆1t1 , where f (t1 ) = t1 .
Chapter 2
Dynamic Systems on Time Scales Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively.
2.1. Structure of Dynamic Systems on Time Scales Definition 2.1.1. The first order linear dynamic equation y∆ = p(t)y,
t ∈ T,
(2.1.1)
is called regressive if p ∈ R . Theorem 2.1.2. Suppose that p ∈ R , t0 ∈ T and y0 ∈ R. Then the unique solution of the IVP y∆ = p(t)y, y(t0 ) = y0 , (2.1.2) is given by y(t) = e p (t,t0)y0 ,
t ∈ T.
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Svetlin G. Georgiev
Proof. Let y be a solution of the IVP (2.1.2). We have y∆ (t)e p (σ(t), t0)
= = =
p(t)e p (σ(t), t0)y(t) p(t) e p (t, t0) + µ(t)e∆ p (t, t0) y(t)
p(t) (e p (t, t0) + µ(t)( p)(t)e p (t, t0)) y(t) µ(t)p(t) e p (t, t0)y(t) = p(t) 1 − 1 + µ(t)p(t) p(t) = e p (t, t0)y(t) 1 + µ(t)p(t) = −( p)(t)e p (t, t0)y(t)
= −e∆ p (t, t0)y(t),
t ∈ T,
whereupon y∆ (t)e p (σ(t),t0) + e∆ p (t,t0)y(t) = 0, or
(ye p (·,t0))∆ (t) = 0,
t ∈ T,
t ∈ T.
Then y(t)e p (t,t0) = c,
t ∈ T,
where c is a constant. Since y(t0 ) = y0 , we get y0 = c, i.e., y(t)e p (t,t0) = y0 ,
t ∈ T.
y(t) = y0 e p (t,t0),
t ∈ T.
Consequently This completes the proof. Definition 2.1.3. If p ∈ R and f : T → R is rd-continuous, then the dynamic equation y∆ = p(t)y + f (t), t ∈ T, (2.1.3) is called regressive.
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Dynamic Systems on Time Scales
Theorem 2.1.4. Suppose that (2.1.3) is regressive. Let t0 ∈ T and x0 ∈ R. The unique solutions of the IVPs x∆ = −p(t)xσ + f (t),
x(t0 ) = x0 ,
t ∈ T,
(2.1.4)
and x∆ = p(t)x + f (t),
x(t0 ) = x0 ,
t ∈ T,
(2.1.5)
are given by x(t) = e p (t,t0)x0 + and x(t) = e p (t,t0)x0 +
Z t
Z t t0
e p (t, τ) f (τ)∆τ,
t ∈ T,
(2.1.6)
e p (t, σ(τ)) f (τ)∆τ,
t ∈ T,
(2.1.7)
t0
respectively. Proof. 1. Consider x(t), defined by (2.1.6). We will prove that it satisfies (2.1.4). We have
x(t0 ) = e p (t0 ,t0 )x0 = x0 ,
and ∆
x (t) = ( p) (t)e p (t,t0)x0 +
Z t t0
( p) (t)e p (t, τ) f (τ)∆τ
+e p (σ(t),t) f (t) Z t p(t) p(t) e p (t,t0)x0 − e p (t, τ) f (τ)∆τ = − 1 + µ(t)p(t) 1 + µ(t)p(t) t0 f (t) + , t ∈ T, 1 + µ(t)p(t)
82
Svetlin G. Georgiev xσ (t) = e p (σ(t),t0)x0 +
Z σ(t) t0
e p (σ(t), τ) f (τ)∆τ
= e p (t,t0)x0 (1 + ( p) (t)µ(t)) + +
Z t
e p (σ(t), τ) f (τ)∆τ t0 Z σ(t)
e p (σ(t), τ) f (τ)∆τ
t
t 1 1 e p (t,t0)x0 + e p (t, τ) f (τ)∆τ 1 + µ(t)p(t) 1 + µ(t)p(t) t0 +µ(t)e p (σ(t),t) f (t) Z t 1 1 = e p (t,t0)x0 + e p (t, τ) f (τ)∆τ 1 + µ(t)p(t) 1 + µ(t)p(t) t0 µ(t) + f (t), t ∈ T. 1 + µ(t)p(t)
Z
=
Therefore, x∆ (t) + p(t)xσ(t)
=
=
t p(t) p(t) e p (t,t0 )x0 − e p (t,τ) f (τ)∆τ 1 + µ(t)p(t) 1 + µ(t)p(t) t0 f (t) + 1 + µ(t)p(t) Z t p(t) p(t) + e p (t,t0 )x0 + e p (t,τ) f (τ)∆τ 1 + µ(t)p(t) 1 + µ(t)p(t) t0 µ(t)p(t) + f (t) 1 + µ(t)p(t) f (t), t ∈ T.
Z
−
Now we multiply the equation (2.1.4) by e p (t,t0) and we get e p (t,t0)x∆ (t) + p(t)e p (t,t0)xσ (t) = e p (t,t0) f (t),
t ∈ T,
or (e p (·,t0)x)∆ (t) = e p (t,t0) f (t),
t ∈ T,
which we integrate from t0 to t and we obtain e p (t,t0 )x(t) − e p (t0,t0 )x(t0 ) =
Z t t0
e p (τ,t0 ) f (τ)∆τ,
t ∈ T,
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Dynamic Systems on Time Scales or e p (t,t0)x(t) = x0 +
Z t t0
e p (τ,t0) f (τ)∆τ,
t ∈ T,
or x(t) = x0 e p (t,t0) + = x0 e p (t,t0) +
Z t
t0 Z t t0
e p (t0 , τ)e p (t,t0) f (τ)∆τ e p (t, τ) f (τ)∆τ,
t ∈ T.
2. Let now x be defined by (2.1.7). Then x∆ (t) = p(t)e p (t,t0)x0 + p(t)
Z t t0
e p (t, σ(τ)) f (τ)∆τ
+e p (σ(t), σ(t)) f (t) = p(t)x(t) + f (t),
t ∈ T.
Now we multiply the equation (2.1.5) by e p (t,t0) and we get e p (t,t0)x∆ (t) − p(t)e p (t,t0)x(t) = e p (t,t0 ) f (t),
t ∈ T,
or p(t) 1 e p (t,t0 )x∆ (t) − e p (t,t0)x(t) 1 + µ(t)p(t) 1 + µ(t)p(t) 1 = e p (t,t0) f (t), t ∈ T, 1 + µ(t)p(t) or e p (σ(t),t0)x∆ (t) + e∆ p (t,t0 )x(t) = e p (σ(t),t0) f (t), or
(e p (·,t0)x)∆ (t) = e p (σ(t),t0) f (t),
t ∈ T,
t ∈ T,
which we integrate from t0 to t and we obtain e p (t,t0)x(t) − e p (t0 ,t0 )x(t0 ) =
Z t t0
e p (σ(τ),t0) f (τ)∆τ,
t ∈ T,
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Svetlin G. Georgiev or e p (t,t0)x(t) = x0 +
Z t t0
e p (t0 , σ(τ)) f (τ)∆τ,
t ∈ T,
or x(t) = x0 e p (t,t0) + = x0 e p (t,t0) +
Z t
t0 Z t t0
e p (t,t0)e p (t0 , σ(τ)) f (τ)∆τ e p (t, σ(τ)) f (τ)∆τ,
t ∈ T.
This completes the proof. Suppose that A is a m × n-matrix on T, A = (ai j )1≤i≤m,1≤ j≤n, shortly A = (ai j ), ai j : T → R, 1 ≤ i ≤ m, 1 ≤ j ≤ n. Definition 2.1.5. We say that A is differentiable on T if each entry of A is differentiable on T and A∆ = a∆i j .
Example 2.1.6. Let T = Z, t +1 t2 + t A(t) = , 2t − 3 2t 2 − 3t + 2
t ∈ T.
We will find A∆ (t), t ∈ T. We have σ(t) = t + 1, a11 (t) = t + 1, a12 (t) = t 2 + t, a21 (t) = 2t − 3, a22 (t) = 2t 2 − 3t + 2,
t ∈ T.
Dynamic Systems on Time Scales Then a∆11 (t) = 1, a∆12 (t) = σ(t) + t + 1 = t +1 +t +1 = 2t + 2, a∆21 (t) a∆22 (t)
= 2, = 2(σ(t) + t) − 3 = 2(t + 1 + t) − 3
= 4t + 2 − 3
t ∈ Tκ .
= 4t − 1, Therefore ∆
A (t) =
1 2t + 2 2 4t − 1
,
t ∈ T.
Example 2.1.7. Let T = 2N0 , 3 t + t t+1 t t+2 A(t) = 1 −t 2 t 2 + t , 1 2 t3 We will find A∆ (t), t ∈ T. We have
σ(t) = 2t, a11 (t) = t 3 + t, t +1 , t +2 a13 (t) = t,
a12 (t) =
a21 (t) = 1, a22 (t) = −t 2 , a23 (t) = t 2 + t, a31 (t) = 1, a32 (t) = 2, a33 (t) = t 3 ,
t ∈ T.
t ∈ T.
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Svetlin G. Georgiev
Then a∆11 (t) = (σ(t))2 + tσ(t) + t 2 + 1 = (2t)2 + 2t 2 + t 2 + 1 = 4t 2 + 3t 2 + 1 = 7t 2 + 1, t + 2 − (t + 1) a∆12 (t) = (t + 2)(σ(t) + 2) 1 , = 2(t + 1)(t + 2) a∆13 (t) = 1, a∆21 (t) = 0, a∆22 (t) = −(σ(t) + t) = −(2t + t) = −3t,
a∆23 (t) = σ(t) + t + 1 = 2t + t + 1 = 3t + 1, a∆31 (t) a∆32 (t) a∆33 (t)
= 0, = 0, = (σ(t))2 + tσ(t) + t 2 = 4t 2 + 2t 2 + t 2 = 7t 2 ,
t ∈ T.
Therefore
7t 2 + 1 A∆ (t) = 0 0
Example 2.1.8. Let T = N20 ,
A(t) =
1 2(t+1)(t+2)
−3t 0
t2 + 1 2
1 t+1
3t
1 3t + 1 , 7t 2
,
t ∈ T.
t ∈ T.
87
Dynamic Systems on Time Scales We will find A∆ (t), t ∈ T. We have σ(t) =
2 √ t +1 ,
a11 (t) = t 2 + 1, 1 a12 (t) = , t +1 a21 (t) = 2, a22 (t) = 3t,
t ∈ T.
Then a∆11 (t) = σ(t) + t √ 2 t +1 +t = √ = t +2 t +1 +t √ = 2t + 2 t + 1, 1 a∆12 (t) = − (t + 1)(σ(t) + 1) 1 √ = − (t + 1)(( t + 1)2 + 1) 1 √ = − , (t + 1)(t + 2 t + 2) a∆21 (t) = 0, a∆22 (t) = 3,
t ∈ T.
Therefore A∆ (t) =
! √ 1 √ 2t + 2 t + 1 − (t+1)(t+2 t+2) , 0 3
Exercise 2.1.9. Let T = 2Z, A(t) = Find A∆ (t), t ∈ T.
t3 t2 2t + 4 t − 1
,
t ∈ T.
t ∈ T.
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Svetlin G. Georgiev
Answer. ∆
A (t) =
3t 2 + 6t + 4 2t + 2 2 1
Definition 2.1.10. Define
t ∈ T.
,
Aσ = aσi j .
Theorem 2.1.11. If A is differentiable on Tκ , then Aσ(t) = A(t) + µ(t)A∆(t),
t ∈ Tκ .
Proof. We have Aσ (t) = =
aσi j (t)
ai j (t) + µ(t)a∆i j (t)
= (ai j (t)) + µ(t) a∆i j (t) = A(t) + µ(t)A∆(t),
t ∈ Tκ .
Below we suppose that B = (bi j )1≤i≤m,1≤ j≤n, bi j : T → R, 1 ≤ i ≤ m, 1 ≤ j ≤ n. Theorem 2.1.12. Let A and B be differentiable on Tκ . Then (A + B)∆ = A∆ + B∆
on Tκ .
Proof. We have (A + B)(t) = (ai j (t) + bi j (t)), (A + B)∆ (t) = a∆i j (t) + b∆i j (t) = a∆i j (t) + b∆i j (t) = A∆ (t) + B∆ (t),
t ∈ Tκ .
This completes the proof. Theorem 2.1.13. Let α ∈ R and A be differentiable on Tκ . Then (αA)∆ = αA∆
on Tκ .
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Dynamic Systems on Time Scales Proof. We have (αai j )∆ (t) = αa∆i j (t) = α a∆i j (t)
(αA)∆ (t) =
= αA∆ (t),
t ∈ Tκ .
This completes the proof. Theorem 2.1.14. Let m = n and A, B be differentiable on Tκ . Then (AB)∆(t) = A∆ (t)B(t) + Aσ(t)B∆(t) = A∆ (t)Bσ(t) + A(t)B∆(t), Proof. We have n
∑ aik (t)bk j (t)
(AB)(t) =
k=1
!
t ∈ Tκ .
t ∈ T.
,
Then (AB)∆ (t)
=
=
n
∑ aik bk j k=1 n
∑ (aik bk j )
!∆
∆
(t)
(t)
k=1
!
! ∆ σ ∆ ∑ aik (t)bk j (t) + aik(t)bk j (t) n
=
k=1 n
= = =
∑
a∆ik (t)bk j (t)
k=1 ∆
n
=
This completes the proof.
n
+
∑
aσik (t)b∆k j (t)
k=1
A (t)B(t) + Aσ(t)B∆ (t) ! n ∆ σ ∆ ∑ aik (t)bk j (t) + aik(t)bk j (t) k=1
=
!
∑
a∆ik (t)bσk j (t)
k=1 ∆ σ
!
n
+
∑
aik (t)b∆k j (t)
k=1
A (t)B (t) + A(t)B∆(t),
t ∈ Tκ .
!
!
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Example 2.1.15. Let T = 2N0 , t t −1 A(t) = , 2 3t + 1
B(t) =
1 t t +1 t −1
,
Then
1 t (AB)(t) = t +1 t −1 2 t + t − 1 2t 2 − 2t + 1 = 3t 2 + 4t + 3 3t 2 − 1 = C(t) t t −1 2 3t + 1
= (ci j (t)),
t ∈ T.
We have σ(t) = 2t, a11 (t) = t, a12 (t) = t − 1,
a21 (t) a22 (t) b11 (t) b12 (t)
= 2, = 3t + 1, = 1, = t,
b21 (t) b22 (t) c11 (t) c12 (t)
= t + 1, = t − 1, = t 2 + t − 1, = 2t 2 − 2t + 1,
c21 (t) = c22 (t) =
3t 2 + 4t + 3, 3t 2 − 1, t ∈ T.
t ∈ T.
Dynamic Systems on Time Scales Then a∆11 (t) = a∆12 (t) =
1, 1,
a∆21 (t) a∆22 (t) b∆11 (t) b∆12 (t)
0, 3, 0, 1,
= = = =
b∆21 (t) = b∆22 (t) = c∆11 (t) =
1, 1, σ(t) + t + 1
= = ∆ c12 (t) = =
2t + t + 1 3t + 1, 2(σ(t) + t) − 2 2(2t + t) − 2
c∆21 (t)
= = = =
c∆22 (t) = = =
Therefore ∆
(AB) (t) =
6t − 2, 3(σ(t) + t) + 4 3(2t + t) + 4 9t + 4, 3(σ(t) + t) 3(2t + t) 9t, t ∈ T.
3t + 1 6t − 2 9t + 4 9t
,
t ∈ T.
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Also, A∆ (t)B(t) = = σ
A (t) = = Aσ (t)B∆(t) = = ∆
σ
∆
A (t)B(t) + A (t)B (t) = =
1 t t +1 t −1 t + 2 2t − 1 , 3t + 3 3t − 3 σ(t) σ(t) − 1 2 3σ(t) + 1 2t 2t − 1 , 2 6t + 1 2t 2t − 1 0 1 2 6t + 1 1 1 2t − 1 4t − 1 , 6t + 1 6t + 3 t + 2 2t − 1 2t − 1 4t − 1 + 3t + 3 3t − 3 6t + 1 6t + 3 3t + 1 6t − 2 , t ∈ T. 9t + 4 9t 1 1 0 3
Consequently (AB)∆ (t) = A∆ (t)B(t) + Aσ(t)B∆(t),
t ∈ T.
Next,
1 σ(t) B (t) = σ(t) + 1 σ(t) − 1 1 2t = , 2t + 1 2t − 1 σ
Dynamic Systems on Time Scales 93 1 1 1 2t A∆ (t)Bσ(t) = 0 3 2t + 1 2t − 1 2t + 2 4t − 1 = , 6t + 3 6t − 3 t t −1 0 1 ∆ A(t)B (t) = 2 3t + 1 1 1 t − 1 2t − 1 = , 3t + 1 3t + 3 2t + 2 4t − 1 t − 1 2t − 1 A∆ (t)Bσ(t) + A(t)B∆(t) = + 6t + 3 6t − 3 3t + 1 3t + 3 3t + 1 6t − 2 = , t ∈ T. 9t + 4 9t Therefore (AB)∆ (t) = A∆ (t)Bσ(t) + A(t)B∆(t), Exercise 2.1.16. Let T = 3Z, 2 t +1 t −2 A(t) = , 2t − 1 t + 1
B(t) =
t ∈ T.
t 2t + 1 t t −1
,
Prove (AB)∆ (t) = A∆ (t)Bσ(t) + A(t)B∆(t), Theorem 2.1.17. Let m = n and A−1 exists on T. Then σ (Aσ )−1 = A−1 on T.
t ∈ Tκ .
Proof. For any t ∈ T we have
A(t)A−1(t) = I,
t ∈ T.
Then Aσ (t) A−1 whereupon
−1
(Aσ ) This completes the proof.
σ
(t) = I,
(t) = A−1
σ
(t),
t ∈ T.
t ∈ T.
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Example 2.1.18. Let T = lN0 , l > 0, t +1 t +2 A(t) = , 1 t +3 Then σ
A (t) = = (Aσ )−1 (t) =
σ(t) + 1 σ(t) + 2 1 σ(t) + 3
1 A (t) = t(t + 3) + 1 whereupon A−1
σ
(t) =
Consequently
=
t +l +1 t +l +2 , 1 t +l +3 1 t + l + 3 −t − l − 2 , −1 t +l +1 (t + l)(t + l + 3) + 1
Next, −1
t ∈ T.
t + 3 −t − 2 −1 t +1
,
t ∈ T,
1 σ(t) + 3 −σ(t) − 2 −1 σ(t) + 1 σ(t)(σ(t) + 3) + 1 1 t + l + 3 −t − l − 2 , −1 t +l +1 (t + l)(t + l + 3) + 1 (Aσ )−1 (t) = (A−1 )σ(t),
Exercise 2.1.19. Let T = 2N0 and t +2 A(t) = t2 + 1
1 t+1 1 t+2
t ∈ T.
t ∈ T.
t ∈ T.
,
t ∈ T.
Prove that (Aσ )−1 (t) = (A−1 )σ (t),
t ∈ T.
Theorem 2.1.20. Let m = n, A be differentiable on Tκ and A−1 , (Aσ )−1 exist on T. Then ∆ −1 A−1 = −A−1 A∆ (Aσ ) −1
= − (Aσ )
A∆ A−1
on Tκ .
95
Dynamic Systems on Time Scales Proof. We have I = AA−1
on T,
whereupon, using Theorem 2.1.14, we get O = I∆ = (AA−1)∆ = A∆ (A−1 )σ + A(A−1 )∆ = A∆ (Aσ)−1 + A(A−1 )∆ = A∆ A−1 + Aσ (A−1 )∆ on Tκ . Hence, A(A−1 )∆ = −A∆ (Aσ)−1 ,
Aσ (A−1 )∆ = −A∆ A−1
on Tκ ,
and (A−1 )∆ = −A−1 A∆ (Aσ )−1 ,
(A−1 )∆ = −(Aσ )−1 A∆ A−1
on Tκ .
This completes the proof. Exercise 2.1.21. Let m = n, A and B be differentiable on Tκ , B−1 and (Bσ )−1 exist on T. Prove (AB−1 )∆ = A∆ − AB−1 B∆ (Bσ)−1 σ = A∆ − AB−1 B∆ B−1 on Tκ .
Definition 2.1.22. We say that the matrix A is rd-continuous on T if each entry of A is rd-continuous. The class of such rd-continuous m × n matrix-valued functions on T is denoted by
Crd = Crd (T) = Crd (T, R m×n ). Below we suppose that A and B are n × n matrix-valued functions.
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Definition 2.1.23. We say that the matrix A is regressive with respect to T provided I + µ(t)A(t) is invertible for all t ∈ Tκ . The class of such regressive and rd-continuous functions is denoted, similar to the scalar case, by R = R (T) = R (T, Rn×n).
Theorem 2.1.24. The matrix-valued function A is regressive if and only if the eigenvalues λi of A are regressive for all 1 ≤ i ≤ n. Proof. Let t ∈ T, j ∈ {1, . . ., n} be arbitrarily chosen and λ j (t) is an eigenvalue corresponding to the eigenvector y(t). Then (1 + µ(t)λ j (t))y(t) = Iy(t) + µ(t)λ j (t)y(t) = Iy(t) + µ(t)A(t)y(t) = (I + µ(t)A(t))y(t), whereupon it follows the assertion. This completes the proof. Example 2.1.25. Let T = 3N0 and t +2 1 A(t) = , t2 + 1 t
t ∈ T.
Consider the equation t + 2 − λ(t) 1 det = 0, t2 + 1 t − λ(t)
t ∈ T.
We have (λ(t) − t)(λ(t) − t − 2) − t 2 − 1 = 0,
t ∈ T,
or (λ(t))2 − 2(t + 1)λ(t) + t 2 + 2t − t 2 − 1 = 0,
t ∈ T,
or (λ(t))2 − 2(t + 1)λ(t) + 2t − 1 = 0,
t ∈ T.
97
Dynamic Systems on Time Scales Therefore λ1,2 (t) = t + 1 ±
q
(t + 1)2 − 2t + 1
p t 2 + 2t + 1 − 2t + 1 p = t + 1 ± t 2 + 2, t ∈ T,
= t +1± and
1 + 3λ1,2 (t) = 0 p 1 + 3 t + 1 ± t2 + 2 = 0
⇐⇒
⇐⇒ p 3t + 4 = ∓3 t 2 + 2
2
2
9t + 24t + 16 = 9t + 18 12t = 1.
⇐⇒
⇐⇒
Consequently 1 + µ(t)λ1,2 (t) 6= 0,
t ∈ T,
i.e., the matrix A is regressive. Theorem 2.1.26. Let A be 2 × 2 matrix-valued function. Then A is regressive if and only if trA + µdetA is regressive. Here trA denotes the trace of the matrix A. Proof. Let A(t) =
a11 (t) a12 (t) a21 (t) a22 (t)
,
t ∈ T.
Then
µ(t)a11(t) µ(t)a12(t) I + µ(t)A(t) = + µ(t)a21(t) µ(t)a22(t) 1 + µ(t)a11(t) µ(t)a12(t) = , t ∈ T. µ(t)a21(t) 1 + µ(t)a22(t) 1 0 0 1
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We get det(I + µ(t)A(t)) = (1 + µ(t)a11(t))(1 + µ(t)a22(t)) − (µ(t))2 a12 (t)a21(t) = 1 + µ(t)a22(t) + µ(t)a11(t) + (µ(t))2 a11 (t)a22(t) − (µ(t))2 a12 (t)a21(t) = 1 + µ(t)(tr(A))(t) + (µ(t))2 (detA)(t) = 1 + µ(t) ((trA)(t) + µ(t)(detA)(t)). (2.1.8) 1. Let A is regressive. Then det(I + µ(t)A(t)) 6= 0,
t ∈ Tκ .
Hence and (2.1.8), we obtain 1 + µ(t) ((trA)(t) + µ(t)(detA)(t)) 6= 0,
t ∈ Tκ ,
(2.1.9)
i.e., trA + µdetA is regressive. 2. Let trA + µdetA is regressive. Then (2.1.9) holds. Hence and (2.1.8), we conclude that A is regressive. This completes the proof.
Definition 2.1.27. Assume that A and B are regressive on T. Then we define A ⊕ B, A, A B by (A ⊕ B)(t) = A(t) + B(t) + µ(t)A(t)B(t), ( A)(t) = −(I + µ(t)A(t))−1A(t),
(A B)(t) = (A ⊕ ( B))(t), respectively.
t ∈ T,
99
Dynamic Systems on Time Scales Example 2.1.28. Let T = 2N and 1 t A(t) = , 2 3t
B(t) =
t 1 2t 3
,
t ∈ T.
Here σ(t) = t + 2, µ(t) = 2, t ∈ T. Then (A ⊕ B)(t)
= =
= = (B ⊕ A)(t)
= =
= =
A(t) + B(t) + µ(t)A(t)B(t) 1 t t 1 + 2 3t 2t 3 1 t t 1 +2 2 3t 2t 3 1 +t 1 +t t + 2t 2 1 + 3t +2 2(1 + t) 3(1 + t) 2t + 6t 2 2 + 9t 1 + 3t + 4t 2 3 + 7t , 2 + 6t + 12t 2 7 + 21t A(t) + B(t) + µ(t)B(t)A(t) 1 t t 1 + 2 3t 2t 3 t 1 1 t +2 2t 3 2 3t 1 +t 1 +t t + 2 t 2 + 3t +2 2(1 + t) 3(1 + t) 2t + 6 2t 2 + 9t 1 +t 1 +t 2t + 4 2t 2 + 6t + 2 + 2t 3 + 3t 4t + 12 4t 2 + 18t
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= I + µ(t)B(t)
= =
det(I + µ(t)B(t))
= =
(I + µ(t)B(t))−1
= =
( B)(t)
= = = =
(A B)(t)
= = =
=
5 + 3t 1 + 7t + 2t 2 , 14 + 6t 3 + 21t + 4t 2 1 0 t 1 +2 0 1 2t 3 1 + 2t 2 , 4t 7
7 + 14t − 8t
7 + 6t, 1 7 −2 −4t 1 + 2t 7 + 6t 7 2 − 7+6t 7+6t , 4t 1+2t − 7+6t 7+6t
−(I + µ(t)B(t))−1B(t) 1 7 −2 t − −4t 1 + 2t 2t 7 + 6t 1 3t 1 − 2t 2t + 3 7 + 6t 1 3t − 7+6t − 7+6t , 2t+3 2t − 7+6t − 7+6t
(A ⊕ ( B))(t)
A(t) + ( B)(t) + µ(t)A(t)( B)(t) 3t 1 − 7+6t − 7+6t 1 t + 2t+3 2t 2 3t − 7+6t − 7+6t 1 3t − 7+6t − 7+6t 1 t +2 2t 2t+3 2 3t − 7+6t − 7+6t ! 2 7+3t 7+6t 14+10t 7+6t
+2
+
6t +7t−1 7+6t 18t 2 +19t−3 7+6t
−2t 2 −3t 7+6t −6t 2 −6t 7+6t 7+3t 7+6t 14+10t 7+6t
=
=
1 3
−2t 2 −3t−1 7+6t −6t 2 −9t−2 7+6t
6t 2 +7t−1 7+6t 18t 2 +19t−3 7+6t
−4t 2 −6t 7+6t −12t 2 −12t 7+6t 7−3t−4t 2 7+6t −12t 2−2t+14 7+6t
!
!
−4t 2−6t−2 7+6t −12t 2−18t−4 7+6t 2t 2 +t−3 7+6t 6t 2 +t−7 7+6t
!
,
! t ∈ T.
101
Dynamic Systems on Time Scales Exercise 2.1.29. Let T = 2N0 , 1 1 A(t) = , 2 −1
B(t) =
3 4 1 0
t ∈ T.
,
Find ( A)(t), Answer. ( A)(t) =
1−4t 1−t−4t 2 2 1−t−4t 2
!
1−t 1−t−4t 2 −1−3t 1−t−4t 2
(A ⊕ B)(t), (A ⊕ B)(t) =
,
t ∈ T.
4 + 4t 3 + 5t
5 + 4t −1 + 8t
Theorem 2.1.30. (R , ⊕) is a group. Proof. Let A, B,C ∈ (R , ⊕). Then (I + µA)−1 ,
(I + µB)−1 ,
(I + µC)−1
exist and I + µ(A ⊕ B) = I + µ(A + B + µAB)
= I + µA + µB + µ2 AB
= I + µA + (I + µA)µB = (I + µA)(I + µB). Therefore (I + µ(A ⊕ B))−1 exists. Also, O⊕A = A⊕O = A.
Next, A ⊕ (−(I + µA)−1 A) = A − (I + µA)−1 A −µ(I + µA)−1 A2
= A − (I + µA)−1 (I + µA)A
= A−A
= O,
,
t ∈ T.
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i.e., the additive inverse of A under the addition ⊕ is −(I + µA)−1 A. Note that I + µ(−(I + µA)−1 A) = (I + µA)−1 (I + µA) −(I + µA)−1 µA
= (I + µA)−1 and then −(I + µA)−1 A ∈ R . Also, (A ⊕ B) ⊕C = (A ⊕ B) +C + µ(A ⊕ B)C
= A + B + µAB +C + µ(A + B + µAB)C
= A + B + µAB +C + µAC + µBC + µ2 ABC, A ⊕ (B ⊕C) = A + (B ⊕C) + µA(B ⊕C)
= A + B +C + µBC + µA(B +C + µBC)
= A + B +C + µBC + µAB + µAC + µ2 ABC. Consequently (A ⊕ B) ⊕C = A ⊕ (B ⊕C), i.e., in (R , ⊕) the associative law holds. This completes the proof. With A we will denote the conjugate matrix of A, with AT we will denote T the transpose matrix of A and A∗ = A is the conjugate transpose of A.
Theorem 2.1.31. Let A and B be regressive. Then 1. A∗ is regressive, 2. A∗ = ( A)∗ , 3. (A ⊕ B)∗ = B∗ ⊕ A∗ .
Proof. Since A is regressive, there exists (I + µA)−1 . 1. We have (I + µA)(I + µA)−1 = I
=⇒
(I + µA)(I + µA)−1
=⇒
(I + µA)
−1 ∗
∗
= I
(I + µA ) = I.
103
Dynamic Systems on Time Scales Therefore (I + µA∗ )−1
and (I + µAT )−1
exist and (I + µA∗ )−1 = (I + µAT )−1 = Consequently A∗ is regressive.
∗ (I + µA)−1 , T (I + µA)−1 .
2. We have ( A)∗ = − (I + µA)−1 A
∗
= −A∗ (I + µA)−1
= −A∗ (I + µA∗ )−1
∗
= A∗ .
3. We have (A + B)∗ = (A + B + µAB)∗ = A∗ + B∗ + µB∗ A∗ = B∗ + A∗ + µB∗ A∗ = B∗ ⊕ A∗ . This completes the proof.
Definition 2.1.32 (Matrix Exponential Functional). Let A ∈ R and t0 ∈ T. The unique solution of the IVP Y ∆ = A(t)Y,
Y (t0 ) = I,
is called the matrix exponential function. It is denoted by eA (·,t0).
Theorem 2.1.33. Let A, B ∈ R and t, s, r ∈ T. Then
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1. e0 (t, s) = I, eA (t,t) = I, 2. eA (σ(t), s) = (I + µ(t)A(t))eA(t, s), ∗ 3. eA (s,t) = e−1 A (t, s) = e A∗ (t, s),
4. eA (t, s)eA(s, r) = eA (t, r), 5. eA (t, s)eB(t, s) = eA⊕B (t, s) if eA (t, s) and B commute. Proof.
1. Consider the IVP Y ∆ = O,
Y (s) = I.
Then its unique solution is eO (t, s) = I. Now we consider the IVP Y ∆ = A(t)Y,
Y (s) = I.
By the definition of eA (·, s) we obtain eA (s, s) = I. 2. By Theorem 2.1.11 and the definition of eA (·, s), we have eA (σ(t), s) = eA (t, s) + µ(t)e∆A(t, s) = eA (t, s) + µ(t)A(t)eA(t, s) = (I + µ(t)A(t))eA(t, s). 3. Let
∗ Z(t) = e−1 A (t, s) .
Dynamic Systems on Time Scales Then
∆
Z (t) =
∆ ∗ −1 eA (t, s)
∗ −1 ∆ = − e−1 (t, s)e (t, s) (e (σ(t), s)) A A A ∗ −1 = − e−1 (t, s)A(t)e (t, s) (e (σ(t), s)) A A A ∗ = − A(t) (eA (σ(t), s))−1 ∗ = − A(t) ((I + µ(t)A(t))eA(t, s))−1 ∗ −1 −1 = − A(t)eA (t, s) (I + µ(t)A(t)) ∗ = eA−1 (t, s) − (I + µ(t)A(t))−1 A(t) ∗ = e−1 A (t, s)( A)(t) = ( A)∗ (t) (eA (t, s))∗
= ( A∗ ) (t)Z(t).
Therefore Z(t) = e A∗ (t, s), i.e., or
∗ e A∗ (t, s) = e−1 A (t, s) , ∗ e−1 A (t, s) = e A∗ (t, s).
4. Let Z(t) = eA (t, s)eA(s, r).
105
106
Svetlin G. Georgiev Then Z ∆ (t) = e∆A (t, s)eA(s, r) = A(t)eA(t, s)eA(s, r) = A(t)Z(t), Z(r) = eA (r, s)eA(s, r) = e−1 A (s, r)eA(s, r) = I = eA (r, r). Therefore eA (t, s)eA(s, r) = eA (t, r).
5. Let Z(t) = eA (t, s)eB(t, s). Then Z ∆ (t) = e∆A (t, s)eσB(t, s) + eA(t, s)e∆B(t, s) = A(t)eA (t, s)(I + µ(t)B(t))eB(t, s) +B(t)eA (t, s)eB(t, s) = A(t)(I + µ(t)B(t))eA(t, s)eB(t, s) + B(t)eA(t, s)eB(t, s) = (A(t) + B(t) + µ(t)A(t)B(t))eA(t, s)eB(t, s) = (A ⊕ B)(t)Z(t),
Z(s) = eA (s, s)eB(s, s) = I. Consequently
eA⊕B (t, s) = eA (t, s)eB(t, s). This completes the proof. Theorem 2.1.34. Let A ∈ R and t, s ∈ T. Then (eA (s,t))t∆ = −eA (s, σ(t))A(t).
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107
Proof. We have (eA (s,t))t∆ = = =
e−1 A (t, s)
∆
t ∆ ∗ e A∗ (t, s) t
∗ (e A∗ (t, s))t∆
= (( A∗ )(t)e A∗ (t, s))∗ ∗ = − (I + µ(t)A∗ (t))−1 A∗ (t)e A∗ (t, s) ∗ = −A∗ (t) (I + µ(t)A∗ (t))−1 e A∗ (t, s)
= (−A∗ (t)e A∗ (σ(t), s))∗ = −e∗ A∗ (σ(t), s)A(t)
= −eA (s, σ(t))A(t). This completes the proof.
Theorem 2.1.35 (Variation of Constants). Let A ∈ R and f : T → Rn be rdcontinuous. Let also, t0 ∈ T and y0 ∈ Rn . Then the IVP y∆ (t) = A(t)y + f (t),
y(t0 ) = y0 ,
(2.1.10)
has unique solution y : T → Rn and this solution is given by y(t) = eA (t,t0)y0 + Proof.
Z t t0
eA (t, σ(τ)) f (τ)∆τ.
1. Let y be given by (2.1.11). Then Z t y(t) = eA (t,t0) y0 + eA (t0, σ(τ)) f (τ)∆τ , t 0 Z t y∆ (t) = A(t)eA(t,t0) y0 + eA (t0 , σ(τ)) f (τ)∆τ t0
+eA (σ(t),t0)eA (t0, σ(t)) f (t)
= A(t)y(t) + f (t), y(t0 ) = y0 , i.e., y satisfies the IVP (2.1.10).
(2.1.11)
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2. Let y˜ be another solution of the IVP (2.1.10). Let v(t) = eA (t0 ,t)y(t). ˜ Then y(t) ˜ = eA (t,t0)v(t) and A(t)eA(t,t0)v(t) + f (t) = A(t)y(t) ˜ + f (t) = y˜∆ (t) = A(t)eA(t,t0)v(t) + eA (σ(t),t0)v∆ (t), whereupon eA (σ(t),t0)v∆ (t) = f (t) or v∆ (t) = eA (t0, σ(t)) f (t). Hence, v(t) = v(t0 ) + = y0 +
Z t
Z t t0
t0
eA (t0, σ(τ)) f (τ)∆τ
eA (t0 , σ(τ)) f (τ)∆τ
and y(t) ˜ = eA (t0 ,t)v(t) Z t = eA (t0 ,t) y0 + eA (t0 , σ(τ)) f (τ)∆τ , t0
i.e., y˜ = y, where y is given by (2.1.11). This completes the proof. Theorem 2.1.36. Let A ∈ R and C be n × n differentiable matrix. If C is a solution of the dynamic equation C∆ = A(t)C −Cσ A(t), then C(t)eA(t, s) = eA (t, s)C(s).
(2.1.12)
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109
Proof. We fix s ∈ T. Let F(t) = C(t)eA(t, s) − eA (t, s)C(s). Then F ∆ (t) = C∆ (t)eA(t, s) +C(σ(t))e∆A(t, s) − e∆A (t, s)C(s) = (A(t)C(t) −C(σ(t))A(t))eA (t, s)
+C(σ(t))A(t)eA(t, s) − A(t)eA(t, s)C(s)
= A(t)C(t)eA(t, s) − A(t)eA(t, s)C(s)
= A(t) (C(t)eA (t, s) − eA (t, s)C(s))
= A(t)F(t),
F(s) = C(s)eA (s, s) − eA(s, s)C(s) = C(s) −C(s) = 0,
i.e., F ∆ (t) = A(t)F(t),
F(s) = 0.
Hence and Theorem 2.1.35, we get F(t) = 0 and (2.1.12) holds. This completes the proof. Corollary 2.1.37. Let A ∈ R and C is an n × n constant matrix that commutes with A. Then C commutes with eA . Proof. We have Cσ = C
and C∆ = A(t)C −CA(t).
Hence and Theorem 2.1.36, it follows that C commutes with eA . This completes the proof. Corollary 2.1.38. Let A ∈ R be a constant matrix. Then A commutes with eA . Proof. We apply Theorem 2.1.36 for C = A. This completes the proof.
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Theorem 2.1.39 (Variation of Constants). Let A ∈ R and f : T → Rn be rdcontinuous. Let also, t0 ∈ T and x0 ∈ Rn . Then the initial value problem x∆ = −A∗ (t)xσ + f (t),
x(t0 ) = x0 ,
(2.1.13)
has a unique solution x : T → Rn and this solution is given by x(t) = e A∗ (t,t0)x0 +
Z t t0
e A∗ (t, τ) f (τ)∆τ.
Proof. The equation (2.1.13) can be rewritten in the form x∆ (t) = −A∗ (t) x(t) + µ(t)x∆ (t) + f (t)
= −A∗ (t)x(t) − µ(t)A∗(t)x∆ (t) + f (t),
whereupon (I + µ(t)A∗ (t))x∆(t) = −A∗ (t)x(t) + f (t), and x∆ (t) = −A∗ (t)(I + µ(t)A∗(t))−1x(t) + (I + µ(t)A∗(t))−1 f (t) = ( A∗ )(t)x(t) + (I + µ(t)A∗ (t))−1 f (t).
Hence and Theorem 2.1.35, we obtain x(t) = e A∗ (t,t0)x0 +
Z t
e A∗ (t, σ(τ))(I + µ(τ)A∗ (τ))−1 f (τ)∆τ
t0
∗ e∗A (σ(τ),t) (I + µ(τ)A(τ))−1 f (τ)∆τ
= e A∗ (t,t0)x0 +
t0 Z t
= e A∗ (t,t0)x0 +
Z t
= e A∗ (t,t0) + = e A∗ (t,t0) + This completes the proof.
Z t
t0 Z t t0
t0
(I + µ(τ)A(τ))−1eA (σ(τ),t)
(eA (τ,t))∗ f (τ)∆τ e A∗ (t, τ) f (τ)∆τ.
∗
f (τ)∆τ
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Dynamic Systems on Time Scales
Theorem 2.1.40 (Liouville’s Formula). Let A ∈ R be a 2 × 2 matrix-valued function and assume that X is a solution of the equation X ∆ = A(t)X. Then X satisfies Liouville’s formula detX(t) = etrA+µdetA (t,t0)detX(t0 ),
t ∈ T.
Proof. By Theorem 2.1.26, it follows that trA + µdetA is regressive. Let A(t) =
a11 (t) a12 (t) a21 (t) a22 (t)
,
X(t) =
x11 (t) x21 (t)
x12 (t) x22 (t)
.
Then
x∆11 (t) x∆21 (t)
x∆12 (t) x∆22 (t)
= =
whereupon
∆ x (t) ∆11 x12 (t) ∆ x21 (t) ∆ x22 (t)
a11 (t) a21 (t)
a12(t) a22(t)
x11 (t) x21 (t)
a11 (t)x11(t) + a12(t)x12(t) a21 (t)x11(t) + a22(t)x21(t)
= = = =
x12 (t) x22 (t)
a11 (t)x12(t) + a12(t)x22(t) a21 (t)x12(t) + a22(t)x22(t)
a11 (t)x11(t) + a12 (t)x21 (t) a11 (t)x12(t) + a12 (t)x22 (t) a21 (t)x11(t) + a22 (t)x21 (t) a21 (t)x12(t) + a22 (t)x22 (t).
,
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Then detX(t)
=
∆
=
(detX) (t)
= =
=
=
=
=
x11 (t)x22(t) − x12(t)x21(t), x∆11 (t)x22(t) + xσ11(t)x∆22(t)
−x∆12 (t)x21(t) − xσ12(t)x∆21(t) ∆ σ x11 (t) xσ12 (t) x11 (t) x∆12 (t) det + det x21 (t) x22 (t) x∆21 (t) x∆22 (t) a11 (t)x11(t) + a12 (t)x21(t) a11 (t)x12(t) + a12(t)x22(t) det x21(t) x22(t) x11 (t) + µ(t)x∆11(t) x12 (t) + µ(t)x∆12(t) +det x∆21(t) x∆22(t) a11(t)x11(t)x22(t) + a12(t)x21(t)x22(t) −a11 (t)x12(t)x21(t) − a12 (t)x21(t)x22(t) ∆ x11 (t) x12 (t) x11(t) x∆12 (t) +det + µ(t)det x∆21 (t) x22 (t) x∆21(t) x∆22 (t) x11(t) x12(t) a11 (t)detX(t) + det x∆21(t) x∆22(t) ∆ x11 (t) x∆12(t) +µ(t)det x∆21 (t) x∆22(t)
a11 (t)detX(t) x11(t) x12(t) +det a21 (t)x11(t) + a22 (t)x21(t) a21 (t)x12(t) + a22(t)x22(t) a11 (t)x11(t) + a12(t)x21(t) a11 (t)x12(t) + a12(t)x22(t) +µ(t)det x∆21 (t) x∆22 (t) a11 (t)detX(t) + a21 (t)x11(t)x12(t) + a22 (t)x11(t)x22(t) −a21 (t)x11(t)x12(t) − a22 (t)x21(t)x12(t)
113
Dynamic Systems on Time Scales +µ(t) a11 (t)det +a12 (t)det =
=
=
x11 (t) x∆21 (t)
x21 (t) x∆21 (t)
x12(t) x∆22(t) x22(t) x∆22(t)
a11 (t)detX(t) + a22(t)detX(t) x11 (t) x12 (t) +µ(t) a11 (t)det a21 (t)x11(t) + a22(t)x21(t) a21 (t)x12(t) + a22(t)x22(t) x21 (t) x22 (t) +a12 (t)det a21 (t)x11(t) + a22(t)x21(t) a21 (t)x12(t) + a22(t)x22(t)
trA(t)detX(t) +µ(t) a11 (t) a21 (t)x12(t)x11(t) + a22(t)x22(t)x11(t) −a21 (t)x11(t)x12(t) − a22(t)x12(t)x21(t) +a12 (t) a21 (t)x21(t)x12(t) + a22 (t)x22(t)x21(t) −a21 (t)x11(t)x22(t) − a22(t)x21(t)x22(t)
trA(t)detX(t) + µ(t) (a11 (t)a22(t)detX(t) − a12(t)a21(t)detX(t))
=
trA(t)detX(t) + µ(t)detA(t)detX(t)
=
(trA(t) + µ(t)detA(t))detX(t),
i.e., (detX)∆(t) = (trA(t) + µ(t)detA(t))detX(t). Therefore detX(t) = etrA+µdetA (t,t0 )detX(t0). This completes the proof. Example 2.1.41. Let T = lZ, l > 0. Consider the IVP 3t 4t + 1 1 0 ∆ X (t) = X(t), X(0) = , 3 2 +t 1 1 Here
3t 4t + 1 A(t) = 3 2 +t µ(t) = l, t ∈ T.
,
t ∈ T,
t > 0.
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Then detA(t) = 3t(2 + t) − 3(4t + 1) = 6t + 3t 2 − 12t − 3
= 3t 2 − 6t − 3,
trA(t) = 3t + 2 + t
= 2 + 4t, trA(t) + µ(t)detA(t) = 2 + 4t + l(3t 2 − 6t − 3)
= 3lt 2 + (4 − 6l)t + 2 − 3l,
t ∈ T.
Let f (t) = 3lt 2 + (4 − 6l)t + 2 − 3l,
t ∈ T.
Then, using Liouville’s formula, we get detX(t) = e f (t, 0)detX(0) = e f (t, 0) t−l
=
∏(1 + µ(s) f (s))
s=0 t−l
=
1 + l 3ls2 + (4 − 6l)s + 2 − 3l
t−l
3l 2 s2 + (4l − 6l 2 )s − 3l 2 + 2l + 1 ,
s=0
=
∏ ∏
s=0
t ∈ T,
t > 0.
Exercise 2.1.42. Let A ∈ R be a n × n matrix-valued function and assume that X is a solution of the equation X ∆ = A(t)X. Prove that X satisfies Liouville’s formula detX(t) = etrA+µdetA (t,t0)detX(t0 ), Hint. Use Theorem 2.1.40 and induction.
t ∈ T.
Dynamic Systems on Time Scales
115
2.2. Constant Coefficients In this section we suppose that A is a n × n constant matrix and A ∈ R . Let t0 ∈ T. Consider the equation x∆ = Ax. (2.2.1) Theorem 2.2.1. Let λ, ξ be an eigenpair of A. Then x(t) = eλ (t,t0)ξ,
t ∈ Tκ ,
is a solution of the equation (2.2.1). Proof. We have Aξ = λξ. Then x∆ (t) = e∆λ (t,t0)ξ = λeλ (t,t0)ξ = eλ (t,t0)(λξ) = eλ (t,t0)Aξ = A (eλ (t,t0)ξ) = Ax(t),
t ∈ Tκ .
This completes the proof. Example 2.2.2. Consider the system ∆ x1 (t) = −3x1 − 2x2 x∆2 (t) = 3x1 + 4x2 . Here A= Then
−3 −2 3 4
.
−3 − λ −2 0 = det 3 4−λ = (λ − 4)(λ + 3) + 6 = λ2 − λ − 6
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Svetlin G. Georgiev
and λ1 = 3,
λ2 = −2.
The considered system is regressive for any time scale for which −2 ∈ R . Note that 1 −2 ξ1 = , ξ2 = −3 1 are eigenvalues corresponding to λ1 and λ2 , respectively. Therefore x(t) = c1 e3 (t,t0)ξ1 + c2 e−2 (t,t0)ξ2 1 −2 = c1 e3 (t,t0) + c2 e−2 (t,t0) , −3 1 where c1 and c2 are real constants, is a solution of the considered system for any time scale for which −2 ∈ R . Example 2.2.3. Consider the system ∆ x1 (t) = x1 (t) − x2 (t) x∆ (t) = −x1 (t) + 2x2 (t) − x3 (t) ∆2 x3 (t) = −x2 (t) + x3 (t).
Here
Then
1 −1 0 A = −1 2 −1 . 0 −1 1
0 = det(A − λI) 1 − λ −1 0 = det −1 2 − λ −1 0 −1 1 − λ
= −(λ − 1)2 (λ − 2) + (λ − 1) + (λ − 1)
= (λ − 1) (−(λ − 1)(λ − 2) + 2) = (λ − 1) −λ2 + 3λ
= −λ(λ − 1)(λ − 3).
117
Dynamic Systems on Time Scales Therefore λ1 = 0,
λ2 = 1,
λ3 = 3.
Note that the matrix A is regressive for any time scale and 1 1 1 ξ1 = 1 , ξ2 = 0 , ξ3 = −2 1 −1 1
are eigenvalues corresponding to λ1 , λ2 and λ3 , respectively. Consequently x(t) = c1 ξ1 + c2 e1 (t,t0)ξ2 + c3 e3 (t,t0)ξ3 1 1 1 = c1 1 + c2 e1 (t,t0) 0 + c3 e3 (t,t0) −2 , 1 −1 1
where c1 , c2 and c3 are constants, is a general solution of the considered system. Example 2.2.4. Consider the system ∆ x (t) = −x1 (t) + x2 (t) + x3 (t) ∆1 x2 (t) = x2 (t) − x3 (t) + x4 (t) ∆ (t) = 2x (t) − 2x (t) x 3 4 ∆3 x4 (t) = 3x4 (t).
Here
Then
−1 0 A= 0 0
1 1 0 1 −1 1 . 0 2 −2 0 0 3
0 = det(A − λI) −1 − λ 1 1 0 0 1 − λ −1 1 = det 0 0 2 − λ −2 0 0 0 3−λ = (λ + 1)(λ − 1)(λ − 2)(λ − 3)
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Svetlin G. Georgiev
and λ1 = −1,
λ2 = 1,
λ3 = 2,
λ4 = 3.
The matrix A is regressive for any time scale for which −1 ∈ R . Note that 1 0 0 0 0 1 0 0 ξ1 = 0 , ξ2 = 0 , ξ3 = 1 and ξ4 = 0 0 0 0 1
are eigenvectors corresponding to λ1 , λ2 , λ3 and λ4 , respectively. Consequently x(t) = c1 e−1 (t,t0)ξ1 + c2 e1 (t,t0)ξ2 + c3 e2 (t,t0)ξ3 + c4 e5 (t,t0)ξ4 1 0 0 1 = c1 e−1 (t,t0) 0 + c2 e1 (t,t0) 0 0 0 0 0 0 0 +c3 e2 (t,t0 ) 1 + c4 e3 (t,t0) 0 , 0 1
where c1 , c2 , c3 and c4 are real constants, is a general solution of the considered system. Exercise 2.2.5. Find a general solution of the system ∆ x1 (t) = x2 (t) x∆2 (t) = x1 (t). Answer. x(t) = c1 e1 (t,t0)
1 1
+ c2 e−1 (t,t0)
where c1 , c2 ∈ R, for any time scale for which −1 ∈ R . Theorem 2.2.6. Assume that A ∈ R . If x(t) = u(t) + iv(t),
t ∈ Tκ ,
1 −1
,
Dynamic Systems on Time Scales
119
is a complex vector-valued solution of the system (2.2.1), where u(t) and v(t) are real vector-valued functions on T, then u(t) and v(t) are real vector-valued solutions of the system (2.2.1) on T. Proof. We have x∆ (t) = A(t)x(t) = A(t) (u(t) + iv(t)) = A(t)u(t) + iA(t)v(t) = u∆ (t) + iv∆ (t),
t ∈ Tκ .
Equating real and imaginary parts, we get u∆ (t) = A(t)u(t), v∆ (t) = A(t)v(t),
t ∈ Tκ .
This completes the proof. Example 2.2.7. Consider the system ∆ x1 (t) = x1 (t) + x2 (t) x∆2 (t) = −x1 (t) + x2 (t). Here A=
1 1 −1 1
.
Then 0 = det(A − λI) 1−λ 1 = det −1 1 − λ = (λ − 1)2 + 1
= λ2 − 2λ + 1 + 1
= λ2 − 2λ + 2, whereupon
λ1,2 = 1 ± i.
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Svetlin G. Georgiev
Note that ξ=
1 i
is an eigenvector corresponding to the eigenvalue λ = 1 + i. We have 1 x(t) = e1+i (t,t0) i 1 = e1 (t,t0) cos 1 (t,t0) + i sin 1 (t,t0) 1+µ 1+µ i ! !! cos 1 (t,t0) i sin 1 (t,t0) 1+µ 1+µ = e1 (t,t0) + i cos 1 (t,t0) − sin 1 (t,t0) 1+µ 1+µ ! ! sin 1 (t,t0) cos 1 (t,t0) 1µ 1+µ + ie1 (t,t0) . = e1 (t,t0) − sin 1 (t,t0 ) cos 1 (t,t0) 1µ
1+µ
Consequently e1 (t,t0)
cos
1 1+µ
− sin
!
(t,t0)
1 1+µ
(t,t0)
and e1 (t,t0)
are solutions of the considered system. Therefore ! cos 1 (t,t0) 1+µ x(t) = c1 e1 (t,t0) + c2 e1 (t,t0) − sin 1 (t,t0) 1+µ
sin cos
1 1+µ 1 1+µ
sin cos
(t,t0) (t,t0)
1 1+µ 1 1+µ
where c1 , c2 ∈ R, is a general solution of the considered system. Example 2.2.8. Consider the system ∆ x1 (t) = x2 (t) x∆ (t) = x3 (t) ∆2 x3 (t) = 2x1 (t) − 4x2 (t) + 3x3 (t).
Here
0 1 0 A = 0 0 1 . 2 −4 3
!
(t,t0) (t,t0 )
!
,
121
Dynamic Systems on Time Scales Then 0 = det(A − λI) −λ 1 0 = det 0 −λ 1 2 −4 3 − λ = −λ2 (λ − 3) + 2 − 4λ
= − λ3 − 3λ2 + 4λ − 2
= −(λ − 1)(λ2 − 2λ + 2), whereupon λ1 = 1, Note that
1 ξ1 = 1 1
λ2,3 = 1 ± i.
1 and ξ2 = 1 + i 2i
are eigenvectors corresponding to the eigenvalues λ1 = 1 and λ2 = 1 + i, respectively. Note that 1 1 e1+i(t,t0 ) 1 + i = e1 (t,t0 ) cos 1 (t,t0 ) + isin 1 (t,t0 ) 1 + i 1+µ 1+µ 2i 2i cos 1 (t,t0 ) sin 1 (t,t0 ) 1+µ 1+µ (1 + i) cos 1 (t,t0 ) + i (1 + i) sin 1 (t,t0 ) = e1 (t,t0 ) 1+µ 1+µ 2icos 1 (t,t0 ) 2isin 1 (t,t0 ) 1+µ 1+µ isin 1 (t,t0 ) cos 1 (t,t0 ) 1+µ 1+µ (1 + i) cos 1 (t,t0 ) + (−1 + i) sin 1 (t,t0 ) = e1 (t,t0 ) 1+µ 1+µ 2icos 1 (t,t0 ) −2 sin 1 (t,t0 ) 1+µ 1+µ cos 1 (t,t0 ) sin 1 (t,t0 ) 1+µ 1+µ cos 1 (t,t0 ) + sin 1 (t,t0 ) 1 (t,t0 ) − sin 1 (t,t0 ) = e1 (t,t0 ) cos 1+µ + i 1+µ 1+µ 1+µ −2 sin 1 (t,t0 ) 2 cos 1 (t,t0 )
1+µ
1+µ
.
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Svetlin G. Georgiev
Consequently cos 1 (t,t0) 1 1+µ 1 (t,t0 ) − sin 1 (t,t0) x(t) = e1 (t,t0) c1 1 + c2 cos 1+µ 1+µ 1 −2 sin 1 (t,t0) 1+µ sin 1 (t,t0) 1+µ cos 1 (t,t ) + sin 1 (t,t ) +c3 0 0 , 1+µ 1+µ 2 cos 1 (t,t0) 1+µ
where c1 , c2 , c3 ∈ R, is a general solution of the considered system. Exercise 2.2.9. Find a general solution of the system ∆ x1 (t) = x1 (t) − 2x2 (t) + x3 (t) x∆ (t) = −x1 (t) + x3 (t) ∆2 x3 (t) = x1 (t) − 2x2 (t) + x3 (t).
Theorem 2.2.10 (Putzer’s Algorithm). Let A ∈ R be a constant n × n matrix and t0 ∈ T. If λ1 , λ2 , . . ., λn are the eigenvalues of A, then n−1
eA (t,t0) =
∑ rk+1(t)Pk ,
k=0
where
is the solution of the IVP λ1 0 0 1 λ2 0 r ∆ = 0 1 λ3 .. .. .. . . . 0 0 0
r1 (t) r(t) = ... rn (t) ... ... ... .. .
0 0 0 .. .
. . . λn
r,
r(t0) =
1 0 0 .. . 0
,
P0 = I, Pk+1 = (A − λk+1 I)Pk ,
0 ≤ k ≤ n − 1.
(2.2.2)
Dynamic Systems on Time Scales
123
Proof. Since A is regressive, we have that all eigenvalues of A are regressive. Therefore the IVP (2.2.2) has unique solution. We set n−1
X(t) =
∑ rk+1(t)Pk.
k=0
We have P1 = (A − λ1 I)P0 = (A − λ1 I),
P2 = (A − λ2 I)P1
= (A − λ2 I)(A − λ1 I), .. .
Pn = (A − λn I)Pn−1 = (A − λn I) . . .(A − λ1 I) = 0.
Therefore X ∆ (t) = X ∆ (t) − AX(t) =
n−1
∆ (t)Pk , ∑ rk+1
k=0 n−1
n−1
∆ (t)Pk − A ∑ rk+1(t)Pk ∑ rk+1
k=0
k=0
n−1
∆ = r1∆ (t)P0 + ∑ rk+1 (t)Pk k=1 n−1
−A ∑ rk+1 (t)Pk k=0
124
Svetlin G. Georgiev n−1
= λ1 r1 (t)P0 + ∑ (rk (t) + λk+1rk+1 (t))Pk k=1 n−1
− ∑ rk+1(t)APk k=1 n−1
=
∑
rk (t)Pk + λ1 r1 (t)P0 k=1 n−1 n−1 + ∑ λk+1 rk+1(t)Pk − ∑ rk+1 (t)APk
=
k=1 n−1
n−1
k=0
∑
∑
rk (t)Pk + λk+1 rk+1(t)Pk k=1 k=0 n−1 − ∑ rk+1(t)APk k=0 n−1
=
n−1
∑ rk (t)Pk − ∑ (A − λk+1I)rk+1(t)Pk
k=1 n−1
=
k=0 n−1
∑ rk (t)Pk − ∑ rk+1(t)Pk+1
k=1
k=0
= −rn (t)Pn
= 0,
t ∈ Tκ .
Also, n−1
X(t0 ) =
∑ rk+1(t0)Pk k=0
= r1 (t0 )P0 = I. This completes the proof. Example 2.2.11. Consider the system ∆ x1 (t) = 2x1 (t) + x2 (t) + 2x3 (t) x∆ (t) = 4x1 (t) + 2x2 (t) + 4x3 (t) ∆2 x3 (t) = 2x1 (t) + x2 (t) + 2x3 (t), t ∈ Tκ .
125
Dynamic Systems on Time Scales Here
2 1 2 A = 4 2 4 . 2 1 2
Then
0 = det(A − λI) 2−λ 1 2 = det 4 2−λ 4 2 1 2−λ
= −(λ − 2)3 + 8 + 8 + 4(λ − 2) + 4(λ − 2) + 4(λ − 2)
= −(λ − 2)3 + 12(λ − 2) + 16
= − λ3 − 6λ2 + 12λ − 8 − 12λ + 24 − 16 = − λ3 − 6λ2
= −λ2 (λ − 6), whereupon
λ1 = 0,
λ2 = 0,
λ3 = 6.
Consider the IVPs r1∆ (t) = 0, r2∆ (t) r3∆ (t)
r1 (t0 ) = 1,
= r1 (t),
r2 (t0 ) = 0,
= r2 (t) + 6r3(t),
r3 (t0 ) = 0.
We have r1 (t) = 1, r2∆ (t) = 1,
t ∈ Tκ
r2 (t0) = 0.
Then r2 (t) = t − t0 ,
t ∈ Tκ ,
and r3∆ (t) = t − t0 + 6r3 (t),
r3 (t0) = 0.
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Therefore r3 (t) =
Z t t0
e6 (t, σ(τ))(τ − t0)∆τ,
t ∈ Tκ .
Next, P0 P1
1 0 0 = 0 1 0 , 0 0 1 = (A − λ1 I)P0
= AP0 = A
2 1 2 = 4 2 4 , 2 1 2 P2 = (A − λ1 I)(A − λ2 I) = A2 I
2 = A
P3
2 1 2 = 4 2 4 2 1 2 12 6 = 24 12 12 6 = 0.
2 1 2 4 2 4 2 1 2 12 24 , 12
Therefore eA (t,t0) = r1 (t)P0 + r2 (t)P1 + r3 (t)P2 1 0 0 2 1 2 = 0 1 0 + (t − t0 ) 4 2 4 0 0 1 2 1 2 12 Z t e6 (t, σ(τ))(τ − t0)∆τ 24 + t0 12
6 12 12 24 , 6 12
127
Dynamic Systems on Time Scales and
x1 (t) c1 x2 (t) = eA (t,t0) c2 , x3 (t) c3
where c1 , c2 , c3 ∈ R, is a general solution of the considered system. Exercise 2.2.12. Using Putzer’s algorithm, find eA (t,t0), where 1. A= 2.
1 2 −1 3
,
1 −1 1 A= 1 0 2 . −1 1 1
2.3. Advanced Practical Problems Problem 2.3.1. Let T = 3N0 , A(t) =
t 2 +2 t+1
4t − 1
2
t + 3t 3t
!
t ∈ T.
,
Find A∆ (t), t ∈ T. Answer. ∆
A (t) =
3t 2 +4t−2 (t+1)(3t+1)
4
Problem 2.3.2. Let T = 3N0 , t + 10 t 2 − 2t + 2 A(t) = , t t2 + t + 1
4t + 3 3
B(t) =
!
,
t ∈ T.
t −2 t +1 t2 t3 − 1
Prove (AB)∆ (t) = A∆ (t)Bσ(t) + A(t)B∆(t),
t ∈ Tκ .
,
t ∈ T.
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Svetlin G. Georgiev
Problem 2.3.3. Let T = 3N0 and 2 t + 2t + 2 A(t) = 1
t +1 t2 + 2
t+1
,
t ∈ T.
Prove that (Aσ )−1 (t) = (A−1 )σ (t), Problem 2.3.4. Let T = N20 and 1 −1 A(t) = , 0 1
B(t) =
t ∈ T.
2 0 −1 1
t ∈ T.
,
Find (A ⊕ B)(t), Answer.
t ∈ T.
√ √ 6( √t + 1) −(2√ t + 2) , −(2 t + 2) 2 t +3
t ∈ T.
Problem 2.3.5. Find a general solution of the system ∆ x1 (t) = 2x1 (t) + 3x2 (t) x∆2 (t) = x1 (t) + 4x2 (t). Answer. x(t) = c1 e1 (t,t0) where c1 , c2 ∈ R.
3 −1
+ c2 e5 (t,t0)
Problem 2.3.6. Find a general solution of the system ∆ x1 (t) = −x1 (t) − x2 (t) − x3 (t) x∆ (t) = x1 (t) − x2 (t) + 3x3 (t) ∆2 x3 (t) = x1 (t) − x2 (t) + 4x3 (t).
1 1
,
Problem 2.3.7. Using Putzer’s algorithm, find eA (t,t0), where 1. A=
2 3 1 −4
,
Dynamic Systems on Time Scales 2.
−1 2 3 A= 1 1 −4 . 1 −1 2
129
Chapter 3
Functionals Suppose that T is a time scale with forward jump operator, backward jump operator, delta differentiation operator and nabla differentiation operator σ, ρ, ∆ and ∇, respectively. Let a, b ∈ T, a < b, and [a, b] is a time scale interval.
3.1. Definition for Functionals Definition 3.1.1 (Functional). By a functional, we mean a correspondence which assigns a definite(real) number to each function belonging to some class.
1 ([a, b]). The formula Example 3.1.2. Let y ∈ Crd
J(y) =
Z b a
1 defines a functional on Crd ([a, b]).
2 y∆ (t) ∆t
1 Example 3.1.3. Let y ∈ Crd ([a, b]), F ∈ C ([a, b] × R × R). The expression
J(y) =
Z b a
1 ([a, b]). defines a functional on Crd
F t, y(t), y∆(t) ∆t
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Svetlin G. Georgiev
Example 3.1.4. Let y ∈ Cld1 ([a, b]). Then Z b 2 ∇ J(y) = y(t) − 2 y (t) ∇t a
defines a functional on Cld1 ([a, b]). Definition 3.1.5. Let X be a linear normed space with a norm k · k. 1. The functional J : X → R is said to be continuous at the point y1 ∈ X if for each ε > 0 there exists a δ = δ(ε) > 0 such that |J(y) − J(y1 )| < ε provided that ky − y1 k < δ. 2. The functional J : X → R is said to be continuous on X if it is continuous at each point of X. 3. The functional J : X → R is said to be linear if J(α1 y1 + α2 y2 ) = α1 J(y1 ) + α2 J(y2 ) for any α1 , α2 ∈ R and for any y1 , y2 ∈ X.
Example 3.1.6. For y ∈ Crd ([a, b]), the expression J(y) =
Z b
y(t)∆t
a
defines a linear functional on Crd ([a, b]). Example 3.1.7. Let f ∈ Cld ([a, b]). For y ∈ Cld ([a, b]), the expression J(y) =
Z b
f (t)y(t)∇t
a
defines a linear functional on Cld ([a, b]).
133
Functionals
3.2. Self-Adjoint Second Order Matrix Equations Suppose that R, P ∈ Crd (T, Rn×n ) be such that R(t) is symmetric and invertible, P(t) is symmetric for each t ∈ T. Consider the second order matrix equation ∆ R(t)Y ∆ + P(t)Y σ = 0. (3.2.1)
1 Definition 3.2.1. We call Y ∈ Crd (T, Rn×n ) a solution of the equation (3.2.1) provided ∆ RY ∆ + PY σ (t) = 0 2
holds for all t ∈ Tκ .
Remark 3.2.2. If Y ∈ Crd (T, Rn×n ) is a solution of the equation (3.2.1), then 1 RY ∆ ∈ Crd (Tκ , Rn×n ).
Theorem 3.2.3 (Wronskian’s Identity). Let Y and Y1 be solutions of (3.2.1). Then T Y T RY1∆ − Y ∆ RY1 is a constant.
Proof. We have
and
∆ RY ∆ = −PY σ , T ∆ Y∆ R = − (Y σ )T P, ∆ T Y T RY1∆ − Y ∆ RY1 =
∆ T − Y ∆ RY1 ∆ T T = (Y σ ) RY1∆ + Y ∆ RY1∆ T ∆ T − Y ∆ R Y1σ − Y ∆ RY1∆ Y T RY1∆
∆
= b (Y σ )T (−PY1σ ) + (Y σ )T PY1σ
= 0. This completes the proof.
134
Svetlin G. Georgiev Now we set Y Z= RY ∆
κ
and S =
on T
0 R−1 −P −µPR−1
on T.
Theorem 3.2.4. Y solves (3.2.1) if and only if Z solves Z ∆ = S(t)Z Proof.
1. Let Y solves (3.2.1). Then ∆ Y ∆ Z (t) = (t) RY ∆ Y ∆ (t) ∆ RY ∆ (t)
= = S(t)Z(t) =
Y ∆ (t) −P(t)Y σ (t)
on Tκ .
(3.2.2)
!
,
0 R−1 (t) −P(t) −µ(t)P(t)R−1(t)
Y ∆ (t) = −P(t)Y(t) − µ(t)P(t)Y ∆ (t) Y ∆ (t) = −P(t) Y (t) + µ(t)Y ∆ (t) Y ∆ (t) = , t ∈ Tκ . −P(t)Y σ (t)
Y (t) R(t)Y ∆ (t)
Therefore Z solves (3.2.2). 2. Let Z solves (3.2.2). Then, using the computations in the previous point, we have ! Y ∆ (t) Y ∆ (t) ∆ = , t ∈ Tκ , −P(t)Y σ (t) RY ∆ (t) whereupon
RY ∆
∆
(t) = −P(t)Y σ (t),
i.e., Y solves (3.2.1). This completes the proof.
t ∈ Tκ ,
135
Functionals Theorem 3.2.5. S is a regressive matrix on T. Proof. We have
0 R−1 (t) I + µ(t)S(t) = + µ(t) −P(t) −µ(t)P(t)R−1(t) I µ(t)R1 (t) = , −µ(t)P(t) I − (µ(t))2 P(t)R−1(t) I 0 0 I
det(I + µ(t)S(t)) = 1 − (µ(t))2 det(P(t)R−1(t)) + (µ(t))2 det(P(t)R−1(t)) = 1,
t ∈ T.
This completes the proof. Therefore, for any choice of a 2n × 2n matrix Za , the problem Z ∆ = S(t)Z,
Z(a) = Za ,
has a unique solution. From here, using Theorem 3.2.4, for any choice of 2n × 2n matrices Ya and Ya0 , the problem R(t)Y ∆
∆
+ P(t)Y σ = 0,
Y (a) = Ya ,
Y ∆ (a) = Ya0 ,
has a unique solution Y . Let Y˜ be the unique solution of the initial value problem R(t)Y ∆
∆
+ P(t)Y σ = 0,
Y (a) = 0,
Y ∆ (a) = R−1 (a).
(3.2.3)
Definition 3.2.6. This Y˜ is called the principal solution of the equation (3.2.1).
Theorem 3.2.7. The matrices ˜ = RY˜ ∆Y˜ −1 (t) and D(t) ˜ = Y˜ Y˜ σ −1 R−1 (t), Q(t)
are symmetric matrices.
t ∈ Tκ ,
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Svetlin G. Georgiev
Proof. By Theorem 3.2.3, the Wronskian identity, we have T T Y˜ T RY˜ ∆ − Y˜ ∆ RY˜ (t) = Y˜ T RY˜ ∆ − Y˜ ∆ RY˜ (a) = Y˜ T RY˜ ∆ (a) = Y˜ T RR−1 (a) = Y˜ T (a)
= 0, Therefore
t ∈ Tκ .
T Y˜ T RY˜ ∆ (t) = Y˜ ∆ RY˜ (t),
t ∈ Tκ ,
i.e., the matrix Y˜ T RY˜ ∆ is a symmetric matrix. Hence, T Y˜ T RY˜ ∆Y˜ −1 (t) = Y˜ ∆ R (t),
and
−1 ∆ T RY˜ ∆Y˜ −1 (t) = Y˜ T Y˜ R (t) T = RY˜ ∆Y˜ −1 (t).
˜ Consequently Q(t), t ∈ Tκ , is a symmetric matrix. Note that −1 RY˜ σY˜ (t) = R Y˜ + µY˜ ∆ Y˜ −1 (t) = R(t) + µ(t) RY˜ ∆Y˜ −1 (t) ˜ = R(t) + µ(t)Q(t), t ∈ Tκ .
From here, since R and Q˜ are symmetric matrices on Tκ , we conclude that RY˜ σY˜ −1 and D˜ are symmetric matrices on Tκ . This completes the proof. Definition 3.2.8. Let Y be a solution of the equation (3.2.1). We say that Y is a conjoined solution of (3.2.1) provided Y T RY ∆ is symmetric.
Definition 3.2.9. Two conjoined solutions Y and Y1 of (3.2.1) are called normalized if T Y T RY1∆ − Y ∆ RY1 = I on Tκ .
137
Functionals Definition 3.2.10. The solution Y˜1 of the problem R(t)Y ∆
∆
+ P(t)Y σ = 0,
Y (a) = −I,
Y ∆ (a) = 0
is called associated solution of (3.2.1) at a. Theorem 3.2.11. Y˜ and Y˜1 are normalized solutions. Proof. By Theorem 3.2.3, the Wronskian identity, we have
T T Y˜1T RY˜1∆ (a) − Y˜1∆ RY˜1 (a) T T = − Y˜1∆ RY˜1 (a) T T = Y˜1∆ R (a)
= RY˜1∆ (a) = 0, T Y˜1T RY˜1∆ (t) = Y˜1∆ RY˜1 (t), t ∈ Tκ , T T Y˜ T RY˜ ∆ (a) − Y˜ ∆ RY˜ (a) T = Y˜ T (a)R(a)Y˜ ∆ (a) T ˜ = Y˜ ∆ (a)R(a)Y(a) Y˜ T RY˜
∆
= 0, T (a) = Y˜ ∆ RY˜ (t),
t ∈ Tκ ,
T T Y˜ RY˜1∆ − Y˜ ∆ RY˜1 (t) = Y˜ RY˜1∆ − Y˜ ∆ RY˜1 (a) T = − Y˜ ∆ RY˜1 (a) T = Y˜ ∆ R (a) = I,
This completes the proof.
t ∈ Tκ .
138
Svetlin G. Georgiev Now we consider the matrix Riccati equation Q∆ + P(t) + QT (R(t) + µ(t)Q)−1 Q = 0.
(3.2.4)
Theorem 3.2.12. If (3.2.1) has a conjoined solution Y such that Y (t) is invertible for all t ∈ [a, b], then Q defined by Q(t) = R(t)Y ∆ (t)Y −1 (t),
t ∈ [a, b],
(3.2.5)
is a symmetric solution of (3.2.4). Conversely, if (3.2.4) has a symmetric solution on [a, b]κ, then there exists a conjoined solution Y of (3.2.1) such that Y (t) is invertible for all t ∈ [a, b] and (3.2.5) holds. Proof. 1. Let Y be a conjoined solution of the equation (3.2.1) such that Y (t) is invertible for all t ∈ [a, b]. Let also, Q be defined by (3.2.5). By Theorem 3.2.7, it follows that Q(t) is symmetric for all t ∈ [a, b]κ. Also, Q∆ (t) = = = = = and
∆ RY ∆Y −1 (t) ∆ ∆ RY ∆ (t) (Y σ )−1 (t) + RY ∆ (t) Y −1 (t) ∆ RY ∆ (t) (Y σ )−1 (t) − RY ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t) −P(t)Y σ (t) (Y σ )−1 (t) − RY ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t) −P(t) − RY ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t), t ∈ [a, b]κ
Q∆ (t) + P(t) + QT (t) (R(t) + µ(t)Q(t))−1 Q(t) = −P(t) − R(t)Y ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t) +P(t) + R(t)Y ∆ (t)Y −1 (t)
× R(t) + µ(t)R(t)Y ∆ (t)Y −1 (t)
−1
R(t)Y ∆ (t)Y −1 (t)
139
Functionals −1
= −R(t)Y ∆ (t) (Y σ )
(t)Y ∆ (t)Y −1 (t)
+R(t)Y ∆ (t)Y −1 (t) I + µ(t)Y ∆ (t)Y −1 (t)
×R−1 (t)R(t)Y ∆ (t)Y −1 (t) = −R(t)Y ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t)
−1
+R(t)Y ∆ (t)Y −1 (t) I + (Y σ (t) −Y (t))Y −1 (t)
−1
Y ∆ (t)Y −1 (t)
= −R(t)Y ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t) −1 +R(t)Y ∆ (t)Y −1 (t) Y σ (t)Y −1 (t) Y ∆ (t)Y −1 (t)
= −R(t)Y ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t)
+R(t)Y ∆ (t)Y −1 (t)Y (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t)
= −R(t)Y ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t) +R(t)Y ∆ (t) (Y σ )−1 (t)Y ∆ (t)Y −1 (t)
= 0,
t ∈ [a, b]κ.
2. Let Q(t), t ∈ [a, b]κ, be a symmetric solution of the equation (3.2.4). Let t0 ∈ [a, b] and Y be the solution of the IVP Y ∆ = R−1 (t)Q(t)Y,
Y (t0 ) = I.
(3.2.6)
Then Y is a fundamental solution of the equation Y ∆ = R−1 (t)Q(t)Y and hence it is invertible on [a, b]. Also, RY ∆
∆
(t) = (QY )∆ (t) = Q∆ (t)Y σ (t) + Q(t)Y ∆ (t) = −P(t) − Q(t) (R(t) + µ(t)Q(t))−1 Q(t) Y σ (t) +Q(t)R−1 (t)Q(t)Y(t)
= −P(t)Y σ (t) − Q(t) (R(t) + µ(t)Q(t))−1 Q(t)Y σ (t) +Q(t)R−1 (t)Q(t)Y(t)
140
Svetlin G. Georgiev = −P(t)Y σ (t) − Q(t) (R(t) + µ(t)Q(t))−1 Q(t)Y σ (t) +Q(t) (R(t) + µ(t)Q(t))−1
× (R(t) + µ(t)Q(t))R−1 (t)Q(t)Y(t)
= −P(t)Y σ (t) + Q(t) (R(t) + µ(t)Q(t))−1 × (R(t) + µ(t)Q(t))R−1 (t)Q(t)Y(t) − Q(t)Y σ (t)
= −P(t)Y σ (t) + Q(t) (R(t) + µ(t)Q(t))−1
× Q(t)Y(t) + µ(t)Q(t)R−1(t)Q(t)Y(t) − Q(t)Y σ (t)
= −P(t)Y σ (t),
t ∈ [a, b]κ.
In the last equation we have used that Y σ (t) −Y (t) = µ(t)R−1(t)Q(t)Y(t), t ∈ [a, b]κ. Since Y T RY ∆ (t0 ) = Q(t0 ), we extend Y to a conjoined solution of (3.2.1). This completes the proof. Theorem 3.2.13 (Piconi’s Identity). Let α ∈ Rn and Y and Y1 be normalized joined solutions of the equation (3.2.1) such that Y is invertible on [a, b]. We put Q = RY ∆Y −1
on [a, b] and D = Y (Y σ )−1 R−1
on [a, b]κ.
Let t ∈ [a, b]κ and y : [a, b] → Rn be differentiable at t. Then we have at t ∆ T yT Qy + 2αT Y −1 y − αT Y −1Y1 α = y∆ Ry∆ − (yσ )T Pyσ T T T − Ry∆ − Qy − Y −1 α D Ry∆ − Qy − Y −1 α .
Proof. We have at t Y ∆Y −1 = R−1 Q, = RY ∆ , T Y T Q = Y ∆ R, QY
Q∆ = =
Functionals ∆ RY ∆Y −1 ∆ ∆ RY ∆ (Y σ )−1 + RY ∆ Y −1
= −PY σ (Y σ )−1 − RY ∆ (Y σ )−1 Y ∆Y −1
= −P − RY ∆ (Y σ )−1 Y ∆Y −1
= −P − QY (Y σ )−1 Y ∆Y −1
= −P − QY (Y σ )−1 R−1 RY ∆Y −1
= −P − QY (Y σ )−1 R−1 Q,
R−1 Q = Y ∆Y −1 , and Y −1Y1
and
∆
= − (Y σ )−1 Y ∆Y −1Y1 + (Y σ )−1 Y1∆
= − (Y σ )−1 R−1 QY1 + (Y σ )−1 R−1 RY1∆ T −1 = − (Y σ ) R−1 Y −1 Y T QY1 T + (Y σ )−1 R−1 Y −1 Y T RY1∆ T T ∆ = (Y σ )−1 R−1 Y −1 Y RY1 −Y T QY1 T T ∆ T = (Y σ )−1 R−1 Y −1 Y RY1 − Y ∆ RY1 T −1 = (Y σ ) R−1 Y −1 ,
∆ T yT Qy + Ry∆ − Qy Y (Y σ )−1 R−1 Ry∆ − Qy T T T = y∆ Qy + (yσ ) Q∆ yσ + (yσ ) Qy∆ T T + y∆ RY (Y σ )−1 y∆ − y∆ RY (Y σ )−1 R−1 Qy −yT QY (Y σ )−1 y∆ + yT QY (Y σ )−1 R−1 Qy
141
142
Svetlin G. Georgiev =
T
Qy − (yσ )T Pyσ − (yσ )T QY (Y σ )−1 R−1 Qyσ + (yσ )T Qy∆ T T + y∆ RY (Y σ )−1 y∆ − y∆ RY (Y σ )−1 R−1 Qy y∆
−yT QY (Y σ )−1 y∆ + yT QY (Y σ )−1 R−1 Qy T = y∆ Qy − (yσ )T Pyσ − (yσ )T QY (Y σ )−1 R−1 Qyσ T + (yσ )T Qy∆ + y∆ R Y σ − µY ∆ (Y σ )−1 y∆ T −1 − y∆ R Y σ − µY ∆ (Y σ ) R−1 Qy −1
−1
−yT QY (Y σ ) y∆ + yT QY (Y σ ) R−1 Qy T T T = y∆ Ry∆ − (yσ ) Pyσ + (yσ ) Qy∆ T −µ y∆ RY ∆ (Y σ )−1 y∆ T −yT QY (Y σ )−1 y∆ + µ y∆ RY ∆ (Y σ )−1 R−1 Qy
+yT QY (Y σ )−1 R−1 Qy − (yσ )T QY (Y σ )−1 R−1 Qyσ
=
t T T y∆ Ry∆ − (yσ ) Pyσ + (yσ ) Qy∆
T − (yσ )T QY (Y σ )−1 R−1 Qyσ − µy∆ + y QY (Y σ )−1 y∆ T + µy∆ + y QY (Y σ )−1 R−1 Qy T T T = y∆ Ry∆ − (yσ ) Pyσ + (yσ ) Qy∆ T
−1
σ T
σ −1
− (yσ ) QY (Y σ )
T
R−1 Qyσ − (yσ ) QY (Y σ )
−1 ∆
−1
+ (y ) QY (Y ) R Qy T T T = y∆ Ry∆ − (yσ ) Pyσ + (yσ ) Qy∆ − (yσ )T Q Y σ − µY ∆ (Y σ )−1 y∆
+ (yσ )T QY (Y σ )−1 R−1 Q (y − yσ ) T = y∆ Ry∆ − (yσ )T Pyσ T −1 −1 +µ (yσ ) Q Y ∆ (Y σ ) −Y (Y σ ) Y ∆Y −1 y∆
y
Functionals =
=
=
=
T y∆ Ry∆ − (yσ )T Pyσ + (yσ )T Q µY ∆ (Y σ )−1 −1 −Y (Y σ ) µY ∆ Y −1 y∆ T y∆ Ry∆ − (yσ )T Pyσ + (yσ )T Q (Y σ −Y ) (Y σ )−1 −1 −Y (Y σ ) (Y σ −Y )Y −1 y∆ T y∆ Ry∆ − (yσ )T Pyσ T −1 −1 + (yσ ) Q I −Y (Y σ ) − I +Y (Y σ ) y∆ T y∆ Ry∆ − (yσ )T Pyσ .
Therefore at t we get
∆ T yT Qy + 2αT Y −1 y − αT Y −1Y1 α − y∆ Ry∆ + (yσ )T Pyσ ∆ T = yT Qy − y∆ Ry∆ + (yσ )T Pyσ ∆ ∆ +2αT Y −1 y − αT Y −1Y1 α T = − Ry∆ − Qy Y (Y σ )−1 R−1 Ry∆ − Qy −2αT (Y σ )−1 Y ∆Y −1 y + 2αT (Y σ )−1 y∆ T −1 −αT (Y σ ) R−1 Y −1 α T = − Ry∆ − Qy Y (Y σ )−1 R−1 Ry∆ − Qy −2αT (Y σ )−1 R−1 Qy + 2αT (Y σ )−1 y∆ T −αT (Y σ )−1 R−1 Y −1 α
T = − Ry∆ − Qy Y (Y σ )−1 R−1 Ry∆ − Qy +2αT (Y σ )−1 R−1 Ry∆ − Qy T −αT (Y σ )−1 R−1 Y −1 α T T −1 = − Ry∆ − Qy − Y −1 α Y (Y σ ) R−1 T × Ry∆ − Qy − Y −1 α .
This completes the proof.
143
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Svetlin G. Georgiev
3.3. Jacobi’s Condition We will start this section with the following useful result. 1 ([a, b]), y : [a, b] → Rn , and Theorem 3.3.1 (Quadratic Functional). Let y ∈ Crd ∆ 2 Ry∆ + Pyσ (t) = 0, t ∈ [a, b]κ .
Then
Z ρ(b)
y∆
a
T
T Ry∆ − (yσ ) Pyσ (t)∆t = yT Ry∆ (ρ(b)) − yT Ry∆ (a). 2
Proof. On [a, b]κ we have yT Ry∆
∆
Hence, Z ρ(b) a
∆ T
y
∆
σ T
= (yσ )T Ry∆
∆
+ y∆
T
Ry∆
T T = (yσ ) (−Pyσ ) + y∆ Ry∆ T = y∆ Ry∆ − (yσ )T Pyσ . σ
Ry − (y ) Py
(t)∆t = = =
This completes the proof.
∆ yT Ry∆ (t)∆t a t=ρ(b) T ∆ y Ry (t) t=a yT Ry∆ (ρ(b)) − yT Ry∆ (a).
Z ρ(b)
We will consider a quadratic functional of the form
F (y) =
Z b
y∆
a
for y ∈ C p1 ([a, b]), y : [a, b] → Rn .
T
T Ry∆ − (yσ ) Pyσ (t)∆t
Functionals
145
Definition 3.3.2. We will say that the functional F is positive definite and we will write F > 0 if
F (y) > 0 for all nontrivial y ∈ C p1([a, b]), y : [a, b] → Rn. Theorem 3.3.3. A sufficient condition for F > 0 is that there exist normalized conjoined solutions Y and Y1 of (3.2.1) with Y invertible on [a, b] and Y (Y σ )−1 R−1 is positive definite on [a, ρ(b)]. Proof. Let y ∈ C p1 ([a, b]), y : [a, b] → Rn , y(a) = y(b) = 0.
1. Suppose that a is right-scattered. Then σ(a) > a and µ(a) > 0. Now, applying Piccone’s identity for α = 0, we get Z b T y∆ Ry∆ − (yσ )T Pyσ (t)∆t F (y) = a
=
Z σ(a) a
=
=
=
≥
y∆
Ry∆ − (yσ )T Pyσ (t)∆t
T Ry∆ − (yσ ) Pyσ (t)∆t σ(a) T y∆ Ry∆ − (yσ )T Pyσ (a)µ(a) Z b T + y∆ Ry∆ − (yσ )T Pyσ (t)∆t σ(a) T y∆ Ry∆ − (yσ )T Pyσ (a)µ(a) Z b ∆ T + yT Qy + Ry∆ − Qy D Ry∆ − Qy (t)∆t σ(a) T y∆ Ry∆ − (yσ )T Pyσ (a)µ(a) + yT Qy (b) − yT Qy (σ(a)) Z b T + Ry∆ − Qy D Ry∆ − Qy (t)∆t σ(a) T ∆ ∆ σ T σ y Ry − (y ) Py (a)µ(a) + yT Qy (b) − yT Qy (σ(a)) +
Z b
T
y∆
T
146
Svetlin G. Georgiev σ (yσ )T Ryσ µ−1 − (yσ )T Pyσ µ − (yσ )T RY ∆Y −1 yσ (a) ∆ σ −1 σ T = (yσ ) Rµ−1 − Pµ − RY σ µ−1 + µ RY ∆ (Y ) y (a) = (yσ )T Rµ−1 − Pµ − RY σ µ−1 − µPY σ (Y σ )−1 yσ (a) = (yσ )T Rµ−1 − Pµ − Rµ−1 + Pµ yσ (a) =
= 0.
In particular, by above computations, we get
F (y) =
Z b σ(a)
T Ry∆ − Qy D Ry∆ − Qy (t)∆t.
Hence, if F (y) = 0, using that P is a positive definite matrix, we conclude that Ry∆ − Qy (t) = 0, t ∈ [σ(a), ρ(b)], or
or Note that
Ry∆ (t) = (Qy)(t),
y∆ (t) = Y ∆Y −1 y (t),
t ∈ [σ(a), ρ(b)], t ∈ [σ(a), ρ(b)].
1 + µY ∆Y −1 = I + (Y σ −Y )Y −1 = I +Y σY −1 − I = Y σY −1 6= 0. Therefore Y ∆Y −1 is a regressive matrix. Then the IVP y∆ (t) = Y ∆Y −1 y (t), t ∈ [σ(a), ρ(b)], y(b) = 0,
has unique solution y(t) = 0, t ∈ [a, b]. Therefore F > 0.
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Functionals
2. Let a is right-dense. Then there exists a right-dense sequence {tm }m∈N ⊂ [a, b] such that tm → a, m → ∞. Let T αm = − Y ∆ Ry (tm), m ∈ N. Now we apply Picone’s identity for α = αm and we get Z b tm
=
≥ = = = =
y∆
T
T Ry∆ − (yσ ) Pyσ (t)∆t
T Qy + 2αT Y −1 y − αT Y −1Y α ∆ (t)∆t y 1 m m m T T T R + tbm Ry∆ − Qy − Y −1 αm D Ry∆ − Qy − Y −1 αm (t)∆t ∆ Rb T T −1 T −1 tm y Qy + 2αmY y − αmY Y1 α m (t)∆t t=b yT Qy + 2αTmY −1 y − αTmY −1Y1 αm t=tm − αTmY −1Y1 αm (b) − yT Qy + 2αTmY −1 y − αTmY −1Y1 αm (tm ) − αTmY −1Y1 αm (b) T + −yT Qy + 2yT RY ∆Y −1 y + yT RY ∆Y −1Y1 Y ∆ Ry (tm ) − αTmY −1Y1 αm (b) T + yT Qy + yT QY1 Y ∆ Ry (tm ). Rb
tm
(3.3.1)
Since Y and Y1 are normalized, we have Y T RY1∆ − I = = Because Y is conjoined, we have Y∆ Therefore
T
Y∆ Y
T
∆ T
RY1 RYY −1Y1 .
RY = Y T RY ∆ .
Y T RY1∆ − I = Y T RY ∆Y −1Y1 = Y T QY1
and then QY1 = Y T
−1
Y T RY1∆ − I .
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Svetlin G. Georgiev Hence and (3.3.1), using that Q is symmetric, we obtain Z b
T Ry∆ − (yσ ) Pyσ (t)∆t = − αTmY −1Y1 αm (b) tm −1 T ∆ ∆ T + yT Qy + yT Y T Y RY1 − I Y Ry (tm ) = − αTmY −1Y1 αm (b) T −1 ∆ T + yT Qy + yT RY1∆ Y ∆ R − Y T Y R y (tm ) = − αTmY −1Y1 αm (b) T + yT Qy + yT RY1∆ Y ∆ Ry − yT Qy (tm) T = − αTmY −1Y1 αm (b) + yT RY1∆ Y ∆ Ry (tm). y∆
T
Consequently
T Ry∆ − (yσ ) Pyσ (t)∆t m→∞ tm ≥ − lim αTm Y −1Y1 (b)αm + yT RY1∆ (tm)αm m→∞ = − lim yT RY ∆ (tm) Y −1Y1 (b)αm + yT RY1∆ (tm)αm
F (y) =
lim
Z b
y∆
T
m→∞
= 0.
Now we suppose that F (y) = 0. Let T zm = Ry∆ − Qy − Y −1 αm → Ry∆ − Qy = z,
as m → ∞,
uniformly. Let k ∈ N and m ≥ k. Then Z b tm
From here,
Z zTm Dzm (t)∆t ≥
b
tk
0 = ≥
lim
Z b
m→∞ tm
Z b tk
zTm Dzm (t)∆t.
zTm Dzm (t)∆t
zT Dz (t)∆t
149
Functionals and 0 = ≥
lim
k→∞
lim
lim
Z b
m→∞ tm Z b T
k→∞ tk
≥ 0.
zTm Dzm
z Dz (t)∆t
(t)∆t
Therefore z = 0 and hence, y = 0. We conclude that F is positive definite. This completes the proof.
Definition 3.3.4. We say that the equation (3.2.1) is disconjugate on [a, b] if the principal solution Y˜ of (3.2.1) satisfies Y˜
is invertible on [a, b],
Y˜ Y˜ σ
−1
R−1
is positive definite on [a, b].
Theorem 3.3.5 (Jacobi’s Condition). F > 0 if and only if (3.2.1) is disconjugate. Proof. 1. Let (3.2.1) is disconjugate. Let Y˜ is the principal solution of (3.2.1) and Y˜1 is the associated solution of (3.2.1). Then Y˜ and Y˜1 are normalized. We apply Theorem 3.3.3 and we conclude that F > 0. 2. Let F > 0. Assume that (3.2.1) is not disconjugate. Then there exists a point t0 ∈ [a, b] with exactly one of the following properties. Case 1. t0 ∈ (a, b], Y˜ (t) is invertible on (a,t0) and Y˜ (t0 ) is singular. Case 2. t0 ∈ (a, b]κ such that Y˜ (t) is invertible for all t ∈ (a, b) and ˜ 0 ) = Y˜ Y˜ σ −1 R−1 (t0 ) D(t
is not positive definite.
150
Svetlin G. Georgiev Let d ∈ Rn \{0} be such that
Y˜ (t0 )d = 0
in case 1, and
˜ (t0 ) ≤ 0 d T Dd
in case 2. We set
y(t) =
Y˜ (t)d for t ≤ t0 0 otherwise
Since Y˜ is the principal solution, we have y(a) = Y˜ (a)d = 0, and using the definition of y, we have y(b) = 0. Also, y(t) 6= 0 for
C p1([a, b]) σ
all t ∈ (a,t0).
Note that y ∈ except for case 2 with t0 right-dense. We have ∆ that y (t) and y (t) are zero for all t > t0 . Hence, T T y∆ Ry∆ − (yσ ) Pyσ (t) = 0 for all t > t0 . Using Theorem 3.3.1, we obtain F (y) =
Z σ(t0 )
y∆
a
=
Z t 0 a
+ = = = = =
T
Z σ(t0 )
Ry∆ − (yσ)T Pyσ (t)∆t
T Ry∆ − (yσ ) Pyσ (t)∆t
y∆
t0
y∆
T
T
Ry∆ − (yσ )T Pyσ (t)∆t
T T y Ry∆ (t0 ) + y∆ Ry∆ − (yσ) Pyσ (t0 )µ(t0 ) T
∆ yT Ry∆ (t0 ) + yT Ry∆ (t0 )µ(t0 ) σ d T Y˜ T RY˜ ∆ d (t0 ) + yT Ry∆ (t0 ) − yT Ry∆ (t0 ) d T Y˜ T RY˜ ∆ d (t0 ) − yT Ry∆ (t0 )
0.
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151
This is a contradiction. Now we consider case 2 with right-dense t0 . There exists d ∈ Rn such that d T R(t0 )d < 0. Assume that F > 0. (a) Let t0 is left-scattered. Then t0 6= b and there there exists a sequence {tm}m∈N ⊂ [a, b] be a right-dense sequence for t0 . For m ∈ N we set ( √tm −t d if t ∈ [t0 ,tm ] tm −t0 ym (t) = 0 otherwise. Then ym (a) = 0, ym (b) = 0, tm − t0 ym (t0 ) = √ d tm − t0 √ = tm − t0 d 6= 0,
and ym ∈ C p1 ([a, b]). Therefore 0
< =
F (ym )
Z σ(tm )
y∆
t0
= =
→
Z σ(tm )
T
Ry∆ − (yσ )T Pyσ (t)∆t
dT d tm − σ(t) tm − σ(t) √ R(t) √ − dT √ P(t) √ d ∆t tm − t0 tm − t0 tm − t0 tm − t0 t0 Z σ(tm ) 1 R(t)∆t d dT tm − t0 t0 Z σ(t ) m (t − σ(t))2 m T −d P(t)∆t d tm − t0 t0 d T R(t0 )d < 0,
m → ∞.
This is a contradiction. (b) Let t0 is left-dense. For m ∈ N, we consider t−t ∗ ∗ m √t0 −tm∗ d if t ∈ [tm,t0 ) √tm −t d if t ∈ (t0 ,tm ] ym (t) = t −t m 0 0 otherwise,
152
Svetlin G. Georgiev if t0 6= b, and ym (t) =
(
∗ √t−tm ∗ d t0 −tm
if t ∈ [tm∗ ,t0) 0 otherwise,
if t0 = b, where {tm∗ }m∈N ⊂ [a, b] is a strictly increasing sequence with limm→∞ tm∗ = t0 . Again we assume F > 0. Then, if t0 6= b, we have 0
0
T
∆ σ T ˜ σ ˜ Ry − (y ) Py (t)∆t
for all nontrivial y ∈ C p1 ([a, b]) with y(a) = y(b) = 0. For such y we also have
F (y) =
Z b
>
T
∆
σ T
σ
y Ry − (y ) Py (t)∆t Z b T ∆ ∆ σ T ˜ σ ˜ y Ry − (y ) Py (t)∆t a
≥
∆
a
0,
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155
i.e., F > 0 and the equation (3.2.1) is disconjugate. This completes the proof.
Chapter 4
Linear Hamiltonian Dynamic Systems Let T be a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Throughout this chapter, we denote by I the 2n × 2n matrix O I I= . −I O
4.1. Linear Symplectic Dynamic Systems Definition 4.1.1 (Symplectic Matrix). A 2n × 2n matrix A is called symplectic if A∗ I A = I .
Example 4.1.2. Consider the matrix 2 0 A= 3 0
0 2 0 1 0 2 . 0 27 0 3 0 7
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Svetlin G. Georgiev
Then
0 0 0 0 IA = −1 0 0 −1 3 0 0 3 = −2 0 0 −1
2 0 3 0 1 0 ∗ A = 2 0 7 2 0 2 0 2 0 3 0 1 0 A∗ JA = 2 0 7 2 0 2 0 0 0 0 0 = −1 0 0 −1
Therefore A is symplectic.
0 2 0 2 0 1 0 1 07 2 0 3 0 2 0 0 0 3 0 7 7 0 2 0 7 , −2 0 0 −2
1 0 0 0
0 3 , 0 7 7 0 3 0 0 2 0 3 3 0 7 0 −2 0 −2 0 7 0 −1 0 −2 1 0 0 1 . 0 0 0 0
Theorem 4.1.3. Let A, B, C and D be n × n matrices. Then the matrix A B C D is symplectic if and only if A∗C and B∗ D are Hermitian and A∗ D −C∗ B = I. Proof. The matrix
A B C D
159
Linear Hamiltonian Dynamic Systems is sympelctic if and only if ∗ A C∗ O I A B I= B∗ D∗ −I O C D
O I −I O
= =
A∗C = C∗ A,
A∗ C ∗ B∗ D∗
C D −A −B
A∗C −C∗ A A∗ D −C∗ B B∗ C − D∗ A B∗ D − D∗ B
B∗ D = D∗ B,
iff
iff
A∗ D −C∗ B = I.
This completes the proof. Theorem 4.1.4. Let A, B, C and D be n × n matrices. Then A B C D is symplectic if and only if it is invertible with
A B C D
−1
=
D∗ −B∗ −C∗ A∗
.
Proof. We have D∗ −B∗ A B D∗ A − B∗ C D∗ B − B∗ D = . C D −C∗ A + A∗C −C∗ B + A∗ D −C∗ A∗ Hence, using Theorem 4.1.3, we conclude that A B C D is symplectic if and only if A∗C = C∗ A,
B∗ D = D∗ B,
A∗ D −C∗ B = I,
D∗ A − B∗C = I,
160 if and only if
Svetlin G. Georgiev
D∗ A − B∗ C D∗ B − B∗ D ∗ ∗ −C A + A C −C∗ B + A∗ D
=
I O O I
if and only if
D∗ −B∗ −C∗ A∗
A B C D
=
I O O I
.
This completes the proof. Exercise 4.1.5. Let A, B, C and D be n × n matrices. Prove that the matrix A B C D is symplectic if and only if the matrices AB∗ and CD∗ are Hermitian and AD∗ − BC∗ = I. Hint. Use that
I O O I
= =
A B C D
D∗ −B∗ −C∗ A∗
AD∗ − BC∗ BA∗ − AB∗ ∗ ∗ CD − DC −CB∗ + DA∗
and then apply Theorem 4.1.4. Definition 4.1.6. A 2n × 2n matrix-valued function S is called symplectic with respect to T if S∗ (t)I + I S(t) + µ(t)S? (t)JS(t) = O,
t ∈ T.
If S is symplectic, then the system z∆ = S(t)z is called a linear symplectic dynamic system.
(4.1.1)
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161
Theorem 4.1.7. If S is symplectic with respect to T, then I + µS is symplectic. Proof. We have (I + µS)∗ J (I + µS) = (I + µS∗ )J (I + µS) = (I + µS∗ )(J + µJS) = J + µJ S + µS∗ J + µ2 S∗ J S = J + µ (J S + S∗ J + µS∗ J S) = J. This completes the proof. Theorem 4.1.8. If S is symplectic with respect to T, then it is regressive. Proof. Since S is symplectic with respect to T, then I + µS is symplectic. Hence and Theorem 4.1.4, it follows that I + µS is invertible. Therefore S is regressive. This completes the proof. Corollary 4.1.9. Let S ∈ Crd is symplectic with respect to T, t0 ∈ T, z0 ∈ R2n . Then the IVP z∆ = S(t)z, z(t0) = z0 , has unique solution z : T → R2n . Theorem 4.1.10. Let A, B, C and D be n × n matrices and A B S= . C D Then S is symplectic with respect to T if and only if (I + µA)∗C −C∗ (I + µA) = O, (I + µA)∗ + A∗ − µC∗ B = O,
B∗ (I + µD) − (I + µD)∗ B = O.
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Svetlin G. Georgiev
Proof. We have ∗
S = and
A∗ C ∗ B∗ D∗
∗ A C∗ O I O S∗ J + JS + µS∗ JS = + B ∗ D∗ −I O −I ∗ A C∗ O I A B +µ ∗ ∗ B D −I O C D
=
=
=
= =
I O
A C
B D
D −C∗ A∗ C + −A −B −D∗ B∗ ∗ ∗ A C C D +µ B∗ D∗ −A −B C −C∗ A∗ + D −A − D∗ B∗ − B ∗ A C −C∗ A A∗ D −C∗ B +µ B∗ C − D∗ A B∗ D − D∗ B C −C∗ D + A∗ −(D + A∗ )∗ B∗ − B A∗C −C∗ A A∗ D −C∗ B +µ −(A∗ D −C∗ B)∗ B∗ D − D∗ B C + µA∗C − µC∗ A −C∗ D + µA∗ D + A∗ − µC∗ B − (A∗ + D + µA∗ D − µC∗ B)∗ B∗ − B + µB∗ D − µD∗ B (I + µA)∗C −C∗ (I + µA) (I + µA)∗ D + A∗ − µC∗ B . − ((I + µA∗ )D + A∗ − µC∗ B)∗ B∗ (I + µD) − (I + µD)∗ B
From here, we conclude that S is symplectic with respect to T if and only if (I + µA)∗C −C∗ (I + µA) = 0, (I + µA)∗ + A∗ − µC∗ B = 0,
B∗ (I + µD) − (I + µD)∗ B = 0. This completes the proof.
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Linear Hamiltonian Dynamic Systems
4.2. Hamiltonian Systems Definition 4.2.1. A 2n × 2n matrix A is called Hamiltonian if A∗ J + J A = O.
Theorem 4.2.2. Let A, B, C and D be n × n matrices. Then the matrix A B C D is Hamiltonian if and only if B and C are Hermitian and D = −A∗ . Proof. We have
A B C D
∗
=
A∗ C ∗ B∗ D∗
and
A C
B D
∗
J +J
A C
B D
= = =
Hence,
O O O O
=
A∗ B∗
C∗ D∗
−C∗ −D∗
A∗ B∗
−C∗ +C −D∗ − A
O I −I O C + −A A∗ + D . B∗ − B
+
D −B
−C∗ +C A∗ + D −D∗ − A B∗ − B
O −I
I O
A C
B D
if and only if C∗ = C,
B∗ = B and A∗ = −D.
This completes the proof. Definition 4.2.3. A 2n × 2n matrix-valued function H is called Hamiltonian with respect to T if H is Hamiltonian and the matrix I − µ(t)H (t)M ∗ M
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Svetlin G. Georgiev
is invertible for all t ∈ T, where
M=
O O I O
.
In this case the system z∆ = H (t) (M ∗ M zσ + M M ∗ z)
(4.2.1)
is called linear Hamiltonian dynamic system.
Remark 4.2.4. Let
H=
A B C D
,
where A, B, C and D are n × n matrices. Then I O A ∗ I − µ(t)H (t)M M = − µ(t) O I C O I O O × O O I O I O A = − µ(t) O I C I O A = − µ(t) O I C I − µ(t)A O = . −µ(t)C I
B D B D
O O
I O O O
Hence and Theorem 4.2.2, H is Hamiltonian with respect to T if and only if B∗ = B,
C∗ = C,
D = −A∗
and I − µA is invertible for all t ∈ T. Now we suppose that A B H= −C D
Linear Hamiltonian Dynamic Systems is Hamiltonian with respect to T. Let now u z= , v where u, v : T → Rn . Then
z
σ
=
,
σ I O u O O vσ σ u = , 0 O O u ∗ MM z = O I v u = , v σ u 0 M ∗ M zσ + M M ∗ z = + 0 v σ u = , v σ A B u ∗ σ ∗ H (M M z + M M z) = −C D v σ Au + Bv = −Cuσ + Dv Auσ + Bv = , −Cuσ − A∗ v ∆ u x∆ = v∆ Auσ + Bv = , −Cuσ − A∗ v
M ∗ M zσ =
i.e., we get the system
uσ vσ
u∆ = Auσ + Bv v∆ = −Cuσ − A∗ v.
165
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Svetlin G. Georgiev
This completes the remark.
4.3. Conjoined Bases Let a, b ∈ T, a < ρ(b). Suppose that A, B and C satisfy the following conditions A, B,C : [a, b] → Rn×n , (I − µA)−1 B
and C
are
A, B,C ∈ Crd ([a, b]),
(4.3.1)
exists on [a, b]κ,
(4.3.2)
on [a, b]κ.
symmetric
(4.3.3)
Note that (I − µA)−1 ∈ Crd ([a, b]). Consider the system ∆ x = Axσ + Bu u∆ = −Cxσ − AT u on [a, b]κ. Let
H=
A B −C −AT
(4.3.4)
.
Then the system (4.3.4) we can rewrite in the form ∆ σ x x x ∗ ∗ =H M M +M M . u u u Note that H is Hamiltonian, and ∗
I − µH M M
A B = I −µ −C −AT A O = I −µ −C O I − µA O = . −C I
I O O O
Hence, using the condition (4.3.2), we conclude that I − µH M ∗ M is invertible on [a, b]κ. Therefore the system (4.3.4) is a Hamiltonian system.
Linear Hamiltonian Dynamic Systems
167
Definition 4.3.1. By a solution of (4.3.4), we mean a pair (x, u) with x, u ∈
Crd1 ([a, b]), satisfying the system (4.3.4) on [a, b]κ.
We use a usual agreement that the vector-valued solutions of (4.3.4) are denoted by small letters and the n × n matrix-valued solutions by capital ones. ˜ U) ˜ Theorem 4.3.2 (Wronskian Identity). For any two solutions (X,U) and (X, of (4.3.4) we have X T U˜ −U T X˜ is a constant on [a, b]. Proof. We have X T U˜ −U t X˜
∆
σ U˜ + X T U˜ ∆ ∆ − U T X˜ σ −U T X˜ ∆ T = X ∆ U˜ + (X σ )T U˜ ∆ T − U ∆ X˜ σ −U T X˜ ∆ =
XT
∆
= (AX σ + BU)T U˜ + (X σ )T −CX˜ σ − AT U˜ T − −CX σ − AT U X˜ σ −U T AX˜ σ + BU˜
T = (X σ ) AT U˜ +U T BU˜ T T − (X σ ) CX˜ σ − (X σ ) AT U˜ T
+ (X σ ) CX˜ σ +U T AX˜ σ −U T AX˜ σ −U T BU˜
= 0 on Tκ . This completes the proof. Let A˜ = (I − µA)−1
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Svetlin G. Georgiev
and (x, u) is a solution of the system (4.3.4). Then x∆ = Axσ + Bu = Ax + µAx∆ + Bu, u∆ = −Cxσ − AT u
= −Cx − µCx∆ − AT u
on Tκ .
Hence, (I − µA)x∆ = Ax + Bu
on Tκ
and ˜ + ABu ˜ x∆ = AAx on Tκ . Then ˜ + ABu) ˜ u∆ = −Cx − µC(AAx − AT u ˜ ˜ + AT )u = −(C + µCAA)x − (µCAB
˜ − µA + µA)x − (µCAB ˜ + AT )u = −CA(I ˜ − (µCAB ˜ + AT )u on Tκ . = −CAx
Therefore the system (4.3.4) can be rewritten in the form ∆ ˜ + ABu ˜ x = AAx ∆ ˜ − (µCAB ˜ + AT )u on Tκ . u = −CAx Let w=
x u
,
S=
˜ ˜ AA AB ˜ + AT ) −CA˜ −(µCAB
Then (4.3.5) takes the form w∆ = Sw.
(4.3.5)
.
Linear Hamiltonian Dynamic Systems
169
We have ST
=
ST J
= =
JS
= =
T
S J + JS
= =
T
S JS
= =
AT A˜ T −A˜ T C , BA˜ T − µBA˜ T C + A AT A˜ T −A˜ T C O I BA˜ T − µBA˜ T C + A −I O T T T ˜ ˜ A C A A , µBA˜ T C + A BA˜ T ˜ ˜ O I AA AB ˜ + AT −I O −C A˜ − µC AB ˜ + AT −C A˜ − µC AB , ˜ ˜ −AA −AB ˜ + AT A˜ T C AT A˜ T −C A˜ − µC AB + ˜ ˜ µBA˜ T C + A BA˜ T −AA −AB T T T T ˜ +A A˜ C −C A˜ A A˜ − µC AB , ˜ ˜ µBA˜ T C + A − AA BA˜ T − AB ˜ ˜ A˜ T C AT A˜ T AA AB ˜ + AT ) µBA˜ T C + A BA˜ T −C A˜ −(µC AB ˜ + AT ˜ − AT A˜ T µC AB ˜ − AT A˜ T C A˜ A˜ T C AB A˜ T C AA . ˜ + AAA ˜ − BA˜ T C A˜ ˜ − BA˜T AT µBA˜ T C AA AAB
Note that ˜ − µAT A˜ T CA˜ A˜ T C −CA˜ + µA˜ T CAA ˜ − (I + µAT A˜ T )CA˜ = A˜ T C(I + µAA)
˜ − µA + µA) − (I − µAT + µAT )A˜ T CA˜ = A˜ T CA(I
= 0, ˜ + AT ) + µA˜ T CAB ˜ − µ2 AT A˜ T CAB ˜ − µAT A˜ T AT A A˜ T − (µCAB ˜ − AT + µA˜ T CAB) ˜ = AT A˜ T (I − µAT ) − µ(I + µAT A˜ T )(AB T
˜ − AT − µA˜ T CAB ˜ = AT − µ(I − µAT + µAT )A˜ T CAB
= 0, ˜ + µ2 BA˜ T CAA ˜ + µAAA ˜ − µBA˜ T CA˜ µBA˜ T C + A − AA ˜ − µA + µA) − µBA˜ T CA˜ + A − A = µBA˜ T CA(I
= 0,
170
Svetlin G. Georgiev ˜ + µAAB ˜ − µBA˜ T AT BA˜ T − AB ˜ = BA˜ T (I − µAT ) − (I − µA)AB = B−B
= 0. Therefore
ST J + JS + µST JS = O and the system (4.3.5) is symplectic. Since S is symplectic with respect to T, by Theorem 4.1.8, it follows that it is regressive. Hence and Corollary 4.1.9, we conclude that the IVP for the system (4.3.5) with initial condition x(s) = x,
u(s) = u,
s ∈ [a, b], x, u ∈ Rn , has unique solution on [a, b]. For a solution (X,U) of (4.3.4) we have X σ = X + µX ∆ = X + µAX σ + µBU
on [a, b]κ,
whereupon (I − µA)X σ = X + µBU
on [a, b]κ,
or ˜ + µABU ˜ X σ = AX
on [a, b]κ,
and U σ = U + µU ∆ = U − µCX σ − µAT U
= −µCX σ + (I − µAT )U ˜ + µABU) ˜ = −µC(AX + (A˜ T )−1U
˜ + (−µ2CAB ˜ + (A˜ T )−1 )U = −µCAX i.e., we get the system σ ˜ + µABU ˜ X = AX σ ˜ + (−µ2CAB ˜ + (A˜ T )−1 )U U = −µCAX
on [a, b]κ,
on [a, b]κ.
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171
From the last system we obtain µCX σ +U σ = (A˜ T )−1U
on [a, b]κ,
U = µA˜ T CX σ + A˜ T U σ
on [a, b]κ,
or and ˜ = X σ − µ2 AB ˜ A˜ T CX σ − µAB ˜ A˜ T U σ AX
on [a, b]κ,
or X = (A˜ −1 − µ2 BA˜ T C)X σ − µBA˜ T U σ
on [a, b]κ.
In this way we obtain the system X = (A˜ −1 − µ2 BA˜ T C)X σ − µBA˜ T U σ U = µA˜ T CX σ + A˜ T U σ on [a, b]κ. Definition 4.3.3. A solution (X,U) of the system (4.3.4) is called a basis, if rank(X T U T ) = n at some, and hence at any, t ∈ [a, b]. Definition 4.3.4. A solution (X,U) of (4.3.4) is called a conjoined solution, if X T U −U T X = O
on [a, b].
˜ U) ˜ are called normalized, if Definition 4.3.5. Two conjoined bases (X,U), (X, X T U˜ −U T X˜ = I
on [a, b].
Definition 4.3.6. The unique solution of the IVP for the system (4.3.4) with initial condition X(a) = O, U(a) = I, is called the principal solution at a.
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Definition 4.3.7. The unique solution (X,U) of the system (4.3.4) with initial condition X(a) = −I, U(a) = O, is called the associated solution at a.
Definition 4.3.8. Together, the principal solution at a and the associated solution at a of the system (4.3.4), is called the special normalized bases of (4.3.4) at a.
Remark 4.3.9. The principal solution and the associated solution of (4.3.4) at a form normalized conjoined bases of (4.3.4). ˜ U) ˜ be any solutions of the system (4.3.4). Then Theorem 4.3.10. Let (X,U), (X, they form normalized conjoined bases of (4.3.4) if and only if the matrix X X˜ U U˜ is symplectic. In this case the following identities hold on [a, b]κ. ˜ X X˜ T , U U˜ T are symmetric, 1. X T U, X˜ T U, 2. ˜ T = UX ˜ T −U X˜ T = I, X U˜ T − XU 3. ˜ X σU˜ T − X˜ σU T = A, 4. ˜ X˜ σ X T − X σ X˜ T = µAB, 5. ˜ U˜ σU T −U σU˜ T = µCA,
173
Linear Hamiltonian Dynamic Systems 6. ˜ − A˜ T U σ X˜ T − U˜ σ X T = µ2CAB
−1
.
˜ U) ˜ form a normalized conjoined bases of the system Proof. Let (X,U), (X, (4.3.4). Then X˜ T U˜ − U˜ T X˜ = O on [a, b],
X T U −U T X = O,
X T U˜ −U T X˜ = I
Let
S=
X X˜ U U˜
T
XT X˜ T
Then S = and ST I S
on [a, b].
XT UT = ˜T ˜ T XT UT X U = T ˜ ˜T XT U T X U −U X = ˜ T U − U˜ T X X O I = −I O = I,
UT U˜ T
(4.3.6) (4.3.7)
.
X X˜ O I ˜ −I O U U U U˜ −X −X˜ X T U˜ −U T X˜ X˜ T U˜ − U˜ T X˜
(4.3.8)
where we have used (4.3.6), (4.3.7). Therefore S is symplectic. If we suppose that S is symplectic, by (4.3.8), we get (4.3.6) and (4.3.7). 1. By (4.3.6), it follows that the matrices X T U and X˜ T U˜ are symmetric matrices. Since the matrix S is symplectic, by Exercise 4.1.5, it follows that the matrices X X˜ T and U U˜ T are symmetric. 2. Again by Exercise 4.1.5, we obtain that ˜ T = I, X U˜ T − XU whereupon ˜ T −U T X˜ = I. UX
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3. We have, using 1, 2, X σU˜ T − X˜ σU T
=
T ˜ + µABU ˜ AX U˜ ˜ U˜ U T − A˜ X˜ + µAB
˜ U˜ T + µABU ˜ U˜ T = AX ˜ UU ˜ T − µAB ˜ T −A˜ XU ˜ U˜ T − XU ˜ ˜ T ) + µAB(U ˜ T) = A(X U˜ T − UU ˜ = A. 4. By 2, we have ˜ T −U X˜ T = I UX and X˜ σ X T − X σ X˜ T
˜ U)X ˜ T = (A˜ X˜ + µAB ˜ + µABU) ˜ −(AX X˜ T
˜ UX ˜ T + µAB ˜ T = A˜ XX ˜ X˜ T − µABU ˜ X˜ T −AX
˜ UX ˜ T −U X˜ T ) = µAB( ˜ = µAB. 5. We have U˜ σU T −U σU˜ T
˜ + (A˜ T )−1 )UU ˜ T + (−µ2CAB ˜ T = −µCA˜ XU ˜ U˜ T + (−µ2CAB ˜ + (A˜ T )−1 )U U˜ T +µCAX ˜ U˜ T − XU ˜ T) = µCA(X ˜ = µCA.
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6. We have U σ X˜ T − U˜ σ XT
˜ X˜ T + −µ2CAB ˜ + A˜ T −1 U X˜ T = −µCAX ˜ + A˜ T −1 UX ˜ T − −µ2CAB ˜ T +µCA˜ XX ˜ − A˜ T −1 UX ˜ T −U X˜ T = µ2CAB ˜ − A˜ T −1 . = µ2CAB
This completes the proof.
Theorem 4.3.11. Let (X,U) be a conjoined basis of (4.3.4). Then there ex˜ U) ˜ of (4.3.4) such that (X,U) and (X, ˜ U) ˜ are ists another conjoined basis (X, normalized. Proof. Let t0 ∈ [a, b]. Since (X,U) is a conjoined basis of (4.3.4) we have that (X T X +U T U)−1 (t0 ) ˜ U) ˜ be the solution of the equation (4.3.4) with initial condition exists. Let (X, ˜ 0) X(t −U(t0 )(X T X +U T U)−1 (t0 ) = . ˜ 0) U(t X(t0 )(X T X +U T U)−1 (t0 ) Then ˜ 0) − U˜ T (t0)X(t ˜ 0) X˜ T (t0 )U(t −1 T −1 = − X T X +U T U (t0)U T (t0)X(t0 ) X T X +U T U (t0 ) −1 T −1 + X T X +U T U (t0)X T (t0 )U(t0) X T X +U T U (t0 ) −1 T = X T X +U T U (t0 ) X T (t0 )U(t0) −U T (t0)X(t0 ) ×(X T X +U T U)−1 (t0 ) = O.
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˜ U) ˜ is a conjoined basis of (4.3.4). Hence and Theorem 4.3.2, we Therefore (X, get X T U˜ −U T X˜ is a constant and this constant is ˜ 0) −U T (t0)X(t ˜ 0) X T (t0 )U(t
−1 −1 = X T (t0 )X(t0) X T X +U T U (t0 ) +U T (t0 )U(t0) X T X +U T U (t0 ) −1 T T T T = X X +U U (t0 ) X X +U U (t0)
= I.
This completes the proof. Theorem 4.3.12. Let t ∈ [a, b]κ and (X,U) be a conjoined solution of (4.3.4) with KerX σ (t) ⊂ KerX(t). Then Ker(X σ )T ⊆ Ker(µBA˜ T )(t). Proof. By Theorem 4.3.11, it follows that there exists another conjioined so˜ U) ˜ of (4.3.4) such that (X,U) and (X, ˜ U) ˜ are normalized. Let lution (X, σ T α ∈ Ker(X ) (t). Then (X σ )T (t)α = 0. Hence, using Theorem 4.3.10, we get X σ (t) X˜ σ
T
(t)α =
X˜ σ (t) (X σ )T (t)α
= 0.
Therefore Ker X˜ σ
T
Consequently X(t) X˜ σ
⊆ Ker(X σ )(t) ⊂ KerX(t). T
(t)α = 0.
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By Theorem 4.3.10, 4, we obtain ˜ µAB Then
T
(t) = X X˜ σ
T
˜ σ)T (t). (t) − X(X
˜ T (t)α = X(X˜ σ )T (t)α − X(X ˜ σ)T (t)α (µAB) = 0.
˜ T . This completes the proof. Consequently α ∈ Ker(µAB) Definition 4.3.13. By M † we denote the Moore-Penrose generalized inverse of the matrix M, i.e., the unique matrix satisfying 1. MM † M = M, 2. M † MM † = M † , 3. MM † and M † M are symmetric.
Definition 4.3.14. A conjoined basis (X,U) of the system (4.3.4) is said to have no focal point in (a, b], provided that X is invertible for all dense points t ∈ (a, b] and ˜ ≥ 0 on [a, b]κ. KerX σ ⊆ KerX, D = X(X σ )T AB Definition 4.3.15. The system (4.3.4) is called disconjugate on [a, b], if the principal solution of (4.3.4) at a has no focal point in (a, b].
4.4. Riccati Equations In this section we will use the notations from the previous sections of this chapter. We consider the Riccati operator ˜ + BQ) on [a, b]κ. R(Q) = Q∆ +C + AT Q + (Qσ + µC)A(A
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We have ˜ + BQ) µR(Q) = µQ∆ + µC + µAT Q + µ(Qσ + µC)A(A σ T σ ˜ + BQ) = Q − Q + µC + µA Q + µ(Q + µC)A(A
˜ = Qσ − Q + µC + µAT Q + (Qσ + µC)A(µA + µBQ) σ T σ ˜ = Q − Q + µC + µA Q + (Q + µC)A(µA − I + I + µBQ) σ T σ −1 ˜ ˜ = Q − Q + µC + µA Q − (Q + µC)AA
˜ + µBQ) +(Qσ + µC)A(I ˜ + µBQ) − Q = µAT Q + (Qσ + µC)A(I ˜ + µBQ) = −(I − µAT )Q + (Qσ + µC)A(I
˜ + µBQ) = −(A˜ T )−1 Q + (Qσ + µC)A(I
on [a, b]κ,
˜ + µBQ) − Q, µA˜ T R(Q) = A˜ T (Qσ + µC)A(I
on [a, b]κ.
whereupon
The next result gives a criterion for existence of a symmetric solution of the Riccati equation. Theorem 4.4.1. The Riccati matrix equation ˜ + BQ) = 0 R(Q) = Q∆ +C + AT Q + (Qσ + µC)A(A
(4.4.1)
has a symmetric solution Q on [a, b] with I + µBQ nonsingular and (I + µBQ)−1B ≥ 0 on [a, b]κ if and only if the system (4.3.4) has a conjoined basis ˜ ≥ 0 on [a, b]κ. (X,U) such that X is invertible on [a, b] with D = X (X σ )−1 AB −1 Moreover, in this case, Q = UX on [a, b] and D = (I + µBQ)−1 B on [a, b]κ. Proof. 1. Let (4.3.4) has a conjoined basis on [a, b] with X invertible on [a, b]. Let Q = −UX −1 on [a, b]. We have that Q is symmetric and on
Linear Hamiltonian Dynamic Systems [a, b]κ Q∆ = (UX −1 )∆ = U(X −1 )∆ +U ∆ (X −1 )σ = −UX −1 X ∆ (X σ )−1 +U ∆ (X σ)−1
= (−UX −1 X ∆ +U ∆ )(X σ )−1
= (−QX ∆ +U ∆ )(X −1 )σ = −Q(AX σ + BU) −CX σ − AT U (X −1 )σ
=
−QAX σ − QBU −CX σ − AT U (X −1 )σ
= −QA −C − (QB + AT )U(X −1 )σ = −QA −C − (QB + AT )(µA˜ T CX σ + A˜ T U σ )(X −1)σ
= −QA −C − (QB + AT )(µA˜ T C + A˜ T (UX −1)σ ) = −QA −C − (QB + AT )(µA˜ T C + A˜ T Qσ)
= −QA −C − (QB + AT )A˜ T (µC + Qσ ), where we have used U = µA˜ T CX σ + A˜ T U σ
on [a, b]κ.
Next, I + µBQ = I + µBUX −1 = XX −1 + µBUX −1 = (X + µBU)X −1
on [a, b]κ.
Since ˜ + µABU ˜ X σ = AX
on [a, b]κ,
we get ˜ + µBU) on [a, b]κ, X σ = A(X and X + µBU = A˜ −1 X σ
on [a, b]κ.
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Svetlin G. Georgiev Therefore I + µBQA˜ −1 = X σ X −1
on [a, b]κ.
Hence, (I + µBQ)−1 B = (A˜ −1 X σ X −1 )−1 B ˜ = X(X σ )−1 AB = D ≥ 0 on [a, b]κ. 2. Let Q is a symmetric solution of (4.4.1) on [a, b]κ with I + µBQ nonsingular and (I + µBQ)−1 B ≥ 0 on [a, b]κ. Note that ˜ + BQ) = A( ˜ A˜ −1 + µA + µBQ) I + µA(A ˜ − µA + µA + µBQ) = A(I
˜ + µBQ) on [a, b]κ. = A(I
˜ + BQ) is regressive on [a, b]κ. Consequently the Therefore the matrix A(A IVP ˜ + BQ)X, X(s) = I, X ∆ = A(A for some s ∈ [a, b], has a unique solution X and it is nonsingular. We set U = QX. Then ˜ + BQ)X X ∆ = A(A ˜ = A(AX + BQX) ˜ = A(AX + BU) σ ˜ = A(A(X − µX ∆ ) + BU) σ ˜ = A((AX + BU) − µAX ∆ )
σ ˜ ˜ ∆ = A(AX + BU) − µAAX σ ˜ ˜ − µA − I)X ∆ = A(AX + BU) + A(I
σ ˜ ˜ − µA)X ∆ − AX ˜ ∆ = A(AX + BU) + A(I σ ˜ = A(AX + BU − X ∆ ) + X ∆ on [a, b]κ,
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whereupon X ∆ = AX σ + BU
on [a, b]κ.
(4.4.2)
Next, U ∆ = (QX)∆ = Qσ X ∆ + Q∆ X ˜ + BQ))X = Qσ X ∆ + (−C − AT Q − (Qσ + µC)A(A ˜ + BQ)X = Qσ X ∆ −CX − AT QX − (Qσ + µC)A(A
= Qσ X ∆ −CX − AT U − (Qσ + µC)X ∆
= −CX − AT U − µCX ∆
= −C(X + µX ∆ ) − AT U = −CX σ − AT U
on [a, b]κ.
Also, we have that rank(X T U T ) = n and X T U = X T QX,
U T X = X T QX,
whereupon X T U −U T X = 0. Therefore (X,U) is a conjoined basis of the system (4.3.4) with X invertible on [a, b]. By (4.4.2), we obtain µX ∆ = µAX σ + µBU
on [a, b]κ,
or X σ − X = µAX σ + µBU
on [a, b]κ,
(I − µA)X σ = X + µBU
on [a, b]κ,
or or ˜ + µBU) X σ = A(X ˜ + µBUX −1 X) = A(X ˜ + µBQ)X = A(I
on [a, b]κ.
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Svetlin G. Georgiev Therefore (X σ)−1 = X −1 (I + µBQ)−1 A˜ −1
on [a, b]κ,
and ˜ D = X(X σ )−1 AB ˜ = XX −1 (I + µBQ)−1 A˜ −1 AB = (I + µBQ)−1 B ≥ 0 on [a, b]κ. This completes the proof.
4.5. Picone’s Identity Definition 4.5.1. The equation x∆ = Axσ + Bu is referred to as equation of motion, while we call u∆ = −Cxσ − AT u Euler’s equation.
Definition 4.5.2. (x, u) is called admissible if it satisfies the equation of motion. Lemma 4.5.3. Let (x, u) be admissible, Q is symmetric on [a, b] and differentiable on [a, b]κ. We set z = u − Qx,
˜ D = B − µBA˜ T (Qσ + µC)AB on [a, b]κ.
Then
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183
(xT QX)∆ + (xσ )T Cxσ − uT Bu + zT Dz = xT (A˜ T R(Q) − µQBA˜ T R(Q))x +2uT (µBA˜ T R(Q))x on [a, b]κ
(4.5.1)
and x + µDz − µ2 BA˜ T R(Q)x = (A˜ −1 − µBA˜ T (Qσ + µC))xσ
(4.5.2)
on [a, b]κ.
Proof. Firstly we will prove (4.5.1). We have (xT Qx)∆ + (xσ )T Cxσ − uT Bu + zT Dz
= (xσ )T (Q∆ xσ + Qx∆ ) + (x∆ )T Qx + (xσ )T Cxσ − uT Bu + zT Dz
= (xσ )T Q∆ xσ + (xσ )T Qx∆ + (x∆ )T Qx + (xσ )T Cxσ − uT Bu + zT Dz ˜ + µABu) ˜ = xT A˜ T + µuT BA˜ T Q∆ (Ax ˜ + xT A˜ T + µuT BA˜ T QA(Ax + Bu) T T T T ˜ + x A + u B A Qx ˜ + µABu) ˜ + xT A˜ T + µuT BA˜ T C(Ax − uT Bu
+(uT − xT Q)D(u − Qx) ˜ + AT A˜ T Q + A˜ T CA˜ + QDQ x = xT A˜ T Q∆ A˜ + A˜ T QAA ˜ + A˜ T QAB ˜ + µA˜ T CAB ˜ − QD u +xT µA˜ T Q∆ AB ˜ + BA˜ T Q + µBA˜ T CA˜ − DQ x +uT µBA˜ T Q∆ A˜ + µBA˜ T QAA ˜ + µBA˜ T QAB ˜ + µ2 BA˜ T CAB ˜ − B + D u, +uT µ2 BA˜ T Q∆ AB
on [a, b]κ. Note that
˜ A˜ = I + µAA ˜ = I + µAA, ˜ µAA˜ = µAA = A˜ − I,
on [a, b]κ.
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Hence, we get
˜ + AT A˜ T Q + A˜ T C A˜ + QDQ x xT A˜ T Q∆A˜ + A˜ T QAA ˜ + A˜ T QAA ˜ + AT A˜ T Q + A˜ T C = xT A˜ T Q∆ + µA˜ T Q∆ AA ˜ +µA˜ T CAA˜ + QBQ − µQBA˜ T (Qσ + µC)ABQ x ˜ + µA˜ T CAA˜ = xT A˜ T Q∆ + A˜ T C + A˜ T AT Q + A˜ T QAA ˜ − A˜ T QAA ˜ +A˜T Qσ AA
˜ +QBQ − µQBA˜T (Qσ + µC)ABQ x = xT A˜ T Q∆ + A˜ T C + A˜ T AT Q + A˜ T (Qσ + µC)AA˜ ˜ +QBQ − µQBA˜T (Qσ + µC)ABQ x ˜ + BQ) = xT A˜ T Q∆ + A˜ T C + A˜ T AT Q + A˜ T (Qσ + µC)A(A ˜ ˜ −A˜T (Qσ + µC)ABQ + QBQ − µQBA˜ T (Qσ + µC)ABQ x ˜ = xT A˜ T R(Q) − µQBA˜T (Qσ + µC)ABQ
˜ + µQBA˜ T (Qσ + µC)AA ˜ −µQBA˜T (Qσ + µC)AA ˜ −A˜T (Qσ + µC)ABQ + QBQ x ˜ + BQ) = xT A˜ T R(Q) − µQBA˜T (Qσ + µC)A(A
−µQBA˜T Q∆ − µQBA˜ T C − µQBA˜ T AT Q +µQBA˜T Q∆ + µQBA˜ T C + µQBA˜ T AT Q ˜ +µQBA˜T (Qσ + µC)AA ˜ −A˜T (Qσ + µC)ABQ + QBQ x = xT A˜ T R(Q) − µQBA˜T R(Q)
+QBA˜T (Qσ − Q) + µQBA˜ T C + µQBA˜ T AT Q ˜ +µQBA˜T (Qσ + µC)AA ˜ −A˜T (Qσ + µC)ABQ + QBQ x = xT A˜ T R(Q) − µQBA˜T R(Q)
+QBA˜T (Qσ + µC) − QBA˜ T Q + µQBA˜ T AT Q +QBA˜T (Qσ + µC)A˜ − QBA˜T (Qσ + µC) ˜ −A˜T (Qσ + µC)ABQ + QBQ x
Linear Hamiltonian Dynamic Systems =
=
and
xT A˜ T R(Q) − µQBA˜ T R(Q) − QBA˜ T Q + QBA˜ T Q
˜ −QBQ + QBA˜ T (Qσ + µC)A˜ − A˜ T (Qσ + µC)ABQ + QBQ x xT A˜ T R(Q) − µQBA˜ T R(Q) x, on [a, b]κ,
˜ + BA˜ T Q + µBA˜ T CA˜ − DQ x uT µBA˜ T Q∆ A˜ + µBA˜ T QAA = uT µBA˜ T R(Q) − µBA˜ T Q∆ − µBA˜ T C − µBA˜ T AT Q ˜ + BQ) + µBA˜ T Q∆ A˜ −µBA˜ T (Qσ + µC)A(A ˜ + BA˜ T Q + µBA˜ T CA˜ − DQ x +µBA˜ T QAA ˜ + µ2 BA˜ T CAA ˜ = uT µBA˜ T R(Q) + µ2 BA˜ T Q∆ AA
˜ + BQ) −µBA˜ T AT Q − µBA˜ T (Qσ + µC)A(A ˜ + BA˜ T Q +µBA˜ T QAA ˜ −BQ + µBA˜ T (Qσ + µC)ABQ x ˜ − µBA˜ T QAA ˜ = uT µBA˜ T R(Q) + µBA˜ T Qσ AA ˜ − B(A˜ T − I)Q +µ2 BA˜ T CAA ˜ + BQ) −µBA˜ T (Qσ + µC)A(A
˜ + BA˜ T Q +µBA˜ T QAA ˜ −BQ + µBA˜ T (Qσ + µC)ABQ x ˜ = uT µBA˜ T R(Q) + µBA˜ T (Qσ + µC)AA
and
˜ + BQ) −µBA˜ T (Qσ + µC)A(A ˜ +µBA˜ T (Qσ + µC)ABQ x = uT µBA˜ T R(Q) x on [a, b]κ,
˜ + A˜ T QAB ˜ + µA˜ T CAB ˜ − QD u T xT µA˜ T Q∆ AB ˜ + BA˜ T Q + µBA˜ T CA˜ − DQ x = uT µBA˜ T Q∆ A˜ + µBA˜ T QAA = uT µBA˜ T R(Q) x on [a, b]κ,
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Svetlin G. Georgiev
and ˜ + µBA˜ T QAB ˜ + µ2 BAC ˜ AB ˜ −B+D u uT µ2 BA˜ T Q∆ AB ˜ + µBA˜ T QAB ˜ + µ2 BA˜ T CAB ˜ −B+D u = uT µBA˜ T (Qσ − Q)AB
˜ − µBA˜ T QAB ˜ + µBA˜ T QAB ˜ + µ2 BA˜ T CAB ˜ −B+D = uT µBA˜ T Qσ AB ˜ + µ2 BA˜ T CAB ˜ − B + B − µBA˜T (Qσ + µC)AB ˜ u = uT µBA˜ T Qσ AB
= 0 on [a, b]κ. Consequently
(xT Qx)∆ + (xσ )T Cxσ − uT Bu + zT Dz = xT A˜ T R(Q) − µQBA˜ T R(Q) x +2uT µBA˜ T R(Q) x on [a, b]κ.
Now we will prove (4.5.2). We have
x + µDz − µ2 BA˜ T R(Q)x ˜ (u − Qx) = x + µ B − µBA˜ T (Qσ + µC)AB ˜ + BQ) x −µ2 BA˜ T Q∆ +C + AT Q + (Qσ + µC)A(A ˜ = x + µBu − µBQx − µ2 BA˜ T (Qσ + µC)ABu 2 ˜T σ ˜ +µ BA (Q + µC)ABQx −µBA˜ T (Qσ − Q + µC)x − µ2 BA˜ T AT Qx ˜ −µ2 BA˜ T (Qσ + µC)AAx ˜ −µ2 BA˜ T (Qσ + µC)ABQx
˜ = −µ2 BA˜ T (Qσ + µC)A(Ax + Bu) 2 ˜T T +x + µBu − µBQx − µ BA A Qx −µBA˜ T (Qσ − Q + µC)x
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= −µ2 BA˜ T (Qσ + µC)x∆ − µBA˜ T (Qσ + µC)x +µBA˜ T Qx + x + µBu − µBQx − µ2 BA˜ T AT Qx
= −µBA˜ T (Qσ + µC)xσ +µBA˜ T (I − µAT )Qx − µBQx + x + µBu
= −µBA˜ T (Qσ + µC)xσ + µBQx − µBQx + x + µBu = −µBA˜ t (Qσ + µC)xσ + x + µBu
˜ + µABu) ˜ = −µBA˜ T (Qσ + µC)xσ + A˜ −1 (Ax = −µBA˜ T (Qσ + µC)xσ + A˜ −1 xσ = A˜ −1 − µBA˜ T (Qσ + µC) xσ on [a, b]κ.
This completes the proof.
Lemma 4.5.4. For any two matrices V and W we have KerV ⊆ KerW
iff W = WV †V
iff W † = V †VW † .
Proof. Let W = WV †V . Let also, x ∈ KerV be arbitrarily chosen. Then V x = 0. Hence, W x = WV †V x = 0. Therefore x ∈ KerW . Because x ∈ KerV was arbitrarily chosen and for it we get that it is an element of KerW , we conclude that KerV ⊆ KerW. Now we suppose that KerV ⊆ KerW. Let x be such that W T x is defined and there exists x = V d1 + d2
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Svetlin G. Georgiev
with d2 ∈ KerV T . Therefore W T x = V T (V d1 + d2 ) = V T V d1 +V T d2 = V T V d1 = V T VV †V d1 T = V T VV † V d1
= V T (V † )T V T V d1 = V T (V † )T W T x = (WV †V )T x.
Therefore W = WV †V. Note that W † = (WV †V )† = V † (V † )†W † = V †VW † . This completes the proof. Lemma 4.5.5. Suppose that X ∆ = AX σ + BU
(4.5.3)
and KerX σ ⊆ KerX hold at t ∈ [a, b]κ. Then X = X(X σ )† X σ ,
X † = (X σ )† X σ X †
(4.5.4)
and ˜ = µAB. ˜ µX σ(X σ )† AB Proof. We apply Lemma 4.5.4 for W = X and V = X σ and we get (4.5.4). By (4.5.3), we get ˜ + µABU ˜ X σ = AX
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and Ker(X σ )T ⊆ Ker(µBA˜ T ) at t. Hence and Lemma 4.5.4, we obtain µBA˜ T = µBA˜ T (X σ )†T (X σ )T , whereupon ˜ = µAB. ˜ µX σ(X σ )† AB This completes the proof. Lemma 4.5.6. Let t ∈ [a, b]κ. Suppose that X,U : [a, b] → Rn×n are matrices such that KerX σ ⊆ KerX and X ∆ = AX σ + BU hold at t. Let also, U and Q be differentiable at t with QX = UX † X symmetric on [t, σ(t)]. 1. If t is right-scattered, or if t is right-dense with X nonsingular at t, then R(Q)X = (U ∆ +CX σ + AT U)X † X. 2. If U ∆ = −CX σ − AT U
at t,
and XTU = UT X
at t,
then ˜ D = X(X σ )T A˜ = B − µBA˜ T (Qσ + µC)AB is symmetric at t and X T A˜ T R(Q)X = µBA˜ T R(Q)X = 0
190
Svetlin G. Georgiev at t. Moreover, if t is right-scattered, or if t is right-dense with X nonsingular at t, then we have R(Q)X = 0
at t,
or R(Q) = 0
at t,
respectively. Proof.
1. Let Z = U − QX.
Then QX = U − Z and ZX † X = (U − QX)X † X
= (U −UX † X)X † X
= UX † X −UX † XX † X = UX † X −UX † X = 0 at t.
Then R(Q)X = =
˜ + BQ) X Q∆ +C + AT Q + (Qσ + µC)A(A ˜ + BQ) XX † X Q∆ +C + AT Q + (Qσ + µC)A(A
= Q∆ XX † X +CXX † X + AT QXX † X ˜ + BQ)XX † X +(Qσ + µC)A(A
= (QX)∆X † X − Qσ X ∆ X † X +CXX † X
+AT (U − Z)X † X ˜ +(Qσ + µC)A(AX + BU − BZ)X † X
Linear Hamiltonian Dynamic Systems = (U − Z)∆ X † X − Qσ X ∆ X † X +CXX † X +AT (U − Z)X † X ˜ +(Qσ + µC)A(AX + BU − BZ)X † X
= U ∆ X † X − Z ∆ X † X − Qσ X ∆ X † X +CXX † X ˜ +AT UX † X + (Qσ + µC)A(AX + BU)X † X = U ∆ X † X − Z ∆ X † X − Qσ X ∆ X † X +CXX † X +AT UX † X + (Qσ + µC)X ∆ X † X
= U ∆ X † X − Z ∆ X † X − Qσ X ∆ X † X +CXX † X
+AT UX † X + Qσ X ∆ X † X +C(X σ − X)X † X
= U ∆ X † X − Z ∆ X † X + AT UX † X +CX σ X † X = U ∆ +CX σ + AT U X † X − Z ∆ X † X.
If t is right-scattered, then µ(t) > 0 and Z∆ X †X
= = = = =
1 σ (Z − Z)X † X µ 1 σ † Z X X µ 1 σ σ † σ † Z (X ) X X X µ 1 (ZX † X)σ X † X µ 0.
If t is right-dense with X nonsingular, then X † = X −1 and Z = ZX −1 X = ZX † X = 0. Therefore
R(Q)X = U ∆ +CX σ + AT U X † X.
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Svetlin G. Georgiev
2. Note that = X T UX † X
X T QX
= U T XX † X = U T X. Then ˜ D = X(X σ )† AB ˜ = −µ2 BA˜ T C + A˜ −1 X σ − µBA˜ T U σ (X σ )† AB ˜ + B − µBA˜ T U σ (X σ )† AB ˜ = −µ2 BA˜ T CAB 2 ˜T ˜ = −µ BA CAB + B
˜ −µBA˜ T (X σ )†T (X σ )T Qσ X σ (X σ )† AB ˜ = B − µBA˜ T (Qσ + µC)AB, i.e., D is symmetric. Note that XT Z = 0 and ˜ ˜ AX = X σ − µABU
˜ = X σ (X σ )† X σ − µX σ (X σ)† ABU σ σ † σ ˜ = X (X ) (X − µABU)
˜ = X σ (X σ )† AX.
Now we apply the computations of 1 and we obtain X T A˜ T R(Q)X = −X T A˜ T Z ∆ X † X = −X T A˜ T (X σ )†T (X σ )T Z ∆ X † X = −X T A˜ T (X σ )†T (X T Z)∆ X † X +X T A˜ T (X σ )†T (X ∆ )T ZX † X
= 0
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193
and µBA˜ T R(Q)X = −µBA˜ T (X σ )†T (X T Z)∆ X † X +µBA˜ T (X σ )†T (X ∆ )T ZX † X = 0. If t is right-scattered, then, using 1, we get R(Q)X = 0 at t. If t is right-dense with X nonsingular at T , using 1, we obtain R(Q) = 0 at t. This completes the proof. Below we suppose the following. ˜ U) ˜ be normalized conjoined basis of (4.3.4) such that (A) Let (X,U) and (X, KerX σ ⊆ KerX holds on [a, b] and X is nonsingular at all dense points of (a, b]. We define the matrix ˜ − X † X)U T Q = UX † + (UX † X˜ − U)(I
on [a, b].
˜ U) ˜ is a normalized conjoined basis, we have Because (X,U) and (X, X T U −U T X = 0,
X˜ T U˜ − U˜ T X˜ = 0 on [a, b],
and X T U˜ −U T X˜ = I
on [a, b].
Note that ˜ − X † X)U T X QX = UX † X + (UX † X˜ − U)(I
on [a, b].
(4.5.5)
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Svetlin G. Georgiev
Since T (UX † X˜ − U˜ )(I − X † X)U T X ˜ T = X T U(I − X † X)(UX †X˜ − U)
˜ T = U T X(I − X † X)(UX †X˜ − U) = U T (X − XX † X)(UX † X˜ − U˜ )T = O on [a, b].
Therefore QX = UX † X
on [a, b].
Next, Q = UX † + (UX † X˜ − U˜ )(U T − X † XU T ) ˜ T − UU ˜ T −UX † XX ˜ † XU T = UX † +UX † XU ˜ † XU T +UX T ˜ T −UX † XXU ˜ = −UU +UX † ˜ T + UX ˜ † XU T +UX † XU
˜ T −UX † XX ˜ † XU T + UX ˜ † XU T = −UU ˜ T) +UX † (I + XU ˜ T −UX † XX ˜ † XU T + UX ˜ †U T = −UU +UX † X U˜ T on [a, b]. By Theorem 4.3.10, 1, we have that U U˜ T is symmetric. Then (U U˜ T )T
= U U˜ T ˜ T, = UU
hence, ˜ T UU
T
U U˜ T = U U˜ T ˜ T. = UU =
T
Linear Hamiltonian Dynamic Systems
195
Next, again by Theorem 4.3.10, 1, we have that X X˜ T is symmetric and X X˜ T
T
= X X˜ T ˜ T, = XX
and UX † X˜ X † XU T
T
T = UX † X X˜ T X † U T ˜ T (X † )T U T = UX † XX ˜ † X)T U T = UX † X(X ˜ † XU T on [a, b]. = UX † XX
Therefore QT
˜ T )T − UX † XX ˜ † XU T T = −(UU ˜ † XU T T + UX † X U˜ T T + UX ˜ T −UX † XX ˜ † XU T = −UU ˜ † XU T +UX † X U˜ T + UX
= Q on [a, b], i.e., Q is symmetric on [a, b]. Note that Q is differentiable on [a, b]κ and if t is right-dense, then X is nonsingular at t and Q = UX −1 at t. Also, if (X,U) has no focal point in (a, b], then (A) holds. Now we consider the functional
F0(x, u) =
Z b a
uT Bu − (xσ )T Cxσ (t)∆t.
Theorem 4.5.7. Suppose that the hypothesis (A) holds. Let an admissible (x, u) with x(a) ∈ ImX(a) be given. Then x ∈ ImX on [a, b] and
F0 (x, u) = (x(b))T Q(b)x(b) − (x(a))T Q(a)x(a) +
Z b a
and
zT Dz (t)∆t,
x + µDz = A˜ −1 − µBA˜ T (Qσ + µC) xσ
on [a, b]κ,
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Svetlin G. Georgiev
where Q is given by (4.5.5), z = u − Qx = u −UX † x on [a, b], and ˜ D = B − µBA˜ T (Qσ + µC)AB on [a, b]κ. ˜ U) ˜ be normalized conjoined basis of (4.3.4) satisProof. Let (X,U) and (X, fying (A). We have that Q is differentiable on [a, b]κ. We will prove that x(t) ∈ ImX(t) for all t ∈ [a, b]. Define the statement E(t) = {x(t) ∈ ImX(t),
t ∈ [a, b]}.
We have 1. E(a) holds. 2. Let t is right-scattered and E(t) holds. Then by Theorem 4.3.12, it follows that E σ (t) holds. 3. Let t is right-dense and E(t) holds. Then X is nonsingular at t and there exists a neighbourhood V of t such that X is invertible at any s ∈ V , s > t. Therefore d(s) = (X(s))−1x(s) ∈ Rn and x(s) = X(s)d(s) ∈ ImX(s). Therefore E(s) holds for all s ∈ V , s > t. 4. Let t be left-dense and E(s) is true for all s ∈ [a,t). Then X is nonsingular at t. Hence, d(t) = (X(t))−1x(t) ∈ Rn and x(t) = X(t)d(t) ∈ ImX(t).
Linear Hamiltonian Dynamic Systems Consequently x ∈ ImX on [a, b]. By Lemma 4.5.6, 2, it follows that ˜ D = X(X σ )† AB ˜ = B − µBA˜ T (Qσ + µC)AB is symmetric at t and X T A˜ T R(Q)X = µBA˜ T R(Q)X = 0 on [a, b]κ. By Lemma 4.5.3, it follows that
and Hence,
x + µDz = A˜ −1 − µBA˜ T (Qσ + µC) xσ
on [a, b]κ,
∆ xT Qx + (xσ )T Cxσ − uT Bu + zT Dz = 0 on [a, b]κ.
F0 (x, u) =
Z b
=
Z b
a
a
uT Bu − (xσ )T Cxσ (t)∆t
Z b ∆ xT Qx (t)∆t + (zT Dz)(t)∆t a
= (xT Qx)(b) − (xT Qx)(a) + Let now t ∈ [a, b] and
Z b a
x(t) = X(t)d(t)
for some d(t) ∈ Rn . Then Qx(t) = QXd(t) = UX † Xd(t) = UX † x(t). Therefore z = u − Qx = u −UX † x This completes the proof.
on [a, b].
(zT Dz)(t)∆t.
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Svetlin G. Georgiev
4.6. “Big” Linear Hamiltonian Systems We define 2n × 2n matrices A∗ , B∗ , C∗ as follows O O O O O O ∗ ∗ ∗ A = , B , C = O A O B O C Then I − µA
∗
= =
I O O I
−µ
I O O I − µA
O O O A
on [a, b].
on [a, b].
Hence, I − µA∗ is invertible on [a, b]κ with A˜ ∗ = (I − µA∗ )−1 I O = O A˜
on [a, b]κ.
Also, A∗ , B∗ , C∗ ∈ Crd ([a, b]), B∗ and C∗ are symmetric. For n × n matrices X˜ and U˜ we define the matrices O I I O ∗ ∗ X = , U = on [a, b]. X X˜ U U˜ Consider the ”big” linear Hamiltonian system ( (X ∗ )∆ = A∗ (X ? )σ + B∗U ∗ (U ∗ )∆ = −C∗ (X ∗ )σ − (A∗ )T U ∗
on [a, b]κ.
(4.6.1)
Suppose that the hypothesis (A) hold. ˜ U) ˜ are normalized conjoined bases of the sysTheorem 4.6.1. (X,U) and (X, ∗ ∗ tem (4.3.4) if and only if (X ,U ) is a conjoined basis of the system (4.6.1). ˜ U) ˜ are normalized conjoined bases of the sysProof. 1. Let (X,U) and (X, tem (4.3.4). Then T X U −U T X = O X˜ T U˜ − U˜ T X˜ = O (4.6.2) T ˜ X U −U T X˜ = I on [a, b].
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199
Then ∗ T
∗
∗ T
(X ) U − (U ) X
∗
=
=
= = =
O XT I O I X˜ T U U˜ I UT O I − O U˜ T X X˜ XTU X T U˜ I + X˜ T U X˜ T U˜ T U X I −U T X˜ − U˜ T X U˜ T X˜ X T U −U T X X T U˜ − I +U T X˜ I + X˜ T U − U˜ T X X˜ T U˜ − U˜ T X˜ O O O O O.
Therefore (X ∗ ,U ∗ ) is a conjoined basis of (4.6.1). 2. Let (X ∗ ,U ∗ ) is a conjoined basis of (4.6.1). Then, using the computations in the previous step, we get (4.6.2). This completes the proof. Let Q = UX † + UX † X˜ − U˜ I − X † X U T , Q˜ = X † + X † X˜ I − X † X U T , ˜ † X Q˜ −X † XX ∗ Q = . Q˜ T Q By the previous section, we have that Q is symmetric with QX = UX † X
on [a, b].
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Svetlin G. Georgiev
Also, we have ˜ †X X † XX
T
=
X †X
T
= X † X X˜
X † X˜
T
T
X† † T
˜ T X = X † XX T = X † X˜ X † X ˜ † X. = X † XX
T
Therefore ∗ T
(Q )
˜ †X − X † XX Q˜ T
=
T
˜ † X Q˜ −X † XX Q˜ T Q ∗ = Q on [a, b], =
Q˜ Q
!
i.e., Q∗ is symmetric on [a, b]. Next, ˜ † X Q˜ −X † XX O I ∗ ∗ Q X = Q˜ T Q X X˜ ˜ ˜ † X + Q˜ X˜ QX −X † XX = . QX Q˜ T + QX˜ Consider ˜ = X † X + X † X(I ˜ − X † X)U T X. QX Since ˜ − X † X)U T X X † X(I
T
˜ T = X T U(I − X † X)(X † X) ˜ T = U T X(I − X † X)(X † X) ˜ T = U T (X − XX † X)(X † X) ˜ T = U T (X − X)(X † X) = 0,
Linear Hamiltonian Dynamic Systems we get ˜ = X † X. QX Next, ˜ + X + Q˜ X˜ = −X † XX ˜ † X + X † + X † X(I ˜ − X † X)U T X˜ −X † XX ˜ † X + X † X˜ + X † X(I ˜ − X † X)(I ˜ − X † X)(X T U˜ − I) = −X † XX ˜ † X + X † X˜ − X † X˜ + X † XX ˜ †X = −X † XX ˜ − X † X)X T U˜ +X † X(I
˜ − X † X)X T U, ˜ = X † X(I and using that
(I − X † X)X T
T
= X(I − X † X)
= X − XX † X
= X −X = 0, we get ˜ † X + Q˜ X˜ = O. −X † XX Also, Q˜ T
T +U X † X˜ −UX † X˜ X † X ˜ †X = (X † )T +U X˜ T (X † )T −UX † XX ˜ T − I)(X † )T −UX † XX ˜ †X = (X † )T + (UX
=
X†
T
˜ † X)T −UX † XX ˜ †X = U(X ˜ † X −UX † XX ˜ † X, = UX
201
202 QX˜
QX˜ + Q˜ T
Svetlin G. Georgiev ˜ †X = UX † X˜ + UX † X˜ − U˜ −UX † XX +U˜ X † X U T X˜
˜ † X + UX ˜ †X = UX † X˜ + UX † X˜ − U˜ −UX † XX × X T U˜ − I = UX † X˜ + UX † X˜ + Q˜ T − U˜ X T U˜ − I = UX † X˜ −UX † X˜ − Q˜ T + U˜ ˜ + UX † X˜ + Q˜ T − U˜ X T U, = U˜ + UX † X˜ + Q˜ T − U˜ X T U˜ ˜ † X −UX † XX ˜ † X − U˜ X T U˜ = U˜ + UX † X˜ + UX = U˜ + UX † X˜ I − X † X X T + U˜ X † X − I X T U˜ ˜ = U.
Consequently ∗
∗
QX =
X †X O QX U˜
.
Note that
whereupon
˜ † X − X † XX ˜ †X 0 = X † XX ˜ † X − X † XX † XX ˜ †X = X † XX = X † X˜ − XX † X˜ X † X, X˜ − XX † X˜ = O.
Now we will prove that ∗ †
(X ) =
−X T X˜ I
X† O
.
203
Linear Hamiltonian Dynamic Systems We have X
∗
−X † X˜ I
X† O
X
∗
= =
I X˜
O X
I X˜
O X
= =
O X
O XX † X I X˜
I X˜
−X † X˜ I X †X O
O I
X† O
O I X X˜
= X ∗, −X † X˜ X † −X † X˜ X † X∗ I O I O † † ˜ O I −X X X −X † X˜ X † = X X˜ I O I O † † ˜ I O −X X X = I O −XX † X˜ + X˜ XX † I O −X † X˜ X † = I O O XX † −X † X˜ X † XX † = I O † † −X X˜ X = . I O
Note that X∗
−X † X˜ X † I O
= = =
−X † X˜ I
X† O
X
∗
= =
−X † X˜ X † I O I O −XX † X˜ + X˜ XX † I O , O XX † −X † X˜ X † O I I O X X˜ X †X O . O I O I X X˜
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Svetlin G. Georgiev
Therefore X
∗
−X † X˜ I
X† O
and
−X † X˜ X † I O
X∗
are symmetric. Consequently ∗ †
(X ) =
−X † X˜ I
X† O
I O U U˜
.
Hence, ∗
∗ †
U (X ) X
∗
=
X †X O O I
X †X O UX † X U˜ † X X O = . QX U˜ =
From here, Q∗ X ∗ = U ∗ (X ∗ )† X ∗ . Moreover, Q∗ is differentiable on [a, b]κ and Q˜ = X −1 at dense points of [a, b]. ˜ U) ˜ be normalized conjoined bases of Theorem 4.6.2. Let (X,U) and (X, (4.3.4). Then Ker(X ∗ )σ ⊆ KerX ∗ if and only if KerX σ ⊆ KerX. Proof.
1. Let
Ker(X ∗ )σ ⊆ KerX ∗ . x x σ ∗ σ Suppose that x ∈ KerX . Then ∈ Ker(X ) . Hence, ∈ 0 0 KerX ∗ and x ∈ KerX. Because x ∈ KerX σ was arbitrarily chosen and for it we get that it is an element of KerX, we conclude that KerX σ ⊆ KerX.
Linear Hamiltonian Dynamic Systems
205
2. Let
KerX σ ⊆ KerX.
x Suppose that ∈ Ker(X ∗ )σ . Then x ∈ KerX σ and hence, x ∈ KerX. 0 x Therefore ∈ KerX ∗ and 0 Ker(X ∗ )σ ⊆ KerX ∗ .
This completes the proof. Let D∗ = X ∗ (X ∗σ)† A˜ ∗ B∗ . We have ∗ †σ
(X )
X ∗ (X ∗σ )†
A˜ ∗ B∗
X ∗ (X ∗σ )† A˜ ∗ B∗
−(X σ )† X˜ σ (X σ)† = , I O O I −(X σ )† X˜ σ (X σ )† = X X˜ I O I O = , −X(X σ )† X˜ σ + X˜ X(X σ)† I O O O = O A˜ O B O O = , ˜ O AB O O I O = ˜ O AB −X(X σ )† X˜ σ + X˜ X(X σ)† O O = ˜ O X(X σ )† AB O O = , O D
i.e., ∗
D =
O O O D
.
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Svetlin G. Georgiev
Also,
−(X σ )† X˜ σ (X σ )† X σ (Q˜ σ )T O O µC∗ = , O µC −(X σ )† X˜ σ (X σ )† X σ (Q∗ )σ + µC∗ = (Q˜ σ )T O O + O µC −(X σ )† X˜ σ (X σ )† X σ = (Q˜ σ )T ∗ σ ∗ ˜∗ ∗ ((Q ) + µC ) A B −(X σ )† X˜ σ (X σ )† X σ = (Q˜ σ )T ˜ O Q˜ σ AB = ˜ O (µC + Qσ ) Q˜ σ AB (Q∗ )σ
A˜ ∗ B∗ A˜ ∗
T T
=
= = =
I O
O A˜ T
O O
O B
O O
O BA˜ T
T
,
I O
O A˜ T
Q˜ σ Qσ
Q˜ σ Qσ
,
Q˜ σ µC + Qσ Q˜ σ µC + Qσ ,
,
O O
O ˜ AB
,
((Q∗ )σ + µC∗ ) A˜ ∗ B∗ ˜ O O O Q˜ σ AB = µ ˜ O BA˜ T O (µC + Qσ ) Q˜ σ AB O O = µ ˜ O BA˜ T (µC + Qσ )Q˜ σ AB O O = , ˜ O µBA˜ T (µC + Qσ )Q˜ σ AB T B∗ − µB∗ A˜ ∗ ((Q∗ )σ + µC∗ ) A˜ ∗ B∗ O O = T σ σ ˜ O B − µBA˜ (µC + Q )Q˜ AB O O = O D µB∗ A˜ ∗
=
D∗ .
Linear Hamiltonian Dynamic Systems Now we define ∗
x =
α x
,
∗
u =
0 u
207
on [a, b].
Theorem 4.6.3. (x∗ , u∗ ) is admissible if and only if (x, u) is admissible. Proof. We have ∗
BU
∗
= =
∗ σ
(X ) ∗
∗ σ
A (X )
= = =
Then ∗ ∆
(X )
= = = =
O O O B
I U O O , BU BU˜ O I , X σ X˜ σ O O O Xσ O A O O . AX σ AX˜ σ
O U˜
I X˜ σ
O O X ∆ X˜ ∆ A∗ (X ∗ )σ + B∗U ∗ O O O O + AX σ AX˜ σ BU BU˜ O O . AX σ + BU AX˜ σ + BU˜
Hence, (x∗ , u∗ ) is admissible if and only if (x, u) is admissible. This completes the proof. Exercise 4.6.4. Let (x∗ , u∗ ) is admissible. Prove that x∗ ∈ ImX ∗ if and only if α +U T x ∈ ImX T
on [a, b].
208
Svetlin G. Georgiev Let Q∗1
=
O O O Q
.
Theorem 4.6.5 (Extended Piconi’s Identity). Suppose that hypothesis (A) holds. Let be given an admissible (x, u) and a constant α ∈ Rn with α + U T (a)x(a) ∈ ImX T (a). Then α +U T x ∈ ImX T on [a, b],
F0 (x, u) =
+
α x(b) Z b
T
Q∗1 (b)
α x(b)
−
α x(a)
T
Q∗1 (a)
α x(a)
(zT Dz)(t)∆t
a
and
x + µDz = A˜ −1 − µBA˜ T (Qσ + µC) xσ − µBA˜ T (Q˜ σ )T α
on [a, b]κ, where
z = u − Qx − Q˜ T α on [a, b].
Proof. Since (x, u) is admissible, we have that (x∗ , u∗ ) is admissible. Therefore α +U T x ∈ ImX T and from here, x∗ ∈ ImX ∗ on [a, b]. Let z∗ = u∗ − Q∗ x∗ . We have z
∗
˜ † X Q˜ −X † XX = − Q˜ T Q † † ˜ ˜ Xα − Qx X XX = T u − Q˜ α − Qx † † ˜ ˜ Xα − Qx X XX = . z 0 u
α x
209
Linear Hamiltonian Dynamic Systems Then ∗ T
∗ ∗
(u ) B u
0 0 = (0, u ) 0 B 0 T = (0, u ) Bu T
= uT Bu,
∗ σ T
∗
∗ σ
((x ) ) C (x )
0 u
0 0 α = (α , (x ) ) 0 C xσ 0 = (αT , (xσ)T ) Cxσ T
σ T
= (xσ )T Cxσ , whereupon
uT Bu − (xσ )T Cxσ = (u∗ )T B∗ u∗ − ((x∗ )σ)T C∗ (x∗ )σ . Also, ∗ T
(x )
Q∗1 x∗
(z∗ )T Dz∗
O O = (α , x ) O Q 0 T T = (α , x ) Qx T
T
α x
= xT Qx, ˜ † Xα T − xT Q˜ T , zT = X † XX † † ˜ ˜ Xα − Qx O O X XX × O D z = zT Dz.
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Svetlin G. Georgiev
Therefore
F0 (x, u) =
uT Bu − (xσ )T Cxσ (t)∆t
Z b a
= F0 (x∗ , u∗ ) Z b ∆ = (x∗ )T Q∗1 x∗ + (z∗ )T D∗ z∗ (t)∆t a ∗ T
= (x ) (b)Q∗1(b)x∗ (b) − (x∗ )T (a)Q∗1 (a)x∗ (a) Z b + (z∗ )T D∗ z∗ (t)∆t a T T α α α α ∗ ∗ = Q1 (b) − Q1 (a) x(b) x(a) x(b) x(a) Z b zT Dz (t)∆t. + a
Note that ∗
∗ ∗
x + µD z
α x
O O = +µ O D α 0 = +µ x Dz α = . x + µDz
˜ ˜ † Xα − Qx X † XX z
Hence, applying Theorem 4.5.7, we get α ∗ ∗ ∗ x + µD z = . A˜ −1 − µBA˜ T (Qσ + µC) xσ
(4.6.3)
Linear Hamiltonian Dynamic Systems
211
Observe that (A˜ ∗ )−1 − µB∗ (A˜ ∗ )T (Q∗ )σ + µC∗ I O O O I O = −µ O A˜ −1 O B O A˜ T ˜ † X σ Q˜ σ O O − X † XX × +µ O C Q˜ σT Qσ I O O O = −µ O A˜ −1 O BA˜ T ˜ †X σ − X † XX Q˜ σ × Q˜ σT Qσ + µC O O I O −µ = O A˜ −1 BA˜ T Q˜ σT BA˜ T (Qσ + µC) I O = , −µBA˜ T Q˜ σT A˜ −1 − µBA˜ T (Qσ + µC) (A˜ ∗ )−1 − µB∗ (A˜ ∗ )T (Q∗ )σ + µC∗ (x∗ )σ I O α = −µBA˜ T Q˜ σT A˜ −1 − µBA˜ T (Qσ +µ C) xσ α = . A˜ −1 − µBA˜ T (Qσ +µ C) xσ − µBA˜ T Q˜ σT α
From the last equality and from (4.6.3), we obtain T −1 A˜ ∗ − µB∗ A˜ ∗ (Q∗ )σ + µC∗ (x∗ )σ = x∗ + µD∗ z∗ .
This completes the proof.
4.7. Positivity of Quadratic Functionals Theorem 4.7.1 (Sufficient Condition for Positive Definiteness). Suppose that there exists a conjoined basis of the system (4.3.4) with no focal point in (a, b]. Then the quadratic functional F0 is positive definite.
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Svetlin G. Georgiev
Proof. Suppose that (X,U) is a conjoined basis of the system (4.3.4) such that X is nonsingular at right-dense points of [a, b] and KerX σ ⊆ KerX,
˜ ≥ 0. D = X(X σ )† AB
˜ U) ˜ By Theorem 4.3.11, it follows that there exists another conjoined basis (X, ˜ ˜ of (4.3.4) such that (X,U) and (X, U) are normalized. Let (x, u) be admissible with x(a) = x(b) = 0. Then x(a) ∈ ImX(a) and by Theorem 4.5.7 we get that x(t) ∈ ImX(t) for all t ∈ [a, b]. 1. Suppose that a is right-scattered. Then, using Theorem 4.5.7, we obtain
F0(x, u) = =
Z b
Z σ(a) a
a
uT Bu − (xσ )T Cxσ (t)∆t
uT Bu − (xσ )T Cxσ (t)∆t
uT Bu − (xσ )T Cxσ (t)∆t σ(a) = µ(a) uT Bu − (xσ )T Cxσ (a) +
≥
= =
=
Z b
+(x(b))T Q(b)x(b) − (xσ(a))T Qσ (a)xσ (a) Z b + zT Dz (t)∆t σ(a) µuT Bu − (xσ )T (Qσ + µC)xσ (a) ˜ + µBu) (a) µuT Bu − (x + µBu)T A˜ T (Qσ + µC)A(x ˜ µuT (B − µBA˜ T (Qσ + µC)AB)u ˜ + 2µBu) (a) −xT A˜ T (Qσ + µC)A(x µzT Dz (a)
≥ 0.
Now we assume that F0 (x, u) = 0. Then, using Theorem 4.5.7, we get Z b a
(zT Dz)(t)∆t = 0.
Linear Hamiltonian Dynamic Systems
213
Since D is positive definite, we obtain zT Dz = 0 on [σ(a), b]κ and Dz = 0
on [σ(a), b]κ.
2. Now we suppose that a is right-dense. We take a sequence {am }m∈N ⊆ (a, b]κ such that am → a, as m → ∞. Since x ∈ ImX on [a, b], we have (I − X † X)U T x = (I − X † X)U T Xd
= (I − X † X)X T Ud = 0
for some d ∈ Rn . Hence, ˜ = X † x on (a, b] Qx and Qx = UX † x on (a, b]. Also, ˜ † XU T x = xT UX † X(X ˜ † )T U T x xT UX † XX = xT UX † X X˜ T X T (X † )T U T x = xT QX X˜ T (X † )T U T x ˜ T (X † )T U T x = xT QT XX ˜ † XU T x. = xT QT XX Note that ˜ T − U˜ T ), QT = (X † )T U T +U(I − X † X)(X † XU ˜ T − U˜ T )X˜ QT X˜ = (X † )T U T X˜ +U(I − X † X)(X † XU = (X † )T U T X˜ = (X † )T (X T U˜ − I).
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Svetlin G. Georgiev Therefore xT UX † X˜ X † XU T x = xT (X † )T (X T U˜ − I)X † XU T x ˜ † XU T x = xT (X † )T X T UX = = =
−xT (X † )T X † XU T x ˜ † XU T x − xT (X † )T X T (X † )T U T x xT XX †UX T † ˜ † x XX UX XU T x − xT (X † )T U T x ˜ † Xαm − xT Qx, xT XX †UX
where αm = U T x(tm). Now we apply Theorem 4.6.5 on the interval [am , b] with α = αm and and we get Z b
am
zm = u − Qx − Q˜ T αm
(uT Bu − (xσ )T Cxσ )(t)∆t
=
T αm αm Q∗ (b) x(b) x(b) T αm αm Q∗ (a) − x(am) x(am)
=
=
=
=
=
Z b
(zTmDzm )(t)∆t am ˜ † Xαm )(b) −(αTmX † XX
+ ≥
on [a, b]
˜ − αTm X † XX ˜ † Xαm (am) − xT Qx + 2αTmQx
˜ † Xαm )(b) −(αTmX † XX ˜ − xT UX † X˜ X † XU T x (am) − xT Qx − 2xT U Qx
˜ † Xαm )(b) −(αTmX † XX − xT Qx − 2xT UX † x − xT UX † X˜ X † XU T x (am)
˜ † Xαm )(b) −(αTmX † XX ˜ † XU T x (am) − xT Qx − 2xT Qx − xT UX † XX
˜ † Xαm )(b) −(αTmX † XX − −xT Q∗ − xT UX † X˜ X † XU T x (am) ˜ † Xαm )(b) −(αTmX † XX ˜ † Xαm (am ). + xT UX † XX
Linear Hamiltonian Dynamic Systems Now, using that αm → 0, as m → ∞, we get
F0(x, u) =
Z b
lim
m→∞ am
uT Bu − (xσ )T Cxσ (t)∆t
˜ † Xαm )(b) lim −(αTm X † XX m→∞ ˜ † Xαm (am) xT UX † XX
≥
= 0. If F0 (x, u) = 0, using lim
Z b
m→∞ am
(zTm Dzm)(t)∆t = 0,
where as m → ∞,
zm → z = u − Qx,
holds uniformly on [ak , b] for any k ∈ N. We fix k ∈ N. Observe that Z b am
(zTmDzm )(t)∆t
Z b
≥
ak
(zTmDzm )(t)∆t
for any m ≥ k. Therefore lim
Z b
m→∞ am
(zTmDzm)(t)∆t
≥
Z b
(zT Dz)(t)∆t
ak
and hence, 0 = ≥
lim
Z b
m→∞ am Z b T
(zTm Dzm)(t)∆t
(z Dz)(t)∆t
a
≥ 0. Consequently zT Dz = 0 on [a, b]κ. Therefore Dz = 0
on [σ(a), b]κ.
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Svetlin G. Georgiev
We have
x = A˜ −1 − µBA˜ T (Qσ + µC) xσ
on [σ(a), b]κ.
(4.7.1)
Now we will show that x = 0 on [a, b]. Consider the statement L(t) = {x(t) = 0 on [σ(a), b]}. 1. We have that L(b) is true because x(b) = 0. 2. Let t is left-scattered and L(t) is true. Then ρ(t) is right-scattered and using (4.7.1), we get L(ρ(t)) is true. 3. Let t ∈ (σ(a), b] is left-dense and L(t) is true. Since X is nonsingular at t, there exists a neighbourhood V of t such that X is nonsingular on V . Because KerX σ (t) ⊆ KerX(t), we get that X σ (t) is nonsingular. Therefore Dz = 0
⇐⇒
Bu = BUX −1x
on V,
and we obtain ˜ + BUX −1 )x on V. x∆ = A(A Note that ˜ + BUX −1 ) = A(I ˜ − µA + µA + µBUX −1 ) I + µA(A ˜ + µBUX −1 ) = A(I ˜ + µBU)X −1 = A(X = X σ X −1 . Therefore ˜ + BUX −1 ) A(A is regressive on V . Hence, the IVP ∆ ˜ x = A(τ)(A(τ) + B(τ)U(τ)X −1(τ))x, x(t) = 0
τ ∈ V,
has unique solution x = 0 on V . Therefore L(s) is true for s ∈ V , s < t.
Linear Hamiltonian Dynamic Systems
217
4. Let t ∈ [σ(a), b) is right-dense and L(s) is true for s ∈ (t, b]. Since x is continuous at t, we have x(t) = lim x(s) s→t
= 0, i.e., L(t) holds. Therefore L(t) is true for all t ∈ [σ(a), b] and since x(a) = 0, we have x = 0 on [a, b]. Consequently F0 > 0. This completes the proof.
Chapter 5
The First Variation Let T be a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively.
5.1. The Dubois-Reymond Lemma In this section it will be deducted several versions of the fundamental lemma of Dubois-Reymond of variational calculus on time scales. Lemma 5.1.1. Let g ∈ Crd ([a, b]), g : [a, b] → Rn . Then Z b a
1 gT (t)η(t)∆t = 0 for all η ∈ Crd ([a, b]),
η : [a, b] → Rn ,
η(a) = η(b) = 0, (5.1.1)
holds if and only if g(t) ≡ 0 on [a, b]κ. Proof.
(5.1.2)
1. Let (5.1.2) holds. Then it follows (5.1.1).
2. Suppose that (5.1.1) holds. Assume that g(t1 ) 6= 0 for some t1 ∈ [a, b]κ. Without loss of generality, we suppose that g(t1) > 0. Since g ∈ Crd ([a, b]), there exists [t2,t3] ⊂ [a, b]κ such that t1 ∈ [t2,t3] and g(t) > 0
220
Svetlin G. Georgiev for t ∈ [t2 ,t3 ]. Consider the function (t − t2 )(t3 − t) .. for t ∈ [t2 ,t3 ] . (t − t2 )(t3 − t) η(t) = 0 .. . for [a, b]\[t2,t3 ]. 0 Then
0 =
Z b a n
=
∑
gT (t)η(t)∆t
Z t3
i=1 t2
gi (t)(t − t2 )(t3 − t)∆t
> 0. This is a contradiction. Consequently g(t) ≡ 0 on [a, b]κ. This completes the proof. Lemma 5.1.2 (Dubois-Reymond Lemma). Let g ∈ Crd ([a, b], g : [a, b] → Rn . Then Z b a
1 ([a, b]), η : [a, b] → Rn, η(a) = η(b) = 0, gT (t)η∆ (t)∆t = 0 for all η ∈ Crd
(5.1.3)
holds if and only if g(t) ≡ c on [a, b]κ Proof.
for some c ∈ Rn .
1. Let (5.1.4) holds. Then Z b a
gT (t)η∆(t)∆t =
Z b
cT η∆ (t)∆t
a
= cT
Z b
η∆ (t)∆t
a
= cT (η(b) − η(a)) = 0.
(5.1.4)
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The First Variation 2. Let (5.1.3) holds. We set G(t) =
Z t
g(τ)∆τ,
a
t ∈ [a, b],
c=
G(b) . b−a
Let η(t) = G(t) − (t − a)c,
t ∈ [a, b].
Then η∆ (t) = G∆ (t) − c = g(t) − c,
η(a) = G(a) = 0,
η(b) = G(b) − (b − a)c = G(b) − G(b)
= 0. Hence, Z b
gT (t)η∆ (t)∆t
Z b
gT (t) (g(t) − c)∆t
Z b
(g(t) − c)T (g(t) − c)∆t + cT
Z b
|g(t) − c|2∆t + cT
=
Z b a
|g(t) − c|2∆t + cT G(b) − cT c(b − a)
=
Z b
|g(t) − c|2∆t + cT (G(b) − c(b − a))
=
Z b
|g(t) − c|2∆t.
0 =
a
=
a
=
a
=
a
a
a
Z b a
Z b a
(g(t) − c)∆t
g(t)∆t − cT c(b − a)
Therefore g(t) ≡ c on [a, b]. This completes the proof.
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Svetlin G. Georgiev
Lemma 5.1.3. Let g ∈ Crd ([a, b]), g : [a, b] → Rn . Then Z b
2
gT (t)η∆ (t)∆t = 0
(5.1.5)
a
2 for all η ∈ Crd ([a, b]), η : [a, b] → Rn , η(a) = η(b) = 0, η∆ (a) = η∆ (b) = 0, if and only if g(t) = c0 + c1 σ(t), t ∈ [a, b]κ, (5.1.6)
for some c0 , c1 ∈ Rn . Proof. 2 ([a, b]),η : [a, b] → Rn , η(a) = η(b) = 0, Let (5.1.6) holds. Suppose that η ∈ Crd ∆ ∆ η (a) = η (b) = 0. Then
Z b
2 cT0 + cT1 σ(t) η∆ (t)∆t a t=b Z b = cT0 + cT1 t η∆ (t) − cT1 η∆ (t)∆t t=a a t=b = −cT1 η(t)
2
gT (t)η∆ (t)∆t =
a
Z b
t=a
= 0, i.e., (5.1.5) holds.
Suppose that (5.1.5) holds. Let c0 , c1 ∈ Rn be such that ( R R b s (g(ξ) − c0 − c1 σ(ξ))∆ξ∆s = 0 Rab a a (g(ξ) − c0 − c1 σ(ξ))∆ξ = 0. We take η(t) =
Z tZ s a
a
(g(ξ) − c0 − c1 σ(ξ)) ∆ξ∆s,
t ∈ [a, b].
223
The First Variation 2 Then η ∈ Crd ([a, b]) and
η(a) = 0, η(b) =
Z bZ s a
a
(g(ξ) − c0 − c1 σ(ξ))∆ξ∆s
= 0, ∆
η (t) =
Z t a
(g(ξ) − c0 − c1 σ(ξ)) ∆ξ,
t ∈ [a, b],
η∆ (a) = 0, η∆ (b) =
Z b a
(g(ξ) − c0 − c1 σ(ξ))∆ξ
= 0. Hence, Z b a
gT (t) − cT0 − cT1 σ(t) (g(t) − c0 − c1 σ(t))∆t
= =
Z b
2 gT (t) − cT0 − cT1 σ(t) η∆ (t)∆t
a
Z b a
−cT1
= = =
2
gT (t)η∆ (t)∆t − cT0 Z b
Z b
2
η∆ (t)∆t
a
2
σ(t)η∆ (t)∆t
a
Z b t=b η∆ (t)∆t −cT1 tη∆ (t) +cT1 t=a a t=b cT1 η(t) t=a
0.
Therefore g(t) ≡ c0 + c1 σ(t) on [a, b]κ. This completes the proof. Lemma 5.1.4. Let α, β ∈ Crd ([a, b]), α, β : [a, b] → Rn . Then Z b a
αT (t)h(t) + βσT (t)h∆(t) ∆t = 0
(5.1.7)
1 ([a, b]), h : [a, b] → Rn , h(a) = h(b) = 0, if and only if for all h ∈ Crd 1 β ∈ Crd ([a, b]) β∆ (t) = α(t),
t ∈ [a, b].
(5.1.8)
224
Svetlin G. Georgiev 1. Let (5.1.8) holds. Then
Proof. Z b
σT
T
∆
α (t)h(t) + β (t)h (t) ∆t =
a
Z b
αT (t)h(t)∆t
a
Z b
+
βσT (t)h∆(t)∆t
a
=
Z b a
t=a
Z b
− =
t=b αT (t)h(t)∆t + βT (t)h(t) β∆T (t)h(t)∆t
a
Z b a
αT (t)h(t)∆t −
Z b
αT (t)h(t)∆t
a
= 0. 2. Let (5.1.7) holds. We set A(t) =
Z t
α(s)∆s,
t ∈ [a, b].
a
Then Z b
Z b
A∆T (t)h(t)∆t t=b Z b = AT (t)h(t) − AσT (t)h∆(t)∆t
αT (t)h(t)∆t =
a
a
t=a
= −
Z b
a
AσT (t)h∆(t)∆t.
a
Therefore 0 =
Z b
=
Z b
=
Z b
a
a
αT (t)h(t) + βσT (t)h∆(t) ∆t
−AσT (t) + βσT (t) h∆ (t)∆t
(−Aσ + βσ )T (t)h∆(t)∆t.
a
Hence and Lemma 5.1.2, we obtain −Aσ (t) + βσ(t) = cσ ,
t ∈ [a, b],
The First Variation
225
for some c ∈ Rn . Consequently β(t) = A(t) + c,
t ∈ [a, b],
1 ([a, b]) and for some c ∈ Rn . From here, we conclude that β ∈ Crd
β∆ (t) = α(t),
t ∈ [a, b].
This completes the proof.
5.2. The Variational Problem Consider the variational problem
L (y) =
Z b a
L(t, yσ (t), y∆(t))∆t −→ min,
y(a) = α, y(b) = β,
(5.2.1)
where a, b ∈ T with a < b, α, β ∈ Rn , n ∈ N, and L : T × R2n → R is a C 2 1 ([a, b]) we define function in the last two variables, L = L(t, x, v). For f ∈ Crd the norm k f k = max | f σ (t)| + max | f ∆ (t)|. t∈[a,b]κ
t∈[a,b]κ
Definition 5.2.1. The function L is called Lagrangian. 1 ([a, b]) with y(a) ˆ = α, y(b) ˆ = β, is called a Definition 5.2.2. A function yˆ ∈ Crd (weak) local minimum of (5.2.1) if there exists δ > 0 such that
L (y) ˆ ≤ L (y) 1 ([a, b]) with y(a) = α, y(b) = β and for all y ∈ Crd
ky − yk ˆ < δ. If L (y) ˆ < L (y) for all such y 6= yˆ is said to be proper.
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Svetlin G. Georgiev
1 Definition 5.2.3. An η ∈ Crd ([a, b]) is called an admissible variation provided
η(a) = η(b) = 0.
For an admissible variation η, we define a function Φ : R → R as follows Φ(ε) = Φ(ε, y, η) = L (y + εη),
ε ∈ R.
Definition 5.2.4. The first and second variation of the variational problem (5.2.1) are defined by ˙ ¨ L1(y, η) = Φ(0) and L2 (y, η) = Φ(0), respectively.
Theorem 5.2.5 (Necessary Condition). If yˆ is a local minimum of (5.2.1), then
L1 (y, ˆ η) = 0 and L2 (y, ˆ η) ≥ 0 for all admissible variations η. 1 ([a, b]κ) for any Proof. Let η be an admissible variation. Then yˆ + εη ∈ Crd ε ∈ R, and
(yˆ + εη) (a) = y(a) ˆ + εη(a) = y(a) ˆ = α, (yˆ + εη) (b) = y(b) ˆ + εη(b) = y(b) ˆ = β. Then, there exists δ > 0 such that 1 L (y) ≥ L (y) ˆ for all y ∈ Crd ([a, b]κ)
The First Variation
227
with y(a) = α, y(b) = β and ky − yk ˆ < δ. Let ε ∈ R be such that |ε|
0
for all nontrivial admissible variations η, then yˆ is a proper weak local minimum of the problem (5.2.1). Proof. Since
L2(y, ˆ η) > 0, there exists an ε > 0 with
L2(y, ˆ η) ≥ 2ε(b − a)kηk2 . Because L is a C2 -function in x and v, there exists a δ > 0 such that |L00 (t, z) − L00 (t, z˜)| < ε, whenever z, z˜ ∈ R2n with |z − z˜| < δ. Let y ∈ C 1 ([a, b]κ) be such that y 6= y, ˆ y(a) = α, y(b) = β, and ky − yk ˆ < δ. We put η = y − y. ˆ
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Svetlin G. Georgiev
Note that η is a nontrivial admissible variation. By Taylor’s formula, we get that there exists a ζ ∈ (0, 1) such that for z=
yˆσ + ζησ yˆ∆ + ζη∆
T
,
z˜ =
yˆσ yˆ∆
,
and |z(t) − z˜(t)| ≤ ζkηk < kηk < δ holds for any t ∈ [a, b]κ, and =
L (y) − L (y) ˆ = Φ(1) − Φ(0)
1¨ dotPhi(0) + Φ(ζ) 2 1¨ Φ(ζ) 2 1 ¨ 1¨ ¨ Φ(0) + Φ(ζ) − Φ(0) = 2 2 1¨ 1 ¨ ¨ ≥ Φ(0) − |Φ(ζ) − Φ(0)| 2 2 Z T ησ (t) 1¨ 1 b ησ (t) 00 00 = Φ(0) − ∆t L (t, z(t)) − L (t, z˜(t)) η∆ (t) η∆ (t) 2 2 a =
≥
= = >
1 b kηkεkηk∆t 2 a 1 ε(b − a)kηk2 − ε(b − a)kηk2 2 b−a 2 ε kηk 2 0.
ε(b − a)kηk2 −
Z
Therefore yˆ is a proper local minimum of (5.2.1). This completes the proof. By Theorem 5.2.5 and Theorem 5.2.6, we get R L1(y, η) = ab Lx (t, yσ(t), y∆(t))ησ(t) +Lv (t, yσ(t), y∆(t)) ∆t,
(5.2.2)
229
The First Variation and R b
L2(y, η) =
a
(ησ )T (t)Lxx(t, yσ(t), y∆(t))ησ(t)
+2 (ησ )T (t)Lxv(t, yσ(t), y∆ (t))η∆(t) T + η∆ (t)Lvv(t, yσ(t), y∆(t))η∆ (t) ∆t.
(5.2.3)
Example 5.2.7. Let n = 1, T = 2N0 . Consider the variational problem Z 8
t 2 + (yσ (t))2 + y∆ (t)
L (y) =
1
Here
3
∆t −→ min,
y(1) = y(8) = 0.
L(t, x, v) = t 2 + x2 + v3 . Then Lx (t, x, v) = 2x, Lv (t, x, v) = 3v2 , Lxx (t, x, v) = 2, Lxv (t, x, v) = 0, Lvv (t, x, v) = 6v. The equations (5.2.2) and (5.2.3) take the form 2 2yσ (t)ησ(t) + 3 y∆ (t) η∆ (t) ∆t, 1 Z 8 2 L2 (y, η) = 2 (ησ (t))2 + 6y∆ (t) η∆ (t) ∆t,
L1 (y, η) =
Z 8 1
respectively.
Example 5.2.8. Let n = 2, T = 2Z. Consider the variational problem
L (y) =
Z 10 0
tyσ1 (t) + (yσ2 (t))2 + y∆1 (t)
y1 (0) = 1,
y2 (0) = 3,
y1 (0) = 4,
y2 (0) = 5.
y∆2 (t)
2
∆t −→
min,
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Svetlin G. Georgiev
Here L(t, x1 , x2 , v1 , v2 ) = tx1 + x22 + v1 v22 . We have Lx1 (t, x1 , x2 , v1 , v2 ) = t, Lx2 (t, x1 , x2 , v1 , v2 ) = 2x2 , Lv1 (t, x1 , x2 , v1 , v2 ) = v22 , Lv) 2 (t, x1 , x2 , v1 , v2 ) = 2v1 v2 , Lx1 x1 (t, x1 , x2 , v1 , v2 ) = 0, Lx1 x2 (t, x1 , x2 , v1 , v2 ) = 0, Lx2 x2 (t, x1 , x2 , v1 , v2 ) = 2, Lx1 v1 (t, x1 , x2 , v1 , v2 ) = 0, Lx1 v2 (t, x1 , x2 , v1 , v2 ) = 0, Lx2 v1 (t, x1 , x2 , v1 , v2 ) = 0, Lx2 v2 (t, x1 , x2 , v1 , v2 ) = 0, Lv1 v1 (t, x1 , x2 , v1 , v2 ) = 0, Lv1 v2 (t, x1 , x2 , v1 , v2 ) = 2v2 , Lv2 v2 (t, x1 , x2 , v1 , v2 ) = 2v1 . Then Lx t, yσ(t), y∆(t) ησ (t) + Lv t, yσ (t), y∆(t) η∆ (t) σ η∆ (t) 2 η1 (t) σ ∆ ∆ ∆ 1 = (t, 2y2 (t)) + y2 (t) , 2y1 (t)y2 (t) ησ2 (t) η∆2 (t) 2 = tησ1 (t) + 2yσ2 (t)ησ2 (t) + y∆2 (t) η∆1 (t) + 2y∆1 (t)y∆2 (t)η∆2 (t)
and the equation (5.2.2) takes the form
L1(y, η) =
Z 10
tησ1 (t) + 2yσ2 (t)ησ2 (t) + y∆2 0 +2y∆1 (t)y∆2 (t)η∆2 (t) ∆t.
2
η∆1 (t)
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The First Variation Next, σ
∆
Lxx t, y (t), y (t) = and hence,
0 0 0 2
(ησ (t))T Lxx t, yσ(t), y∆(t) ησ (t) = (ησ1 (t), ησ2 (t)) =
(ησ1 (t), ησ2 (t))
= 2 (ησ2 (t))2 ,
,
0 0 0 2
0 2ησ2 (t)
ησ1 (t) ησ2 (t)
and σ
∆
Lxv t, y (t), y (t) = 2 (ησ (t))T Lxv t, yσ(t), y∆(t) η∆ (t) = 0, 0 2y∆2 (t) σ ∆ Lvv t, y (t), y (t) = , 2y∆2 (t) 2y∆1 (t) η∆ (t) = =
0 0 0 0
,
T
Lvv t, yσ(t), y∆(t) η∆ (t) ∆ 0 2y∆2 (t) η1 (t) ∆ ∆ η1 (t), η2 (t) 2y∆2 (t) 2y∆1 (t) η∆2 (t) 2y∆2 (t)η∆2 (t) η∆1 (t), η∆2 (t) 2y∆2 (t)η∆1 (t) + 2y∆1 (t)η∆2 (t)
= 2y∆2 (t)η∆1 (t)η∆2 (t) + 2y∆2 (t)η∆1 (t)η∆2 (t) 2 +2y∆1 (t) η∆2 (t) 2 = 4y∆2 (t)η∆1 (t)η∆2 (t) + 2y∆1 (t) η∆2 (t) .
The equation (5.2.3) takes the form
L2(y, η) =
Z 10 0
2 (ησ2 (t))2 + 4y∆2 (t)η∆1 (t)η∆2 (t) + 2y∆1 (t) η∆2 (t)
2
∆t.
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Svetlin G. Georgiev
Example 5.2.9. Let n = 1, T = N0 . Consider the variational problem Z 2 2 L (y) = tyσ (t)y∆(t) + y∆ (t) ∆t −→ min, −2
y(2) = 1,
y(−2) = 4.
We have L(t, x, v) = txv + v2 , Lx (t, x, v) = tv, Lv (t, x, v) = tx + 2v, Lxx (t, x, v) = 0, Lxv (t, x, v) = t, Lvv (t, x, v) = 2. The equations (5.2.2) and (5.2.3) take the form
L1(y, η) =
Z 2
L2(y, η) =
Z 2
ty∆ (t)ησ(t) + tyσ (t)2y∆(t) η∆ (t) ∆t,
−2
−2
= 2
Z 2
= 2
Z 2
−2
−2
2tησ (t)η∆(t) + 2η∆ (t) ∆t
t=2 tησ (t)η∆ (t)∆t + 2η(t)
t=−2
tησ (t)η∆ (t)∆t,
respectively. Exercise 5.2.10. Let n = 1, T = N20 . Write the equations (5.2.2) and (5.2.3) for the variational problem
L (y) =
Z 49 0
tyσ (t)y∆ (t)∆t −→ min,
y(0) = y(49) = 1.
Answer.
L1(y, η) =
Z 49
L2(y, η) = 2
0
t y∆ (t)ησ(t) + yσ (t)η∆ (t) ∆t,
Z 49 0
tησ (t)η∆ (t)∆t.
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The First Variation
5.3. The Euler-Lagrange Equation Theorem 5.3.1. If yˆ is a local minimum of (5.2.1), then the Euler-Lagrange equation L∆v (t, yˆσ(t), yˆ∆(t)) = Lx (t, yˆσ(t), yˆ∆(t)), t ∈ [a, b]κ, (5.3.1) holds, where
∆ (Lv ) ∆t Proof. Let yˆ be a local minimum of (5.2.1) and η be an admissible variation. Then L∆v =
0 = Φ(0) = L1 (y, ˆ η) =
Z b a
Lx (t, yˆσ(t), yˆ∆(t))ησ(t)∆t
Z b
+ Lv (t, yˆσ(t), yˆ∆(t))η∆(t)∆t a Z t t=b σ ∆ = Lx s, yˆ (s), yˆ (s) ∆s η(t) t=a a Z b Z t σ ∆ ∆ − Lx s, yˆ (s), yˆ (s) η (t) ∆t a
a
Lv t, yˆσ(t), yˆ∆(t) η∆ (t)∆t a Z b Z t = − Lx s, yˆσ (s), yˆ∆(s) η∆ (t) ∆t +
Z b
+
Z b
a
a
a
Lv t, yˆσ(t), yˆ∆(t) η∆ (t)∆t
Hence and Lemma 5.1.2, we get
Lx (t, yˆσ(t), yˆ∆(t)) = L∆v (t, yˆσ(t), yˆ∆(t)),
t ∈ [a, b]κ.
This completes the proof. Definition 5.3.2. The equations (5.3.1) are called the Euler-Lagrange equations.
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Svetlin G. Georgiev
Example 5.3.3. Let n = 1. We will find the solution of the problem Z bq 2 1 + (y∆ (t)) ∆t −→ min, y(a) = 0, y(b) = 1. a
Here L(t, x, v) = Then
p
1 + v2 .
Lx (t, x, v) = 0, √
Lv (t, x, v) = Therefore
whereupon
and
v . 1 + v2
∆
∆
q y (t) = 0, 2 ∆ 1 + (y (t)) q
y∆ (t) 2
= c,
1 + (y∆ (t))
y∆ (t) = c1 ,
t ∈ [a, b]κ,
t ∈ [a, b]κ,
y(t) = c1 t + c2 ,
t ∈ [a, b]κ.
Here c, c1 and c2 are constants. Note that y(a) = c1 a + c2 = 0, y(b) = c1 b + c2 = 1. Then c1 = Consequently
1 , b−a
y(t) =
t −a , b−a
c2 = −
a . b−a
t ∈ [a, b]κ.
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The First Variation
Example 5.3.4. Let n = 1 and T = 2N0 . We will find the solution of the variational problem Z 8 1
Here
t 2 + yσ (t) y∆ (t)∆t −→ min,
y(1) = y(8) = 1.
σ(t) = 2t, L(t, x, v) = (t 2 + x)v. Then Lx (t, x, v) = v, Lv (t, x, v) = x + t 2 . Using the Euler-Lagrange equation, we get t 2 + yσ (t) or
∆
= y∆ (t),
t 2 + y(t) + ty∆ (t) or
∆
= y∆ (t),
σ(t) + t + y∆ (t) + ty∆ (t) or ty∆ (t)
∆
∆
t ∈ [1, 8]κ,
= y∆ (t),
t ∈ [1, 8]κ,
= −3t,
t ∈ [1, 8]κ.
ty∆ (t) = −t 2 + c1 ,
t ∈ [1, 8]κ,
c1 , t
t ∈ [1, 8]κ.
Hence, or
y∆ (t) = −t + Therefore y(t) − y(1) = − or
t ∈ [1, 8]κ,
Z t 1
τ∆τ + c1
Z t ∆τ 1
τ
Z t 1 τ=t ∆τ y(t) = 1 − τ2 +c1 , 3 τ=1 1 τ
,
t ∈ [1, 8]κ, t ∈ [1, 8]κ,
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Svetlin G. Georgiev
or
1 1 y(t) = 1 − t 2 + + c1 3 3
or y(t) =
4 1 2 − t + c1 3 3
Z t ∆τ
τ
1
Z t ∆τ
τ
1
t ∈ [1, 8]κ,
,
t ∈ [1, 8]κ.
,
Here c1 is a constant which will be determined by the condition y(8) = 1. We have Z 8 4 64 ∆τ − + c1 = 1, 3 3 1 τ or c1
Z 8 ∆τ
or
= 21,
τ
1
21 c1 = R 8 ∆τ . 1 τ
Consequently
Rt
4 − t 3 21 1 ∆τ + R 8 ∆ττ , y(t) = 3 1 τ
t ∈ [1, 8]κ.
Example 5.3.5. Let T = 3Z. We will find the solution of the variational problem Z 9 0
2 2 − y∆1 (t) + yσ2 (t)y∆1 (t) + tyσ1 (t) + y∆2 (t) ∆t −→ min,
y1 (0) y2 (0)
=
9 1
,
y1 (9) y2 (9)
=
−1 1
Here L(t, x1 , x2 , v1 , v2 ) = −v21 + x2 v1 + tx1 + v22 . Then Lx1 (t, x1 , x2 , v1 , v2 ) = t, Lx2 (t, x1 , x2 , v1 , v2 ) = v1 , Lv1 (t, x1 , x2 , v1 , v2 ) = x2 − 2v1 , Lv2 (t, x1 , x2 , v1 , v2 ) = 2v2 .
.
237
The First Variation Then the Euler-Lagrange equations take the form ( ∆ yσ2 (t) − 2y∆1 (t) = t ∆ 2 y∆2 (t) = y∆1 (t), t ∈ [0, 9]. Let
1 3 f (t) = t 2 − t, 2 2
t ∈ [0, 9].
Then 3 1 (σ(t) + t) − 2 2 1 3 = (t + 3 + t) − 2 2 = t.
f ∆ (t) =
Therefore
or
or
yσ2 (t) − 2y∆1 (t) = 12 t 2 − 32 t + c1 2y∆2 (t) = y1 (t) + c2 , t ∈ [0, 9], 2
y2 (t) + 3y∆2 (t) − 4y∆2 (t) = 12 t 2 − 23 t + c1 y1 (t) = 2y∆2 (t) − c2 , t ∈ [0, 9], 2
4y∆2 (t) − 3y∆2 (t) − y2 (t) = − 12 t 2 + 32 t + c1 y1 (t) = 2y∆2 (t) − c2 , t ∈ [0, 9].
Here c1 and c2 are constants. Consider the equation 2 1 3 4y∆2 (t) − 3y∆2 (t) − y2 (t) = − t 2 + t + c1 , 2 2
t ∈ [0, 9].
(5.3.2)
Note that y2h (t) = a1 e− 1 (t, 0) + a2e1 (t, 0), 4
t ∈ [0, 9],
is a solution of the corresponding homogeneous equation of the equation (5.3.2). Here a1 and a2 are constants. We will search a solution of the equation (5.3.2) in the form y2 (t) = a1 (t)e− 1 (t, 0) + a2 (t)e1 (t, 0), 4
t ∈ [0, 9],
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Svetlin G. Georgiev
where a1 , a2 ∈ C 1 ([0, 9]) will be determined below. We have y∆2 (t) = a∆1 (t)e− 1 (t + 3, 0) + a∆2 (t)e1 (t + 3, 0) 4
1 − a1 (t)e− 1 (t, 0) + a2 (t)e1 (t, 0), 4 4
t ∈ [0, 9].
We want a∆1 (t)e− 1 (t + 3, 0) + a∆2 (t)e1 (t + 3, 0) = 0,
t ∈ [0, 9].
1 y∆2 (t) = − a1 (t)e− 1 (t, 0) + a2 (t)e1(t, 0), 4 4
t ∈ [0, 9].
4
Then
Hence, 2 1 y∆2 (t) = − a∆1 (t)e− 1 (t + 3, 0) + a∆2 (t)e1 (t + 3, 0) 4 4 1 + a1 (t)e− 1 (t, 0) + a2 (t)e1(t, 0), t ∈ [0, 9], 4 16
and 2
4y∆2 (t) − 3y∆2 (t) − y2 (t) = −a∆1 (t)e− 1 (t + 3, 0) + 4a∆2 (t)e1 (t + 3, 0) 4
1 + a1 (t)e− 1 (t, 0) + 4a2(t)e1 (t, 0) 4 4 3 + a1 (t)e− 1 (t, 0) − 3a2(t)e1 (t, 0) 4 4 −a1 (t)e− 1 (t, 0) − a2 (t)e1 (t, 0) 4
= or
1 2 3 t − t + c1 , 2 2
t ∈ [0, 9],
1 3 −a∆1 (t)e− 1 (t + 3, 0) + 4a∆2 (t)e1 (t + 3, 0) = t 2 − t + c1 , 4 2 2 t ∈ [0, 9]. Thus we get the system ( a∆1 (t)e− 1 (t + 3, 0) + a∆2 (t)e1(t + 3, 0) = 0 4 −a∆1 (t)e− 1 (t + 3, 0) + 4a∆2 (t)e1(t + 3, 0) = 21 t 2 − 32 t + c1 , t ∈ [0, 9]. 4
239
The First Variation Consequently a∆2 (t) =
1 5
1 2 3 t − t + c1 e1 (0,t + 3), 2 2
t ∈ [0, 9],
and we take 1 a2 (t) = 5
Z 9 1 0
3 τ − τ + c1 e1 (0, τ + 3)∆τ, 2 2 2
t ∈ [0, 9],
and a∆1 (t) = −a∆2 (t)e1 (t + 3, 0)e− 1 (0,t + 3) 4 1 1 2 3 = − t − t + c1 e1 (0,t + 3)e1 (t + 3, 0)e− 1 (0,t + 3) 4 5 2 2 1 1 2 3 t − t + c1 e− 1 (0,t + 3), t ∈ [0, 9]. = − 4 5 2 2 We take 1 a1 (t) = − 5
Z t 0
1 2 3 τ − τ + c1 e− 1 (0, τ + 3)∆τ, 4 2 2
t ∈ [0, 9].
Therefore a1 (t)e− 1 (t, 0) + a2 (t)e1 (t, 0) 4 Z 1 t 1 2 3 = − τ − τ + c1 e− 1 (t, τ + 3)∆τ 4 5 0 2 2 Z t 1 2 3 1 τ − τ + c1 e1 (t, τ + 3)∆τ + 5 0 2 2 Z t 1 1 2 3 = τ − τ + c1 e1 (t, τ + 3) − e− 1 (t + 3, τ + 3) ∆τ, 4 5 0 2 2
t ∈ [0, 9],
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Svetlin G. Georgiev
and y2 (t) = a1 e− 1 (t, 0) + a2e1 (t, 0) 4 Z t 1 1 2 3 + τ − τ + c1 e1 (t, τ + 3) − e− 1 (t, τ + 3) ∆τ, 4 5 0 2 2 1 y∆2 (t) = − a1 e− 1 (t, 0) + a2e1 (t, 0) 4 4 1 1 2 3 t − t + c1 + e1 (t + 3,t + 3) − e− 1 (t + 3,t + 3) 4 5 2 2 Z t 1 2 3 1 1 τ − τ + c1 e1 (t, τ + 3) + e− 1 (t, τ + 3) ∆τ, + 5 0 2 2 4 4 1 y1 (t) = − a1 e− 1 (t, 0) + 2a2e1 (t, 0) 4 2 Z 2 t 1 2 3 1 + τ − τ + c1 e1 (t, τ + 3) + e− 1 (t, τ + 3) ∆τ − c2 , 5 0 2 2 4 4 t ∈ [0, 9]. For the constants a1 , a2 , c1 and c2 we have the system − 12 a1 + 2a2 − c2 = 0 a1 + a2 = 1 1 − 2 a1 e− 41 (9, 0) + 2a2e1(9, 0) 2R9 1 2 3 1 (9, τ + 3) ∆τ − c2 = −1 + τ − τ + c e (9, τ + 3) + e 1 1 1 5 0 2 2 4 −4 a1 e− 1 (9, 0) + a2e1 (9, 0) R4 + 1 2 1 τ2 − 3 τ + c1 e1 (9, τ + 3) − e 1 (9, τ + 3) ∆τ = 1. − 5 0 2 2 4
Exercise 5.3.6. Let T = 3N0 . Find the solution of the variational problem. Z 27 q 2 q 2 ∆ ∆ 1 + y1 (t) + 1 + y2 (t) ∆t −→ min, 1
y1 (1) y2 (1)
=
0 0
,
Answer. y1 (t) = y2 (t) =
y1 (27) y2 (27)
t −1 , 26
=
t ∈ [1, 27].
1 1
.
241
The First Variation
5.4. Legendre’s Condition We introduce the following notations. P = Lvv (·, yˆσ, yˆ∆ ),
Q = Lxx (·, yˆσ, yˆ∆ ),
R = Lxv (·, yˆσ, yˆ∆ ), 1 α† = if α ∈ R\{0}, α
0† = 0.
Theorem 5.4.1 (Legendre’s Condition). If yˆ is a weak local solution of the variational problem (5.2.1), then P(t) + µ(t) R(t) + RT (t) + µ(t)Q(t) + (µ(σ(t)))† P(σ(t)) ≥ 0, 2
t ∈ [a, b]κ .
2
Proof. Let s ∈ [a, b]κ . 1. Suppose that s < σ(s) < t. We take γ ∈ Rn arbitrarily and define η : [a, b] → Rn as follows. p γ µ(s) if t = σ(s) η(t) = 0 otherwise. We have η(a) = η(b) = 0. Therefore η is an admissible variation. Note that η∆ (s) = = =
η(σ(s)) − η(s) µ(s) p γ µ(s) µ(s) γ p µ(s)
242
Svetlin G. Georgiev and η(σ(σ(s))) − η(σ(s)) µ(σ(s)) p γ µ(s) = − . µ(σ(s))
η∆ (σ(s)) =
Next, η∆ (t) = 0 for all t ∈ [a, b]\{s, σ(s)}. Thus Z σ(s) a
=
=
= = =
T (ησ )T Qησ + 2 (ησ )T Rη∆ + η∆ Pη∆ (t)∆t T µ(s) (ησ )T Qησ + 2 (ησ )T Rη∆ + η∆ Pη∆ (s) √ γ √ √ µ(s) γT µQγ µ + 2γT µR √ µ T γ γ + √ P √ (s) µ µ 1 T T T µ(s) µγ Qγ + 2γ Rγ + γ Pγ (s) µ T 2 T T γ Pγ + µ γ Qγ + γ µRγ + γµRT γ (s) γT P(s) + µ(s) R(s) + RT (s) + µ(s)Q(s) γ.
(a) Let σ(s) is right-dense. Then Z b σ(s)
(ησ )T Qησ + 2 (ησ )T Rη∆ + η∆
T
Pη∆ (t)∆t = 0.
243
The First Variation (b) Let σ(s) is right-scattered. Then Z b T (ησ )T Qησ + 2 (ησ )T Rη∆ + η∆ Pη∆ (t)∆t σ(s)
Pη∆ (t)∆t σ(s) T T T = µ(σ(s)) (ησ ) Qησ + 2 (ησ ) Rη∆ + η∆ Pη∆ (σ(s)) =
Z σ(σ(s))
= µ(σ(s)) =
T
T
(ησ ) Qησ + 2 (ησ ) Rη∆ + η∆
T
µ(s) γT P(σ(s))γ (µ(σ(s)))2
µ(s) T γ P(σ(s))γ. µ(σ(s))
Consequently Z b T (ησ )T Qησ + 2 (ησ )T Rη∆ + η∆ Pη∆ (t)∆t a T T γ P(s) + µ(s) R(s) + R (s) + µ(s)Q(s) γ if σ(σ(s)) = σ(s) γT P(s) + µ(s) R(s) + RT (s) + µ(s)Q(s) γ = + µ(s) γT P(σ(s))γ if σ(σ(s)) > σ(s) µ(σ(s)) T = γ P(s) + µ(s) R(s) + RT (s) + µ(s)Q(s) + (µ(σ(s)))† γ. Hence and Theorem 5.2.5, we conclude that γT P(s) + µ(s) R(s) + RT (s) + µ(s)Q(s) + (µ(σ(s)))† γ ≥ 0.
2. Let s is right-dense.
(a) Let s is left-scattered. Therefore there exists a decreasing sequence {sk}k∈N ⊂ [a, b] such that lim sk = s.
k→∞
For γ ∈ Rn and k ∈ N, we define ( γ √sks−t−s if t ∈ [s, sk] k ηk (t) = 0 otherwise.
244
Svetlin G. Georgiev Note that ηk is an admissible variation. We have Z s T (ησ ) Qησ + 2(ησ )T Rη∆ + (η∆ )T Pη∆ (t)∆t a
= µ(ρ(s)) (ησ )T Qησ + 2(ησ )T Rη∆ + (η∆ )T Pη∆ (ρ(s)) sk − s T = µ(ρ(s)) (sk − s)γT Q(ρ(s))γ + 2 γ R(ρ(s))γ µ(ρ(s)) sk − s T + γ P(ρ(s))γ (µ(ρ(s)))2 1 T P(ρ(s)) γ = (sk − s)γ 2R(ρ(s)) + µ(ρ(s))Q(ρ(s)) + µ(ρ(s))
and Z b
(ησ )T Qησ + 2(ησ )T Rη∆ + (η∆ )T Pη∆ (t)∆t s Z sk (sk − σ(t))2 T sk − σ(t) T 1 T γ Q(t)γ − 2 γ R(t)γ + γ P(t)γ ∆t. = sk − s sk − s sk − s s
Hence and Theorem 5.2.5, letting k → ∞, we get γT P(s)γ ≥ 0, i.e., γT P(s) + µ(s) R(s) + RT (s) + µ(s)Q(s) + (µ(σ(s)))† γ ≥ 0. (b) Assume that s is left-dense. Then there exists an increasing sequence {zk }k∈N ⊂ [a, b] such that lim zk = s.
k→∞
For γ ∈ Rn and k ∈ Rn we define s −t k γ √sk −s if t ∈ [s, sk] t−z ηk (t) = γ √s−zk if t ∈ [zk , s] k 0 otherwise.
The First Variation
245
Then ηk is an admissible variation. We have Z s
T (ησ ) Qησ + 2(ησ)T Rη∆ + (η∆ )T Pη∆ (t)∆t
a
+ =
Z b s
Z s
T (ησ ) Qησ + 2(ησ )T Rη∆ + (η∆ )T Pη∆ (t)∆t
(zk − σ(t))2 T σ(t) − zk T 1 T γ Q(t)γ − 2 γ R(t)γ + γ P(t)γ ∆t s − zk s − zk s − zk zk Z sk (sk − σ(t))2 T sk − σ(t) T 1 T + γ Q(t)γ − 2 γ R(t)γ + γ P(t)γ ∆t. sk − s sk − s sk − s s
Hence, using Theorem 5.2.5 and letting k → ∞, we obtain γT P(s)γ ≥ 0 and γT P(s) + µ(s) R(s) + RT (s) + µ(s)Q(s) + (µ(σ(s)))† γ ≥ 0.
This completes the proof.
5.5. Jacobi’s Condition Theorem 5.5.1. Suppose that P(t) and P(t) + µ(t)R(t) are Then
L2(y, ˆ η) = where
invertible for all t ∈ [a, b]κ. (5.5.1)
Z b a
ξ = Pη∆ + RT ησ ,
T (ησ ) Cησ + ξT Bξ (t)∆t,
B = P−1 ,
C = Q − RP−1 RT .
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Svetlin G. Georgiev
Proof. We have
L2(y, ˆ η) =
Z b
(ησ )T Qησ + 2 (ησ )T Rη∆ + η∆
a
=
Z b
=
Z b
T
Pη∆ (t)∆t
T T (ησ ) C + RBRT ησ + 2 (ησ ) RBPη∆ a ∆ T ∆ + Pη B Pη (t)∆t Z b T T = (ησ ) Cησ + Pη∆ + RT ησ B Pη∆ + RT ησ (t)∆t a
a
(ησ )T Cησ + ξT Bξ (t)∆t.
This completes the proof.
We introduce the following notations. A = −P−1 RT ,
B = P−1 ,
C = Q − RP−1 RT .
Then ξ = Pη∆ + RT ησ −1
Bξ = P ξ
⇐⇒
= η∆ + P−1 RT ησ = η∆ − Aησ , i.e., η∆ = Aησ + Bξ. Theorem 5.5.2 (Jacobi’s Condition). Assume that (5.5.1) holds. Then L2 is positive definite if and only if the linear Hamiltonian system ∆ η = A(t)ησ + B(t)ξ ξ∆ = C(t)ησ − AT (t)ξ is disconjugate on [a, b].
247
The First Variation Proof. We have I − µA = I + µP−1 RT
= P−1 P + µRT
=
(P + µR)P−1
T
is invertible. Then the assertion follows from the results in the previous section. This completes the proof. Theorem 5.5.3. Let R is symmetric, differentiable and invertible, and P + µR is invertible for all t ∈ [a, b]κ. Then L2 is positive definite if and only if the linear Hamiltonian system ∆ η = P˜ −1 (t)ξ σ ˜ ξ∆ = Q(t)η is disconjugate on [a, b], where Q˜ = Q − R∆ ,
P˜ = P + µR.
Proof. We have
Hence,
∆ T ηT Rη = (ησ )T R∆ ησ + (ησ )T Rη∆ + η∆ Rη T = (ησ )T R∆ ησ + 2 (ησ )T Rη∆ − µ η∆ Rη∆ .
L2(y, η) =
Z b
=
Z b
∆ ((ησ )T Qησ + ηT Rη − (ησ )T R∆ ησ a T T +µ η∆ Rη∆ + η∆ Pη∆ (t)∆t Z b T = (ησ )T (Q − R∆ )ησ + η∆ (µR + P)η∆ (t)∆t a
T ˜ σ (ησ ) Qη + η∆
a
T
˜ ∆ (t)∆t, Pη
whereupon, using the results in the previous chapter, it follows the assertion. This completes the proof.
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Svetlin G. Georgiev
5.6. Advanced Practical Problems Problem 5.6.1. Let n = 1, T = 3N0 . Write the equations (5.2.2) and (5.2.3) for the variational problem
L (y) =
Z 27
t 2 + y∆ (t)
1
Answer.
2
∆t −→ min,
L1(y, η) = 2
Z 27
L2(y, η) = 2
Z 27
y(1) = y(27) = 3.
y∆ (t)η∆ (t)∆t,
1
1
2 η∆ (t) ∆t.
Problem 5.6.2. Write the Euler-Lagrange equations for the following variational problems. 3 R 1. 010 tyσ (t)y∆(t) + y∆ (t) ∆t −→ min, y(0) = 1, y(10) = −2, T = Z, R 2. 116 tyσ1 (t)yσ2 (t) + y∆1 (t)y∆2 (t) ∆t −→ min, y1 (1) 0 y1 (16) 1 = , = , T = 2 N0 . y2 (1) −1 y2 (16) 1 Problem 5.6.3. Check Legendre’s condition for the following variational problems. 2 R 1. 19 (yσ (t))2 − 3 y∆ (t) ∆t −→ min, y(1) = y(9) = 1,
2.
T = 3 N0 .
R 20 σ ∆ (t) + yσ (t) y∆ (t) 3 + t 2 ∆t −→ min, y (t)y 0 y(0) = 1,
y(20) = 3,
T = 2Z.
Problem 5.6.4. Check Euler’s condition for the following variational problems.
1.
2.
R 10 −1
The First Variation 3 4 t 2 yσ (t) − 3(t + 1) y∆ (t) ∆t −→ min, y(−1) = 0,
y(10) = 25,
249
T = Z.
7 3 2 R8 σ ∆ 3 σ ∆ y (t) + y (t) + y (t) y (t) + t + t − 2t + 1 ∆t −→ min, 1 y(1) = 0,
y(8) = 4,
T = 2 N0 .
Problem 5.6.5. Check the conditions of Theorem 5.5.3 for the following variational problems. 3 2 R 1. 127 t 2 − 7yσ (t) + t y∆ (t) − t ∆t −→ min, y(1) = 0,
2.
y(27) = 2,
T = 3 N0 .
R8 σ ∆ (t) + tyσ (t) y∆ (t) 2 + t ∆t −→ min, y (t) + y 1 y(1) = 1,
y(8) = −3,
T = 2 N0 .
Chapter 6
Higher Order Calculus of Variations Suppose that T is time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let also, the time scale T has sufficiently many points such that all computations to make sense and assume that T is such that σ(t) = a1 t + a0 for some a1 ∈ R+ and a0 ∈ R, for t ∈ T. Such time scales exist. For instance, T = Z, T = qN0 . Let a, b ∈ T, a < b, be such that all computations on [a, b] to have a sense. Note that it is impossible to have points which are simultaneously left-dense and right-scattered, or simultaneously leftscattered and right-dense.
6.1. Statement of the Variational Problem In this chapter we investigate the variational problem. Z ρr−1 (b) r r−1 r−1 r L (y) = L t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ∆t −→ min, a
y(a) r−1 y∆ (a)
= y0a , .. . =
yr−1 a ,
(6.1.1)
y ρr−1 (b) = y0b , r−1 y∆
ρr−1 (b)
(6.1.2) =
where y : [a, b] → Rn , yla , ylb ∈ Rn , l ∈ {0, . . ., r}, n, r ∈ N.
yr−1 b ,
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Svetlin G. Georgiev
Definition 6.1.1. The function L is called Lagrangian. Assume that the Lagrangian L(t, u0 , u1 , . . ., ur ) of the problem (6.1.1), (6.1.2) is continuous and has continuous partial derivatives with respect to u0 , u1 , . . ., ur , r ≥ 1, and continuous partial delta derivative with respect to t up to order r + 1. For a subset A ⊆ T we define n o 2r 2r C 2r (A) = y : T → R : y∆ is continuous on Tκ .
Let y ∈ C 2r([a, b]).
Definition 6.1.2. We say that yˆ ∈ C 2r ([a, b]) is a (weak) local minimum of the problem (6.1.1), (6.1.2) provided there exists a δ > 0 such that
L (y) ˆ ≤ L (y) for all y ∈ C 2r ([a, b]) satisfying (6.1.2) and ky − yk ˆ < δ, where
∞
l r−l kyk = ∑ yσ ∆ , ∞
l=0
kyk∞ = sup |y(t)|. t∈[a,b]κ
r
If L (y) ˆ < L (y) for all such y 6= yˆ is said to be proper. Definition 6.1.3. We say that η ∈ C 2r([a, b]) is an admissible variation for the problem (6.1.1), (6.1.2) if η(a) = 0, η ρr−1 (b) = 0, .. . r−1 ∆r−1 η (a) = 0, η∆ ρr−1 (b) = 0.
Higher Order Calculus of Variations Lemma 6.1.4. Let f be defined on a, ρ2r (b) and it is continuous. Then Z ρ2r−1 (b)
r
f (t)ησ (t)∆t = 0
253
(6.1.3)
a
for every admissible variation η if and only if f (t) = 0 for all t ∈ a, ρ2r (b) . Proof. If f (t) = 0 for all t ∈ a, ρ2r (b) ,2rthenthe assertion is true. Let (6.1.3) holds. Suppose that there exists t0 ∈ a, ρ (b) such that f (t0 ) > 0. 1. Let t0 = a. Since f is continuous on a, ρ2r(b) , there exists a δ > 0 such that f (t) > 0 for t ∈ [a, a + δ). We define η as follows. h2r+2 (t, a)h2r+2(a + δ,t) if t ∈ [a, a + δ) η(t) = 0 otherwise. We have that η ∈ C 2r ([a, b]) and r−1
η(a) = . . . = η∆
r−1 (a) = η ρr−1 (b) = . . . = η∆ ρr−1 (b) = 0,
i.e., η is an admissible variation. Then 0 =
Z ρ2r−1 (b)
r
f (t)ησ (t)∆t
a
=
Z a+δ
r
f (t)ησ (t)∆t
a
> 0, which is a contradiction. 2. Suppose that t0 6= a and t0 is dense. Then there exists a δ > 0 such that (t0 − δ,t0 + δ) ⊂ [a, b] and f (t) > 0 for t ∈ (t0 − δ,t0 + δ). Define h2r+2 (t,t0)h2r+2(t0 + δ,t) if t ∈ (t0 − δ,t0 + δ) η(t) = 0 otherwise. Then 0 =
Z ρ2r−1 (b)
r
f (t)ησ (t)∆t
a
=
Z t0 +δ t0 −δ
> 0,
r
f (t)ησ (t)∆t
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Svetlin G. Georgiev which is a contradiction.
3. Suppose that t0 6= a and t0 is right-scatttered. Then all points t, t ≥ t0 , are r isolated. Define η such that ησ (t0 ) = 1 and zero elsewhere. Therefore 0 =
Z ρ2r−1 (b)
r
f (t)ησ (t)∆t
a
=
Z σ(t0 )
r
f (t)ησ (t)∆t
t0
r
= µ(t0 ) f (t0)ησ (t0 ) > 0, which is a contradiction. This completes the proof.
6.2. Euler’s Equation Theorem 6.2.1. If yˆ is a weak local minimum for the problem (6.1.1), (6.1.2), then yˆ satisfies Euler’s equation r
∑ (−1) l=0
l
1 a1
t ∈ a, ρ2r (b) .
(l−1)l 2
l r r−1 l−1 l L∆ul t, yˆσ (t), yˆσ ∆ (t), . . ., yˆσ∆ (t), yˆ∆ (t) = 0, (6.2.1)
Proof. Let
Φ(ε) = L (yˆ + εη),
ε ∈ R.
This function has a minimum at ε = 0. Therefore ˙ Φ(0) =0
255
Higher Order Calculus of Variations and 0=
Z ρr−1 (b) r
∑ Lul
a
l=0
Z ρr (b) r
=
∑ Lul
a
+
r−l l r r−1 r−1 r t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ ∆ (t)∆t
l=0
Z ρr−1 (b) r ρr (b)
r−l l r r−1 r−1 r t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ ∆ (t)∆t
∑ Lul
l=0
r−l l σr σr−1∆ σ∆r−1 ∆r L t, y (t), y (t), . . ., y (t), y (t) ησ ∆ (t)∆t u ∑ l
Z ρr (b) r
=
r−l l r r−1 r−1 r t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ ∆ (t)∆t
a
l=0
r r r−1 r−1 r +µ (ρr (b)) ∑ Lul ρr (b), yσ (ρr (b)), yσ ∆ (ρr (b)), . . ., yσ∆ (ρr (b)) , y∆ (ρr (b)) l=0
σr−l ∆l
×η
(ρr (b)) .
The last equality we integrate by parts and we get 0=
Z ρr (b) a
r
−∑
r r r−1 r−1 r Lu0 t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ (t)∆t
Z ρr (b)
l=1 a
r−1 l−1 r r−l r−1 r L∆ul t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ ∆ σ (t)∆t
r−l l−1 t=ρr (b) r r−1 r−1 r + ∑ Lul t, yσ (t), yσ ∆(t), . . ., yσ∆ (t), y∆ (t) ησ ∆ (t) r
t=a
l=1
r
r
+µ (ρr (b)) ∑ Lul ρr (b), yσ (ρr (b)), yσ l=0 σr−l ∆l
×η
(ρr (b)) .
r−1∆
(ρr (b)) , . . ., yσ∆
r−1
r (ρr (b)), y∆ (ρr (b))
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Svetlin G. Georgiev
Note that 2
ησ (a) = ησ (a) + µ(a)ησ∆(a) = η(a) + µ(a)η∆ (a) + µ(a)a1η∆σ (a) 2 = a1 µ(a) η∆ (a) + µ(a)η∆ (a)
= 0, σ3
2
2
η (a) = ησ (a) + µ(a)ησ ∆ (a) 2
= µ(a)a21η∆σ (a)
= µ(a)a21 η∆σ (a) + µ(a)η∆σ∆ (a) 2 2 = µ(a)a21 η∆ (a) + µ(a)η∆ (a) + µ(a)a1 η∆ σ (a) 2 3 = (µ(a))2 a31 η∆ (a) + µ(a)η∆ (a) = 0,
and so on, r−1
ησ
(a) = 0,
and r−2
ησ∆
r−2
(a) = a1 η∆ σ (a) r−2 r−1 = a1 η∆ (a) + µ(a)η∆ (a) = 0,
σ2 ∆r−3
η
r−3 2
(a) = a21 η∆ σ (a) r−3 r−3 = a21 η∆ σ (a) + µ(a)η∆ σ∆ (a) r−3 r−2 = a21 η∆ (a) + µ(a)η∆ (a) r−2 +a1 µ(a)η∆ σ (a) r−2 r−1 = a31 µ(a) η∆ (a) + µ(a)η∆ (a) = 0,
and so on,
r−l ∆l
ησ
(a) = 0,
l ∈ {1, . . ., r}.
Higher Order Calculus of Variations Next, note that ρ(b),
ρ2 (b),
. . ., ρr−1 (b)
are right-scattered points. We have η∆ ρr−1 (b) = 0
and hence,
η ρr−1 (b) − η ρr−2 (b) , 0= ρr−1 (b) − ρr−2 (b)
whereupon Now, using that we get
η ρr−2 (b) = 0.
2 η∆ ρr−1 (b) = 0,
η∆ ρr−1 (b) − η∆ ρr−2 (b) 0 = ρr−1 (b) − ρr−2 (b) η ρr−2 (b) − η ρr−3 (b) = − r−2 (ρ (b) − ρr−3 (b))(ρr−1 (b) − ρr−2 (b)) and and so on,
η ρr−3 (b) = 0, η(ρ(b)) = 0.
Also, 2 0 = η∆ ρr−1 (b) , η∆ ρr−1 (b) − η∆ ρr−2 (b) 0 = ρr−1 (b) − ρr−2 (b) η∆ ρr−2 (b) = − r−1 , ρ (b) − ρr−2 (b) and
η∆ ρr−2 (b) = 0
257
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Svetlin G. Georgiev
and so on, η∆ (ρ(b)) = 0. Hence, r−2 ∆
0 = ησ
(ρr (b)) r−2
∆σ = ar−2 1 η
= a1r−2 η∆ = 0 and so on,
r−l ∆l
ησ
(ρr (b)) = 0,
(ρr (b)) ρ2 (b) l ∈ {1, . . ., r}.
Consequently 0=
Z ρr (b) a
r
−∑
r r r−1 r−1 r Lu0 t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ (t)∆t
Z ρr (b)
l=1 a
0=
r σr σr−1∆ σ∆r−1 ∆r L t, y (t), y (t), . . ., y (t), y (t) ησ (t)∆t ∑ ul
Z ρr (b) r a
r−1 l−1 r r−l r−1 r L∆ul t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ ∆ σ (t)∆t,
l=0
r
1 a l=1 1
−∑
Z ρr (b) a
r−1+1 l−1 r r−l r−1 r ∆ L∆ul t, yσ (t), yσ ∆ (t), . . ., yσ∆ (t), y∆ (t) ησ (t)∆t.
The last equation we integrate by parts and so on, using Lemma 6.1.4, we get (6.2.1). This completes the proof. Example 6.2.2. Consider the variational problem
L (y) =
Z ρ(b) a
2 2 L t, yσ (t), yσ∆ (t), y∆ (t) ∆t −→ min,
y(a) = y0a ,
y(ρ(b)) = y0b ,
y∆ (a) = y1a ,
y∆ (ρ(b)) = y1b .
259
Higher Order Calculus of Variations Then Euler’s equation is 2 2 0 = Lu0 t, yˆσ (t), yˆσ∆(t), yˆ∆ (t) 2 2 −L∆u1 t, yˆσ (t), yˆσ∆(t), yˆ∆ (t) 1 2 2 2 + L∆u2 t, yˆσ (t), yˆσ∆ (t), yˆ∆ (t) , a1 Example 6.2.3. Consider the variational problem
L (y) =
t ∈ a, ρ4 (b) .
3 2 2 3 L t, yσ (t), yσ ∆ (t), yσ∆ (t), y∆ (t) ∆t −→ min,
Z ρ2 (b) a
y(a) = y0a , y∆ (a) = y1a , 2
y∆ (a) = y2a ,
y(b) = y1b , y∆ (b) = y1b , 2
y∆ (b) = y2b .
Then Euler’s equation is 3 2 2 3 0 = Lu0 t, yσ (t), yσ ∆ (t), yσ∆ (t), y∆ (t) 3 2 2 3 −L∆u1 t, yσ (t), yσ ∆ (t), yσ∆ (t), y∆ (t) 1 2 3 2 2 3 + L∆u2 t, yσ (t), yσ ∆ (t), yσ∆ (t), y∆ (t) a1 3 2 2 3 1 3 − 3 L∆u3 t, yσ (t), yσ ∆ (t), yσ∆ (t), y∆ (t) , a1
t ∈ a, ρ6 (b) .
Example 6.2.4. Let T = 3N0 . Consider the variational problem
L (y) =
Z 81 1
2
2
tyσ (t)yσ∆(t)y∆ (t)∆t −→ min,
y(1) = 0,
y(81) = 0,
∆
y∆ (81) = −1.
y (1) = 1,
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Svetlin G. Georgiev
Here a1 = 3,
r = 2,
L(t, u0 , u1 , u2 ) = tu0 u1 u2 . Then Lu0 (t, u0, u1 , u2 ) = tu1 u2 , Lu1 (t, u0, u1 , u2 ) = tu0 u2 , Lu2 (t, u0, u1 , u2 ) = tu0 u1 , L∆u1 (t, u0, u1 , u2 ) = u0 u2 , L∆u2 (t, u0, u1 , u2 ) = u0 u1 , 2
L∆u2 (t, u0, u1 , u2 ) = 0. The Euler’s equation is 2
2
2
tyσ∆ (t)y∆ (t) − yσ (t)y∆ (t) = 0,
t ∈ [1, 3].
Exercise 6.2.5. Write Euler’s equation for the following variational problem Z ρ2 (b) 3 2 3 t 2 yσ (t) + yσ∆ − 10ty∆ (t) ∆t −→ min . a
6.3. Advanced Practical Problems Problem 6.3.1. Write Euler’s equation for the following variational problems. R 2 3 3 1. aρ (b) t + yσ (t) + y∆ (t) ∆t −→ min, R 2 3 2 3 2. aρ (b) t + yσ (t) + 2yσ ∆ (t) − 7y∆ (t) ∆t −→ min, R 3 4 3 3. aρ (b) t 2 + tyσ (t) + t 3 yσ∆ (t) ∆t −→ min, R 2 2 4. aρ(b) t − t 3 yσ (t) + t 4 y∆ (t) ∆t −→ min, R 4 5 4 3 2 5 5. aρ (b) tyσ (t) + yσ∆ (t) + yσ ∆ (t) − t 3 y∆ (t) ∆t −→ min.
Chapter 7
Double Integral Calculus of Variations Let T1 and T2 be time scales with forward jump operators and delta differentiation operators σ1 , σ2 and ∆1 , ∆2 , respectively. With Crd we denote the set of functions f (x, y) on T1 × T2 with the following properties. 1. f is rd-continuous in x for fixed y. 2. f is rd-continuous in y for fixed x. 3. If (x0 , y0 ) ∈ T1 × T2 with x0 right-dense or maximal and y0 right-dense or maximal, then f is continuous at (x0 , y0 ). 4. If x0 and y0 are both left-sided, then the limit f (x, y) exists(finite) as (x, y) approaches (x0 , y0 ) along any path in {(x, y) ∈ T1 × T2 : x < x0 , (1)
y < y0 } .
By Crd we denote the set of all continuous functions for which both the ∆1 partial derivative and the ∆2 -partial derivative exist and are of the class Crd .
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Svetlin G. Georgiev
7.1. Statement of the Variational Problem Let E ⊂ T1 × T2 be a set of type ω and let Γ be its positively oriented fence. Suppose that a function L(x, y, u, p, q),
(x, y) ∈ E
[
and (u, p, q) ∈ R3 ,
Γ
is given, it is continuous together with its partial delta derivatives of the first and second order with respect to x, y and partial usual derivatives of the first and second order with respect to u, p, q. Consider the functional
L (u) =
Z Z
E
L(x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))∆1 x∆2 y
whose domain of definition D(L ) consists of functions u ∈ ing the ”boundary conditions”
S Crd(1)(E Γ)
u = g(x, y) on Γ,
(7.1.1) satisfy-
(7.1.2)
where g is a fixed function defined and continuous on Γ. Definition 7.1.1. We call functions u ∈ D(L ) admissible. (1)
Definition 7.1.2. The functions η ∈ Crd (E admissible variations. (1)
If f ∈ Crd (E
S
S
Γ) and η = 0 on Γ, are called
Γ), we define the norm
kfk =
supS | f (x, y)| + sup f ∆1 (x, σ2 (y))
(x,y)∈E
Γ
(x,y)∈E
+ sup f ∆2 (σ1 (x), y) . (x,y)∈E
Definition 7.1.3. A function uˆ ∈ D(L ) is called a weak local minimum of L provided there exists a δ > 0 such that
L (u) ˆ ≤ L (u)
Double Integral Calculus of Variations for all u ∈ D(L ) with
263
ku − uk ˆ < δ.
If
L (u) ˆ < L (u) for all such u 6= u, ˆ then uˆ is said to be proper weak local minimum.
7.2. First and Second Variation For a fixed element u ∈ D(L ) and a fixed admissible variation η, we define Φ : R → R as follows. Φ(ε) = L (u + εη). Definition 7.2.1. The first and second variation L at the point u are defined by ˙ ¨ L1(u, η) = Φ(0) and L2 (u, η) = Φ(0), respectively.
Theorem 7.2.2 (Necessary Condition). If uˆ ∈ D(L ) is a local minimum of L , then L1(u, η) = 0 and L2(u, η) ≥ 0 for all admissible variations η. Proof. Assume that L has a local minimum at uˆ ∈ D(L ). Let η be an arbitrary admissible variation. Then ˙ ¨ Φ(0) = L1 (u, ˆ η) and Φ(0) = L2 (u, ˆ η). By Taylor’s formula, we get 1 ¨ 2 ˙ , Φ(ε) = Φ(0) + Φ(0)ε + Φ(α)ε 2! where |α| ∈ (0, |ε|). If |ε| is sufficiently small, then kuˆ + εη − uk ˆ = |ε|kηk
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Svetlin G. Georgiev
will be as small as we please. Hence, from the definition of a local minimum, we obtain L (uˆ + εη) ≥ L (u), ˆ i.e., Φ(ε) ≥ Φ(0). Therefore Φ has a local minimum for ε = 0. From here, ˙ Φ(0) = 0, or, equivalently,
L1(u, ˆ η) = 0. ˙ Since Φ(0) = 0, we have 1¨ 2 . Φ(ε) − Φ(0) = Φ(α)ε 2 ¨ Therefore Φ(α) ≥ 0 for all ε whose absolute values is sufficiently small. Letting ¨ is continuous, we get ε → 0 and using that α → ∞, as ε → 0, and Φ ¨ Φ(0) ≥ 0, or, equivalently,
L2(u, ˆ η) ≥ 0. This completes the proof. Theorem 7.2.3 (Sufficient Condition). Let uˆ ∈ D(L ) be such that
L1(u, ˆ η) = 0 for all admissible variations η. If L2 (u, η) ≥ 0 for all u ∈ D(L ) and all admissible variations η, then L has an absolute minimum at the point u. ˆ If L2 (u, η) ≥ 0 for all u in some neighbourhood of the point uˆ and all admissible variations η, then the functional L has a local minimum at u. ˆ
Double Integral Calculus of Variations
265
Proof. For the function Φ we have ˙ Φ(1) = Φ(0) + Φ(0) +
1 ¨ Φ(α), 2!
α ∈ (0, 1).
(7.2.1)
Note that Φ(1) = L (uˆ + η), ˙ Φ(0) = L1 (u, ˆ η)
Φ(0) = L (u), ˆ
= 0, 2 d ¨ L (uˆ + εη) Φ(α) = 2 dε ε=α 2 d = L ( u ˆ + αη + βη) dβ2 β=0 = L2 (uˆ + αη, η). Hence and (7.2.1), we obtain
L (uˆ + η) = L (u) ˆ +
1 L2(uˆ + αη, η) 2!
for all admissible variations η, where α ∈ (0, 1). Suppose that L2 (u, η) ≥ 0 for all u ∈ D(L ) and all admissible variations η. If u ∈ D(L ), then putting η = u − u, ˆ we get
L (u) ≥ L (u). ˆ Then L has an absolute minimum at the point u. ˆ Now we suppose that L2(u, η) ≥ 0 for all u in some neighbourhood of the point uˆ and all admissible variations η. There exists r > 0 such that for u ∈ D(L ) and ku − uk ˆ < r, we have L2 (u, η) ≥ 0 for all admissible variations η. We take such an element u and we put η = u − u. ˆ Then 1 2
L (u) = L (u) ˆ + L2 (uˆ + αη, η).
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Note that kuˆ + αη − uk ˆ = kαηk
= |α|kηk
≤ kηk
= ku − uk ˆ
< r. Hence,
L2(uˆ + αη, η) ≥ 0, and, then
L (u) ≥ L (u). ˆ This completes the proof. By Theorem 7.2.2 and Theorem 7.2.3, it follows that
L1 (u, η) =
and
Z Z
E
Lu (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))η(σ1 (x), σ2 (y))
+L p (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))η∆1 (x, σ2 (y)) +Lq (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))ηδ2 (σ1 (x), y) ∆1 x∆2 y, (7.2.2) Luu (x,y,u(σ1(x),σ2 (y)),u∆1 (x,σ2 (y)),u∆2 (σ1(x),y)) (η(σ1(x),σ2(y)))2 E 2 +L pp (x,y,u(σ1(x),σ2(y)),u∆1 (x,σ2 (y)),u∆2 (σ1 (x),y)) η∆1 (x,σ2 (y)) 2 +Lqq (x,y,u(σ1(x),σ2(y)),u∆1 (x,σ2(y)),u∆2 (σ1 (x),y)) η∆2 (σ1(x),y) ∆ ∆ 2 1 +2Lup (x,y,u(σ1(x),σ2(y)),u (x,σ2 (y)),u (σ1 (x),y))η(σ1(x),σ2 (y))η∆1 (x,σ2(y)) +2Luq(x,y,u(σ1(x),σ2 (y)),u∆1 (x,σ2 (y)),u∆2 (σ1(x),y))η(σ1(x),σ2 (y))η∆2(σ1 (x),y) +2L pq (x,y,u(σ1(x),σ2(y)),u∆1 (x,σ2 (y)),u∆2 (σ1 (x),y))η∆1(x,σ2 (y))η∆2 (σ1(x),y) ∆1 x∆2 y. (7.2.3)
L2 (u,η) =
Z Z
Example 7.2.4. Let L(x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) 2 = x + y + u(σ1 (x), σ2 (y)) + u∆1 (x, σ2 (y)) 3 + u∆2 (σ1 (x), y) .
Double Integral Calculus of Variations
267
Here L(x, y, u, p, q) = x + y + u + p2 + q3 . Then Lu (x, y, u, p, q) = 1, L p (x, y, u, p, q) = 2p, Lq (x, y, u, p, q) = 3q2 , Luu (x, y, u, p, q) = 0, L pp (x, y, u, p, q) = 2, Lqq (x, y, u, p, q) = 6q, Luq (x, y, u, p, q) = 0, Lup (x, y, u, p, q) = 0, L pq (x, y, u, p, q) = 0. Therefore the equations (7.2.2) and (7.2.3) take the form Z Z L1(u, η) = η(σ1 (x), σ2 (y)) + 2u∆1 (x, σ2 (y))η∆1 (x, σ2 (y)) E 2 +3 u∆1 (σ1 (x), y) η∆2 (σ1 (x), y) ∆1 x∆2 y, Z Z 2 2 L2(u, η) = 2 η∆1 (x, σ2 (y)) + 6u∆2 (σ1 (x), y) η∆2 (σ1 (x), y) ∆1 x∆2 y. E
Exercise 7.2.5. Write the equations (7.2.2) and (7.2.3) for the following functionals. 1. L(x, y, u, p, q) = xy + u2 + upq, 2. L(x, y, u, p, q) = xyupq, 3. L(x, y, u, p, q) = x2 + u2 p2 + u2 q2 , 4. L(x, y, u, p, q) = x2 + yupq, 5. L(x, y, u, p, q) = (x − yu)2 + (y + pq)2 .
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7.3. Euler’s Condition Let E is an ω-type subset of T1 × T2 and Γ be the positively oriented fence of E. We set E σ = {(x, y) ∈ E : (σ1 (x), σ2 (y)) ∈ E} . Lemma 7.3.1 (Dubois-Reymond’s Lemma). If M(x, y) is continuous on E with Z Z M(x, y)η(σ1 (x), σ2 (y))∆1 x∆2 y = 0
S
Γ
E
for every admissible variation η, then M(x, y) = 0
for all (x, y) ∈ E σ .
Proof. Assume the contrary. Without loss of generality, we suppose that (x0 , y0 ) ∈ E σ is such that M(x0 , y0 ) > 0. The continuity of M(x, y) ensures that M(x, y) is positive in a rectangle Ω = [x0 , x1 ) × [y0 , y1 ) ⊂ E for some points x1 ∈ T1 , y1 ∈ T2 such that σ1 (x0 ) ≤ x1
and σ2 (y0 ) ≤ y1 .
We set
(x − x0 )2 (x − σ1 (x1 ))2 (y − y0 )(y − σ2 (y1 ))2 for (x, y) ∈ Ω 0 for (x, y) ∈ E\Ω. S (1) We have that η ∈ Crd (E Γ), η = 0, i.e., η is an admissible variation. We Γ have that η(x, y) =
Z Z
E
M(x, y)η(σ1 (x), σ2 (y))∆1 x∆2 y =
Z Z
Ω
M(x, y)η(σ1 (x), σ2 (y))∆1 x∆2 y
> 0, which is a contradiction. This completes the proof.
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Double Integral Calculus of Variations
Theorem 7.3.2 (Euler’s Necessary Condition). Suppose that an admissible function uˆ provides a local minimum for L and the function uˆ has continuous partial delta derivatives of the second order. Then uˆ satisfies the Euler-Lagrange equation 0 = Lu (x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) −L∆p 1 (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) −L∆q 2 (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
(7.3.1)
for (x, y) ∈ E σ .
Proof. Since uˆ is a local minimum for L , by Theorem 7.2.2, it follows that
L1(u, ˆ η) = 0 for all admissible variations η. Hence and (7.2.2), applying integration by parts and Green’s formula, we get 0 = L1 (u, ˆ η) Z Z = Lu (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) E
×η(σ1 (x), σ2 (y))∆1x∆2 y
+L p (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) ×η∆1 (x, σ2 (y)) =
+Lq (x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))η∆2 (σ1 (x), y) ∆1 x∆2 y
Z Z
E
Lu (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
×η(σ1 (x), σ2 (y))∆1x∆2 y Z Z + L p (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) E
×η∆1 (x, σ2 (y))
+Lq (x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
×η∆2 (σ1 (x), y)∆1 x∆2 y
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Svetlin G. Georgiev =
Z Z
E
Lu (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
×η(σ1 (x), σ2 (y))∆1 x∆2 y Z Z ∂ L p (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) + ∆ x E 1 ×η(x, σ2 (y)) ∂ + Lq (x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) ∆2 y ×η(σ1 (x), y) ∆1 x∆2 y Z Z − L∆p 1 (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) E +L∆q 2 (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
×η(σ1 (x), σ2 (y))∆1 x∆2 y
=
Z Z
E
Lu (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
×η(σ1 (x), σ2 (y))∆1 x∆2 y Z Z ∂ ∆1 L p (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) − E ∆1 x +L∆q 2 (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) ×η(σ1 (x), σ2 (y)) ∆1 x∆2 y Z + L p (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) Γ
×η(x, σ2 (y))∆2y
=
−Lq (x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) ×η(σ1 (x), y)∆1x
Z Z
E
Lu (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
×η(σ1 (x), σ2 (y))∆1 x∆2 y Z Z + L∆p 1 (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y)) E ∆2 +Lq (x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y))
×η(σ1 (x), y)∆1x∆2 y.
Double Integral Calculus of Variations
271
From here and from Lemma 7.3.1, we get (7.3.1). This completes the proof. Example 7.3.3. Let T1 = 2N0 , T2 = 3N0 . Consider the variational problem
L (y) =
Z Z
E
x2 y3 u(2x, 3y)u∆1 (x, 3y)u∆2 (2x, y)∆1 x∆2 y −→ min,
where E = {(x, y) ∈ T1 × T2 : 1 ≤ x ≤ 8,
1 ≤ y ≤ 27}.
Here L(x, y, u, p, q) = x2 y3 upq, σ1 (x) = 2x, σ2 (y) = 3y,
x ∈ T1 ,
y ∈ T2 .
Then Lu (x, y, u, p, q) = x2 y3 pq, L p (x, y, u, p, q) = x2 y3 uq, Lq (x, y, u, p, q) = x2 y3 up, L∆p 1 (x, y, u, p, q) = (σ1 (x) + x)y3 uq = (2x + x)y3 uq = 3xy3 uq, L∆q 2 (x, y, u, p, q) = x2 (σ2 (y))2 + yσ2 (y) + y2 up = x2 (3y)2 + y(3y) + y2 up = x2 (9y2 + 3y2 + y2 )up = 13x2 y2 up. The Euler-Lagrange equation takes the form x2 y3 u∆1 (x, 3y)u∆2 (2x, y) − 3xy3 u(2x, 3y)u∆2 (2x, y) −13x2 y2 u(2x, 3y)u∆1 (x, 3y) = 0,
(x, y) ∈ E.
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Svetlin G. Georgiev
Exercise 7.3.4. Write the Euler-Lagrange equation for the variational problem L (y) =
Z Z
E
2 x2 + y2 u(σ1 (x),σ2(y)) + u∆1 (x,σ2 (y)) u∆2 (σ1(x),y) ∆1 x∆2 y −→ min,
in the cases 1. T1 = Z, T2 = 2Z, 2. T1 = T2 = 3Z, 3. T1 = 2N0 , T2 = 3Z, 4. T1 = N0 , T2 = 3N0 , 5. T1 = T2 = N30 .
7.4. Advanced Practical Problems Problem 7.4.1. Write the equations (7.2.2) and (7.2.3) for the following functionals. 1. L(x, y, u, p, q) = x + p + q + u2 − (y + 2p)2 , 2. L(x, y, u, p, q) = (x2 + p)3 + yuq, 3. L(x, y, u, p, q) = (x − p)2 + (y + q)3 , 4. L(x, y, u, p, q) = (x + y + u + p + q)2 , 5. L(x, y, u, p, q) = (x − u − p)2 − q4 . Problem 7.4.2. Write the Euler-Lagrange equation for the variational problem Z Z 2 3 2 4 L (y) = x − 3xy + y + y (u(σ1 (x), σ2 (y))) ∆1 x∆2 y −→ min, E
in the cases
1. T1 = Z, T2 = 2Z, 2. T1 = T2 = 3Z,
Double Integral Calculus of Variations 3. T1 = 2N0 , T2 = 3Z, 4. T1 = N0 , T2 = 3N0 , 5. T1 = T2 = N30 .
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Chapter 8
Noether’s Second Theorem Suppose that T is time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let also, the time scale T has sufficiently many points such that all computations to make sense and assume that T is such that σ(t) = a1t + a0 for some a1 ∈ R+ and a0 ∈ R, for any t ∈ T. Let a, b ∈ T, a < b, be such that all computations on [a, b] to have a sense. For a subset A ⊆ T k (A) we denote the space of all functions y : [a, b] → R such and k ∈ N with Crd r r k k that y∆ , r ∈ {1, . . ., k}, exist on Aκ and y∆ is rd-continuous on [a, b]κ .
8.1. Invariance under Transformations We will start with the following useful lemma. Lemma 8.1.1 (Higher Order Fundamental Lemma of the Calculus of Variations). Let f 0 , f 1 , . . ., f m ∈ Crd ([a, b]). If ! Z m−1 ρ
a
(b)
m
(t) ∆t = 0,
l=0
2m for all η ∈ Crd ([a, b]) such that
η(a) = 0, .. . m−1
η∆
m−l ∆l
∑ fl (t)ησ
(a) = 0,
η ρm−1 (b) = 0, m−1
η∆
ρm−1 (b) = 0,
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Svetlin G. Georgiev
then m
∑ (−1) l=0
l
1 a1
l(l−1) 2
l
m
fl∆ (t) = 0,
t ∈ [a, b]κ .
Proof. We will use mathematical induction. 1. Let m = 0. Then the assertion follows from Lemma 5.1.1. 2. Assume that the assertion is true for some m ∈ N.
3. We will prove the assertion for m + 1. We have, using the proof of Theorem 6.2.1, 0
=
Z ρm (b)
m+1
Z ρm (b)
m
∑
a
=
l=0
∑
a
=
m
∑
a
m+1−l ∆l
fl (t)ησ
Z ρm (b)
m+1
fm+1 (t)η∆
m
m+1
!
(t) ∆t
(t) ∆t !
m+1−l ∆l
∑ fl (t)ησ
a
l=0
!
(t) ∆t
(t) + fm+1 (t)η∆
l=0
Z ρm (b) a
=
m+1−l ∆l
fl (t)ησ
l=0
Z ρm (b)
+
m+1−l ∆,
fl (t)ησ
(t) ∆t
t=ρm (b) m + fm+1 (t)η∆ (t) t=0
−
=
Z ρm (b) a
Z ρm (b) a
−
1 a1
mσ
∆ fm+1 (t)η∆ m
(t)∆t
σm+1−l ∆l
∑ fl (t)η l=0
m Z
a
ρm (b)
!
(t) ∆t m
∆ fm+1 (t)ησ∆ (t)∆t
!
(t) ∆t
277
Noether’s Second Theorem =
Z ρm (b)
=
Z ρm−1 (b)
a
m−1
∑
m−1
∑
σm+1−l ∆l
fl (t)η
(t) +
l=0
fm (t) −
fm (t) −
1 a1
m
1 a1
m ∆ fm+1 (t) ησ∆ (t) ∆t
m Z
a
ρm (b)
∆ fm+1 (t)
σ∆m
η
!
(t) ∆t
!
1 m ∆ m fm+1 (t) ησ∆ (t) ∆t a ρm (b) 1 l=0 ! m Z ρm−1 (b) m−1 1 ∆ σ∆m σm+1−l ∆l (t) + fm (t) − fm+1 (t) η (t) ∆t ∑ fl (t)η a1 a l=0 ! m m−1 1 m σm+1−l ∆l ∆ σ∆m +µ (ρ (b)) ∑ fl (t)η (t) + fm (t) − fm+1 (t) η (t) m a1 t=ρ (b) l=0 ! m Z ρm−1 (b) m−1 m+1−l l m 1 ∆ (t) ησ∆ (t) ∆t. ∑ fl (t)ησ ∆ (t) + fm (t) − a1 fm+1 a l=0 +
=
(t) +
l=0
a
=
m+1−l ∆l
fl (t)ησ
!
Z ρm−1 (b)
m−1
∑
m+1−l ∆l
fl (t)ησ
(t) +
fm (t) −
Hence and the induction hypothesis, we get l(l−1) 2 l 1 fl∆ (t) 0 = ∑ (−1) a 1 l=0 m(m−1) m 2 1 1 m ∆ +(−1) fm+1 (t) fm(t) − a1 a1 m(m+1) m+1 2 l m+1 1 l = ∑ (−1) fl∆ (t), t ∈ [a, b]κ . a1 l=0 m−1
l
This completes the proof.
Suppose that y = (y1 , y2 , . . ., yn ). We consider the transformations t = t, yk+1 (t) = yk (t) + ∑kj=1 T k j (p j )(t), where
m
m−(l+1) ∆l
T k j (p j ) = ∑ gklj pσj l=0
k ∈ {1, . . ., n},
(8.1.1)
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Svetlin G. Georgiev
1 and gklj ∈ Crd ([a, b]), gklj : [a, b] → R, l ∈ {0, . . ., m}, j ∈ {1, . . ., r}, k ∈ m ([a, σm (b)]), {1, . . ., n}, which depend on arbitrary functions p1 , p2 , . . ., pr ∈ Crd pl : [a, b] → R, l ∈ {1, . . ., r} and their delta derivatives up to order m. Consider the variational problem
L (y) =
Z b a
L t, yσ (t), y∆(t) ∆t −→
extremize,
(8.1.2)
where L : [a, b] × R2n → R, L = L(t, u, v), is a given function that is continuous and has continuous partial derivatives with respect to u and v, and continuous partial delta derivative with respect to t. Definition 8.1.2. The functional L is called invariant under the transformations (8.1.1) if and only if Z b Z b σ ∆ L t, y (t), y (t) ∆t = L t, yσ (t), y∆ (t) ∆t. a
a
Theorem 8.1.3 (Necessary Condition of Invariance). If the functional L is invariant under the transformations (8.1.1), then σ R 0 = ∑nk=1 ab Lyσk t, yσ (t), y∆(t) ∑rj=1 T k j (p j ) (t) (8.1.3) ∆ +Ly∆ t, yσ(t), y∆(t) ∑rj=1 T k j (p j ) (t) ∆t. k
Proof. Since L is invariant under the transformations (8.1.1), we have Z b L t, yσ(t), y∆(t) ∆t a !σ Z b r σ 1j = L t, y1 (t) + ε ∑ T (p j ) (t), . . ., a
yσn (t) + ε
j=1
r
∑ T n j(p j ) j=1
y∆1 (t) + ε
r
∑ T 1 j (p j)
j=1
y∆n (t) + ε
r
∑ T n j (p j)
j=1
!σ
!∆
!∆
(t),
(t), (t) ∆t
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Noether’s Second Theorem
for any ε ∈ R. Differentiating the last equality with respect to ε and then taking ε = 0, we get (8.1.3). This completes the proof.
8.2. Noether’s Second Theorem without Transformations of Time Define Ek (L) = Lσyk − L∆yk , where L∆y∆ = k
k = 1, . . ., n,
∆ L ∆. ∆t yk
Theorem 8.2.1. If the functional L is invariant under the transformations (8.1.1), then n
m
∑ ∑ (−1)
l
k=1 l=0
1 a1
l(l+1) 2
gklj
σ
Ek (L)
∆l
= 0.
Proof. By Theorem 8.1.3, we have n
0 =
∑
Z b
σ
∆
Lyσk t, y (t), y (t)
k=1 a
+Ly∆ t, yσ (t), y∆(t) k
r
r
∑T
kj
(p j )
j=1
∑ T k j (p j ) j=1
!∆
!σ
(t)
(t) ∆t.
We fix j ∈ {1, . . ., r}. By the arbitrariness of the functions p1 , p2 , . . ., pr , we can suppose that ph ≡ 0 for h 6= j, h ∈ {1, . . ., r}, and p j (a) = 0, .. . m−1
p∆j
σ−1 ∆m
pj
(a) = 0, (a) = 0,
p j (b) = 0, m−1
p∆j
(b) = 0,
σ−1 ∆m
pj
(b) = 0.
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Therefore n
∑
0 =
Z b
Lyσk t, yσ(t), y∆(t)
k=1 a
+Ly∆ t, yσ(t), y∆(t) k
σ k j T (p j ) (t)
∆ k j T (p j ) (t) ∆t.
Integrating by parts the last equality, we get
Z b σ ∆ ∆ σ ∆ σ t, y (t), y (t) − L ∆ t, y (t), y (t) L yk ∑ y n
0 =
k
a
k=1
× T k j (p j )
σ
(t)∆t
t=b +Lyk t, yσ (t), y∆(t) T k j (p j )(t) . t=a
By the proof of Theorem 8.2.1, we obtain t=b T k j (p j )(t) = 0, t=a
Therefore n
0 =
∑
Lyσk t, yσ(t), y∆(t) − L∆y∆ t, yσ (t), y∆(t)
Z b
k=1 a
k
× T k j (p j ) or
n
∑
Z b
k=1 a
σ
(t)∆t,
Ek (L) t, yσ(t), y∆(t)
or n
0 =
k ∈ {1, . . ., n}.
∑
Z b
k=1 a
Ek (L) t, yσ (t), y∆(t)
m
×
∑
σ k j T (p j ) (t)∆t = 0,
l=0
m−(l+1) ∆l gklj pσj
!σ
(t)∆t,
281
Noether’s Second Theorem or 0 =
Z b m
n
∑ ∑ Ek (L)
t, yσ(t), y∆(t)
a l=0 k=1
×
gklj
σ 1 l a1
m−l ∆l
pσj
(t)∆t.
Hence and Lemma 8.1.1, we obtain m
n
∑ ∑ (−1) l=0 k=1
l
1 a1
or n
m
∑ ∑ (−1)
k=1 l=0
This completes the proof.
l
l(l−1)
2
1 a1
σ 1 l Ek (L) gklj a1
l(l+1) 2
gklj
σ
Ek (L)
∆l
!∆l
= 0,
= 0.
8.3. Noether’s Second Theorem with Transformations of Time In this section we suppose that the Lagrangian L is defined for all t ∈ R, not only for t from the time scale T, and L is continuous and has continuous partial derivative with respect to t. Consider the transformations t = t + ∑rj=1 H j (p j )(t), yk (t) = yk (t) + ∑rj=1 T k j (p j )(t),
k ∈ {1, . . ., n},
where m
H j (p j ) =
l=0 m
T k j (p j ) =
m−(l+1) ∆l
∑ fl j pσj
m−(l+1) ∆l
∑ gklj pσj
l=0
, ,,
(8.3.1)
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Svetlin G. Georgiev
1 fl j , gklj ∈ Crd ([a, b]), k ∈ {1, . . ., n}, l ∈ {0, . . ., m}, j ∈ {1, . . ., r}, which de2m pend on arbitrary functions p1 , . . ., pr , ps ∈ Crd ([a, σm(b)]), and their partial derivatives up to order m. Assume that the map r
t 7−→ α(t) = t + ∑ H j (p j )(t) j=1
1 is strictly increasing Crd function and its image is again a time scale T with forward jump operator σ and delta differentiation operator ∆.
Definition 8.3.1. The functional L is called invariant under the transformations 1 ([a, b]) we have (8.3.1) if and only if for all y ∈ Crd Z b a
Z b L t, yσ(t), y∆(t) ∆t = L t, yσ (t), y∆ (t) ∆t. a
Theorem 8.3.2 (Noether’s Second Theorem with Transformations of Time). If L is invariant under the transformations (8.3.1), then 0 =
l ∑nk=1 ∑m l=1 (−1)
+
for j ∈ {1, 2, . . ., r}.
l(l+1) 1 a1
2
gklj
σ
∆l
Ek (L) ∆l ∆ σ ∆ fl j Lt − L − yk Ly∆ − µLt k
Proof. Let q 6= 0. We define
v e q. L(t, s, y, q, v) = L s − µ(t)q, y, q
1 For s(t) = t and for any y ∈ Crd ([a, b]), y : [a, b] → Rn , we have
L t, yσ(t), y∆(t) = e L t, sσ(t), yσ(t), s∆(t), y∆(t) .
(8.3.2)
Noether’s Second Theorem Therefore, for s(t) = t, we get
L (y) =
Z b
=
Z b
a
a
L t, yσ (t), y∆(t) ∆t
e L t, sσ(t), yσ(t), s∆(t), y∆(t) ∆t
= L (s, y)
and using that L is invariant under the transformations (8.3.1), we get
L (s, y) =
a
=
L t, yσ (t), y∆(t) ∆t
Z b
L t, (y ◦ σ) (t), y∆ (t) ∆t
Z α(b) α(a)
L α(t), (y ◦ σ ◦ α)(t), y∆ (α(t)) α∆ (t)∆t a ! Z b (y ◦ α)∆ (t) = L α(t), (y ◦ α ◦ σ) (t), α∆ (t)∆t ∆ (t) α a =
Z b
! (y ◦ α)∆ (t) = L α (t) − µ(t)α (t), (y ◦ α) (t), α∆ (t)∆t ∆ (t) α a Z b e = L t, ασ(t), (y ◦ α)σ (t), α∆(t), (y ◦ α)∆ (t) ∆t Z b
σ
σ
∆
a
Let
e (α, y ◦ α). = L
H(t, y(t)) = α(t), r
T k (t, y(t)) = yk (t) + ∑ T k j (p j )(t), j=1
T
=
Hence, for s(t) = t, we have
k ∈ {1, . . ., n},
T 1 , T 2 , . . ., T n .
(α(t), (y◦ α)(t)) = (t, y(t)) = (H(t, y(t)), T(t, y(t))) = (H(s(t), y(t)), T(s(t), y(t))).
283
284
Svetlin G. Georgiev
Therefore, for s(t) = t, we obtain e (s, y) = L e (H(s, y), T (s, y)). L
e is invariant on Consequently L e = (s, y) : s(t) = t, U
1 y ∈ Crd ([a, b])
under the group of state transformations
(s, y) = (H(s, y), T(s, y)) in the sense of Definition 8.1.2. Hence and Theorem 8.2.1, we get l(l+1) σ ∆l 2 1 k ˜ L 0 = g E k lj a1 l(l+1) ∆l 2 σ l 1 e + ∑m (−1) f E ( L) , l j s l=0 a1 l ∑nk=1 ∑m l=0 (−1)
where Note that
Es ( e L) = e Lsσ − e L∆s∆ .
e Lsσ t, sσ(t), yσ(t), s∆(t), y∆(t) y∆ (t) ∆ s (t) = Lt sσ (t) − µ(t)s∆(t), yσ(t), ∆ s (t) and e Ls∆ t, sσ(t), yσ(t), s∆(t), y∆(t) y∆ (t) σ ∆ σ = L s (t) − µ(t)s (t), y (t), ∆ s (t) n ∆ yk (t) y∆ (t) σ ∆ σ − ∑ ∆ Ly∆ s (t) − µ(t)s (t), y (t), ∆ k s (t) k=1 s (t) ∆ y (t) −Lt sσ (t) − µ(t)s∆(t), yσ(t), ∆ µ(t)s∆(t). s (t)
(8.3.3)
Noether’s Second Theorem
285
Hence, for s(t) = t, we obtain Es ( e L) t, sσ(t), yσ(t), s∆(t), y∆(t) = Lt t, yσ(t), y∆(t) − L t, yσ(t), y∆(t) − ∑nk=1 y∆k (t)Ly∆ t, yσ(t), y∆(t) k ∆ −µ(t)Lt t, yσ(t), y∆(t) , Ek ( e L) t, sσ(t), yσ(t), s∆(t), y∆(t) = Ek (L) t, yσ (t), y∆(t)
(8.3.4)
(8.3.5)
for k ∈ {1, . . ., n}. We substitute (8.3.4) and (8.3.5) into (8.3.3) and we get (8.3.2). This completes the proof.
8.4. Noether’s Second Theorem-Double Delta Integral Case Let T1 and T2 be two time scales with forward jump operators and delta differentiation operators σ1 , σ2 and ∆1 , ∆2 , respectively. Let E ⊆ T1 × T2 be an ω-type set and let Γ be its positively fence and E σ = {(x, y) ∈ E : (σ1 (x), σ2 (y)) ∈ E} . Suppose that L(x, y, u, p, q), (x, y) ∈ E Γ, (u, p, q) ∈ R3n , be given. Let also, L is continuous together with its partial delta derivatives of first and second order with respect to x and y, and partial usual derivatives of the first and second order with respect to u, p, q. Consider the variational problem S
L (u)
=
Z Z
E
L x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y) ∆1 x∆2 y
−→ extremize,
where the set of admissible functions is n [ (1) D = u : E Γ → Rn , u ∈ Crd ,
o u = g on Γ ,
g is a fixed function defined and continuous on Γ. With ρ1 and ρ2 we denote the backward jump operators of T1 and T2 , respectively. Suppose that T1 and T2
286
Svetlin G. Georgiev
be such that σ1 (ρ1 (x)) = x, σ2 (ρ2 (y)) = y,
x ∈ T1κ ,
y ∈ T2κ .
Let u(x, y) = (u1 (x, y), u2 (x, y), . . ., un (x, y)). Consider the transformations x = x, y = y, uk (x, y) = uk (x, y) + T k (p)(x, y),
(8.4.1) k ∈ {1, . . ., n},
where T k (p)(x, y) = ak0 (x, y)p(x, y) +ak1 (x, y)p∆1 (x, y) +ak2 (x, y)p∆2 (x, y),
k ∈ {1, . . ., n},
where a0 , a1 and a2 are C 1 functions and p has continuous partial delta derivatives of the first and second order . Note that the transformations (8.4.1) depend on an arbitrary continuous function p and the partial derivatives of p. Definition 8.4.1. The functional L is called invariant under the transformations (8.4.1) if and only if Z Z
=
E
L x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y) ∆1 x∆2 y Z Z
E
L x, y, u(σ1 (x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y) ∆1 x∆2 y.
We introduce the following notations. T k (pσ )(x, y) = T k (σ1 (x), σ2 (y)), T k (pσ1 )(x, y) = T k (σ1 (x), y), T k (pσ2 )(x, y) = T k (x, σ2 (y)),
k ∈ {1, . . ., n},
x ∈ T1 ,
y ∈ T2 .
287
Noether’s Second Theorem
Theorem 8.4.2 (Necessary Condition of Invariance). If the functional L is invariant under the transformations (8.4.1), then RR 0 = ∑nk=1 E Luσk T k (pσ) + Lu∆1 T k∆1 (pσ2 ) k (8.4.2) +Lu∆2 T k∆2 (pσ1 ) ∆1 x∆2 y. k
Proof. Let ε ∈ R. Since L is invariant under the transformations (8.4.1), we have Z Z L x, y, u(σ1(x), σ2 (y)), u∆1 (x, σ2 (y)), u∆2 (σ1 (x), y) ∆1 x∆2 y E Z Z = L x, y, u1(x, y) + εT 1 (p)(x, y), . . ., un (x, y) + εT n (p)(x, y), E
∆
u1 1 (x, σ2 (y)) + εT 1∆1 (p)(x, σ2 (y)), . . ., u∆n 1 (x, σ2 (y)) + εT n∆1 (p)(x, y), ∆ u1 2 (σ1 (x), y) + εT 1∆2 (p)(x, σ2 (y)), . . ., u∆n 2 (σ1 (x), y) + εT n∆2 (p)(x, y) ∆1 x∆2 y.
We differentiate the last equality with respect to ε and then letting ε → 0, we get (8.4.2). This completes the proof. Define
Eˆk (L) = Luσ − L∆∆11 − L∆∆22 , k
where L∆∆11 = uk
uk
∂ L∆ , ∆1 uk 1
uk
k ∈ {1, . . ., n},
L∆∆22 = uk
∂ L∆ . ∆2 uk 2
Theorem 8.4.3. If L is invariant under the transformations (8.4.1), then n
∑ k=1
Proof. We take p such that
Z Z
E
Eˆk (L)T k (pσ )∆1 x∆2 y = 0.
p(x, σ2 (y)) Γ p(σ1 (x), y) Γ ∆1 p (ρ1 (x), σ2 (y)) Γ ∆1 p (x, y) Γ ∆2 p (x, y) Γ p∆2 (σ1 (x), ρ2 (y)) Γ
=
0,
=
0,
=
0,
=
0,
=
0,
=
0.
(8.4.3)
288
Svetlin G. Georgiev
We fix l ∈ {1, . . ., n}. Then, using Green’s formula, we get Z Z l∆1 σ2 l∆2 σ1 L ∆1 T (p ) + L ∆2 T (p ) ∆1 x∆2 y ul
E
Z Z
=
ul
∆1 l σ2 Lul ∆1 T (p ) + L
ul
E
L∆∆11 T l (pσ ) + L∆∆22 T l (pσ ) ∆1 x∆2 y ul ul E Z L ∆1 T l (pσ2 )∆2 y − L ∆2 T l (pσ1 )∆1 x ul ul Γ Z Z ∆ ∆ − L ∆11 T l (pσ ) + L ∆22 T l (pσ ) ∆1 x∆2 y ul ul E Z Z ∆ ∆ − L ∆11 T l (pσ ) + L ∆22 T l (pσ ) ∆1 x∆2 y. Z Z
−
=
=
ul
ul
E
Hence,
∆2 ! l σ1 ∆1 x∆2 y ∆2 T (p )
l∆2 σ1 l∆1 σ2 T (p ) + L T (p ) ∆1 x∆2 y ∆ ∑ E u∆l 1 ul 2 l=1 ∆2 ∆1 n RR l σ l σ = − ∑l=1 E L ∆1 T (p ) + L ∆2 T (p ) ∆1 x∆2 y. n
Z Z
L
ul
ul
By Theorem 8.4.2, we have that n
∑
k=1
Z Z
E
Luσk T k (pσ )∆1 x∆2 y
n
=
−∑
k=1
Z Z
+L
∆ uk 2
E
L
∆
uk 1
T k∆1 (pσ )
T k∆2 (pσ ) ∆1 x∆2 y.
Hence and (8.4.4), we get (8.4.3). This completes the proof. Theorem 8.4.4. For any k ∈ {1, . . ., n}, we have Z Z
k
E
σ
qT (p )∆1 x∆2 y
Proof. We take p such that p(σ1 (x), y) = 0, Γ
=
Z Z
qak0 − (qak1 )∆1 −(qak2 )∆2 pσ ∆1 x∆2 y. E
p(x, σ2 (y)) = 0. Γ
(8.4.4)
Noether’s Second Theorem
289
Note that Z Z
E
qT k (pσ )∆1 x∆2 y
=
Z Z
E
qak0 (x, y)p(σ1 (x), σ2 (y))∆1 x∆2 y
Z Z
E
Z Z
E
+ +
qak1 (x, y)p∆1 (x, σ2 (y))∆1 x∆2 y qak2 (x, y)p∆2 (σ1 (x), y)∆1 x∆2 y.
Observe that Z Z
=
E
qak1(x,y)p∆1 (x,σ2 (y)) + qak2(x,y)p∆2 (σ1 (x),y) ∆1 x∆2 y Z Z ∆1 ∆2 qak1 (x,y)p(x,σ2(y)) + qak2(x,y)p(σ1(x),y) ∆1 x∆2 y E Z Z ∆1 ∆2 − qak1 (x,y) p(σ1(x),σ2(y)) + qak2 (x,y) p(σ1(x),σ2 (y)) ∆1 x∆2 y E
=
=
−qak2 (x,y)p(σ1(x),y) ∆1 x + qak1 (x,y)p(x,σ2(y)) ∆2 y Γ Z Z ∆1 ∆2 − qak1 (x,y) p(σ1(x),σ2(y)) + qak2 (x,y) p(σ1(x),σ2 (y)) ∆1 x∆2 y E Z Z ∆1 ∆2 qak1 (x,y) p(σ1(x),σ2(y)) + qak2 (x,y) p(σ1(x),σ2 (y)) ∆1 x∆2 y. − Z Z
E
Therefore Z Z
k
E
σ
qT (p )∆1 x∆2 y =
This completes the proof.
Z Z
qak0 (x, y)p(σ1 (x), σ2 (y))∆1 x∆2 y Z Z ∆1 − qak1 (x, y) p(σ1 (x), σ2 (y)) E ∆2 p(σ1 (x), σ2 (y)) ∆1 x∆2 y. + qak2 (x, y) E
We define ∆1 ∆2 Tek (q) = qak0 − qak1 − qak2 ,
k ∈ {1, . . ., n}.
Theorem 8.4.5 (Noether’s Second Theorem without Transformation of Time). If the functional L is invariant under the transformations (8.4.1), then n
∑ Tek (Eˆk (L)) = 0
k=1
on E σ .
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Svetlin G. Georgiev
Proof. By Theorem 8.4.3 and Theorem 8.4.4, we get n
∑ k=1
Z Z
E
Eˆk (L)T k (pσ )∆1 x∆2 y
n
=
∑ k=1
=
0.
Z Z
E
Tek Eˆk (L) pσ ∆1 x∆2 y
Hence and Lemma 7.3.1, we obtain n
∑ Tek
k=1
This completes the proof.
Eˆk (L) = 0 on E σ .
References [1] R. Agarwal and M. Bohner, Quadratic functionals for second order matrix equations on time scales. Nonlinear Analysis, 33(1998), pp. 675-692 [2] M. Bohner, Calculus of Variation on Time Scales, Dynamic Systems and Applications, 13(2004), pp. 339-349. [3] M. Bohner and G. Sh. Guseinov, Double integral calculus of variations on time scales. Computers and Mathematics with Applications, 54(2007), pp. 45-57. [4] M. Bohner and A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications, Birkh¨auser, Boston, 2001. [5] M. Bohner and S. Georgiev, Multivariable Dynamic Calculus on Time Scales, Springer, 2017. [6] R. Ferreira, A. Malinowska and D. Torres, Optimality conditions for the calculus of variations with higher-oredr delta derivatives. Applied Mathematic Letters, 24(2011), pp. 87-92. [7] R. Hilscher, Linear Hamiltonian Systems on Time Scales: Positivity of Quadratic Functionals. Mathematical and Computer Modelling, 32(2000), pp. 507-527.
Author Contact Information Svetlin G. Georgiev Sorbonne University Paris, France Email: [email protected]
Index C p1(T), 14 Crd (T), 14 Crd1 (T), 14 R , 96 R (T, Rn×n), 96 R (T), 96 Λn , 30 Additive Inverse Under ⊕, 21 Admissible Functions, 262 Admissible Pair, 182 Admissible Variation, 226, 252, 262 Associated Solution, 137, 172 Backward Jump Operator, 31 Basis, 153, 171 Boundary, 59 Broken Line, 73 Chain Rule, 10, 11 Circle Dot Multiplication, 21 Circle Minus Substraction, 24 Circle Plus Addition, 21, 24 Component of a Set, 73 Conjoined Solution, 136, 171 Connected Set, 73 Continuous Function, 56 Continuous Functional, 132
Curve, 66, 73 closed, 66 final point, 66 initial point, 66 Jordan, 68 length, 69 nonrectifiable, 69 oriented, 67 parameter, 67 parametric equations, 67 rectifiable, 69 simple, 68 Cylindrical Transformation, 23 Darboux ∆-Integral, 50 lower, 49 upper, 49 Darboux ∆-Sum lower, 49 upper, 49 Delta Derivative, 5 Delta Differentiable Function, 5 Dense, 33, 59 Dense Point, 2 Derivative of the Inverse, 12 Disconjugate Equation, 149 Disconjugate System, 177 Domain, 73
296
Index
Dual Version of Induction Principle, 15 Dubois-Reymond’s Lemma, 268 Equation of Motion, 182 Euler’s Equation, 182 Euler’s Necessary Condition, 269 Euler-Lagrange Equation, 234, 269 First Variation, 226, 263 Forward Jump Operator, 1, 2, 30 Functional, 131 Generalized Exponential Function, 25 Generalized Square, 22 Graininess, 33 Graininess Function, 3 Green’s Formula, 74 Hamiltonian Dynamic System, 164 Hamiltonian Matrix, 163, 164 Harmonic Numbers, 3 Higher-Dimensional Time Scale, 30 Higher-Order Delta Derivative, 9 Higher-Order Fundamental Lemma of the Calculus of Variations, 276 Higher-Order Hilger Derivative, 9 Hilger Alternative Axis, 19 Hilger Complex Number, 19 Hilger Derivative, 5 Hilger Differentiable Function, 5 Hilger Imaginary Circle, 19 Hilger Purely Imaginary Number, 20 Hilger Real Axis, 19 Hilger Real Part, 20 Hyperbolic Functions, 29
Improper Integral, 19 Induction Principle, 14 Invariance Under Transformations, 278, 282, 286 Isolated, 33 strictly, 33 Isolated Point, 2 Jacobi’s Condition, 246 Jordan ∆-Measurable, 59, 60 Jordan ∆-Measure, 60 inner, 60 outer, 60 Jordan Curve, 68 Lagrangian, 225, 252 Left-Dense, 33 Left-Dense Point, 2 Left-Scattered, 33 strictly, 33 Left-Scattered Point, 2 Legendre’s Condition, 241 Leibniz’s Formula, 9, 44, 46 Line Delta Integral first kind, 70 second kind, 71 Line Segment horizontal, 73 vertical, 73 Linear Functional, 132 Liouville’s Formula, 111 Lipschitz Condition, 57 Lipschitz Constant, 57 Matrix Exponential Function, 103 Moore-Penrose Generalized Inverse Matrix, 177
Index Noether’s Second Theorem with Transformations of Time, 282 Noether’s Second Theorem without Transformation of Time, 290 Normalized Bases, 171 Normalized Solutions, 136 Partial Delta Derivative, 37 higher-order, 43 second-order, 42 Partial Hilger Derivative, 37 Partial Nabla Derivative, 45 Partition inner, 62 Partition of Interval, 48 Partition of Rectangle, 48 Piconi’s Identity, 140, 196 Polygonal Path, 73 Positive Definite Functional, 145 Positively Oriented Fence, 74 set of type ω, 74 Pre-Antiderivative, 16 Pre-Differentiable Function, 13 Principal Solution, 135, 172 Proper Weak Local Minimum, 225, 263 Putzer’s Algorithm, 123 Quadratic Functional, 144 Rd-Continuous Function, 14 Rd-Continuous Matrix, 95 Refinement of ∆-Partition, 50 Regressive Dynamic Equation, 80 Regressive Equation, 79
297
Regressive Function, 24 Regressive Group, 24 Regressive Matrix, 96 Regulated Function, 12 Riccati Equation, 138 Riccati Operator, 178 Riemann ∆-Integrable, 62 Riemann ∆-Integral, 53 Riemann Multiple ∆-Integral, 62 Right-Dense, 32 Right-Dense Point, 2 Right-Scattered, 32 strictly, 32 Right-Scattered Point, 2 Scattered, 59 Second Delta Derivative, 9 Second Hilger Derivative, 9 Second Variation, 226, 263 Semigroup Property, 25 Set of Type ω, 74 Special Normalized Bases, 172 Sufficient Condition for Positive Definiteness, 212 Symplectic Dynamic System, 161 Symplectic Matrix, 157, 161 Time Scale, 1 Trigonometric Functions, 29 Weak Local Minimum, 225, 252, 263