Topics in partial differential equations [First edition.] 9789352741021

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PARTIAL DIFFERENTIAL EQUATIONS

ISBN 978-93-5274-102-1

9 789352 741021

PARTIAL DIFFERENTIAL EQUATIONS

TOPICS IN

PARTIAL DIFFERENTIAL EQUATIONS

Salient features of the present edition : H It has detailed theory supplemented with well explained examples. H It has adequate number of unsolved problems of all types in exercises. H It has working rules for solving problems before exercises. H It has hints of tricky problems after relevant exercises.

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Comprehensive Differential Equations and Calculus of Variations (For B.A./B.Sc. II) Comprehensive Differential Equations and Differential Geometry (For B.A./B.Sc. II) Comprehensive Abstract Algebra (For B.A./B.Sc. III) Comprehensive Discrete Mathematics (For B.A./B.Sc. III, B.C.A., M.C.A.) Comprehensive Business Mathematics (For B.Com. I, B.T.M.) Comprehensive Business Statistics (For B.Com. II, B.B.A., B.I.M.) A Textbook of Pharmaceutical Mathematics Vol. I (For B.Pharma.) A Textbook of Pharmaceutical Mathematics Vol. II (For B.Pharma.) A Textbook of Quantitative Techniques (For M.B.A.)

TOPICS IN

PARTIAL DIFFERENTIAL EQUATIONS

By PARMANAND GUPTA B.Sc. (Hons.), M.Sc. (Delhi) M.Phil (KU), Pre. Ph.D. (IIT Delhi) Associate Professor of Mathematics Former Head of Department of Mathematics Indira Gandhi National College, Ladwa Kurukshetra University, Haryana

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CONTENTS Chapter

Pages

1. Partial Differential Equations ......................................................................................... 1–12 1.1. Introduction ................................................................................................................................. 1 1.2. Definition of a Partial Differential Equation ........................................................................... 1 1.3. Order of a Partial Differential Equation .................................................................................. 1 1.4. Linear Partial Differential Equation ........................................................................................ 1 1.5. Notation ....................................................................................................................................... 2 1.6. Formation of a Partial Differential Equation ........................................................................... 2 1.7. Formation of a Partial Differential Equation by Elimination of Arbitrary Constants ......... 2 1.8. Formation of a Partial Differential Equation by Elimination of Arbitrary Functions ......... 7

2. Partial Differential Equations of the First Order (Equations Linear in p and q) .... 13–28 2.1. Introduction ............................................................................................................................... 13 2.2. Solution of a Partial Differential Equation ............................................................................ 13 2.3. Complete Solution ..................................................................................................................... 13 2.4. Particular Solution ................................................................................................................... 14 2.5. Singular Solution ...................................................................................................................... 14 2.6. General Solution ....................................................................................................................... 14 2.7. Lagrange Linear Equation ....................................................................................................... 15 2.8. Solution of Lagrange Linear Equation ................................................................................... 15

3. Partial Differential Equations of the First Order (Equations Non-linear in p and q) .......................................................................................................................... 29–59 3.1. Introduction ............................................................................................................................... 29 3.2. Special Type I : Equations Containing Only p and q ............................................................. 29 3.3. Special Type II : Equations of the Form z = px + qy + g(p, q) ............................................... 34 3.4. Special Type III : Equations Containing Only z, p and q ...................................................... 39 3.5. Special Type IV : Equations of the Form f1(x, p) = f2(y, q) ..................................................... 44 3.6. Use of Transformations ............................................................................................................ 50 3.7. Charpit’s General Method of Solution .................................................................................... 53

4. Homogeneous Linear Partial Differential Equations with Constant Coefficients . 60–80 4.1. Introduction ............................................................................................................................... 60 4.2. Partial Differential Equations of Second and Higher Order ................................................ 60 4.3. Homogeneous Linear Partial Differential Equations with Constant Coefficients .............. 60 4.4. Some Theorems ......................................................................................................................... 61 4.5. General Solution of Homogeneous Linear Partial Differential Equation f(D, D′)z = 0 with Constant Coefficients ....................................................................................................... 62

(v)

Chapter

Pages

4.6. General Solution of Homogeneous Linear Partial Differential Equation f(D, D′)z = F(x, y) with Constant Coefficients ......................................................................... 66 4.7. Particular Integral of f(D, D′)z = F(x, y) .................................................................................. 66 4.8. Particular Integral When F(X, Y) is Sum or Difference of Terms of the Form xmyn ........... 66 4.9. Particular Integral When F(x, y) is of the Form φ(ax + by) ................................................... 68 4.10. General Method of Finding Particular Integral ..................................................................... 75

5. Non-homogeneous Linear Partial Differential Equations with Constant Coefficients .................................................................................................................... 81–97 5.1. Introduction ............................................................................................................................... 81 5.2. Non-homogeneous Linear Partial Differential Equations with Constant Coefficients ..... 81 5.3. Reducible and Irreducible Non-homogeneous Linear Partial Differential Equations with Constant Coefficients ................................................................................................................ 81 5.4. General Solution of Reducible Non-homogeneous Linear Partial Differential Equation f(D, D′)z = 0 with Constant Coefficients ................................................................................. 82 5.5. General Solution of Irreducible Non-homogeneous Linear Partial Differential Equation f(D, D′)z = 0 With Constant Coefficients ................................................................................. 85 5.6. General Solution of Non-homogeneous Linear Partial Differential Equation With Constant Coefficients ................................................................................................................ 88 5.7. Particular Integral of f(D, D′)z = F(x, y) .................................................................................. 88 5.8. Particular Integral When F(x, y) is Sum or Difference of Terms of the Form xmyn ............ 88 5.9. Particular Integral When F(x, y) is of the Form eax+by ........................................................... 91 5.10. Particular Integral When F(x, y) is of the Form sin (ax + by) or cos(ax + by) ...................... 93 5.11. Particular Integral When F(x, y) is of the Form eax+by V(x, y) ............................................... 95

6. Partial Differential Equations Reducible to Equations with Constant Coefficients ................................................................................................. 98–103 6.1. Introduction ............................................................................................................................... 98 6.2. Reducible Linear Partial Differential Equations with Variable Coefficients ...................... 98 6.3. Solution of Reducible Linear Partial Differential Equations with Variable Coefficients .. 98

7. Monge’s Methods ...................................................................................................... 104–118 7.1. Introduction ............................................................................................................................. 104 7.2. Partial Differential Equation of Second Order .................................................................... 104 7.3. Intermediate Integral ............................................................................................................. 104 7.4. Monge’s Methods ..................................................................................................................... 104 7.5. Monge’s Method of Solving Rr + Ss + Tt = V ........................................................................ 105 7.6. Monge’s Method of Solving Rr + Ss + Tt + U(rt – s2) = V .................................................... 113

(vi)

PREFACE The present book on ‘‘Partial Differential Equations’’ has been written as a textbook according to the latest guidelines and syllabus in Mathematics issued by the U.G.C. for various universities. The text of the book has been prepared with the following salient features: (i) The language of the book is simple and easy to understand. (ii) Each topic has been presented in a systematic, simple, lucid and exhaustive manner. (iii) A large number of important solved examples properly selected from the previous university question papers have been provided to enable the students to have a clear grasp of the subject and to equip them for attempting problems in the university examination without any difficulty. (iv) Apart from providing a large number of examples, different type of questions in ample quantity have been provided for a thorough practice to the students. (v) A large number of ‘notes’ and ‘remarks’ have been added for better understanding of the subject. A serious effort has been made to keep the book free from mistakes and errors. In fact no pains have been spared to make the book interesting and useful. Suggestions and comments for further improvement of the book will be welcomed. —AUTHOR

(vii)

SYMBOLS Greek Alphabets A B Γ D E Z H Θ

α β γ δ ε ζ η θ ∃

Alpha Beta Gamma Delta Epsilon Zeta Eta Theta there exists

I K Λ M N Ξ O Π

ι κ λ μ ν ξ ο π V

Iota Kappa Lambda Mu Nu Xi Omicron Pi for all

P Σ T Y Φ X Ψ Ω

ρ σ τ υ ϕ χ ψ ω

Rho Sigma Tau Upsilon Phi Chi Psi Omega

Metric Weights and Measures LENGTH 10 millimetres 10 centimetres 10 decimetres 10 metres 10 decametres 10 hectometres

CAPACITY 10 millilitres 10 centilitres 10 decilitres 10 litres 10 decalitres 10 hectolitres

= 1 centilitre = 1 decilitre = 1 litre = 1 decalitre = 1 hectolitre = 1 kilolitre

VOLUME 1000 cubic centimetres = 1 centigram 1000 cubic decimetres = 1 cubic metre

AREA 100 square metres 100 ares 100 hectares

= 1 are = 1 hectare = 1 square kilometre

WEIGHT 10 milligrams 10 centigrams 10 decigrams 10 grams 10 decagrams 10 hectograms 100 kilograms 10 quintals

ABBREVIATIONS kilometre km metre m centimetre cm millimetre mm kilolitre kl litre l millilitre ml

tonne quintal kilogram gram are hectare centiare

= = = = = =

= = = = = = = =

1 centimetre 1 decimetre 1 metre 1 decametre 1 hectometre 1 kilometre

1 centigram 1 decigram 1 gram 1 decagram 1 hectogram 1 kilogram 1 quintal 1 metric ton (tonne)

( ix)

t q kg g a ha ca

1

Partial Differential Equations

1.1. INTRODUCTION Partial differential equations arise in applied mathematics and mathematical physics when the functions involved depend on two or more independent variables. The use of partial differential equation is enormous as compared to that of ordinary differential equations. In the present chapter, we shall learn the method of solving various types of partial differential equations. 1.2. DEFINITION OF A PARTIAL DIFFERENTIAL EQUATION An equation containing one or more partial derivatives of an unknown function of two or more independent variables is called a partial differential equation. The following are some of the examples of partial differential equations : 1.

∂z ∂z +3 = 5z + tan ( y − 3x ) ∂x ∂y

2. xz

3. ( y 2 + z 2 )

∂z ∂z − xy = − xz ∂x ∂y

4. x

∂2 z

∂2 z ∂2 z −2 2 =0 ∂x∂y ∂y

6. x 2

5. 2

∂x 2

−3

∂z ∂z + yz = xy ∂x ∂y

F GG H

FG IJ IJ H K JK

∂z ∂z ∂z + 3y = 2 z − x2 ∂x ∂y ∂y

2

2 ∂2 z ∂z ∂z 2 ∂ z − y −y +x = 0. 2 2 ∂y ∂x ∂x ∂y

1.3. ORDER OF A PARTIAL DIFFERENTIAL EQUATION The order of a partial differential equation is defined as the order of the highest partial derivative occurring in the partial differential equation. For the partial differential equations (1–6) given above, the order of the first four equations are one each and the order of the last two equations are two each. 1.4. LINEAR PARTIAL DIFFERENTIAL EQUATION A partial differential equation is said to be linear if the dependent variable and its partial derivatives occur only in the first degree and are not multiplied together. A partial differential equation which is not linear is called non-linear. Out of partial differential equations (1–6) given above the first, fifth and sixth equations are linear and others are non-linear. 1

2

PARTIAL DIFFERENTIAL EQUATIONS

∂z + 5 y = 7 is not a linear partial differential equation ∂x ∂z are multiplied together. because the dependent variable z and its partial derivative ∂x

The partial differential equation z

1.5. NOTATION If z = f(x, y) be a function of two independent variables x and y, then we shall use the following notation : ∂z = p, ∂x

∂z = q, ∂y

∂2 z = r, ∂x 2

∂2 z = s, ∂x∂y

∂2 z = t. ∂y 2

1.6. FORMATION OF A PARTIAL DIFFERENTIAL EQUATION There are two ways of forming partial differential equations depending on the given relation between variables. A relation between variables may contain arbitrary constants and arbitrary functions. The elimination of arbitrary constants (or functions) give rise to a partial differential equation. 1.7.

FORMATION OF A PARTIAL DIFFERENTIAL EQUATION BY ELIMINATION OF ARBITRARY CONSTANTS

Let z be a function of two independent variables x and y defined by f(x, y, z, a, b) = 0,

...(1)

where a and b are arbitrary constants. Differentiating (1) partially w.r.t. x and y, we get

∂f ∂f ∂z ∂f ∂f ∂z + = 0 and + =0 ∂x ∂z ∂x ∂y ∂z ∂y ⇒ and

∂f ∂f +p =0 ∂x ∂z

...(2)

∂f ∂f +q =0 ∂y ∂z

...(3)

In general, a and b may be eliminated from (1), (2), (3) and we get an equation of the form g(x, y, z, p, q) = 0. This is the required partial differential equation. The order of this equation shall be one. Remark 1. If the number of arbitrary constants is less than the number of independent variables, then the elimination of arbitrary constants shall usually give rise to more than one differential equation of order one. For example, if z = λx + y, then we have differential equations p =

z− y and q = 1. x

2. If the number of arbitrary constants is greater than the number of independent variables, then the elimination of arbitrary constants shall give rise to a partial differential equation of order usually greater than one.

3

PARTIAL DIFFERENTIAL EQUATIONS

ILLUSTRATIVE EXAMPLES Example 1. Form partial differential equations by eliminating arbitrary constants from the following relations : (i) z = (x + a)(y + b) (ii) z = ax2 + by2 + ab 2 2 (iv) z = aebx sin by. (iii) z = (x + a)(y + b) Sol. (i) We have z = (x + a)(y + b) ...(1) ⇒ z = xy + ay + bx + ab Differentiating z partially w.r.t. x and y, we get ∂z = y(1) + 0 + b(1) + 0 ∂x ∂z = x(1) + a(1) + 0 + 0 ∂y

and

...(2) ...(3)

∂z ∂z –y (3) ⇒ a = –x ∂x ∂y Putting the values of a and b in (1), we get

(2)

⇒

b=

FG H

IJ FG KH

IJ K

∂z ∂z −x y+ −y ∂y ∂x (ii) We have z = ax2 + by2 + ab. Differentiating (1) partially w.r.t. x and y, we get z= x+

∂z = 2ax + 0 + 0 ∂x

...(2)

p 2x Putting the values of a and b in (1), we get

(2)

⇒

p = 2ax ⇒ a =

z=

or

or

z=

q = 2by ⇒ b =

⇒

FG p IJ x + FG q IJ y + FG p IJ FG q IJ H 2x K H 2 y K H 2x K H 2 y K 2

2

or

z=

...(2) and

∂z = ( x 2 + a )( 2 y + 0) ∂y

⇒

p = 2x(y2 + b)

⇒

y2 + b =

p 2x

(3)

⇒

q = 2y(x2 + a)

⇒

x2 + a =

q 2y

z=

q p . or pq = 4xyz. 2 y 2x

q . 2y

...(1)

(2)

Putting the values of y2 + b and x2 + a in (1), we get

...(3)

px qy pq + + 2 2 4 xy

4xyz = 2px2y + 2qxy2 + pq. (iii) We have z = (x2 + a)(y2 + b). Differentiating (1) partially w.r.t. x and y, we get ∂z = ( y 2 + b)( 2x + 0) ∂x

...(1)

∂z = 0 + 2by + 0 ∂y

and (3)

∂z ∂z . ∂x ∂y

...(3)

4

PARTIAL DIFFERENTIAL EQUATIONS

(iv) We have z = aebx sin by. Differentiating (1) partially w.r.t. x and y, we get

and (2) (3)

∂z = ( a sin by ) . be bx ∂x

...(2)

∂z = ( ae bx ) . b cos by ∂y

...(3)

p = abebx sin by q = abebx cos by

⇒ ⇒

(4) ⇒

p = bz

⇒

FG IJ H K

p p = tan y q z

∴

...(4) ...(5)

p = tan by. q

Dividing (4) by (5), we get Also,

...(1)

b=

p . z p = q tan

or

py . z

Example 2. Find a partial differential equation by eliminating a, b and c from x2 a2 x2

+

y2

+

b2 y2

z2 c2

= 1.

z2

= 1. a 2 b2 c 2 Differentiating (1) partially w.r.t. x and y, we get

Sol. We have

+

+

...(1)

∂z 2x 2 z ∂z +0+ 2 = 0 or c 2 x + a 2 z =0 2 ∂x a c ∂x

and

0+

2y 2

+

...(2)

2z ∂z ∂z = 0 or c 2 y + b2 z =0 2 ∂y ∂ y c

...(3)

b Differentiating (2) partially w.r.t. y, we get

F GH

0 + a2 z

I JK

∂2z ∂z ∂z + =0 ∂y∂x ∂y ∂x

⇒

z

∂ 2z ∂z ∂z + = 0. ∂y∂x ∂x ∂y

Example 3. Find the partial differential equation of all planes which are at a constant distance ‘a’ from the origin. Sol. Let lx + my + nz = a ...(1) be the equation of a plane where l, m, n are d.c.’s of the normal to the plane. Differentiating (1) partially w.r.t. x and y, we get l(1) + 0 + n

∂z =0 ∂x

(2) ⇒ l + np = 0 (3) ⇒ m + nq = 0 Also l2 + m2 + n2 = 1 ∴ (– np)2 + (– nq)2 + n2 = 1

...(2) or or

and l = – np m = – nq

0 + m(1) + n

∂z =0 ∂y

...(3)

5

PARTIAL DIFFERENTIAL EQUATIONS

(p2 + q2 + 1) n2 = 1

or

p + q2 + 1 p , m = − nq = − 2 p + q2 + 1

l = − np = −

∴

1

n=

or

(Assuming n > 0)

2

q 2

p + q2 + 1

Putting the values of l, m and n in (1), we get −

px p2 + q 2 + 1

−

qy p2 + q 2 + 1

+

z p2 + q 2 + 1

=a

z = px + qy + a p2 + q 2 + 1 .

or

Example 4. Find the differential equation of the family of spheres of radius 7 with centres on the plane x – y = 0. Sol. Let (a, a, b) be any point on the plane x – y = 0. ∴ With centre at (a, a, b), the equation of the sphere of radius 7 is (x – a)2 + (y – a)2 + (z – b)2 = 49 ...(1) ∴ (1) represents a family of spheres where a and b are arbitrary constants. Differentiating (1) partially w.r.t. x and y, we get 2(x – a) + 0 + 2(z – b) p = 0 ...(2) and 0 + 2(y – a) + 2(z – b) q = 0 ...(3) (2) ⇒ x – a = – (z – b)p and (3) ⇒ y – a = – (z – b)q ∴ (1) ⇒ (z – b)2p2 + (z – b)2q2 + (z – b)2 = 49 ...(4) ⇒ (p2 + q2 + 1) (z – b)2 = 49 x−y (2) – (3) ⇒ 2(x – y) = – 2(z – b) (p – q) ⇒ z − b = − p−q ∴

(4)

⇒

2

(p + q

2

F x − y IJ + 1) G H p − qK

2

= 49

(p2 + q2 + 1)(x – y)2 = 49(p – q)2. Example 5. Show that the differential equation of all cones which have their vertex at the origin is px + qy = z. Sol. The equation of the family of cones is the homogeneous equation ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0 ...(1) where a, b, c, f, g, h are arbitrary constants. Differentiating (1) partially w.r.t. x and y, we get 2ax + 0 + 2czp + 2fyp + 2g(z.1 + xp) + 2hy = 0 ...(2) and 0 + 2by + 2czq + 2f(yq + z.1) + 2gxq + 2hx = 0 ...(3) (2) ⇒ ax + gz + hy + p(cz + fy + gx) = 0 ...(4) (3) ⇒ by + fz + hx + q(cz + fy + gx) = 0 ...(5) Multiplying (4) by x and (5) by y and adding, we get ax2 + gxz + hxy + by2 + fyz + hxy + (px + qy) (cz + fy + gx) = 0 or ax2 + by2 + fyz + gzx + 2hxy + (px + qy) (cz + fy + gx) = 0 Using (1), we get – (cz2 + fyz + gzx) + (px + qy) (cz + fy + gx) = 0 ⇒ (cz + fy + gx) (– z + px + qy) = 0 ⇒ px + qy = z. or

6

PARTIAL DIFFERENTIAL EQUATIONS

WORKING RULES FOR SOLVING PROBLEMS Rule I.

For a given relation involving variables and arbitrary constants, the relation is differentiated partially w.r.t. independent variables and arbitrary constants are eliminated to get the corresponding partial differential equation. Rule II. If the number of arbitrary constants is less than the number of independent variables, then the elimination of arbitrary constants shall usually give rise to more than one differential equation of order one. Rule III. If the number of arbitrary constants is equal to the number of independent variables, then the elimination of arbitrary constants shall give rise to one differential equation of order one. Rule IV. If the number of arbitrary constants is greater than the number of independent variables, then the elimination of arbitrary constants shall give rise to a differential equation of order usually greater than one.

TEST YOUR KNOWLEDGE Form partial differential equations by eliminating arbitrary constants from the following relations (Q. no. 1–10) : 2. z = ax + (1 – a)y + b 1. az + b = a2x + y 3. z = ax + by + ab 4. z = ax + a2y2 + b

x2

y2

6. z = axe y +

1 2 2y a e +b 2

5.

2z =

7.

8. ax2 + by2 + cz2 = 1 z = xy + y x 2 − a 2 + b 2 2 10. z = ax + by + cxy. z = ax + bxy + cy Form a partial differential equation by eliminating a and b from the equation (x – a)2 + (y – b)2 + z2 = k2. Find the partial differential equation of planes having equal x and y intercepts. Find the differential equation of all spheres of fixed radius and having their centres in the xy-plane. Find the differential equation of all spheres whose centre lies on z-axis.

9. 11. 12. 13. 14.

a

2

+

b2

Answers 1.

∂z ∂z . =1 ∂x ∂y

4.

∂z ∂z = 2y ∂y ∂x

7.

∂z ∂z ∂z ∂z =x +y ∂x ∂y ∂x ∂y

2 8. ∂z ∂z + z ∂ z = 0 ∂x ∂y ∂x∂y

∂2 z

∂z ∂z ∂2 z +y − xy ∂x ∂y ∂x∂y

10.

12.

2.

FG IJ H K

∂x

2

= 0,

∂2 z ∂y

2

∂z ∂z − =0 ∂x ∂y

2

∂z ∂z + =1 ∂x ∂y

5. 2z = x

= 0, z = x

13. z2

3. z = x

∂z ∂z +y ∂x ∂y

F F ∂z I F ∂z I GG GH ∂x JK + GH ∂y JK H 2

6.

2

FG IJ H K

∂z ∂z ∂z =x + ∂y ∂x ∂x

9. x

+ 1 = k2 14. x

2

∂z ∂z +y = 2z ∂x ∂y

11. z 2

I JJ K

∂z ∂z ∂z ∂z +y + . ∂x ∂y ∂x ∂y

F F ∂z I F ∂z I GG GH ∂x JK + GH ∂y JK H 2

∂z ∂z −y = 0. ∂y ∂x

2

I JJ K

+ 1 = k2

7

PARTIAL DIFFERENTIAL EQUATIONS

1.8. FORMATION OF A PARTIAL DIFFERENTIAL EQUATION BY ELIMINATION OF ARBITRARY FUNCTIONS Let u and v be independent functions of three variables x, y, z and let f(u, v) = 0 ...(1) be an arbitrary relation between u and v. Regarding z as a function of x, y and differentiating (1) partially w.r.t. x, we get ∂f ∂u ∂u ∂z ∂f ∂v ∂v ∂z + + =0 + ∂u ∂x ∂z ∂x ∂v ∂x ∂z ∂x ∂f ∂u ∂u ∂f ∂v ∂v ⇒ + +p =0 +p ∂u ∂x ∂z ∂v ∂x ∂z ∂f ∂f ∂v ∂v ∂u ∂u ⇒ +p ...(2) =− +p ∂u ∂v ∂x ∂z ∂x ∂z Similarly, differentiating (1) partially w.r.t. y, we get

FG H

IJ K IJ K

FG IJ H K IJ FG FG H H K FG IJ FG IJ H K H K F ∂v + q ∂v IJ FG ∂u + q ∂u IJ ∂ f ∂f =−G H ∂y ∂z K H ∂y ∂z K ∂u ∂v Eliminating f using (2) and (3), we get F ∂v + p ∂v IJ FG ∂u + p ∂u IJ = − FG ∂v + q ∂v IJ FG ∂u + q ∂u IJ −G H ∂x ∂z K H ∂x ∂z K H ∂y ∂z K H ∂y ∂z K FG ∂v + p ∂v IJ FG ∂u + q ∂u IJ = FG ∂v + q ∂v IJ FG ∂u + p ∂u IJ ⇒ H ∂x ∂z K H ∂y ∂z K H ∂y ∂z K H ∂x ∂z K F ∂u ∂v − ∂u ∂v IJ p + FG ∂u ∂v − ∂u ∂v IJ q = ∂u ∂v − ∂u ∂v ⇒ G H ∂y ∂z ∂z ∂y K H ∂z ∂x ∂x ∂z K ∂x ∂y ∂y ∂x

...(3)

Pp + Qq = R, ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v where and R = P= − , Q= − − . ∂y ∂z ∂z ∂y ∂z ∂x ∂x ∂z ∂x ∂y ∂y ∂x This is the required partial differential equation. The order of this equation is one. ⇒

Remark. The functions u and v are said to be independent if u/v is not merely a constant.

ILLUSTRATIVE EXAMPLES Example 1. Form partial differential equations by eliminating arbitrary function from the following relations : (i) z = f(x2 + 2y2) (iii) f(x2 + y2, z – xy) = 0

(ii) f(x2 + y2 + z2) = ax + by + cz (iv) f(x2 + y2 + z2, z2 – 2xy) = 0.

Sol. (i) We have z = f(x2 + 2y2). Differentiating (1) partially w.r.t. x and y, we get ∂z ∂ = p = f ′( x 2 + 2 y 2 ) ( x 2 + 2 y 2 ) = 2x f ′( x 2 + 2 y 2 ) ∂x ∂x ∂z ∂ = q = f ′( x 2 + 2 y 2 ) ( x 2 + 2 y 2 ) = 4 y f ′( x 2 + 2 y 2 ) ∂y ∂y

and Dividing we get

p x = or 2py − qx = 0. q 2y

...(1)

8

PARTIAL DIFFERENTIAL EQUATIONS

(ii) We have f(x2 + y2 + z2) = ax + by + cz. Differentiating (1) partially w.r.t. x and y, we get

...(1)

f ′( x 2 + y 2 + z 2 ) . 2x + 0 + 2 z

...(2)

f ′( x 2 + y 2 + z 2

...(3)

and

or or

∂z I ∂z FG = a .1 + 0 + c J H K ∂x ∂x F ∂z I ) . G 0 + 2 y + 2z J = 0 + b . 1 + c ∂∂yz H ∂y K

(2)

⇒

2 f ′( x 2 + y 2 + z 2 ) . ( x + pz) = a + cp

...(4)

(3)

⇒

2 f ′( x 2 + y 2 + z 2 ) . ( y + qz) = b + cq

...(5)

Dividing (4) by (5), we get x + pz a + cp = or ( x + pz) (b + cq) = ( y + qz) (a + cp) y + qz b + cq bx + cxq + bpz + czpq = ay + cyp + aqz + czpq (bz – cy)p + (cx – az)q = ay – bx. (iii) We have f(x2 + y2, z – xy) = 0. Let u = x2 + y2 and v = z – xy. ∴ (1) ⇒ f(u, v) = 0 Differentiating (2) partially w.r.t. x, we get

FG H

...(1) ...(2)

IJ K

∂f ∂u ∂f ∂v ∂v ∂z =0 + + ∂u ∂x ∂v ∂x ∂z ∂x ∂f ∂f . 2x + ( − y + 1 . p) = 0 ∂u ∂v

⇒

∂f ∂f + ( p − y) =0 ∂u ∂v Differentiating (2) partially w.r.t. y, we get

...(3)

2x

⇒

FG H

IJ K

∂f ∂u ∂f ∂v ∂v ∂z + + =0 ∂u ∂y ∂v ∂y ∂z ∂y

⇒ ⇒ Eliminating ⇒

∂f ∂f . 2y + ( − x + 1 . q) = 0 ∂u ∂v 2y

∂f ∂f + (q − x ) =0 ∂u ∂v

...(4)

2x p − y * ∂f ∂f =0 from (3) and (4), we get and 2y q − x ∂u ∂v 2x(q – x) – 2y(p – y) = 0 or xq – yp = x2 – y2.

*Why this step. If ax + by = 0 and cx + dy = 0, then

∴ Eliminating x, y, we get −

b d =− a c

x b x d =− and =− . y a y c

or ad − bc = 0 or

a b = 0. c d

9

PARTIAL DIFFERENTIAL EQUATIONS

(iv) We have f(x2 + y2 + z2, z2 – 2xy) = 0. Let u = x2 + y2 + z2 and v = z2 – 2xy. ∴ (1) ⇒ f(u, v) = 0 Differentiating (1) partially w.r.t. x, we get

FG H

IJ K

FG H

...(1) ...(2)

IJ K

∂f ∂u ∂u ∂f ∂v ∂v + + p =0 p + ∂u ∂x ∂z ∂v ∂x ∂z

∂f ∂f ( 2x + 2zp ) + ( − 2 y + 2zp ) = 0 ∂u ∂v ∂f ∂f ( x + zp ) + ( zp − y ) =0 ⇒ ∂u ∂v Differentiating (1) partially w.r.t. y, we get

⇒

IJ K

FG H

...(3)

IJ K

FG H

∂f ∂u ∂ u ∂ f ∂v ∂v + + q + q =0 ∂u ∂y ∂z ∂ v ∂y ∂z

⇒ ⇒

∂f ∂f ( 2 y + 2zq ) + ( − 2x + 2zq ) = 0 ∂u ∂v ∂f ∂f ( y + zq) + ( zq − x) =0 ∂u ∂v

Eliminating

...(4)

x + zp zp − y ∂f ∂f and from (3) and (4), we get = 0. ∂u ∂v y + zq zq − x

⇒ (x + zp)(zq – x) – (y + zq)(zp – y) = 0 2 ⇒ xzq – x + z2pq – xzp – yzp + y2 – z2pq + yzq = 0 ⇒ (x + y)zp – (x + y)zq = y2 – x2 ⇒ z(p – q) = y – x. Example 2. Form partial differential equations by eliminating arbitrary functions from the following relations : (i) z = xφ(y) + yψ(x) (ii) z = f(x2 – y) + g(x2 + y) (iii) x = f(z) + g(y) (iv) z = f(y + ax) + g(y + bx), a ≠ b. Sol. (i) We have z = xφ(y) + yψ(x). ...(1) Differentiating (1) partially w.r.t. x and y, we get ∂z ∂z = xφ ′( y) + ψ ( x) . 1 = φ( y ) . 1 + yψ ′ ( x ) ...(2) and ...(3) y ∂ ∂x Differentiating (2) w.r.t. y, we get ∂2 z = φ ′( y ) + ψ ′ ( x ) . 1 ∂y∂x

FG H

IJ K

...(4)

FG H

IJ K

∂2 z 1 ∂z 1 ∂z = − ψ ( x) + − φ( y) ∂y∂x x ∂y y ∂x

⇒ ⇒

xy

⇒

xy

∂2 z ∂z ∂z =x +y − ( xφ( y) + yψ ( x)) ∂y∂x ∂x ∂y ∂ 2z ∂z ∂z =x +y − z. ∂y∂x ∂x ∂y

(Using (2) and (3))

10

PARTIAL DIFFERENTIAL EQUATIONS

(ii) We have z = f(x2 – y) + g(x2 + y). Differentiating partially w.r.t. x and y, we get

...(1)

∂ ∂ ( x 2 − y ) + g ′( x 2 + y ) (x 2 + y) ∂x ∂x ∂ ∂ q = f ′( x 2 − y) ( x 2 − y) + g ′( x 2 + y) ( x 2 + y) ∂y ∂y p = 2x f ′(x2 – y) + 2x g′(x2 + y)

p = f ′( x 2 − y )

and

or

...(2) ...(3)

(2) ⇒ (3) ⇒ q = – f ′(x2 – y) + 1 . g′(x2 + y) Differentiating (4) w.r.t. x, we get r = 2x f ″(x2 – y) . 2x + 2.1 f ′(x2 – y) + 2x g″(x2 + y) . 2x + 2.1 g′(x2 + r = 4x2 (f ″(x2 – y) + g″(x2 + y)) + 2 (f ′(x2 – y) + g′(x2 + y)) Differentiating (5) w.r.t. y, we get t = – f ″(x2 – y) . (– 1) + g″(x2 + y) . 1 or t = f ″(x2 – y) + g″(x2 + y) ∴

(6)

r = 4 x 2t + 2

⇒

⇒

x

∂ 2z 2

= 4x 3

∂ 2z 2

FG p IJ H 2x K +

y) ...(6)

...(7)

(Using (4) and (7))

∂z . ∂x

∂x ∂y (iii) We have x = f(z) + g(y). Differentiating (1) partially w.r.t. x and y, we get 1 = f ′(z) p + 0 ...(2) and 0 = f ′(z) q + g′(y) Differentiating (2) and (3) w.r.t. x, we get 0 = f ″(z) p . p + f ′(z) r ...(4) and 0 = f ″(z) p . q + f ′(z) s + 0 (4) ⇒ f ″(z) p2 = – f ′(z) r (5) ⇒ f ″(z) pq = – f ′(z) s Dividing, we get

...(4) ...(5)

...(1) ...(3) ...(5)

p r = or ps − qr = 0. q s

(iv) We have z = f(y + ax) + g(y + bx). Differentiating (1) partially w.r.t. x and y, we get p = f ′(y + ax) . a + g′(y + bx) . b and q = f ′(y + ax) . 1 + g′(y + bx) . 1 Differentiating (2) partially w.r.t. x and y, we get r = f ″(y + ax) a2 + g″(y + bx) b2 and s = f ″(y + ax) a . 1 + g″(y + bx) b . 1 Differentiating (3) w.r.t. y, we get t = f ″(y + ax) . 1 + g″(y + bx) . 1 (4) ⇒ a2 f ″(y + ax) + b2 g″(y + bx) – r = 0 (5) ⇒ a f ″(y + ax) + b g″(y + bx) – s = 0 (6) ⇒ f ″(y + ax) + g″(y + bx) – t = 0

...(1) ...(2) ...(3) ...(4) ...(5) ...(6)

11

PARTIAL DIFFERENTIAL EQUATIONS

Eliminating f ″(y + ax) and g″(y + bx) from these three equations, we get

(a2

b 2)

a2 a

b2 b

1

1

r s =0 t

⇒ (a – b) r – – s+ – ab2) t = 0 ⇒ r – (a + b)s + abt = 0. Example 3. The equation of any cone with vertex at P(x0 , y0 , z0) is of the form f

(a2b

FG x − x Hz−z

y − y0 = 0. z − z0

,

y − y0 = 0. z − z0

0

Find the differential equation. f

Sol. We have

Fx−x GH z − z

0

IJ K

,

0

0

I JK

...(1)

x − x0 y − y0 and v = . z − z0 z − z0 ∴ (1) ⇒ f(u, v) = 0 Differentiating (2) partially w.r.t. x, we get u=

Let

F GH

I JK

F GH

I JK

x − x0 ∂z y − y0 ∂z ∂f 1 − 0 ∂f − + − =0 2 ∂u z − z0 ( z − z0 ) ∂x ∂v ( z − z0 ) 2 ∂x

F GH

I JK

F GH

...(2)

I JK

...(3)

I JK

...(4)

x − x0 y − y0 ∂f 1 ∂f + −p =0 −p 2 ∂u z − z0 ∂ v ( z − z0 ) ( z − z0 ) 2

⇒

Differentiating (2) partially w.r.t. y, we get

F GH

I JK

F GH

I JK

x − x0 ∂z y − y0 ∂z ∂f ∂f 1 − 0 − + − =0 2 ∂u ∂v z − z0 ( z − z0 ) 2 ∂y ( z − z0 ) ∂y

F GH

I JK

F GH

⇒

x − x0 y − y0 ∂f ∂f 1 + −q =0 −q 2 ∂u ∂v z − z 0 ( z − z0 ) ( z − z0 )2

Eliminating

∂f ∂f and from (3) and (4), we get ∂u ∂v x − x0 1 −p z − z0 ( z − z 0 )2 x − x0 −q ( z − z0 )2

⇒

z − z0 − p ( x − x 0 ) − q (x − x0 )

−p

y − y0

( z − z0 )2 =0 y − y0 1 −q z − z0 ( z − z0 )2 − p ( y − y0 ) z − z0 − q ( y − y0 )

=0

⇒

[ z − z0 − p ( x − x0 )] [ z − z0 − q ( y − y0 )] − pq ( x − x0 ) ( y − y0 ) = 0

⇒

( z − z0 ) 2 − p ( x − x0 ) ( z − z0 ) − ( z − z0 ) q ( y − y0 ) = 0

⇒

p(x – x0) + q(y – y0) = z – z0.

12

PARTIAL DIFFERENTIAL EQUATIONS

WORKING RULES FOR SOLVING PROBLEMS Rule I.

For a given relation involving variables and arbitrary functions, the relation is differentiated partially w.r.t. independent variables and arbitrary functions are eliminated to get the corresponding partial differential equation. Rule II. If the number of arbitrary functions is less than the number of independent variables, then the elimination of arbitrary functions shall give rise to a differential equation of order one. Rule III. If the number of arbitrary functions is equal to the number of independent variables, then the elimination of arbitrary functions shall give rise to a differential equation of order usually greater than one.

TEST YOUR KNOWLEDGE Find partial differential equation by eliminating arbitrary functions from the following relations : 2. z = f(x2 – y2)

1. z = f(x + ky) 3.

f(x2

+

y2

+

z2)

=x+y+z

4. z = x + y + f(xy)

5. z = xy + f(x2 + y2)

6. z = f(xy/z)

7. f(x + y + z) = xyz

8. z = (x + y) f(x2 – y2)

9. z = f(x) +

ey

g(x)

10. z = f(xy) + g(x + y) 12. f(x + y + z, x2 + y2 – z2) = 0

11.

z = f(xy) + g(x/y)

13.

z = f(x cos α + y sin α – at) + g(x cos α + y sin α + at).

Answers 1. q = kp

2. yp + xq = 0

5. py – qx =

y2

–

x2

3. (y – z) p + (z – x) q = x – y

6. px – qy = 0

7. x(y – z)p + y(z – x)q = z(x – y) 10. x(y – x) r – (y2 – x2) s + y(y – x)t + (p – q) (x + y) = 0 12. p(y + z) – (x + z)q = x – y

8. yp + xq = z 9. t – q = 0 2 2 11. x r – y t + xp – yq = 0 13.

∂2 z

∂x 2

+

∂2 z

∂y 2

=

1 ∂2 z

a 2 ∂t2

.

Hint 12.

u = x + y + z, v =

x2

+

y2

Diff. w.r.t. x, we get ⇒ Similarly, Eliminating

–

z2

FG H

4. px – qy = x – y

⇒

f(u, v) = 0

IJ K

FG H

IJ K

∂f ∂u ∂u ∂z ∂f ∂v ∂v ∂z + + + =0 ∂u ∂x ∂z ∂x ∂v ∂x ∂z ∂x

∂f ∂f (1 + 1 . p) + (2 x − 2 zp) = 0 ∂u ∂v ∂f ∂f (1 + 1 . q) + (2 y − 2 zq) = 0 ∂u ∂v ∂f ∂f , we get 1 + p 2 x − 2 zp = 0 . , 1 + q 2 y − 2 zq ∂u ∂v

Partial Differential Equations of the First Order

2

(Equations Linear in p and q)

2.1. INTRODUCTION In the last chapter, we studied the methods of forming partial differential equations. The next step is to solve partial differential equations. Solving a partial differential equation means to find a function which satisfies the given partial differential equation. A function satisfying a partial differential equation is called its solution (or integral). In the present chapter, we shall confine ourselves to the solution of partial differential equations of first order and at the same time linear in p and q. 2.2. SOLUTION OF A PARTIAL DIFFERENTIAL EQUATION A solution of a partial differential equation is a relation between the variables by means of which the partial derivatives are derived there from the given partial differential equation is satisfied. A solution of a partial differential equation is also called an integral of the equation. In the context of partial differential equations of first order, there are four types of solutions. These are : (i) Complete solution (ii) Particular solution (iii) Singular solution (iv) General solution. 2.3. COMPLETE SOLUTION Let z be a function of two independent variables x and y defined by f(x, y, z, a, b) = 0 where a and b are arbitrary constants. Differentiating (1) partially w.r.t. x and y, we get ∂f ∂f +p = 0 ...(2) ∂x ∂z

and

∂f ∂f +q =0 ∂y ∂z

...(1)

...(3)

Eliminating a and b from (1), (2) and (3), we get an equation of the form g(x, y, z, p, q) = 0. This is a partial differential equation of first order. 13

14

PARTIAL DIFFERENTIAL EQUATIONS

In (1), the number of arbitrary constants is two which is equal to the number of independent variables in g(x, y, z, p, q) = 0. The function f(x, y, z, a, b) = 0 is called the complete solution of the equation g(x, y, z, p, q) = 0. For example z = (x + a)(y + b) is a complete solution of the equation pq = z. 2.4. PARTICULAR SOLUTION A solution obtained by giving some particular values to the arbitrary constants in the complete solution of a partial differential equation of first order is called a particular solution of the concerned equation. For example z = (x + 1)(y + 4) is a particular solution of the equation pq = z. 2.5. SINGULAR SOLUTION Let f(x, y, z, a, b) = 0 be the complete solution of a partial differential equation g(x, y, z, p, q) = 0. The relation between x, y and z obtained by eliminating the arbitrary constants a and b ∂f ∂f = 0, = 0 is called the singular solution of the between the equations f ( x, y, z, a, b) = 0, ∂a ∂b equation g(x, y, z, p, q) = 0, provided it satisfies this equation. This solution represents the envelope of the surfaces represented by the complete solution of the given partial differential equation. The singular solution may or may not be contained in the complete solution of the equation. For example, z = ax + by – (a2 + b2) is the complete solution of the partial differential equation z = px + qy – (p2 + q2). Let f(x, y, z, a, b) = z – ax – by + a2 + b2. ∂f ∂f = − x + 2a, = − y + 2b ∂a ∂b Eliminating a and b from the equations,

∴

z – ax – by + a2 + b2 = 0, – x + 2a = 0,

– y + 2b = 0, we get z =

x 2 + y2 . 4

This also satisfy the given equation. ∴

z=

x 2 + y 2 is the singular solution of the equation z = px + qy – (p2 + q2). 4

2.6. GENERAL SOLUTION Let f(x, y, z, a, b) = 0 be the complete solution of a partial differential equation g(x, y, z, p, q) = 0. Let b = φ(a). ∴ f(x, y, z, a, φ(a)) = 0 is a one-parameter family of the surfaces of g(x, y, z, p, q) = 0. The relation between x, y and z obtained by eliminating the arbitrary constant a between the ∂f equations f ( x, y, z, a, φ(a)) = 0 and = 0 is called the general solution of the equation ∂a g(x, y, z, p, q) = 0, provided it satisfies this equation. This solution represents the envelope of the surfaces represented by the equation f(x, y, z, a, φ(a)) = 0.

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

15

If b = φ(a), where φ is an arbitrary function, then the elimination of a between the ∂f equations f ( x , y , z , a , φ( a )) = 0, and = 0 is not possible. Thus the general solution of the equa∂a ∂f tion g(x, y, z, p, q) = 0 is written as the set of equations f ( x , y , z , a , φ( a )) = 0, = 0 , where φ is ∂a any arbitrary function. We know that if u and v be independent functions of x, y, z and f(u, v) = 0

...(1)

be an arbitrary function of u and v, then Pp + Qq = R

...(2)

where

P=

∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v − , Q= − , R= − . ∂y ∂z ∂z ∂y ∂z ∂x ∂x ∂z ∂x ∂y ∂y ∂x

(2) is a partial differential equation of first order. Thus f(u, v) = 0 is a solution of the equation Pp + Qq = R. Since f(u, v) = 0 contains an arbitrary function ‘f ’, it is the general solution of the equation (2). 2.7. LAGRANGE LINEAR EQUATION We know that a partial differential equation of first order involves only the first order partial derivatives of the dependent variable (z) w.r.t. the independent variables (x and y). Thus an equation of first order involves x, y, z, p, q and may also involve powers of partial derivatives p and q. In particular, a partial differential equation of first order and at the same time linear in p and q is of the form Pp + Qq = R where P, Q, R are functions of x, y, z. This type of a partial differential equation is called a Lagrange linear equation. For example 4xp + 6y2q = x2 + y2 + z2 is a Lagrange linear equation. 2.8. SOLUTION OF LAGRANGE LINEAR EQUATION Let

Pp + Qq = R

...(1)

be a Lagrange linear equation where P, Q, R are functions of the dependent variable z and independent variables x and y. The system of equations

dx dy dz = = P Q R

...(2)

is called the Lagrange system of ordinary differential equations for the equation (1). Let u = C1 and v = C2 be two independent solutions of the equations (2). Let f(u, v) = 0, be an arbitrary function of u and v. Differentiating (3) partially w.r.t. x and y, we get

IJ FG IJ FG H K H K ∂f F ∂u ∂u I ∂f F ∂v ∂v I G + q J + G + qJ = 0 ∂u H ∂y ∂z K ∂v H ∂y ∂z K

∂f ∂u ∂u ∂f ∂v ∂v + + p + p =0 ∂u ∂x ∂z ∂v ∂x ∂z

and

...(3)

16

PARTIAL DIFFERENTIAL EQUATIONS

Eliminating

⇒ ⇒ ⇒

∂f ∂f , we get and ∂u ∂v

∂u ∂u +p ∂x ∂z ∂u ∂u +q ∂y ∂z

∂v ∂v +p ∂x ∂z = 0 ∂v ∂v +q ∂y ∂z

FG ∂u + p ∂uIJ FG ∂v + q ∂vIJ − FG ∂v + p ∂vIJ FG ∂u + q ∂uIJ = 0 H ∂x ∂z K H ∂y ∂z K H ∂x ∂z K H ∂y ∂z K FG ∂u ∂v − ∂u ∂vIJ p + FG ∂u ∂v − ∂u ∂vIJ q + ∂u ∂v − ∂u ∂v = 0 H ∂z ∂y ∂y ∂z K H ∂x ∂z ∂z ∂x K ∂x ∂y ∂y ∂x FG ∂u ∂v − ∂u ∂vIJ p + FG ∂u ∂v − ∂u ∂vIJ q = ∂u ∂v − ∂u ∂v H ∂y ∂z ∂z ∂y K H ∂z ∂x ∂x ∂z K ∂x ∂y ∂y ∂x

...(4)

∴ (3) is a solution of the equation (4).

Taking differentials of u = C1 and v = C2, we get

and

∂u ∂u ∂u dx + dy + dz = 0 ∂x ∂y ∂z

...(5)

∂v ∂v ∂v dx + dy + dz = 0 ∂x ∂y ∂z

...(6)

Since u and v are independent functions, we have dx ∂u ∂v ∂u ∂v − ∂y ∂z ∂z ∂y

=

dy ∂u ∂v ∂u ∂v − ∂z ∂x ∂x ∂z

=

dz

...(7)

∂u ∂v ∂u ∂v − ∂x ∂y ∂y ∂x

Using (2) and (7), we have ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v − − − ∂y ∂z ∂z ∂y ∂x ∂y ∂y ∂x ∂ z ∂ x ∂ x ∂ z = = = λ (say ) P Q R

∴

∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v − = λP, − = λQ, − = λR . ∂y ∂z ∂z ∂y ∂z ∂x ∂x ∂z ∂x ∂y ∂y ∂x

∴ Putting these values in (4), we get λPp + λQq = λR

or

Pp + Qq = R,

which is the given partial differential equation. ∴ If u = C1 and v = C2 be two independent solutions of the system of differential dx dy dz = = equations , then any arbitrary function f(u, v) of u and v is a solution of the P Q R Lagrange linear equation Pp + Qq = R. The solution f(u, v) = 0 is the general solution of the equation Pp + Qq = R. In particular, for arbitrary constants a and b, the solution u = av + b is a complete solution of the equation Pp + Qq = R. Remark. The Lagrange system of ordinary differential equations

dx dy dz = = for the partial P Q R

differential equation Pp + Qq = R is also known as the auxiliary system of equations or simply as the auxiliary equations of the equation Pp + Qq = R.

17

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

WORKING STEPS FOR SOLVING Pp + Qq = R Step I.

Indentify the functions P, Q and R.

dx dy dz = = . P Q R Step III. Find two independent solutions u = C1 and v = C2 of the system given in step II. Step IV. Write f(u, v) = 0 and call it the general solution of the given equation. Step II. Form the system :

Type I. In this type, we shall consider the solution of the equation Pp + Qq = R for dx dy dz = = gives an equation in which the equality of two factors of the auxiliary equations P Q R the variables whose differentials are involved. Two independent solutions of the auxiliary equations are calculated in this manner.

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following Lagrange linear equations : (i) 2p + 5q = 1

(ii) y2p + x2q = x2y2z2

(iv) zp = x (v) (x2 + 2y2) p – xyq = xz. Sol. (i) We have 2p + 5q = 1. Here P = 2, Q = 5, R = 1 ∴ Auxiliary equations are i.e.,

(iii)

y2z p + xzq = y 2 x

dx dy dz . = = P Q R

dx dy dz = = 2 5 1 Taking the first two fractions of (1), we get 5dx – 2dy = 0

...(1)

...(2) ...(3) Integrating (2), we get 5x – 2y = C1 Taking the last two fractions of (1), we get dy – 5dz = 0 ...(4) Integrating (4), we have y – 5z = C2 ...(5) From (3) and (5), the general solution of the given equation is f(5x – 2y, y – 5z) = 0, where f is any arbitrary function. (ii) We have y2p + x2q = x2y2z2. Here P = y2, Q = x2, R = x2y2z2. ∴ Auxiliary equations are i.e.,

dx dy dz = = . P Q R dx y

2

=

dy x

2

=

dz 2 2 2

x y z

Taking the first two fractions of (1), we get x2dx – y2dy = 0 or 3x2dx – 3y2dy = 0

...(1)

...(2)

18

PARTIAL DIFFERENTIAL EQUATIONS

Integrating (2), we get x3 – y3 = C1 ...(3) 2 –2 Taking the last two fractions of (1), we get 3y dy – 3z dz = 0 ...(4) 3 –1 Integrating (4), we get y + 3z = C2 ...(5) From (3) and (5), the general solution of the given equation is f(x3 – y3, y3 + 3z–1) = 0, where f is any arbitrary function. (iii) We have

y2 z p + xzq = y 2 . x

y2z , Q = xz , R = y 2 x dx dy dz ∴ Auxiliary equations are . = = P Q R P=

Here

xdx

dy dz ...(1) = y z xz y 2 Taking the first two fractions of (1), we get x2dx = y2dy or 3x2dx – 3y2dy = 0 ...(2) Integrating (2), we have x3 – y3 = C1 ...(3) Taking the first and last fractions of (1), we get xdx = zdz or 2xdx – 2zdz = 0 ...(4) 2 2 Integrating (4), we have x – z = C2 ...(5) From (3) and (5), the general solution of the given equation is f(x3 – y3, x2 – z2) = 0, where f is any arbitrary function. (iv) We have zp = x. ∴ zp + 0.q = x Here P = z, Q = 0, R = x.

i.e.,

2

∴ Auxiliary equations are i.e.,

=

dx dy dz . = = P Q R

dx dy dz = = z 0 x Second fraction of (1) implies dy = 0

...(1)

∴ y = C1 ...(2) Taking the first and third fractions of (1), we get xdx = zdz or 2xdx – 2zdz = 0 ...(3) 2 2 Integrating (3), we have x – z = C2 ...(4) 2 2 From (2) and (4), the general solution of the given equation is f(y, x – z ) = 0, where f is any arbitrary function. (v) We have (x2 + 2y2) p – xyq = xz. Here P = x2 + 2y2, Q = – xy, R = xz. ∴ Auxiliary equations are i.e.,

dx dy dz = = . P Q R

dx x 2 + 2y 2

=

dy dz = − xy xz

...(1)

19

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Taking the first two fractions of (1), we get

x + 2y

dx x 2 + 2 y 2 = dy − xy

⇒ 2x

⇒

(2)

⇒

⇒

2

2x

FG IJ H K

=

dy . − xy

dx 2x2 =− − 4y dy y

dx 2 + x2 = − 4y dy y

Let z = x2. ∴

dx 2

...(2)

FG IJ H K

dz 2 +z = − 4y dy y

This is a linear differential equation of order one.

∴

I.F. = e

z

zy 2 =

( − 4 y ) y 2 dy + C1

z

2 dy y

= e 2 log y = y 2

Taking the last two fractions of (1), we get

or x2y2 + y 4 = C1

dy dz + =0 y z

...(3) ...(4)

Integrating (4), we get

log y + log z = log C2 or yz = C2 ...(5) 2 2 4 From (3) and (5), the general solution of the given equation is f(x y + y , yz) = 0, where f is any arbitrary function. Type II. In this type, we shall consider the solution of the equation Pp + Qq = R for

dx dy dz gives an equation in = = P Q R the variables whose differentials are involved. Another independent solution of the auxiliary equations is found by using the first solution. Example 2. Find the general solution of the following Lagrange linear equations : (i) p + 2q = 5z + tan (y – 2x) (ii) yp + xq = xyz2 (x2 – y2) 2 2 4 (iii) xz(z + xy) p – yz(z + xy) q = x (iv) z (p – q) = z2 + (x + y)2. Sol. (i) We have p + 2q = 5z + tan (y – 2x). which the equality of two factors of the auxiliary equations

dx dy dz . = = 1 2 5z + tan ( y − 2x )

...(1)

Taking the first two fractions of (1), we get dy – 2dx = 0 Integrating (2), we have y – 2x = C1 Taking the last two fractions of (1) and using (3), we have

...(2) ...(3)

∴ Auxiliary equations are

dy dz − =0 2 5z + tan C1 1 1 y − log|5z + tan C1 |= C2 2 5 5y – 2 log | 5z + tan (y – 2x) | = 10C2

...(4)

Integrating (4), we have or

...(5)

20

PARTIAL DIFFERENTIAL EQUATIONS

From (3) and (5), the general solution of the given equation is f(y – 2x, 5y – 2 log | 5z + tan (y – 2x) |) = 0, where f is any arbitrary function. (ii) We have yp + xq = xyz2 (x2 – y2). ∴ Auxiliary equations are

dx dy dz = = . 2 y x xyz ( x 2 − y 2 )

...(1)

Taking the first two fractions of (1), we get 2xdx – 2ydy = 0 Integrating (2), we have x2 – y2 = C1

...(2) ...(3)

Taking the last two fractions of (1) and using (3), we have ydy − Integrating (4), we have

y2 1 − 2 C1

Fz I =C GH − 1 JK

dz C1 z 2

=0

−1

2

y2 1 + = C2 2 2 z( x − y 2 )

⇒

...(4)

...(5)

From (3) and (5), the general solution of the given equation is

F GH

f x2 − y2 ,

I JK

y2 1 + = 0 , where f is any arbitrary function. 2 z(x 2 − y 2 )

(iii) We have xz(z2 + xy) p – yz(z2 + xy) q = x4. ∴ Auxiliary equations are

dx xz ( z 2 + xy )

=

Taking the first two fractions of (1), we get Integrating (2), we have

dy − yz ( z 2 + xy )

=

dz x4

.

dx dy + =0 x y

log x + log y = log C1 or xy = C1

...(1) ...(2) ...(3)

dx dz = 3 Taking the first and third fractions of (1) and using (3), we get 2 z( z + C 1 ) x

x3dx – (z3 + C1z) dz = 0

or Integrating (4), we have

F GH

I JK

...(4)

x4 z 4 C1 z 2 − + = C2 4 4 2

x4 – z4 – 2xyz2 = 4C2 ...(5) 4 4 2 From (3) and (5), the general solution of the given equation is f(xy, x – z – 2xyz ) = 0, where f is any arbitrary function. (iv) We have z(p – q) = z2 + (x + y)2. ⇒ zp – zq = z2 + (x + y)2.

or

∴ Auxiliary equations are

dx dy dz . = = 2 z − z z + ( x + y )2

Taking the first two fractions of (1), we have dx + dy = 0 Integrating, we have x + y = C1

...(1) ...(2) ...(3)

21

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Taking the last two fractions of (1) and using (3), we have dy +

zdz 2

z + C1

2

=0

1 1 log| z 2 + C 12 |= C 2 or y + log ( z 2 + ( x + y ) 2 ) = C 2 2 2 From (3) and (5), the general solution of the given equation is

Integrating (4), we get y +

FG H

f x + y, y +

e

1 log z2 + (x + y)2 2

...(4) ...(5)

jIJK = 0 ,

where f is any arbitrary function. Type III. In this type, we shall consider the solution of the equation Pp + Qq = R by using the formula : dx dy dz P1dx + Q1dy + R1dz = = = , P Q R P1P + Q1Q + R1 R

where P1, Q1, R1 are some functions of x, y and z. If for some choice of P1, Q1, R1, the sum P1P + Q1Q + R1R is zero, then we have P1dx + Q1dy + R1dz = 0. We integrate this equation to get one solution of the auxiliary equations. P1, Q1, R1 are called multipliers. By using different set of multipliers or by using two fractions of the auxiliary equations, we find another independent solution of the auxiliary equations. Example 3. Find the general solution of the following Lagrange linear equations : (i) x(y2 – z2)p + y(z2 – x2)q = z(x2 – y2) (ii) x(y2 + z) p – y(x2 + z) q = z(x2 – y2) (iii) (y2 + z2)p – xyq + xz = 0 (iv) (4y – 3z)p + (4x – 2z)q = 2y – 3x. 2 2 2 2 Sol. (i) We have x(y – z ) p + y(z – x ) q = z(x2 – y2). dx

∴ Auxiliary equations are Taking

2

2

x( y − z )

=

dy 2

2

y( z − x )

=

dz

...(1)

2

z( x − y 2 )

1 1 1 , , as multipliers, each fraction of (1) x y z 1 1 1 1 1 1 dx + dy + dz dx + dy + dz x y z x y z = = 2 2 2 2 2 2 0 ( y − z ) + (z − x ) + (x − y )

1 1 1 dx + dy + dz = 0 x y z Integrating, we get log |x| + log |y| + log |z| = log C1 |xyz| = C1 or xyz = ± C1 Taking x, y, z as multipliers, each fraction of (1) ∴

or

=

xdx + ydy + zdz 2

2

2

2

2

2

2

2

2

x ( y − z ) + y (z − x ) + z (x − y )

=

...(2)

xdx + ydy + zdz 0

∴ xdx + ydy + zdz = 0 or 2xdx + 2ydy + 2zdz = 0 Integrating, we get x2 + y2 + z2 = C2 ...(3) From (2) and (3), the general solution of the given equation is f(xyz, x2 + y2 + z2) = 0, where f is any arbitrary function.

22

PARTIAL DIFFERENTIAL EQUATIONS

(ii) We have

x(y2 + z) p – y(x2 + z) q = z(x2 – y2).

∴ Auxiliary equations are

dx dy dz = = x( y 2 + z) − y( x 2 + z) z( x 2 − y 2 )

...(1)

Taking x, y, – 1 as multipliers, each fraction of (1) =

xdx + ydy + (−1) dz xdx + ydy − dz = 0 x 2 ( y 2 + z) − y 2 ( x 2 + z) − z( x 2 − y 2 )

∴ 2xdx + 2ydy – 2dz = 0 Integrating, we get x2 + y2 – 2z = C1 Taking

...(2)

1 1 1 , , as multiplier, each fraction of (1) x y z 1 1 1 1 1 1 dx + dy + dz dx + dy + dz x y z x y z = = 2 0 y + z − x2 − z + x2 − y2

∴

1 1 1 dx + dy + dz = 0 x y z

Integrating, we get log |x| + log |y| + log |z| = log C2 or |xyz| = C2 or xyz = ± C2 ...(3) 2 2 From (2) and (3), the general solution of the given equation is f(x + y – 2z, xyz) = 0, where f is any arbitrary function. (iii) We have (y2 + z2) p – xyq = – xz. ∴ Auxiliary equations are

dx y2 + z2

=

dy dz = − xy − xz

Taking the last two fractions of (1), we get

...(1)

dy dz − =0 y z

Integrating, we have log |y| – log |z| = log C1 or

y = C1 or z

y = ± C1 z

...(2)

Taking x, y, z as multipliers, each fraction of (1) =

xdx + ydy + zdz 2

2

2

xy + xz − xy − xz

2

=

xdx + ydy + zdz 0

∴ 2xdx + 2ydy + 2zdz = 0 Integrating, we have x2 + y2 + z2 = C2 ...(3) 2 2 2 From (2) and (3), the general solution of the given equation is f(y/z, x + y + z ) = 0, where f is any arbitrary function. (iv) We have (4y – 3z) p + (4x – 2z) q = 2y – 3x. ∴ Auxiliary equations are

dx dy dz = = 4 y − 3z 4 x − 2z 2 y − 3 x

...(1)

23

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Taking a, b, c as multipliers, each fraction of (1)

adx + bdy + cdz a( 4 y − 3z ) + b( 4 x − 2z ) + c( 2 y − 3x ) Let a(4y – 3z) + b(4x – 2z) + c(2y – 3x) = 0 (2) ⇒ (4b – 3c) x + (4a + 2c) y + (– 3a – 2b) z = 0 Let 4b – 3c = 0, 4a + 2c = 0, – 3a – 2b = 0 ⇒ b : c = 3 : 4, a : c = 1 : – 2, a : b = – 2 : 3 ⇒ a:b:c=–2:3:4 ∴ (2) is true for a = – 2, b = 3, c = 4 ∴ Each fraction of (1) =

...(2)

− 2dx + 3dy + 4dz − 2dx + 3dy + 4dz = − 2( 4 y − 3z ) + 3( 4 x − 2z ) + 4( 2 y − 3x ) 0 ∴ – 2dx + 3dy + 4dz = 0 ...(3) Integrating, we have – 2x + 3y + 4z = C1 Also, (2) ⇒ 4(ay + bx) – 3(az + cx) + 2(cy – bz) = 0 Let ay + bx = 0, az + cx = 0, cy – bz = 0 ∴ a : b = – x : y, a : c = – x : z, b : c = y : z ∴ a:b:c=–x:y:z ∴ (2) is true for a = – x, b = y, c = z. ∴ Each fraction of (1) =

− xdx + ydy + zdz − xdx + ydy + zdz = − x ( 4 y − 3 z ) + y( 4 x − 2 z ) + z( 2 y − 3 x ) 0 ∴ – 2xdx + 2ydy + 2zdz = 0 ...(4) Integrating, we get – x2 + y2 + z2 = C2 From (3) and (4), the general solution of the given equation is f(– 2x + 3y + 4z, – x2 + y2 + z2) = 0, where f is any arbitrary function. Type IV. In this type, we shall consider the solution of the equation Pp + Qq = R by using the formula : =

dx dy dz P1dx + Q1dy + R1dz = = = , P Q R P1P + Q1Q + R1 R

where P1, Q1, R1 are some functions of x, y and z. If for some choice of P1, Q1, R1, the sum P1dx + Q1dy + R1dz is exact differential of a factor of P1P + Q1Q + R1R, then the quotient P1dx + Q1dy + R1dz is equated with a suitable fraction of auxiliary equations to get one solution P1P + Q1Q + R1 R of the auxiliary equations. By using different set of multipliers or by using two fractions of the auxiliary equations, we find another independent solution of the auxiliary equations.

Example 4. Find the general solution of the following Lagrange linear equations : (i) (x2 – yz) p + (y2 – zx) q = z2 – xy (iii) p cos (x + y) + q sin (x + y) = z +

(ii) (y + z) p + (z + x) q = x + y 1 z

(iv) (y2 + yz + z2) p + (z2 + zx + x2) q = x2 + xy + y2.

24

PARTIAL DIFFERENTIAL EQUATIONS

Sol. (i) We have (x2 – yz) p + (y2 – zx) q = z2 – xy. dx

∴ Auxiliary equations are

2

x − yz

=

dy 2

y − zx

=

dz

...(1)

2

z − xy

Taking 1, 1, 1 and x, y, z as multipliers, each fraction of (1) = ⇒ ⇒ ⇒

1. dx + 1. dy + 1. dz 2

2

2

x − yz + y − zx + z − xy

=

xdx + ydy + zdz 3

x − xyz + y 3 − xyz + z 3 − xyz

dx + dy + dz xdx + ydy + zdz = 2 x + y + z − yz − zx − xy ( x + y + z)( x 2 + y 2 + z 2 − yz − zx − xy) 2

2

( x + y + z ) d( x + y + z ) =

1 ( 2xdx + 2 ydy + 2zdz ) 2

1 d( x 2 + y 2 + z 2 ) = 0 2 1 1 ( x + y + z) 2 − ( x 2 + y 2 + z 2 ) = C 1 2 2 xy + yz + zx = C1

( x + y + z ) d( x + y + z ) −

Integrating, we get or

...(2)

Taking 1, – 1, 0 and 0, 1, – 1 as multipliers, each fraction of (1) =

dx − dy + 0 0 + dy − dz = ( x 2 − yz) − ( y 2 − zx) + 0 0 + ( y 2 − zx) − ( z 2 − xy) dx − dy dy − dz = 2 2 x − y + z( x − y) y − z 2 + x( y − z)

⇒ ⇒

2

dx − dy dy − dz = ( x − y) ( x + y + z) ( y − z) ( y + z + x)

⇒

d( x − y) d( y − z) − =0 x− y y− z

Integrating, we get log | x – y | – log | y – z | = log C2

x− y = C2 or y− z

or

x− y = ± C2 y−z

...(3)

FG H

From (2) and (3), the general solution of the given equation is f xy + yz + zx, where f is any arbitrary function. (ii) We have

IJ K

x−y = 0, y−z

(y + z) p + (z + x) q = x + y.

∴ Auxiliary equations are

dx dy dz = = y+z z+x x+y

...(1)

Taking 1, – 1, 0 and 0, 1, – 1 as multipliers, each fraction of (1) = ⇒

dx − dy + 0 0 + dy − dz = ( y + z) − (z + x ) + 0 0 + (z + x ) − (x + y)

dx − dy dy − dz = − (x − y) − ( y − z)

⇒

d( x − y) d( y − z) − =0 x− y y−z

...(2)

25

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Integrating, we get log | x – y | – log | y – z | = log C1

x− y = C1 or y−z

⇒

x− y = ± C1 y−z

...(3)

Taking 1, – 1, 0 and 1, 1, 1 as multipliers, each fraction of (1) and (2)

∴

dx − dy + 0 dx + dy + dz = ( y + z) − ( z + x) + 0 ( y + z) + ( z + x) + ( x + y) d( x − y ) d( x + y + z ) or 2 d( x − y) + d( x + y + z) = 0 = x− y x+ y+ z − ( x − y ) 2( x + y + z )

Integrating, we get 2 log | x – y | + log | x + y + z | = log C2 ⇒ (x – y)2 | x + y + z | = C2 ⇒ (x – y)2 (x + y + z) = ± C2 From (3) and (4), the general solution of the given equation is

FG x − y , (x − y) Hy−z

f

2

IJ K

(x + y + z) = 0 ,

where f is any arbitrary function.

(iii) We have p cos (x + y) + q sin (x + y) = z +

1 . z

dx dy dz = = 1 cos ( x + y ) sin ( x + y ) z+ z Taking 1, 1, 0 and 1, – 1, 0 as multipliers, each fraction of (1)

∴ Auxiliary equations are

dx + dy + 0 dx − dy + 0 = cos ( x + y ) + sin ( x + y ) + 0 cos ( x + y ) − sin ( x + y ) + 0

= (1) and (2) ⇒

1

⇒

Integrating, we get

Also, ⇒

2z

.

2

2 z +1 1 2

FG H

dt 2 sin (t + π/4)

dz − cos ec t +

IJ K

π dt = 0 4

log|z 2 + 1|− log tan

FG H

(2)

⇒

...(2)

, where t = x + y.

FG H

IJ K

π 1 t+ = log C1 2 4

( z 2 + 1)1/ 2 = C1 x+y π tan + 2 8

⇒

...(1)

zdz d( x + y) = z 2 + 1 cos ( x + y) + sin ( x + y) 1 2 zdz dt = = . 2 2 z + 1 cos t + sin t

⇒

...(4)

IJ K

cos ( x + y ) − sin ( x + y ) d( x + y ) = d( x − y ) cos ( x + y ) + sin ( x + y )

cos t − sin t dt − d( x − y) = 0, where t = x + y. cos t + sin t

...(3)

26

PARTIAL DIFFERENTIAL EQUATIONS

Integrating, we get ⇒

log| cos t + sin t | – (x – y) = log C2 | cos t + sin t | ey–x = C2

bcos (x + y) + sin ( x + y)g e

⇒

y− x

...(4)

= ± C2

From (3) and (4), the general solution of the given equation is

F G (z + 1) fG , bcos (x + y) + sin (x + y)g e GG tan FGH x + y + π IJK H 2 8 2

1/ 2

y−x

I JJ = 0, JJ K

where f is any arbitrary function. (iv) We have (y2 + yz + z2) p + (z2 + zx + x2) q = x2 + xy + y2. dx

∴ Auxiliary equations are

2

y + yz + z

2

=

dy 2

z + zx + x

2

=

dz 2

x + xy + y 2

...(1)

Taking 1, – 1, 0 and 0, 1, – 1 as multipliers, each fraction of (1) =

dx − dy + 0 0 + dy − dz = 2 2 2 2 ( y + yz + z ) − ( z + zx + x ) + 0 0 + ( z + zx + x 2 ) − ( x 2 + xy + y 2 ) 2

dx − dy

⇒

2

2

y − x + yz − zx

=

dy − dz 2

z − y 2 + zx − xy

⇒

dx − dy dy − dz = ( y − x ) ( y + x + z) (z − y) (z + y + x )

⇒

d( x − y ) d( y − z ) = − (x − y) − ( y − z)

Integrating, we get

⇒

d( y − z ) d( x − y ) − =0 y−z x−y

log | y – z | – log | x – y | = log C1

y−z y−z = C1 or = ± C1 x−y x−y

⇒

...(2)

Taking 1, – 1, 0 and 1, 0, – 1 as multipliers, each fraction of (1) =

dx − dy 2

2

2

2

( y + yz + z ) − ( z + zx + x )

=

dx − dz 2

( y + yz + z 2 ) − ( x 2 + xy + y 2 )

⇒

dx − dy dx − dz = y 2 − x 2 + yz − zx z 2 − x 2 + yz − xy

⇒

dx − dy dx − dz = ( y − x)( y + x + z) ( z − x)( z + x + y)

⇒

dx − dy dx − dz = − (x − y) − (x − z)

⇒

d( x − z ) d( x − y ) − =0 x−z x−y

Integrating, we get log | x – z | – log | x – y | = log C2. ⇒

x−z = C2 or x− y

x−z = ± C2 x− y

...(3)

27

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

From (2) and (3), the general solution of the given equation is f f is any arbitrary function.

 y  z , x  z   0, where  x  y x  y

TEST YOUR KNOWLEDGE Find the general solution of the following Lagrange linear equations: 1. (i) p + q = sin x (ii) ap + bq = c (iii) p tan x + q tan y = tan z

(iv) yzp + zxq = xy (vi) x2p + y2q = z2

(v) xp + yq = z

(viii) y2p – xyq = x(z – 2y)

(vii) (x – a) p + (y – b)q = z – c 2. (i) p – 2q =

3x2

sin (y + 2x)

(ii) p + 3q = z + cot (y – 3x) (iv) y3q – xy2p = axz

(iii) zp – zq = x + y (v) xyp + y2q + 2x2 – xyz = 0

(vi) (p – q) (x + y) = z

(vii) z(xy + z2)(px – qy) = x4

(viii) xzp + yzq = xy

3. (i) (z – y)p + (x – z)q = y – x

 1  1  p   1  1 q  1  1  z y  x z y x  b  c  yzp   c  a  zxq   a  b  xy (v)   a   b   c 

(iii)

(vii) x(y2 – z2)p – y(z2 + x2)q = z(x2 + y2) 4. (i)

(x2

–

y2

–

z2)p

+ 2xyq = 2xz

(ii) x(y – z)p + y(z – x)q = z(x – y) (iv) x2(y – z)p + y2(z – x)q = z2(x – y) (vi) z(x + y)p + z(x – y)q = x2 + y2 (viii) (x – y)p + (x + y)q = 2xz (ii) (1 + y)p + (1 + x)q = z

(iii) (x2 – y2 – yz)p + (x2 – y2 – zx)q = z(x – y) (iv) xzp + yzq = xy (v) x(x + y)p – y(x + y)q + (x – y)(2x + 2y + z) = 0

1

6 1

6

(vi) y( x  y )  az p  x ( x  y )  az q  z( x  y ) (vii) xp + zq + y = 0

(viii) (x2 + y2)p + 2xyq = (x + y)z.

Answers 1. (i) f(x – y, z + cos x) = 0

 sin x , sin x   0  sin y sin z   x y (v) f  ,  = 0  y z  x  a , y  b   0 (vii) f   y  b z  c (iii) f

2. (i) f(2x + y, x3 sin (2x + y) – z) = 0 (iii) f(x + y, 2(x + y)x – z2) = 0

(ii) f(bx – ay, cy – bz) = 0 (iv) f(x2 – y2, x2 – z2) = 0 (vi) f

 1  1 , 1  1   0  y x z y

(viii) f(x2 + y2, yz – y2) = 0

1  ax  (iv) f  xy, log| z|  0 3 y  

6

(ii) f y  3x , x  log| z  cot ( y  3x )|  0 2

28

PARTIAL DIFFERENTIAL EQUATIONS

(v) f

F x , x − log GH y

z−

2x y

I =0 JK

(vi) f(x + y, x – (x + y) log | z |) = 0

3. (i) f(x + y + z, x2 + y2 + z2) = 0

Fy ,x + y +z I =0 GH z JK z F x −y I JK = 0 (iii) f G z − x + y , z H (v) f b xy, ( x + y)( x + y + z)g = 0 F I =0 x (vii) f G y + z , JK H e 2

2

1 1 1 + + =0 x y z

IJ K

(vi) f(2xy – z2, x2 – y2 – z2) = 0

FH x + y + 2I F (ii) f G (1 + x ) − (1 + y ) , JK = 0 H z Fx I (iv) f G , xy − z J = 0 Hy K F x + y , x − y − 2azIJ = 0 (vi) f G H z K F y , x + yI = 0 . (viii) f G H x − y z JK 2

2

−1

( y/ x )

2

2

2

2

IJ K

2 2 − 2 tan (viii) f x + y − log| z |, ( x + y ) e

(vii) f(x2 + y2 + z2, x/yz) = 0 2

FG H

(iv) f xyz ,

(v) f(ax2 + by2 + cz2, a2x2 + b2y2 + c2z2) = 0

2

x =0 y

(ii) f(x + y + z, xyz) = 0

(iii) f(x + y + z, xyz) = 0

4. (i) f

FG H

(viii) f xy − z 2 ,

(vii) f(xy, x4 –2xyz2 – z4) = 0

2

2

2

tan −1 ( y / z )

2

2

Hints 3. (vi) Try x, – y, – z and y, x, – z as multipliers. (vii) Try x, y, z and 1/x, – 1/y, – 1/z as multipliers. 4. (i) Try x, y, z as multipliers. (ii) Try 1, 1, 0 as multipliers. (iii) Try 1, – 1, 0 and x, – y, 0 as multipliers. (iv) Try 1/x, 1/y, 0 as multipliers. (v) Try 1, 1, 0 and 1, 1, 1 as multipliers. (vi) Try 1, 1, 0 and x, – y, 0 as multipliers.

FG IJ H K FG IJ H K

y 1 dy + − 2 dz d( y/ z) dx 0 + zdy − ydz z z = = = (vii) We have 2 x z2 + y2 1 + ( y/ z)2 y 1+ z

∴

log| x | − tan −1

y = log C . z

(viii) Try 1, 1, 0 and 1, – 1, 0 as multipliers.

IK = 0

Partial Differential Equations of the First Order

3

(Equations Non-linear in p and q) 3.1. INTRODUCTION By now we have learnt the method of solving first order partial differential equations which are linear in partial derivatives p and q. A partial differential equation of first order need not be linear in p and q. In the present chapter, we shall study the methods of solving such equations. In the first part, we shall study the method of solving some special types of equations which can be solved easily by methods other than the general method. In the second part, we shall take up Charpit’s general method of solution. 3.2. SPECIAL TYPE I : EQUATIONS CONTAINING ONLY p AND q Let g(p, q) = 0 be a partial differential equation of first order and containing only p and q. Let z = ax + φ(a) y + c be the complete solution of (1), where φ(a) is some function of a.

...(1) ...(2)

∂z ∂z = a and q = = φ(a) ∂x ∂y ∴ (1) ⇒ g(a, φ(a)) = 0. ∴ Complete solution of (1) is z = ax + φ(a) y + c, where g(a, φ(a)) = 0 and a, c are arbitrary constants. To find the singular solution, let f(x, y, z, a, c) = z – ax – φ(a) y – c (2)

⇒

p=

∴ Using f(x, y, z, a, c) = 0,

∂f ∂f = 0, = 0, the singular solution is given by eliminat∂a ∂c

ing a and c from the equations : z – ax – φ(a) y – c = 0, – x – φ′(a) y = 0, – 1 = 0 This is impossible, because – 1 ≠ 0. ∴ There is no singular solution. To find the general solution, let c = ψ(a), where ψ is any arbitrary function. 29

30

PARTIAL DIFFERENTIAL EQUATIONS

∂f = 0, the general solution is given by ∂a z – ax – φ(a) y – ψ(a) = 0, – x – φ′(a) y – ψ′(a) = 0.

∴ Using f(x, y, z, a, ψ(a)) = 0,

ILLUSTRATIVE EXAMPLES Example 1. Solve the following partial differential equations : (i) p2 + q2 = λ2 (ii) p2 – q2 = k2 2 (iii) p = 2q + 1 (iv) p2 + 6p + 2q + 4 = 0. 2 2 Sol. (i) We have p + q = λ2. This equation is of the form g(p, q) = 0. Let z = ax + φ(a) y + c be the complete solution of (1), where φ(a) is some function of a. (2) ∴ (1)

⇒

p=

...(1) ...(2)

∂z ∂z = a and q = = φ(a) ∂x ∂y

2 2 2 2 2 ⇒ a + ( φ( a )) = λ or φ( a ) = ± λ − a

φ(a) = λ2 − a 2

Let

and − λ ≤ a ≤ λ.

∴ The complete solution is z = ax + λ2 − a 2 y + c , where a and c are arbitrary constants and – λ ≤ a ≤ λ. There is no singular solution. To find the general solution, let f(x, y, z, a, c) = z – ax – ∴ Using f(x, y, z, a, ψ(a)) = 0,

∂f = 0, the general solution is given by the equations ∂a

z − ax − λ2 − a 2 y − ψ(a) = 0,

−x+

a 2

λ − a2

y − ψ ′ (a) = 0,

where ψ is any arbitrary function. (ii) We have p2 – q2 = k2. This equation is of the form g(p, q) = 0. Let z = ax + φ(a) y + c be the complete solution of (1), where φ(a) is some function of a. (2)

⇒

p=

λ2 − a 2 y − c and c = ψ(a).

...(1) ...(2)

∂z ∂z = a and q = = φ(a) ∂x ∂y

∴ (1) ⇒ a 2 − (φ(a)) 2 = k 2 or φ( a) = ± a 2 − k 2 Let

φ(a) =

a 2 − k 2 and a2 ≥ k2.

∴ The complete solution is z = ax + a 2 − k 2 y + c , where a and c are arbitrary constants and a2 ≥ k2. There is no singular solution. To find the general solution, let

31

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

f ( x , y , z , a , c ) = z − ax − a 2 − k 2 y − c and c = ψ(a).

∴ Using f(x, y, z, a, ψ(a)) = 0,

∂f = 0, the general solution is given by the equations ∂a

z − ax − a 2 − k 2 y − ψ(a) = 0, − x −

a a 2 − k2

y − ψ ′ (a) = 0,

where ψ is any arbitrary function. (iii) We have p = 2q2 + 1. This equation is of the form g(p, q) = 0. Let z = ax + φ(a) y + c be the complete solution of (1), where φ(a) is some function of a.

∴

⇒

p=

(1)

⇒

a = 2 ( φ( a )) 2 + 1 or φ( a ) = ±

φ(a) =

...(2)

∂z ∂z = a and q = = φ(a) ∂x ∂y

(2)

Let

...(1)

a −1 . 2

a −1 and a ≥ 1. 2 a−1 y + c , where a and c are arbitrary 2

∴ The complete solution is z = ax +

constants and a ≥ 1. There is no singular solution. To find the general solution, let f ( x , y , z , a , c ) = z − ax −

∴ Using f(x, y, z, a, ψ(a)) = 0,

a −1 y − c and c = ψ( a ). 2

∂f = 0, the general solution is given by the equations: ∂a

a−1 1 y − ψ(a) = 0 , − x − y − ψ ′ (a) = 0 , 2 2 2 a−1 where ψ is any arbitrary function. (iv) We have p2 + 6p + 2q + 4 = 0. z − ax −

This equation is of the form g(p, q) = 0. Let z = ax + φ(a) y + c be the complete solution of (1) where φ(a) is some function of a (2)

∴ (1) ⇒

a2 + 6a + 2φ(a) + 4 = 0 or

∴ The complete solution is z = ax − constants.

...(2)

∂z ∂z = a and q = = φ(a) ∂x ∂y

p=

⇒

Fa GH 2

2

φ( a ) = −

I JK

...(1)

Fa GH 2

2

I JK

+ 3a + 2

+ 3a + 2 y + c , where a and c are arbitrary

32

PARTIAL DIFFERENTIAL EQUATIONS

There is no singular solution. To find the general solution, let

f ( x , y , z , a , c ) = z − ax + ∴ Using f ( x , y , z , a , ψ( a )) = 0,

z − ax +

Fa GH 2

2

Fa GH 2

2

I JK

+ 3a + 2 y − c and c = ψ( a ).

∂f = 0, the general solution is given by the equations : ∂a

I JK

+ 3a + 2 y − ψ(a) = 0, − x + (a + 3)y − ψ ′ (a) = 0,

where ψ is any arbitrary function.

WORKING STEPS FOR SOLVING g(p, q) = 0 Step I.

Take complete solution as z = ax + φ(a) y + c where a and c are arbitrary constants.

∂z ∂z = a, q = = φ( a ). ∂x ∂y Step III. Substitute the values of p and q in g(p, q) = 0 and find the value of φ(a) in terms of a. Put the value of φ(a) in the complete solution z = ax + φ(a) y + c. Step IV. For general solution take f(x, y, z, a, c) = z – ax – φ(a)y – c, and c = ψ(a). Differentiate ‘f ’ partially w.r.t. a and write the general solution as f(x, y, z, ∂f a, ψ(a)) = 0, = 0, i.e. z – ax – φ(a) y – ψ(a) = 0, – x – φ′(a) y – ψ′(a) = 0, where ∂a ψ is any arbitrary function. Step V. Equation of the form g(p, q) = 0 has no singular solution.

Step II. Find p =

TEST YOUR KNOWLEDGE Solving the following partial differential equations : 1. p2 + q2 = 16

2. p = q2

3. p2 – q2 = 1

4. p + q = pq

5. p2 + p = q2

6. p + q + pq = 0

7. p = 9.

eq

p2q3

=1

8. p2 + q2 = npq 10. pq = k.

Answers 1.

2 C.S. z = ax − 16 − a y + c , where a and c are arbitrary constants and – 4 ≤ a ≤ 4.

S.S. No singular solution. G.S. z − ax − 16 − a2 − ψ (a) = 0, − x + 2.

a 16 − a2

y − ψ ′ (a) = 0, where ψ is any arbitrary function.

C.S. z = ax + a y + c , where a and c are arbitrary constants and a ≥ 0. S.S. No singular solution.

33

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

G.S. z − ax − a y − ψ (a) = 0, − x − 3.

1 2 a

y − ψ ′ (a) = 0, where ψ is any arbitrary function.

C.S. z = ax + a 2 − 1 y + c , where a and c are arbitrary constants and |a| ≥ 1. S.S. No singular solution. 2 G.S. z − ax − a − 1 y − ψ(a) = 0, − x −

4.

a 2

a −1

y − ψ ′ (a) = 0 , where ψ is any arbitrary function.

a y + c , where a and c are arbitrary constants and a ≠ 1. a −1

C.S. z = ax +

S.S. No singular solution. a 1 y − ψ ( a ) = 0, − x + y − ψ ′( a ) = 0, where ψ is any arbitrary function. a −1 ( a − 1)2

G.S. z − ax − 5.

C.S. z = ax + a 2 + a y + c , where a and c are arbitrary constants and a ∈ R – (– 1, 0). S.S. No singular solution. 2 G.S. z − ax − a + a y − ψ (a) = 0, − x −

6.

C.S. z = ax −

2a + 1 2 a2 + a

y − ψ ′(a) = 0, where ψ is any arbitrary function.

a y + c , where a and c are arbitrary constants and a ≠ – 1. a+1

S.S. No singular solution. G.S. z − ax + 7.

a 1 y − ψ ( a ) = 0, − x + y − ψ ′( a ) = 0, where ψ is any arbitrary function. a +1 ( a + 1)2

C.S. z = ax + y log a + c, where a and c are arbitrary constants and a > 0. S.S. No singular solution. G.S. z − ax − y log a − ψ ( a ) = 0, − x −

8.

C.S. z = ax +

FG H

IJ K

y − ψ ′( a ) = 0, where ψ is any arbitrary function. a

a n + n 2 − 4 y + c, where a and c are arbitrary constants. 2

S.S. No singular solution. G.S. z − ax −

FG H

arbitrary function. 9.

IJ K

FG H

IJ K

a 1 n + n 2 − 4 y − ψ( a ) = 0, − x − n + n 2 − 4 y − ψ ′( a ) = 0, where ψ is any 2 2

−2/ 3 y + c , where a and c are arbitrary constants and a ≠ 0. C.S. z = ax + a

S.S. No singular solution. G.S. z − ax − a −2/ 3 y − ψ ( a ) = 0, − x + 10.

C.S. z = ax +

2 −5 / 3 a y − ψ ′( a ) = 0, where ψ is any arbitrary function. 3

k y + c , where a and c are arbitrary constants and a ≠ 0. a

S.S. No singular solution. G.S. z − ax −

k k y − ψ( a ) = 0, − x + 2 y − ψ ′( a ) = 0, where ψ is any arbitrary function. a a

34

PARTIAL DIFFERENTIAL EQUATIONS

3.3. SPECIAL TYPE II : EQUATIONS OF THE FORM z = px + qy + g(p, q) Consider the equation Let be a solution of (1).

z = px + qy + g(p, q). z = ax + by + c

...(1) ...(2)

∂z ∂z = a and q = =b ∂x ∂y ∴ (1) ⇒ ax + by + c = ax + by + g(a, b) ⇒ c = g(a, b) ∴ (2) ⇒ z = ax + by + g(a, b). ∴ Complete solution of (1) is z = ax + by + g(a, b), where a and b are arbitrary constants. To find the singular solution, let f(x, y, z, a, b) = z – ax – by – g(a, b). (2)

∵

⇒

p=

Using f(x, y, z, a, b) = 0,

a and b from the equations :

∂f ∂f = 0, = 0 , the singular solution is given by eliminating ∂a ∂b

z − ax − by − g(a, b) = 0, − x −

∂g ∂g = 0, − y − = 0 , provided it satisfies the given ∂a ∂b

equation. To find the general solution, let b = ψ(a), where ψ is any arbitrary function. ∂f ∴ Using f ( x, y, z, a, ψ (a)) = 0, = 0, the general solution is given by ∂a z – ax – ψ(a)y – g(a, ψ(a)) = 0, – x – ψ′(a) y – g′′(a, ψ(a)) = 0.

Remark. The partial differential equation z = px + qy + g(p, q) is analogous to the Clairaut’s equation z = px + f(p). The equation z = px + qy + g(p, q) is known as extended Clairaut’s equation.

ILLUSTRATIVE EXAMPLES Example 1. Solve the following partial differential equations : (i) z = px + qy + pq (ii) z = px + qy + p2q2 (iii) z = px + qy + 4 1 + p 2 + q 2

(iv) z = px + qy + log (pq).

Sol. (i) We have z = px + qy + pq. ...(1) This equation is of the form z = px + qy + g(p, q). ∴ Complete solution of (1) is z = ax + by + ab, where a and b are arbitrary constants. To find the singular solution, let f(x, y, z, a, b) = z – ax – by – ab ∴ ∴

∂f ∂f = – x – b and =–y–a ∂a ∂b f(x, y, z, a, b) = 0 ⇒ z – ax – by – ab = 0 ∂f =0 ∂a ∂f =0 ∂b

...(2)

⇒

–x–b=0

...(3)

⇒

–y–a=0

...(4)

35

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Putting the values of a and b from (3) and (4) in (2), we get z – (– y) x – (– x) y – (– y)(– x) = 0 or z + xy = 0 and it also satisfies (1). ∴ Singular solution is z + xy = 0. To find the general solution, let b = ψ(a). ∂f = 0, the general solution is given by the equations : ∴ Using f ( x, y, z, a, ψ (a)) = 0, ∂a z – ax – ψ(a) y – a ψ(a) = 0, – x – ψ′(a) y – ψ(a) – aψ′(a) = 0, where ψ is any arbitrary function. (ii) We have z = px + qy + p2q2. ...(1) This equation is of the form z = px + qy + g(p, q). ∴ Complete solution of (1) is z = ax + by + a2b2, where a and b are arbitrary constants. To find the singular solution, let f(x, y, z, a, b) = z – ax – by – a2b2 ∂f ∂a f(x,y, z, a, b) ∂f ∂a ∂f ∂b

∴ ∴

=0

⇒

∂f = – y – 2a2b ∂b z – ax – by – a2b2 = 0

=0

⇒

– x – 2ab2 = 0

...(3)

=0

⇒

– y – 2a2b = 0

...(4)

= – x – 2ab2 and

x 2 y ,a b = − , 2 2

ab 2 = −

∴

( ab)3 =

FG IJ H K

4 x x ( ab)b = − ⇒ b=− . 2 2 xy

∴ Similarly,

a=−

Fy I z+G J H 2x K 2

∴ ⇒

z=−

1/3

3

4/3

Fx I x+G J H 2yK 2

Fy I GH 2x JK 2

xy , 4

1/ 3

=−

ab =

FG xy IJ H4K

x 2/ 3

1/ 3

Fx I =−G J H 2y K 2

21/ 3 y1/ 3

...(2)

1/ 3

.

1/ 3

1/3

y−

FG xy IJ H4K

2/3

= 0 ⇒ z + x 2/3 y2/ 3

FG 1 H2

1/3

+

1 2

1/3

−

IJ K

1 =0 2.2 1/3

x 2/ 3 y 2/ 3 .

2 This gives the singular solution, because it also satisfies (1). To find the general solution, let b = ψ(a).

∂f = 0, the general solution is given by the equations : ∂a z – ax – ψ(a) y – a2(ψ(a))2 = 0, – x – ψ′(a) y – 2a (ψ(a))2 – 2a2ψ(a) ψ′(a) = 0, where ψ is any arbitrary function.

∴ Using f ( x, y, z, a, ψ (a)) = 0,

(iii) We have

z = px + qy + 4 1 + p2 + q 2 .

This equation is of the form z = px + qy + g(p, q). ∴ Complete solution of (1) is z = ax + by + 4 1 + a 2 + b 2 , where a and b are arbitrary constants.

...(1)

36

PARTIAL DIFFERENTIAL EQUATIONS

To find the singular solution, let f ( x , y , z , a , b) = z − ax − by − 4 1 + a 2 + b 2 ∂f 4a =−x− ∂a 1 + a 2 + b2

∴

f(x, y, z, a, b) = 0

∴

∂f =0 ∂a

⇒

–x–

∂f =0 ∂b

⇒

–y–

∴

1 + a 2 + b2 =

(3)

a=− x

⇒

Similarly,

4a 1 + a 2 + b2 4b 1 + a 2 + b2

...(2)

=0

...(3)

=0

...(4)

16( a 2 + b 2 ) 1 + a 2 + b2

16 − x 2 − y 2 = 16 −

∴

∴

z – ax – by – 4 1 + a 2 + b 2 = 0

⇒

Using (3) and (4), we get x 2 + y 2 =

∂f 4b =−y− ∂b 1 + a 2 + b2

and

16( a 2 + b 2 ) 2

1+a +b

2

=

16 1 + a 2 + b2

4 16 − x 2 − y 2

1 + a 2 + b2 4 x =− . =− 4 4 16 − x 2 − y 2

x 16 − x 2 − y 2

.

y

b=−

16 − x 2 − y 2

Putting the values of a and b in (2), we get z+

or

z=

x2 2

16 − x − y

2

16 − x 2 − y 2 2

16 − x − y

2

+

y2 2

16 − x − y

= 16 − x 2 − y 2

2

−

16 16 − x 2 − y 2

=0

or x2 + y2 + z2 = 16.

This gives the singular solution, because it also satisfies (1). To find the general solution, let b = ψ(a). Using f ( x, y, z, a, ψ (a)) = 0,

∂f = 0, the general solution is given by the equations : ∂a

z − ax − ψ (a)y − 4 1 + a 2 + (ψ(a))2 = 0 , − x − ψ ′ (a)y −

is any arbitrary function. (iv) We have z = px + qy + log (pq). This equation is of the form z = px + qy + g(p, q).

4(a + ψ(a) ψ ′ (a)) 1 + a 2 + (ψ(a))2

= 0, where ψ

...(1)

37

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

∴ Complete solution of (1) is z = ax + by + log (ab), where a and b are arbitrary constants. To find the singular solution, let f(x, y, z, a, b) = z – ax – by – log (ab) ∂f 1 =−x− ∂a a

∴

f(x, y, z, a, b) = 0

∴

∂f 1 =−y− ∂b b

and

⇒ z – ax – by – log (ab) = 0

...(2)

∂f =0 ∂a

⇒

–x–

1 =0 a

...(3)

∂f =0 ∂b

⇒

–y–

1 =0 b

...(4)

1 x

∴ (3)

⇒

a=−

∴ (2)

⇒

z + 1 + 1 − log

and (4)

1 =0 xy

⇒

b=−

1 y

⇒ z + 2 + log xy = 0.

This is the singular solution, because it also satisfies (1). To find the general solution, let b = ψ(a). ∂f = 0, the general solution is given by the equations : ∂a 1 ψ ′ (a) = 0, z – ax – ψ(a)y – log a – log ψ(a) = 0, – x – ψ′(a)y – − a ψ(a)

Using f ( x, y, z, a, ψ (a)) = 0,

where ψ is any arbitrary function. Example 2. Show that the complete integral of z = px + qy – 2p – 3q represents all possible planes through the point (2, 3, 0). Also find the envelope of all planes represented by the complete integral, i.e., find the singular solution. Sol. We have

z = px + qy – 2p – 3q.

...(1)

This equation is of the form z = px + qy + g(p, q). ∴ Complete solution of (1) is z = ax + by – 2a – 3b, where a and b are arbitrary constants. This represents a family of planes each passing through (2, 3, 0) because 0 = a(2) + b(3) – 2a – 3b for all constants a and b. To find the singular solution, let f(x, y, z, a, b) = z – ax – by – 2a – 3b ∂f = 0 ⇒ − x − 2 = 0, ∂a

∂f =0 ⇒ − y −3=0 ∂b

∴ z – a (– 2) – b (– 3) – 2a – 3b = 0 or z = 0. It also satisfies (1). ∴ The singular solution is

z = 0.

38

PARTIAL DIFFERENTIAL EQUATIONS

WORKING STEPS FOR SOLVING z = px + qy + g(p, q) Step I.

Take complete solution as z = ax + by + g(a, b), where a and b are arbitrary constants.

Step II. For singular solution, take f(x, y, z, a, b) = z – ax – by – g(a, b). Find Eliminate a and b from the equations : f = 0,

∂f ∂f and . ∂a ∂b

∂f ∂f = 0, = 0. This gives the ∂a ∂b

singular solution. Step III. For general solution, take b = ψ(a), where ψ is any arbitrary function. The ∂f equations : f = 0, = 0 constitute the general solution. ∂a

TEST YOUR KNOWLEDGE Solve the following partial differential equations (Q. No. 1–10) : 2. z = px + qy + p2 + q2

1. z = px + qy + 5pq 3. z = px + qy +

p2

–

q2

4. z = px + qy – 2p – 3q

5. z = px + qy + 3(pq)1/3

6. z = px + qy + p/q

7. z = px + qy + 2 pq

8. z = px + qy – 2

9. z = px + qy + p2 + pq + q2 11.

10. z = px + qy +

pq αp2 + βq 2 + 1 .

pq Show that the complete integral of the equation z = px + qy + pq − p − q represents a family of planes such that the algebraic sum of the intercepts on the three coordinates axes is unity.

12.

Show that the complete integral of the equation z = px + qy +

p2 + q 2 + 1 represents a family of

planes each at a unit distance from the origin.

Answers 1.

C.S. z = ax + by + 5ab S.S. 5z + xy = 0 G.S. z – ax – ψ(a) y – 5aψ(a) = 0, x + 5ψ(a) + (y + 5a) ψ′(a) = 0

2.

C.S. z = ax + by + a2 + b2 S.S. x2 + y2 + 4z = 0 G.S. z – ax – ψ(a) y – a2 – (ψ(a))2 = 0, x + 2a + (y + 2ψ(a)) ψ′(a) = 0

3.

C.S. z = ax + by + a2 – b2 S.S. x2 – y2 + 4z = 0 G.S. z – ax – ψ(a) y – a2 + (ψ(a))2 = 0, x + 2a + (y – 2ψ(a)) ψ′(a) = 0

4.

C.S. z = ax + by – 2a – 3b S.S. z = 0 G.S. z – ax – ψ(a) y + 2a + 3ψ′(a) = 0, x + (y – 3) ψ′(a) – 2 = 0

39

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

5.

C.S. z = ax + by + 3(ab)1/3 S.S. xyz – 1 = 0 G.S. z – ax – ψ(a) y – 3(a ψ(a))1/3 = 0, x + ψ′(a) y +

6.

ψ ( a ) + aψ ′( a ) ( a ψ ( a ))2/ 3

=0

C.S. z = ax + by + a/b S.S. xz + y = 0 G.S. z – ax – ψ(a) y –

7.

1 a ψ ′( a ) a − = 0, x + ψ′(a) y + =0 ψ ( a ) ψ( a ) ( ψ ( a ))2

C.S. z = ax + by + 2 ab S.S. (x – z)(y – z) = 1 G.S. z – ax – ψ(a) y – 2 aψ( a ) = 0, x + ψ′(a) +

8.

ψ( a ) + aψ ′( a ) a ψ( a )

=0

C.S. z = ax + by – 2 ab S.S. (x – z)(y – z) = 1 G.S. z – ax – ψ(a) y + 2 aψ( a ) = 0, x + ψ′(a) y –

9.

ψ( a ) + aψ ′( a ) a ψ( a )

=0

C.S. z = ax + by + a2 + ab + b2 S.S. x2 + y2 – xy + 3z = 0 G.S. z – ax – ψ(a) y – a2 – aψ(a) – (ψ(a))2 = 0, x + (y + a + 2ψ(a)) ψ′(a) + 2a + ψ(a) = 0

10.

C.S. z = ax + by + αa 2 + βb2 + 1 S.S.

x2 y2 + + z2 = 1 α β

G.S. z – ax – ψ(a) y –

αa 2 + β( ψ( a ))2 + 1 = 0, x + ψ′(a) y +

aα + βψ ( a ) ψ ′( a ) αa 2 + β( ψ ( a ))2 + 1

= 0.

Hint 7.

S.S. We have z – ax – by – 2 ab = 0, x = − Now

x – z = x – (ax + by + 2 ab ) = −

Similarly, y – z = −

b a

and

y=−

a . b

b b a +a +b − 2 ab = − a a b

b a

a . b

3.4. SPECIAL TYPE III : EQUATIONS CONTAINING ONLY z, p AND q Let g(z, p, q) = 0 be a partial differential equation of first order and containing only z, p and q.

...(1)

40

PARTIAL DIFFERENTIAL EQUATIONS

Let z = G(u) where u = x + ay be a solution of (1) where a is an arbitrary constant. dz ∂u dz dz dz ∂u dz dz = .1= and q = = .a=a ∴ p= du ∂x du du du ∂y du du

FG H

IJ K

dz dz =0 ,a du du This is an ordinary differential equation of first order. The solution of this equation, say f(x, y, z, a, b) = 0 gives the complete solution of (1) where a and b are arbitrary constants. The singular solution is obtained by eliminating a and b from the equations :

∴ (1) ⇒

g z,

∂f ∂f = 0, = 0 , provided it satisfies the given equation. ∂a ∂b Let b = φ(a) where φ is an arbitrary function. The general solution is given by the equations :

f(x, y, z, a, b) = 0,

∂f = 0. ∂a

f(x, y, z, a, φ(a)) = 0,

ILLUSTRATIVE EXAMPLES Example 1. Solve the following partial differential equations : (i) p2 + q2 = 4z (ii) z2 (p2 + q2 + 1) = 1 2 (iii) p(1 – q ) = q(1 – z). Sol. (i) We have p2 + q2 = 4z. This equation is of the form g(z, p, q) = 0. Let z = G(u), where u = x + ay be a solution of (1).

p=

∴

FG dz IJ + FG a dz IJ H du K H du K F dz I (1 + a ) G J H du K 2

∴ (1) ⇒ ⇒

2

dz

⇒

z

Integrating, we get ⇒

1/ 2

2

= 4z 2

= 4z =

2 z=

2 1 + a2

∂z dz ∂u dz ∂z dz ∂u dz = = and q = = =a . ∂x du ∂x du ∂y du ∂y du

⇒ 2

1 + a2 2u 1 + a2

2 dz = z 1/ 2 2 du 1+a du +b

z = 2( x + ay ) + b 1 + a 2

4(1 + a2)z = 4(x + ay + c)2, where 2c = b 1 + a 2 ⇒ (1 + a2) z – (x + ay + c)2 = 0. This is the complete solution. Let f(x, y, z, a, c) = (1 + a2) z – (x + ay + c)2 ⇒

...(1)

41

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

∂f ∂f = 2az – 2(x + ay + c)y and = – 2(x + ay + c) ∂a ∂c f(x, y, z, a, c) = 0 ⇒ (1 + a2)z – (x + ay + c)2 = 0

∴ ∴

∂f =0 ⇒ 2az – 2(x + ay + c) = 0 ∂a ∂f =0 ⇒ – 2(x + ay + c) = 0 ∂c (4) ⇒ x + ay + c = 0 ∴ (3) ⇒ 2az – 2(0)y = 0 ⇒ 2az = 0 ⇒ z = 0. This is the singular solution, because z = 0 also satisfies (1).

...(2) ...(3) ...(4)

∂f = 0, the general solution is given by ∂a (1 + a2) z – (x + ay + φ(a))2 = 0, 2az – 2(x + ay + φ(a)) (y + φ′(a)) = 0, where φ is any arbitrary function. (ii) We have z2(p2 + q2 + 1) = 1. This equation is of the form g(z, p, q) = 0. Let z = G(u), where u = x + ay be a solution of (1).

Let c = φ(a). Using f(x, y, z, a, φ(a)) = 0,

p=

∴ ∴ (1) ⇒ ⇒

z2

∂z dz ∂u dz ∂z dz ∂u dz = = and q = = =a . ∂x du ∂x dx ∂y du ∂y du

F F dz I + F a dz I + 1I GH GH du JK GH du JK JK = 1 F dz IJ = 1 (1 + a ) G H du K z 2

2

2

2

2

−1

dz 1 = du 1 + a2

Let z

dz =

∴

1− z

Integrating, we get

− 1 − z2 =

⇒

...(1)

− 1 + a2

2

dz ±1 = du 1 + a2

⇒

1 − z2 z

1 − z2 z

.

du 1 + a2

u 1 + a2

+b

1 − z 2 = u + c, where c = b 1 + a 2

⇒ (1 + a2) (1 – z2) = (x + ay + c)2. This is the complete solution. Let f(x, y, z, a, c) = (1 + a2)(1 – z2) – (x + ay + c)2 ∴ ∴

∂f ∂f = 2a(1 – z2) – 2(x + ay + c)y and = – 2(x + ay + c) ∂a ∂c f(x, y, z, a, c) = 0 ⇒ (1 + a2) (1 – z2) – (x + ay + c)2 = 0 ∂f =0 ∂a

⇒

2a(1 – z2) – 2(x + ay + c)y = 0

...(2) ...(3)

42

PARTIAL DIFFERENTIAL EQUATIONS

(4) ⇒

∂f =0 ∂c x + ay + c = 0

⇒ – 2(x + ay + c) = 0

∴ (3) ⇒

2a(1 – z2) – 2(0)y = 0

⇒

2a(1 – z2) = 0

⇒

...(4)

1 – z2 = 0 ⇒ z2 – 1 = 0.

This is the singular solution, because z2 – 1 = 0 also satisfies (1). ∂f = 0, the general solution is given by ∂a (1 + a2)(1 – z2) – (x + ay + φ(a))2 = 0, 2a(1 – z2) – 2(x + ay + φ(a))(y + φ′(a)) = 0,

Let c = φ(a). Using f(x, y, z, a, c) = 0,

where φ is any arbitrary function. p(1 – q2) = q(1 – z).

(iii) We have

...(1)

This equation is of the form g(z, p, q) = 0. Let z = G(u), where u = x + ay be a solution of (1).

p=

∴ ∴ (1) ⇒

F GH

∂z dz ∂u dz ∂z dz ∂u dz = . = and q = = =a ∂x du ∂x dx ∂y du ∂y du

FG IJ IJ = a dz (1 − z ) H K K du

dz dz 1 − a2 du du

2

dz =0 du

⇒ 1 − a2

or

FG dz IJ H du K

...(2)

2

= a(1 – z)

...(3)

(2) ⇒ z = c. This is not a complete solution because it does not contain two arbitrary constants. (3)

FG dz IJ H du K

⇒

=

1 − a + az a2

.

1 − a + az dz = du a

Let ∴

2

(1 – a + az)–1/2 dz =

Integrating, we get

du a

(1 − a + az) 1/2 u = +b (1 / 2) a a

⇒

2 1 − a + az = u + ab

⇒

4(1 – a + az) = (x + ay + c)2, where c = ab.

This is the complete solution. Let

f(x, y, z, a, c) = 4(1 – a + az) – (x + ay + c)2

43

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

∂f = 4(– 1 + z) – 2(x + ay + c)y and ∂a

∴

∂f = – 2(x + ay + c) ∂c

f(x, y, z, a, c) = 0 ⇒ 4(1 – a + az) – (x + ay + c)2 = 0

∴

...(4)

∂f =0 ∂a

⇒

4(– 1 + z) – 2(x + ay + c)y = 0

...(5)

∂f =0 ∂c

⇒

– 2(x + ay + c) = 0

...(6)

(6)

⇒

x + ay + c = 0

∴ (5)

⇒

∴ (4)

4(– 1 + z) – 2(0)y = 0 ⇒ 4(– 1 + z) = 0

⇒

–1+z=0

⇒

(0)2

4(1 – a + a(1)) –

⇒

z=1

=0

4 = 0, which is impossible.

⇒

∂f = 0, the ∂a general solution is given by 4(1 – a + az) – (x + ay + φ(a))2 = 0, – 4 + 4z – 2(x + ay + φ(a)) (y + φ′(a)) = 0, where φ is any arbitrary function.

∴ There is no singular solution. Let c = φ(a). Using f(x, y, z, a, φ(a)) = 0 and

WORKING STEPS FOR SOLVING g(z, p, q) = 0 Step I.

Take z = G(u), where u = x + ay. dz dz and q = a , the given equation reduces to an ordinary du du differential equation of first order. Let its solution be f(x, y, z, a, b) = 0. This gives the complete solution of the given equation.

Step II. By putting p =

∂f ∂f = 0, = 0. ∂a ∂b Step IV. For general solution, take b = φ(a), where φ is any arbitrary function. The ∂f = 0 constitute the general solution. equations : f = 0, ∂a

Step III. For singular solution, eliminate a and b from the equations : f = 0,

TEST YOUR KNOWLEDGE Solve the following partial differential equations : 1. p2 + q2 = z 3.

p2

+ pq = 4z

7.

+

q3

4. pz = 1 + q2 6. 9(p2z + q2) = 4

5. z = pq p3

2. z2(p2 + q2 + 2) = 1

= 3pqz, z > 0

9. 4(1 + z3) = 9z4 pq

8. p3 + q3 = 27z 10. q2 = z2p2(1 – p2).

44

PARTIAL DIFFERENTIAL EQUATIONS

Answers 1.

C.S. 4(1 + a2) z = (x + ay + c)2 S.S. z = 0 G.S. 4(1 + a2) z – (x + ay + φ(a))2 = 0, 8az – 2(x + ay + φ(a))(y + φ′(a)) = 0

2.

C.S. (1 + a2)(1 – 2z2) = 4(x + ay + c)2 S.S. 1 – 2z2 = 0 G.S. (1 + a2)(1 – 2z2) – 4(x + ay + φ(a))2 = 0, a(1 – 2z2) – 4(x + ay + φ(a))(y + φ′(a)) = 0

3.

C.S. (1 + a)z = (x + ay + c)2 S.S. z = 0 G.S. (1 + a)z – (x + ay + φ(a))2 = 0, z – 2(x + ay + φ(a))(y + φ′(a)) = 0

4.

C.S. z2 – z z 2 − 4 a 2 + 4 a 2 log ( z +

z 2 − 4 a 2 ) = 4( x + ay + b )

S.S. There is no singular solution. 2 2 2 G.S. z2 – z z − 4 a + 4 a log ( z +

az z 2 − 4a 2

5.

FG H

+ 2a log z + z 2 − 4a 2

z 2 − 4 a 2 ) − 4( x + ay + φ( a )) = 0,

IJ − K (z +

4a3 z 2 − 4a 2 ) z 2 − 4a 2

– y – φ′(a) = 0.

C.S. 4az = (x + ay + b)2 S.S. z = 0 G.S. 4az – (x + ay + φ(a))2 = 0, 2z – (x + ay + φ(a)) (y + φ′(a)) = 0.

6.

C.S. (z + a2)3 = (x + ay + b)2 S.S. No singular solution. G.S. (z + a2)3 – (x + ay + φ(a))2 = 0, 3a(z + a2)2 – (x + ay + φ(a))(y + φ′(a)) = 0

7.

C.S. (1 + a3) log z = 3a(x + ay) + b S.S. No singular solution. G.S. (1 + a3) log z – 3a(x + ay) – φ(a) = 0, 3a2 log z – 3x – 6ay – φ′(a) = 0.

8.

C.S. (1 + a3) z2 = 8(x + ay + b)3 S.S. z = 0 G.S. (1 + a3) z2 – 8(x + ay + φ(a))3 = 0, a2z2 – 8(x + ay + φ(a))2(y + φ′(a)) = 0.

9.

C.S. a(1 + z3) = (x + ay + b)2 S.S. z3 + 1 = 0 G.S. a(1 + z3) – (x + ay + φ(a))2 = 0, 1 + z3 – 2(x + ay + φ(a))(y + φ′(a)) = 0

10.

C.S. z2 = (x + ay + b)2 + a2 S.S. z = 0 G.S. z2 – (x + ay + φ(a))2 – a2 = 0, (x + ay + φ(a))(y + φ′(a)) + a = 0.

3.5. SPECIAL TYPE IV : EQUATIONS OF THE FORM f1(x, p) = f2(y, q) Consider the equation

f1(x, p) = f2(y, q).

...(1)

Let each side of (1) be equal to a. ∴

f1(x, p) = a

...(2)

f2(y, q) = a

...(3)

45

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Solving (2) for p, let Solving (3) for q, let

p = F1(x, a) q = F2(y, a).

∂z ∂z dx + dy = p dx + q dy ∂x ∂y dz = F1(x, a) dx + F2(y, a) dy

Since z is a function of x and y, we have dz = ∴ Integrating, we get

z=

z

F1(x, a) dx +

z

F2 (y, a) dy + b.

z

This represents the complete solution of the given equation. To find the singular solution, let f(x, y, z, a, b) = z – ∴ Using f(x, y, z, a, b) = 0,

z

a and b from the equations : z–

FH

F1 ( x , a ) dx −

z

F1 ( x , a ) dx −

z

F2 ( y , a ) dy − b .

∂f ∂f = 0, = 0, the singular solution is given by eliminating ∂a ∂b

z

F2 ( y , a ) dy − b = 0,

z

IK

∂ F1 ( x , a ) dx + F2 ( y , a ) dy = 0 and − 1 = 0. ∂a This is impossible, because – 1 ≠ 0. ∴ There is no singular solution. To find the general solution, let b = φ(a), where φ is an arbitrary function. Using ∂f = 0, the general solution is given by the equations : ∂a ∂ − F1 (x,a)dx − F2 (y, a)dy − φ′ (a) = 0. F1 (x, a)dx − F2 (y, a)dy − φ(a) = 0, ∂a

f(x, y, z, a, φ(a)) = 0, z–

z

z

FH

z

z

IK

ILLUSTRATIVE EXAMPLES Example 1. Solve the following partial differential equations : (i) (iii)

(ii) py + qx + pq = 0

p − q + 3x = 0

p2y

(1 +

x 2)

=

qx2

(iv) px + q = p2.

Sol. (i) We have

p + 3x = q .

...(1)

This equation is of the form f1(x, p) = f2(y, q). Let each side of (1) be equal to a. ∴

...(2)

p + 3x = a

(2) ⇒ Now ∴

q =a

p = (a – 3x)2 and (3) ⇒ q = a2 dz = p dx + q dy dz = (a – 3x)2 dx + a2 dy

Integrating, we get

z=

⇒

z=

z

( a − 3x ) 2 dx +

z

a 2 dy + b.

(a − 3x)3 + a 2 y + b. −9

...(3)

46

PARTIAL DIFFERENTIAL EQUATIONS

This is the complete solution. There is no singular solution. To find the general solution, let f(x, y, z, a, b) = z +

1 (a – 3x)3 – a2y – b and b = φ(a). 9

∂f = 0. ∂a The general solution is given by the equations :

∴ Using

f(x, y, z, a, φ(a)) = 0,

1 1 2 (a – 3x)3 – a2y – φ(a) = 0, (a − 3x) − 2ay − φ′ (a) = 0, 9 3 where φ is any arbitrary function. (ii) We have py + qx + pq = 0.

z+

p q =− x+p y This equation is of the form f1(x, p) = f2(y, q). Let each side of (1) be equal to a. py + q(x + p) = 0 ⇒

⇒

p =a x+p

∴

...(2)

Now

ax and 1−a dz = p dx + q dy

∴

dz =

(2)

p=

⇒

−

q =a y

...(1)

...(3)

(3) ⇒ q = – ay

ax dx – ay dy 1−a

z

ax dx − 1− a

z

ay dy +

Integrating, we get

z=

⇒

z=

a x 2 ay 2 b . − + 1− a 2 2 2

⇒

2z =

a x 2 − ay 2 + b . 1− a

b . 2

This is the complete solution. There is no singular solution. To find the general solution, let f(x, y, z, a, b) = 2z –

a x 2 + ay 2 − b a −1

and b = φ(a).

∂f =0, ∂a The general solution is given by the equations :

∴ Using

f(x, y, z, a, φ(a)) = 0,

2

2z − tion.

x a + y 2 – φ′ (a) = 0 , where φ is any arbitrary funcx 2 + ay 2 − φ(a) = 0, − 2 (1 − a) 1− a

(iii) We have ⇒

p2y (1 + x2) = qx2. 1 + x2 x2

p2 =

q y

...(1)

47

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

This equation is of the form f1(x, p) = f2(y, q). Let each side of (1) be equal to a. 1 + x2

∴ (2)

x

2

p2 = a

ax

p=±

⇒

Let

1 + x2

and

Now

1 + x2 dz = p dx + q dy

∴

dz = z=

(3) ⇒ q = ay

ax

dx + ay dy

1 + x2

z

a 2

2x 1+ x

2

dx + a

z

y dy +

(Assuming a ≥ 0)

b . 2

ay 2 b + ⇒ 2z = 2 a 2 2 This is the complete solution. There is no singular solution. z=

⇒

...(3)

ax

p=

Integrating, we get

q =a y

...(2)

a 1 + x2 +

1 + x 2 + ay 2 + b.

To find the general solution, let f(x, y, z, a, b) = 2z – 2 a 1 + x 2 – ay2 – b and b = φ(a). ∂f = 0 , the general solution is given by the equations : ∂a 1 – ay2 – φ(a) = 0, – 2 . . 1 + x 2 – y2 – φ′(a) = 0 2 a

∴ Using f(x, y, z, a, φ(a)) = 0, 2z – 2 a 1 + x 2

2z – 2 a 1 + x 2 – ay2 – φ(a) = 0,

or

1 + x 2 + ay 2 + a φ′ (a) = 0,

where φ is any arbitrary function. (iv) We have

px + q = p2. p2 – px = q

⇒

...(1) f1(x, p) = f2(y, q).

This equation is of the form

Let each side of (1) be equal to a. p2 – px = a

∴ (2) Let

⇒

...(2)

p2 – px – a = 0 ⇒ p = p=

FH

FH

1 x ± x 2 + 4a 2

1 x + x 2 + 4a 2

Now

dz = p dx + q dy

∴

dz =

FH

q=a

IK

IK

1 x + x 2 + 4a dx + a dy 2

IK

...(3)

48

PARTIAL DIFFERENTIAL EQUATIONS

Integrating, we get

z=

LM MN

2 x 2 1 x x + 4a 4a + + log x + x 2 + 4a 4 2 2 2

FH

IK

OP PQ + ay + b

1 2 x + x x 2 + 4a + a log x + x 2 + 4a + ay + b . 4 This is the complete solution. There is no singular solution. To find the general solution, let

or

z=

f(x, y, z, a, b) = z –

1 2 ( x + x x 2 + 4a ) – a log (x + 4

∴ Using f(x, y, z, a, φ(a)) = 0, z–

x 2 + 4a ) – ay – b and b = φ(a).

∂f = 0 , the general solution is given by the equations : ∂a

1 2 ( x + x x 2 + 4a ) – a log ( x + x 2 + 4a ) – ay – φ(a) = 0, 4

1 x 4 − . – log ( x + x 2 + 4a ) – a . x + x 2 + 4a 4 2 x 2 + 4a

F GG 0 + H 2

4 x2

I J – y – φ′(a) = 0 + 4a JK

1 2 (x + x x 2 + 4a ) – a log (x + x 2 + 4a ) – ay – φ(a) = 0, 4 x 2a 2 + log (x + x + 4a ) + + y + φ′ φ′(a) = 0, 2 2 2 x + 4a (x + x + 4a ) x 2 + 4a where φ is any arbitrary function.

or

z–

WORKING STEPS FOR SOLVING f1(x, p) = f2(y, q) Step I. Take each side of f1(x, p) = f2(y, q) equal to a. Step II. Solve equations for p and q. Let p = F1(x, a), q = F2(y, a). Write z = p dx + q dy and substitute the values of p and q. Integrate this equation to get the complete solution. Step III. Equation of the form f1(x, p) = f2(y, q) has no singular solution. Step IV. Take the complete solution as f(x, y, z, a, b) = 0. Put b = φ(a). The general ∂f = 0. solution is given by the equations : f(x, y, z, a, φ(a)) = 0, ∂a

TEST YOUR KNOWLEDGE Solve the following partial differential equations : 1. p – q = x2 + y2

2. x(1 + y) p = y(1 + x) q

3. pq = xy

4. q = xyp2

5.

x2p2

=

q2y

7. yp = 2yx + log q 9. p2 – x = q2 – y

6. q(p – cos x) = cos y 8.

p + q = 2x

10. p – 3x2 = q2 – y.

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Answers 1.

1 3 ( x − y 3 ) + a( x + y ) + b 3 S.S. No singular solution

C.S. z =

1 3 ( x − y3 ) − a( x + y) − φ(a) = 0, x + y + φ′(a) = 0 3 2. C.S. z = a log xy + a(x + y) + b S.S. No singular solution G.S. z – a log xy – a(x + y) – φ(a) = 0, log xy + x + y + φ′(a) = 0 3. C.S. 2z = ax2 + y2/a + b S.S. No singular solution G.S. 2z – ax2 – y2/a – φ(a) = 0, x2 – y2/a2 + φ′(a) = 0 G.S. z –

4.

2 C.S. 2 z = 4 ax + ay + b S.S. No singular solution

G.S. 2 z − 4 ax − ay2 − φ(a) = 0, 2 x/a + y2 + φ′(a) = 0 5.

C.S. z =

a log x + 2 ay + b

S.S. No singular solution G.S. z – 6.

a log x − 2 ay − φ( a ) = 0, log x + 2 y + 2 a φ′( a ) = 0

1 sin y + b a S.S. No singular solution

C.S. z = ax + sin x +

G.S. z – ax – sin x – 7.

e ay +b a S.S. No singular solution

C.S. z = x2 + ax +

G.S. z – x2 – ax – 8.

1 1 sin y – φ(a) = 0, x – 2 sin y + φ′(a) = 0 a a

e ay ( ay − 1) e ay – φ(a) = 0, x + + φ′(a) = 0 a a2

1 ( 2x − a )3 + a 2 y + b 6 S.S. No singular solution

C.S. z =

1 ( 2x − a )3 − a 2 y − φ( a ) = 0 , 4ay – (2x – a)2 + 2φ′(a) = 0 6 C.S. 3z = 2(x + a)3/2 + 2(y + a)3/2 + b S.S. No singular solution

G.S. z – 9.

G.S. 3z – 2(x + a)3/2 – 2(y + a)3/2 – φ(a) =0, 3 x + a + 3 y + a + φ′(a) = 0 10.

2 ( a + y )3/ 2 + b 3 S.S. No singular solution

C.S. z = ax + x3 +

G.S. z – ax – x3 –

2 ( a + y )3/ 2 – φ(a) = 0, x + 3

a + y + φ′(a) = 0.

49

50

PARTIAL DIFFERENTIAL EQUATIONS

3.6. USE OF TRANSFORMATIONS At times the use of transformations helps a lot in changing a partial differential equation to a much simpler form. Remark. Keeping in view the scope of the present book, we are restricting ourselves only to the finding of complete solutions of partial differential equations which are reducible to the form g(P, Q) = 0, ∂Z ∂Z ,Q= where P = . ∂X ∂Y

ILLUSTRATIVE EXAMPLES Example 1. Find the complete solution of the following differential equations with the help of transformations : (i) x2p2 + y2q2 = z (ii) x2p2 + y2q2 = 4z2 m n 2l (iii) pq = x y z (iv) (1 – x2) yp2 + x2q = 0. Sol. (i) We have x2p2 + y2q2 = z. ...(1) (1)

⇒

Fz GH x

x2 z

−1/ 2

⇒

−1

FG ∂z IJ + y FG ∂z IJ H ∂x K z H ∂y K F z ∂z I ∂z I +G J ∂x K H y ∂y JK 2

2

2

−1/ 2 −1

=1 2

=1

...(2)

dX = x–1 dx, dY = y–1 dy, dZ = z–1/2 dz.

Let X, Y, Z be new variables such that ∴

2

By using integration, we have X = log x, Y = log y, Z = 2 z P=

∴

∂Z dZ ∂z dx 1 ∂z z −1/ 2 ∂z = = . . . . x = −1 ∂X dz ∂x dX ∂x x z ∂x

∂Z dZ ∂z dy 1 ∂z z −1/2 ∂z = . . = . . y = −1 . ∂Y dz ∂y dY ∂y y z ∂y ∴ (2) ⇒ P2 + Q2 = 1 This equation is of the form g(P, Q) = 0. Let Z = aX + φ(a)Y + c be the complete solution of (3), where φ(a) is some function of a.

and

Q=

(4)

⇒

P=

...(3) ...(4)

∂Z ∂Z = a and Q = = φ(a) ∂X ∂Y

∴ (3) ⇒ a2 + (φ(a))2 = 1 or φ(a) = ± 1 − a 2 Let

φ(a) =

1 − a 2 , – 1 ≤ a ≤ 1.

∴ The complete solution of (3) is Z = aX +

1 − a 2 Y + c.

∴ The complete solution of (1) is 2 z = a log x + arbitrary constants. (ii) We have (1)

⇒

x

2

z

2

FG ∂z IJ H ∂x K

x2p2 + y2q2 = 4z2. 2

+

y

2

z

2

FG ∂z IJ H ∂y K

2

=4

1 − a 2 log y + c, where a and c are

...(1)

51

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Fz GH x

⇒

−1 −1

∂z ∂x

I +Fz JK GH y 2

−1 −1

∂z ∂y

I JK

2

=4

...(2)

Let X, Y, Z be new variables such that dX = x–1 dx, dY = y–1 dy, dZ = z–1 dz ∴ By using integration, we have X = log x, Y = log y, Z = log z P=

∴

∂Z dZ ∂z dx 1 ∂z z −1 ∂z = . . = . . x = −1 ∂X dz ∂x dX z ∂x x ∂x

∂Z dZ ∂z dy 1 ∂z z −1 ∂z = . . = . . y = −1 . ∂Y dz ∂y dY z ∂y y ∂y ∴ (2) ⇒ P2 + Q2 = 4 This equation is of the form g(P, Q) = 0. Let Z = aX + φ(a)Y + c be the complete solution of (3), where φ(a) is some function of a.

and

Q=

(4) ⇒

P=

...(3) ...(4)

∂Z ∂Z = a and Q = = φ(a) ∂X ∂Y

∴ (3) ⇒ a2 + (φ(a))2 = 4 or φ(a) = ± 4 − a 2 Let

φ(a) =

4 − a 2 , – 2 ≤ a ≤ 2.

∴ The complete solution of (3) is Z = aX +

4 − a 2 Y + c.

∴ The complete solution of (1) is log z = a log x + arbitrary constants. (iii) We have

pq = xmynz2l.

z − 2l

⇒

m n

x y

pq = 1

⇒

Fz GH x

−l m

∂z ∂x

4 − a 2 log y + c, where a and c are

I Fz JK GH y

−l n

∂z ∂y

I JK

...(1) =1

...(2)

Let X, Y, Z be new variables such that dX = xm dx, dY = yn dy, dZ = z–l dz ∴

By using integration, we have X =

∴ and

xm + 1 yn + 1 z1 − l , Y= , Z= . m+1 n+1 1− l

P=

∂Z dZ ∂z dx ∂z 1 z − l ∂z = = z− l . . . . m = m ∂X dz ∂x dX ∂x x x ∂x

Q=

∂Z dZ ∂z dx ∂z 1 z − l ∂z = = z− l = n . . . n ∂Y dz ∂y dY ∂y y y ∂y

∴ (2) ⇒ PQ = 1 This equation is of the form g(P, Q) = 0. Let Z = aX + φ(a) Y + c be the complete solution of (3), where φ(a) is some function of a. (4) ⇒

P=

∂Z ∂Z = a and Q = = φ(a) ∂X ∂Y

...(3) ...(4)

52

PARTIAL DIFFERENTIAL EQUATIONS

∴ (3) ⇒

a φ(a) = 1

or

φ(a) = 1/a

∴ The complete solution of (3) is Z = aX +

∴ The complete solution of (1) is

1 Y + c. a

z1 − l a 1 = xm+ 1 + y n + 1 + c , where a and 1− l m +1 (n + 1) a

c are arbitrary constants and a ≠ 0. (1 – x2) yp2 + x2q = 0.

(iv) We have (1)

1 − x2

⇒

x

F GG Hx/

⇒

2

1 1 − x2

FG ∂z IJ H ∂x K I ∂z J ∂x JK

2

+

1 ∂z =0 y ∂y

+

1 ∂z =0 y ∂y

2

Let X, Y, Z be new variables such that dX = ∴

...(1)

...(2) x

1 − x2

dx , dY = y dy, dZ = dz

By using integration, we have X= −

1 (1 − x 2 ) 1/2 y2 . = − 1 − x2 , Y = , Z = z. 2 1/ 2 2

2 ∂Z dZ ∂z dx ∂z 1 − x 1 ∂z = . . = 1 . . = P= 2 ∂x ∂X dz ∂x dX ∂x x x/ 1− x

∴

and

Q= ∴ (2)

⇒

∂Z dZ ∂z dy ∂z 1 1 ∂z . . . = . = = 1. ∂Y dz ∂y dY ∂y y y ∂y

P2 + Q = 0

...(3)

This equation is of the form g(P, Q) = 0. Let

Z = aX + φ(a) Y + c

...(4)

be the complete solution of (3), where φ(a) is some function of a. (4)

P=

⇒

∴ (3)

⇒

∂Z ∂Z = a and Q = = φ(a) ∂X ∂Y

a2 + φ(a) = 0 or φ(a) = – a2

∴ The complete solution of (3) is Z = aX – a2Y + c. ∴ The complete solution (1) is z = − a 1 − x 2 − constants.

a2 2 y + c , where a and c are arbitrary 2

53

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

TEST YOUR KNOWLEDGE Find 1. 3. 5. 7. 9.

the complete solution of the following partial differential equations : 2. pq = xm yn zl p2x + q2y = z p2 – q2 = z 4. p2 + q2 = z p2y (1 + x2) = qx2 6. zy2p = x(y2 + z2q2) 2 2 2 2 2 8. x4p2 + y2zq – z2 = 0 z (p /x + q /y ) = 1 m 2m l n 2n lm/(m–n) p sec x + z q cosec y = z 10. z2p2y + 6zpxy + 2zqx2 + 4x2y = 0.

Answers 1. 3.

z = a x ± 1 − a2

2 z = ax ± a 2 − 1 y + c, |a| ≥ 1

2 5. z = a 1 + x +

7. z2 = ax2 ± 9. 10.

y + c, – 1 ≤ a ≤ 1

1 2 2 a y +c 2

1 − a 2 y2 + c, – 1 ≤ a ≤ 1

2.

2z1 − ( l / 2 ) a 1 xm + 1 + y n + 1 + c, a ≠ 0 = 2−l m +1 (n + 1) a

4. 2 z = ax ± 1 − a 2 y + c, – 1 ≤ a ≤ 1 6. z2 = ax2 ±

a − 1 y2 + c, a ≥ 1

8. xy log z + ay = (a2 – 1) x + cxy

m 1/ n a m−n z ( m − n − l )/( m − n ) = ( 2x + sin 2x ) + (1 − a ) (2y – sin 2y) + c m−n−l 4 4

z 2 = ax 2 −

Fa GH 2

2

I JK

+ 3a + 2 y 2 + c.

3.7. CHARPIT’S GENERAL METHOD OF SOLUTION If the given partial differential equation is not of any of the given special types, then the given equation is solved by using Charpit’s general method. Let f(x, y, z, p, q) = 0 ...(1) be the given partial differential equation of first order and non-linear in p and q. Since z is a function of x and y, we have dz = p dx + q dy ...(2) The procedure is to first find an equation involving x, y, z, p, q. Let F(x, y, z, p, q) = 0 ...(3) be the required equation involving x, y, z, p, q. The equations (1) and (3) are solved to find the values of p and q. The values of p and q are substituted in (2) and is then integrated to get the desired result. Differentiating (1) and (3) partially w.r.t. x and y, we get

∂f ∂f ∂f ∂p ∂f ∂q + p+ + =0 ∂x ∂z ∂p ∂x ∂q ∂x

...(4)

∂f ∂f ∂f ∂p ∂f ∂q + q+ + =0 ∂y ∂z ∂p ∂y ∂q ∂y

...(5)

∂F ∂F ∂F ∂p ∂F ∂q + p+ + =0 ∂x ∂z ∂p ∂x ∂q ∂x

...(6)

54

PARTIAL DIFFERENTIAL EQUATIONS

∂F ∂F ∂F ∂p ∂F ∂q + q+ + =0 ∂y ∂z ∂p ∂y ∂q ∂y Multiplying (4) by

∂F ∂f , (6) by and subtracting, we get ∂p ∂p

FG H

IJ K

FG H

IJ K

...(8)

FG H

IJ FG K H

IJ K

...(9)

∂f ∂F ∂F ∂f ∂f ∂F ∂F ∂f ∂f ∂F ∂F ∂f ∂q − − + − p+ =0 ∂x ∂p ∂x ∂p ∂z ∂p ∂z ∂p ∂q ∂p ∂q ∂p ∂x Multiplying (5) by

∂F ∂F , (7) by and subtracting, we get ∂q ∂q

∂f ∂F ∂F ∂f ∂f ∂F ∂F ∂f ∂f ∂F ∂F ∂f ∂p − =0 − + − q+ ∂y ∂q ∂y ∂q ∂z ∂q ∂z ∂q ∂p ∂q ∂p ∂q ∂y Adding (8) and (9), we get

FG H

...(7)

IJ K

FG H

IJ K

∂f ∂F ∂F ∂f ∂f ∂F ∂F ∂f ∂f ∂F ∂F ∂f ∂f ∂F ∂F ∂f − + − p+ − + − q =0 ∂x ∂p ∂x ∂p ∂z ∂p ∂z ∂p ∂y ∂q ∂y ∂q ∂z ∂q ∂z ∂q

⇒

F∵ GH

∂q ∂2 z ∂2 z ∂p = = = ∂x ∂x∂y ∂y∂x ∂q

FG ∂f + p ∂f IJ ∂F + FG ∂f + q ∂f IJ ∂F + FG − p ∂f − q ∂f IJ ∂F H ∂x ∂z K ∂p H ∂y ∂z K ∂q H ∂p ∂q K ∂z F ∂f IJ ∂F + FG − ∂f IJ ∂F = 0 + G− H ∂p K ∂x H ∂q K ∂y

I JK

...(10)

This is a Lagrange equation of first order with independent variables x, y, z, p, q and dependent variable F. The auxiliary equations of (10) are dp dq dz dx dy dF ...(11) = = = = = ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f 0 +p +q −p −q − − ∂x ∂z ∂y ∂z ∂p ∂q ∂p ∂q Any of the integrals of (11) satisfy (10). If any such integral involve p or q or both, it can be taken as the assumed relation (3). Simpler the integral involving p or q or both, that is derived from (11), the easier will be the solution of (1). Substituting these values of p and q in dz = p dx + q dy and then integrating, we find the required solution.

ILLUSTRATIVE EXAMPLES Example 1. Find the complete solution of the following partial differential equations by using Charpit’s method : (i) z = px + qy + p2 + q2 (ii) z2 = pqxy (iii) px + qy = pq (iv) (p2 + q2) y = qz 2 (v) p = (qy + z) . Sol. (i) We have z = px + qy + p2 + q2. ...(1) 2 2 Let f(x, y, z, p, q) = z – px – qy – p – q . ∂f ∂f ∂f ∂f ∂f = − p, = − q, = 1, = − x − 2 p, = − y − 2q ∴ ∂x ∂y ∂z ∂p ∂q

55

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

Charpit’s auxiliary equations are dp ∂f ∂f +p ∂x ∂z

=

dq ∂f ∂f +q ∂y ∂z

=

dz dx dy = = . ∂f ∂f ∂f ∂f −p −q − − ∂p ∂q ∂p ∂q

dp dq dz dx dy = = = = − p + p(1) − q + q(1) − p( − x − 2 p) − q( − y − 2q ) − x − 2 p − y − 2q

⇒

dp dq dz dx dy = = = = 2 2 − − − − 2q 0 0 x 2 p y px + qy + 2 p + 2q

⇒

First fraction implies dp = 0. Let p = a. Similarly, let q = b. Now dz = p dx + q dy ∴ dz = a dx + b dy Integrating, we get z = ax + by + c. Putting the value of z in (1), we get ax + by + c = ax + by + a2 + b2 or c = a2 + b2. ∴ z = ax + by + a2 + b2. This is the complete solution. (ii) We have z2 = pqxy. ...(1) 2 Let f(x, y, z, p, q) = z – pqxy.

∂f ∂f ∂f ∂f ∂f = − pqy, = − pqx, = 2 z, = − qxy, = − pxy ∂x ∂y ∂z ∂p ∂q Charpit’s auxiliary equations are

∴

dp dq dz dx dy = = = = ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f +p +q −p −q − − ∂x ∂z ∂y ∂z ∂p ∂q ∂p ∂q

⇒

dy dp dq dz dx = = = = − ( − pxy ) − pqy + p( 2z ) − pqx + q( 2z ) − p( − qxy ) − q( − pxy ) − ( − qxy )

⇒ (2) ⇒

dp dq dz dx dy = = = = 2 pz − pqy 2qz − pqx 2 pqxy qxy pxy xdp + pdx ydq + qdy = x ( 2 pz − pqy ) + p( qxy ) y( 2qz − pqx ) + q( pxy )

x dp + p dx y dq + q dy = 2xpz 2 yqz

⇒ Integrating, we get ⇒

⇒

⇒

d( xp ) d( yq ) = xp yq

log xp = log yq + a xp = yqb2

Solving (1) and (3) for p and q, we get p = Now

...(2)

dz = p dx + q dy ∴ dz =

...(3) (Putting a = log b2)

bz z and q = . x by

bz z dx + dy x by

dz b 1 = dx + dy z x by

56

PARTIAL DIFFERENTIAL EQUATIONS

Integrating, we get

log z = b log x +

1 log y + log c. b

∴ z = cxb y1/b This is the complete solution. (iii) We have px + qy = pq. Let f(x, y, z, p, q) = px + qy – pq. ∂f ∂f ∂f ∂f ∂f = p, = q, = 0, = x − q, = y− p ∴ ∂x ∂y ∂z ∂p ∂q Charpit’s auxiliary equations are

...(1)

dp dq dz dx dy = = = = ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f +p +q −p −q − − ∂x ∂z ∂y ∂z ∂p ∂q ∂p ∂q

dp dq dz dx dy = = = = p + p( 0) q + q( 0) − p( x − q ) − q( y − p ) − ( x − q ) − ( y − p )

⇒ ⇒ (2)

⇒

dp dq dz dx dy = = = = p q − px − qy + 2 pq q − x p − y

...(2)

dp dq = p q

...(3)

⇒ log p = log q + log a ⇒ p = aq

Solving (1) and (3), we get p = 0, q = 0 or p = ax + y, q =

ax + y . a

Case I. p = 0, q = 0 ∴ dz = p dx + q dy ⇒ dz = 0 dx + 0 dy = 0 ⇒ z = c. This is not a complete solution, because it does not contain two arbitrary constants. Case II. p = ax + y, q = ∴

ax + y a

dz = p dx + q dy ⇒ dz = (ax + y) dx + =

FG ax + y IJ dy H a K

ax + y ax + y ( a dx + dy ) = d ( ax + y ) a a

(ax + y)2 + b. 2a This is the complete solution.

Integrating, we get z =

(iv) We have (p2 + q2) y = qz. Let f(x, y, z, p, q) = p2y + q2y – qz.

∂f ∂f ∂f ∂f ∂f = 0, = p2 + q 2 , = − q, = 2 py, = 2qy – z ∂x ∂y ∂z ∂p ∂q Charpit’s auxiliary equations are

∴

dp dq dz dx dy = = = = ∂f ∂f ∂f ∂f ∂f ∂f ∂f ∂f +p +q −p −q − − ∂x ∂z ∂y ∂z ∂p ∂q ∂p ∂q

...(1)

57

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

⇒

dp dq dz dx dy = 2 = = = 2 0 + p( − q ) p + q + q( − q ) − p( 2 py ) − q( 2qy − z ) − 2 py − ( 2qy − z )

⇒

dz dx dy dp dq = 2 = = = 2 2 − 2 py z − 2qy − pq p − 2 p y − 2q y + qz

(2) ⇒

dp dq dp dq = 2 ⇒ = − pq p −q p

p2 q2 a + = 2 2 2

Integrating, we get

Solving (1) and (3), we get

⇒

or p2 + q2 = a a2 y2 z

2

z dz – ay dy =

2

2 2

az − a y

dx ⇒

1 d ( az 2 − a 2 y 2 ) = az 2 − a 2 y 2 dx 2a

⇒

...(3)

and q =

dz = p dx + q dy ⇒ dz =

∴ ⇒

⇒ pdp + qdq = 0

a−

p=

...(2)

ay . z

az 2 − a 2 y 2 z

d

dx +

ay dy z

F z I − d F ay I = GH 2 JK GH 2 JK 2

2

d( az 2 − a 2 y 2 ) az 2 − a 2 y 2

az 2 − a 2 y 2 dx

= 2adx

Integrating, we get 2 az 2 − a 2 y 2 = 2ax + 2b ⇒ az2 – a2y2 = (ax + b)2. This is the complete solution. (v) We have p = (qy + z)2. Let f(x, y, z, p, q) = (qy + z)2 – p ∂f = 0, ∂x

∴

∂f = 2(qy + z)q, ∂y

...(1) ∂f = 2(qy + z), ∂z

∂f = – 1, ∂p

∂f = 2(qy + z) y ∂q

Charpit’s auxiliary equations are

dp ∂f ∂f +p ∂x ∂z

=

dq dz dx dy = = = ∂f ∂f ∂f ∂f ∂f ∂f − − +q −p −q ∂y ∂z ∂p ∂q ∂p ∂q

dp dq dz dx = = = 0 + p2(qy + z) 2q(qy + z) + q2(qy + z) − p(− 1) − q2 y(qy + z) − (− 1) dy = − 2 y(qy + z) dp dq dz dx dy = = ⇒ = = ...(2) 2 p(qy + z) 4 q(qy + z) p − 2 yq(qy + z) 1 − 2 y(qy + z) dp dy dp dy = + (2) ⇒ ⇒ =0 2 p(qy + z) − 2 y(qy + z) p y Integrating, we get log | p | + log | y | = log C ⇒ | py | = C ⇒ py = ± C ⇒ py = a (Putting a = ± C) ∴ p = a/y ⇒

58

PARTIAL DIFFERENTIAL EQUATIONS

a = (qy + z)2. y

Putting the value of p in (1), we get qy + z =

⇒

a y

⇒ qy =

a −z y

dz = pdx + qdy ⇒ dz =

∴ ⇒

ydz = adx +

⇒

d(yz) = adx +

Integrating, we get yz = ax +

F GH

a y a y

a .

I JK

− z dy

⇒ q=

F GH

a y

3/ 2

−

I JK

z y

a a z dx + 3 / 2 − dy y y y

⇒ ydz + zdy = adx +

a y

dy

dy

y 1/2 + b ⇒ yz = ax + 2 ay + b. 1 /2

This is the complete solution. WORKING STEPS FOR USING CHARPIT’S METHOD Step I.

Shift all terms of the given equation to the left side and denote the left side by f(x, y, z, p, q).

∂f ∂f ∂f ∂f ∂f , , , . ∂x ∂y ∂z ∂p ∂q Step III. Write the Charpit’s auxiliary equations and substitute the values of partial derivatives of f and simplify. Step IV. Select any two fractions so that the resulting integral is the simplest relation involving at least one of p and q. This relation and the given equation are solved to find the values of p and q. Step V. Put the values of p and q in the equation dz = p dx + q dy and integrate. This gives the complete solution of the given equation.

Step II. Find

TEST YOUR KNOWLEDGE Find the complete solution of the following partial differential equations by using Charpit’s method : 2. q = 3p2 1. q = px + q2 2 2 2 2 3. p – y q = y – x 4. pxy + pq + qy = yz 5. 2(z + px + qy) = yp2 6. 2z + p2 + qy + 2y2 = 0 2 8. 2xz – px2 – 2qxy + pq = 0 7. q = px + p 2 9. p(1 + q ) + (b – z)q = 0 10. (p2 + q2)x = pz.

Answers 1. z = (a – a2) log x + ay + b 3. z =

x a2 − x 2 a2 x a2 sin − 1 − + − y+b a y 2 2

2. z = ax + 3a2y + b 4. (z – ax) (y + a)a = bey

59

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

5. yz – 7. z = –

ax a2 + =b y 4 y2

6. y2((a – x)2 + 2z + y2) = b

LM MN

x x2 + 4 a x2 ± + a log ( x + 4 4

OP PQ

x2 + 4 a ) + ay + b

8. z = ay + b(x2 – a) 10.

az2

–

a2x2

= (ay +

9. 2 c( z − b) − 1 = x + cy + a b)2.

Hints y( z − ax ) . a+y

4. p = a ⇒ q =

∴ dz = p dx + q dy ⇒ dz = a dx + 5. Charpit’s auxiliary equations implies Also dz =

a y

2

F GH

dx + −

I JK

y( z − ax ) dy a+y

⇒

dz − a dx y = dy . z − ax a+y

dp dy = or py2 = a 4 p − 2y

z ax a2 − 3 + dy y y 2y4

⇒ (y dz + z dy) – a

F 1 dx − x dy I − a GH y J y K 2y

2

2

3

1 ( 2z + 2 y 2 + ( a − x )2 ) dy y Multiplying by 2y2, we get (2y2 dz + 4yz dy) = (2y2(a – x) dx – 2y(a – x)2 dy) – 4y3 dy.

6. dz = (a – x) dx –

dy = 0 .

Homogeneous Linear Partial Differential Equations with Constant Coefficients

4

4.1. INTRODUCTION Till now we have been discussing the methods of solving partial differential equations of the first order. A partial differential equation of the first order involves, only the first order partial derivatives (p and q) of the dependent variable z. Now we shall consider the solution of partial differential equations of order higher than one. 4.2. PARTIAL DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER We know that the order of a partial differential equation is the order of the highest partial derivative occurring in the given partial differential equation. For example,

∂3 z ∂x 3

+2

∂ 2 z ∂z + = x2 + y is a partial differential equation of order 3. For ∂x∂y ∂y

∂ ∂ are denoted by D(or Dx) and D′(or Dy) respectively. Thus, and ∂x ∂y the above differential equation can also be written as (D 3 + 2DD′ + D′)z = x 2 + y or as (D x3 + 2Dx Dy + Dy)z = x2 + y. the sake of simplicity,

Remark. DD′z stands for

∂z ∂z ∂z ∂z ∂2 z . . and not for . The product is denoted as (Dz)(D′z). ∂ ∂ ∂ x y x ∂y ∂x∂y

4.3. HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS We know that a partial differential equation is called linear if the dependent variable and its partial derivatives occur only in the first degree and are not multiplied together. A linear partial differential equation of order n is of the form

FA GH

0

∂n z ∂x

n

+ A1

∂n z ∂x

n−1

∂y

+ ......+ A n

∂n z ∂y

n

I + FB JK GH

0

∂ n − 1z ∂x

n−1

+ ......+ Bn − 1

FG H

∂ n − 1z ∂y

n−1

I JK

IJ K

∂z ∂z +N + Pz = F(x, y), ∂x ∂y where the coefficients A0, A1, ...... N, P are constants or functions of x and y. If the coefficients A0, A1, ......, N, P are all constants then such a differential equation is called a linear partial differential equation with constant coefficients. + ...... + M

60

61

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

In a linear partial differential equation, the orders of various partial derivatives occurring in the equation may or may not be equal. In case the orders of all partial derivatives involved in the equation are same then it is called a homogeneous linear partial differential equation and otherwise it is called a non-homogeneous linear partial differential equation. A homogeneous linear partial differential equation with constant coefficients is of the ∂n z ∂n z ∂n z + A 1 n − 1 + ......+ A n = F( x, y) , form A 0 n ∂x ∂x ∂y ∂y n where A0, A1, ......, An are all constants. Consider the following partial differential equations : (i) 4

∂2 z ∂2 z ∂2 z + + 5 = ex ∂x∂y ∂y 2 ∂x 2

(ii)

∂2 z ∂2 z − = cos x cos 2y ∂x 2 ∂x∂y

∂2 z ∂z + – z = e–x ∂x∂y ∂y ∂x (vi) r – s + 2q – z = x2y2 (viii) r + (y/x)s = 15xy2

(iii) (D3 – 2D2D′ – DD′2 + 2D′3)z = sin x

(iv)

(v) (2DD′ + D′2 – 3D)z = 3 cos (3x – 2y) (vii) xyr + x2s – yt = x3ey

∂2 z 2

+

(ix) yt + xs + q = 8yx2 + 9y2.

Differential equations (i), (ii) and (iii) are homogeneous linear partial differential equations with constant coefficients. Differential equations (iv), (v) and (vi) non-homogeneous linear partial differential equation with constant coefficients. Differential equations (vii) and (viii) are homogeneous linear partial differential equations with variable coefficients. Differential equation (ix) is a non-homogeneous linear partial differential equation with variable coefficients. In the present chapter, we shall consider the methods of solving homogeneous linear partial differential equations with constant coefficients. 4.4. SOME THEOREMS Let f(D, D′)z = F(x, y) be a linear partial differential equation with constant coefficients. ∴ The function f(D, D′) is of the form

FA GH

0

∂n z ∂x

n

+ A1

∂n z ∂x

n−1

∂y

+ ......+ A n

∂n z ∂y

n

I + FB JK GH

0

∂ n − 1z ∂x

n−1

+ ......+ Bn − 1

∂ n − 1z ∂y

n−1

I JK

FG H

+ ...... + M

∂z ∂z +N ∂x ∂y

IJ + Pz K

where A0, A1, ......, N, P are all constants. Theorem 1. Let f(D, D′)z = 0 be a linear partial differential equation with constant coefficients. If u1, u2, ..., um be m solutions of f(D, D′)z = 0, then prove that m

∑cu i

i=1

i

is also a solution of f(D, D′′)z = 0.

Proof. ui is a solution of f(D, D′) z = 0 for 1 ≤ i ≤ m. ∴ f(D, D′) ui = 0 for 1 ≤ i ≤ m

62

PARTIAL DIFFERENTIAL EQUATIONS

F I GG ∑ c u JJ = ∑ f (D, D′ ) (c u ) = ∑ c f (D, D′ )u = ∑ c (0) = 0. H K m

Now f(D, D′)

m

m

i i

i=1

i i

m

i

i=1

i

i=1

i

i=1

m

∑ cu

∴

i i

is also a solution of the equation F(D, D′)z = 0.

i=1

Theorem 2. Let f(D, D′)z = F(x, y) be a linear partial differential equation with constant coefficients. If u is a solution of f(D, D′)z = 0 and v is a solution of f(D, D′) z = F(x, y), then prove that u + v is a solution of f(D, D′) z = F(x, y). Proof. u is a solution of f(D, D′)z = 0. ∴ f(D, D′)u = 0 ...(1) v is a solution of f(D, D′)z = F(x, y). ∴ f(D, D′)v = F(x, y) ...(2) Now f(D, D′) (u + v) = f(D, D′)u + f(D, D′)v = 0 + F(x, y) (Using (1), (2)) = F(x, y). ∴ u + v is a solution of f(D, D′)z = F(x, y). 4.5. GENERAL SOLUTION OF HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATION f(D, D′′)z = 0 WITH CONSTANT COEFFICIENTS f(D, D′)z = A 0

∂n z

A0

FG D IJ H D′ K

n

+ A1

FG D IJ H D′ K

n−1

+ A1

∂n z

∂n z

=0 ...(1) ∂x n ∂x n − 1∂y ∂y n be a homogeneous linear partial differential equation with constant coefficients, where A0 ≠ 0. (1) can also be written as (A0Dn + A1Dn–1D′ + ...... + AnD′n) z = 0 ...(2) Let (2) be equivalent to A0[(D – m1D′)(D – m2D′) ...... (D – mnD′)] z = 0 ...(3) Treating D and D′ as variables, the equations (2) and (3) implies Let

+ ......+ A n

+ ...... + An = A 0

FG D − m IJ FG D − m IJ ...... FG D − m IJ H D ′ K H D′ K H D ′ K 1

2

n

∴ m1, m2, ......, mn are roots of the equation A0mn + A1mn–1 + ...... + An = 0. ...(4) The equation (4) is called the auxiliary equation of (2). This equation can be obtained by putting D equal to m and D′ equal to ‘1’ in the operator of equation (2) and equating it to zero. The roots m1, m2, ......, mn of the auxiliary equation may or may not be distinct. Case I. Roots are distinct. ∴ m1 ≠ m2 ≠ m3 ≠ ...... ≠ mn. Equation (3) shows that for 1 ≤ i ≤ n, the solution of (D – miD′) z = 0 is a solution of (3) and hence of (1). (D – miD′) z = 0 ⇒ p – mi q = 0 dx dy dz = = ⇒ ⇒ dy + mi dx = 0, dz = 0 1 − mi 0 ⇒ y + mix = c1, z = c2

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

63

(Putting c2 = φi(c1)) ⇒ z = φi(y + mix). ∴ z = φ1(y + m1x), z = φ2(y + m2x), ...... , z = φn(y + mn x) are solutions of (1). ∴ z = φ1(y + m1x) + φ2(y + m2x) + ...... + φn(y + mnx) is also a solution of (1). Since this solution contains n arbitrary functions φ1, φ2, ......, φn, this solution represents the general solution of the given equation. Case II. Roots are not distinct. Let m1 = m2 ≠ m3 ≠ ...... ≠ mn. In this case, the solution of (1) can be written as z = (φ1 + φ2)(y + m1x) + ...... + φn(y + mnx). (∵ m1 = m2) This solution contains only n – 1 arbitrary functions φ1 + φ2, φ3, ......, φn. ∴ This is not a general solution. Using m1 = m2 in (3), we get, A0 [(D – m1D′)2 (D – m3D′) ...... (D – mnD′)] z = 0 ...(5) Equation (5) shows that for 3 ≤ i ≤ n, the solution of (D – miD′)z = 0 is also a solution of (5) and hence of (1). ∴ z = φi(y + mix), 3 ≤ i ≤ n is a solution of (1). ∴ z = φ3(y + m3x) + ...... + φn(y + mnx) ...(6) is a solution of (1). The solution of (D – m1D′)2z = 0 is also a solution of (5) and hence of (1). (D – m1D′)2 z = 0 ⇒ (D – m1D′)(D – m1D′) z = 0 ...(7) ...(8) Let (D – m1D′) z = u. ∴ (7) ⇒(D – m1D′)u = 0 ⇒ u = ψ1(y + m1x), where ψ1 is arbitrary. ∴ (7) ⇒ p – m1q = ψ1(y + m1x) ⇒

dx dy dz = = 1 − m1 ψ 1 ( y + m1 x)

...(9)

(9) ⇒ dy + m1dx = 0 ⇒ y + m1x = c Taking the first and third fractions of (9), we get dx =

dz ψ 1 (c)

or dz = ψ1(c) dx

⇒ z = ψ1(c) x + d ⇒ z = xψ1(y + m1x) + ψ2(y + m2x) ...(10) (Putting d = ψ2(c)) Combining (6) and (10), z = xψ1(y + m1x) + ψ2(y + m2x) + φ3(y + m3x) + ...... + φn(y + mnx) is also a solution of (1). Since this solution contains n arbitrary functions ψ1, ψ2, φ3, ......, φn , this solution represents the general solution of the given equation. Remarks 1. If the root m1 of the auxiliary equation is repeated r times, then the corresponding part of the general solution is φ1(y + m1x) + xφ2(y + m1x) + ...... + xr–1φr(y + m1x), where φ1, φ2, ......, φr are arbitrary functions. 2. The auxiliary equation of a homogeneous linear partial differential equation with constant coefficients is obtained by putting D = m and D′ = 1 in the operator of the equation and then equating it to zero.

Exceptional Case. If A0 = 0, A1 ≠ 0, then equation (2) becomes (A1Dn–1D′ + ...... + AnD′n) z = 0 or D′(A1Dn – 1 + A2Dn–2D′ + ... + AnD′ n–1)z = 0 ∴ The solution of D′z = 0 is also a solution of (1).

...(1)

64

PARTIAL DIFFERENTIAL EQUATIONS

∂z = 0 ⇒ z = φ(x), where φ is arbitrary. ∂y Similarly, if D′r is a factor of the operator of the equation then the corresponding part of the general solution is z = φ1(x) + yφ2(x) + ...... + yr–1φr(x), where φ1, φ2, ......, φr are arbitrary functions. D′z = 0 ⇒

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations : ∂2z ∂2z ∂2z − 8 + 15 =0 (ii) 2r + 5s + 2t = 0 ∂x∂y ∂x 2 ∂y 2 (iii) (D3 – 6D2D′ + 11DD′ 2 – 6D′ 3) z = 0 (iv) (D4 + D′ 4 – 2D2D′ 2) z = 0 (v) (D3D′ 2 + D2D′ 3) z = 0.

(i)

∂2 z ∂2 z ∂2 z −8 + 15 2 = 0 . 2 ∂x∂y ∂x ∂y 2 2 ⇒ (D – 8DD′ + 15D′ ) z = 0

Sol. (i) We have

...(1)

By putting D = m and D′ = 1 in the operator of (1), the auxiliary equation of (1) is m2 – 8m + 15 = 0. ∴ m = 3, 5 ∴ The general solution of the given equation is z = φ1(y + 3x) + φ2(y + 5x), where φ1, φ2 are arbitrary functions. (ii) We have 2r + 5s + 2t = 0. ∴ (2D2 + 5DD′ + 2D′2) z = 0 ...(1) The auxiliary equation of (1) is 2m2 + 5m + 2 = 0. ∴ m = – 1/2, – 2 ∴ The general solution of the given equation is

FG H

IJ K

1 x + φ 2 (y − 2x) , where φ1, φ2 are arbitrary functions. 2 (iii) We have (D3 – 6D2D′ + 11DD′2 – 6D′3)z = 0. ...(1) 3 2 The auxiliary equation of (1) is m – 6m + 11m – 6 = 0. ∴ m = 1, 2, 3 ∴ The general solution of the given equation is z = φ1(y + x) + φ2(y + 2x) + φ3(y + 3x), where φ1, φ2, φ3 are arbitrary functions. (iv) We have (D4 + D′4 – 2D2D′2)z = 0 ...(1) The auxiliary equation of (1) is m4 + 1 – 2m2 = 0. ∴ (m2 – 1)2 = 0 or m = 1, 1, – 1, – 1. ∴ The general solution of the given equation is φ2(y + x) + φ3(y – x) + xφ φ4(y – x), z = φ1(y + x) + xφ where φ1, φ2, φ3, φ4 are arbitrary functions. (v) We have (D3D′2 + D2D′3)z = 0. ...(1) (1) ⇒ D′2D2(D + D′)z = 0 The part of general solution corresponding to the factor D′2 is φ1(x) + yφ2(x). z = φ1 y −

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

The auxiliary equation of D2(D + D′)z = 0 is m2(m + 1) = 0 ∴ m = 0, 0, – 1 ∴ The part of general solution corresponding to the factor D2(D + D′) is φ3(y + 0.x) + xφ4(y + 0.x) + φ5(y + (– 1)x). ∴ The general solution of the given equation is φ2(x) + φ3(y) + xφ φ4(y) + φ5(y – x), z = φ1(x) + yφ where φ1, φ2, φ3, φ4, φ5 are arbitrary functions. WORKING STEPS FOR SOLVING PROBLEMS Step I. Express the given equation in terms of D and D′. Step II. Put D = m and D′ = 1 in the operator of the equation and equate it to zero. This is the auxiliary equation of the given equation. Solve the auxiliary equation. Step III. If mi is a distinct root then the corresponding part of the general solution is φ(y + mix). If mi is a root repeated r times then the corresponding part of the general solution is φ1(y + mi x) + xφ2(y + mi x) + ...... + xr–1φr(y + mi x). Step IV. If D′r is a factor of the operator of the equation then we put D = m and D′ = 1 in the other factor of the operator and the part of the general solution corresponding to D′r is taken as φ1(x) + yφ2(x) + ...... + yr–1φr(x).

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. r = a2t 3.

2.

F ∂ z − 2 ∂ z − ∂ zI = 0 GH ∂x ∂x∂y ∂y JK 2

2

2

2

∂2 z ∂x

2

−

∂2 z

∂y2

=0

4. (D2 – 4DD′ + 4D′2) z = 0

2

6. (D3 + 2D2D′ – DD′2 – 2D′3) z = 0 8. (D3 + D2D′ – 6DD′2) z = 0

5. 25r – 40s + 16t = 0 7. (D3 – 3DD′2 + 2D′3) z = 0 9. (D3 – 2D2D′) z = 0

10.

∂3 z

∂3 z

+6

∂3 z

=0 ∂x3 ∂x∂y2 ∂y3 12. (D2D′ – 3DD′2 + 2D′3) z = 0 14. (D′3D + D′4) z = 0.

11. (D3 – 3D2D′ + 3DD′2 – D′3) z = 0 13. (D4 – 2D3D′ + 2DD′3 – D′4) z = 0

−7

Answers 1. z = φ1(y + ax) + φ2(y – ax) 3. z = φ1(y + (1 +

FG H

z= z= z= z= z=

FG H

IJ K

2 )x)

4. z = φ1(y + 2x) + xφ2(y + 2x)

4 4 6. x + xφ 2 y + x 5 5 φ1(y + x) + xφ2(y + x) + φ3(y – 2x) 8. 10. φ1(y) + xφ2(y) + φ3(y + 2x) φ1(y + x) + xφ2(y + x) + x2φ3(y + x) 12. φ1(y – x) + φ2(y + x) + xφ3(y + x) + x2φ4(y + x) φ1(x) + yφ2(x) + y2φ3(x) + φ4(y – x).

5. z = φ1 y + 7. 9. 11. 13. 14.

IJ K

2. z = φ1(y + x) + φ2(y – x)

2 )x) + φ2(y + (1 –

z = φ1(y + x) + φ2(y – x) + φ3(y – 2x) z = φ1(y) + φ2(y + 2x) + φ3(y – 3x) z = φ1(y + x) + φ2(y + 2x) + φ3(y – 3x) z = φ1(x) + φ2(y + x) + φ3(y + 2x)

65

66

PARTIAL DIFFERENTIAL EQUATIONS

4.6. GENERAL SOLUTION OF HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATION f(D, D′′)z = F(x, y) WITH CONSTANT COEFFICIENTS Let f(D, D′) z = F(x, y) ...(1) be a homogeneous linear partial differential equation with constant coefficients. Let u be the general solution of f(D, D′) z = 0. ∴ f(D, D′) u = 0 ...(2) Let v be a particular integral i.e., a particular solution of f(D, D′) z = F(x, y). ∴ f(D, D′) v = F(x, y) ...(3) Now, f(D, D′)(u + v) = f(D, D′)u + f(D, D′)v = 0 + F(x, y) = F(x, y) (Using (2) and (3)) ∴ u + v is a solution of f(D, D′)z = F(x, y). Since u is the general solution of the equation f(D, D′)z = 0, it contains arbitrary functions equal in number to its order. ∴ The solution u + v of the equation f(D, D′)z = F(x, y) also contains as many arbitrary functions as the order of f(D, D′)z = F(x, y). ∴ u + v is the general solution of the equation f(D, D′)z = F(x, y). The general solution u of the equation f(D, D′)z = 0 is called the complementary function (C.F.) of the equation f(D, D′)z = F(x, y). ∴ The general solution of the equation f(D, D′′)z = F(x, y) is obtained by adding the general solution of the equation f(D, D′′)z = 0 to any particular integral of the equation f(D, D′′)z = F(x, y). 4.7. PARTICULAR INTEGRAL OF f(D, D′)z = F(x, y) Let f(D, D′)z = F(x, y) be a homogeneous linear partial differential equation with constant coefficients. The inverse operator f(D, D′) ∴

1 of the operator f(D, D′) is defined by the identity f (D, D′ )

FG 1 F(x, y)IJ = F(x, y) K H f (D, D′ )

1 F(x, y) is a particular integral of the equation (1) because f (D, D′ ) f(D, D′)

∴

...(1)

FG 1 F(x, y)IJ = F(x, y) = R.H.S. of (1). H f (D, D′ ) K

1 F(x, y) is a particular integral of the equation f(D, D′)z = F(x, y). f(D, D′ )

4.8. PARTICULAR INTEGRAL WHEN F(X, Y) IS SUM OR DIFFERENCE OF TERMS OF THE FORM xmyn If F(x, y) is sum or difference of terms of the form x my n, then the particular integral 1 F(x, y) of the differential equation f (D, D′)z = F(x, y) is obtained by expanding f (D, D′ ) 1 in an infinite series in ascending powers of either D or D′. The particular integrals f (D, D′ ) obtained in the above mentioned two methods may not be identical. Any of the two particular

67

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

integrals may be used. If m < n, then it is advisible to expand D and in case n < m, then we should expand

1 in ascending powers of f (D, D′ )

1 in ascending powers of D′. f (D, D′ )

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations: (i) (2D2 – 5DD′ + 2D′2)z = 24(y – x) (ii) r + (a + b)s + abt = xy. Sol. (i) We have (2D2 – 5DD′ + 2D′2) z = 24(y – x). ...(1) 2 The A.E. of (1) is 2m – 5m + 2 = 0. ∴ m = 1/2, 2. 1 ∴ C.F. = φ 1 y + x + φ 2 ( y + 2 x) . 2 1 Now, P.I. = 24(y – x) 2D 2 − 5DD′ + 2D′ 2

FG H

IJ K

F F GH GH F 1 + F 5D ′ − D ′ GH GH 2D D

I I 24( y − x) JK JK I + ......I ( y − x) 24 = JK JK 2D 12 F 5 I F x y − x IJ + 30 (1) ( y − x) + (1) + 0J = 12 G G = K H 2 6K D 2D D H

1 5D′ D′ 2 1 − − 2 = 2D 2D 2 D 2

−1

2

2

2

2

3

3

x3 = 6 x2 y + 3x3 . 6 ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is 1 z = φ 1 y + x + φ 2 (y + 2x) + 6x 2 y + 3x 3 , 2 where φ1 and φ2 are arbitrary functions. (ii) We have r + (a + b) s + abt = xy.

= 6 x 2 y − 2 x 3 + 30 .

FG H

IJ K

∂2 z ∂2 z ∂2 z + ( a + b ) + ab = xy ∂x∂y ∂x 2 ∂y 2 ⇒ (D2 + (a + b) DD′ + ab D′2) z = xy The A.E. of (1) is m2 + (a + b) m + ab = 0. ∴ m = – a, – b ∴ C.F. = φ1(y – ax) + φ2(y – bx).

⇒

Now,

P.I. = =

1 D 2 + (a + b) DD′ + D′ 2

F GH

(xy) =

...(1)

1 D2

I JK

LM1 + F (a + b) D′ + D′ I OP J D D K PQ MN GH

D′ D′ 2 1 1 − ( a + b ) − 2 + ...... ( xy) D D2 D

2

2

−1

( xy)

68

PARTIAL DIFFERENTIAL EQUATIONS

=

1

LM xy − (a + b) (x) − 0OP = yFG x IJ − (a + b) x . 24 D N Q H 6K 3

4

D2 ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is 1 3 a+b 4 x , z = φ1(y – ax) + φ2(y – bx) + x y − 6 24 where φ1 and φ2 are arbitrary functions.

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. (D2 + 3DD′ + 2D′2) z = 2x + 3y 2. (D2 – 2DD′ + D′2) z = 12xy 2 2 4. (D2 – a2D′2) z = x2 3. (D – DD′ – 6D′ ) z = xy 2 2 2 5. (D – 6DD′ + 9D′ ) z = 12x + 36xy 6. (D2 – 2DD′ – 15D′2)z = 12xy.

Answers 7 3 3 2 x + x y 6 2 1 3 1 4 x 3. z = φ1(y – 2x) + φ2(y + 3x) + x y + 6 24 5. z = φ1(y + 3x) + xφ2(y + 3x) + 10x4 + 6x3y 1. z = φ1(y – x) + φ2(y – 2x) –

2. z = φ1(y + x) + xφ2(y + x) + 2x3y + x4

1 4 x 12 6. z = φ1(y – 3x) + φ2(y + 5x) + x4 + 2x3y.

4. z = φ1(y + ax) + φ2(y – ax) +

4.9. PARTICULAR INTEGRAL WHEN F(x, y) IS OF THE FORM φ(ax + by) Theorem. If f(D, D′) be a homogeneous function of D and D′ of degree n, then the particular integral of f(D, D′)z = φ(ax + by) is given by

zz z

1 ...... φ(v) dv dv ...... dv, where v = ax + by, provided f(a, b) ≠ 0. f(a, b) Proof. We have Drφ(ax + by) = arφ(r)(ax + by), D′sφ(ax + by) = bsφ(s)(ax + by) r and D D′sφ(ax + by) = arbsφ(r+s)(ax + by). ∴ f(D, D′) φ(ax + by) = f(a, b) φ(n) (ax + by) ( ∵ f(D, D′) is a homogeneous function of degree n) Dividing both sides by non-zero constant f(a, b), we get φ(ax + by) = φ (n) (ax + by) f (a, b) ∴ By definition of the inverse operator f(D, D′), we have f(D, D′)

⇒

1 φ(ax + by) φ (n) (ax + by) = f (D, D′ ) f (a, b) 1 1 φ (n) (v) = φ(v), where v = ax + by. f (D, D′ ) f (a, b)

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

69

Replacing φ(n)(v) by φ(v), we get 1 1 φ(v) = ...... φ(v) dv dv ...... dv , where v = ax + by. f (D, D′ ) f (a, b) On the right side, the function φ(v) is to be integrated n times w.r.t. v, which is also the degree of the homogeneous function f(D, D′).

zz z

zz z

1 1 ...... φ(v) dv dv ...... dv , φ(ax + by) = f(D, D′ ) f(a, b) where v = ax + by, provided f(a, b) ≠ 0. Exceptional Case. If f(a, b) = 0, then bD – aD′ must be a factor of f(D, D′) because ∴ P.I. =

ba – ab = 0. Let the factor bD – aD′ be repeated r times, where r ≥ 1. The value of 1 xr φ(ax + by). r φ(ax + by) is given by (bD − aD′ ) br r ! r 1 x φ(ax + by) = r φ(ax + by). ∴ (bD − aD′ )r b r!

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations : (i) (2D2 – 5DD′ + 2D′ 2) z = 5(y – x) (ii) r + s – 2t = (2x + y)1/2 3 2 2 (iii) (D – 4D D′ + 4DD′ ) z = 6 sin (3x + 2y) (iv) (D3 – 6D2D′ + 11DD′ 2 – 6D′ 3) z = e5x+6y Sol. (i) We have (2D2 – 5DD′ + 2D′ 2)z = 5(y – x). ...(1) The A.E. of (1) is 2m2 – 5m + 2 = 0. ∴ m = 1/2, 2

FG H

Also ∴ ∴

IJ K

1 x + φ 2 ( y + 2 x) . 2 1 5( y − x) P.I. = 2 2D − 5DD′ + 2D′ 2 y – x = ax + by ⇒ a = – 1, b = 1. f(D, D′) = 2D2 – 5DD′ + 2D′2. f(a, b) = f(– 1, 1) = 2(– 1)2 – 5(– 1)(1) + 2(1)2 = 9 ≠ 0 1 P.I. = 5v dv dv, where v = y – x f (a, b)

C.F. = φ 1 y +

∴

(2)

⇒

z

zz

1 5 v3 5 v2 .5. dv = . = ( y − x) 3 . 9 2 9 6 54 Using G.S. = C.F. + P.I., the general solution of the given equation is

=

FG H

z = φ1 y +

IJ K

1 5 x + φ 2 (y + 2x) + (y − x)3 , 2 54

where φ1, φ2 are arbitrary functions. (ii) We have r + s – 2t = (2x + y)1/2. ⇒

∂2 z ∂2 z ∂2 z + − 2 = (2x + y)1/2 ∂x 2 ∂x∂y ∂y 2

...(2)

70

PARTIAL DIFFERENTIAL EQUATIONS

⇒ (D2 + DD′ – 2D′2) z = (2x + y)1/2 The A.E. of (1) is m2 + m – 2 = 0. ∴ m = – 2, 1 ∴ C.F. = φ1(y – 2x) + φ2(y + x). P.I. =

(2)

1 (2x + y)1/2 D + DD′ − 2D′ 2

...(2)

2

2x + y = ax + by ⇒ a = 2, b = 1. f(D, D′) = D2 + DD′ – 2D′2. f(a, b) = f(2, 1) = (2)2 + (2)(1) – 2(1)2 = 4 ≠ 0.

Also, ∴ ∴

...(1)

⇒

P.I. =

1 f (a, b)

z

zz

v1/2 dv dv , where v = 2x + y

1 2v3 / 2 1 2 v5 / 2 1 (2x + y)5/2. = dv = . 4 3 4 3 5/2 15 ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is 1 (2x + y)5/ 2 , where φ1, φ2 are arbitrary functions. z = φ1(y – 2x) + φ2(y + x) + 15 (iii) We have (D3 – 4D2D′ + 4DD′2) z = 6 sin (3x + 2y). ...(1) The A.E. of (1) is m3 – 4m2 + 4m = 0. ∴ m = 0, 2, 2 ∴ C.F. = φ1(y + 0.x) + φ2(y + 2x) + xφ3(y + 2x) = φ1(y) + φ2(y + 2x) + xφ3(y + 2x) 1 6 sin (3x + 2y) ...(2) P.I. = 3 2 D − 4D D′ + 4DD′ 2 3x + 2y = ax + by ⇒ a = 3, b = 2. Also, f(D, D′) = D3 – 4D2D′ + 4DD′2. ∴ f(a, b) = f(3, 2) = (3)3 – 4(3)2(2) + 4(3)(2)2 = 3 ≠ 0. =

∴ (2)

⇒

P.I. =

1 f (a, b)

zz

zzz

6 sin v dv dv dv, where v = 3x + 2y

1 .6 − cos v dv dv = 2 3 = 2 cos v = 2 cos (3x + 2y).

=

z

(− sin v) dv

Using G.S. = C.F. + P.I., the general solution of the given equation is φ3(y + 2x) + 2 cos (3x + 2y), z = φ1(y) + φ2(y + 2x) + xφ where φ1, φ2, φ3 are arbitrary functions. ∴

(iv) We have (D3 – 6D2D′ + 11DD′2 – 6D′3) z = e 5 x + 6 y . The A.E. of (1) is m3 – 6m2 + 11m – 6 = 0 ∴ m = 1, 2, 3 ∴ C.F. = φ1(y + 1 . x) + φ2(y + 2x) + φ3(y + 3x). P.I. =

1 e5 x + 6 y D − 6D D′ + 11DD′ 2 − 6D′ 3 3

2

...(1)

...(2)

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

71

5x + 6y = ax + by ⇒ a = 5, b = 6. Also, f(D, D′) = D3 – 6D2D′ + 11DD′2 – 6D′3 ∴ f(a, b) = f(5, 6) = (5)3 – 6(5)2(6) + 11(5)(6)2 – 6(6)3 = – 91 ≠ 0. 1 e v dv dv dv, where v = 5x + 6y ∴ (2) ⇒ P.I. = f (a, b) 1 1 1 v 1 5x + 6 y e v dv dv = − e v dv = − e =− e . = − 91 91 91 91 ∴ Using G.S. = C.F. +P.I., the general solution of the given equation is

zz

zzz

z

z = φ1(y + x) + φ2(y + 2x) + φ3(y + 3x) −

1 5x + 6y e , 91

where φ1, φ2, φ3 are arbitrary functions. Example 2. Find the general solution of the following partial differential equations : (i) (D3 – 4D2D′ + 4DD′2)z = 4 sin (2x + y) (ii) (D3 – 7DD′ 2 – 6D′ 3)z = x2 + xy2 + y3 + cos (x – y) (iii) (D3 – 4D2D′ + 5DD′ 2 – 2D′ 3) z = ey+2x + (y + x)1/2 (iv) (D3 – 7DD′ 2 – 6D′ 3)z = sin (x + 2y) + e3x+y. Sol. (i) We have (D3 – 4D2D′ + 4DD′ 2) z = 4 sin (2x + y). ...(1) 3 2 The A.E. of (1) is m – 4m + 4m = 0. ∴ m = 0, 2, 2 ∴ C.F. = φ1(y + 0.x) + φ2(y + 2x) + xφ3(y + 2x). 1 4 sin (2x + y) D − 4D D′ + 4DD′ 2 2x + y = ax + by ⇒ a = 2, b = 1.

P.I. =

3

2

...(2)

Also, f(D, D′) = D3 – 4D2D′ + 4DD′2. ∴ f(a, b) = f(2, 1) = (2)3 – 4(2)2(1) + 4(2)(1)2 = 0. D3 – 4D2D′ + 4DD′2 = D(D2 – 4DD′ + 4D′2) = D(D – 2D′)2 1 1 ∴ (2) ⇒ P.I. = . 4 sin (2x + y) (∵ D = 2, D′ = 1 ⇒ D – 2D′ = 0) 2 (D − 2D′ ) D 1 4 (− cos (2 x + y)) 1 . 4 sin ( 2x + y ) = dx* = 2 2 (D − 2D′ ) (D − 2D′ ) 2

z

=–2

1 cos (2x + y) (D − 2D′ ) 2

=–2.

x2

cos (2x + y) (1) 2 2 ! = – x2 cos (2x + y).

( ∵ bD – aD′ = (1)D – 2D′ = D – 2D′)

∴ Using G.S. = C.F. + P.I., the general solution of the given equation is φ 3(y + 2x) – x 2 cos (2x + y), where φ1, φ2, φ3 are arbitrary functions. z = φ 1(y) + φ 2(y + 2x) + xφ (ii) We have (D3 – 7DD′2 – 6D′3) z = x2 + xy2 + y3 + cos (x – y). ...(1) *Alternatively,

1 1 4 sin (2 x + y) = D 2

z

4 sin v dv , where v = 2x + y

= – 2 cos v = – 2 cos (2x + y).

72

PARTIAL DIFFERENTIAL EQUATIONS

The A.E. of (1) is m3 – 7m – 6 = 0. ⇒ m = – 1, – 2, 3 ∴ C.F. = φ1(y + (– 1)x) + φ2(y + (– 2)x) + φ3(y + 3x). 1 P.I. = 3 (x2 + xy2 + y3 + cos (x – y)) D − 7DD′ 2 − 6D′ 3 = Now,

1 1 (x2 + xy2 + y3) + 3 cos (x – y). 2 3 D − 7DD′ − 6D′ D − 7DD′ 2 − 6D′ 3 3

1 (x2 + xy2 + y3) D − 7DD′ 2 − 6D′ 3 3

=

1 D

3

LM1 − F 7 D′ MN GH D LM1 + F 7 D′ MN GH D

D′ 3 +6 3 D

2 2

D′ 3 D3

I OP (x + xy + y ) JK PQ I + ......OP (x + xy + y ) JK PQ −1

2

2

3

=

1 D3

=

1 7 6 ( x 2 + xy 2 + y 3 ) + 5 (2 x + 6 y) + 6 (6) D3 D D 5

=

+6

2

Fx + x y =G H 60 24 Also

2

4

2

+

2

I F JK GH

2

I JK

3

F I GH JK

x 3 y3 x6 x5 y x6 +7 + + 36 6 360 20 720

5 6 1 5 7 5 1 4 2 1 3 3 x + x + x y+ x y + x y . 72 60 20 24 6

1 cos (x – y) D − 7DD′ 2 − 6D′ 3 1 cos (x – y) = (D + D′ ) (D + 2D′ ) (D − 3D′ ) 1 1 . = cos (x – y) D + D′ (D + 2D′ ) (D − 3D′ ) (∵ D = 1, D′ = – 1 ⇒ D + D′ = 0) 1 1 = . cos v dv dv , where v = x – y D + D′ (1 + 2(− 1)) (1 − 3(− 1)) 3

zz

FG H

IJ K

=

1 1 1 1 . . cos ( x − y) (– cos v) = D + D′ − 4 D + D′ 4

=

1 1 1 1 . cos (x – y) = – . cos (x – y) (Note this step) 4 D + D′ 4 ( − D − D′ )

=– =

1 x′ . cos (x – y) 4 (− 1) 1 1 !

( ∵ bD – aD′ = (– 1)D – (1)D′ = – D – D′)

1 x cos (x – y) 4

5 6 1 5 7 5 1 4 2 1 3 3 x x + x + x y+ x y + x y + cos ( x − y ) . 72 60 20 24 6 4 ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is

∴

P.I. =

73

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

z = φ 1(y – x) + φ 2(y – 2x) + φ 3(y + 3x) + +

5 6 1 5 7 5 1 4 2 1 x + x + x y + x y + x3y3 72 60 20 24 6

x cos (x − y) , where φ1, φ2, φ3 are arbitrary functions. 4 (iii) We have (D3 – 4D2D′ + 5DD′2 – 2D′3) z = ey+2x + (y + x)1/2. ...(1) 3 2 The A.E. of (1) is m – 4m + 5m – 2 = 0. ∴ m = 1, 1, 2 ∴ C.F. = φ1(y + 1 . x) + xφ2(y + 1 . x) + φ3(y + 2x). 1 (e y + 2 x + ( y + x) 1/2 ) ∴ P.I. = 3 2 D − 4D D′ + 5DD′ 2 − 2D′ 3 1 1 e y + 2x + ( y + x) 1/2 . = 2 2 (D − D′ ) (D − 2D′ ) (D − D′ ) (D − 2D′ ) 1 1 1 . e2 x + y e y + 2x = Now, D − 2D′ (D − D′ ) 2 (D − D′ ) 2 (D − 2D′ ) (∵ D = 2, D′ = 1 ⇒ D – 2D′ = 0) 1 1 . e v dv dv , where v = 2x + y = D − 2D′ (2 − 1) 2

zz

1 x1 e2 x + y = e 2 x + y = xe 2 x + y . D − 2D′ (1) 1 1 ! ( ∵ bD – aD′ = (1)D – 2D′ = D – 2D′) 1 1 1 ( y + x) 1/2 = . ( x + y) 1/2 Also (D − D′ ) 2 (D – 2D′ ) (D − D′ ) 2 D − 2D′ (∵ D = 1, D′ = 1 ⇒ D – D′ = 0) 1 1 1/2 v dv , where v = x + y = . (D − D′ ) 2 1 − 2(1)

=

z

=

1 1 v3 / 2 2 . − =− . ( x + y) 3 / 2 2 3/2 3 (D − D′ ) 2 (D − D′ )

= −

x2 2 . 2 ( x + y ) 3/ 2 3 (1) 2 !

( ∵ bD – aD′ = (1)D – (1)D′ = D – D′)

1 2 x ( x + y )3/ 2 3 1 ∴ P.I. = xe 2 x + y − x 2 ( x + y) 3 / 2 . 3 ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is 1 2x + y − x 2 (x + y)3/ 2 , φ2(y + x) + φ3(y + 2x) + xe z = φ1(y + x) + xφ 3 where φ1, φ2, φ3 are arbitrary functions.

= −

(iv) We have (D3 – 7DD′ 2 – 6D′3) z = sin (x + 2y) + e3x+y. The A.E. of (1) is m3 – 7m – 6 = 0. ∴ m = – 1, – 2, 3 ∴ C.F. = φ1(y + (– 1)x) + φ2(y + (– 2)x) + φ3(y + 3x).

...(1)

74

PARTIAL DIFFERENTIAL EQUATIONS

1 (sin ( x + 2 y) + e 3 x + y ) D − 7DD′ 2 − 6D′ 3 1 1 e3 x + y . sin ( x + 2 y) + = (D + D′ )(D + 2D′ )(D − 3D′ ) (D + D′ )(D + 2D′ )(D − 3D′ ) 1 sin ( x + 2 y) Now, (D + D′ )(D + 2D′ )(D − 3D′ ) 1 sin v dv dv dv , where v = x + 2y = (1 + 2) (1 + 4) (1 − 6) 1 1 1 1 − cos v dv dv = cos ( x + 2 y) . sin v dv = =– (– cos v) = – 75 75 75 75 1 e3 x + y Also (D + D′ )(D + 2D′ )(D − 3D′ ) 1 1 . e3 x + y = (∵ D = 3, D′ = 1 ⇒ D – 3D′ = 0) D − 3D′ (D + D′ ) (D + 2D′ ) 1 1 . e v dv dv , where v = 3x + y = D − 3D′ (3 + 1) (3 + 2)

P.I. =

3

zzz

zz

z

zz

1 x1 e 3x + y 1 3x + y . e = 1 . D − 3D′ 20 (1) 1 ! 20

=

( ∵ bD – aD′ = (1)D – 3D′ = D – 3D′)

1 xe 3 x + y 20 ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is

=

z = φ1(y – x) + φ2(y – 2x) + φ3(y + 3x) −

1 1 cos (x + 2y) + xe 3x + y , 75 20

where φ1, φ2, φ3 are arbitrary functions.

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. (D2 + 3DD′ + 2D′2) z = 2x + 3y

2. (D2 + 2DD′ + D′2) z = e2x + 3y

3. (D2 – 2DD′ + D′2) z = ex+2y

4. (D3 – 4D2D′ + 4DD′2) z = cos (2x + 3y)

5. s =

ex+y

6. 4r – 4s + t = 16 log (x + 2y)

7. 2r – s – 3t = 5ex/ey

8. (D2 – 5DD′ + 4D′2) z = sin (4x + y)

9. (D2 – 2aDD′ + a2D′2) z = g(y + ax) 11.

(D3

13.

(D3 – 4D2D′ + 4DD′2) z = cos (2x + y)

–

2D2D′

–

DD′2

+

2D′3)

z=

10. (2D2 – 5DD′ + 2D′2) z = 5 sin (2x + y)

ex+y

12. (D3 – 4D2D′ + 4DD′2) z = sin (y + 2x) 14. (D2 – 3DD′ + 2D′2) z = e2x–y + ex+y + cos (x + 2y).

Answers 1. z = φ1(y – x) + φ2(y – 2x) +

1 (2 x + 3 y)3 240

3. z = φ1(y + x) + xφ2(y + x) + e x + 2 y

2. z = φ1(y – x) + xφ2(y – x) +

e2 x + 3 y 25

75

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

4. z = φ1(y) + φ2(y + 2x) + xφ3(y + 2x) –

1 sin (2 x + 3 y) 32

5. z = φ1(x) + φ2(y) + ex+y 1 1 2 6. z = φ 1 y + x + xφ 2 y + x + 2 x log ( x + 2 y) 2 2 1 3 x− y 7. z = φ 1 ( y − x) + φ 2 y + x + xe 8. z = φ1(y + x) + φ2(y + 4x) – x cos (4 x + y) 3 2

FG H

IJ K

FG H

FG H

IJ K

IJ K

9. z = φ1(y + ax) + xφ2(y + ax) +

x2 g ( y + ax) 2

11.

z = φ1(y – x) + φ2(y + x) + φ3(y + 2x) –

12.

z = φ1(y) + φ2(y + 2x) + xφ3(y + 2x) –

13. 14.

FG H

10. z = φ 1 y +

1 x+y xe 2

IJ K

1 5 x + φ 2 ( y + 2 x) − x cos (2 x + y) 2 3

x2 cos (2 x + y) 4 x2 sin (2 x + y) z = φ1(y) + φ2(y + 2x) + xφ3(y + 2x) + 4 1 2x − y 1 e − xe x + y − cos ( x + 2 y) . z = φ1(y + x) + φ2(y + 2x) + 12 3

4.10. GENERAL METHOD OF FINDING PARTICULAR INTEGRAL Let f(D, D′)z = F(x, y) ...(1) be a homogeneous linear partial differential equation of order n with constant coefficients. Let m1, m2, ......, mn be the roots of the auxiliary equation of (1). Here some of the roots may be repeated. P.I. =

1 1 F( x, y) = F( x, y) f (D, D′ ) (D − m1D′ )(D − m2 D′ ) ...... (D − mn D′ )

We first study the method of evaluating

1 g( x, y) for some constant m and D − mD′

function g(x, y). Let

z=

1 g( x, y) D − mD′

(D – mD′) z = g(x, y) p – mq = g(x, y)

∴ ⇒

∴ Lagrange auxiliary equations are (2)

⇒

(2)

⇒

dy −m dz dx = g( x, y)

dx =

dx dy dz = = 1 − m g( x, y)

⇒ dy + mdx = 0 ⇒ y + mx = a

z

⇒ z = g( x , a − mx ) dx

⇒

dz = g(x, a – mx) dx

∴

1 g(x, y) = g(x, a − mx) dx D − mD′

z

...(2)

...(3)

76

PARTIAL DIFFERENTIAL EQUATIONS

After evaluation of right side, the constant a is replaced by y + mx. Now we come to the evaluation of a particular integral of (1). 1 F( x, y) P.I. of (1) = (D − m1D′ )(D − m2 D′ ) ...... (D − mn D′ ) 1 1 1 F( x, y) . ...... = D − m1D′ D − m2 D′ D − mn D′ This right side is evaluated by repeated application of the formula (3).

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations: (i) (D2 + 2DD′ + D′ 2) z = 2 cos y – x sin y (ii)

∂2z ∂ 2 z 4x y − 4 = 2 − 2 2 2 ∂x ∂y y x

∂2z ∂2z − = tan3 x tan y – tan x tan3 y ∂x 2 ∂y 2 (iv) (D2 + DD′ – 6D′2) z = x2 sin (x + y). ...(1) Sol. (i) We have (D2 + 2DD′ + D′2) z = 2 cos y – x sin y. 2 The A.E. of (1) is m + 2m + 1 = 0. ∴ m = – 1, – 1 ∴ C.F. = φ1(y + (– 1)x) + xφ2(y + (– 1)x) = φ1(y – x) + xφ2(y – x). 1 (2 cos y − x sin y) P.I. = 2 D + 2DD′ + D′ 2 1 (2 cos y − x sin y) = (D + D′ ) (D + D′ ) 1 [2 cos (a + x) − x sin (a + x)] dx = D + D′ (D + D′ = D – mD′ ⇒ m = – 1. ∴ y = a – mx = a + x) 1 [2 sin (a + x) − {− x cos (a + x) + sin (a + x)}] = D + D′ 1 1 [sin y + x cos y] [sin (a + x) + x cos (a + x)] = = D + D′ D + D′

(iii)

z

=

z

(sin ( a + x ) + x cos ( a + x )) dx

(D + D′ = D – mD′ ⇒ m = – 1. ∴ y = a – mx = a + x) = – cos (a + x) + x sin (a + x) + cos (a + x) = x sin (a + x) = x sin y. ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is φ2(y – x) + x sin y, where φ1, φ2 are arbitrary constants. z = φ1(y – x) + xφ (ii) We have ⇒

∂2 z ∂2 z 4 x y − 4 = 2 − 2. 2 2 ∂x ∂y y x 4x y (D2 – 4D′2) z = 2 − 2 y x

...(1)

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

77

The A.E. of (1) is m2 – 4 = 0. ∴ m = – 2, 2 ∴ C.F. = φ1(y + (– 2)x) + φ2(y + 2x) = φ1(y – 2x) + φ2(y + 2x). P.I. = =

1 2

D − 4D′ 1 D + 2D′

=

1 D + 2D′

=

1 D + 2D′

2

F 4x − y I = F 4x − y I 1 J GH y x JK (D + 2D′ )(D − 2D′ ) GH y x K F 4 x − a − 2 x I dx GH (a − 2 x) x JK

z z FGH z FGH

2

2

2

2

2

2

(D – 2D′ = D – mD′ ⇒ m = 2 ∴ y = a – mx = a – 2x)

I JK

−

2a − 4 x − 2a a 2 − 2 + dx 2 x ( a − 2 x) x

−

a 2 2a 2 + − 2 + dx 2 a − 2 x (a − 2 x) x x

I JK

LM OP N Q Llog y + y + 2x + y + 2x + 2 log xOP 1 = y x D + 2D′ MN Q LMlog y + 2 log x + 2x + y + 3OP 1 = D + 2D′ N y x Q F I 2x a + 2x + + 3J dx = G log (a + 2 x) + 2 log x + a + 2x x H K =

=

a a 1 log (a − 2 x) + + + 2 log x a − 2x x D + 2D′

z z FGH

(D + 2D′ = D – mD′ ⇒ m = – 2. ∴ y = a – mx = a + 2x)

log (a + 2 x) + 2 log x +

= log (a + 2 x) . x −

z

IJ K

2x a + + 5 dx a + 2x x

2 . x dx + 2 (log x) . x − a + 2x +

z

z

2 . x dx x

2x dx + a log x + 5 x a + 2x

= x log (a + 2x) + (2x + a) log x + 3x = x log y + y log x + 3x. Using G.S. = C.F. + P.I., the general solution of the given equation is z = φ1(y – 2x) + φ2(y + 2x) + x log y + y log x + 3x, where φ1, φ2 are arbitrary functions. (iii) We have

∂2 z ∂2 z − = tan3 x tan y – tan x tan3 y. ∂x 2 ∂y 2

⇒ (D2 – D′2) z = tan3 x tan y – tan x tan3 y The A.E. of (1) is m2 – 1 = 0. ∴ m = – 1, 1 ∴ C.F. = φ1(y + (– 1)x) + φ2(y + 1.x) = φ1(y – x) + φ2(y + x).

...(1)

78

PARTIAL DIFFERENTIAL EQUATIONS

P.I. =

1 (tan3 x tan y – tan x tan3 y) D 2 − D′ 2

1 (tan3 x tan y – tan x tan3 y) (D + D′ )(D − D′ ) 1 [tan 3 x tan (a − x) − tan x tan 3 (a − x)] dx = D + D′ (D – D′ = D – mD′ ⇒ m = 1. ∴ y = a – mx = a – x) =

=

1 D + D′

=

1 D + D′

z z LMz N

tan x tan (a − x) [sec 2 x − 1 − sec 2 (a − x) + 1] dx tan (a − x) . tan x sec 2 x dx − –

LM N

1 tan 2 x 1 − + tan ( a x ) . = D + D′ 2 2 + tan x . =

OP Q

tan x . tan (a − x) sec 2 (a − x) dx

sec 2 ( a − x) tan 2 x dx

tan 2 (a − x) 1 − 2 2

z

sec 2 x tan 2 (a − x) dx

1 tan (a − x) tan 2 x + tan x tan 2 (a − x) 2(D + D′ )

− =

z

z

z

(sec 2 x (sec 2 (a − x) − 1) − sec 2 (a − x) (sec 2 x − 1) dx

1 tan (a − x) tan 2 x + tan x tan 2 (a − x) 2(D + D′ ) −

z

(sec 2 (a − x) − sec 2 x) dx

=

1 [tan y tan2 x + tan x tan2 y + tan y + tan x] 2(D + D′ )

=

1 [tan y sec2 x + tan x sec2 y] 2(D + D′ )

=

1 2

z

LM N

OP Q

OP Q

[tan (a + x) sec2 x + tan x sec2 (a + x)] dx (D + D′ = D – mD′ ⇒ m = – 1. ∴ y = a – mx = a + x)

z

z

1 tan (a + x) tan x − sec 2 (a + x) tan x dx + tan x sec 2 (a + x) dx 2 1 = tan y tan x. 2 ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is

=

z = φ1(y – x) + φ2(y + x) +

OP Q

1 tan x tan y, where φ1, φ2 are arbitrary functions. 2

OP Q

79

HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

(iv) We have (D2 + DD′ – 6D′2) z = x2 sin (x + y). The A.E. of (1) is m2 + m – 6 = 0. ∴ m = – 3, 2 ∴ C.F. = φ1(y + (– 3)x) + φ2(y + 2x) = φ1(y – 3x) + φ2(y + 2x). P.I. = =

=

1 x2 sin (x + y) D 2 + DD′ − 6D′ 2

z

1 1 x 2 sin ( x + a − 2 x) dx x2 sin (x + y) = D + 3D′ (D + 3D′ )(D − 2D′ ) (D – 2D′ = D – mD′ ⇒ m = 2. ∴ y = a – mx = a – 2x) 1 D + 3D′

z

LM N 1 Lx = D + 3D′ MN =

...(1)

x 2 sin (a − x) dx

1 x 2 cos (a − x) − D + 3D′ 2

z

2 x cos (a − x) dx

cos (a − x) − (− 2 x sin (a − x) +

OP Q

z

2 sin (a − x) dx)

=

1 x 2 cos (a − x) + 2 x sin (a − x) − 2 cos (a − x) D + 3D′

=

1 ( x 2 − 2) cos (2 x + y − x) + 2 x sin (2 x + y − x) D + 3D′

=

1 ( x 2 − 2) cos ( x + y) + 2 x sin ( x + y) D + 3D′

=

=

z z

OP Q

[( x 2 − 2) cos ( x + a + 3 x) + 2 x sin ( x + a + 3 x)] dx

(D + 3D′ = D – mD′ ⇒ m = – 3. ∴ y = a – mx = a + 3x) [( x 2 − 2) cos (4 x + a) + 2 x sin (4 x + a)] dx

= (x2 – 2)

sin (4 x + a) − 4

z

2x .

z

sin (4 x + a) dx + 4

z

2 x sin (4 x + a) dx

=

1 2 3 ( x − 2) sin (4 x + a) + 4 2

=

1 2 3 cos (4 x + a) −x − ( x − 2) sin (4 x + a) + 4 2 4

=

1 2 3 3 ( x − 2) sin (4 x + a) − x cos (4 x + a) + sin (4 x + a) 4 8 32

Fx GH 4 Fx =G H4

2

=

2

x sin (4 x + a) dx

LM N

I JK 13 I 3 sin ( x + y) − x cos ( x + y) . − J 32 K 8

−

z

1. −

13 3 sin (4 x + y − 3 x) − x cos (4 x + y − 3 x) 32 8

cos (4 x + a) dx 4

OP Q

80

PARTIAL DIFFERENTIAL EQUATIONS

∴ Using G.S. = C.F. + P.I., the general solution of the given equation is

Fx z = φ (y – 3x) + φ (y + 2x) + G H4

2

1

2

where φ1, φ2 are arbitrary functions.

−

I JK

13 3 sin (x + y) − x cos (x + y) , 32 8

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. 3. 5. 6.

r + s – 6t = y cos x 2. (D2 + DD′ – 6D′2) z = y sin x 4. (D2 – DD′ – 2D′2) z = (y – 1) ex r – s – 2t = (2x2 + xy – y2) sin xy – cos xy 3 2 3 y (D – 3DD′ – 2D′ ) z = cos (x + 2y) – e (3 + 2x) (D3 + 2D2D′ – DD′2 – 2D′3) z = (y + 2) ex.

Answers 1. z = φ1(y + 2x) + φ2(y – 3x) – y cos x + sin x 3. z = φ1(y + 2x) + φ2(y – x) + sin xy

2. z = φ1(y + 2x) + φ2(y – 3x) – y sin x – cos x 4. z = φ1(y + 2x) + φ2(y – x) + yex

1 sin (x + 2y) + xey 27 6. z = φ1(y + x) + φ2(y – x) + φ3(y – 2x) + yex. 5. z = φ1(y – x) + xφ2(y – x) + φ3(y + 2x) +

Non-homogeneous Linear Partial Differential Equations with Constant Coefficients

5

5.1. INTRODUCTION From the last chapter, we have been solving linear partial differential equations with constant coefficients. In that chapter we found the general solution of only such equations in which the orders of all partial derivatives involved in the equation were same. In other words, we solved only homogeneous linear partial differential equations with constant coefficients. In the present chapter, we shall learn the methods of finding general solution of linear partial differential equations which are not homogeneous. 5.2.

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL CONSTANT COEFFICIENTS

EQUATIONS WITH

A linear partial differential equation with constant coefficients is called a non-homogeneous linear partial differential equation with constant coefficients if the orders of partial derivatives occurring in the equation are not equal. For example, the following partial differential equations are all non-homogeneous linear partial differential equations with constant coefficients : ∂z ∂ 2 z − = e2 x + 3 y ∂x ∂y 2 (ii) (2DD′ + D′2 – 3D′)z = 3 cos (3x – 2y)

(i)

(iii) (D – 2D′ + 5)(D2 + D′ + 3)z = e 3 x − 4 y sin (x – 2y) 5.3.

REDUCIBLE AND IRREDUCIBLE NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

Let f(D, D′) z = F(x, y) be a non-homogeneous linear partial differential equation with constant coefficients. This equation is called reducible if f(D, D′) can be resolved into factors each of which is of the first degree in D and D′. For example, (D2 – D′2 + 3D – 3D′)z = sin x is a reducible non-homogeneous linear partial differential equation with constant coefficients because D2 – D′2 + 3D – 3D′ = (D – D′) (D + D′ + 3). A non-homogeneous linear partial differential equation with constant coefficients is called irreducible if it is not reducible. For example (2D2 – D′2 + D) z = x2 – y is an irreducible non-homogeneous linear partial differential equation with constant coefficients because (2D2 – D′2 + D) cannot be resolved into linear factors in D and D′. 81

82 5.4.

PARTIAL DIFFERENTIAL EQUATIONS

GENERAL SOLUTION OF REDUCIBLE NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATION f(D, D′)z = 0 WITH CONSTANT COEFFICIENTS

Let f(D, D′) z = (a1D + b1D′ + c1) ...... (anD + bnD′ + cn) z = 0 ...(1) be a reducible non-homogeneous linear partial differential equation with constant coefficients. The factors on the left side of (1) may or may not be distinct. Case I. Factors are distinct Equation (1) shows that for 1 ≤ i ≤ n, the solution of (aiD + biD′ + ci) z = 0 is a solution of (1). ⇒ ai p + bi q = – ciz (aiD + biD′ + ci) z = 0 ⇒ (2)

aidy – bidx = 0

⇒

Also,

(2)

⇒

dx dy dz = = ai bi − ci z

...(2)

⇒ ai y – bix = λ

c dz = − i dx z ai − c x / ai

⇒ z = μe i This is true only when ai ≠ 0.

⇒ log z = –

ci x + log μ ai

⇒ z = e − ci x / ai φ i (ai y − bi x)

(Putting μ = φi(λ))

If bi ≠ 0, then by taking IInd and IIIrd fractions of (2), we can show that z = e − ci y / bi φ i (ai y − bi x) is a solution of (1). ∴ z = e − c1 x / a1 φ 1 (a1 y − b1 x) , z = e − c2 x / a2 φ 2 (a2 y − b2 x) , ......, z = e − cn x / an φ n (an y − bn x) are solutions of (1). ∴ z = e − c 1x /a 1 φ 1 (a 1y − b 1x) + e − c 2 x / a 2 φ 2 (a 2 y − b 2 x) + ...... + e − c n x / a n φ n (a n y − b n x) is also a solution of (1). Since this solution contains n arbitrary functions φ1, φ2, ......, φn, this solution represents the general solution of the given equation. Here we have assumed that a1, a2, ......, an are all non-zero constants. Case II. Factors are not distinct Let the first two factors be same and all others distinct. In this case, the solution of (1) can be written as z = e − c1 x / a1 (φ 1 + φ 2 )( a1 y − b1 x) + ...... + e − cn x / an φ n ( an y − bn x) . This solution contains only n – 1 arbitrary functions φ1, φ2, φ3, ......, φn. ∴ This is not a general solution. (1) ⇒ (a1D + b1D′ + c1)2 (a3D + b3D′ + c3) ...... (anD + bnD′ + cn) z = 0

...(3)

Equation (3) shows that for 3 ≤ i ≤ n, the solution of (aiD + biD′ + ci) z = 0 is also a solution of (3) and hence of (1). ∴ z = e − ci x / ai φ i ( ai y − bi x) , 3 ≤ i ≤ n is a solution of (1), provided ai ≠ 0, 3 ≤ i ≤ n. ∴

z = e − c3 x / a3 φ 3 (a3 y − b3 x) + ...... + e − cn x / an φ n ( an y − bn x)

...(4)

is a solution of (1). The solution of (a1D + b1D′ + c1)2 z = 0 is also a solution of (3) and hence of (1). (a1D + b1D′ + c1)2 z = 0 ⇒ (a1D + b1D′ + c1)(a1D + b1D′ + c1) z = 0

...(5)

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

Let

(a1D + b1D′ + c1) z = u

∴ (5) ⇒

(a1D + b1D′ + c1) u = 0

83 ...(6)

− c x/a u = e 1 1 ψ 1 (a1 y − b1 x)

⇒

where ψ1 is arbitrary. We are assuming that a1 ≠ 0. ∴ (6) ⇒

a1p + b1q + c1z = e − c1 x / a1 ψ 1 (a1 y − b1 x) a1p + b1q = e − c1 x / a1 ψ 1 (a1 y − b1 x) – c1z

⇒

dx dy dz = = − c x/ a 1 1 a1 b1 e ψ 1 (a1 y − b1 x) − c1 z

⇒

...(7)

(7)

⇒ a1dy – b1dx = 0 ⇒ a1y – b1x = c Taking the first and third fractions of (7), we get dz 1 − c1x / a1 = (e ψ 1 (a1 y − b1 x) − c1 z) dx a1 dz c1 1 − c1x / a1 z= e + ψ 1 (a1 y − b1 x) dx a1 a1

⇒

This is a linear equation. ∴

z I.F. = e

∴ Solution of (8) is z · e c1 x / a1 = =

z

1 a1

c1 dx a1

= e c1 x / a1

1 − c1 x / a1 e ψ 1 ( a1 y − b1 x ) . e c1 x / a1 dx a1

z

ψ 1 (c) dx =

xψ 1 (c) +d a1

x ψ 1 (a1 y − b1 x) + d a1

⇒

ze c1 x / a1 =

⇒

ze c1x / a1 = x φ 1 (a1 y − b1 x) + φ 2 (a1 y − b1 x)

FG Taking φ H

∴ Combining (4) and (9),

...(8)

1

=

IJ K

1 ψ 1 and d = φ 2 (c) a1

z = e − c1 x / a1 (x φ1(a1y – b1x) + φ2(a1y – b1x))

...(9)

– c x/a z = e 1 1 (x φ1(a1y – b1x) + φ2(a1y – b1x))

+ e − c 3 x /a 3 φ 3 (a 3 y − b 3 x) + .. .. .. + e − c n x / a n φ n (a n y − b n x) . Since this solution contains n arbitrary functions φ1, φ2, ......, φn, this solution represents the general solution of the given equation. Remark. If the factor a1x + b1y + c1 is repeated r times, then the corresponding part of the general solution is e− c1 x / a1 (φ1(a1 y − b1x) + x φ 2 (a1 y − b1x) + ...... + xr–1 φr(a1y – b1x)), where φ1, φ2, ......, φr are arbitrary functions.

84

PARTIAL DIFFERENTIAL EQUATIONS

WORKING STEPS FOR SOLVING PROBLEMS Step I. Express f(D, D′) as the product of linear factors in D and D′. Step II. Corresponding to each non-repeated factor aD + bD′ + c, the part of G.S. is e − cx / a φ(ay − bx) , if a ≠ 0 or e − cy / b φ(ay − bx) if b ≠ 0.

Step III. Corresponding to each repeated factor aD + bD′ + c, r times, the part of G.S. is e − cx/a

r

∑

x i − 1 φ i (ay − bx) , if a ≠ 0 or e − cy/b

i=1

r

∑

y i − 1 yi (ay − bx) , if b ≠ 0.

i=1

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of following partial differential equations : (i) t + s + q = 0 (ii) DD′(D – 2D′ – 3) z = 0 (iii) (DD′ + aD + bD′ + ab)z = 0 (iv) r + 2s + t + 2p + 2q + z = 0. Sol. (i) We have t + s + q = 0. ⇒ (D′2 + DD′ + D′) z = 0 ⇒ (0.D + 1.D′ + 0)(D′ + D + 1) z = 0 ...(1) ∴ The general solution of (1) is z = e–0y/1 φ1(0y – 1.x) + e–1.x/1 φ2(1.y – 1.x) or z = φ1(– x) + e–x φ2(y – x), where φ1 and φ2 are arbitrary functions.

Remark. φ1(– x) is also a function of x and can also be written as ψ(x) for some arbitrary function ψ.

(ii) We have DD′(D – 2D′ – 3) z = 0. ...(1) ⇒ (1.D + 0.D′ + 0)(0.D + 1.D′ + 0)(D – 2D′ – 3)z = 0 ∴ The general solution of (1) is z = e–0.x/1 φ1(1.y – 0.x) + e–0.y/1 φ2(0.y – 1.x) + e3x/1 φ3(1.y + 2.x) or z = φ1(y) + φ2(– x) + e3x φ3(2x + y), where φ1, φ2, φ3 are arbitrary functions. (iii) We have (DD′ + aD + bD′ + ab) z = 0. ⇒ (D + b)(D′ + a) z = 0 ⇒ (1.D + 0.D′ + b)(0.D + 1.D′ + a)z = 0 ...(1) ∴ The general solution of (1) is z = e–bx/1 φ1(1.y – 0.x) + e–ay/1 φ2(0.y – 1.x) or z = e–bx φ1(y) + e–ay φ2(– x), where φ1 and φ2 are arbitrary functions. (iv) We have ⇒ ⇒ ⇒

(D2

r + 2s + t + 2p + 2q + z = 0

+ 2DD′ + [(D +

D′2

D′)2

+ 2D + 2D′ + 1) z = 0

+ 2(D + D′) + 1] z = 0 (D + D′ + 1)2 z = 0

∴ The general solution of (1) is z = e–1.x/1 (φ1(1.y – 1.x) + x φ2(1.y – 1.x)) or

z = e–x (φ φ1(y – x) + x φ2(y – x)),

where φ1, φ2 are arbitrary functions.

...(1)

85

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

5.5.

GENERAL SOLUTION OF IRREDUCIBLE NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATION f(D, D′)z = 0 WITH CONSTANT COEFFICIENTS

Let f(D, D′) z = 0 ...(1) be an irreducible non-homogeneous linear partial differential equation with constant coefficients. Let z = ceax+by. r s ∴ D D z = Dr Ds(ceax+by) = carbseax+by = ar br ceax+by ∴ f(D, D′) z = f(a, b) ceax+by ax+by ∴ z = ce is a solution of (1) if f(a, b) = 0 and the constant c is arbitrary. Let (ai, bi) be one of infinitely many pairs satisfying f(ai, bi) = 0. ∴

z = ci e ai x + bi y is a solution of (1) for each i.

∴ The general solution of (1) is z =

∞

∑ce i

a i x + bi y

, where f(ai, bi) = 0.

i=1

Remark. For the above irreducible partial differential equation (1), the general solution has been written in terms of arbitrary constants.

WORKING STEPS FOR SOLVING PROBLEMS Step I.

Express the given equation in the form f(D, D′) g(D, D′) z = 0, where f(D, D′) is expressible as a product of linear factors in D and D′ and g(D, D′) is irreducible. Step II. Write the solution of f(D, D′)z = 0 in terms of arbitrary functions. Step III. Write the solution of g(D, D′)z = 0 in terms of arbitrary constants. Step IV. Add both general solutions to get the general solution of the given equation. Example 2. Find the general solution of the following partial differential equations : (i) (D2 + D + D′) z = 0

(ii) (2D4 – 3D2D′ + D′2) z = 0

(iii) (D + 2D′ – 3)(D2 + D′)z = 0

(iv) (D′ + 3D)2 (D2 + 5D + D′) z = 0.

Sol. (i) We have (D2 + D + D′) z = 0.

...(1)

D2 + D + D′ is irreducible Let z = ceax+by be a solution of (1). ∴

(a2 + a + b) ceax+by = 0 a2 + a + b = 0 ⇒ b = – (a2 + a)

⇒

z = ce ax − ( a

∴

∞

∴ The general solution of (1) is z =

∑

2

+ a) y

, where a and c are arbitrary constants.

ciea ix − (a i

i=1

2

+ a i )y

, where ai and ci are arbitrary

constants. (ii) We have (2D4 – 3D2D′ + D′2) z = 0. ⇒

(2D2

–

D′)(D2

– D′) z = 0

...(1)

86

PARTIAL DIFFERENTIAL EQUATIONS

The factor 2D2 – D′ is irreducible. Let z = ceax+by be a solution of (2D2 – D′) z = 0. ∴ ∴

(2a2 – b) ceax+by = 0 ⇒ 2a2 – b = 0 or b = 2a2 ax + 2 a z = ce

2

y

∴ Corresponding to 2D2 – D′, the part of general solution of (1) is

∞

∑c e i

2

ai x + 2ai y

.

i =1

The factor D2 – D′ is also irreducible. Let z = c′ e a ′ x + b′ y be a solution of (D2 – D′) z = 0. ∴ ⇒

(a′2 – b′) c′ e a′ x + b′ y = 0 a′2 – b′ = 0 or b′ = a′2

∴

a′x + a′ z = c′ e

2

y

∴ Corresponding to D2 – D′, the part of general solution of (1) is

∞

∑c ′e i

a′ i x + a′ i 2 y

i=1

∴ The general solution of (1) is

z=

∞

∑c e i

a i x + 2a i 2 y

∞

+

i=1

∑c ′e i

ai ′x + ai ′2 y

, where ai, ci,

i=1

ai′, ci′ are arbitrary constants. (iii) We have (D + 2D′ – 3)(D2 + D′) z = 0. The factor D + 2D′ – 3 is linear. ∴ Corresponding to D + 2D′ – 3, the part of general solution of (1) is

...(1)

e − ( − 3 ) x /1 φ1(1.y – 2.x) i.e. e3xφ1(y – 2x). The factor D2 + D′ is irreducible. Let z = ceax+by be a solution of (D2 + D′) z = 0. ∴ (a2 + b) ceax+by = 0. ⇒ a2 + b = 0 or b = – a2 ∴

ax − a z = ce

2

y

∴ Corresponding to D2 + D′, the part of general solution of (1) is

∞

∑c e i

2

ai x − ai y

.

i =1 ∞

∴ The general solution of (1) is z = e3x φ1(y – 2x) +

∑c e i

i=1

2

a ix − a i y

, where φ1 is arbitrary

function and ai, ci are arbitrary constants. (iv) We have (D′ + 3D)2 (D2 + 5D + D′) z = 0. The factor (3D + D′)2 is linear repeated. ∴ Corresponding to (3D + D′)2, the part of general solution of (1) is e–0.x/3 (φ1(3.y – 1.x) + xφ2(3.y – 1.x)), i.e., φ1(3y – x) + xφ2(3y – x). 2 The factor D + 5D + D′ is irreducible. Let z = ceax+by be a solution of (D2 + 5D + D′) z = 0. ∴ (a2 + 5a + b) ceax+by = 0

...(1)

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

87

a2 + 5a + b = 0 or b = – (a2 + 5a)

⇒

ax − ( a z = ce

∴

2

+ 5a ) y

∴ Corresponding to D2 + 5D + D′, the part of general solution of (1) is

∞

∑c e i

2

ai x − ( ai + 5a i ) y

.

a i x − (a i 2 + 5a i )y

,

i =1

∴ The general solution of (1) is z = φ1(3y – x) + x φ2(3y – x) +

∞

∑c e i

i=1

where φ1, φ2 are arbitrary functions and ai, ci are arbitrary constants.

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. (D + D′ – 1)(D + 2D′ – 2)z = 0

2. (D + 1)(D + D′ – 1)z = 0

3. (D2 – DD′ – 2D)z = 0

4. (D2 – D′2 + D – D′)z = 0

5. (D – D′ – 1)(D – D′ – 2) z = 0

6. (D2 – DD′ + D′ – 1)z = 0

7. s + p – q – z = 0

8. (D2 – DD′ – 2D′2 + 2D + 2D′)z = 0

9.

∂2 z

∂x2

+

∂2 z ∂2 z −6 2 =0 ∂x∂y ∂y

10. (D – 3D′ – 2)2 z = 0

11.

(D – D′2) z = 0

12. (2D2 – D′2 + D) z = 0

13.

(D2

14. (D – 2D′ – 1)(D – 2D′2 – 1)z = 0

15.

(2D – 3D′ + 5)2 (D2 + 3D′) z = 0.

+ DD′ + D + D′ + 1) z = 0

Answers 1. z = exφ1(y – x) + e2xφ2(y – 2x) or 3. z = 5. z = 7. z =

z = eyφ1(y – x) + e2yφ2(y – φ1(y) + e2xφ2(y + x) exφ1(y + x) + e2xφ2(y + x) exφ1(y) + e–yφ2(– x)

2. z = e–xφ1(y) + exφ2(y – x) 2x) 4. z = φ1(y + x) + e–xφ2(y – x) 6. z = exφ1(y) + e–xφ2(y + x) 8. z = φ1(y – x) + e–2xφ2(y + 2x) 10. z = e2x (φ1(y + 3x) + xφ2(y + 3x))

9. z = φ1(y + 2x) + φ2(y – 3x) ∞

11.

z=

∑c e i

∞

2

bi x + bi y

12.

i =1 ∞

13.

z=

∑c e i

ai x + bi y

∑ ce i

i= 1

, where ai2 + aibi + ai + bi + 1 = 0

i=1

∞

14.

z = ex φ(y + 2x) +

∑c e i

2

(1 + 2bi ) x + bi y

i =1

∞

15.

z = e–5x/2 (φ1(2y + 3x) + xφ2(2y + 3x)) +

∑c e i

i=1

ai x − ai 2 y / 3

.

ai x + bi y

, where 2ai2 – bi2 + ai = 0

88 5.6.

PARTIAL DIFFERENTIAL EQUATIONS

GENERAL SOLUTION OF NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS

Let

f(D, D) z = F(x, y)

...(1)

be a non-homogeneous linear partial differential equation with constant coefficients. Let 

u be the general solution of f(D, D) z = 0. f(D, D) u = 0

...(2)

Let v be a particular integral of f(D, D) z = F(x, y).  Now

f(D, D) v = F(x, y)

...(3)

f(D, D) (u + v) = f(D, D) u + f(D, D) v = 0 + F(x, y) = F(x, y).

 u + v is a solution of f(D, D) z = F(x, y). Since u is the general solution of the equation f(D, D) z = 0, the solution u + v of the equation f(D, D) z = F(x, y) is the general solution of the equation f(D, D) z = F(x, y). The general solution u of the equation f(D, D) z = 0 is called the complementary function (C.F.) of the equation f(D, D) z = F(x, y).  The general solution of the equation f(D, D¢) z = F(x, y) is obtained by adding the general solution of the equation f(D, D¢)z = 0 to any particular integral of the equation f(D, D¢)z = F(x, y). 5.7. PARTICULAR INTEGRAL OF f(D, D¢)z = F(x, y) Let

f(D, D) z = F(x, y)

...(1)

be a non-homogeneous linear partial differential equation with constant coefficients. Since, f(D, D)

 1 F(x, y)"# = F(x, y), the function 1 F(x, y) is a particular f (D, D ) ! f (D, D ) $

integral of the equation f(D, D) z = F(x, y).

5.8. PARTICULAR INTEGRAL WHEN F(x, y) IS SUM OR DIFFERENCE OF TERMS OF THE FORM xmyn If F(x, y) is sum or difference of the terms of the form xmyn, then the particular integral

1 1 F( x, y) of the differential equation f(D, D) z = F(x, y) is obtained by expanding f (D, D ) f (D, D ) in an infinite series in ascending powers of either D or D. The particular integrals obtained in the above mentioned two method may not be identical. Any one of the two particular integrals may be used.

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations : (i) (D2 – D – 1)z = x2y (ii) (D2 – D2 – 3D + 3D)z = xy. 2 2 Sol. (i) We have (D – D – 1) z = x y. ...(1)

89

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

D2 – D′ – 1 is irreducible. ∞

C.F. =

∴

∑c e i

a i x + bi y

i =1 ∞

C.F. =

∴

∑c e i

, where ai2 – bi – 1 = 0 or bi = ai2 – 1

2

a i x + ( ai − 1) y

.

i =1

1 1 x2 y = − x2 y D 2 − D′ − 1 1 + (D′ − D 2 ) = – (1 + (D′ – D2))–1 x2y = – (1 – (D′ – D2) + (D′ – D2)2 – ......) x2y = – (1 – D′ + D2 – 2D′D2 + ......) x2y = – (x2y – x2 + 2y – 4 + 0 + ......) = – x2y + x2 – 2y + 8. ∴ Using G.S. = C.F. + P.I., the general solution of (1) is

P.I. =

z=

∞

∑c e i

a i x + (a i 2 − 1)y

− x 2 y + x 2 − 2y + 8 , where ai and ci are arbitrary constants.

i =1

(ii) We have (D2 – D′2 – 3D + 3D′) z = xy. ...(1) D2 – D′2 – 3D + 3D′ = (D – D′) (D + D′ – 3) = (1.D – 1.D′ + 0)(1.D + 1.D′ – 3) ∴ C.F. = e– 0.x/1 φ1(1.y + 1.x) + e– (– 3)x/1 φ2(1.y – 1.x) = φ1(y + x) + e3xφ2(y – x). 1 xy P.I. = (D − D′ )(D + D′ − 3)

FG FG IJ IJ xy KK H H D′ 1 F I F D D′ + FG D + D′ IJ + ......IJ xy 1+ + ......J G1 + + =– G KH 3 3 H3 3K 3D H D K 1 F D′ I F D D′ + 2DD′ + ......IJ xy + ......J G1 + + 1+ =– G KH 3 3 9 K D 3D H 1 F ′ D′ D′ I + + + ......J xy =– G1 + D3 + D3′ + 2DD K 3D H 9 D 3 I 1 F y x 2 x x xy + + + + + + 0 + ......J =– G 3 3 9 2 3 3D H K x I 1 F x y xy x 2x x + + + + + =– G 3H 2 3 6 9 6 6 JK x 1 F x y xy x 2 I + + + + xJ . =– G 3H 2 3 3 6 9 K Using G.S. = C.F. + P.I., the general solution of (1) is 1 F x y xy x x z = φ (y + x) + e φ (y – x) − G + + + 3H 2 3 3 6 =

FG H

1 D′ 1− D D

IJ K

−1

.

1 D D′ + 1− −3 3 3

−1

2

2

∴

2

2

2

2

1

where φ1, φ2 are arbitrary functions.

3

2

3

3x

2

2

2

3

+

I JK

2 x , 9

90

PARTIAL DIFFERENTIAL EQUATIONS

Alternative method of finding P.I. P.I. =

1 xy (D − D′ )(D + D′ − 3)

FG IJ xy H K F 1 + D + D′ + FG D + D′ IJ + ...I xy 1 =– G 3 H 3 K JK 3(D − D′ ) H 1 FG1 + D + D′ + 2DD′ + ...IJ xy =– K 3 3 9 3(D − D′ ) H 1 FG xy + y + x + 2 IJ =– F D′ I H 3 3 9 K 3DG1 − J H DK 1 F =– G1 − DD′ IJK FGH xy + 3y + 3x + 29 IJK 3D H I F y x 2I 1 F D′ D′ =– + + ...J G xy + + + J 1+ G 3D H D D K H 3 3 9K I 1 F y x 2 1 F 1I xy + + + + G x + J + 0 + ...J =– G 3D H 3 3 9 DH 3K K y x 2I 1 F 1 F 1I =– xy + + + J − x+ J G G 3D H 3 3 9 K 3D H 3K 1 F x y xy x 2x I 1 F x x I =– G − G + + + + J 3H 2 3 6 9K 3H 6 6 JK 1 F x y xy x x 2 I =– G + + + + xJ . 3H 2 3 3 6 9 K =–

1 D + D′ 1− 3(D − D′ ) 3

−1

2

−1

2

2

2

2

2

2

2

3

2

3

Remark. In solving problems, the second method is found comparatively easier and straight forward.

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. r – s + p = 1

2. (D – D′ – 1)(D – D′ – 2) z = x

3. s + p – q = z + xy

4. (D2 – DD′ – 2D′2 + 2D + 2D′) z = xy

5. (D2 – D′2 + D + 3D′ – 2) z = x2y

6. D(D + D′ – 1)(D + 3D′ – 2) z = x2 – 4xy + 2y2.

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

91

Answers 1. z = φ1(y) + e–x φ2(y + x) + x

2. z = ex φ1(y + x) + e2x φ2(y + x) +

3. z = ex φ1(y) + e–y φ2(x) – xy – y + x + 1

x 2 y xy x3 3 x 2 x − − + − 4 4 12 8 2 1 2 3 3 21 x y + xy + x 2 + y + 3 x + z = e–2x φ1(y + x) + ex φ2(y – x) – 2 2 2 4 1 (2 x3 − 12 x2 y + 12 xy2 − 21x2 + 24 xy + 3 x) . z = φ1(y) + ex φ2(y – x) + e2x φ3(y – 3x) + 12

4. z = φ1(y – x) + e–2x φ2(y + 2x) + 5. 6.

x 3 + 2 4

FG H

IJ K

5.9. PARTICULAR INTEGRAL WHEN F(x, y) IS OF THE FORM eax+by Theorem. If f(D, D′′) be a function of D and D′′, then provided f(a, b) ≠ 0.

1 1 e ax + by = e ax + by f(a, b) f(D, D′ )

Proof. We have D r e ax + by = a r e ax + by D s e ax + by = b s e ax + by D r D s e ax + by = a r b s e ax + by .

and ∴

f(D, D′) e ax + by = f(a, b) e ax + by

Dividing both sides by f(a, b), we get f(D, D′) ∴ By definition of the inverse operator

e ax + by = e ax + by f (a, b)

1 e ax + by 1 e ax + by = , we have . f (D, D′ ) f (a, b) f (D, D′ )

Remark 1. The above result is not true for any general function φ(ax + by) of ax + by. 1 1 eax + by = eax + by is applicable only when f(a, b) ≠ 0. The f (D, D′ ) f (a, b) case when f(a, b) = 0 will be discussed a little later. Remark 2. The method

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations : (i) (DD′ + aD + bD′ + ab)z = emx+ny (ii) (D2 + D′ + 4)z = e4x–y. Sol. (i) We have (DD′ + aD + bD′ + ab)z = e mx + ny . ...(1) DD′ + aD + bD′ + ab = (D + b)(D′ + a) = (1.D + 0.D′ + b)(0.D + 1.D′ + a) ∴ C.F. = e–bx/1 φ1(1.y – 0.x) + e– ay/1 φ2(0.y – 1.x) = e– bx φ1(y) + e– ay φ2(– x). 1 e mx + ny P.I. = ...(2) DD′ + aD + bD′ + ab D = m, D′ = n ⇒ DD′ + aD + bD′ + ab = mn + am + bn + ab = (m + b)(n + a) ≠ 0 (Assuming m ≠ – b, n ≠ – a) ∴ (2) ⇒

P.I. =

1 e mx + ny (m + b)(n + a )

92

PARTIAL DIFFERENTIAL EQUATIONS

∴ Using

G.S. = C.F. + P.I., the general solution of (1) is z = e–bx φ1(y) + e–ay φ2(– x) +

where φ1 and φ2 are arbitrary functions. (ii) We have (D2 + D′ + 4) z = e4x–y. D2 + D′ + 4 is irreducible. ∞

C.F. =

∴

∑c e i

a i x + bi y

∞

∑c e i

...(1)

, where ai2 + bi + 4 = 0

i =1

=

e mx + ny , (m + b) (n + a)

2

ai x − ( ai + 4 ) y

.

i =1

1

e4 x − y D + D′ + 4 D = 4, D′ = – 1 ⇒ D2 + D′ + 4 = (4)2 + (– 1) +4 = 19 ≠ 0

P.I. =

∴ (2)

⇒ P.I. =

...(2)

2

1 4x − y e 19

∴ Using G.S. = C.F. + P.I., the general solution of (1) is z =

∞

∑

c i a i x − (a i

2

+ 4)y

i=1

+

1 4x − y , e 19

where ai and ci are arbitrary constants.

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. (D2 – DD′ – 2D) z = e2 x + y

2. (D2 – D′2 + D – D′)z = e2 x + 3 y

3. (D2 – 4DD′ + D – 1) z = e3 x − 2 y

4. (D2 – DD′ – 2D′2 + 2D + 2D′) z = e2 x + 3 y

5. (D2 – D′2 + D + 3D′ – 2)z = e x − y

6. (D3 – 3DD′ + D + 1) z = e2 x + 3 y .

Answers 1.

z = φ1( y) + e2 x φ2 ( y + x) −

3.

z=

∞

∑c e i

ai x + bi y

+

i =1

1 2x + y e 2

1 3x − 2 y e , where ai2 – 4aibi + ai – 1 = 0 35

1 2x + 3 y e 10 1 φ1( y + x) + e x φ2 ( y − x) − e x − y 4

4.

z = φ1( y − x) + e− 2 x φ2 ( y + 2 x) −

5.

z = e− 2 x

6.

z=

∞

∑ ce

i =1

2. z = φ1( y + x) + e− x φ2 ( y − x) −

ai x + bi y

−

1 2x + 3 y e , where ai3 – 3aibi + ai + 1 = 0. 7

1 2x + 3 y e 6

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

93

5.10. PARTICULAR INTEGRAL WHEN F(x, y) IS OF THE FORM sin(ax + by) OR cos(ax + by) Let the given partial differential equation be f(D, D′) z = F(x, y), where F(x, y) = sin (ax + by) (or cos (ax + by)). 1 ∴ P.I. = sin (ax + by) (or cos (ax + by)) f (D, D′ ) This is evaluated by putting D2 = –a2, DD′ = –ab and D′2 = –b2, provided the denominator is not zero.

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations : (i) (D + D′)(D + D′ – 2)z = sin (x + 2y) (ii) (D2 – DD′ – 2D′ 2 + 2D + 2D′)z = e2x+3y + sin (2x + y). Sol. (i) We have (D + D′)(D + D′ – 2) z = sin (x + 2y). ...(1) ⇒ (1 . D + 1 . D′ + 0)(1 . D + 1 . D′ – 2)z = sin (x + 2y) ∴ C.F. = e– 0.x/1 φ1(1.y – 1.x) + e– (– 2)x/1 φ2(1.y – 1.x) ∴ C.F. = φ1(y – x) + e2x φ2(y – x).

1 sin (x + 2y) (D + D′ )(D + D′ − 2) a = 1, b = 2 D2 = – a2 = – (1)2 = – 1, DD′ = – ab = – (1)(2) = – 2, D′2 = – b2 = – (2)2 = – 4.

P.I. = Here ∴

∴ (2)

⇒ P.I. = = =

1 2

(D + D′ ) − 2(D + D′ )

sin (x + 2y)

1 2

2

D + D′ + 2DD′ − 2D − 2D′

sin (x + 2y)

1 sin (x + 2y) − 1 − 4 + 2( − 2) − 2D − 2D′

=– =–

1 sin (x + 2y) 2D + 2D′ + 9 2D + 2D′ − 9 (2D + 2D′ ) 2 − 81

sin (x + 2y)

=−

2D + 2D′ − 9 sin ( x + 2 y) 4D 2 + 8DD′ + 4D′ 2 − 81

=−

2D + 2D′ − 9 sin ( x + 2 y) 4(– 1) − 8(1)(2) − 4(– 4) − 81

=

2D + 2D′ − 9 sin ( x + 2 y) 117

...(2)

94

PARTIAL DIFFERENTIAL EQUATIONS

1 [2 cos (x + 2y) + 4 cos (x + 2y) – 9 sin (x + 2y)] 117 1 =− [6 cos (x + 2y) – 9 sin (x + 2y)] 117 ∴ Using G.S. = C.F. + P.I., the general solution of (1) is 1 z = φ1(y – x) + e2x φ2(y – x) – [6 cos (x + 2y) – 9 sin (x + 2y)], 117 where φ1 and φ2 are arbitrary functions. (ii) We have (D2 – DD′ – 2D′2 + 2D + 2D′) z = e2x+3y + sin (2x + y). ...(1) D2 – DD′ – 2D′2 + 2D + 2D′ = (D + D′)(D – 2D′) + 2(D + D′) = (D + D′)(D – 2D′ + 2) = (1.D + 1.D′ + 0)(1.D – 2D′ + 2) – 0.x/1 ∴ C.F. = e φ1(1.y – 1.x) + e– 2x/1 φ2(1.y – (– 2)x) ∴ C.F. = φ1(y – x) + e–2x φ2(y + 2x). 1 (e 2 x + 3 y + sin (2 x + y)) P.I. = 2 D − DD′ − 2D′ 2 + 2D + 2D′ 1 e2 x+3 y = 2 2 D − DD′ − 2D′ + 2D + 2D′ 1 sin (2 x + y) + 2 D − DD′ − 2D′ 2 + 2D + 2D′ 1 e2 x+3 y = 2 (2) − 2(3) − 2(3) 2 + 2(2) + 2(3) 1 sin (2 x + y) + (− 4) + 2(1) − 2(– 1) + 2D + 2D′ 1 2 x +3 y 1 e + sin (2 x + y) =– 2D + 2D′ 10 1 2 x +3 y D − D′ e + sin (2 x + y) =– 10 2(D 2 − D′ 2 ) 1 2 x +3 y D − D′ e + sin (2 x + y) =– 10 2(− 4 + 1) 1 2 x+3 y 1 =– e − (2 cos (2 x + y) − cos (2 x + y)) 10 6 1 2 x +3 y 1 e − cos (2 x + y) . =– 10 6 ∴ Using G.S. = C.F. + P.I., the general solution of (1) is 1 2x + 3y 1 e z = φ1(y – x) + e–2x φ2(y + 2x) – − cos (2x + y) , 10 6 where φ1, φ2 are arbitrary functions. =−

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. (D2 – DD′ + D′ – 1)z = cos (x + 2y) 3. (D2 – DD′ – 2D) z = sin (3x + 4y) 5. (D – D′2)z = cos (x – 3y)

2. (D – D′ – 1)(D – D′ – 2) z = sin (2x + 3y) 4. (2DD′ + D′2 – 3D′) z = 3 cos (3x – 2y) 6. (D2 + D′)(D – D′ – D′2) z = sin (2x + y).

95

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

Answers 1 sin ( x + 2 y) 2 1 [sin (2 x + 3 y) − 3 cos (2 x + 3 y)] 2. z = ex φ1(y + x) + e2x φ2(y + x) + 10 1 [sin (3 x + 4 y) + 2 cos (3 x + 4 y)] 3. z = φ1(y) + e2x φ2(y + x) + 15 3 [4 cos (3 x − 2 y) + 3 sin (3 x − 2 y)] 4. z = φ1(x) + e3x/2 φ2(2y – x) + 50 1. z = ex φ1(y) + e–x φ2(y + x) +

∞

5. z =

∑c e i

i =1 ∞

6. z =

∑

2

bi x + bi y

ci eai x − ai

2

y

+

1 [sin ( x − 3 y ) + 9 cos ( x − 3 y )] 82 ∞

+

i=1

∑ke i

(bi + bi 2 ) x + bi y

−

i=1

1 [5 sin (2 x + y) − 3 cos (2 x + y)]. 34

5.11. PARTICULAR INTEGRAL WHEN F(x, y) IS OF THE FORM eax+by V(x, y) Let the given partial differential equation be f (D, D′) z = F(x, y), where F(x, y) = eax+by V(x, y). P.I. =

∴

1 eax+by V(x, y) f (D, D′ )

This is evaluated by using the formula :

1 1 e ax + by V(x, y) = e ax + by V(x, y) . f(D, D′ ) f(D + a, D′ + b) Remark. If F(x, y) = eax+by and f(a, b) = 0, then we cannot write

In such a case, we write

1 1 e ax + by . e ax + by as f (a, b) f (D, D′ )

1 1 1 (eax + by . 1) = eax + by eax + by = . 1. f (D + a, D + b) f (D, D′ ) f (D, D′ )

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations: (i) (D2 – 4DD′ + 4D′2 + D – 2D′) z = ex+y (ii) (D + D′ – 1)(D + D′ – 3)(D + D′) z = ex+y sin (2x + y). Sol. (i) We have (D2 – 4DD′ + 4D′2 + D – 2D′) z = ex+y. D2

– 4DD′ +

4D′2

+ D – 2D′ = (D –

2D′)2

...(1)

+ (D – 2D′)

= (D – 2D′)(D – 2D′ + 1) = (1.D – 2D′ + 0)(1.D – 2D′ + 1) ∴

C.F. = e–0.x/1 φ1(1.y – (– 2)x) + e–1.x/1 φ2(1.y – (– 2)x) = φ1(y + 2x) + e–x φ2(y + 2x). P.I. =

1 e x+ y (D − 2D′ ) (D − 2D′ + 1)

...(2)

96

PARTIAL DIFFERENTIAL EQUATIONS

Here a = 1, b = 1. ∴ ∴

D = 1, D′ = 1 ⇒ D – 2D′ = 1 – 2(1) = – 1 ≠ 0 and D – 2D′ + 1 = 1 – 2(1) + 1 = 0 P.I. =

FG H

1 1 ex + y D − 2D′ + 1 D − 2D′

=

IJ K

=

FG H

1 1 ex + y D − 2D′ + 1 1 − 2(1)

IJ K

1 1 e x+ y .1 . = – ex+y (D + 1) − 2(D′ + 1) + 1 D − 2D′ + 1 − 1

= – e x+ y

FG H

1 1 D′ 1 = − e x+ y . 1− 2 D − 2D′ D D

= − ex+y .

FG H

IJ K

IJ K

−1

.1

1 D′ 1 1+2 + ...... . 1 = − e x + y . (1) = − xe x + y . D D D

∴ Using G.S. = C.F. + P.I., the general solution of (1) is z = φ1(y + 2x) + e–x φ2(y + 2x) – xex+y, where φ1, φ2 are arbitrary functions. (ii) We have (D + D′ – 1)(D + D′ – 3)(D + D′) z = ex+y sin (2x + y). ...(1) x + y ⇒ (1.D + 1.D′ – 1)(1.D + 1.D′ – 3)(1.D + 1.D′ + 0)z = e sin (2x + y) ∴ C.F. = e– (– 1)x/1 φ1(1.y – 1.x) + e– (– 3)x/1 φ2(1 · y – 1 · x) + e– 0.x/1 φ3(1 · y – 1 · x) = ex φ1(y – x) + e3x φ2(y – x) + φ3(y – x).

1 e x + y sin (2x + y) (D + D′ − 1)(D + D′ − 3)(D + D′ ) 1 = ex + y sin (2x + y) (D + 1 + D′ + 1 − 1) (D + 1 + D′ + 1 − 3) (D + 1 + D′ + 1) 1 sin (2x + y) = ex + y (D + D′ + 1 ) (D + D′ − 1) (D + D′ + 2) 1 . (D + D′ − 2) = ex + y sin (2x + y) ((D + D′ ) 2 − 1 ) ((D + D′ ) 2 − 4) D + D′ − 2 = ex + y sin (2x + y) 2 2 (D + D′ + 2DD′ − 1 ) (D 2 + D′ 2 + 2DD′ − 4) D + D′ − 2 = ex + y sin (2x + y) (− 4 − 1 + 2(– 1) . 2 . 1 − 1) (− 4 − 1 + 2(− 1) (2 . 1) − 4)

P.I. =

= ex + y

D + D′ − 2 sin (2x + y) − 130

1 x+ y (2 cos (2x + y) + cos (2x + y) – 2 sin (2x + y)) e 130 1 x+ y e =− (3 cos (2x + y) – 2 sin (2x + y)). 130 ∴ Using G.S. = C.F. + P.I., the general solution of (1) is =−

z = ex φ1(y – x) + e3x φ2(y – x) + φ3(y – x) − where φ1, φ2, φ3 are arbitrary functions.

1 e x + y (3 cos (2x + y) – 2 sin (2x + y)), 130

NON-HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1. (D2 – DD′ + D′ – 1) z = ey 3.

(D2

– D′) z =

2. (D2 – DD′ + D′ – 1) z = ex

ex+y

4. (D2 – D′2 – 3D + 3D′) z = xy + ex+2y

5. (D2 + DD′ + D + D′ – 1) z = e–2x (x2 + y2)

6. D(D – 2D′)(D + D′) z = ex+2y (x2 + 4y2) .

Answers 1. z = ex φ1(y) + e–x φ2(y + x) – xey ∞

3. z =

∑c e i

2

ai x + ai y

2. z = ex φ1(y) + e–x φ2(y + x) +

1 x xe 2

− ye x + y

i =1

4. z = φ1(y + x) + e3x φ2(y – x) – ∞

5. z =

∑c e i

ai x + bi y

1 2 1 1 1 3 2 x y − xy − x 2 − x − x − xe x + 2 y 6 9 9 18 27

+ e–2x (x2 + y2 + 6x + 2y + 18), where a2i + aibi + ai + bi – 1 = 0

i =1

6. z = φ1(y) + φ2(y + 2x) + φ3(y – x) –

1 (9 x2 + 36 y2 − 18 x − 72 y + 76) e x + 2 y . 81

97

Partial Differential Equations Reducible to Equations with Constant Coefficients

6

6.1. INTRODUCTION Till now we have been discussing the solution of linear partial differential equations which are with constant coefficients. Now we shall consider the method of solving a particular type of linear partial differential equations with variable coefficients that are capable of reducing to a linear partial differential equations with constant coefficients. 6.2. REDUCIBLE LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH VARIABLE COEFFICIENTS Let f(xD, yD) = F(x, y) ...(1) be a linear partial differential equation with variable coefficients. Here f(xD, yD) is some function of xD and yD such that (1) may be a linear partial differential equation. The following are some of the partial differential equations of the form f(xD, yD) = F(x, y) : (i) x2D2 – y2D2 = x2y (ii) (x2D2 – 2xyDD + y2D2 – xD + 3yD) z = 8(y/x) (iii) x2r – y2t + px – qy = log x. In general, a reducible linear partial differential equation is of the form : a0 x n

n z

 a1 x n  1 y

n z

x n x n  1y This equation can also be written as

 ......  an y n

n z y n

 ......  F( x, y)

(a0 x n D n  a1 x n  1 yD n  1D  ......  an y n D n  ......) z  F( x, y) .

6.3. SOLUTION OF REDUCIBLE LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH VARIABLE COEFFICIENTS Let

f(xD, yD) = (a0 x n D n  a1 x n  1 yD n  1D  ......  an y n D n  ......) z  F( x, y)

be a reducible linear partial differential equation with variable coefficients. Define variables u and v by u = log x and v = log y.  x = eu and y = ev.   Let D1 = and D = . u v 98

...(1)

99

PARTIAL DIFFERENTIAL EQUATIONS REDUCIBLE TO EQUATIONS WITH CONSTANT COEFFICIENTS

∂z ∂z ∂u ∂z 1 ∂z =x =x = ∂x ∂u ∂x ∂u x ∂u ∂ ∂ x or xD = D1 = ∂x ∂u

Now

x

∴ Also, x

F GH

I LM JK N

...(2)

∂ ∂ n − 1z ∂n z ∂ n − 1z x n − 1 n − 1 = x x n − 1 n + (n − 1) x n − 2 n − 1 ∂x ∂x ∂x ∂x n = x

∴

xn

∴

xn

∂n z ∂x n ∂n ∂x n

∂n z ∂x n

+ (n − 1) x n − 1

FG ∂ − (n − 1)IJ x H ∂x K F ∂ − n + 1IJ x = Gx H ∂x K

= x

n−1

n−1

∂n − 1z

OP Q

∂x n − 1

∂n − 1z ∂x n − 1 ∂n − 1

∂x n − 1

xnDn = (xD – n + 1) xn–1Dn–1 xnDn = (D1 – n + 1) xn–1Dn–1.

or or

...(3)

When n = 2, (3) ⇒

x2D2

= (D1 – 1) xD = (D1 – 1)D1

∴

x2D2

= D1(D1 – 1)

When n = 3, (3) ⇒

x3D3

= (D1 – 2) x2D2 = (D1 – 2) D1(D1 – 1)

∴

x3D3 = D1(D1 – 1)(D1 – 2) etc.

Thus, we have

xD = D1, x2D2 = D1(D1 – 1),

Similarly, we have

yD′ = D1′, y2D′2 = D1′(D1′ – 1), y3D′3 = D1′(D1′ – 1)(D1′ – 2) ......

Also,

x3D3 = D1(D1 – 1)(D1 – 2), ......

xyDD′ = D1D1′, xrysDrD′s

and

= D1(D1 – 1) ...... (D1 – r + 1)D1′(D1′ – 1) ...... (D1′ – s + 1).

Substituting these values, the given equation will be reduced to a linear partial differential equation with constant coefficients and with independent variables u and v. This equation is solved by known methods and then the values of u and v are substituted in terms x and y. This represents the general solution of the given equation.

ILLUSTRATIVE EXAMPLES Example 1. Find the general solution of the following partial differential equations : (i) x 2 (ii) x 2

2 ∂2z ∂2z ∂z ∂z 2 ∂ z + + +x +y −z=0 2xy y 2 2 ∂x∂y ∂x ∂y ∂x ∂y 2 ∂2z ∂2z ∂z 2 ∂ z − + + 6y = x3 y4 4xy 4y ∂x∂y ∂y ∂x 2 ∂y 2

(iii) x2r – y2t + px – qy = log x 2 (iv) x

∂2z ∂x

2

+ 2xy

FG H

IJ K

∂2z ∂2z ∂z ∂z + y2 + nz = n x +y + x2 + y2 + x3 . 2 ∂x∂y ∂x ∂y ∂y

100

PARTIAL DIFFERENTIAL EQUATIONS

2 ∂2 z ∂2 z ∂z ∂z 2 ∂ z + 2 xy + y +x +y − z = 0. 2 2 ∂x∂y ∂x ∂y ∂x ∂y ⇒ (x2D2 + 2xyDD′ + y2D′2 + xD + yD′ – 1) z = 0 ...(1) Let u = log x and v = log y. ∴ x = eu and y = ev. 2 Also, xD = D1, yD′ = D1′, x D2 = D1(D1 – 1), xyDD′ = D1D1′, y2D′2 = D1′(D1′ – 1), ∂ ∂ where D1 = and D1′ = . ∂u ∂v ∴ (1) ⇒ (D1(D1 – 1) + 2D1D1′ + D1′(D1′ – 1) + D1 + D1′ – 1) z = 0 ⇒ (D1′2 + 2D1D1′ + D1′2 – 1) z = 0 ⇒ ((D1 + D1′)2 – 1) z = 0 ⇒ (D1 + D1′ – 1)(D1 + D1′ + 1) z = 0 ⇒ (1.D1 + 1.D1′ – 1)(1.D1 + 1.D1′ + 1) z = 0. ∴ z = e– (– 1).u/1 φ1(1.v – 1.u) + e–1.u/1 φ2(1.v – 1.u) = eu φ1(v – u) + e–u φ2(v – u)

Sol. (i) We have

x2

= xφ1(log y – log x) +

FG H

= xφ 1 log ∴ The general solution is z = xψ 1

IJ K

1 φ (log y – log x) x 2

FG H

IJ K

FG IJ H K

FG IJ H K

y y 1 y 1 y + ψ2 + φ 2 log = xψ 1 , (say) x x x x x x

FG y IJ + 1 ψ FG y IJ , where ψ , ψ are arbitrary functions. H xK x H xK 2

1

2

2 ∂2 z ∂2 z ∂z 2 ∂ z − 4 xy + 4 y + 6y = x3 y4 . 2 2 ∂x∂y ∂y ∂x ∂y ⇒ (x2D2 – 4xyDD′ + 4y2D′2 + 6yD′) z = x3y4 ...(1) Let u = log x and v = log y. ∴ x = eu and y = ev. 2 2 Also, xD = D1, yD′ = D1′, x D = D1(D1 – 1), xyDD′ = D1D1′, y2D1′2 = D1′(D1′ – 1),

(ii) We have

where

x2

∂ ∂ and D1′ = . ∂u ∂v ∴ (1) ⇒ (D1(D1 – 1) – 4D1D1′ + 4D1′(D1′ – 1) + 6D1′) z = e3ue4v

D1 =

⇒ ⇒ ⇒ ∴

(D12 – 4D1D1′ + 4D1′2 – D1 + 2D1′) z = e3u+4v ((D1 – 2D1′)2 – (D1 – 2D1′)) z = e3u+4v (D1 – 2D1′)(D1 – 2D1′ – 1) z = e3u+4v (1.D1 – 2D1′ + 0)(1.D1 + (– 2)D1′ – 1) z = e3u + 4v C.F. = e0.u/1 φ1(1.v + 2u) + e1.u/1 φ2(1.v + 2u) = φ1(2u + v) + eu φ2(2u + v) = φ1(2 log x + log y) + xφ2(2 log x + log y) = φ1(log x2y) + xφ2(log x2y) = ψ1(x2y) + xψ2(x2y), say 1 e 3u + 4 v P.I. = ...(2) (D 1 − 2D 1 ′ )(D 1 − 2D 1 ′ − 1)

PARTIAL DIFFERENTIAL EQUATIONS REDUCIBLE TO EQUATIONS WITH CONSTANT COEFFICIENTS

Here

101

a = 3, b = 4 D1 = 3, D1′ = 4 ⇒ D1 – 2D1′ = 3 – 2(4) = – 5 ≠ 0 D1 – 2D1′ – 1 = 3 – 2(4) – 1 = – 6 ≠ 0

and

1 1 3 4 e 3u + 4v = x y . 30 ( − 5)( − 6) ∴ Using G.S. = C.F. + P.I., the general solution of the given equation is

∴ (2)

P.I. =

⇒

ψ2(x2y) + z = ψ1(x2y) + xψ

1 3 4 x y , where ψ1, ψ2 are arbitrary functions. 30

(iii) We have x2r – y2t + px – qy = log x. ⇒ (x2D2 – y2D′2 + xD – yD′) z = log x Let u = log x and v = log y. ∴ x = eu and y = ev Also, xD = D1, yD′ = D1′, x2D2 = D1(D1 – 1), where

...(1)

y2D′2 = D1′(D1′ – 1),

∂ ∂ and D1′ = . ∂u ∂v ∴ (1) ⇒ (D1(D1 – 1) – D1′(D1′ – 1) + D1 – D1′) z = u ⇒ (D12 – D1′2) z = u ⇒ (D1 + D1′)(D1 – D1′) z = u ⇒ (1.D1 + 1.D1′ + 0)(1.D1 + (– 1).D1′ + 0)z = u ∴ C.F. = e– 0.u/1 φ1(1.v – 1 . u) + e– 0.u/1 φ2(1.v – (– 1).u) = φ1(v – u) + φ2(v + u) = φ1(log y – log x) + φ2(log y + log x)

D1 =

FG H

FG IJ H K D ′ I 1 1 F 1− u= G J u D K −D ′ D H F 1 + D ′ − ......I u = 1 (u + 0) = u GH D JK D 6

= φ 1 log P.I. =

=

D 12 1 D1

2

IJ K

y y + φ2(log xy) = ψ 1 + ψ2(xy), (say.) x x

1

2

1

2 1

1

1

−1

2

2 1

2

3

2

1

2

=

1 (log x) 3 . 6

∴ Using G.S. = C.F. + P.I., the general solution of the given equation is z = ψ1

(iv) We have

x2

FG y IJ H xK

+ ψ 2 (yx) +

1 (log x)3 , where ψ1, ψ2 are arbitrary functions. 6

FG H

2 ∂2 z ∂2 z ∂z ∂z 2 ∂ z + 2 xy + y + nz = n x +y 2 2 ∂x∂y ∂x ∂y ∂x ∂y

IJ + x K

2

⇒ (x2D2 + 2xyDD′ + y2D′2 – nxD – nyD′ + n) z = x2 + y2 + x3 Let u = log x and v = log y. ∴ x = eu and y = ev. Also, xD = D1, yD′ = D1′, x2D2 = D1(D1 – 1), xyDD′ = D1D1′, y2D′2 = D1′(D1′ – 1), where D1 =

∂ ∂ and D1′ = . ∂u ∂v

+ y2 + x3. ...(1)

102

PARTIAL DIFFERENTIAL EQUATIONS

(1) ⇒ (D1(D1 – 1) + 2D1D1′ + D1′(D1′ – 1) – nD1 – nD1′ + n) z = e2u ⇒ (D12 + 2D1D1′ + D1′2 – D1 – D1′ – nD1 – nD1′ + n) z = e2u ⇒ ((D1 + D1′)2 – (n + 1)(D1 + D1′) + n) z = e2u ⇒ (D1 + D1′ – 1)(D1 + D1′ – n) z = e2u ⇒ (1.D1 + 1.D1′ – 1)(1.D1 + 1.D1′ – n)z = e2u 1.u/1 ∴ C.F. = e φ1(1.v – 1.u) + enu/1 φ2(1.v – 1.u)

FG H

= eu φ1(v – u) + enu φ2(v – u) = xφ 1 log = xψ1(y/x) + xnψ2(y/x), (say.) P.I. = =

IJ K

FG H

y y + x n φ 2 log x x

+ + + + +

e2v e2v e2v e2v e2v

+ + + + +

e3u e3u e3u e3u e3u.

IJ K

1 ( e 2 u + e 2 v + e 3u ) (D 1 + D 1 ′ − 1) (D 1 + D 1 ′ − n) 1 1 e 2u + e 2v (D 1 + D 1 ′ − 1) (D 1 + D 1 ′ − n) (D 1 + D 1 ′ − 1) (D 1 + D 1 ′ − n)

+

1 e 3u (D 1 + D 1 ′ − 1) (D 1 + D 1 ′ − n)

=

1 1 1 e 2u + e 2v + e 3u (2 + 0 − 1) (2 − 0 − n) (0 + 2 − 1) (0 + 2 − n) (3 + 0 − 1) (3 + 0 − n)

=

1 1 1 x2 + y2 + x3 . 2−n 2−n 2(3 − n)

∴ Using G.S. = C.F. + P.I., the general solution of the given equation is z = xψ 1

FG y IJ + x H xK

n

ψ2

FG y IJ + 1 (x H xK 2 − n

2

+ y2) +

1 x3 , 2(3 − n)

where ψ1 and ψ2 are arbitrary functions.

WORKING STEPS OF SOLVING f(xD, yD′′) = F(x, y) Step I. Put u = log x and v = log y. Step II. Change the whole equation in independent variables u and v by using x = eu and y = ev. We shall get a linear partial differential equation with constant coefficients. Step III. Find the general solution of the equation obtained in step II. Step IV. In the general solution, put u = log x and v = log y. This gives the general solution of the given equation.

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations : 1.

x2

3.

x2

∂2 z ∂x

2

∂2 z ∂x

2

+ 2 xy − y2

∂2 z ∂2 z + y2 2 = 0 ∂x∂y ∂y

∂2 z

∂y2

+x

∂z ∂z −y =0 ∂x ∂y

∂2 z ∂2 z ∂z + 2 y2 2 + 5 y − 2z = 0 ∂x∂y ∂y ∂x ∂y 2 ∂2 z 2 ∂ z − y2 2 = xy 4. x 2 ∂x ∂y 2 2. x

∂2 z 2

− 3 xy

PARTIAL DIFFERENTIAL EQUATIONS REDUCIBLE TO EQUATIONS WITH CONSTANT COEFFICIENTS

5.

x2

∂2 z ∂x

2

− y2

∂2 z ∂y

2

= x2 y

7. x2r – 3xys + 2y2t + px + 2qy = x + 2y 9. yt – q = xy 11.

x2r – xys – 2y2t + xp – 2yq = log

FG y IJ + xψ FG y IJ H xK H xK F yI z = ψ (xy) + ψ G J H xK F yI 1 x y z = ψ (xy) + xψ G J + H xK 2

1. z = ψ 1 3. 5.

2

1

2

1

2

2

7. z = ψ1(xy) + ψ2(x2y) + x + y

z = ψ1(x2y) + ψ 2

12.

z=

∞

∑c x i

i=1

ai bi

y

y 1 − x 2

FG IJ H K

+ x 2 y2

2

+ 2 xy

12. x2r – 4y2t – 4yq – z = x2y2 log y.

Answers 1 ψ 2 ( x2 y) x

2. z = x2 ψ 1 ( xy) +

FG y IJ + xy log x H xK x y F yI F yI 6. z = ψ G J + xψ G J + H xK H x K (m + n)(m + n − 1) F yI x 8. z = ψ (y) + x ψ G J – H x K 9y F yI F y I (x + y ) 10. z = ψ G J + xψ G J + H xK H x K n(n − 1) 4. z = ψ1(xy) + xψ2

1 xy2 log y 2 y 1 1 + (log x)2 log y − (log x)2 x 2 4

9. z = ψ1(x) + y2ψ2(x) + 11.

∂2 z ∂2 z + y2 2 = x m yn , m + n ≠ 0, 1 ∂x∂y ∂x ∂y 3 x 8. x2r + 2xys – xp = 2 y 10. x2r + 2xys + y2t = (x2 + y2)n/2 2 6. x

∂2 z

103

m n

1

1

2

2

3

2

2

2

2

1

2

(16 − 15 log y) , where ai2 – 4bi2 – ai – 1 = 0. 225

2 n/ 2

7

Monge’s Methods

7.1. INTRODUCTION In the last chapter we discussed the methods of solving some special type of linear partial differential equations with variable coefficients which were capable of being reduced to linear partial differential equations with constant coefficients by changing the independent variables. Solving any given partial differential equation with variable coefficients is not an easy task. We are moving in this direction step by step. 7.2. PARTIAL DIFFERENTIAL EQUATION OF SECOND ORDER A partial differential equation of the second order is of the form f(x, y, z, p, q, r, s, t) = 0. It is only in special cases that a partial differential equation of second order can be solved. Monge’s methods are used to solve some particular types of equations of second order. 7.3. INTERMEDIATE INTEGRAL Let

f(x, y, z, p, q, r, s, t) = 0

...(1)

be a partial differential equation of second order. A relation of the form u = φ(v), where u, v are functions of x, y, z, p, q and φ is an arbitrary function, is called an intermediate integral of (1) if the given partial differential equation (1) could be derived by eliminating the arbitrary function φ. 1 log y = φ(x) is an intermediate integral of the equation xys = 1, x 1 ∂p 1 1 1 because differentiating p − log y = φ(x) w.r.t. y, we get − . = 0 or s – = 0 or xys = 1. x ∂y x y xy

For example

p−

Remark. Finding of one or more intermediate integrals of a partial differential equation of second order is the first step in the direction of finding the general solution of the given partial differential equation of second order.

7.4. MONGE’S METHODS Let

Rr + Ss + Tt + U(rt – s2) = V

...(1)

be a partial differential equation of second order, where R, S, T, U, V are functions of x, y, z, p, q. An equation of the form (1) may or may not admit of a solution. Monge’s methods are used to solve any solvable equation of the form (1). 104

105

MONGE’S METHODS

In particular if U = 0, then (1) reduces to Rr + Ss + Tt = V. We shall be considering the cases U = 0 and U ≠ 0 separately. 7.5. MONGE’S METHOD OF SOLVING Rr + Ss + Tt = V Let Rr + Ss + Tt = V ...(1) be a solvable partial differential equation, where R, S, T, V are functions of x, y, z, p, q. Since z is a function of x and y, we have

∂p ∂p ∂2 z ∂2 z dx + dy = 2 dx + dy = r dx + s dy ∂x ∂y ∂y∂x ∂x ∂q ∂q ∂2 z ∂2 z dx + dy = dx + 2 dy = s dx + t dy. dq = ∂x ∂y ∂x∂y ∂y

dp = and

Solving these equations for r and t, we get ∴ (1)

⇒

R

r=

dp − sdy dx

FG dp − sdy IJ + Ss + T FG dq − sdx IJ = V H dx K H dy K

and

t=

dq − sdx . dy

...(2) ⇒ s[R(dy)2 – S dxdy + T(dx)2] = R dydp + T dxdq – V dxdy 2 2 The equations : R(dy) – S dxdy + T(dx) = 0 ...(3) and R dydp + T dxdq – V dxdy = 0 ...(4) are called Monge’s equations. The equation (3) may have either distinct or same factors. Case I. Let R(dy)2 – S dxdy + T(dx)2 = (A1dy + B1dx)(A2dy + B2dx) = 0. In this case we have two distinct systems

OP Q A dy + B dx = 0O Rdydp + Tdxdq − Vdxdy = 0PQ A 1dy + B 1dx = 0 Rdydp + Tdxdq − Vdxdy = 0

and

2

2

...(5) ...(6)

Let system (5) be integrable. Let u = u(x, y, z, p, q) = a and v = v(x, y, z, p, q) = b satisfy the system (5). ∴ u = ψ(v) is an intermediate integral of (1), since u = a, v = b satisfy (2) and hence (1). If system (6) is also integrable, we get another intermediate integral. These intermediate integrals are solved to find the values of p and q in terms of x and y. The values of p and q are substituted in dz = p dx + q dy. This is integrated to get the general solution of (1). In case we get only one intermediate integral or we want to use only one intermediate integral then we express it in the form Pp + Qq = R and use Lagrange’s method to find the general solution of (1). Case II. Let R(dy)2 – Sdx dy + T(dx)2 = (Ady + Bdx)2 = 0 Let u = u(x, y, z, p, q) = a, v = v(x, y, z, p, q) = b satisfy the system Ady + Bdx = 0 Rdydp + Tdxdq – Vdxdy = 0 ∴ u = ψ(v) is an intermediate integral of (1), since u = a, v = b satisfy (2) and hence (1). We express it in the form Pp + Qq = R and use Lagrange’s method to find the general solution of (1). Type I. Equations giving two distinct intermediate integrals and both are used to find the general solution.

OP Q

106

DIFFERENTIAL EQUATIONS AND INTEGRAL TRANSFORMS

WORKING STEPS FOR SOLVING PROBLEMS Step I. Write the given equation in the form Rr + Ss + Tt = V. Step II. Substitute the values of R, S, T, V in the Monge’s equations : R(dy)2 – Sdxdy + T(dx)2 = 0 ...(1) and Rdydp + Tdxdq – Vdxdy = 0 ...(2) Step III. Factorise (1) into two distinct factors. Step IV. Find two intermediate integrals. Solve these to find the values of p and q. Step V. Put p and q in dz = pdx + qdy and integrate to get the general solution of the given equation.

ILLUSTRATIVE EXAMPLES

and

and

Example 1. Find the general solution of the following partial differential equations : (i) r – t cos2 x + p tan x = 0 (ii) xy (r – t) – s(x2 – y2) = qx – py. 2 Sol. (i) We have r – t cos x + p tan x = 0. ...(1)  r – t cos2 x = – p tan x Comparing (1) with Rr + Ss + Tt = V, we get R = 1, S = 0, T = – cos2 x, V = – p tan x. The Monge’s equations are : R(dy)2 – Sdxdy + T(dx)2 = 0 ...(2) Rdydp + Tdxdq – Vdxdy = 0 ...(3) (2)  (dy)2 – cos2  (dx)2 = 0  (dy – cos  dx) (dy + cos  dx) = 0  dy – cos x dx = 0 ...(4) dy + cos  dx = 0 ...(4) (3)  1.dydp + (– cos2 ) dxdq – (– p tan x) dxdy = 0  dydp – cos2  dxdq + p tan x dx dy = 0 ...(5) We consider the system (4) and (5). Integrating (4), we get y – sin x = a (4)  dy = cos x dx. Putting this value of dy in (5), we get cos x dxdp – cos2 x dxdq + p tan x dx cos x dx = 0  cos x dx (dp – cos x dq + p tan x dx) = 0  dp – cos x dq + p tan x dx = 0  (dp + p tan x dx) – cos x dq = 0  Multiplying by sec x, we get (sec x dp + p tan x sec x dx) – dq = 0  d(p sec x) – dq = 0 Integrating, we get p sec x – q = b Let b = (a),  arbitrary.  p sec x – q = (y – sin x) ...(6) Similarly by solving (4) and (5), we get p sec x + q = (y + sin x) ...(7)

107

MONGE’S METHODS

Solving (6) and (7), we get p= and

1 (φ(y – sin x) + ψ(y + sin x)) 2 sec x

1 (φ(y – sin x) – ψ(y + sin x)) 2 Now dz = pdx + qdy. cos x 1 (φ( y – sin x) + ψ ( y + sin x) dx – (φ ( y – sin x) – ψ ( y + sin x)) dy ∴ dz = 2 2 1 1 ⇒ dz = φ( y – sin x) (cos x dx – dy) + ψ ( y + sin x) (cos x dx + dy) 2 2 1 1 ⇒ dz = – φ( y – sin x) d( y – sin x) + ψ( y + sin x) d( y + sin x) 2 2 Integrating, we get 1 1 ψ( y + sin x) d( y + sin x) z = – φ( y – sin x) d( y – sin x) + 2 2 ⇒ z = φ1(y – sin x) + φ2(y + sin x), (say.) This is the general solution of the given equation. Here φ1, φ2 are arbitrary functions. (ii) We have xy(r – t) – s(x2 – y2) = qx – py. ...(1) ⇒ xyr – (x2 – y2) s – xyt = qx – py Comparing (1) with Rr + Ss + Tt = V, we get R = xy, S = – x2 + y2, T = – xy, V = qx – py. The Monge’s equations are : R(dy)2 – Sdxdy + T(dx)2 = 0 ...(2)

q=–

z

and

z

Rdydp + Tdxdq – Vdxdy = 0 (2) ⇒ + (x2 – y2) dxdy – xy(dx)2 = 0 ⇒ (xdy – ydx)(ydy + xdx) = 0 ⇒ xdy – ydx = 0 ydy + xdx = 0 (3) ⇒ xydydp – xy dxdq – (qx – py) dxdy = 0 We consider the system (4) and (5), dy dx = (4) ⇒ xdy = ydx ⇒ y x Integrating, we get log y = log x + log a y y ⇒ log = log a ⇒ = a. x x (5) ⇒ ydp . xdy – xdq . ydx – qdx . xdy + pdy . ydx = 0 ⇒ ydp – xdq – qdx + pdy = 0 ⇒ (ydp + pdy) – (xdq + qdx) = 0 ⇒ d(yp – xq) = 0 ⇒ yp – xq = b Let b = φ(a). ∴ yp – xq = φ(y/x) Now we consider the system (4′) and (5). (4′) ⇒ d(x2 + y2) = 0 ⇒ x2 + y2 = c

...(3)

xy(dy)2

and

...(4) ...(4′) ...(5)

[By using (4)]

...(6)

108

DIFFERENTIAL EQUATIONS AND INTEGRAL TRANSFORMS

(5) ⇒ xdp . ydy – ydq . xdx – qdy . xdx + pdx . ydy = 0 ⇒ xdp + ydq + qdy + pdx = 0 ⇒ (xdp + pdx) + (ydq + qdy) = 0 ⇒ d(xp) + d(yq) = 0 ⇒ d(xp + yq) = 0 ⇒ xp + yq = k. Let k = ψ(c). ∴ xp + yq = ψ(x2 + y2) ∴ We get two intermediate integrals of (1). Solving (6) and (7) for p and q, we get p=

and

Now

xψ ( x 2 + y 2 ) + yφ ( y/ x) x2 + y2 dz = pdx + qdy

∴

dz =

xψ ( x 2 + y 2 ) + yφ ( y/ x) yψ ( x 2 + y 2 ) − xφ ( y/ x) dx + dy . x2 + y2 x2 + y2

⇒

dz =

ψ(x 2 + y2 ) φ( y/ x) ( xdx + ydy) + 2 ( ydx – xdy) 2 2 x +y x + y2

⇒

dz =

ψ(x 2 + y2 ) φ( y/ x) xdy – ydx d( x 2 + y 2 ) – x2 2 (x 2 + y2 ) 1 + ( y/ x) 2

q=

[By using (4′)]

...(7)

yψ ( x 2 + y 2 ) − xφ ( y/ x) . x2 + y2

ψ(x 2 + y2 ) φ( y/ x) d( x 2 + y 2 ) – d ( y/ x) 2 2 2 (x + y ) 1 + ( y/ x) 2 Integrating, we get

⇒

dz =

z

z

ψ(x2 + y2 ) φ ( y/ x) d( x 2 + y 2 ) + – d ( y/ x) . 2 2 2 (x + y ) 1 + ( y/ x) 2 ⇒ z = φ1(x2 + y2) + φ2(y/x), (say). This is the general solution of the given equation. Here φ1, φ2 are arbitrary functions. Type II. Equations giving two distinct intermediate integral and only one is used to find the general solution.

z=

WORKING STEPS FOR SOLVING PROBLEMS Step I. Write the given equation in the form Rr + Ss + Tt = V. Step II. Substitute the values of R, S, T, V in the Monge’s equations : R(dy)2 – Sdxdy + T(dy)2 = 0 ...(1) and Rdydp + Tdxdq – Vdxdy = 0 ...(2) Step III. Factorize (1) into two distinct factors. Step IV. Using any of the factors, find an intermediate integral. Solve this integral by using Lagrange method to get the general solution of the given equation. Example 2. Find the general solution of the following partial differential equations : (i) (r – s)y + (s – t)x + q – p = 0 (ii) (x – y) (xr – xs – ys + yt) = (x + y) (p – q). Sol. (i) We have (r – s)y + (s – t)x + q – p = 0. ⇒ yr + (x – y)s – xt = p – q ...(1) Comparing (1) with Rr + Ss + Tt = V, we get R = y, S = x – y, T = – x, V = p – q.

109

MONGE’S METHODS

and

The Monge’s equations are : R(dy)2 – Sdxdy + T(dx)2 = 0 ...(2) Rdydp + Tdxdq – Vdxdy = 0 ...(3) 2 2 (2) ⇒ y(dy) – (x – y) dxdy – x(dx) = 0 ⇒ (dx + dy)(ydy – xdx) = 0 ⇒ dx + dy = 0 ...(4) and ydx – xdy = 0 ...(4′) (3) ⇒ ydydp – xdx dq – (p – q) dx dy = 0 ...(5) We consider the system (4) and (5). Integrating (4), we get x + y = a. (5) ⇒ ydydp – xdxdq – pdxdy + qdxdy = 0 ⇒ ydpdy + xdq (– dx) + pdy (– dx) + qdxdy = 0 ⇒ ydp + xdq + pdy + qdx = 0 (∵ dy = – dx) ⇒ (ydp + pdy) + (xdq + qdx) = 0 ⇒ d(py) + d(qx) = 0 ⇒ py + qx = b Let b = ψ(a). ∴ py + qx = ψ(x + y) This is a Lagrange linear equation. dx dy dz Auxiliary equations are ...(6) = = y x ψ( x + y) dx dy 1 = (6) ⇒ ⇒ xdx – ydy = 0 ⇒ d( x 2 − y 2 ) = 0 y x 2 ⇒ x2 – y2 = k, where k is arbitrary. dx dz dx dz = ⇒ = (6) ⇒ 2 y ψ ( x + y) x − k ψ x + x2 − k

FH

dz =

⇒

u=x+

x2 − k

⇒

x2 − k

ψ x+

F GG H

du = 1 +

x2 − k

x 2

IK

FH

IK

dx

I J dx = − k JK

x du ψ(u) ∴ (7) ⇒ dz = ψ(u) . ⇒ dz = du u u Integrating, we get ψ(u) du + b . z= u ⇒ z = φ1(u) + b, say

...(7) u 2

x −k

dx

⇒

dx 2

x –k

=

du u

z

⇒

z = φ1 ( x +

x 2 – k ) + φ2 (a), say

⇒ z = φ1 (x + y) + φ2 (x2 – y2). This is the general solution of the given equation. Here φ1, φ2 are arbitrary functions. (ii) We have (x – y) (xr – xs – ys + yt) = (x + y) (p – q). ⇒ (x2 – xy) r – (x2 – y2) s + (xy – y2) t = (x + y) (p – q) ...(1) Comparing (1) with Rr + Ss + Tt = V, we get R = x2 – xy, S = – (x2 – y2), T = xy – y2, V = (x + y) (p – q).

110

and

DIFFERENTIAL EQUATIONS AND INTEGRAL TRANSFORMS

The Monge’s equations are : R(dy)2 – Sdxdy + T(dx)2 = 0 ...(2) Rdydp + Tdxdq – Vdxdy = 0 ...(3) (2) ⇒ (x2 – xy)(dy)2 + (x2 – y2) dxdy + (xy – y2)(dx)2 = 0 ⇒ x(dy)2 + (x + y)dxdy + y(dx)2 = 0 ⇒ (xdy + ydx)(dx + dy) = 0 ⇒ xdy + ydx = 0 ...(4) and dx + dy = 0 ...(4′) 2 2 (3) ⇒ (x – xy) dydp + (xy – y ) dxdq – (x + y)(y + p) dxdy = 0 We consider the system (4) and (5). (4) ⇒ d(xy) = 0 ⇒ xy = a. (5) ⇒ (x – y)dp . xdy + (x – y)dq . ydx – (p – q)dx . xdy – (p – q)dy . ydx = 0 ⇒ (x – y)dp – (x – y)dq – (p – q)dx + (p – q)dy = 0 (∵ xdy = – ydx) ⇒ (x – y)(dp – dq) – (p – q)(dx – dy) = 0

dp − dq dx − dy d( p − q) d ( x − y) = = ⇒ x− y p−q x− y p− q Integrating, let p – q = b(x – y). Let b = ψ(a). ∴ p – q = (x – y) ψ(xy). This is a Lagrange linear equation. ⇒

Auxiliary equations are

dx dy dz = = 1 – 1 ( x − y) ψ ( xy)

...(6)

(6) ⇒ dx = – dy ⇒ x + y = c. Taking yψ(xy), xψ(xy), 1 as multipliers, each fraction of (6) = ∴

yψ ( xy) dx + xψ ( xy) dy + dz yψ ( xy) dx + xψ ( xy) dy + dz = . 0 yψ ( xy) − xψ ( xy) + ( x − y) ψ ( xy)

yψ ( xy) dx + xψ ( xy) dy + dz = 0

⇒ ψ(xy)(ydx + xdy) + dz = 0 ⇒ ψ(xy) d(xy) + dz = 0 Integrating, let φ1(xy) + z = λ Let λ = φ2(c). ∴ φ1(xy) + z = φ2(x + y) ∴ The general solution of the given equation is z = φ2(x + y) – φ1(xy), where φ1 and φ2 are arbitrary functions. Type III. Equations giving two identical intermediate integrals. WORKING STEPS FOR SOLVING PROBLEMS Step I. Write the given equation in the form Rr + Ss + Tt = V. Step II. Substitute the values of R, S, T, V in the Monge’s equations : ...(1) R(dy)2 – Sdxdy + T(dy)2 = 0 and Rdydp + Tdxdq – Vdxdy = 0 ...(2) Step III. Factorise (1) into two identical factors. Step IV. Using one factor, find an intermediate integral. Solve this integral by using Lagrange method to get the general solution of the given equation.

111

MONGE’S METHODS

and

Example 3. Find the general solution of the following partial differential equations : (i) y2r – 2ys + t = p + 6y (ii) (y – x) (q2r – 2pqs + p2t) = (p + q)2 (p – q). Sol. (i) We have y2r – 2ys + t = p + 6y. ...(1) Comparing (1) with Rr + Ss + Tt = V, we get R = y2, S = – 2y, T = 1, V = p + 6y. The Monge’s equations are R(dy)2 – Sdxdy + T(dx)2 = 0 ...(2) Rdydp + Tdxdq – Vdxdy = 0 ...(3) 2 2 2 (2) ⇒ y (dy) + 2y dxdy + (dx) = 0 ⇒ (ydy + dx)2 = 0 ⇒ ydy + dx = 0 ⇒ dx = – ydy ...(4) y2 + a ⇒ y2 + 2x = b, where b = 2a 2 (3) ⇒ y2dydp + 1.dxdq – (p + 6y) dxdy = 0 ⇒ y2dydp – ydydq + y(p + 6y) (dy)2 = 0 ⇒ ydp – dq + (p + 6y) dy = 0 ⇒ (ydp + pdy) – dq + 6ydy = 0 Integrating, py – q + 3y2 = k. Let k = ψ(b). ∴ py – q + 3y2 = ψ(y2 + 2x) This is a Lagrange linear equation.

Integrating, let x = –

...(5)

The auxiliary equations are dx dy dz = = 2 y − 1 ψ( y + 2 x) − 3 y 2

(6)

⇒

dx dy = −1 y

⇒ ydy + dx = 0 ⇒ y2 + 2x = c

dz ψ(c) − 3 y 2 Integrating, let y3 – yψ(c) = z + d.

(6)

⇒

– dy =

...(6)

⇒ (3y2 – ψ(c)) dy = dz

Let d = φ1(c) ψ(y2 + 2x) = z + φ1(y2 + 2x). ∴ y3 – yψ This is the general solution of (1). Here ψ, φ1 are arbitrary functions. (ii) We have (y – x)(q2r – 2pqs + p2t) = (p + q)2 (p – q). 2 ⇒ (y – x)q r – 2pq(y – x)s + p2(y – x)t = (p + q)2 (p – q) ...(1) Comparing (1) with Rr + Ss + Tt = V, we get R = (y – x)q2, S = – 2pq(y – x), T = p2(y – x), V = (p + q)2 (p – q). The Monge’s equations are R(dy)2 – Sdxdy + T(dx)2 = 0 ...(2) and Rdydp + Tdxdq – Vdxdy = 0 ...(3) 2 2 2 2 (2) ⇒ (y – x) q (dy) + 2pq(y – x)dxdy + p (y – x)(dx) = 0 ⇒ (y – x)(qdy + pdx)2 = 0 ⇒ pdx + qdy = 0 ⇒ dz = 0 ⇒ z = a. (3) ⇒ (y – x) q2dydp + p2(y – x) dxdq – (p + q)2 (p – q) dxdy = 0 ⇒ (y – x) [qdp . qdy + pdq . pdx] – (p2 – q2) [pdx . dy + qdy . dx] = 0 ⇒

(y – x) (qdp – pdq) – (p2 – q2) (– dy + dx) = 0 (∵ pdx + qdy = 0)

112

DIFFERENTIAL EQUATIONS AND INTEGRAL TRANSFORMS

qdp – pdq – (p2 – q2)

⇒

q 2d

⇒

FG pIJ + ( p H qK

2

− q2 )

d( x − y) =0 y−x d( x − y) =0 x− y

FG IJ H K

p 1 d( x − y) + d =0 2 q x− y ( p/q) − 1 1 p/q − 1 1 = log b . Integrating, we get log (x – y) + log 2 p/q + 1 2 p− q =b ⇒ (x – y)2 p+ q p− q = ψ( z) Let b = ψ(a). ∴ (x – y)2 p+ q ⇒ (x – y)2 (p – q) – (p + q) ψ(z) = 0 ⇒ p((x – y)2 – φ(z)) – q((x – y)2 + ψ(z)) = 0 This is a Lagrange linear equation. The auxiliary equations are ⇒

(5)

⇒

dx dy dz = = 0 ( x − y) 2 − φ( z) − (( x − y) 2 + ψ ( z)) dz = 0 ⇒ z = c.

Each fraction of (5)

=

Integrating, we get

d( x − y) ( x − y) 2

x + y = – ψ(c) .

x+y–

⇒

...(5)

dx + dy dx − dy = − 2ψ( z) 2( x − y) 2

d(x + y) = – ψ(c)

⇒

...(4)

Let

( x − y) − 1 + d. −1

ψ( z) =d x− y d = φ(c).

ψ(z) = φ(z) . x−y This represents the general solution of (1). Here ψ, φ are arbitrary functions. x+y−

∴

TEST YOUR KNOWLEDGE Find the general solution of the following partial differential equations by using Monge’s method : 1. r = k2t

2. t – r sec4 y = 2q tan y

3. (r – s)x = (t – s)y

4. pt – qs = q3

5. q(1 + q)r – (p + q + 2pq)s + p(1 + p)t = 0

6. xy(t – r) + (x2 – y2)(s – 2) = py – qx

7.

x2r

–

y2t

– 2xp + 2z = 0

9. y2r + 2xys + x2t + px + qy = 0

8. x2r – 2xs + t + q = 0

113

MONGE’S METHODS

Answers 1. z = φ1(y – kx) + φ2(y + kx)

2. z = φ1(tan y – x) + φ2(tan y + x)

z = φ1(x + y) + φ2(y/x) x+ y 5. x = φ1(z) + φ2(x + y + z)

4. y = xz + φ1(z) + φ2(x)

3.

7. zy =

(xy)3/2

6. z = xy + φ1(x2 + y2) + φ2(x/y)

φ1(y/x) + φ2(xy)

8. z = xφ1(y + log x) + φ2(y + log x)

9. z = ψ (y2 – x2) log (y + x) + φ (y2 – x2).

Hint 3. The intermediate integrals are p – q = f(y/x) and xp + yq – z = g(x + y). Solving for p and q, we get

LM FG y IJ OP H xKQ N 1 L q= M z + g(x + y) − xf FGH xy IJK OPQ . x+ y N 1 L 1 L F yI O F yIO z + g( x + y) + yf G J P dx + z + g( x + y) − xf G J P dy dz = M M H K H xKQ x+ y N x Q x+ y N F yI (x + y) dz = zd(x + y) + g(x + y) d(x + y) + f G J (ydx – xdy) H xK p=

and ∴ ⇒ ⇒

( x + y) dz − zd( x + y) ( x + y)2

=

1 z + g( x + y) + yf x+ y

g ( x + y)

( x + y)2

d( x + y) +

f ( y/ x)

1 + ( y/ x)2

d( y/ x) .

7.6. MONGE’S METHOD OF SOLVING Rr + Ss + Tt + U(rt – s2) = V Let Rr + Ss + Tt + U(rt – s2) = V ...(1) be a solvable partial differential equation, where R, S, T, U, V are functions of x, y, z, p, q. Since z is a function of x and y, we have

∂2 z ∂2 z dx + dy = rdx + sdy 2 ∂y∂x ∂x ∂2 z ∂2 z dx + 2 dy = sdx + tdy . and ∂x∂y ∂y dp − sdy dq − sdx and t = Solving these equations for r and t, we get r = . dx dy dp − sdy dq − sdx (dp − sdy) (dq − sdx) + Ss + T − s2 = V ∴ (1) ⇒ R + U dx dy dxdy ⇒ s[R(dy)2 – Sdxdy + T(dx)2 + U(dxdp + dydq)] = Rdydp + Tdxdq + Udpdq – Vdxdy ...(2) The equations : R(dy)2 – Sdxdy + T(dx)2 + U(dxdp + dydq) = 0 ...(3) and Rdydp + Tdxdq + Udpdq – Vdxdy = 0 ...(4) are called Monge’s equations. Here, the equation (3) cannot be factored. Let λ = λ(x, y, z, p, q) be a function such that λ [R(dy)2 – Sdxdy + T(dx)2 + U(dxdp + dydq)] + Rdydp + Tdxdq + Udpdq – Vdxdy be factorisable. ∂p dx + ∂x ∂q dx + dq = ∂x

dp =

FG H

∂p dy = ∂y ∂q dy = ∂y

IJ K

FG H

IJ K

FG H

IJ K

114

DIFFERENTIAL EQUATIONS AND INTEGRAL TRANSFORMS

Let λ [R(dy)2 – Sdxdy + T(dx)2 + U(dxdp + dydq)] + Rdydp + Tdxdq + Udpdq – Vdxdy = (ady + bdx + cdp) (αdy + βdx + γdq) = aα(dy)2 + (aβ + bα) dxdy + bβ(dx)2 + cβdxdp + aγdydq + cαdydp + bγdxdq + cγdpdq = 0 Comparing coefficients, we get aα = λR, aβ + bα = – λS – V, bβ = λT, cβ = λU, aγ = λU, cα = R, bγ = T, cγ = U. Let a = λ, α = R. ∴ aα = λR Also aγ = λU ⇒ γ = U, cα = R ⇒ cR = R ⇒ c = 1, cβ = λU ⇒ 1 . β = λU ⇒ β = λU, bγ = T ⇒ bU = T ⇒ b = T/U. ∴ aβ + bα = – λS – V ⇒ λ(λU) + (T/U) R = – λS – V ...(5) 2 2 ⇒ U λ + SUλ + TR + UV = 0 Let λ1, λ2 be the roots of (5). We have (ady + bdx + cdp)(αdy + βdx + γdq) = 0 ...(6)

FG H

IJ K

T dx + 1 . dp (Rdy + λ 1Udx + Udq) = 0 U ⇒ (λ1Udy + Tdx + Udp) (Rdy + λ1Udx + Udq) = 0 Similarly, taking λ = λ2, we get (λ2Udy + Tdx + Udp) (Rdy + λ2Udx + Udq) = 0 Equations (7) and (8) give four systems of equations:

Taking

λ = λ1, (6) becomes λ 1dy +

OP Q λ Udy + Tdx + Udp = 0 O Rdy + λ Udx + Udq = 0PQ λ Udy + Tdx + Udp = 0O Rdy + λ Udx + Udq = 0PQ Rdy + λ Udx + Udq = 0 O Rdy + λ Udx + Udq = 0PQ λ 1Udy + Tdx + Udp = 0 λ 2 Udy + Tdx + Udp = 0 1

...(7) ...(8) ...(9) ...(10)

2

2

...(11)

1

1

...(12)

2

Subtracting equations of (9), we get (λ1 – λ2) Udy = 0. If λ1 ≠ λ2, then Udy = 0 identically, which is not true. ∴ System (9) does not give intermediate integral. Similarly, we reject system (12). ∴ We have only two systems given below:

OP Rdy + λ Udx + Udq = 0Q λ Udy + Tdx + Udp = 0O Rdy + λ Udx + Udq = 0PQ λ 1Udy + Tdx + Udp = 0

...(10)

2

2

...(11)

1

The equation (5) may have either distinct or same roots. Case I. λ1 ≠ λ2 In this case we have two distinct systems (10) and (11). Let system (10) be integrable. Let u = u(x, y, z, p, q) = a, v = v(x, y, z, p, q) = b satisfy the system (10). ∴ u = ψ(v) is an intermediate integral of (1).

115

MONGE’S METHODS

If system (11) is also integrable, we get another intermediate integral. These intermediate integrals are solved to find the values of p and q in terms of x and y. The values of p and q are substituted in dz = pdx + qdy. This is integrated to get the general solution of (1). In case we get only one intermediate integral or we want to use only one intermediate integral then we express it in the form Pp + Qq = R and use Lagrange’s method to find the solution of (1). Case II. λ1 = λ2 In this case we have identical systems (10) and (11). Let system (10) be integrable. Let u = u(x, y, z, p, q) = a, v = v(x, y, z, p, q) = b satisfy the system (10). ∴ u = ψ(v) is an intermediate integral of (1). We express it in the form Pp + Qq = R and use Lagrange’s method to find the solution of (1). Remark. In case the computation of finding general solution of an equation is difficult, we restrict ourselves to a solution with arbitrary constants.

ILLUSTRATIVE EXAMPLES Example 1. Find the solution of the following partial differential equations : (i) r + 4s + t + rt – s2 = 2 (ii) 3r + s + t + rt – s2 = – 9. Sol. (i) We have r + 4s + t + rt – s2 = 2 ...(1) Comparing (1) with Rr + Ss + Tt + U(rt – s2) = V, we get R = 1, S = 4, T = 1, U = 1, V = 2. λ-quadratic equation is U2λ2 + SUλ + TR + UV = 0. ⇒ λ2 + 4λ + (1 + 2) = 0 ⇒ λ = – 1, – 3. Let λ1 = – 1, λ2 = – 3. First system of equations giving intermediate integral is ...(2) λ1Udy + Tdx + Udp = 0 Rdy + λ2Udx + Udq = 0 ...(3) (2) ⇒ – dy + dx + dp = 0 ...(4) (3) ⇒ dy – 3dx + dq = 0 ...(5) Integrating (4) and (5), we get – y + x + p = a and y – 3x + q = b, where a and b are arbitrary constants. Let b = ψ(a). ∴ y – 3x + q = ψ(– y + x + p) In particular let ψ(– y + x + p) = α(– y + x + p) + β. ∴ y – 3x + q = α(– y + x + p) + β ⇒ αp – q = – (α + 3)x + (α + 1) y – β This is a Lagrange equation. dx dy dz = = The auxiliary equations are ...(6) α − 1 − (α + 3) x + (α + 1) y − β dx (6) ⇒ = – dy ⇒ dx + αdy = 0 ⇒ x + αy = γ. ...(7) α Equating second and third fractions of (6), we get dy dz = (Using (7)) − 1 − (α + 3) (γ − αy) + (α + 1) y − β dz ⇒ dy = 2 − (α + 4α + 1) y + αγ + 3γ + β 2 ⇒ ((α + 4α + 1) y – (αγ + 3γ + β)) dy + dz = 0

116

DIFFERENTIAL EQUATIONS AND INTEGRAL TRANSFORMS

(α2 + 4α + 1)

⇒

y2 – (αγ + 3γ + β) y + z = k 2

y2 – ((α + 3) (x + αy) + β) y + z = φ(γ) [Taking k = φ(γ)] 2 1 2 α2 y + (α + 3) xy + (α + 3) βy = z − φ( x + αy) +α− ⇒ 2 2 1 2 2 ⇒ z = (α + 2α − 1) y + (α + 3) xy + (α + 3) βy + φ(x + αy). 2 This is the general solution of the given equation. Here α, β are arbitrary constants and φ is an arbitrary function.

⇒

(α2 + 4α + 1)

F GH

I JK

Remark. For the above equation, y – 3x + q = ψ(– y + x + p) is an intermediate integral. Since p appears in the argument of the arbitrary function ψ, we cannot find the value of p using this equation and the other intermediate integral of the given equation.

(ii) We have 3r + s + t + rt – s2 = – 9. ...(1) Comparing (1) with Rr + Ss + Tt + U(rt – s2) = V, we get R = 3, S = 1, T = 1, U = 1, V = – 9. λ-quadratic equation is U2λ2 + SUλ + TR + UV = 0. ⇒ λ2 + λ – 6 = 0 ⇒ λ = 2, – 3. Let λ1 = 2, λ2 = – 3. First system of equations giving intermediate integral is λ1Udy + Tdx + Udp = 0 ...(2) Rdy + λ2Udx + Udq = 0 ...(3) (2) ⇒ 2dy + dx + dp = 0 ⇒ 2y + x + p = a (3) ⇒ 3dy – 3dx + dq = 0 ⇒ 3y – 3x + q = b Let a = ψ(b). ∴ 2y + x + p = ψ(3y – 3x + q) In particular, let ψ(3y – 3x + q) = α(3y – 3x + q) + β. ∴ 2y + x + p = α(3y – 3x + q) + β ⇒ p – αq = (3α – 2)y – (3α + 1)x + β This is a Lagrange’s equation. The auxiliary equations are dx dy dz = = ...(4) 1 − α (3α − 2) y − (3α + 1) x + β dy (4) ⇒ dx = ⇒ dy + αdx = 0 ⇒ y + αx = γ. ...(5) −α dz (4) ⇒ dx = (Using (5)) (3α − 2) (γ − αx) − (3α + 1) x + β ⇒ (– (3α2 + α + 1) x + 3αγ – 2γ + β) dx = dz 1 ⇒ z = – (3α2 + α + 1) x2 + (3αγ – 2γ + β) x + k 2 1 ⇒ z = – (3α2 + α + 1) x2 + (3αy + 2α2x – 2y – 2αx + β)x + φ(γ) 2 [(Putting k = φ(γ)] 1 α2 – 5α α – 1) x2 + (3α α – 2) xy + βx + φ(y + αx). ⇒ z= (3α 2

117

MONGE’S METHODS

This is the general solution of the given equation. Here α and β are arbitrary constants and φ is an arbitrary function. Example 2. Find the solution of the following partial differential equations: (i) 5r + 6s + 3t + 2(rt – s2) + 3 = 0 (ii) (q2 – 1) zr – 2pqzs + (p2 – 1) zt + z2(rt – s2) = p2 + q2 – 1. Sol. (i) We have 5r + 6s + 3t + 2(rt – s2) = – 3. ...(1) Comparing (1) with Rr + Ss + Tt + U(rt – s2) = V, we get R = 5, S = 6, T = 3, U = 2, V = – 3. λ-quadratic equation is U2λ2 + SUλ + TR + UV = 0. ⇒ 4λ2 + 12λ + 9 = 0 ⇒ λ = – 3/2, – 3/2. Let λ1 = – 3/2, λ2 = – 3/2. ∴ The system of equations giving intermediate integral is λ1Udy + Tdx + Udp = 0 ...(2) Rdy + λ2Udx + Udq = 0 ...(3) (2) ⇒ – 3dy + 3dx + 2dp = 0 ⇒ – 3y + 3x + 2p = a (3) ⇒ 5dy – 3dx + 2dq = 0 ⇒ 5y – 3x + 2q = b Let a = ψ(b). ∴ – 3y + 3x + 2p = ψ(5y – 3x + 2q) In particular, let ψ(5y – 3x + 2q) = α(5y – 3x + 2q) + β. ∴ – 3y + 3x + 2p = α(5y – 3x + 2q) + β ⇒ 2p – 2αq = 3y – 3x + 5αy – 3αx + β. This is a Lagrange’s equation. The auxiliary equations are dx dy dz = = ...(4) 2 − 2α 3 y − 3 x + 5αy − 3αx + β dy ...(5) (4) ⇒ dx = – α ⇒ dy + αdx = 0 ⇒ y + αx = γ. dx dz = (4) ⇒ (Using (5)) 2 (3 + 5α) (γ − αx) − 3(1 + α) x + β (– (5α2 + 6α + 3)x + 3γ + 5αγ + β) dx = 2dz

⇒ –

⇒ ⇒

–

1 (5α2 + 6α + 3)x2 + (3γ + 5αγ + β) x = 2z + k 2

1 (5α2 + 6α + 3)x2 + ((3 + 5α) (y + αx) + β) x = 2z + φ(γ) 2

[Putting k = φ(γ)]

1 (5α2 + 6α + 3)x2 + (3αx2 + 5α2x2 + 3xy + 5αxy + βx) = 2z + φ(y + αx) 2 1 α2 – 3) x2 + (3 + 5α α) xy + βx = 2z + φ(y + αx). ⇒ (5α 2 This is the general solution of the given equation.

⇒ –

Here α, β are arbitrary constants and φ is an arbitrary function. (ii) We have (q2 – 1) zr – 2pqzs + (p2 – 1) zt + z2(rt – s2) = p2 + q2 – 1. Comparing (1) with Rr + Ss + Tt + U(rt – s2) = V, we get R = (q2 – 1) z, S = – 2pqz, T = (p2 – 1) z, U = z2, V = p2 + q2 – 1. λ-quadratic equation is U2λ2 + SUλ + RT + UV = 0.

...(1)

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DIFFERENTIAL EQUATIONS AND INTEGRAL TRANSFORMS

⇒ z4λ2 – 2pqz3λ + (q2 – 1)(p2 – 1) z2 + z2 (p2 + q2 – 1) = 0 ⇒ z4λ2 – 2pqz3λ + p2q2z2 = 0 ⇒ (zλ – pq)2 = 0 ⇒ λ = pq/z, pq/z. Let λ1 = pq/z, λ2 = pq/z. ∴ The system of equations giving intermediate integral is λ1Udy + Tdx + Udp = 0 ...(2) Rdy + λ2Udx + Udq = 0 ...(3) pq 2 . z dy + (p2 – 1) z dx + z2dp = 0 (2) ⇒ ...(4) z pq 2 . z dx + z 2 dq = 0 (3) ⇒ (q2 – 1) zdy + ...(5) z (4) ⇒ pqdy + (p2 – 1) dx + zdp = 0 ⇒ p(qdy + pdx) – dx + zdp = 0 ⇒ pdz + zdp – dx = 0 ⇒ d(zp) – dx = 0 ⇒ zp – x = a. 2 (5) ⇒ (q – 1) dy + pqdx + zdq = 0 ⇒ q(qdy + pdx) – dy + zdq = 0 ⇒ qdz + zdq – dy = 0 ⇒ d(zq) – dy = 0 ⇒ zq – y = b. Let a = ψ(b) ∴ zp – x = ψ(zq – y). In particular, let ψ(zq – y) = α(zq – y) + β. ∴ zp – x = α(zq – y) + β ⇒ zp – αzq = x – αy + β. This is a Lagrange’s equation. The auxiliary equations are dx dy dz ...(6) = = z − αz x − αy + β dy (6) ⇒ dx = – ⇒ dy + αdx = 0 ⇒ y + αx = γ ...(7) α dx dz = ⇒ [(1 + α2)x – αγ + β] dx = zdz (Using (7)) (6) ⇒ z x − α (γ − αx) + β ⇒

x2 z2 − αγx + βx = +k 2 2 (1 + α2) x2 – 2x(y + αx) x + 2βx = z2 + 2k

(1 + α2)

⇒ ⇒ (1 – α2) x2 – 2αxy + 2βx = z2 + φ(γ) αxy + 2β βx = z2 + φ(y + αx). ⇒ (1 – α2) x2 – 2α This is the general solution of the given equation. Here α, β are arbitrary constants and φ is an arbitrary function.

[Putting 2k = φ(γ)]

TEST YOUR KNOWLEDGE Find the solution of the following partial differential equations by using Monge’s method: 2. 3s – 2(rt – s2) = 2 1. 3s + (rt – s2) = 2 2 4. 2r – 6s + 2t + (rt – s2) = 4. 3. 3r + 4s + t + (rt – s ) = 1

Answers α 2 y + 2xy + βy – φ(x + αy) 2 2 3. x + 3y2 + 2z – 4xy – 2βx = φ (y + αx) 1. z =

5 2 y – 2αxy – βy + φ(αx – 2y) 2 2 4. z = (α + α – 1) x2 + (2α – 2) xy + βx + φ(y + αx)

2. αz =

PARTIAL DIFFERENTIAL EQUATIONS

ISBN 978-93-5274-102-1

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PARTIAL DIFFERENTIAL EQUATIONS