Higher Engineering Mathematics

AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS 513

I

53.3 Solution of partial differential

equations by direct partial

integration

The simplest form of partial differential equations
occurs when a solution can be determined by direct
partial integration. This is demonstrated in the fol-
lowing worked problems.

Problem 1. Solve the differential equation

∂^2 u

∂x^2

= 6 x^2 ( 2 y− 1 )given the boundary condi-

tions that atx=0,

∂u

∂x

=sin 2yandu=cosy.

Since

∂^2 u

∂x^2

= 6 x^2 ( 2 y− 1 )then integrating partially

with respect toxgives:

∂u

∂x

=

∫

6 x^2 (2y−1)dx=(2y−1)

∫

6 x^2 dx

=(2y−1)

6 x^3

3

+f(y)

= 2 x^3 (2y−1)+f(y)

wheref(y) is an arbitrary function.
From the boundary conditions, whenx=0,

∂u

∂x

=sin 2y.

Hence, sin 2y=2(0)^3 (2y−1)+f(y)

from which, f(y)=sin 2y

Now

∂u

∂x

= 2 x^3 (2y−1)+sin 2y

Integrating partially with respect toxgives:

u=

∫

[2x^3 (2y−1)+sin 2y]dx

=

2 x^4

4

(2y−1)+x(sin 2y)+F(y)

From the boundary conditions, whenx=0,
u=cosy, hence

cosy=

(0)^4

2

(2y−1)+(0)sin 2y+F(y)

from which,F(y)=cosy

Hence, the solution of

∂^2 u

∂x^2

= 6 x^2 (2y−1) for the

given boundary conditions is:

u=

x^4

2

(2y−1)+xsiny+cosy

Problem 2. Solve the differential equation:

∂^2 u

∂x∂y

=cos(x+y) given that

∂u

∂x

=2 wheny=0,

andu=y^2 whenx=0.

Since

∂^2 u

∂x∂y

=cos(x+y) then integrating partially

with respect toygives:

∂u

∂x

=

∫

cos(x+y)dy=sin(x+y)+f(x)

From the boundary conditions,

∂u

∂x

=2 wheny=0,

hence

2 =sinx+f(x)

from which, f(x)= 2 −sinx

i.e.

∂u

∂x

=sin(x+y)+ 2 −sinx

Integrating partially with respect toxgives:

u=

∫

[sin(x+y)+ 2 −sinx]dx

=−cos(x+y)+ 2 x+cosx+f(y)

From the boundary conditions,u=y^2 whenx=0,

hence

y^2 =−cosy+ 0 +cos 0+f(y)

= 1 −cosy+f(y)

from which,f(y)=y^2 − 1 +cosy

Hence, the solution of

∂^2 u

∂x∂y

=cos(x+y) is given by:

u=−cos(x+y)+ 2 x+cosx+y^2 − 1 +cosy

Problem 3. Verify that

φ(x,y,z)=

1

√

x^2 +y^2 +z^2

satisfies the partial

differential equation:

∂^2 φ

∂x^2

+

∂^2 φ

∂y^2

+

∂^2 φ

∂z^2

=0.