Mathematical Methods for Physics and Engineering : A Comprehensive Guide

20.3 GENERAL AND PARTICULAR SOLUTIONS

which, when substituted into the PDE (20.9), give
[
A(x, y)

∂p

∂x

+B(x, y)

∂p

∂y

]

df(p)

dp

=0.

This removes all reference to the actual form of the functionf(p) since for

non-trivialpwe must have

A(x, y)

∂p

∂x

+B(x, y)

∂p

∂y

=0. (20.10)

Let us now consider the necessary condition forf(p) to remain constant asx

andyvary; this is thatpitself remains constant. Thus forfto remain constant

implies thatxandymust vary in such a way that

dp=

∂p

∂x

dx+

∂p

∂y

dy=0. (20.11)

The forms of (20.10) and (20.11) are very alike and become the same if we

require that

dx

A(x, y)

=

dy

B(x, y)

. (20.12)

By integrating this expression the form ofpcan be found.

For

x

∂u

∂x

− 2 y

∂u

∂y

=0, (20.13)

find(i)the solution that takes the value 2 y+1on the linex=1, and(ii)a solution that

has the value 4 at the point(1,1).

If we seek a solution of the formu(x, y)=f(p), we deduce from (20.12) thatu(x, y) will
be constant along lines of (x, y)thatsatisfy

dx

x

=

dy

− 2 y

,

which on integrating givesx=cy−^1 /^2. Identifying the constant of integrationcwithp^1 /^2
(to avoid fractional powers), we conclude thatp=x^2 y. Thus the general solution of the
PDE (20.13) is

u(x, y)=f(x^2 y),

wherefis an arbitrary function.
We must now find the particular solutions that obey each of the imposed boundary
conditions. For boundary condition (i) a little thought shows that the particular solution
required is

u(x, y)=2(x^2 y)+1=2x^2 y+1. (20.14)

For boundary condition (ii) some obviously acceptable solutions are

u(x, y)=x^2 y+3,

u(x, y)=4x^2 y,

u(x, y)=4.