Mathematical Methods for Physics and Engineering : A Comprehensive Guide

20.3 GENERAL AND PARTICULAR SOLUTIONS

We could, of course, have takenh(x, y)=y, but this only leads to a solution that

is already contained in (20.25).

Solve

∂^2 u

∂x^2

+2

∂^2 u

∂x∂y

+

∂^2 u

∂y^2

=0,

subject to the boundary conditionsu(0,y)=0andu(x,1) =x^2.

From our general result, functions ofp=x+λywill be solutions provided

1+2λ+λ^2 =0,

i.e.λ=−1 and the equation is parabolic. The general solution is therefore

u(x, y)=f(x−y)+xg(x−y).

The boundary conditionu(0,y) = 0 impliesf(p)≡0, whilstu(x,1) =x^2 yields

xg(x−1) =x^2 ,

which givesg(p)=p+ 1, Therefore the particular solution required is

u(x, y)=x(p+1)=x(x−y+1).

To reinforce the material discussed above we will now give alternative deriva-

tions of the general solutions (20.22) and (20.25) by expressing the original PDE

in terms of new variables before solving it. The actual solution will then become

almost trivial; but, of course, it will be recognised that suitable new variables

could hardly have been guessed if it were not for the work already done. This

does not detract from the validity of the derivation to be described, only from

the likelihood that it would be discovered by inspection.

We start again with (20.20) and change to new variables

ζ=x+λ 1 y, η=x+λ 2 y.

With this change of variables, we have from the chain rule that

∂

∂x

=

∂

∂ζ

+

∂

∂η

,

∂

∂y

=λ 1

∂

∂ζ

+λ 2

∂

∂η

.

Using these and the fact that

A+Bλi+Cλ^2 i=0 fori=1, 2 ,

equation (20.20) becomes

[2A+B(λ 1 +λ 2 )+2Cλ 1 λ 2 ]

∂^2 u

∂ζ∂η

=0.