PDES: GENERAL AND PARTICULAR SOLUTIONS
Each is a valid solution (the freedom of choice of form arises from the fact thatu
is specified at only one point (1,1), and not along a continuum (say), as in boundary
condition (i)). All three are particular examples ofthe general solution, which may be
written, for example, as
u(x, y)=x^2 y+3+g(x^2 y),whereg=g(x^2 y)=g(p) is an arbitrary function subject only tog(1) = 0. For this
example, the forms ofgcorresponding to the particular solutions listed above areg(p)=0,
g(p)=3p−3,g(p)=1−p.
As mentioned above, in order to find a solution of the formu(x, y)=f(p)werequire that the original PDE contains no term inu, but only terms containing
its partial derivatives. If a term inuis present, so thatC(x, y)= 0 in (20.9),
then the procedure needs some modification, since we cannot simply divide out
the dependence onf(p) to obtain (20.10). In such cases we look instead for
a solution of the formu(x, y)=h(x, y)f(p). We illustrate this method in the
following example.
Find the general solution ofx∂u
∂x+2
∂u
∂y− 2 u=0. (20.15)We seek a solution of the formu(x, y)=h(x, y)f(p), with the consequence that
∂u
∂x=
∂h
∂xf(p)+hdf(p)
dp∂p
∂x,
∂u
∂y=
∂h
∂yf(p)+hdf(p)
dp∂p
∂y.
Substituting these expressions into the PDE (20.15) and rearranging, we obtain
(
x
∂h
∂x+2
∂h
∂y− 2 h)
f(p)+(
x∂p
∂x+2
∂p
∂y)
hdf(p)
dp=0.
The first factor in parentheses is just the original PDE withureplaced byh. Therefore, if
hisanysolution of the PDE,however simple, this term will vanish, to leave
(
x
∂p
∂x+2
∂p
∂y)
hdf(p)
dp=0,
from which, as in the previous case, we obtain
x∂p
∂x+2
∂p
∂y=0.
From (20.11) and (20.12) we see thatu(x, y) will be constant along lines of (x, y)that
satisfy
dx
x
=
dy
2,
which integrates to givex=cexp(y/2). Identifying the constant of integrationcwithpwe
findp=xexp(−y/2). Thus the general solution of (20.15) is
u(x, y)=h(x, y)f(xexp(−^12 y)),wheref(p) is any arbitrary function ofpandh(x, y) is any solution of (20.15).