Partial Differential Equations with MATLAB

20 An Introduction to Partial Differential Equations with MATLAB©R

that is, we have managed toseparatethe variablexfrom the variabley.
We say that the equation isseparableand that we haveseparated the
variables. Now, we have a situation where a function ofxequals a function
ofy,thatis,where
f(x)=g(y)

forallvalues ofxandyin the domain of the problem. So chooseanysuch
x-value,x=x 0 .Wethenhave

f(x 0 )=g(y)

for all values ofy,thatis,thatg(y)isaconstantfunction! Then, it follows
thatf(x) is a constant function, as well!
So, at this point, we have

u(x, y)=X(x)Y(y)isasolution⇒

X′

X

=−

Y′

Y

=λ (1.15)

for some real constantλ. Conversely, given any real constantλ, if (1.15) is
satisfied, thenu=XYis a solution of the PDE (why?).
Equation (1.15) actually is two equations:

X′

X

=λ and

Y′

Y

=−λ.

Therefore, we conclude thatu=XYis a solution of the PDE if and only if
XandYsatisfy the ODEs

X′−λX=0 and Y′+λY =0

for the sameλ. The product solutions, thus, are

X(x)=eλx and Y(y)=e−λy

or
u(x, y)=eλ(x−y),

for any real constant,λ. Further, any linear combination of these solutions
is, again, a solution.
As we shall see in the following chapter, although it looks as though we have
found only solutions which are linear combinations of product solutions, in
many cases that will be enough to solve any well-posed problem involving the
given PDE. In the process of solving these initial-boundary-value problems,
we shall find that only certain values ofλwill lead tonontrivialsolutions of
the problem, that is, to solutions other than the zero-function.

Example 2Find all product solutions of the heat equation,ut=uxx.
We letu(x, t)=X(x)T(t) and substitute:

X(x)T′(t)=X′′(x)T(t).