Partial Differential Equations with MATLAB

22 An Introduction to Partial Differential Equations with MATLAB©R

Example 4Separate the PDE (inu(x, y, z)),

ux− 2 uyy+3uz=0.

We letu(x, y, z)=X(x)Y(y)Z(z) and, substituting, get

X′YZ− 2 XY′′Z+3XYZ′=0.

Let’s divide byu=XYZand see what happens:

X′

X

−

2 Y′′

Y

+

3 Z′

Z

=0.

At the very least, we can separate any one of the variables from the other
two. For example, we can write

X′

X

=

2 Y′′

Y

−

3 Z′

Z

=−λ 1 ,

where we have concluded, as before, that each side of the separated equation
must be constant. Now, we immediately get the ODE

X′+λ 1 X=0.

As for the second half, we can rewrite it as

2 Y′′

Y

=

3 Z′

Z

−λ 1 ,

and we have separated the variablesyandz. Hence, we conclude that

2 Y′′

Y

=

3 Z′

Z

−λ 1 =−λ 2

for any realλ 2 ,or

2 Y′′+λ 2 Y=0 and 3Z′+(λ 2 −λ 1 )Z=0.

Therefore,u=XYZis a solution if and only if there exist constantsλ 1 and
λ 2 such thatX,Y andZsatisfy the three ODEs above (with, of course, the
sameλ 1 and the sameλ 2 in each).
We donotwant to give the impression that all linear, homogeneous PDEs
are separable—in fact, “most” are not separable. However, many of the equa-
tions which are important in applicationsareseparable (rather, many of the
simplifications which are made in deriving PDEs are madeso that the re-
sulting PDEs are linear and, often, separable). It is very easy to prove that
a PDE is separable—by separating it! However, it is more difficult toprove
that a PDE isnotseparable.