Partial Differential Equations with MATLAB

Introduction 21

Again, dividing both sides byugives us

T′

T

=

X′′

X

= constant.

For the sake of convenience (we’ll see why later), we call the constant−λ:

T′

T

=

X′′

X

=−λ

or
X′′+λX=0 and T′+λT=0.

So, for each real numberλ, we must solve these two ODEs. Now, the form of
the solution of the first PDE will depend on the sign ofλ(this didnothappen
in the previous example), so we must consider three cases.

Case 1: λ> 0
X=ccos

√

λx+dsin

√

λx,T=e−λt

and
u=e−λt[ccos

√

λx+dsin

√

λx].

Case 2: λ=0
X=cx+d, T=1

and
u=cx+d.

Case 3: λ< 0
X=ce

√

−λx+de−

√

−λx,T=e−λt

and
u=e−λt[ce

√

−λx+de−

√

−λx].

In each case,canddare arbitrary constants. Again, any linear combination
of solutions is a solution.

Example 3Separate the PDE 3uyy− 5 uxxxy+7uxxy=0.
Again, letu=XY:

3 XY′′− 5 X′′′Y′+7X′′Y′=0.

Then, dividing byXYdoesn’t help us, but dividing byXY′gives us

3 Y′′

Y′

=

5 X′′′− 7 X′′

X

=−λ

or
5 X′′′− 7 X′′+λX=0 and 3Y′′+λY′=0.