26 An Introduction to Partial Differential Equations with MATLAB©R
the latter in that it includes theparameterλ. Remember that we solve the
X-ODEfor each real number λ.Foreachλ, the ODE has infinitely many
solutions. Now, though, we need to findwhich of these solutions “survive”
the boundary conditions—that is, we shall see that, for “most” real numbers
λ, the only solution that also satisfies the boundary conditions is the zero-
solution,X(x)≡0. Thus, we need to identify those values ofλfor which
theX-system hasnontrivialsolutions (and, of course, find those solutions).
Then we will solve theT-equation, but only for these values ofλ,andform
the nontrivial product solutions of the PDE and boundary conditions.
These values ofλare calledeigenvalues‖of theX-system, and the corre-
sponding nontrivial solutions are theeigenfunctionsassociated withλ.The
system itself is an example of an ODEeigenvalue problem.
Let’s calculate some eigenthings.
Example 1Find all eigenvalues and eigenfunctions of the eigenvalue problem
y′′+λy=0,y=y(x), 0 <x< 1
y(0) =y(1) = 0.As with ODE initial-value problems, and the boundary-value problems from
the exercises in Section 1.3, we first find the general solution of the ODE,
then apply the boundary conditions. To do this, we sety=erxand find the
characteristic equationr^2 +λ= 0. It now becomes apparent that we need to
treat the casesλ>0,λ=0andλ<0 separately.
Case 1: λ< 0
Ifλ<0, then we can writeλ=−k^2 for some real numberkwithk>0.
Then, the characteristic equationr^2 −k^2 = 0 leads to the two independent
solutionsekxande−kx. However, it turns out that life is much easier if we
use, instead, the functions
y 1 =
ekx+e−kx
2=cosh(kx)and
y 2 =
ekx−e−kx
2= sinh(kx)(see Exercise 22). Then, the general solution is
y=c 1 cosh(kx)+c 2 sinh(kx).‖These eigenvalues are similar to those which we see in Linear Algebra, where the eigen-
values of a matrix are those real numbers for which the matrix equationAv=λvhas a
nontrivial solution. Here, we are looking for real numbers for which thefunctionalequation
Ly=λyhas a nontrivial solution, whereLy=−y′′.