Partial Differential Equations with MATLAB

28 An Introduction to Partial Differential Equations with MATLAB©R

As in Case 1, this forcesc 2 =0except in those cases wherekis a number with
the propertysink= 0. For these latter values ofk, we neednothavec 2 =0;in
fact, there is no restriction onc 2 ,sothetermc 2 sin(kx)survives both boundary
conditions. In other words, these values ofkgive us the eigenvaluesλ=k^2
of the problem; for each suchk, the functions

y=csin(kx)

are the associated eigenfunctions. In practice, we say thattheeigenfunction
isy= sin(kx), realizing that any constant multiple of an eigenfunction is an
eigenfunction (why?).
So the eigenvalues are those numbersλ=k^2 where sink= 0. Therefore,
we have

k=π, 2 π, 3 π,...=nπ, n=1, 2 , 3 ,...(remember:k>0)

and
λ=π^2 , 4 π^2 , 9 π^2 ,...=n^2 π^2 ,n=1, 2 , 3 ,....

We write the eigenvalues as

λn=n^2 π^2 ,n=1, 2 , 3 ,...

and the corresponding eigenfunctions as

yn= sin(nπx),n=1, 2 , 3 ,....

Example 2Do the same for

y′′+λy=0,

y′(0) =y′(3) = 0.

Case 1: λ< 0 ,λ=−k^2 ,k> 0
We have
y=c 1 cosh(kx)+c 2 sinh(kx),

so that
y′=c 1 ksinh(kx)+c 2 kcosh(kx).

(See Exercise 22.) Then,

y′(0) = 0 =c 2 k⇒c 2 =0

and
y′(3) = 0 =c 1 ksinh 3k⇒c 1 =0,

so there are no negative eigenvalues.