32 An Introduction to Partial Differential Equations with MATLAB©R
Therefore (see Exercise 23),f(x) has no roots whenx>0 and the problem
has no negative eigenvalues.
Case 2: λ=0
The general solution here is
y=c 1 x+c 2
and, sincey′=c 1 , the boundary conditions give
y(0) = 0 =c 2
y(1) +y′(1) = 0 = 2c 1 +c 2.Therefore,c 1 =c 2 =0,soλ 0 = 0 is not an eigenvalue.
Case 3: λ> 0 ,λ=k^2 ,k>0.
Here, as usual,
y=c 1 coskx+c 2 sinkx,
so
y′=−c 1 ksinkx+c 2 kcosx.
Then,
y(0) = 0 =c 1
y(1) = 0 =c 2 (sink+kcosk).This system has only the solutionc 1 =c 2 =0unlesskis such that
sink+kcosk=0.Therefore, the eigenvalues correspond to those values ofksatisfying
−k=tank.How do we solve fork? We don’t—because we can’t! However, we can show
that there are infinitely many such values ofk, by looking at the graphs of
y=−kandy=tankfork>0, which we have plotted in Figure 1.2. In
fact, it looks as thoughy=−kintersects each branch ofy=tankexactly
once. Therefore, our eigenvalues correspond to values ofkn,n=1, 2 , 3 ,...,
satisfying
π
2<k 1 <3 π
2
3 π
2<k 2 <5 π
2
..
.
(2n−1)π
2<kn<
(2n+1)π
2