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y1 (x) = sin x, y2(x) = sin2x, ¦ ¦ ¦, yn (x) = sin nx, ¦ ¦ ¦, (24)
remembering that multiples of these functions are also eigenfunctions.
Now let us suppose that X < 0. If we let X = — j2, then Eq. (18) becomes
y" — j2y = 0^
Chapter 10. Partial Differential Equations and Fourier Series
The characteristic equation for Eq. (25) is r2 — j2 = 0 with roots r = ±j, so its general solution can be written as
ó = c1 cosh jx + c2 sinh jx. (26)
We have chosen the hyperbolic functions cosh jx and sinh jx as a fundamental set of solutions rather than the exponential functions exp (jx) and exp (—jx) for convenience in applying the boundary conditions. The first boundary condition requires that c1 = 0 and then the second boundary condition gives c2 sinh jn = 0. Since j = 0, it follows that sinh jn = 0, and therefore we must have c2 = 0. Consequently, ó = 0 and there are no nontrivial solutions for ê < 0. In other words, the problem (18), (19) has no negative eigenvalues.
Finally, consider the possibility that ê = 0. Then Eq. (18) becomes
y" = 0, (27)
and its general solution is
ó = c1 x + c2. (28)
The boundary conditions (19) can only be satisfied by choosing c1 = 0 and c2 = 0, so there is only the trivial solution ó = 0 in this case as well. That is, k = 0 is not an eigenvalue.
To summarize our results: We have shown that the problem (18), (19) has an infinite sequence of positive eigenvalues kn = n2for n = 1, 2, 3,... and that the corresponding eigenfunctions are proportional to sin nx. Further, there are no other real eigenvalues. There remains the possibility that there might be some complex eigenvalues; recall that a matrix with real elements may very well have complex eigenvalues. In Problem 17 we outline an argument showing that the particular problem (18), (19) cannot have complex eigenvalues. Later, in Section 11.2, we discuss an important class of boundary value problems that includes (18), (19). One of the important properties of this class of problems is that all their eigenvalues are real.
In later sections of this chapter we will often encounter the problem
y" + ky = 0, y(0) = 0, y( L) = 0, (29)
which differs from the problem (18), (19) only in that the second boundary condition is imposed at an arbitrary point x = L rather than at x = ï. The solution process is exactly the same as before up to the step where the second boundary condition is applied. For the problem (29) this condition requires that
c2sin j L = 0 (30)
rather than Eq. (22), as in the former case. Consequently, jL must be an integer
multiple of ï, so j = nn/L, where n is a positive integer. Hence the eigenvalues and
eigenfunctions of the problem (29) are given by
kn = ï2ï 2/L2, yn (x) = sin(nï x/L), n = 1, 2, 3,.... (31)
As usual, the eigenfunctions yn (x) are determined only up to an arbitrary multiplicative constant. In the same way as for (18), (19) you can show that the problem (29) has no eigenvalues or eigenfunctions other than those in Eq. (31).
The problems following this section explore to some extent the effect of different boundary conditions on the eigenvalues and eigenfunctions. A more systematic discussion of two-point boundary and eigenvalue problems appears in Chapter 11.
10.2 Fourier Series
In each of Problems 1 through 10 either solve the given boundary value problem or else show that it has no solution.
1. ó' + ó = 0, 0, ó'(ï) = 1
2. ó' + 2ó = 0, ó'(0) = 1 ó'(ï) = 0
3. ó' + ó = 0, ó(0) = 0, y(L) = 0
4. ó' + ó = 0, ó'(0) = 1, y(L) = 0
5. ó' + ó = õ , 0, ó(ï) = 0