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Boundary value problems with higher order differential equations can also occur; in them the number of boundary conditions must equal the order of the differential equation. As a rule, the order of the differential equation is even, and half the boundary conditions are given at each end of the interval. It is also possible for a single boundary condition to involve values of the solution and/or its derivatives at both boundary points; for example,
y(0) — y(L) = 0.
Chapter 11. Boundary Value Problems and Sturm-Liouville Theory
The following example involves one boundary condition of the form (15) and is therefore more complicated than the problems in Section 10.1.
Find the eigenvalues and the corresponding eigenfunctions of the boundary value problem
/ + ky = 0, (18)
7(0) = 0, /(1) + y(\) = 0. (19)
One place where this problem occurs is in the heat conduction problem in a bar of unit length. The boundary condition at x = 0 corresponds to a zero temperature there. The boundary condition at x = 1 corresponds to a rate of heat flow that is proportional to the temperature there, and units are chosen so that the constant of proportionality is 1 (see Appendix A of Chapter 10).
The solution of the differential equation may have one of several forms, depending
on k, so it is necessary to consider several cases. First, if ê = 0, the general solution of
the differential equation is
y = qx + C2. (2°)
The two boundary conditions require that
c2 = 0, 2c1 + c2 = 0, (21)
respectively. The only solution of Eqs. (21) is c1 = c2 = 0, so the boundary value problem has no nontrivial solution in this case. Hence ê = 0 is not an eigenvalue.
If ê > 0, then the general solution of the differential equation (18) is
y = c1 sin Vk x + c2 cos VI x, (22)
where Vk > 0. The boundary condition at x = 0 requires that c2 = 0; from the boundary condition at x = 1 we then obtain the equation
c1 (sin Vk + Vk cos Vk) = 0.
For a nontrivial solution y we must have c1 = 0, and thus k must satisfy
sinVk + Vk cos VI = 0. (23)
Note that if k is such that cos Vk = 0, then sin Vk = 0, and Eq. (23) is not satisfied. Hence we may assume that cos Vk = 0; dividing Eq. (23) by cos Vk, we obtain
Vk = — tan Vk. (24)
The solutions of Eq. (24) can be determined numerically. They can also be found approximately by sketching the graphs of f (Vk) = Vk and g(Vk) = — tan Vk for Vk > 0 on the same set of axes, and identifying the points of intersection of the two curves (see Figure 11.1.1). The point Vk = 0 is specifically excluded from this argument because the solution (22) is valid only for Vk = 0. Despite the fact that the curves intersect there, k = 0 is not an eigenvalue, as we have already shown. The first three positive solutions of Eq. (24) are Jk[ = 2.029, yk^ = 4.913, and yk^ = 7.979. As can be seen from Figure 11.1.1, the other roots are given with reasonable accuracy
11.1 The Occurrence of Two-Point Boundary Value Problems
by — 1)n/2 for n = 4, 5,..., the precision of this estimate improving as
n increases. Hence the eigenvalues are
k = 4.116, k2 = 24.14,
1 2 22 (25)
k3 = 63.66, kn = (2n — 1)2n 2/4 for n = 4, 5,....
Finally, since c2 = 0, the eigenfunction corresponding to the eigenvalue kn is
Ôï (x, kn) = kn sin^k~nx; n = 1, 2,..., (26)
where the constant kn remains arbitrary.
Next consider k < 0. In this case it is convenient to let k = —/ã so that ã > 0. Then Eq. (14) becomes
/ — ãó = 0, (27)
and its general solution is
ó = c1 sinh „/ã x + c2 cosh „/ã x, (28)
where „/ã > 0. Proceeding as in the previous case, we find that ã must satisfy the equation
-/ã = — tanh „/ã. (29)
From Figure 11.1.2 it is clear that the graphs of f(*/fi) = and g(*/fi) = — tanh^/ã
intersect only at the origin. Hence there are no positive values of „/ã that satisfy
Eq. (29), and hence the boundary value problem (18), (19) has no negative eigenvalues.
FIGURE 11.1.1 Graphical solution of \/k = — tan -fk.
Chapter 11. Boundary Value Problems and Sturm Liouville Theory