Question 11.1.1: Find the eigenvalues and the corresponding eigenfunctions of......

The solution of the differential equation may have one of several forms, depending on λ, so it is necessary to consider several cases.

Case I: First, if λ = 0, the general solution of the differential equation (18) is

y=c_{1}x+c_{2}. (20)

The two boundary conditions (19) require that

c_{2}=0,\quad2c_{1}+c_{2}=0, (21)

respectively. The only solution of equations (21) is c_{1} = c_{2} = 0, so the boundary value problem has no nontrivial solution in this case. Hence λ = 0 is not an eigenvalue.

Case II: If λ > 0, then the general solution of the differential equation (18) is

y=c_{1}\sin\left({\sqrt{\lambda}}x\right)+c_{2}\cos\left({\sqrt{\lambda}}x\right), (22)

where {\sqrt{\lambda}} > 0. The boundary condition at x = 0 requires that c_{2} = 0; from the boundary condition at x = 1 we then obtain the equation

c_{1}\left(\sin\sqrt{\lambda}+\sqrt{\lambda}\cos\sqrt{\lambda}\right)=0.

For a nontrivial solution y we must have c_{1}\neq0, and thus λ must satisfy

\sin\sqrt{\lambda}+\sqrt{\lambda}\cos\sqrt{\lambda}=0. (23)

Note that if λ is such that \cos{\sqrt{\lambda}} = 0, then \sin{\sqrt{\lambda}} ≠ 0, and equation (23) is not satisfied. Hence we may assume that \cos{\sqrt{\lambda}} ≠ 0; dividing equation (23) by \cos{\sqrt{\lambda}}, we obtain

{\sqrt{\lambda}}=-\tan{\sqrt{\lambda}}. (24)

The solutions of equation (24) cannot be determined exactly; they must be approximated numerically. They can also be estimated by sketching the graphs of f({\sqrt{\lambda}}) = {\sqrt{\lambda}} and g({\sqrt{\lambda}}) = −\tan{\sqrt{\lambda}} for {\sqrt{\lambda}} > 0 on the same set of axes and identifying the points of intersection of the two curves (see Figure 11.1.1). The point {\sqrt{\lambda}} = 0 is specifically excluded from this argument because the solution (22) is valid only for {\sqrt{\lambda}} ≠ 0. Despite the fact that the curves intersect there, λ = 0 is not an eigenvalue, as we have already shown. The first four positive solutions of equation (24) are {\sqrt{\lambda_{1}}} ≅ 2.02876, {\sqrt{\lambda_{2}}} ≅ 4.91318, {\sqrt{\lambda_{3}}} ≅ 7.97867, and {\sqrt{\lambda_{4}}} ≅ 11.08554. Figure 11.1.1 shows that the roots can be approximated with reasonable accuracy by {\sqrt{\lambda_{n}}} ≅ (2n − 1)π/2 for n = 5, 6, . . . , the precision of this estimate improving as n increases. Hence the eigenvalues are

\lambda_{1}\cong4.11586,\quad\lambda_{2}\cong24.13934,\quad\lambda_{3}\cong63.65911,\quad\lambda_{4}\cong122.88916,\\\lambda_{n}\cong\frac{(2n-1)^{2}\pi^{2}}{4},\quad\mathrm{for}\,n=5,6,\ \dots\,. (25)

Finally, since c_{2} = 0, the eigenfunction corresponding to the eigenvalue \lambda_{n} is

\phi_{n}(x,\lambda_{n})=k_{n}\sin\biggl(\sqrt{\lambda_{n}}\,x\biggr);\quad n=1,2,\ \ldots\,, (26)

where the constant k_{n} remains arbitrary.

Case III: Next consider λ < 0. In this case let λ = −μ so that μ > 0. Then equation (14)

T^{\prime}+\lambda T=0. (14)

becomes

y^{\prime\prime}-\mu y=0, (27)

and its general solution is

y=c_{1}\sinh(\sqrt{\mu}x)+c_{2}\cosh(\sqrt{\mu}x), (28)

where \sqrt{\mu} > 0. Proceeding as in the previous case, we find that μ must satisfy the equation

{\sqrt{\mu}}=-\operatorname{tanh}{\sqrt{\mu}}. (29)

From Figure 11.1.2 it is clear that the graphs of f(\sqrt{\mu}) = \sqrt{\mu} and g(\sqrt{\mu}) = −\tanh{\sqrt{\mu}} intersect only at the origin. Hence there are no positive values of \sqrt{\mu} that satisfy equation (29), and hence the boundary value problem (18), (19) has no negative eigenvalues.

Finally, it is necessary to consider the possibility that λ may be complex. It is possible to show by direct calculation that the problem (18), (19) has no complex eigenvalues. However, in Section 11.2 we consider in more detail a large class of problems that includes this example. One of the things we show there is that every problem in this class has only real eigenvalues. Therefore, we omit the discussion of the nonexistence of complex eigenvalues here. Thus all the eigenvalues and eigenfunctions of the problem (18), (19) are given by equations (25) and (26).