What does it mean if there arent any eigenvalues and eigenfunctions?

Question

I have been trying to solve this problem of eigenvalues:

$X''(x)+\lambda X(x)=0 , X(1)=0,X'(1)=0$ however I cannot find any eigenvalues or eigenfunctions.

What does this mean for $X(x)$?

Answer 1

Assuming $\lambda \neq 0$, the characteristic equation for $X'' + \lambda X = 0$ is $a^2 + \lambda = 0$, which has solutions $a = \pm \sqrt{-\lambda}$, and the general solution is $X(x) = C_1 e^{i \sqrt{\lambda} x} + C_2 e^{-i \sqrt{\lambda} x}$ (I could split this into the cases when $\lambda$ is positive, negative or complex, but it's not going to matter for the next step).

We can then put in the given initial conditions to solve for the coefficients, and we get:

$$\begin{eqnarray} X(1) & = & C_1 e^{i \sqrt{\lambda}} + C_2 e^{-i \sqrt{\lambda}} \\ \implies C_1 e^{i \sqrt{\lambda}} + C_2 e^{-i \sqrt{\lambda}} & = & 0 \\ X'(1) & = & i C_1 e^{i \sqrt{\lambda}} - i C_2 e^{-i \sqrt{\lambda}} \\ \implies i C_1 e^{i \sqrt{\lambda}} - i C_2 e^{-i \sqrt{\lambda}} & = & 0 \end{eqnarray}$$

and a quick bit of linear algebra will show that the solution to these two equations is $C_1 = C_2 = 0$, and hence $X(x) = 0$ is the solution to the given DE.

If, on the other hand, $\lambda = 0$, then the DE becomes $X'' = 0$ and its general solution is $X(x) = C_1 x + C_2$, but it's pretty easy to see that $X(1) = X'(1) = 0$ also gives $X(x) = 0$ as the solution.