Eigenvalue problem with Robin boundary conditions at both ends

Question

From Walter A. Strauss's Partial Differential Equations:

Consider the eigenvalue problem with Robin boundary conditions at both ends:
$$ -X'' = \lambda X, \qquad X'(0) - a_0 X(0) = 0, \qquad X'(l) - a_1 X(l)=0 $$ a) Show that $\lambda=0$ is an eigenvalue if and only if $a_0+a_1=-a_0a_1 l$

b) Find the eigenfunctions corresponding to the zero eigenvalue

Hint: First solve ODE for $X(x)$. The solutions are not sines or cosines.

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I was able to do part a) if $\lambda=0$ then $a_0+a_1=-a_0a_1 l$. But the rest of part a) and part b) I am unable to do. Can someone please help me to finish this problem?

Answer 1

If $\lambda=0$ is an eigenvalue, then $X(x)$ will be of the form $X(x) = mx + c$. Plug this into the boundary conditions, solve for $m$ and $c$, and voilà! The converse for part (a) is just as easy - if the condition is satisfied, then show there exist $m$ and $c$ so that $X(x)$ satisfies the boundary conditions.