# eigen value problem with Robin Boundary Conditions at both ends

This is a problem from the book Partial Differential Equations by Walter.A.Strauss.
Consider the eigen value problem with Robin Boundary Conditions at both ends:
$-X''=\lambda X$
$X'(0)-a_0X(0)=0$
$X'(l)-a_1X(l)=0$

a)Show that $\lambda=0$ is an eigenvalue if and only if $a_0+a_1=-a_0a_1 l$
b)Find the eigen functions corresponding to the zero eigenvalue
(Hint: First solve ODE for X(x).The solutions are not sines or cosines).

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.

Haven't though through: What happens if you work with $Y(x) = X(x) - \left((x/l)a_0X(0) + ((l-x)/l)a_1 X(l)\right)$? – Eric Towers Jun 17 '14 at 16:24
@EricTowers : When $\lambda=0$ the solution is $X(x)=Ax+B$. Plugging boundary conditions I get two equations as $A-a_0B=0$ and $A+a_1(Al+B)=0$. Then in order to find the **eigen function ** do I have to solve one of $A-a_0B=0$ or $A+a_1(Al+B)=0$. When I use $A-a_0B=0$ I get $A=a_0B$. Then how can I find B and how does it become a function of x – clarkson Jun 18 '14 at 1:24
@EricTowers : Also I don't understand what you have given as $Y(x) = X(x) - \left((x/l)a_0X(0) + ((l-x)/l)a_1 X(l)\right)$ – clarkson Jun 18 '14 at 1:30
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.