Sturm-Liouville system variable bounded in interval

Question

I am solving the following Sturm-Liouville system problem

$1.$ $X''(x) + \lambda X(x)=0, 0<x<2 $

$ 2.$ $X'(0)=0 , X(2) =0$

So I have considered the cases when $ \lambda <0 , \lambda=0$ .Now I am at the 3rd case when $\lambda >0$ then the solutions is $X(x)= c_1 \cos(\lambda x)+c_2 \sin(\lambda x )$ .

From $X'(0) =0$ we get that $c_2=0$ and from $X(2)=0 \implies c_1=0 $ is one solution,the other is $\cos(\lambda x)=0$.

Now when $\cos(\lambda x)=0$,how should I proceed with the solution considering $0<x<2$ ?

Answer 1

Now you have $\cos{2\lambda}=0$, from the homogeneous boundary condition, so $2\lambda = \frac{\pi}{2}+n\pi$, for $n \in \mathbb{Z}$

Answer 2

For $\lambda \gt 0$, general solution of the given Sturm-Liouville problem is $$X(x)= c_1 \cos(\sqrt \lambda x)+c_2 \sin(\sqrt \lambda x )$$ Now $X'(0)=0 , X(2) =0 \quad$ gives $c_2=0$ and $c_1\cos(2\sqrt \lambda )=0$

Now if $c_1=0$, then we get the trivial solution, so we take $c_1 \neq 0 $ and

$\cos(2\sqrt \lambda )=0\implies 2\sqrt \lambda=(2n+1)\frac{\pi}{2}\quad$ where $n=0,\pm1,\pm2,. . . $

So $\sqrt \lambda = (2n+1)\frac{\pi}{4} \implies \lambda= (2n+1)^2 \frac{\pi^2}{16}\quad$ where $n=0,\pm1,\pm2,. . . $

Hence the general solution is $$X(x)= c_1 \cos((2n+1)\frac{\pi}{4} x),\qquad \text{where} \quad n=0,\pm1,\pm2,. . . $$