# Sturm-Liouville Problem: Finding eigenvalues and eigenfunctions

I am trying to find the eigenvalues and eigenfunctions of the following Sturm-Liouville problem:

$$\begin{cases} -u''=\lambda u, \ &x\in (a, b), \\ u(a)=u(b)=0, \ &b>a.\\ \end{cases}$$

I already checked the eigenvalues for when $$\lambda<0$$ and $$\lambda=0$$ and got that the only there are only trivial results for those two cases. However, for $$\lambda>0$$, found a general solution of $$u_{gen}(x)=c_1\sin((\sqrt\lambda)x) + c_2\cos((\sqrt\lambda)x)$$ and then plugging in initial conditions to get $$u(a)=c_1\sin((\sqrt\lambda)a) + c_2\cos((\sqrt\lambda)a)$$ and $$u(b)=c_1\sin((\sqrt\lambda)b) + c_2\cos((\sqrt\lambda)b)$$

Then I put these values into a matrix, took the determinant and ended up with $$\sin(\sqrt\lambda a)\cos(\sqrt\lambda b)-\sin(\sqrt\lambda b)\cos(\sqrt\lambda a)=0$$ which eventually simplified to $$\tan(\sqrt\lambda a)=\tan(\sqrt\lambda b)$$ which led me to an eigenvalue of $$\lambda_n = \left( \frac{n\pi }{ (b-a) } \right)^2$$

Here is where I am getting stuck. I am not completely sure how to find the eigenfunctions for this problem, and I feel like the answer is already in my grasp, I just have no idea what it is. In examples we did in class, the initial conditions worked out so that in order to not have a trivial solution, one of the c values was being multiplied by a sin function, meaning the sin function had to be zero. Can anyone help me figure out what the eigenfunction is?

Oooh, you could have made it so much easier on yourself by starting from $$u_{gen}=c_1\sin\left(\sqrt{\lambda}(x-a)\right)+c_2\cos\left(\sqrt{\lambda}(x-a)\right)$$ Then you have $$u(a)=c_2=0$$ And $$u(b)=c_1\sin\left(\sqrt{\lambda}(b-a)\right)=0$$ So for a nontrivial solution $$\sqrt{\lambda}(b-a)=n\pi$$ So $$\lambda=\frac{n^2\pi^2}{(b-a)^2}$$ And then $$u_n(x)=c_1\sin\left(\frac{n\pi(x-a)}{b-a}\right)$$ If we want to normalize to unity, let $$\int_a^bc_1^2\sin^2\left(\frac{n\pi(x-a)}{b-a}\right)dx=\frac12c_1^2(b-a)=1$$ So we get the normalized eigenfunctions $$u_n(x)=\sqrt{\frac2{b-a}}\sin\left(\frac{n\pi(x-a)}{b-a}\right)$$
Any non-trivial solution $$u$$ of $$-u''=\lambda u$$ with $$u(a)=0$$ will be a non-constant multiple of the solution where $$u(a)=0,u'(a)=1$$. That solution is $$\varphi_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}(x-a))}{\sqrt{\lambda}}.$$ The limiting form where $$\lambda=0$$ is also correct. In this case $$\varphi_{0}(x)=x-a$$. So, the solutions $$u_{\lambda}$$ where $$u_{\lambda}(a)=0=u_{\lambda}(b)$$ are the solutions $$\varphi_{\lambda}$$ for which $$\lambda$$ satisifes the algebraic equation $$\frac{\sin(\sqrt{\lambda}(b-a))}{\sqrt{\lambda}}=0.$$ $$\lambda=0$$ is not a solution because the limiting equation where $$\lambda=0$$ is $$b-a=0$$, which is not valid. So, the solutions $$\lambda$$ are $$\sqrt{\lambda}(b-a)=n\pi \\ \lambda = \frac{n^2\pi^2}{(b-a)^2},\;\; n=1,2,3,\cdots.$$ These are the eigenvalues.