# Eigenvalue problem with Robin Boundary Conditions

Solve the eigenvalue problem with Robin boundary conditions: $$\phi''(x)+\lambda\phi(x)=0,$$ $$\phi(0)+\phi'(0)=0,$$ $$\phi(\pi)+\phi'(\pi)=0$$ I worked out that both $$\lambda=0$$ and $$\lambda<0$$ aren't eigenvalues, and for $$\lambda>0$$ with $$\lambda=+p^2$$ I found that $$c_1+c_2=0$$ and $$c_1\cos(\pi p)+c_2p\cos(\pi p)=0$$. Something cool was supposed to happen and I'm not sure if I'm doing it right.  So continuing with this I got that $$p=1$$ and therefore $$\lambda=1$$.

• I think you made a mistake with applying your boundary conditions $\phi(x) = c_1 \cos(p x) + c_2 \sin(p x)$, $\phi'(x) = p[ -c_1 \sin(p x) + c_2 \cos(p x)]$, would lead to $c_1 + p c_2 = 0$ and similar for the other one. Jun 8, 2020 at 18:16

Every non-zero solution of $$\phi''+\lambda\phi =0, \\ \phi(0)+\phi'(0)=0$$ can be normalized so that $$\phi(0)-\phi'(0)=1$$ by scaling the solution. If you could not do that, then the solution $$\phi$$ would satisfy $$\phi(0)=\phi'(0)=0$$, which is satisfied only by the $$0$$ function. So, start by solving $$\phi''+\lambda \phi=0$$ subject to $$\phi(0)+\phi'(0)=0,\;\;\; \phi(0)-\phi'(0)=1, \\ \implies \phi(0)=1/2,\;\; \phi'(0)=-1/2.$$ This has the unique solution $$\phi_{\lambda}(x) = \frac{1}{2}\cos(\sqrt{\lambda}x)-\frac{1}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}x).$$ In order for $$\lambda$$ to be an eigenvalue it will necessary for $$\phi_{\lambda}(\pi)+\phi_{\lambda}'(\pi)=0$$, which gives the eigenvalue equation $$\frac{1}{2}\cos(\sqrt{\lambda}\pi)-\frac{1}{2\sqrt{\lambda}}\sin(\sqrt{\lambda}\pi)-\frac{\sqrt{\lambda}}{2}\sin(\sqrt{\lambda}\pi)-\frac{1}{2}\cos(\sqrt{\lambda}\pi) = 0 \\ \left(1+\lambda\right) \frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}}=0.$$ The eigenvalues are $$\lambda=-1 \mbox{ and } \lambda = 1^2,2^2,3^2,\cdots .$$ The corresponding eigenfunctions are $$\phi_{-1}=\frac{1}{2}\cosh(x)-\frac{1}{2}\sinh(x)=e^{-x},\\ \phi_{n^2}=\frac{1}{2}\cos(nx)-\frac{1}{2n}\sin(nx),\;\; n=1,2,3,\cdots.$$