Consider the following ODE $$y''+a\delta (x)y+\lambda y=0$$ subject to the initial conditions $$y(\pm\pi )=0$$ (1) Show that there is a set of eigenvalues $$\tan (\pi \sqrt{\lambda })=\frac{2\sqrt{\lambda }}{a}$$ (2) Investigate if the condition $$\lambda =-u^{2}$$ where u is a positive number, is possible.
I tried Laplace transform but noticed that the initial value is not satisfying. Then I tried to separate cases where x is zero or non-zero, but it didn't lead to the answer.
Keep trying: So I noticed $$\int_{-\epsilon }^{\epsilon }[y''+a\delta (x)y+\lambda y]dx=\int_{-\epsilon }^{\epsilon }y''dx+\int_{-\epsilon }^{\epsilon }a\delta (x)ydx+\int_{-\epsilon }^{\epsilon }\lambda ydx$$ $$=y'(\epsilon )-y'(-\epsilon )+ay(0)+\lambda \int_{-\epsilon }^{\epsilon } ydx$$ Then I took $$\lim_{\epsilon \rightarrow 0^{+}}$$ and obtain the following equation $$y'(0^{+})-y'(0^{-})+ay(0)=0$$ Next I tried to solve the ODE on the interval $$(-\infty ,0)\cup (0,\infty )$$ (in this way the delta function is evaluated to be zero) So the ODE becomes $$y''+\lambda y=0$$ which has the solution form$$y=c_{1}\sin \sqrt{\lambda }x+c_{2}\cos \sqrt{\lambda }x$$ But after I substitute the initial conditions $$y(\pm\pi )=0$$ I got $$c_{1}=c_{2}=0$$ I am wondering what went wrong here...
