Solving $\sin(\sqrt{\lambda}L) + \beta \cos(\sqrt{\lambda}L)\sqrt{\lambda} = 0$

Question

I'm working with the ODE $$-\frac{d^2u}{dx^2}=\lambda u$$ and trying to find eigenvalues and eigenfunctions corresponding the boundary conditions $$u(0)=0, u(L)+\beta \frac{du}{dx}(L)=0$$ Assuming that $\lambda >0$ and applying $u(0)=0$, we get that $c_2=0$. Applying the second condition, we get $$\sin(\sqrt{\lambda}L) + \beta \cos(\sqrt{\lambda}L)\sqrt{\lambda} = 0 \text{ (Eq. 1)}$$

My guess is that there is no solution for $\lambda$ to satisfy Eq. 1. But how do I show it?

Answer 1

In your first equation you can divide with $\cos (\sqrt{\lambda} L)$ and get

$$ \tan (\sqrt{\lambda} L)= - \beta \sqrt{\lambda}, $$

Define $f(x) = \tan (\sqrt{x} L)+ \beta \sqrt{x}$ and notice,

$$\lim _{x\rightarrow^{+} \frac{(\pi +4n\pi)^2}{4L^2} }f(x)=-\infty$$ and that

$$\lim _{x\rightarrow^{-} \frac{(\pi +4(n+1)\pi)^2}{4L^2} }f(x)=\infty$$

Therefore in each interval $(\frac{(\pi +4n\pi)^2}{4L^2}, \frac{(\pi +4(n+1)\pi)^2}{4L^2})$ by the intermediate value theorem you get an eigenvalue.