Show that $\lambda=\mu^2$ is an eigenvalue of the BVP $$-u''(x)=\lambda u(x), 0<x<1$$ $$u(0)=0, u(1)=u'(1),$$

provided that $\mu$ is a root of the equation $\tan\mu=\mu$. By considering the graphs of $\tan\mu$ and $\mu$, show that there are an infinite number of eigenvalues $\lambda_n$ and that $\lambda_n\to(n+\frac{1}{2})^2\pi^2$ as $n\to\infty$.


I consider 3 cases: when $\lambda<0, \lambda=0, \lambda>0$.

CASE I: Suppose $\lambda<0$. We write $\lambda=-\mu^2 (\mu\neq 0)$ then the ODE has general solution

$$u(x)=Ae^{\mu x}+Be^{-\mu x}$$

The first BC implies that $A+B=0$ so $u(x)=A(e^{\mu x}-e^{-\mu x})$ and the second BC gives $A\mu(e^{\mu}+e^{-\mu})=A(e^{\mu}-e^{-\mu})\to\mu=\tanh(\mu)$.

Then $u(x)=A(e^{\tanh(\mu)x}-e^{-\tanh(\mu)x})$ for some constant $A$. I am not sure how to continue from here.

CASE II: Suppose $\lambda=0$. Then $u''=0$ which has general solution $u(x)=A+Bx$. The BCs imply that $A=0$ only without saying anything about $B$...

CASE III: Suppose $\lambda>0$. We write $\lambda=\omega^2$ and we have the general solution $$u(x)=A\sin(\omega x)+B\cos(\omega x)$$ The first BC implies that $B=0$ while the second BC implies $\tan\omega=1$, but I think I'm suppose to arrive at $\tan\omega=0$ to satisfy the question being asked.

Some hints to continue the argument for the three cases (and notification to any mistakes) will be greatly appreciated.


Case I: $\mu=0$ is the only solution of $\mu=\tanh\mu$.

Case II: $\mu=0$ is an eigenvalue and $u(x)=x$ an eigenfunction.

Case III: the boundary condition at $x=1$, when imposed on $u(x)=\sin(\omega\,x)$ gives $$ \sin\omega=\omega\cos\omega\implies\omega=\tan\omega. $$ There is a solution on each interval $(2\,n\,\pi,2\,n\,\pi+1)$


Case 1: You correctly identified that a solution can only exist if $\mu = \tanh \mu$. As it happens, there exists no $\mu \neq 0$ such that $\mu = \tanh \mu$! (You can see this if you sketch the graphs.) So case 1 gives you no solutions.

Case 2: Your solution $u = Bx$ is fine. Don't worry about not being able to determine $B$ - just by looking at the equations and boundary conditions, you can see that if $u = x$ is a solution then $u = Bx$ must also be a solution for any constant $B$.

Case 3: Right idea, but you forgot a factor of $\omega$ when differentiating $u$ to get $u(1)$. The correct condition on $\omega$ is $ \tan \omega = \omega$, not $\tan \omega = 1$. Now draw graphs of $\tan \omega$ and $\omega$ to see where the solutions to $ \tan \omega = \omega$ are. Note that solutions to $ \tan \omega = \omega$ with $\omega$ very large lie close to the vertical asymptotes of the graph of $\tan \omega$, and you can see this from your graph. This will enable you to complete the question.


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