An "unusual" PDE eigenvalue problem

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Question: (Strauss Partial Differential Equations: An Introduction, Ch. 4.3, Ex. 12) "Consider the unusual eigenvalue problem"

$$-v_{xx}=\lambda v \\ v_x(0)=v_x(\ell)=\frac{v(\ell)-v(0)}{\ell}$$

for $0<x<\ell$.

  1. Get an equation for the positive eigenvalues $\lambda>0$.
  2. Reduce it to the form $\gamma\sin(\gamma)\cos(\gamma)=\sin^2(\gamma)$ for $\gamma=\frac12\ell\sqrt\lambda$.
  3. Find the eigenvalues by finding $\gamma$.
  4. Write the eigenfunctions for the positive eigenvalues.

Problem: For $\lambda=\beta^2>0$, we have $v_{xx}+\beta^2v=0$ or

$$v(x)=A\cos(\beta x)+B\sin(\beta x) \\ v'(x)=\beta B\cos(\beta x)-\beta A\sin(\beta x)$$

Then

$$\beta B=\beta B\cos(\beta\ell)-\beta A\sin(\beta\ell)=\frac{A\cos(\beta\ell)+B\sin(\beta\ell)-A}{\ell}$$

Which is totally an equation whose solutions $\beta$ give $\lambda=\beta^2$. This solves problem (1).

Now suppose $\beta=\frac2\ell\gamma$. I could reduce the above equation to

$$\gamma B+(A-2\gamma B)\sin^2(\gamma)=(B-2\gamma A)\sin(\gamma)\cos(\gamma)$$

but it doesn't seem possible to go further. Can't do (2). Well, suppose by some black magic I manage to reduce it. Now I'm tasked with solving it. Mathematica suggests for (3) that $\gamma=n\pi$ for $n\in\mathbb Z$, but how can I explicitly derive it?

Finally, for (4), can I even solve for the coefficients $A,B$ with the boundary conditions given?

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Applying the boundary conditions to $v = A\sin(\beta x) + B\cos(\beta x)$ we get the two equations

$$\matrix{v_x(0) = v_x(\ell) &\implies& A[1-\cos(\beta \ell)] + B\sin(\beta \ell) = 0\\v(\ell)-v(0) = \ell v_x(0) &\implies& A[\sin(\beta \ell)-\ell\beta] = B[1-\cos(\beta \ell)]}$$

Combinding them, to get ridd of $A$ and $B$, gives us

$$[1-\cos(\beta \ell)]^2 + [\sin(\beta \ell)-\ell\beta]\sin(\beta \ell) = 0$$

Using $\tan(x/2) = \frac{1-\cos(x)}{\sin(x)}$ we can simply this down to

$$\sin^2(\beta\ell/2) = \frac{\beta\ell}{2}\sin(\beta\ell/2)\cos(\beta\ell/2)$$

Thus $\gamma = \frac{\beta\ell}{2}$ are the zeros of the function

$$f(x) = \sin(x)^2-x\sin(x)\cos(x)$$

which is the same equation as mentioned in the problem.