Let $A:L^{2}[0,\pi]\rightarrow L^{2}[0,\pi]$ be defined by $$ A\phi(x)=\int_{0}^{\pi}K\left(x,\varepsilon\right)\phi(\varepsilon)d\varepsilon, $$ where the kernel is $$ K=\begin{cases} \frac{\sin\left(kx\right)\sin\left[k\left(L-\varepsilon\right)\right]}{k\sin\left(k\pi\right)}, & 0\leq x\leq\varepsilon\leq\pi,\\ \frac{\sin\left(k\varepsilon\right)\sin\left[k\left(L-x\right)\right]}{k\sin\left(k\pi\right)}, & 0\leq\varepsilon\leq x\leq\pi. \end{cases} $$ Note that $k$ is a non-integer wave speed, and $K$ is the Green's function solutions to the harmonic oscillator equation (though I don't think this should have any bearing on the techniques we apply to solve the problem).
Solve $A\phi(x)=\lambda\phi(x),$ for the eigenfunctions $\phi$ and eigenvalues $\lambda$.
Normally, I would take the derivative of both sides one or two times, applying the Leibniz integral rule as needed. However, I am confused because the kernel is defined differently over $[0,x]$ and $[x,\pi]$. Even when the integral is split, each part of the summation would contain a $sin$ function in terms of $x$. Therefore, repeatedly taking the derivative does not appear to 'get rid' of the integral. So, I am not able to reduce the equation to an ODE, which is my goal.
Any suggestions? or reference material I should check out?
Thanks!
You already know that your Green's function $K(x|\xi)$ solves $K''(x|\xi) + k^2 K(x|\xi) = -\delta(x-\xi)$, and hence \begin{align*} \psi'' + k^2\psi = -f \iff \psi = \int_{0}^{\pi}K(x|\xi)f(\xi) \, \mathrm{d}\xi =: Af \end{align*} To push this into the form $A\phi = \lambda \phi$ you want to take $\psi = \lambda \phi$, and $f = \phi$ to obtain the equivalent ode \begin{align*} \phi'' + k^2\phi = -\frac{\phi}{\lambda} \iff \phi'' + \gamma \phi = 0 \end{align*} where $\gamma := k^2 + \frac{1}{\lambda}$.
We didn't have to differentiate under the kernel because we could use properties of the Green's function instead. See Duffy's book 'Green's functions with Applications' for more info.