Let me first write what is asked:
Given the operator $L = \dfrac{d}{dx^2} + k^2$ over the space of twice-differentiable functions:
- Prove that $L$ is linear.
- Find the Green function for $L$, under boundary conditions $G(0,y) = G\left(\dfrac{\pi}{2k},y\right)=0$.
- Solve the boundary problem \begin{align} \begin{cases} &Lf(x) = p(x)=x\\ &f(0) = f\left( \dfrac{\pi}{2k} \right) = 0 \end{cases} \end{align} and prove that the solution can be written as \begin{align} f(x) = \int_{0}^{\dfrac{\pi}{2k}}G(x,y)p(y)dy. \end{align}
My attempted solution is this (the first subquestion is trivial), starting from (2) \begin{align} LG(x,y) = \delta(x-y) = \begin{cases} 0, \ x \neq y\\ \infty, \ x = y \end{cases}. \end{align}
- For $x \in [0,y)$ we have \begin{align} &G(x,y) = A(y) \cos (kx) + B(y) \sin (kx)\\ &G_{-}(0,y) = 0 \implies G_{-}(0,y) = A(y) \implies G(x,y) = B(y) \sin (kx), \ x \in [0,y). \end{align}
- For $x \in (y, \dfrac{\pi}{2k} ]$ we have \begin{align} &G_{+}(x,y) = C(y) \cos(kx) + D(y) \sin(kx)\\ &G_{+}\left( \dfrac{\pi}{2k}, y \right) =0 \implies G_{+} \left( \dfrac{\pi}{2k}, y \right) = D(y) \end{align} so \begin{align} G_{+}(x,y) = C(y) \cos(kx), \ x \in (y, \dfrac{\pi}{2k}]. \end{align}
- For $x=y$ we have continuity, so: \begin{align} B(y) \sin(ky) = C(y) \cos(ky) \end{align} and by integrating over $\left[ y-\epsilon, y+\epsilon \right]$: \begin{align} kB(y)\cos(ky) + kC(y) \sin(ky) = -1 \end{align} gives us \begin{align} \begin{cases} B(y) &= -\dfrac{\cos(ky)}{k}\\ C(y) &= -\dfrac{\sin(ky)}{k} \end{cases} \end{align} and therefore \begin{align} G(x,y) = \begin{cases} -\dfrac{1}{k} \cos(ky) \sin(kx), \ x \in [0,y)\\ -\dfrac{1}{k} \sin(ky) \cos(kx), \ x \in (y, \dfrac{\pi}{2k}]. \end{cases} \end{align}
For (3), my work was \begin{align} L f(x) = x &\implies \left( \dfrac{d^2}{dx^2} + k^2 \right)f(x) = x\\ &\implies \dfrac{d^2}{dx^2}(f(x)) + k^2 f(x) = x\\ &\implies f''(x) + k^2 f(x) = x. \end{align} To solve this is pretty much trivial. The homogeneous equation is $f_{hom} = c_1 \cos(kx) + c_2 \sin (kx) $ and by using any known method for arriving at the full solution (I did a simple polynomial substitution) we get that $f_{par} = \dfrac{1}{k^2}x$.
At any rate we get \begin{align} f(x) = c_1 \cos(kx) + c_2 \sin(kx) + \dfrac{1}{k^2}x \end{align} and by applying the boundary conditions we get \begin{align} &f(0) = 0 \implies c_1 = 0\\ &f\left( \dfrac{\pi}{2k} \right) = 0 \implies c_2 = - \dfrac{\pi}{2k^3}. \end{align} I am stuck at how am I supposed to prove that the Green function found above can be connected to the solution of my problem for \begin{align} f(x) = \int_{0}^{\dfrac{\pi}{2k}}G(x,y)p(y)dy. \end{align}