Solve the given differential equation by using Green's function method

Question

I am really struggling with the concept and handling of the Green's function. I have to solve the given differential equation using Green's function method $$\frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');\qquad y(0)=y(L)=0$$

Answer 1

For these sorts of problems, knowing the properties of the solution you are looking for is just as important as computing the solution. Generally, we assume that $y$ is continuous, $y$ is differentiable except at $x'$, and that \begin{align} \lim_{\epsilon\to0} \int_{x'-\epsilon}^{x'+\epsilon} y'' +k^{2}y \, \mathrm{d}y = \lim_{\epsilon\to0} \int_{x'-\epsilon}^{x'+\epsilon} \delta(x-x')\, \mathrm{d}y. \end{align} This latter condition implies that \begin{align} \lim_{\epsilon\to0} y'(x'+\epsilon) - y'(x'-\epsilon) = 1 \end{align} We'll write this condition as "$y'(x^{'}_{+}) -y'(x^{'}_{-}) = 1$."

For $x<x'$, our DE reads $y''+k^{2}y = 0$ with $y(0) = 0$, giving $y(x) = A\sin(kx)$.

For $x>x'$, our DE reads $y''+k^{2}y = 0$ with $y(L) = 0$, giving $y(x) = B\sin(k(x-L))$.

Since $y$ is continuous, $A\sin(kx') = B\sin(k(x'-L))$, so that \begin{align} y(x) = \begin{cases} A\sin(kx), & x<x', \\ A\frac{\sin(kx')}{\sin(k(x'-L))}\sin(k(x-L)), & x\ge x' \end{cases} \end{align} The condition $y'(x^{'}_{+}) -y'(x^{'}_{-}) = 1$ specializes to \begin{align} \frac{Ak\sin(kx')\cos(k(x'-L))}{\sin(k(x'-L))} - Ak\cos(kx') = 1, \end{align} so that \begin{align} A = \frac{1}{k\left(\sin(kx')\cot(k(x'-L)) - \cos(kx') \right)} \end{align}