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 $\frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');y(0)=y(L)=0$

Answer 1

The idea is this: For $x < x'$ you can see that you have a solution of $y''+k^{2}y=0$ with $y(0)=0$. For $x > x'$ you have a solution of $y''+k^{2}y=0$ with $y(L)=0$. To get the delta function at $x=x'$, you have to stitch the two solutions (the one to the left and the one to the right) in such a way that the they agree at $x=x'$ and the derivative jumps by a value of $1$. That way the second derivative, i.e. the derivative of the the first derivative, gives a delta function.

So, solve for two functions $\phi$ and $\psi$ that satisfy $y''+k^{2}y=0$, and satisfy $$ \phi(0)=0,\;\; \phi'(0)=1,\;\;\;\;\; \psi(L)=0,\;\;\psi'(L)=1 $$ I'll let you find such solutions; they're multiples of $\sin(kx)$ and $\sin(k(L-x))$, respectively. Then find $A$, and $B$ (which will depend on $x'$) such that $$ A\phi(x')-B\psi(x')=0,\\ A\phi'(x')-B\psi'(x')=-1. $$ This is a $2\times 2$ linear system has a solution if the determinant satisfies $$ \phi(x')\psi'(x')-\phi'(x')\psi(x') \ne 0. $$ This determinant (which is the Wronskian) is constant in $x'$ because $$ \frac{d}{dx}(\phi(x)\psi'(x)-\phi'(x)\psi(x))=\phi\psi''-\phi''\psi= -k^{2}\phi\psi+k^{2}\phi\psi = 0. $$ The only time you can't solve this equation is when the $\psi$ and $\phi$ are linearly-dependent, which is exactly when there is a non-trivial solution of $f''+k^{2}f=0$ with $f(0)=f(L)=0$. In other words, you can solve the equation for a unique $A$ and $B$ if $k$ is not an eigenvalue of the equation.