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
$\displaystyle \frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');\qquad y(0)=y(L)=0$
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
$\displaystyle \frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');\qquad y(0)=y(L)=0$
The "Green's function" for a given differential operator, D(f), is a function, G(x,x'), satisfying D(G)= 0 as long as $\displaystyle x\ne x'$ and such that $\displaystyle \lim_{x'\to x^+} G(x,x')- \lim_{x'\to x^-} G(x, x')= 1$.
Functions satisfying $\displaystyle y''+ k^2y= 0$ are all of the form y(x)= C cos(kx)+ D sin(kx). So we must have $\displaystyle G(x, x')= C_1cos(kx)+ D_1(kx)$ for some constants (with respect to x, they may depend on x') for $\displaystyle x\le x'$ and $\displaystyle G(x, x')= C_2 cos(kx)+ D_2 sin(kx)$ for $\displaystyle x\ge x'$. So the problem is to determine $\displaystyle C_1$, $\displaystyle C_2$, $\displaystyle D_1$, and $\displaystyle D_2$.
If x= 0 then $\displaystyle x\le x'$ because x' must also be between 0 and L so we must have $\displaystyle y(0)= C_1 cos(0)+ D_1 sin(0)= C_1= 0$.
Similarly of x= L then $\displaystyle x\ge x'$ so we must have $\displaystyle y(L)= C_2 cos(L)+ D_2 sin(L)= 0$. That gives $\displaystyle C_2= tan(L) D_2$
So far, we have $\displaystyle G(x, x')= D_1sin(x)$ if $\displaystyle x\le x'$ and $\displaystyle G(x, x')= D_2(tan(L)cos(x)+ sin(x))$.
At x= x' those must both be true: $\displaystyle G(x', x')= D_1 sin(x')= D_2(tan(L)cos(x')+ sin(x'))$.
The derivative, from the left, at x= x', is $\displaystyle D_1 cos(x')$ and, from the right, $\displaystyle D_2(cos(x')- tan(L)sin(x'))$ so we must have $\displaystyle D_1 cos(x')- D_2(cos(x')- tan(L)sin(x'))= 1$.
That gives two equation to solve for $\displaystyle D_1$ and $\displaystyle D_2$ in terms of x'.