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

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

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

2. ## Re: Solve the given differential equation by using Green's function method

The "Green's function" for a given differential operator, D(f), is a function, G(x,x'), satisfying D(G)= 0 as long as $x\ne x'$ and such that $\lim_{x'\to x^+} G(x,x')- \lim_{x'\to x^-} G(x, x')= 1$.

Functions satisfying $y''+ k^2y= 0$ are all of the form y(x)= C cos(kx)+ D sin(kx). So we must have $G(x, x')= C_1cos(kx)+ D_1(kx)$ for some constants (with respect to x, they may depend on x') for $x\le x'$ and $G(x, x')= C_2 cos(kx)+ D_2 sin(kx)$ for $x\ge x'$. So the problem is to determine $C_1$, $C_2$, $D_1$, and $D_2$.

If x= 0 then $x\le x'$ because x' must also be between 0 and L so we must have $y(0)= C_1 cos(0)+ D_1 sin(0)= C_1= 0$.

Similarly of x= L then $x\ge x'$ so we must have $y(L)= C_2 cos(L)+ D_2 sin(L)= 0$. That gives $C_2= tan(L) D_2$

So far, we have $G(x, x')= D_1sin(x)$ if $x\le x'$ and $G(x, x')= D_2(tan(L)cos(x)+ sin(x))$.

At x= x' those must both be true: $G(x', x')= D_1 sin(x')= D_2(tan(L)cos(x')+ sin(x'))$.

The derivative, from the left, at x= x', is $D_1 cos(x')$ and, from the right, $D_2(cos(x')- tan(L)sin(x'))$ so we must have $D_1 cos(x')- D_2(cos(x')- tan(L)sin(x'))= 1$.

That gives two equation to solve for $D_1$ and $D_2$ in terms of x'.

3. ## Re: Solve the given differential equation by using Green's function method

I think $C_{2}=-D_{2}tan(L)$
What is the role of delta function in the right hand side then ?