dirac delta function

[logistic_guy]

Solve the differential equation by Dirac delta function to obtain Green's function.

\(\displaystyle \frac{d^2y}{dx^2} + k^2y = f(x)\)

\(\displaystyle a < x < b\)

 

[logistic_guy]

[imath]\large g(x,s) =\begin{cases} \sin k(x - a) & a \leq s < x\\[2ex] \sin k(b - x) & x < s \leq b\end{cases}[/imath]

 

[logistic_guy]

[imath]\large g(x,s) =\begin{cases} \sin k(x - a) & a \leq s < x\\[2ex] \sin k(b - x) & x < s \leq b\end{cases}[/imath]

This should be written as:

[imath]\large g(x,s) =\begin{cases} A\sin k(s - a) & a \leq s < x\\[2ex] B\sin k(b - s) & x < s \leq b\end{cases}[/imath]

 

[logistic_guy]

[imath]\large g(x,s) =\begin{cases} A\sin k(s - a) & a \leq s < x\\[2ex] B\sin k(b - s) & x < s \leq b\end{cases}[/imath]

\(\displaystyle A\) and \(\displaystyle B\) in this system are in fact not constants. They are functions of \(\displaystyle x\). Therefore, they should be written as \(\displaystyle A(x)\) and \(\displaystyle B(x)\). I wrote them without the variable \(\displaystyle x\), just to save some ink and also the system looks more clean like this!

The goal is to find these two functions, \(\displaystyle A\) and \(\displaystyle B\). We will use the two properties of continuity, those are:

\(\displaystyle g(x^-) - g(x^+) = 0\)
\(\displaystyle g'(x^+) - g'(x^-) = 1\)

Let us apply the properties.

\(\displaystyle A\sin k(x - a) - B\sin k(b - x) = 0\)
\(\displaystyle -kB\cos k(b - x) - kA\cos k(x - a) = 1\)

 

[logistic_guy]

\(\displaystyle A\sin k(x - a) - B\sin k(b - x) = 0\)
\(\displaystyle -kB\cos k(b - x) - kA\cos k(x - a) = 1\)

Let us solve for \(\displaystyle A\) in the first equation.

\(\displaystyle A = \frac{B\sin k(b - x)}{\sin k(x - a)}\)

 

[logistic_guy]

\(\displaystyle A = \frac{B\sin k(b - x)}{\sin k(x - a)}\)

Substitute this result in the second equation.

\(\displaystyle -kB\cos k(b - x) - k\left(\frac{B\sin k(b - x)}{\sin k(x - a)}\right)\cos k(x - a) = 1\)

Solve for \(\displaystyle B\).

\(\displaystyle -kB\cos k(b - x)\sin k(x - a) - kB\sin k(b - x)\cos k(x - a) = \sin k(x - a)\)


\(\displaystyle B\bigg[k\cos k(b - x)\sin k(x - a) + k\sin k(b - x)\cos k(x - a)\bigg] = -\sin k(x - a)\)


\(\displaystyle B = -\frac{\sin k(x - a)}{k\cos k(b - x)\sin k(x - a) + k\sin k(b - x)\cos k(x - a)}\)

We will use this identity.

\(\displaystyle \sin A \cos B + \sin B \cos A = \sin(A + B)\)

Then,

\(\displaystyle B = -\frac{\sin k(x - a)}{k\sin k(b - a)}\)

And

\(\displaystyle A = \bigg[-\frac{\sin k(x - a)}{k\sin k(b - a)}\bigg]\frac{\sin k(b - x)}{\sin k(x - a)}\)

Or

\(\displaystyle A = -\frac{\sin k(b - x)}{k\sin k(b - a)}\)

 

[logistic_guy]

[imath]\large g(x,s) =\begin{cases} A\sin k(s - a) & a \leq s < x\\[2ex] B\sin k(b - s) & x < s \leq b\end{cases}[/imath]

Substitute \(\displaystyle A\) and \(\displaystyle B\) in the function above.

[imath]\large g(x,s) =\begin{cases} \left(-\frac{\sin k(b - x)}{k\sin k(b - a)}\right)\sin k(s - a) & a \leq s < x\\[2ex] \left(-\frac{\sin k(x - a)}{k\sin k(b - a)}\right)\sin k(b - s) & x < s \leq b\end{cases}[/imath]

 

[logistic_guy]

[imath]\large g(x,s) =\begin{cases} \left(-\frac{\sin k(b - x)}{k\sin k(b - a)}\right)\sin k(s - a) & a \leq s < x\\[2ex] \left(-\frac{\sin k(x - a)}{k\sin k(b - a)}\right)\sin k(b - s) & x < s \leq b\end{cases}[/imath]

Simplify.

[imath]\large g(x,s) =\begin{cases} -\frac{\sin k(s - a)\sin k(b - x)}{k\sin k(b - a)} & a \leq s < x\\[2ex] -\frac{\sin k(x - a)\sin k(b - s)}{k\sin k(b - a)} & x < s \leq b\end{cases}[/imath]

And this was the \(\displaystyle \color{pink} \bold{fourth}\) Green's function. We learnt to find Green's function in \(\displaystyle 4\) different ways. We also learnt how to use it to solve a differential equation. If you completely followed the four ways I made, you would be able to build a solid foundation in the very deep topic of \(\displaystyle \color{green} \bold{Green \ functions!}\)