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]

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

 

[logistic_guy]

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

This should be written as:

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

 

[logistic_guy]

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

$\displaystyle A$ 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]

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

Substitute $\displaystyle A$ and $\displaystyle B$ in the function above.

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

 

[logistic_guy]

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

Simplify.

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

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 $\displaystyle \color{green} \bold{Green \ functions!}$