dirac delta function - 2

[logistic_guy]

Solve the differential equation the hard way 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]

$\large g(x,s) =\begin{cases} A\cos kx + B\sin kx & a \leq s < x\\[2ex] C\cos kx + D\sin kx & x < s \leq b\end{cases}$

 

[logistic_guy]

$\large g(x,s) =\begin{cases} A\cos kx + B\sin kx & a \leq s < x\\[2ex] C\cos kx + D\sin kx & x < s \leq b\end{cases}$

This should be written as:

$\large g(x,s) =\begin{cases} A\cos ks + B\sin ks & a \leq s < x\\[2ex] C\cos ks + D\sin ks & x < s \leq b\end{cases}$

 

[logistic_guy]

$\large g(x,s) =\begin{cases} A\cos ks + B\sin ks & a \leq s < x\\[2ex] C\cos ks + D\sin ks & x < s \leq b\end{cases}$

As I said before, $\displaystyle A, B, C, \text{and} \ D$ are not constants, but rather functions of $\displaystyle x$. And the goal is find them to get Green's function.

Four unknowns need $\displaystyle 4$ equations. We can get two of them from boundary conditions and two from continuity conditions. That is FOUR.

$\displaystyle A\cos ka + B\sin ka = 0$
$\displaystyle C\cos kb + D\sin kb = 0$
$\displaystyle A\cos kx + B\sin kx - C\cos kx - D\sin kx = 0$
$\displaystyle -kC\sin kx + kD\cos kx + kA\sin kx - kB\cos kx = 1$

 

[logistic_guy]

We will use a calculator to solve this system. This calculator does not accept functions so we will replace them with letters.

$\displaystyle F = \cos ka$
$\displaystyle G = \sin ka$
$\displaystyle H = \cos kb$
$\displaystyle L = \sin kb$
$\displaystyle Y = \cos kx$
$\displaystyle Z = \sin kx$

 

[logistic_guy]

If I solve this systems of equations, I get:

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

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

$\displaystyle C = -\frac{\sin kb\sin k(x-a)}{k\sin k(b - a)}$

$\displaystyle D = \frac{\cos kb\sin k(x-a)}{k\sin k(b - a)}$

 

[logistic_guy]

Our Green function so far:


$\Large g(x,s) =\begin{cases} \left(\frac{\sin ka\sin k(b-x)}{k\sin k(b - a)}\right)\cos ks + \left(-\frac{\cos ka\sin k(b-x)}{k\sin k(b - a)}\right)\sin ks & a \leq s < x\\[3ex] \left(-\frac{\sin kb\sin k(x-a)}{k\sin k(b - a)}\right)\cos ks + \left(\frac{\cos kb\sin k(x-a)}{k\sin k(b - a)}\right)\sin ks & x < s \leq b \end{cases}$

 

[logistic_guy]

With a little simplification, we get:


$\Large g(x,s) =\begin{cases} \left(\frac{\sin k(b-x)}{k\sin k(b - a)}\right)(\sin ka\cos ks - \cos ka \sin ks) & a \leq s < x\\[3ex] \left(\frac{\sin k(x-a)}{k\sin k(b - a)}\right)(-\sin kb\cos ks + \cos kb\sin ks) & x < s \leq b \end{cases}$

 

[logistic_guy]

$\Large g(x,s) =\begin{cases} \left(\frac{\sin k(b-x)}{k\sin k(b - a)}\right)(\sin ka\cos ks - \cos ka \sin ks) & a \leq s < x\\[3ex] \left(\frac{\sin k(x-a)}{k\sin k(b - a)}\right)(-\sin kb\cos ks + \cos kb\sin ks) & x < s \leq b \end{cases}$

We simplify the expression above to get:

$\Large g(x,s) =\begin{cases} \left(\frac{\sin k(b-x)}{k\sin k(b - a)}\right)\sin k(a - s) & a \leq s < x\\[3ex] \left(\frac{\sin k(x-a)}{k\sin k(b - a)}\right)\sin k(s - b) & x < s \leq b \end{cases}$


Or


$\Large g(x,s) =\begin{cases} -\frac{\sin k(s - a)\sin k(b-x)}{k\sin k(b - a)} & a \leq s < x\\[3ex] -\frac{\sin k(x-a)\sin k(b - s)}{k\sin k(b - a)} & x < s \leq b \end{cases}$

Which matches the result of this post:

green function

Find Green's function for the following differential equation: \frac{d^2y}{dx^2} + k^2y = f(x) a < x < b Then use it to find the solution of: \frac{d^2y}{dx^2} + 4y = x y(1) = 5 y(5) = 1 1 < x < 5 Solve the differential equation again with your favorite normal method. Does your answer...

www.freemathhelp.com

Which means we have found the Green's function successfully the hard way!