dirac delta function

logistic_guy

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

d2ydx2+k2y=f(x)\displaystyle \frac{d^2y}{dx^2} + k^2y = f(x)

a<x<b\displaystyle a < x < b
 
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g(x,s)={sink(xa)as<xsink(bx)x<sb\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}
 
g(x,s)={sink(xa)as<xsink(bx)x<sb\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:

g(x,s)={Asink(sa)as<xBsink(bs)x<sb\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}
 
Last edited:
g(x,s)={Asink(sa)as<xBsink(bs)x<sb\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}
A\displaystyle A and B\displaystyle B in this system are in fact not constants. They are functions of x\displaystyle x. Therefore, they should be written as A(x)\displaystyle A(x) and B(x)\displaystyle B(x). I wrote them without the variable x\displaystyle x, just to save some ink and also the system looks more clean like this!

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

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

Let us apply the properties.

Asink(xa)Bsink(bx)=0\displaystyle A\sin k(x - a) - B\sin k(b - x) = 0
kBcosk(bx)kAcosk(xa)=1\displaystyle -kB\cos k(b - x) - kA\cos k(x - a) = 1
 
Asink(xa)Bsink(bx)=0\displaystyle A\sin k(x - a) - B\sin k(b - x) = 0
kBcosk(bx)kAcosk(xa)=1\displaystyle -kB\cos k(b - x) - kA\cos k(x - a) = 1
Let us solve for A\displaystyle A in the first equation.

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