Solving wave equation using Green's function Watch

roshanhero
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I am currently self studying/ self teaching myself this great book Mathematical methods of physics by Mathews and Walker and must tell you that I am really struggling with the explanation of the solution of wave equation given on the book using Green's function from the start to the last, so suggest me the technique used in here to solve this problem
\nabla^{2}\phi-\frac{1}{c^{2}}\frac{\partial^2 \phi}{\partial t^2}=\delta(x-x')\delta(t-t')
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steve44
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Quick review: First you meant:

(\nabla^2 \phi - {1 \over c^2} {\partial^2 \phi \over \partial t^2 }) = \delta (x-x') \delta (t-t')

Easiest case:

({\partial^2 \over \partial x^2} - {1 \over c^2} {\partial^2 \over \partial t^2}) u(x,t) = 0

separating by parts, we can write the solution in the form

u(x,t) = f(x) e^{i\omega_0 t}

where

({d^2 \over d x^2} + k_0^2) f(x) =0

and k_0 = \omega_0 / c.

If we would like to include a source of wave disturbance, we must describe an inhomogeneous equation

({d^2 \over d x^2} + k_0^2) f(x) = \rho(x)

In solving

\mathcal{L} (x) f(x) = \rho(x)

where \mathcal{L} (x) is a linear differential operator. We show f(x) has the simple integral representation

f(x) = \int_{-\infty}^\infty G(x-x_0) \rho(x_0) d x_0.

where \mathcal{L} (x) G(x-x_0) = \delta (x-x_0).

This is easily done by direct substitution

\mathcal{L} (x) f(x) =
= \mathcal{L} (x) \int_{-\infty}^\infty G(x-x_0) \rho(x_0) d x_0
= \int_{-\infty}^\infty \mathcal{L} (x) G(x-x_0) \rho(x_0) d x_0
= \int_{-\infty}^\infty \delta (x-x_0) \rho(x_0) d x_0 = \rho (x)

Knowing the Green's function G(x-x_0) we can in principle find f(x).

Let's solve the inhomogeneous equation

({d^2 \over d x^2} + k_0^2) G(x - x_0) = \delta (x - x_0).

We solve this with point source \delta (x), the general solution is then obtained by substituting x \mapsto x- x_0 in G(x),

({d^2 \over d x^2} + k_0^2) G(x) = \delta (x)

This equation is solved via Fourier transform, \mathcal{F}, which transforms the differential equation into an algebraic equation. Works the following way,

\mathcal{F} ([{d^2 \over d x^2} + k_0^2] G (x)) = \mathcal{F} (\delta (x)) = {1 \over \sqrt{2 \pi}}

or

\int_{-\infty}^\infty e^{ik x} ([{d^2 \over d x^2} + k_0^2] G (x)) dx = (-k^2 + k_0^2) \tilde{G} (k) = {1 \over \sqrt{2 \pi}}

So

\tilde{G} (k) = - {1 \over \sqrt{2 \pi}} {1 \over k^2 - k_0^2} .

The contour integral to find G(x) is then subtle. The integrand has poles at k = \pm k_0 on the real axis. Probably have questions already... Do you know about Jordan's Lemma? The Cauchy principal value? By the way doing this off the top my head. I'll will think about the general situation later.
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