Greens function of 1-d forced wave equation

Question

[ORIGINAL PROBLEM] You are given hat the Green's function $g(x,t,\xi, \phi)$ is

$\frac{\partial^2g}{\partial t^2} - \frac{\partial^2g}{\partial x^2}=\delta(t-\tau)\delta(x-\xi)$

with

$g(x,0,\xi, \tau)= \frac{\partial }{\partial t} g(x,0,\xi, \phi)=0$

Show that the fourier transform in x of the Green's function is given by

$G(x,t,\xi, \phi)=\frac{e^{ik\xi}sink(t-\tau)H(t-\tau)}{k} $ where H(x) is the Heaviside function.

I get that

$\frac{\partial^2 \tilde{g}}{\partial t^2}-k^2 \tilde{g} = \delta(t-\tau)e^{-ik\xi} $

so

$\tilde{g}=Ae^{kt}+Be^{-kt}+C.F$

but I cannot see what the complementary function should be or where to proceed from here.[ORIGINAL PROBLEM]

I would like a full worked solution for all of this question and in return I am offering a bounty.

Answer 1

Part a) After performing a Fourier Transform we obtain:

$$ \partial^2 _t G+k^2 G= \delta( t - \tau) \exp(i k \xi)$$

using the opposite signed Fourier Transform(please note:you made a sign mistake on the $k^2G$ term), then let $ G(x,t) = \exp(ik \xi) f(x,t)$ so we only have:

$$ \partial^2 _t f+k^2 f= \delta( t - \tau) $$

Now break it into two cases, the homogenous and the non-homogenous. The homogenous case is solving

$$\partial^2_T f = - k^2 f $$ which has solutions cosine and sine, but the initial conditions gives only sine. So we get:

$$f = A \sin( k T) = A \sin(k(t - \tau))$$

Notice that $T=0 \implies t-\tau =T=0$ on the homogenous case. Now vary the constant (i.e suppose $A$ is a function of $t$ as well). It's easy to see, using the fact that

$$ \int \delta(t -\tau) = H(t -\tau ) $$

we get

$$f = \frac{\sin(k(t-\tau)) H(t-\tau)}{k} $$

Hence:

$$ G(k,t) = \frac{ e^{ik \xi}\sin(k(t-\tau)) H(t-\tau)}{k} $$

Details of the variation:

$$ f' = A' \sin(kT) + kA \cos(kT) $$

$$ f'' = A'' \sin(kT) + 2k A' \cos(kT) + k^2 \sin(kT) $$

I won't do the full analysis version of this, but you can formalize this. So we see for $\epsilon >0$, we let $T \in ( -\epsilon, \epsilon)$ Thus

$$ f'' \approx k A'\quad \& \quad f \approx 0 \implies k A' = \delta(T)$$ $$ \implies A(t) = \int \frac{ \delta(T)}{k}$$

Part b) We take the inverse transform:

$$ g(x,T) = \mathcal{F}^{-1} (G) = \frac{1}{2 \pi} \int_\mathbb{R} G(k,T) \exp(-i xk) dk = \int_{-\infty}^\infty \frac{H(T) \sin(k T) \exp( i k( \xi-x) ) }{2 \pi k} dk$$

Use the identity:

$$ 2\sin(a)(\cos(b) + \sin(b)) = \sin(a-b) + \sin(a+b) +\cos(a-b) - \cos(a+b)$$

Then using the fact you're given allows you to write where $\sigma = \xi -x$:

$$g(\sigma,T) = \frac{1}{4} H(T) \left ( \text{sgn}(T - \sigma) + \text{ sgn} (T + \sigma) \right)$$

Rewrite this as

$$ g(\sigma,T) = \frac{1}{2} H( T - |\sigma| ) $$

Part c) Since we based everything off $h(x,t) = \delta(T) \delta ( \sigma)$, and we now have $h(x,t) = \delta(x)$... I'll leave this to you.

Answer 2

To avoid contradiction, I use the notation mentioned in the problem. As your work, $G$ solves the following IVP of $t$: $$ G_{tt}(k,t)+k^2G(k,t)=\delta(t-\tau)e^{-ik\xi}\\ G(k,0)=0,G_t(k,0)=0 $$ A quick way to solve this equation is to use Laplace's transform in $t$. Let $\mathcal{L}:t\rightarrow s$ be the Laplace transform operator and $\mathcal{L}[G]=\widehat{G}$. Then, by the differential property, the equation about $G$ can be rewritten by: $$ s^2\widehat{G}(k,s)+k^2\widehat{G}(k,s)=e^{-\tau s-ik\xi}\\ \widehat{G}(k,s)=\dfrac{e^{-\tau s-ik\xi}}{s^2+k^2} $$ Note that is form implicit the initial data that $G(k,0)=G_t(k,0)=0$. Then, apply the inverse Laplace's transform and convolution formula, we have: $$ G(k,t)=\dfrac{\int_0^t\delta(s-\tau)e^{-ik\xi}\sin(k(t-s))ds}{k}=\dfrac{e^{-ik\xi}H(t-\tau)\sin(k(t-\tau))}{k} $$ So I think there is a typo in the book that the exponent of $e$ should be $-ik\xi$ not $ik\xi$. Then (a) solve.

(b) can be check by a simple computation. I think it is trivial to show all the procedure so I just give you a hint: $$ F^{-1}[G]=H(t-\tau)F^{-1}[\dfrac{e^{-ik\xi}\sin(k(t-\tau))}{k}] $$ Note the Euler's formula gives $e^{-ik\xi}=\cos(k\xi)-i\sin(k\xi)$, then plug it in the inverse Fourier's transform, the proposition is follows.

Following the same way about (a) and (b), we can check (c). I also omit the detail procedure and just show you the hint:

The Fourier's transform of $u_{tt}-u_{xx}=\delta(x)$ is $\widehat{u}_{tt}+k^2\widehat{u}=1$ where $\widehat{u}$ is the Fourier's transform of $u$. The solution is $$ \widehat{u}(k,t)=\dfrac{1-\cos(kt)}{k^2} $$ Applying the inverse Fourier's transform, we have: $$ u(x,t)=F^{-1}[\dfrac{1-\cos(kt)}{k^2}]=F^{-1}[-\dfrac{e^{ikt}}{k^2}+\dfrac{1}{k^2}-\dfrac{e^{-ikt}}{k^2}] $$ Using integral and exponential shifting property, we obtain: $$ u(x,t)=(x+t)H(x+t)+(x-t)H(x-t)-2xH(x) $$ $u(x,t)$ at $t=1$ and $t=2$ are trivial to show.