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Re. I. M. Gelfand "Generalized Functions", Vol. 1, Academic Press 1964

Using an unconventional approach which can yield greater intuition, Gelfand on p.113 derives the integral term in D'Alembert's Formula for the solution to the 1D wave equation by convolving the infinite plane's impulse response with the wave equation's second initial condition (initial velocity). However he has set the wave propagation speed to unity (ie. $c = 1$). I am trying to re-derive it for non unity values of c.

I should get the integral term in the following D'Alembert's Formula:

$$u(x,t)= \frac 1{2}[ f(x+t)+ f(x-t) ]+ \frac1{2c} \int_{x-ct}^{x+ct} g(y)dy$$

but I do not get the $1/c$ in the integral term.

My question is: Why do I not get 1/c?

What I have tried follows:

Given the infinite planes impulse response is $E(x,t) = \frac 12 \theta(ct-|x|)$.

Given initial conditions $u(x,0)=0$ and $u_1(x)=u_t(x,t)|_{t=0}$ the response to $u_1$ is

$$u(x,t)=E(x,t)*u_1(x)=\int_{\xi=-\infty}^{\xi=+\infty} E(\xi,t) u_1(x-\xi)d\xi $$

Since $E(x,t)$ is zero for $\xi>+ct$ and also for $\xi<-ct$, and is $\frac12$ elsewhere:

$$u(x,t)=\frac12\int_{\xi=-ct}^{\xi=+ct} u_1(x-\xi)d\xi $$

For a change of variables, let $\eta=x-\xi$. Then $\xi=x-\eta$, and $\frac {d\xi}{d\eta}=-1$, so $d\xi = -d\eta$.

For the limits of integration $\xi=+ct$ becomes $\eta=\xi - ct$ and $\xi=-ct$ becomes $\eta=\xi + ct$.

So

$$u(x,t)=\frac{-1}{2}\int_{\eta=x+ct}^{\eta=x-ct} u_1(\eta)d\eta $$

$$u(x,t)=\frac{1}{2}\int_{x-ct}^{x+ct} u_1(\eta)d\eta $$

PROBLEM--MISSING 1/c !!!!

References:

http://mathworld.wolfram.com/dAlembertsSolution.html
https://en.wikipedia.org/wiki/D%27Alembert%27s_formula
http://www.jirka.org/diffyqs/htmlver/diffyqsse32.html
http://people.uncw.edu/hermanr/pde1/dAlembert/dAlembert.htm

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  • $\begingroup$ I am very familiar with the conventional derivation--which is what is on your wiki link. I am trying to make Gelfand's unconventional derivation work for c in general. $\endgroup$ – user45664 Jul 8 at 3:28
  • $\begingroup$ Good catch--I deleted it. $\endgroup$ – user45664 Jul 8 at 3:52
  • $\begingroup$ Please keep the link. It helps those easily look into the details of the formula if needed. $\endgroup$ – Aaron Stevens Jul 8 at 3:58
  • $\begingroup$ Sorry-deleted it. Here it is en.wikipedia.org/wiki/D%27Alembert%27s_formula $\endgroup$ – user45664 Jul 8 at 4:01
  • $\begingroup$ I already put it back. I was the one who originally put it in. $\endgroup$ – Aaron Stevens Jul 8 at 4:04
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The point is that the response function should read \begin{equation} E(x,t) = \frac{1}{2c} \theta(ct - |x|). \end{equation} You can check this by calculating the second derivatives \begin{equation} E_{xx} = \frac{1}{2c} \delta'(ct - |x|), \end{equation} \begin{equation} E_{tt} = \frac{c}{2} \delta'(ct - |x|), \end{equation} and taking your wave equation, \begin{equation} E_{tt} = c^2 E_{xx}. \end{equation}

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  • $\begingroup$ There seems to be a dimension problem with incorporating 1/c into the impulse response: If for example the response was for a vibrating string E should have dimension [L] but including 1/c would multiply the dimension by [T]/[L] giving dimension [T]. $\endgroup$ – user45664 Jul 8 at 18:14
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    $\begingroup$ The response function does not necessarily have to be of dimension $[L]$, this is required only for $u(x,t)$ which represents the actual displacement of the string. If you look at the expression of $u(x,t)$ as a convolution of $E(x,t)$ and $u_1(x)$ you will see that it works correctly: $E(x,t)$ gives $\frac{T}{L}$, $u_1(x,t)$ gives $\frac{L}{T}$ being a derivative of $u(x,t)$ and $dx$ gives $[L]$, so that you finish with $[L]$ for $u(x,t)$. $\endgroup$ – Ciruzz Broncio Jul 9 at 11:53
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    $\begingroup$ Moreover, you can see it from a more general check: the response function for a wave equation should obey such initial value condition that its time derivative in $t=0$ should be a delta function, $E_t(x,0) = \delta(x)$. Since the delta function has the dimension inverse to that of its argument (in this case $[L]^{-1}$) and the time derivative has $[T]^{-1}$ you finish with $\frac{[T]}{[L]}$ for the response function. $\endgroup$ – Ciruzz Broncio Jul 9 at 12:11
  • $\begingroup$ Is there a source for derivation of this this impulse response without using D'Alembert's formula? Or if you could do it I can write another question about the impulse response for you to answer. $\endgroup$ – user45664 Jul 11 at 20:32
  • $\begingroup$ Unfortunately, in the literature, I have seen only a check (similar to Gelfand's one) whether the wave equation and initial conditions are fulfilled. $\endgroup$ – Ciruzz Broncio Jul 16 at 17:37
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to use the D'Alembert's Formula your function $g(y)$

must be $D_2:=\frac{\partial}{\partial t}f(\xi,t)\bigg|_{t=0}$ so the integral is now:

$$I=\frac{1}{c}\int_{x-ct}^{x+ct}D_2(\xi,t)\,d\xi\tag 1$$

$\Rightarrow$

$$\frac{\partial I}{\partial t}=\frac{1}{2}\left[D_2(x+ct,0)+D_2(x-ct,0)\right] $$

and

$$\frac{\partial^2 I}{\partial t^2}=\frac{c}{2}\left[D_{12}(x+ct,0)+D_{12}(x-ct,0)\right] $$

where :

$D_{12}=\frac{\partial}{\partial t}\,D_2(x+ct,t)\bigg|_{t=0}$

with:

$f(\xi,t)=\frac{1}{2}\theta(ct-|\xi|)\,\sigma(\xi)$

$\Rightarrow$

$D_2(\xi\,,t)=1/2\,\mbox {D} \left( \theta \right) \left( ct- \left| \xi \right| \right) c\sigma \left( \xi \right) $

and

$D_{12}=1/2\, \left( D^{ \left( 2 \right) } \right) \left( \theta \right) \left( ct- \left| \xi \right| \right) {c}^{2}\sigma \left( \xi \right) $

where: $D(\theta)=d\theta/dt$ and $D^2(\theta)=d^2(\theta)/dt^2$

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  • $\begingroup$ I am trying to derive D'Alemberts Formula's integral term via convolution of the velocity initial condition with the impulsively excited plane's impulse response. $\endgroup$ – user45664 Jul 8 at 18:20
  • $\begingroup$ I realized this, I just wanted to write the D'Alembert's Formula solution for your function, it must be valid for every function $f(\xi,t)$ $\endgroup$ – Eli Jul 8 at 19:47

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