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A function $u(x,t)$ is called a traveling wave if it has the form $u(x,t) = f(x - at)$, for some function $f$, called the waveform, and some number $a$, called the wave speed.

a.) Show that if a traveling wave solves the wave equation, and the waveform is not a line, then $a = \pm c$.

b.) Show that for the diffusion equation, there are traveling wave solutions with any speed. What is the general form of the waveform for a given speed $a$?

I am not sure how to proceed with this, any suggestions are greatly appreciated

Attempted solution a.) A wave equation is of the form $$u_{tt} - c^2 u_{xx} = 0$$ Suppose the solution is of the form $u(x,t) = f(x - at)$. Then we have $$u_{tt} = (a)^2 f^{\prime\prime}(x - at), \ \ u_{xx} = f^{\prime \prime}(x - at)$$ Thus from the wave equation we have $$(a)^2 f^{\prime\prime}(x - at) - c^2f^{\prime \prime}(x - at) = 0$$ Hence we must have $a = \pm c$ in order for this PDE to be satisfied.

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    $\begingroup$ For a) the point about the waveform not being a line just means that $f''(y) \neq 0$ for some $y$. Since the given $u$ satisfies the wave equation, just compute $u_{tt}, u_{xx}$ and see what you end up withe. $\endgroup$
    – copper.hat
    Commented May 28, 2017 at 20:30
  • $\begingroup$ @copper.hat I made an edit to my post, am I sort of on the right track for part a.)? $\endgroup$
    – justanewb
    Commented May 29, 2017 at 19:12

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@Wolfgang-1 , regards. I may give partial detail.

For the 1st question :

  • A wave equation is of the form : $$ u_{tt} - c^{2} u_{xx} = 0 $$ You could check that $u(x,t) = f(x-at)$ solves the equation if $a^{2} = c^{2}$ and $f$ is not a line.

I am not familiar with your method. But the 1st question could be solved this way : 'presume' the solution is of the form $ u(x,t) = f(x-at) $, so that

$$ a^{2}f''(\xi) - c^{2}f''(\xi) = 0, \:\: \text{with} \: \xi = x -at $$

from here you can see that the solution $f(x-at)$ must have $a = \pm c$. Note that $f$ must not be a line, if it is then it's 2nd derivative would be 0. This answers the 1st question.

You could try to visualize $u(x,t)= \cos\left[k(x-ct)\right]+ \cos\left[k(x+ct)\right] $ (with small $k$, so that the wave directions is more visible) $\\$ $\\$ $\\$

For the 2nd :

  • A 1D diffusion equation is of the form : $$ u_{t} = K u_{xx}, \:\: K > 0 $$ If you put $u(x,t) = f(x - at)$, then from the PDE you could get

$$ f''(\xi) + \frac{a}{K}f'(\xi) = 0, \:\:\ \ \ \text{with } \:\: \xi = x- at $$

So there are travelling-wave solutions for the PDE, which is achieved by solving this ODE. You can solve this ODE and get the waveform $f(\xi)$, which is a translation of exponential function

$$ u(x,t) = f(\xi) = e^{-\frac{a}{K}(x-at)} $$

Hope this would help. Regards, Arief.

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  • $\begingroup$ For the first question, should I use d'Alembert's formula. I am not famaliar with solving these types of questions. $\endgroup$
    – justanewb
    Commented May 29, 2017 at 18:40
  • $\begingroup$ I am not sure if I answered your first question if you could have a look at my edit $\endgroup$
    – justanewb
    Commented May 29, 2017 at 19:11
  • $\begingroup$ @Wolfgang-1 from your question, the part a.), is to show that if the travelling wave solution $f(x-at)$ is a solution to the wave equation, then $a= \pm c$. If $f(x-at)$ is a solution, it means $u(x,t)=f(x-at)$ fits the wave eqn. To answer the question, you only need to show that $a$ must satisfy $a= \pm c $. What you have done is trying to find the exact solution of the equation, and you have not try to 'presume' the solution is of the form $$u=f(x-at)$$ Thanks. $\endgroup$
    – Redsbefall
    Commented May 29, 2017 at 23:03
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    $\begingroup$ Ok, I understand the solution being $u(x,t) = f(x - ct) + h (c + ct)$ but I don't see how we are going to have the form $F(x - at)$ with $a = \pm c$ $\endgroup$
    – justanewb
    Commented May 30, 2017 at 13:33
  • $\begingroup$ I need some more help if you can $\endgroup$
    – justanewb
    Commented May 30, 2017 at 14:56

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