# Traveling wave solving the wave equation [closed]

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.

• 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. Commented May 28, 2017 at 20:30
• @copper.hat I made an edit to my post, am I sort of on the right track for part a.)? Commented May 29, 2017 at 19:12

@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.

• For the first question, should I use d'Alembert's formula. I am not famaliar with solving these types of questions. Commented May 29, 2017 at 18:40
• I am not sure if I answered your first question if you could have a look at my edit Commented May 29, 2017 at 19:11
• @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. Commented May 29, 2017 at 23:03
• 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 $a = \pm c$ Commented May 30, 2017 at 13:33
• I need some more help if you can Commented May 30, 2017 at 14:56