2

Dalambert's formula for the following BVP $$\hspace{35mm}u_{tt}=C^2u_{xx} ~~-\infty <x<\infty, t>0$$ $$u(x,0)=\phi(x)$$ $$u_t(x,o)=\gamma(x)$$

is that $$u(x,t)=\frac{1}{2}[\phi(x+ct)+\phi(x-ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}\gamma(x)$$

Know my question is that what is the behaviuor of the wave equation when $\phi(x)=0$,

Thanks.

Rosa
  • 1,502
  • What do you meean by behaviour? Exactly the same formula applies. – Mariano Suárez-Álvarez May 02 '13 at 05:33
  • I probably write obvious thing, but it's $$ u(x,t) = \frac 1{2c}\int_{x-ct}^{x+ct} \gamma(\tau)d\tau $$ – Kaster May 02 '13 at 05:41
  • when $\gamma(x)=0$ then the wave has two part, one of them go to the right by speed $c$ and by half height, and another, go the left by speed $c$ and by half height, now I want to know what happen when $\phi(x)=0$ – Rosa May 02 '13 at 05:42

1 Answers1

7

As an example, suppose we want to solve the wave equation $u_{t t}=u_{x x}$ subject to $u(x,0)=0$ and $u_t(x,0)=2x \left/\left(1+x^4\right)\right.$. We can interpret this as follows. We impart an initial velocity to a very long string at rest. The precise initial velocity is given by $g(x)=2x\left/\left(1+x^4\right)\right.$. We might do this by striking the string in two locations with equal, but opposite vertical forces. Here is the graph of this function:

enter image description here

We can use d'Alembert's formula to find the solution analytically.

$$u(x,t) = \frac{1}{2}\int_{x-t}^{x+t} \frac{2s}{1+\left(s^2\right)^2} \, ds=\frac{1}{2}\arctan \left(s^2\right)|_{x-t}^{x+t} = \frac{1}{2}\left(\arctan \left((x+t)^2\right)-\arctan \left((x-t)^2\right)\right)$$

Here's the graph of $\arctan \left(x^2\right)$.

enter image description here

The solution is the superposition of two waves half this size; one is shifted to the left and the other is shifted to the right and reflected. At time $t=0$, the two waves cancel each other out and we get zero - the initial condition. A short time later, a downward hump appears to the left and an upward hump appears to the right. These two waves travel in their respective direction as time progresses. Here's an animation of this process.

enter image description here

I generated the animation with the following Mathematica code.

g[x_] := 2 x/(1 + x^4); 
u[x_, t_] = 1/2 Integrate[g[s], {s, x - t, x + t},
  Assumptions -> {Element[x, Reals], t > 0}];
Animate[Plot[u[x, t], {x, -5, 5},
  PlotRange -> {{-5, 5}, {-1, 1}},
  AspectRatio -> 1/3], {t, 0, 4}]
Mark McClure
  • 30,510
  • Thanks, it was so interseting. – Rosa May 02 '13 at 05:53
  • @Hakhak Glad to hear you find it interesting! In that case, you might consider giving it an upvote, or even an accept. I notice that you've asked 14 questions without a single upvote or accept! – Mark McClure May 02 '13 at 06:00
  • @Hakhak Thanks!! – Mark McClure May 02 '13 at 06:08
  • @MarkMcClure One upvote from my side ...:). Could you tell me how you make this animation? – Srijan May 02 '13 at 07:59
  • @srijan Thanks! I generated the animation with Mathematica and have added the code. If you happen to have access to Mathematica, I lifted this demo out of the "long waves" notebook here: https://sites.google.com/a/unca.edu/mark-mcclure/home/classes/spring-2013/pde – Mark McClure May 02 '13 at 08:14
  • @MarkMcClure Thank you very much. I use mathematica . It is very nice. :) – Srijan May 02 '13 at 08:16