# d'alembert solution Help

• May 14th 2010, 05:46 PM
Niall101
d'alembert solution Help
Hi I have been trying to figure out how this method works.

Searched the forum but got no results. Have attached an example Im trying to do. If someone could show me it would be great.

Or even show me a similar example worked out.

Edit: I am trying another question
u_tt = u_xx
u( x, 0) = 0 = f(x) ?
u_t( x,0) = xe^-(x^2) = g(x)?

So as f(x) = 0 can I just figure out the ingegral of g(x) with limits (x+ct) and (x-tc)

The answer is supposed to be 11 but I just get

1/4 [ -e^(x+t)^2 + e^-(x-t)^2]

I use the formula and as f(x) = 0
• May 15th 2010, 07:08 AM
Jester
Quote:

Originally Posted by Niall101
Hi I have been trying to figure out how this method works.

Searched the forum but got no results. Have attached an example Im trying to do. If someone could show me it would be great.

Or even show me a similar example worked out.

Edit: I am trying another question
u_tt = u_xx
u( x, 0) = 0 = f(x) ?
u_t( x,0) = xe^-(x^2) = g(x)?

So as f(x) = 0 can I just figure out the ingegral of g(x) with limits (x+ct) and (x-tc)

The answer is supposed to be 11 but I just get

1/4 [ -e^(x+t)^2 + e^-(x-t)^2]

I use the formula and as f(x) = 0

The answer can't be 11! If the answer was $u(x,t) = 11$, then $u_t(x,t) = 0$ always and so $u_t(x,0) = 0$ and not $u_t(x,0) = xe^{-x^2} \ne 0$ for all $x$. Your answer (almost) is correct

$
u(x,t) = \frac{1}{4} \left( e^{-(x-t)^2} - e^{-(x+t)^2}\right)
$

It satisfies both the PDE and the initial conditions.
• May 16th 2010, 10:37 AM
Niall101
Hey Thanks for that. Someone on the college forum said they got 11 for the answer which threw me way off.