d'alembert solution Help

Aug 2008
98
8
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.

Thanks in advance!



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
 

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Last edited:

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
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.

Thanks in advance!



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 \(\displaystyle u(x,t) = 11\), then \(\displaystyle u_t(x,t) = 0\) always and so \(\displaystyle u_t(x,0) = 0 \) and not \(\displaystyle u_t(x,0) = xe^{-x^2} \ne 0 \) for all \(\displaystyle x\). Your answer (almost) is correct

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

It satisfies both the PDE and the initial conditions.
 
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Reactions: Niall101
Aug 2008
98
8
Hey Thanks for that. Someone on the college forum said they got 11 for the answer which threw me way off.