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avatar+248 

Question:

Solve the PDE utt = uxx  where u(x,0) = e(-x^2)  ,  ut(x,0) = d/dx(e-x^2) using d'Alembert's solution. Sketch the solutions at t= 0 , 1, 10.

Problem I'm having is that we spent one 50 minute lecture on this, and 40 minutes of the lecture was deriving d'Alembert's general solution to an Initial Value Problem so I'm clueless on how to actually use his formula.

d'Alemberts general soln to an IVP: 

u(x,t) = 1/2[f(x-at)+f(x+at)] + 1/2a $$\int_{x-at}^{x+at} \mathrm{g}{(s)}\,\mathrm{d}s$$  In this case a = 1.

Any help at all is appreciated.

difficulty advanced

Best Answer 

 #5
avatar+19632 
+20

Question:

 u(x,0) = e(-x^2)  ,  ut(x,0) = d/dx(e-x^2)

Solve  using d'Alembert's solution

u(x,t) = 1/2[f(x-at)+f(x+at)] + 1/2a $$\int_{x-at}^{x+at} \mathrm{g}{(s)}\,\mathrm{d}s$$  In this case a = 1.

 

$$u(x,0) = g(x) = e^{-x^2} \qquad
\boxed{
g(x-t) = e^{-(x-t)^2} \qquad
g(x+t) = e^{-(x+t)^2} }$$

$$u_t(x,0) = h(x) = \dfrac{d\left( e^{-x^2} \right)}{dx} \qquad
\boxed{
\int_{x-t}^{x+t} h(\xi) \, d\xi. = \int_{x-t}^{x+t} \dfrac{d\left( e^{-\xi^2} \right)}{d\xi} = \left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}\\
}$$

 

 

 

$$u(x,t) = \frac{1}{2}\left[g(x-t) + g(x+t)\right] + \frac{1}{2} \int_{x-t}^{x+t} h(\xi) \, d\xi.\\\\
u(x,t) = \frac{1}{2}\left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2}\left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}$$

$$u(x,t) = \frac{1}{2} \left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} - e^{-(x-t)^2} \right]\\\\
u(x,t) = \frac{1}{2} \left[ e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} } \right]\\\\
\boxed{ u(x,t) = e^{-(x+t)^2} }\\\\
u(x,0) = e^{-(x)^2}\\
u(x,1) = e^{-(x+1)^2}\\
u(x,10) = e^{-(x+10)^2}$$

heureka  May 6, 2015
 #1
avatar+285 
0

41.024382993532i

kes1968  May 6, 2015
 #2
avatar+26750 
+15

Here f(x-at) = e-(x-t)^2 and g(s) = -2s*e-s^2 so:

 

pde solution

I'll leave you to sketch the solutions.

 

(Note: for f use u(x,0), and for g use ut(x,0)) 

Alan  May 6, 2015
 #3
avatar+248 
0

Thanks a lot Alan, cant thumbs up enough =)

 #4
avatar+26750 
0

You're welcome!

Alan  May 6, 2015
 #5
avatar+19632 
+20
Best Answer

Question:

 u(x,0) = e(-x^2)  ,  ut(x,0) = d/dx(e-x^2)

Solve  using d'Alembert's solution

u(x,t) = 1/2[f(x-at)+f(x+at)] + 1/2a $$\int_{x-at}^{x+at} \mathrm{g}{(s)}\,\mathrm{d}s$$  In this case a = 1.

 

$$u(x,0) = g(x) = e^{-x^2} \qquad
\boxed{
g(x-t) = e^{-(x-t)^2} \qquad
g(x+t) = e^{-(x+t)^2} }$$

$$u_t(x,0) = h(x) = \dfrac{d\left( e^{-x^2} \right)}{dx} \qquad
\boxed{
\int_{x-t}^{x+t} h(\xi) \, d\xi. = \int_{x-t}^{x+t} \dfrac{d\left( e^{-\xi^2} \right)}{d\xi} = \left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}\\
}$$

 

 

 

$$u(x,t) = \frac{1}{2}\left[g(x-t) + g(x+t)\right] + \frac{1}{2} \int_{x-t}^{x+t} h(\xi) \, d\xi.\\\\
u(x,t) = \frac{1}{2}\left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2}\left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}$$

$$u(x,t) = \frac{1}{2} \left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} - e^{-(x-t)^2} \right]\\\\
u(x,t) = \frac{1}{2} \left[ e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} } \right]\\\\
\boxed{ u(x,t) = e^{-(x+t)^2} }\\\\
u(x,0) = e^{-(x)^2}\\
u(x,1) = e^{-(x+1)^2}\\
u(x,10) = e^{-(x+10)^2}$$

heureka  May 6, 2015
 #6
avatar+92781 
0

I'll help with my thumb too :)

Thanks Alan and Heureka   

Melody  May 6, 2015

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