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# Need explanation on how to use d’Alembert’s solution to solve and plot PDE's

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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.

#5
+18777
+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
Sort:

#1
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41.024382993532i

kes1968  May 6, 2015
#2
+26366
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Here f(x-at) = e-(x-t)^2 and g(s) = -2s*e-s^2 so:

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
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Thanks a lot Alan, cant thumbs up enough =)

#4
+26366
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You're welcome!

Alan  May 6, 2015
#5
+18777
+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
#6
+91233
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I'll help with my thumb too :)

Thanks Alan and Heureka

Melody  May 6, 2015

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