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Cauchy Pde Problem!

  • Thread starter mmmboh
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  • #1
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14diu4k.jpg


Could someone tell me where to start? I tried separating variables, which got me no where (plus we haven`t technically learned it), and I tried putting it into a form of D^2U, but I couldn`t figure that out either. Please help.

Thank you.
 

Answers and Replies

  • #2
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Anyone?
 
  • #3
hunt_mat
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Have you tried Fourier transforms?
 
  • #4
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We haven`t learned that. There must be a more simple way. :S
 
  • #5
hunt_mat
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What HAVE you done on your course? Once I know this I can select the appropriate technique.
 
  • #6
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Cauchy problems for the heat and wave equations. Poisson equation, laplace equation, characteristic curves, dirichlet problem, finite difference method, advection in 1d..
 
  • #7
hunt_mat
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Okay have you done characteristics for 3 variables rather than two?

The way to do this in 1d is to use characteristic co-ordinates.

Mat
 
  • #8
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No we`ve only done it for two variables :S...I`ll give it a shot, but can you let me know what substitutions to use please? Do i just introduce a third variable equal to like y-ct or what :S
 
  • #9
hunt_mat
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Hmmm, come to thin about it have you done anything in 2 spacial dimensions? Can you extend what you did on the cauchy problem for the wave eqution?
 
  • #10
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We looked at the Cauchy problem in 1 spacial dimension. Most of what we have done is in two dimensions total.
 
  • #11
hunt_mat
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I think the question is saying that it is radially symmetric, do you agree?
 
  • #12
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Yes that`s what it seems like.
 
  • #13
hunt_mat
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So what is the radially symmetric part of Laplace operator [tex]\nabla^{2}[/tex]
 
  • #14
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In polar coordinates, it is (1/r)(d/dr)(rdu/dr)...Am I suppose to take the laplacian of u, and see something?
 
  • #15
hunt_mat
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Expand that out and denote v=ru, and what equation do you have now?
 
  • #16
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But u(x,y,t) is not in terms of r, can I write u(r,t) instead and use the laplacian on that? If that`s what you mean..or do you mean something else?

I can expand it, and get (1/r)(du/dr)+(d/dr)(du/dr), but there is no u*r in this.

If I let u=v/r, I get v/r^3.
 
Last edited:
  • #17
hunt_mat
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You should end up with the 1D equation for v, solve this in the usual manner and the problem is easily solved.
 
  • #18
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I get v/r^3 = 0 :S...I edited my post if you didn`t see, can you check if that`s what I`m suppose to get? (but I still don`t yet see how this will show that u vanishes for the given condition, or why we can assume u satisfies laplace`s equation).

Thanks for the help btw.
 
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  • #19
hunt_mat
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Just to confirm the equation you should be solving is:

[tex]
\frac{\partial^{2}u}{\partial t^{2}}-c^{2}\left(\frac{\partial^{2}u}{\partial r^{2}}+\frac{2}{r}\frac{\partial u}{\partial r}\right) =0
[/tex]

Use [tex] v(t,r)=ru(r,t)[/tex] to obtain a 1D wave equation, turn that intop characteristic co-ordinates in the usual way and that will give you your solution.

I am off to bed now.
 
  • #20
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So do you mean I should I write u=v/r, and insert that into the equation you wrote? Good night, thanks.

If so, I get vtt=0, hope that is right!
 
  • #21
hunt_mat
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Yes, have you not been doing that?
 
  • #22
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I wasn`t before :S, but now I get I get vtt=0, so I guess I can just integrate that and substitute u in.
 
  • #23
hunt_mat
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as u=u(t,r) so is v=v(t,r)
 
  • #24
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I get u(r,t)=a(r)*r*t+r*b(r), where a,b are some functions of r. I don`t see how this means u vanished for the given condition though :S

Edit: Oops I didn`t differentiate the v.
 
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  • #25
hunt_mat
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Calculate:

[tex]
\frac{\partial^{2}v}{\partial r^{2}}
[/tex]

to get the folowing equation:

[tex]
\frac{\partial^{2}v}{\partial t^{2}}-c^{2}\frac{\partial^{2}v}{\partial r^{2}}=0
[/tex]

Then introduce the co-ordinates:

[tex]
\begin{array}{ccc}
\alpha & = & r+ct \\
\beta & = & r-ct
\end{array}
[/tex]

use the condition you have and this should give you the result you need.
 

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