Solving Cauchy PDEs using the Method of Characteristics

In summary, the conversation discusses different techniques for solving a problem involving the Laplace operator in a radially symmetric situation. The suggested method involves using characteristic co-ordinates and solving a 1D wave equation. The conversation also touches on different equations and variables, with one participant eventually arriving at a potential solution.
  • #1
mmmboh
407
0
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.
 
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  • #2
Anyone?
 
  • #3
Have you tried Fourier transforms?
 
  • #4
We haven`t learned that. There must be a more simple way. :S
 
  • #5
What HAVE you done on your course? Once I know this I can select the appropriate technique.
 
  • #6
Cauchy problems for the heat and wave equations. Poisson equation, laplace equation, characteristic curves, dirichlet problem, finite difference method, advection in 1d..
 
  • #7
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
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
Hmmm, come to thin about it have you done anything in 2 spatial dimensions? Can you extend what you did on the cauchy problem for the wave eqution?
 
  • #10
We looked at the Cauchy problem in 1 spatial dimension. Most of what we have done is in two dimensions total.
 
  • #11
I think the question is saying that it is radially symmetric, do you agree?
 
  • #12
Yes that`s what it seems like.
 
  • #13
So what is the radially symmetric part of Laplace operator [tex]\nabla^{2}[/tex]
 
  • #14
In polar coordinates, it is (1/r)(d/dr)(rdu/dr)...Am I suppose to take the laplacian of u, and see something?
 
  • #15
Expand that out and denote v=ru, and what equation do you have now?
 
  • #16
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.
 
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  • #17
You should end up with the 1D equation for v, solve this in the usual manner and the problem is easily solved.
 
  • #18
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
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
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
Yes, have you not been doing that?
 
  • #22
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
as u=u(t,r) so is v=v(t,r)
 
  • #24
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
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.
 

FAQ: Solving Cauchy PDEs using the Method of Characteristics

How do I determine if a PDE is a Cauchy problem?

A PDE is classified as a Cauchy problem if it is specified on a subset of the domain where the initial conditions are given. This means that the solution can be uniquely determined at a specific point within the domain.

What is the general approach to solving Cauchy PDEs?

The general approach to solving Cauchy PDEs is to first rewrite the problem in a standard form, which involves separating the highest order derivatives. Then, the problem can be solved using various techniques such as separation of variables, method of characteristics, or Fourier transforms.

What are the initial conditions in a Cauchy problem?

The initial conditions in a Cauchy problem refer to the values of the dependent variables and their derivatives at a specific point within the domain. These conditions are used to uniquely determine the solution to the PDE.

How do I know if my solution to a Cauchy PDE is correct?

One way to check if your solution to a Cauchy PDE is correct is to substitute it back into the original PDE and initial conditions. If the solution satisfies the PDE and initial conditions, then it is considered to be a valid solution.

Are there any special techniques for solving nonlinear Cauchy PDEs?

Yes, there are special techniques for solving nonlinear Cauchy PDEs such as the method of characteristics and the inverse scattering transform. These techniques are used to transform the nonlinear PDE into a system of linear PDEs, which can then be solved using standard methods.

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