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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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- Thread starter mmmboh
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- #1

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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.

- #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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- #7

hunt_mat

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

Mat

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- #9

hunt_mat

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- #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]

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- #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.

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

hunt_mat

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- #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.

Thanks for the help btw.

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- #19

hunt_mat

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[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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If so, I get v

- #21

hunt_mat

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Yes, have you not been doing that?

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- #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.

Edit: Oops I didn`t differentiate the v.

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- #25

hunt_mat

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[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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