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**Home Assignment 3 / Re: Problem 6**

« **on:**October 08, 2012, 08:05:08 AM »

Professor can we assume that u is 0 at positive and negative infinity? Thanks!

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Professor can we assume that u is 0 at positive and negative infinity? Thanks!

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Should part (c) and part (d) be: solve the IBVP for x>0 or for all x? Thanks!

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Professor for part (b), is there any restriction on when we can use method of reflection (continuation) to solve IBVP? Thanks!

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Should we consider solutions on different intervals (i.e. x>2t and 0<x<2t) since they differ in part (a) and part (d)?

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Used a different way to do part (d)

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That means in part (c) we don't need to assume that u is even and we will use this assumption in part (d)? Thanks!

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In this question r is always positive right (since it's the distance to the origin)? Should u(r,0) and ut(r,0) be even functions of r? I guess we need additional information about u(r,0) and ut(r,0) so that v can be extended to negative values.

Thanks!

Thanks!

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Professor I have a question for part (c). Does the solution have to be continuous? For example I have f(x) on x>2t, g(x) on -2t<x<2t and h(x) on -3t<x<-2t. Should f(2t) = g(2t) and g(-2t)=h(-2t) (so the overall solution is continuous)?

My problem is that some of the f, g, h involve a constant K and I was wondering if I should use continuity to specify what K is.

Thanks a lot!

My problem is that some of the f, g, h involve a constant K and I was wondering if I should use continuity to specify what K is.

Thanks a lot!

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Part (a) and (b), part (c) and (d) the same questions? Thanks

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OMG I didn't realize there was a correction on Sep 23! I finished the assignment on Friday night and didn't expect that there would be any change to the questions just several hours before the assignment is due....

Professor can you give some consideration to the situation this time?

Professor can you give some consideration to the situation this time?

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Professor I used a similar method but got the final answer completely different (see attached).

In the equation Ï•(0)+Ïˆ(âˆ’6t)=2t, shouldn't we have Ïˆ'(âˆ’6t) since the second initial condition gives du/dt?

In the equation Ï•(0)+Ïˆ(âˆ’6t)=2t, shouldn't we have Ïˆ'(âˆ’6t) since the second initial condition gives du/dt?

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Using polar coordinates would make things much easier (see attachments). However we need to be careful here since arccos gives us the angle from 0 to pi. When y<0 (i.e. theta > pi), it will become 2pi - arccos.

Please let me know if there's anything wrong with my posted solution.

Please let me know if there's anything wrong with my posted solution.

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I followed the normal steps and found the general solutions to both equations. I cannot figure out why one of them do not exist. Is it because some function is not defined? Get lost in part (c)...

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From the auxiliary equations dx/a=dy/b=du/f, we can either express du in terms of dx and integrate over x, or express du in terms of dy and integrate over y. But sometimes these two approaches give different results. Then can we say that the general solution does not exist?

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Thank you for your hint but I still didn't get the point..

For example if the general solution has the form f(x/y), how can I make them continuous at (0,0)? Thanks!!

For example if the general solution has the form f(x/y), how can I make them continuous at (0,0)? Thanks!!