Messages

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Home Assignment Y / Re: HAY--as preparation for TT2

« on: November 13, 2012, 07:05:16 PM »

about condition of a+b+a b l=o and b=-a in Appendix C. It does not cross BOTH branches. They intersect at origin. Maybe b=a?

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Home Assignment Y / Re: HAY--as preparation for TT2

« on: November 12, 2012, 09:34:32 PM »

Thank You!

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Home Assignment Y / Re: HAY--as preparation for TT2

« on: November 11, 2012, 10:28:14 PM »

Where this reasoning came from? (I mean what is connection between sign of eigenvalues and intersection of these two graphs, and why do we take specifically 1/alpha or -1/alpha)

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Home Assignment Y / Re: HAY--as preparation for TT2

« on: November 11, 2012, 09:06:22 PM »

In last year TT2 #2 Solution. I can't understand the following reasons. Why do we need to say that tanh(beta l) intersects -1/
alpha?
 

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Home Assignment 3 / Problem 4

« on: October 07, 2012, 03:32:35 PM »

Is initial Gaussian centered at $0$? Considering opposite I'm getting $M(T)$ and $m(T)$  neither increasing nor decreasing, what seems suspicious to me. If it is centered then $M(T)$ is decreasing... In other words, is $u(0,0)$ or $u(l,0) = max    u(x,t)$ for all $x$

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Home Assignment 3 / Re: Problem 6

« on: October 07, 2012, 10:47:13 AM »

For 6b boundary conditions include u(l,t) or no? So that u(0,t)=u(l,t)=0 (Dirichlet) or u_x(0,t)=u_x(l,t)=0 (Newman). 

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Misc Math / Re: Method of Continuation

« on: October 06, 2012, 09:29:29 AM »

I understood the even/odd "trick" as a classification, what to do in each case, but I can't understand the reason and idea of applying, what results in disability to apply method to another types of equations (not 1dwe)

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Home Assignment 2 / Re: Problem4

« on: September 30, 2012, 02:04:47 PM »

And I believe that's the only condition u have  ,

Not only. u(x,t) must be continuous and have partials up to 3rd degree

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Home Assignment 2 / Problem4

« on: September 30, 2012, 11:33:01 AM »

Are we allowed in part "a" to use initial wave equation for equality proof? ([d^2u/dt^2-d^2u/dx^2=0] in de/dt=dp/dx and de/dx=dp/dt (partials) )