Author Topic: HAY--as preparation for TT2  (Read 9621 times)

Victor Ivrii

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Vitaly Shemet

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Re: HAY--as preparation for TT2
« Reply #1 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?
 

Zarak Mahmud

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Re: HAY--as preparation for TT2
« Reply #2 on: November 11, 2012, 09:51:28 PM »
The only way we can have a negative eigenvalue is if the line $y=-\frac{1}{\alpha}$ intersects $\tanh \beta l$. This can't happen if $\alpha$ is positive. Have you tried drawing the graph?

By the way, for negative eigenvalues the convention is to use $\gamma$ instead of $\beta$.

 
« Last Edit: November 11, 2012, 10:04:44 PM by Zarak Mahmud »

Vitaly Shemet

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Re: HAY--as preparation for TT2
« Reply #3 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)

Zarak Mahmud

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Re: HAY--as preparation for TT2
« Reply #4 on: November 11, 2012, 10:38:16 PM »
It is discussed in lecture 13. If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3

Victor Ivrii

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Re: HAY--as preparation for TT2
« Reply #5 on: November 12, 2012, 04:06:52 AM »
It is discussed in lecture 13. If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3

See  Appendix C and Appendix B.
« Last Edit: November 12, 2012, 09:50:39 PM by Victor Ivrii »

Zarak Mahmud

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Re: HAY--as preparation for TT2
« Reply #6 on: November 12, 2012, 11:44:04 AM »
It is discussed in lecture 13. If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3

See  Appendix C and Appendix B.

You linked to appendix C twice.  :)

Victor Ivrii

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Re: HAY--as preparation for TT2
« Reply #7 on: November 12, 2012, 01:49:49 PM »
It is discussed in lecture 13. If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3

See  Appendix C and Appendix B.

You linked to appendix C twice.  :)

Yes, right. Fixed.

Zarak Mahmud

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Re: HAY--as preparation for TT2
« Reply #8 on: November 12, 2012, 03:49:15 PM »
By the way, I just wanted to remark - the hyperbola dividing the $(\alpha , \beta)$ plane is a great way to keep everything straight. I was getting mixed up with the signs until I start thinking of it in this way.
« Last Edit: November 12, 2012, 04:40:54 PM by Victor Ivrii »

Vitaly Shemet

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Re: HAY--as preparation for TT2
« Reply #9 on: November 12, 2012, 09:34:32 PM »
Thank You!

Vitaly Shemet

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Re: HAY--as preparation for TT2
« Reply #10 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?

Victor Ivrii

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Re: HAY--as preparation for TT2
« Reply #11 on: November 13, 2012, 07:34:20 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?

Yes, fixed (one can see from (3) that it was exactly $\beta=\alpha$).