HAY--as preparation for TT2

[Victor Ivrii]

http://www.math.toronto.edu/courses/apm346h1/20129/HAY.html
http://www.math.toronto.edu/courses/apm346h1/20129/HAY.pdf

You may post solutions immediately

[Vitaly Shemet]

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]

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 $\beta$.

 

[Vitaly Shemet]

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]

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]

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.

[Zarak Mahmud]

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]

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]

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.

[Vitaly Shemet]

Thank You!

[Vitaly Shemet]

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]

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 $\beta=\alpha$).