# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment 4 => Topic started by: Calvin Arnott on October 18, 2012, 04:39:28 PM

Title: Problem 1
Post by: Calvin Arnott on October 18, 2012, 04:39:28 PM
It seems to me that question 1 part c) doesn't ask any question and instead makes a statement. Is there anything I'm missing there?
Title: Re: Problem 1
Post by: Victor Ivrii on October 18, 2012, 06:47:58 PM
It seems to me that question 1 part c) doesn't ask any question and instead makes a statement. Is there anything I'm missing there?

Justify  statements.
Title: Re: Problem 1
Post by: Kun Guo on October 19, 2012, 12:26:33 AM
for part a, should we add a condition that alpha and beta are real? Or they have to be real since we are assuming all eigenvalues are real?
Title: Re: Problem 1
Post by: Victor Ivrii on October 19, 2012, 04:26:48 AM
for part a, should we add a condition that alpha and beta are real? Or they have to be real since we are assuming all eigenvalues are real?

Right, I put it explicitly. Thanks for checking and asking questions (on behalf of everyone).
Title: Re: Problem 1
Post by: James McVittie on October 20, 2012, 09:07:18 AM
What does OX stand for?
Title: Re: Problem 1
Post by: Victor Ivrii on October 20, 2012, 10:42:33 AM
What does OX stand for?

$x$ axis
Title: Re: Problem 1
Post by: Shu Wang on October 20, 2012, 11:53:24 PM
For X_n expression in a), if why don't we have w_n in front of the cosine, instead we have w?

Also for e) can we assume there are no degeneracy in the eigenfunctions/states? Otherwise they would be orthogonal with equal eigenvalue.
Title: Re: Problem 1
Post by: Victor Ivrii on October 21, 2012, 03:17:35 AM
For X_n expression in a), if why don't we have w_n in front of the cosine, instead we have w?
Corrected (btw, it is $\omega$, not $w$

Quote
Also for e) can we assume there are no degeneracy in the eigenfunctions/states? Otherwise they would be orthogonal with equal eigenvalue.

Not sure what you mean.
Title: Re: Problem 1
Post by: Heqian Zhang on October 22, 2012, 04:49:08 PM
Hi, I have a question. To solve this question, should we substituting the Xn given into the initial conditions to justify whether it is right ? Or we just use the initial conditions to get the 2 equation given in the problem?
Title: Re: Problem 1
Post by: Victor Ivrii on October 22, 2012, 05:10:02 PM
Hi, I have a question. To solve this question, should we substituting the Xn given into the initial conditions to justify whether it is right ? Or we just use the initial conditions to get the 2 equation given in the problem?

Since we have two conditions (one at each end) we call them boundary conditions.
Title: Re: Problem 1
Post by: James McVittie on October 24, 2012, 09:30:43 PM
Solution to Problem 1(d)

To show that eigenfunctions corresponding to different eigenvalues are orthogonal, we evaluate the following:

$$(\lambda_{n}-\lambda_{m})\intop_{0}^{l}X_{n}(x)X_{m}(x)dx$$

Notice that we can make a simple substitution, apply the Fundamental Theorem of Calculus using the boundary conditions. Then,

$$(\lambda_{n}-\lambda_{m})(X_{n}(x)X_{m}(x))=X_{n}"(x)X_{m}-X_{n}(x)X"_{m}(x)=(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'$$

Plugging into the original integral, we obtain:

$$\intop_{0}^{l}(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'dx=X_{n}'(l)X_{m}(l)-X_{n}(l)X'_{m}(l)-X_{n}'(0)X_{m}(0)+X_{n}(0)X'_{m}(0)=0$$

Therefore, the eigenfunctions corresponding to different eigenvalues are orthogonal.
Title: Problem 1--not posted
Post by: Victor Ivrii on October 26, 2012, 08:55:06 AM
You may assume that all eigenvalues are real (which is the case).

Justify examples 6--7 of
Lecture 13  (http://www.math.toronto.edu/courses/apm346h1/20129/L13.html)
Consider eignevalue problem with Robin boundary conditions
\begin{align*}
& X'' +\lambda X=0 && 0<x<l,\\[3pt]
& X'(0)=\alpha X(0), \quad X'(l)=-\beta X(l)
\end{align*}
$\alpha, \beta \in \mathbb{R}$.

• (a) Prove that positive eigenvalues are $\lambda_n=\omega_n^2$ and the corresponding eigenfunctions are $X_n$ where $\omega_n>0$ are roots of
\begin{align*}
& \tan (\omega l)= \frac{(\alpha+\beta)\omega}{\omega^2-\alpha\beta};\\
& X_n= \omega_n \cos (\omega_n x) +\alpha \sin (\omega_n x);
\end{align*}
($n=1,2,\ldots$).

Solve this equation graphically.
• (b) Prove that negative eigenvalues if there are any are $\lambda_n=-\gamma_n^2$ and the corresponding eigenfunctions are $Y_n$ where $\gamma_n>0$ are roots of
\begin{align*}
& \tanh (\gamma l )= {-\frac{(\alpha + \beta)\gamma }{\gamma ^2 + \alpha\beta}},\\
& Y_n(x) = \gamma_n \cosh (\gamma_n x) + \alpha \sinh (\gamma_n x).
\end{align*}

Solve  this equation graphically.
• (c) To investigate how many negative eigenvalues are, consider the threshold case of eigenvalue $\lambda=0$: then $X=cx+d$ and plugging into b.c. we have $c=\alpha d$ and $c=-\beta (d+lc)$; this system has non-trivial solution $(c,d)\ne 0$ iff $\alpha+\beta+\alpha \beta l =0$. This hyperbola divides $(\alpha,\beta)$-plane into three zones:

(http://www.math.toronto.edu/courses/apm346h1/20129/L13-1.png)

Check above arguments and justify that in the described zones there are really no, one, two negative eigenvalues respectively.
• (d) Prove</strong> that eigenfunctions corresponding to different eigenvalues are orthogonal:

\int_0^l X_n(x)X_m (x)\,dx =0\qquad\text{as } \lambda_n\ne \lambda_m
\label{eq-ort}

where we consider now all eigenfunctions (no matter corresponding to positive or negative eigenvalues).
• (e) Bonus  Prove that eigenvalues are simple, i.e. all eigenfunctions corresponding to the same eigenvalue are proportional.

We have proof of (d) but I definitely want (a)--(c) and (e) (I have seen that they were solved in what was submitted). In (c) I allow to use the following fact (which is due to simple variational arguments which unfortunately we have no time to study): Since the problem is symmetric (which implies that e.v. are real) and quadratic form in the right-hand expression of (13.29) is monotone increasing with respect to $\alpha,\beta$, so do eigenvalues $\lambda_n =\lambda_n(\alpha,\beta)$.[/list]
Title: Re: Problem 1
Post by: Aida Razi on October 29, 2012, 06:57:50 PM
Part (a) proof:
Title: Re: Problem 1
Post by: Aida Razi on October 29, 2012, 09:46:48 PM
Part (b) proof:
Title: Re: Problem 1
Post by: Fanxun Zeng on December 19, 2012, 10:05:43 PM
Thanks for parts a b d above. As no one has posted part c yet, I just post solution attached.
Title: Re: Problem 1
Post by: Fanxun Zeng on December 20, 2012, 02:12:38 AM
In addition to Part c solution I posted above, here is Problem 1 Part e Bonus solution to prove eigenvalues are simple.
Title: Re: Problem 1
Post by: Victor Ivrii on December 20, 2012, 02:39:30 AM
In each of three zones number of negative eigenvalues stays the same as they can cross to positives only on the borders. So in fact one can go along line $\alpha=\beta$ which intersects all of them. Then things are slightly simpler to analyze.