Author Topic: TT1 problem 2  (Read 939 times)

Victor Ivrii

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TT1 problem 2
« on: March 12, 2015, 07:14:52 PM »
Solve  the eigenvalue problem
\begin{gather*}
X''+\lambda X =0, \qquad x\in (0,l),\\
X(0)=0; \quad X'(l) +\alpha  X(l) =0.
\end{gather*}

a.  For which values of  $\alpha$  is $\lambda=0$  an eigenvalue?
b.  For which values of $\alpha$  are  there  negative eigenvalues? How many?
c.  Find the positive eigenvalues.
d. How do positive eigenvalues depend on $\alpha$?


Note. If  the equations for  eigenvalues cannot be solved explicitly, indicate in which interval they can be found, and write the corresponding eigenfunctions.


Biao Zhang

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Re: TT1 problem 2
« Reply #1 on: March 13, 2015, 07:41:59 AM »
(a)

If $\lambda = 0$ is an eigenvalue than: $X' = 0$ $\implies$ $X'(x) = Ax + B$, and
$X'(x) = A$,$X(0) = 0 = B$. Therefore $X(x) = Ax$.
$$X'(l)+\alpha X(l)=A+\alpha Al=A(1+\alpha l) =0$$
In order for $X(x)$ not to be trivial we must have $A \neq 0$. Therefore $(1 + \alpha l) = 0$
$\implies$ $α = −1/l$.

(b)

 If $\lambda < $0. Let $\lambda = -\beta
^2$ where $\beta  > 0$. Than $X'' - \beta
^2X = 0$, where
$X(x) = A\cosh(\beta x) + B\sinh(\beta x)$. If $ X(0) = A = 0$, than $X(x) = B\sinh(\beta x)$
and $X' = B\beta cosh(\beta x$).
$$X'(l)+\alpha X(l)=B(\beta \cosh(\beta l)+\alpha\sinh(\beta l))$$
For $X(x)$ not to be trivial we must have $B\neq 0$ then $\beta \cosh(\beta l)+\alpha \sinh(\beta l)=0$. Then since $\alpha \neq 0$ we have that $\beta \cos (\beta l)>0$. Therefore
$$-\frac{\beta}{\alpha}=\tanh$$
Then if $\alpha < 0\implies y=\frac{1}{\alpha}$ intersects $\tanh(\beta l)$
$$(\tanh(\beta l))'|_{\beta=0}=l$$
When $\beta =0$. Note $(-\frac{\beta}{\alpha})'=\frac{1}{\alpha}$. Than $l>\frac{1}{\alpha}$ . Therefore the only eigenvalue if found when $\alpha < \frac{1}{l}$.

(c)

If $\lambda =\beta^2>0$ we have:
$$X'(l)+\alpha X(l) = B\beta \cos(\beta l) + \alpha B \sin(\beta l)=0$$
For $X$ to be non-trivial, we have $-\frac{\beta}{\alpha}=\tan (\beta l)$. Note $\alpha \neq 0$. We get:
$$\lambda _m=\frac{\beta^2_n \pi}{2l}+\frac{n\pi}{l}\leq\beta_n\leq\frac{3\pi}{2l}+\frac{(n-1)\pi}{n}$$

(d)

$$\tan(\beta l)'=l\sec^2(\beta l)=\frac{1}{\cos^2(\beta l)}$$
If $\frac{-1}{\alpha}>l$ we have an intersection with the first branch. If $\frac{-1}{l}<\alpha<0$ we have that
$$\frac{\pi}{2l}+\frac{(n-1)\pi}{l}<\beta_n<\frac{3\pi}{2l}+\frac{(n-1)\pi}{n}$$
also $0<\beta_0<\frac{\pi}{2}$.\\
If $\alpha<0$, There is no intersection.
« Last Edit: March 13, 2015, 07:43:31 AM by Biao Zhang »

Victor Ivrii

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Re: TT1 problem 2
« Reply #2 on: March 13, 2015, 11:45:52 AM »
a. correct.
b. We get (some misprints): $\lambda=\beta^2$, $X=A \sinh (\beta x)$ , $A[\beta \cosh (\beta l) + \alpha  \sinh(\beta l) ]=0$, so $A\ne 0$ means $\beta^{-1}\tanh (\beta l)=-1/\alpha $.  Consider graph $z(\beta)=\beta^{-1}\tanh (\beta l)$ as $\beta>0$. Then
$$
z'= \frac{1}{\beta^2(\cosh (\beta l))^2}[-\sinh (\beta l) \cosh(\beta l) + \beta l]<0
$$
because $\sinh (t)>t$, $\cosh (t)>1$ as $t>0$. Note that $z(+\infty)=0$, $\lim_{\beta\to +0}z(\beta)= l$ and $z(\beta)=-1/\alpha$ for some $\beta>0$ if and only if $-1/\alpha \in (0, l)$.

So answer is: $\alpha < -1/l$.

Also note that when $\alpha <0$ increases $-1/\alpha$ increases and therefore $\beta$ decreases; then $\lambda=-\beta^2$ increases. (not required)



c. $\lambda=-\beta^2$, $X=A \sin (\beta x)$, $\beta^{-1}\tan (\beta l)=-1/\alpha$. Let $z(\beta)=\beta^{-1}\tan (\beta l)$; then
$$
z'= \frac{1}{\beta^2(\cos (\beta l))^2}[-\sin (\beta l) \cos (\beta l) + \beta l]>0
$$
because  $|\sin (t)|<t$, $|\cos (t)|\le 1$ as $t>0$. Also note that on each interval $((n-\frac{1}{2})\frac{\pi}, {l}, (n+\frac{1}{2})\frac{\pi}{l})$ this function runs from $-\infty$ to $+\infty$ thus intersecting $-1/\frac{\alpha}$ exactly once, so we have $\beta_n$ on this interval.

d. Note that as $\alpha $ increases $-1/\alpha$ increases and therefore $\beta$ also increases;  then $\lambda=\beta^2$ increases.