### Author Topic: TT2 problem 1  (Read 2867 times)

#### Victor Ivrii

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##### TT2 problem 1
« on: March 12, 2015, 07:13:38 PM »
Consider the eigenvalue problem
\begin{gather}
x^2 X'' + 2 x X' + \lambda X =0, \qquad x \in (\frac{2}{3},\frac{3}{2}),\label{eq-1}\\
X'(\frac{2}{3}) =0 ; \ X'(\frac{3}{2}) =0.\notag
\end{gather}
Assume  $\lambda \ge 0$. Find all the eigenvalues and the corresponding eigenfunctions.

Hint: as (\ref{eq-1}) is Euler equation, look   for elementary  solutions  in the form $x^m$).

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### Re: TT2 problem 1
« Reply #1 on: March 13, 2015, 09:56:25 AM »
Almost everyone did poorly on this problem. Taking coefficient $2$ instead of $1$  in comparison with TT2 of 2012 made it much more difficult (not beyond the reach, but much more difficult). This was not my intention. So I will consider it as a Bonus problem and all other will get weight 25.

Characteristic equation $k(k-1)+2k+\lambda=0$ and therefore making change $t= \ln (3x/2)$ we arrive to equation

\ddot{X} + \dot{X}+\lambda X=0

where over-dot means derivative with respect to $t=\ln \left(\frac{3x}{2}\right)$, $0\le t\le l=\ln (9/4)$. We took $t=\ln \left(\frac{3x}{2}\right)=\ln (x) + \ln (3/2)$ instead just $t=\ln (x)$ to have interval $[0,l]$; here $l=2\ln (3/2)$.

So, $k(k+1)+\lambda=0$. This equation for given $k$ has two rots, one is $k$ and another is $-(k+1)$. So $X=Ae^{kt}+Be^{-(k+1)t}$. Since $X'=x^{-1}\dot{X}$ boundary conditions become $\dot{X}(0)=\dot{X}(l)=0$:

\left\{\begin{aligned}
&A k  - B(k+1)=0,\\
&Ake^{kl}-B(k+1)e^{-(k+1)l}=0
\end{aligned}\right.

Determinant of this system is $k(k+1)\bigl(e^{kl}-e^{-(k+1)l}\bigr)=0$. So either $k(k+1)=0\iff \lambda=0$ or $e^{kl}=e^{-(k+1)l}\iff e^{(2k+1)l}=0\iff 2k+1= \pi ni/l$ with $n=0,1,2,\ldots$ (taking $n=-1,-2,\ldots$ would repeat the same pair $k,-(k+1)$).

So,  $k_n = -\frac{1}{2}+ \frac{\pi ni}{l}$, then $\lambda_n = \frac{1}{4}+\left(\frac{\pi n }{l}\right)^2$.

\lambda_n = \frac{1}{4}+\left(\frac{\pi n }{\ln (9/4)}\right)^2\qquad n=0,1,2,\ldots.

Then $A_n = k_n+1=\frac{1}{2}+ \frac{\pi i n}{l}$, $B_n =k_n=-\frac{1}{2}+ \frac{\pi i n}{l}$ (up to a numerical factor), and up to a numerical factor

X_n = e^{-\frac{t}{2}} \left[\sin \left(\frac{\pi nt}{l}\right) + \frac{\pi n}{l} \cos \left(\frac{\pi nt}{l}\right)\right]   \qquad n=0,1,2,\ldots

and we need to plug $t$ and $l$ here.

We need to test $\lambda=0$, then $X= Ae^{-\frac{t}{2}}+Bte^{-\frac{t}{2}}$ and boundary conditions imply $-\frac{A}{2} +B=0$, $-\frac{A}{2}- \frac{B}{2} l +B=0$ which implies $A=B=0$; so we do not have this eigenvalue.
« Last Edit: March 13, 2015, 07:17:01 PM by Victor Ivrii »