Author Topic: HA4 problem 1  (Read 2876 times)

Victor Ivrii

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HA4 problem 1
« on: February 12, 2015, 05:14:49 PM »
Justify examples 6--7 of  Lecture 11: Consider eignevalue problem with Robin boundary conditions
& X'' +\lambda X=0 && 0<x<l,\label{eq-4.1}\\[3pt]
& X'(0)=\alpha X(0), \qquad X'(l)=-\beta X(l),\label{eq-4.2}
with $\alpha, \beta \in \mathbb{R}$.

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)=
    & X_n= \omega_n \cos (\omega_n x) +\alpha \sin (\omega_n x);
with $n=1,2,\ldots$. 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
    & \tanh (\gamma l )= {-\frac{(\alpha + \beta)\gamma }
    {\gamma   ^2 + \alpha\beta}},\label{eq-4.5}\\
    & Y_n(x) = \gamma_n \cosh (\gamma_n  x) + \alpha \sinh (\gamma_n x).
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.

d.  Prove 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
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.
« Last Edit: February 12, 2015, 05:19:16 PM by Victor Ivrii »

Biao Zhang

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Re: HA4 problem 1
« Reply #1 on: March 19, 2015, 11:21:26 PM »
$$X''+\lambda X=0 \Rightarrow X(x)=C\cos \omega x +D\sin \omega x$$
Let $\lambda = \omega^2$
$$X'(x)=-C\omega \sin \omega x + D \omega \cos \omega x$$
Since $X'(0)=D\omega$ and $X(0)=C$, we have:
$$X'(0)=\alpha X(0)$$
D\omega =\alpha C \rightarrow
D=\frac{\alpha C}{\omega}
$$X'(l)=-c\omega \sin \omega l+D\omega \cos \omega l$$
$$X(l)=c\cos \omega l + D \sin \omega l$$
$$X'(l)=-\beta X(l)$$
$$-C\omega \sin(\omega l)+D\omega \cos(\omega l)=-\beta C\cos(\omega l)-\beta D\sin(\omega l)$$
$$(D\omega + \beta C)\cos(\omega l)=(C\Omega -\beta D)\sin(\omega l)$$
$$(\alpha C+\beta C)\cos(\omega l)=(C \omega -\frac{\alpha C\beta}{\omega})\sin(\omega l)$$
$$\rightarrow (\alpha + \beta)\omega \cos(\omega l)=(\omega^2-\alpha \beta)\sin(\omega l)$$
$$\Rightarrow \tan(\omega l)=\frac{(\alpha + \beta)\omega}{\omega^2-\alpha \beta}$$
Sub to $X(x)=C\cos \omega x +D\sin \omega x$ and By 8
$$X_n=\omega_n \cos(\omega_n x)+\alpha\sin(\omega_n x)$$
$\lambda = -\gamma$,$-X''=-\gamma^2\implies X''-\gamma^2X=0$,\\
char. is $k^2-\gamma^2=0\implies$ \\
$k=\pm \gamma \implies X(x)=Ae^{\gamma x}+Be^{-\gamma x}=A\cosh \gamma x + B\sinh \gamma x$
$$X'(x)=A\gamma \sinh \gamma x +B\gamma  \cosh \gamma x$$
$$X'(0)=B\gamma \textbf{ and } X(0)=A$$
$$X'(0)=\alpha X(0)\implies$$
B\gamma = \gamma A \implies B=\frac{\alpha A}{\gamma}
$$X'(l)=A\gamma \sinh \gamma l + B\gamma \cosh \gamma l$$
$$X(l)=A\cosh \gamma l +B \sinh \gamma l$$
$$A\gamma \sinh \gamma l +B\gamma \cosh \gamma l=-\beta A\cosh \gamma l-\beta B \sinh \gamma l$$
$$(A\gamma +\beta B)\sinh\gamma l = -(\beta A+B\gamma)\cosh \gamma l$$
$$By (9) \implies$$
$$(A\gamma^2+\beta \alpha A)\sinh \gamma l =-(\beta \gamma A+ \alpha A \gamma )\cosh \gamma l $$
$$\frac{\sinh \gamma l}{\cosh \gamma l}=-\frac{(\alpha + \beta)\gamma }{\gamma^2 +\alpha \beta}\implies \tanh(\gamma l)=\frac{(\alpha +\beta)\gamma}{\gamma^2+\alpha \beta}$$
$$X(x)=A\cosh (\gamma x)+B\sinh(\gamma x)$$
By (9)
$$X_n(x)=\cosh(\gamma_n x)+\frac{\alpha}{\gamma_n}\sinh (\gamma_n x)$$
« Last Edit: March 20, 2015, 12:59:04 AM by Biao Zhang »