Toronto Math Forum
APM3462018S => APM346Tests => Quiz4 => Topic started by: Jingxuan Zhang on March 01, 2018, 10:07:07 PM

Note: Since problems for both sections are very similar I suggest to discuss them together, so certain parts of the solutions could be used without repetition
(V.I.)
The only difference that in Wed section condition on the right end are $u_{xx}_{x=l}=u_{xxx}_{x=l}=0$,
This is problem 3 part 1,2 from http://www.math.toronto.edu/courses/apm346h1/20181/PDEtextbook/Chapter4/S4.2.P.html
The associated eigenvalue problem is
\begin{align}X^{iv}\omega^4 X=0\label{1}\\X(0)=X'(0)=0\label{2}\\X(l)=X'(l)=0
\label{3}\end{align}
From (\ref{1}) we write
\begin{equation}X=A\cosh (\omega x) + B\sinh(\omega x)+C\cos(\omega x)+D\sin(\omega x)\label{4}\end{equation}
whence (\ref{2}) implies, if $\omega\neq 0 $,
\begin{equation}A+C=B+D=0\label{5}\end{equation}
and so (\ref{4}) becomes
\begin{equation}X=A(\cosh (\omega x) \cos(\omega x)) + B(\sinh(\omega x)\sin(\omega x))\label{6}\end{equation}
Now the algebraic system in variable of $A,B$ obtained from (\ref{3}) has nontrivial solution if and only if the coefficient matrix is singular, that is:
\begin{equation}\left\begin{array}{cc}\cosh (\omega l) \cos(\omega l)&\sinh(\omega l)\sin(\omega l)\\ \sinh (\omega l) +\sin(\omega l)&\cosh(\omega l)\cos(\omega l)\end{array}\right=22\cosh (\omega l)\cos(\omega l)=0\iff\cosh (\omega l)\cos(\omega l)=1\label{equation}.\end{equation}
The null space of this system is
\begin{equation}(A,B)'=t(\sinh(\omega l)+\sin(\omega l),\cosh (\omega l) \cos(\omega l))',t\in\mathbb{R}\label{8}\end{equation}
and so (\ref{6}) becomes
\begin{equation}X=(\sinh(\omega l)+\sin(\omega l))(\cosh (\omega x) \cos(\omega x))+(\cosh (\omega l) \cos(\omega l))(\sinh(\omega x)\sin(\omega x))\label{9}\end{equation}
up to a scalar multiple. This is the eigenspace.
The graphs are those of $1/\cosh$ and $\cos$, imagined to be in variable of $\omega l$. Their (infinitely many) intersections suffice (\ref{equation}).

There is also $ T(t) $, which has the ODE $ T'' + k\omega^4T = 0 $
$$ \implies T = E\cos(\omega^2\sqrt{k}) + F\sin(\omega^2\sqrt{k}) $$
So the full solution for $ u(x, y) = X(x)T(t) $ with X given in Jingxuan's answer below

Ioanna, please correct expression for $T(t)$.

Sorry, forgot the actual variable so
$$ T(t) = E\cos(\omega^2\sqrt{k}t) + F\sin(\omega^2\sqrt{k}t) $$

Sorry I don't understand why we can assume $\lambda$ to be a positive number ( here $\omega^4$), can someone help?

Sorry I don't understand why we can assume $\lambda$ to be a positive number ( here $\omega^4$), can someone help?
You can derive this directly, but there is a more general way:
1) If $u$, $v$ satisfy b.c., then $\int u^{(4)}v\,dx =\int u v^{(4)}\,dx$; thus operator is selfadjoint, which implies that eigenvalues must be real.
2) Also $\int u^{(4)}u\,dx= \int (u'')^2\,dx> 0$ for $u\ne 0$ (also satisfying boundary conditions); thus operator is selfadjoint and positive, which implies that that eigenvalues must be positive$.