APM346-2018S > Quiz-4

Quiz 4 -- both sections

(1/2) > >>

Jingxuan Zhang:
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/PDE-textbook/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|=2-2\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}).

Ioana Nedelcu:
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

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

Ioana Nedelcu:
Sorry, forgot the actual variable so

$$ T(t) =  E\cos(\omega^2\sqrt{k}t) + F\sin(\omega^2\sqrt{k}t) $$

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

Navigation

[0] Message Index

[#] Next page