# Toronto Math Forum

## APM346-2018S => APM346--Tests => Quiz-4 => Topic started by: Jingxuan Zhang on March 01, 2018, 10:07:07 PM

Title: Quiz 4 -- both sections
Post 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/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
$$X=A\cosh (\omega x) + B\sinh(\omega x)+C\cos(\omega x)+D\sin(\omega x)\label{4}$$
whence (\ref{2}) implies, if $\omega\neq 0$,
$$A+C=B+D=0\label{5}$$
and so (\ref{4}) becomes
$$X=A(\cosh (\omega x) -\cos(\omega x)) + B(\sinh(\omega x)-\sin(\omega x))\label{6}$$

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:
$$\left|\begin{array}{cc}\cosh (\omega l) -\cos(\omega l)&\sinh(\omega l)-\sin(\omega l)\\ \sinh (\omega l) +\sin(\omega l)&\cosh(\omega l)-\cos(\omega l)\end{array}\right|=2-2\cosh (\omega l)\cos(\omega l)=0\iff\cosh (\omega l)\cos(\omega l)=1\label{equation}.$$
The null space of this system is
$$(A,B)'=t(-\sinh(\omega l)+\sin(\omega l),\cosh (\omega l) -\cos(\omega l))',t\in\mathbb{R}\label{8}$$
and so (\ref{6}) becomes
$$X=(-\sinh(\omega l)+\sin(\omega l))(\cosh (\omega x) -\cos(\omega x))+(\cosh (\omega l) -\cos(\omega l))(\sinh(\omega x)-\sin(\omega x))\label{9}$$
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}).
Title: Re: Quiz 4 Thursday
Post by: Ioana Nedelcu on March 01, 2018, 11:36:02 PM
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
Title: Re: Quiz 4
Post by: Victor Ivrii on March 02, 2018, 03:34:47 AM
Ioanna, please correct expression for $T(t)$.
Title: Re: Quiz 4
Post by: Ioana Nedelcu on March 02, 2018, 02:39:27 PM
Sorry, forgot the actual variable so

$$T(t) = E\cos(\omega^2\sqrt{k}t) + F\sin(\omega^2\sqrt{k}t)$$
Title: Re: Quiz 4 -- both sections
Post by: Ruite Xu on March 06, 2018, 04:11:59 PM
Sorry I don't understand why we can assume $\lambda$ to be a positive number ( here $\omega^4$), can someone help?
Title: Re: Quiz 4 -- both sections
Post by: Victor Ivrii on March 06, 2018, 04:22:06 PM
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 self-adjoint, 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 self-adjoint and positive, which implies that that eigenvalues must be positive\$.