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APM3462012 => APM346 Math => Home Assignment 4 => Topic started by: Calvin Arnott on October 17, 2012, 04:46:26 PM

In problem 2, do we have: [K > 0] for [u_{tt} + K u_{xxxx} = 0]?

Oscillations of the beam which with both its ends having fixed positions and fixed directions (bricked into the walls: it is called ''clamped'') are described by an equation
\begin{equation*}
u_{tt} + K u_{xxxx}=0, \qquad 0<x<l
\end{equation*}
with $K>0$ and the boundary conditions
\begin{equation*}
u(0,t)=u_{x}(0,t)=u(l,t)=u_{x}(l,t)=0.
\end{equation*}
 (a) Find equation describing frequencies and corresponding eigenfunctions
(You may assume that all eigenvalues are real and positive).  (b) Solve this equation graphically.
 (c) Prove that eigenfunctions corresponding to different eigenvalues are orthogonal.
 (d) Bonus Prove that eigenvalues are simple, i.e. all eigenfunctions corresponding to the same eigenvalue are proportional.

Solution is attached!

Thanks for a and b. As no one has posted Problem 2 part c yet, I just did using IBP as attached.

In addition to Part c solution I posted above, here is Problem 2 Part d Bonus solution to prove eigenvalues are simple.

Actually in 2c we prove that the problem is symmetric i.e. that for all $X,Y$ satisfying $0$ b.c. $(HX,Y)=(X,HY)$ where $HX=X^{IV}$ and $(.,.)$ is an inner product in $L^2((0,l))$: $(X,y)=\int_0^l X\bar{Y}\,dx$ ($Y=\bar{Y}$ as problem is realvalued). This implies that eigenvalues must be real and $(HX,X)=\ X''\^2$ implies that either eigenvalue is positive or it is $0$ and in the latter case $X''=0$ (but then $X$ is linear and $X(0)=X'(0)=0$ implies that $X=0$ and $0$ is not an eigenvalue).
Now $\lambda=\omega^4$ with real $\omega>0$ and characteristic equation $k^4=\lambda$ returns $k_{1,2}=\pm \omega$, $k_{3,4}=\pm i\omega$ and
$$
X= A\cosh (\omega x)+ B\sinh (\omega x) + C\cos (\omega x)+ D\sin (\omega x)
$$
and apriori multiplicity could be as high as $4$.
Initial conditions $X(0)=0$ and $X'(0)=0$ imply that $A+C=0$, $B+D=0$ and therefore
$$
X= A\bigl(\cosh (\omega x) \cos (\omega x) \bigr) + B\bigl(\sinh (\omega x)  \sin (\omega x)\bigr)
$$
and therefore multiplicity could be at most $2$ (two possibly independent coefficients).
Then $X(l)=X'(l)=0$ rewrites as
\begin{align*}
&A\bigl(\cosh (\omega l) \cos (\omega l) \bigr) + B\bigl(\sinh (\omega l)  \sin (\omega l)\bigr)=0\\[3pt]
&A\bigl(\sinh (\omega l)+ \sin (\omega l) \bigr) + B\bigl(\cosh (\omega l)  \cos (\omega l)\bigr)=0
\end{align*}
and $\omega^4$ is an eigenvalue iff
$$
\left\begin{matrix}
\cosh (\omega l) \cos (\omega l) & \sinh (\omega l)  \sin (\omega l) \\
\sinh (\omega l)+ \sin (\omega l) & \cosh (\omega l)  \cos (\omega l)
\end{matrix}\right=0
$$
which rewrites as
$$
\cosh(\omega l) \cos(\omega (l)=1.
$$
Part (d)
What about simplicity (Fanxung arguments are wrong): we need to prove that there is exactly 1 linearly independent solution which means that the matrix
$$
\begin{pmatrix}
\cosh (\omega l) \cos (\omega l) &\sinh (\omega l)  \sin (\omega l) \\
\sinh (\omega l)+ \sin (\omega l)& \cosh (\omega l)  \cos (\omega l)
\end{pmatrix}
$$
has then rank $1$ (not $0$). However if rank was $0$ then all the elements would be $0$, in particular $\cosh (\omega l) \cos (\omega l)=0$ which is impossible for $\omega l>0$.