Author Topic: Problem 2  (Read 5878 times)

Calvin Arnott

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Problem 2
« on: October 17, 2012, 04:46:26 PM »
In problem 2, do we have: [K > 0] for [u_{tt} + K u_{xxxx} = 0]?

Victor Ivrii

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Re: Problem 2--not solved
« Reply #1 on: October 17, 2012, 05:20:29 PM »
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.
« Last Edit: October 26, 2012, 09:00:26 AM by Victor Ivrii »

Aida Razi

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Re: Problem 2
« Reply #2 on: November 11, 2012, 04:16:18 PM »
Solution is attached!

Fanxun Zeng

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Re: Problem 2
« Reply #3 on: December 19, 2012, 10:30:49 PM »
Thanks for a and b. As no one has posted Problem 2 part c yet, I just did using IBP as attached.
« Last Edit: December 19, 2012, 10:37:05 PM by Peter Zeng »

Fanxun Zeng

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Re: Problem 2
« Reply #4 on: December 20, 2012, 02:10:59 AM »
In addition to Part c solution I posted above, here is Problem 2 Part d Bonus solution to prove eigenvalues are simple.

Victor Ivrii

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Re: Problem 2
« Reply #5 on: December 20, 2012, 02:27:33 AM »
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 real-valued). 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 $\omega l>0$.

« Last Edit: December 20, 2012, 02:30:18 AM by Victor Ivrii »