Author Topic: HA6-P3  (Read 1216 times)

Bruce Wu

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Bruce Wu

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Re: HA6-P3
« Reply #1 on: October 29, 2015, 12:49:01 AM »
a) Following the hint, I made a change of variables $r=x-\frac{l}{2}$. The beam equation and its boundary conditions then become:
$$u_{tt}+Ku_{rrrr}=0,~-\frac{l}{2}<r<\frac{l}{2}\\u(-\frac{l}{2},t)=u_{r}(-\frac{l}{2},t)= u(\frac{l}{2},t)=u_{r}(\frac{l}{2},t)=0$$
Letting $u=T(t)R(r)$, we see that $$T''+K\lambda T=0\\ R''''-\lambda R=0$$
Here we are assuming that $\lambda>0$, so let $\lambda=\omega^{4}$. Solving the corresponding characteristic equations and ODEs we get
$$T=A\cos(\sqrt{K}\omega^{2}t)+B\sin(\sqrt{K}\omega^{2}t)\\ R=Ce^{\omega r}+De^{-\omega r}+E\cos(\omega r)+F\sin(\omega r)$$
Using the 4 boundary conditions, the fact that $T$ is a constant with respect to $r$ and is not identically $0$ we arrive at the system of 4 equations below
$$Ce^{-\omega\frac{l}{2}}+De^{\omega\frac{l}{2}}+E\cos(\omega\frac{l}{2})-F\sin(\omega\frac{l}{2})=0\\
Ce^{-\omega\frac{l}{2}}-De^{\omega\frac{l}{2}}+E\sin(\omega\frac{l}{2})+F\cos(\omega\frac{l}{2})=0\\
Ce^{\omega\frac{l}{2}}+De^{-\omega\frac{l}{2}}+E\cos(\omega\frac{l}{2})+F\sin(\omega\frac{l}{2})=0\\
Ce^{\omega\frac{l}{2}}-De^{-\omega\frac{l}{2}}-E\sin(\omega\frac{l}{2})+F\cos(\omega\frac{l}{2})=0$$
We can convert this into a matrix, and it only has non-trivial solutions when the determinant of the matrix is $0$. This computation is tedious, but the final condition that $\omega$ must satisfy is
$$\cos(\omega l)\cosh(\omega l)=1$$

Bruce Wu

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Re: HA6-P3
« Reply #2 on: October 29, 2015, 12:53:53 AM »
b) One can see a plot of the general form of the function $\cos(x)\cosh(x)$ here http://www.wolframalpha.com/input/?i=cos%28x%29cosh%28x%29
It is a very weird function. The different $\omega_{n}$'s are the values of $\frac{x}{l}$ when the function intersects the horizontal line $y=1$

Bruce Wu

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Re: HA6-P3
« Reply #3 on: October 29, 2015, 01:14:04 AM »
After this I am not quite sure how to continue. Row reducing the matrix to solve for three of $C,D,E,F$ in terms of the other one seems like a daunting task, but without them I cannot find the eigenfunctions. The hint says to consider separately even and odd eigenfunctions, but I'm not certain what forms they should take. The exponentials cannot have any symmetry unless $D=\pm C$ so that they turn into $2C\cosh(\omega r)$ and $2C\sinh(\omega r)$, respectively.
If my logic is correct, then the even eigenfunctions should be $$2C\cosh(\omega r) + E\cos(\omega r)$$
and the odd eigenfunctions $$2C\sinh(\omega r) + F\sin(\omega r)$$
However the relationship between $C,E,F$ is unknown.

Victor Ivrii

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Re: HA6-P3
« Reply #4 on: October 29, 2015, 05:34:44 AM »
1) It would be simpler to keep notation $L$ rather than plug $l/2$ instead.

2) Important: After you got this $4\times 4$-system you can by adding and subtraction reduce it to two $2\times 2$ systems (the first with respect to coefficients at $\cosh,\cos$; the second with respect to coefficients at $\sinh,\sin$).

a) The first system has non-trivial solution off its determinant is $0$: $\Delta_1=0$

b) The second system has non-trivial solution off its determinant is $0$: $\Delta_2=0$

One needs to calculate $\Delta_1$ and $\Delta_2$  and write explicit equations, getting two series of eigenvalues corresponding to even and odd eigenfunctions.  There are few more questions to be explored:

c) Can be it that $\Delta_1=\Delta_2=0$? No philosophy, just calculations!

d) As $\Delta_1=0$ can you find corresponding coefficients (up to a common factor which you can take $=1$)?

e) As $\Delta_2=0$ can you find corresponding coefficients (up to a common factor which you can take $=1$)?

However the relationship between C,E,F is unknown

Chi Ma

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Quiz 4 Solution
« Reply #5 on: October 31, 2015, 01:42:46 PM »
Let $y = x - L$. The problem can be written as follows.

\begin{equation}
u_{tt} + Ku_{yyyy} = 0
\end{equation}

\begin{equation}
u(-L,t) = u_y(-L,t) = u(L,t) = u_y(L,t) = 0 \label{BC}
\end{equation}

Let $u(y,t) = Y(y)T(t)$ and apply separation of  variables.
\begin{equation}
\frac{Y^{(4)}(y)}{Y(y)} = \omega^4 \label{y}
\end{equation}

The general solution to (\ref{y}) is
\begin{equation}
Y(y) = A\cosh(\omega y) + B\sinh(\omega y) + C\cos(\omega y) + D\sin(\omega y)
\end{equation}

Boundary conditions (\ref{BC}) imply

\begin{align}
Y(-L) &= A\cosh(\omega L) - B\sinh(\omega L) + C\cos(\omega L) - D\sin(\omega L) = 0\\
Y(L) &= A\cosh(\omega L) + B\sinh(\omega L) + C\cos(\omega L) + D\sin(\omega L) = 0 \\
Y'(-L) &= -A\sinh(\omega L) + B\cosh(\omega L) + C\sin(\omega L) + D\cos(\omega L) = 0 \\
Y'(L) &= A\sinh(\omega L) + B\cosh(\omega L) - C\sin(\omega L) + D\cos(\omega L) = 0
\end{align}

This $4\times4$ system is equivalent to two $2\times2$ systems:

\begin{equation}
 \left[\begin{array}{cc} \sinh(\omega L) & -\sin(\omega L) \\ \cosh(\omega L) & \cos(\omega L) \end{array} \right] \left[\begin{array}{cc} A \\ C \end{array} \right] \equiv \Delta_2 \left[\begin{array}{cc} A \\ C \end{array} \right] = 0
\end{equation}

\begin{equation}
 \left[\begin{array}{cc} \sinh(\omega L) & \sin(\omega L) \\ \cosh(\omega L) & \cos(\omega L) \end{array} \right] \left[\begin{array}{cc} B \\ D \end{array} \right] \equiv \Delta_1 \left[\begin{array}{cc} B \\ D \end{array} \right] = 0
\end{equation}

For any given $\omega \in {\rm I\!R}$, $\Delta_1=\Delta_2=0$ iff $\omega = 0$. Since $0$ is not an eigenvalue, every eigenvalue corresponds to either an even or an odd eigenfunction. For odd eigenfunctions, the coefficients $[B \space D]'$ belong to the null space of $\Delta_1$. For even eigenfunctions, the coefficients $[A \space C]'$ belong to the null space of $\Delta_2$. Because both of these 2 null spaces are 1-dimensional, eigenvalues are simple. All eigenfunctions corresponding to the same eigenvalue are proportional.

A general odd eigenfunction can be written as follows:

\begin{equation}
Y(y) = B\sin(\omega_{odd} L)\sinh(\omega_{odd} y) - B\sinh(\omega_{odd} L) \sin(\omega_{odd} y)
\end{equation}

where $\omega_{odd}$ is a solution to $\tanh(\omega L) = \tan(\omega L)$.


A general even eigenfunction can be written as follows:

\begin{equation}
Y(y) = A\cos(\omega_{even} L)\cosh(\omega_{even} y) - A\cosh(\omega_{even} L) \cos(\omega_{even} y)
\end{equation}

where $\omega_{even}$ is a solution to $\tanh(\omega L) = - \tan(\omega L)$.
« Last Edit: October 31, 2015, 04:24:52 PM by Chi Ma »

Victor Ivrii

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Re: HA6-P3
« Reply #6 on: November 01, 2015, 07:51:40 AM »
Chi Ma solution is correct. You need to select some $B\ne 0$


 Interesting observation: for $\omega_n \gg 1$ we have $\tanh (\omega_n L)\approx 1$ and therefore $\tan (\omega_nL) \approx \pm 1$ and finally (changing notations) $\omega_{m}' \approx \pi/4+\pi m$, $\omega_{m}'' \approx -\pi/4+\pi m$. Also  we can select
\begin{equation}
X_n =\cos (\omega_n} -\cosh(\omega _n x)\cos (\omega_n L)/\cosh (\omega_nL)
\end{equation}
and similarly to odd eigenfunctions and as $|x|< L$ and $\omega_n \gg 1$ the last term is small (not uniformly as $|x|\to L$!) and thus is of  a boundary layer type.

This is an example of asymptotic studies of large eigenvalues!