### Author Topic: Web Bonus Problem--3  (Read 11331 times)

#### Victor Ivrii

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##### Web Bonus Problem--3
« on: October 26, 2012, 09:14:14 AM »
Oscillations of the beam  (with left end clamped and right end free) 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_{xx}(l,t)=u_{xxx}(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.

Compare with eigenvalues of Problem 2 of HA2

#### Calvin Arnott

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• OK
##### Re: Web Bonus Problem--3
« Reply #1 on: October 27, 2012, 01:12:11 PM »
OK, well, I ended up making this way more detailed than I had planned. We first separate variables. Let: [u(x,t) = X(x)T(t)] in [u_{tt}(x,t) + K u_{xxxx} = 0,   K > 0]

Then: [u_{tt}(x,t) = X(x)T''(t),  u_{xxxx}(x,t) = X''''(x)T(t)]
[u_{tt}(x,t) + K u_{xxxx} = X(x)T''(t) + K X''''(x)T(t) = 0]
[\frac{X''''(x)}{X(x)} = \frac{-T''(t)}{K T(t)} = \lambda]

For some constant, as both sides are independant of the other respective variable. Now, we have by assumptions that [\lambda = c^4 > 0,   K = k^2 > 0]

So we are left with two ODE's in the form: [X''''(x) = c^4 X(x),  T''(t) = -c^4 k^2 T(t)]

Which yield solutions: [X(x) = A \cosh(c x) + B \sinh(c x) + C \cos(c x) + D \sin(c x)]
[T(t) = A \cos(c^2 k t) + B \sin(c^2 k t)]

Note that correct notation is \cos which produces $\cos$ etc

Using the two boundary conditions: [u(0,t) = 0 = u_x(0,t)]

[X(0) = A \cosh(c 0) + B \sinh(c 0) + C \cos(c 0) + D \sin(c 0) = A + C = 0,   A = -C]
[X'(0) = A c \sinh(c 0) + B c \cosh(c 0) - C c \sin(c 0) + D c \cos(c 0) = B c + D c,   D = -B]

As we disregard the case where: [X(0) \ne 0,   X'(x) \ne 0 \implies T(t) â‰¡ 0]

OK, great. We plug into the 3rd and 4th boundary conditions.

[u_{xx}(l,t) = 0 = u_{xxx}(l,t)]

\begin{multline*}
X''(l) = A c^2 \cosh(c l) + B c^2 \sinh(c l) - C c^2 \cos(c l) - D c^2 \sin(c l) =\\
A c^2 \cosh(c l) + B c^2 \sinh(c l) + A c^2 \cos(c l) + B c^2 \sin(c l) = A c^2 (\cosh(c l) + \cos(c l)) + B c^2 (\sinh(c l) + \sin(c l)) = 0
\end{multline*}
There is \multline
[X'''(l) = A c^3 (\sinh(c l) - \sin(c l)) + B c^3 (\cosh(c l) + \cos(c l)) = 0]

Yielding: [A (\cosh(c l) + \cos(c l)) + B (\sinh(c l) + \sin(c l)) = 0]
[A (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0]
[A = -B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))}]
[-B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0]
[-\frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l) + (\cosh(c l) + \cos(c l)))]
[-\sinh(c l)^2 + \sin(c l)^2 + \cosh(c l)^2 + 2 \cosh(c l) \cos(c l) + \cos(c l)^2 =0]

Where we used the facts that [B = 0 \implies X(x) â‰¡ 0,   (\cosh(c l) + \cos(c l)) = 0 \implies B = A = 0]
Which we disregard as we're not so interested in the trivial case.
[ \cosh(c l)^2 - \sinh(c l)^2 + \sin(c l)^2+ \cos(c l)^2 + 2 \cosh(c l) \cos(c l)  = 1 + 1 + 2 \cosh(c l) \cos(c l) = 0]

As: [\cosh^2 - \sinh^2 = 1,   \cos^2 + \sin^2 = 1]

So our eigenvalues are those c which satisfy [\cosh(c l) \cos(c l) = -1 \blacksquare]

Next consider: [\int_0^l X_n(x)X_m(x) dx,   n \ne m,   \lambda_n \ne \lambda_m]

Then [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = \int_0^l \lambda_n X_n(x)X_m(x) -  \lambda_m X_n(x)X_m(x) dx]
[\int_0^l X''''_n(x)X_m(x) - X_n(x)X''''_m(x) dx = (X'''_n(x)X_m(x) - X_n(x)X'''_m(x))_{x=(0,l)} - \int_0^l X'''_n(x)X'_m(x) - X'_n(x)X'''_m(x) dx]
[= 0 - (X''_n(x)X'_m(x) - X_n'(x)X''_m(x))_{x=(0,l)} + \int_0^l X''_n(x)X''_m(x) - X''_n(x)X''_m(x) dx = 0 + 0 = 0]

Using integration by parts, and the fact that our boundary conditions vanish at:

[X(0) = 0, X'(0) = 0, X''(l) = 0, X'''(l) = 0]

Then we have [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = 0,     \lambda_n \ne \lambda_m \ne 0 \implies \int_0^l X_n(x)X_m(x) dx = 0]

And our eigenfunctions are orthogonal â– .

Let our differential operator be [\mathcal{I}, st.  \mathcal{I} X = \lambda X,   \mathcal{I} Y = \lambda Y]

We show that: [\langle \mathcal{I}X,Y\rangle = \langle X,\mathcal{I}^*Y\rangle]

[\langle \mathcal{I}X,Y\rangle = \int_0^l \mathcal{I}X(x)Y(x)^* dx = \int_0^l \lambda X(x)Y(x)^* dx = \int_0^l X(x)\lambda^* Y(x)^* dx = \int_0^l X(x)\mathcal{I}^*Y(x)^* dx = \langle X,\mathcal{I}^*Y\rangle]

As we had by assumption our eigenvalues were real. Then our operator is hermetian, so the eigenfuncitons of different eigenvalues are linearly independant, and the eigenfunctions of the same eigenvalue are linearly dependant.â–
« Last Edit: October 27, 2012, 02:10:00 PM by Calvin Arnott »

#### Victor Ivrii

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##### Re: Web Bonus Problem--3
« Reply #2 on: October 27, 2012, 01:22:56 PM »
Good!

What we got in the case of both clamped ends?
« Last Edit: October 27, 2012, 01:24:35 PM by Victor Ivrii »

#### Calvin Arnott

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• OK
##### Re: Web Bonus Problem--3
« Reply #3 on: October 27, 2012, 02:06:11 PM »
In the clamped case, things get a lot easier! Some modifications to the previous post:

Let: [u(x,t) = X(x)T(t)] in [u_{tt}(x,t) + K u_{xxxx} = 0,   K > 0,    u(0,t) = u_{xx}(0,t)=u(l,t)=u_{xx}(l,t)=0]

Then: [u_{tt}(x,t) = X(x)T''(t),  u_{xxxx}(x,t) = X''''(x)T(t)]
[u_{tt}(x,t) + K u_{xxxx} = X(x)T''(t) + K X''''(x)T(t) = 0]
[\frac{X''''(x)}{X(x)} = \frac{-T''(t)}{K T(t)} = \lambda]

For some constant, as both sides are independant of the other respective variable. Now, we have by assumptions that [\lambda = c^4 > 0,   K = k^2 > 0]

So we are left with two ODE's in the form: [X''''(x) = c^4 X(x),  T''(t) = -c^4 k^2 T(t)]

Which yield solutions: [X(x) = A \cosh(c x) + B \sinh(c x) + C \cos(c x) + D \sin(c x)]
[T(t) = A \cos(c^2 k t) + B \sin(c^2 k t)]

Using the two boundary conditions: [u(0,t) = 0 = u_{xx}(0,t)]

[X(0) = A \cosh(c 0) + B \sinh(c 0) + C \cos(c 0) + D \sin(c 0) = A + C = 0,   A = -C]
[X''(0) = A c^2 \cosh(c 0) + B c^2 \sinh(c 0) - C c^2 \cos(c 0) - D c^2 \sin(c 0) = A c^2 + D c^2,   A = C]
[A = C = -C \implies A = C = 0]

As we disregard the case where: [X(0) \ne 0,   X''(x) \ne 0 \implies T(t) â‰¡ 0]

OK, nice. We plug into the 3rd and 4th boundary conditions, using our new function for X:

[X(x) = B \sinh(c x) + D \sin(c x)]

[u(l,t) = 0 = u_{xx}(l,t)]

[X(l) = B \sinh(c l) + D \sin(c l) = 0]
[X''(l) = B c^2 \sinh(c l) - c^2 D\sin(c l) = B \sinh(c l) - D\sin(c l) = 0]

As c != 0. Adding equations gives us that: [B \sinh(c l) + D \sin(c l) + B \sinh(c l) - D\sin(c l) = 2 B \sinh(c l) = 0]

Now, for real x, sinh(x) has a root only at c = 0, which would yield an eigenvalue of 0, which we disregard by assumption. So:

[2 B \sinh(c l) = B \sinh(c l) = 0 \implies B = 0]

Now we cannot have D = 0, or our solution is trivial. So, resubstituting for B, we're left with only:

[B \sinh(c l) + D \sin(c l) = D \sin(c l) = \sin(c l) = 0]

And our eigenvalues are those c which satisfy [\sin(c l) = 0 \blacksquare]

Which we can provide a closed form for in this case since our solution is so nice:
[\sin(c l) = 0 \implies c = \frac{\pi n}{l} \implies \lambda = \frac{\pi ^4 n^4}{l^4}]

These next two sections need basically no modification, as only which term vanishes in our boundary case during integration by parts changes.

[\int_0^l X_n(x)X_m(x) dx,   n \ne m,   \lambda_n \ne \lambda_m]

Then [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = \int_0^l \lambda_n X_n(x)X_m(x) -  \lambda_m X_n(x)X_m(x) dx]
[\int_0^l X''''_n(x)X_m(x) - X_n(x)X''''_m(x) dx = (X'''_n(x)X_m(x) - X_n(x)X'''_m(x))_{x=(0,l)} - \int_0^l X'''_n(x)X'_m(x) - X'_n(x)X'''_m(x) dx]
[= 0 - (X''_n(x)X'_m(x) - X_n'(x)X''_m(x))_{x=(0,l)} + \int_0^l X''_n(x)X''_m(x) - X''_n(x)X''_m(x) dx = 0 + 0 = 0]

Using integration by parts, and the fact that our boundary conditions vanish at:

[X(0) = 0, X''(0) = 0, X(l) = 0, X''(l) = 0]

Then we have [(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = 0,     \lambda_n \ne \lambda_m \ne 0 \implies \int_0^l X_n(x)X_m(x) dx = 0]

And our eigenfunctions are orthogonal â– .

Let our differential operator be [\mathcal{I}, st.  \mathcal{I} X = \lambda X,   \mathcal{I} Y = \lambda Y]

We show that: [\langle \mathcal{I}X,Y\rangle = \langle X,\mathcal{I}^*Y\rangle]

[\langle \mathcal{I}X,Y\rangle = \int_0^l \mathcal{I}X(x)Y(x)^* dx = \int_0^l \lambda X(x)Y(x)^* dx = \int_0^l X(x)\lambda^* Y(x)^* dx = \int_0^l X(x)\mathcal{I}^*Y(x)^* dx = \langle X,\mathcal{I}^*Y\rangle]

As we had by assumption our eigenvalues were real. Then our operator is hermetian, so the eigenfuncitons of different eigenvalues are linearly independant, and the eigenfunctions of the same eigenvalue are linearly dependant.â–
« Last Edit: October 27, 2012, 07:59:34 PM by Calvin Arnott »

#### Victor Ivrii

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##### Re: Web Bonus Problem--3
« Reply #4 on: October 28, 2012, 09:06:27 AM »
Actually clamped both ends and clamped left-free right end are twins! From general solution
$$X= A\cosh (cx)+ B\cos (cx) +C\sinh (cx) +D\sin (cx)$$
to satisfy conditions on the left end we conclude that $B=-A$, $D=-C$ (really, $X(0)=A+B$ and $c^{-1}X'(0)=C+D$) and therefore
$$X= A\bigl(\cosh (cx)-\cos (cx)\bigr) +C\bigl(\sinh (cx) - \sin (cx) \bigr))$$
and
$$c^{-1}X'= A\bigl(\sinh (cx)+\sin (cx)\bigr) +C\bigl(\cosh (cx) - \cos (cx) \bigr))$$
and taking $x=l$ we get for $X(l)=X'(l)=0$
$$\left| \begin{matrix} \cosh (cl) - \cos (cl) & \sinh (cl) - \sin (cl)\\ \sinh (cl) + \sin (cl) & \cosh (cl) - \cos (cl) \end{matrix}\right|=0$$
which rewrites as
$$[\cosh (cl) - \cos (cl)]^2 - \sinh^2 (cl) + \sin^2 (cl) = 2-2\cosh (cl) \cos (cl)=0$$
and for $X''(l)=X'''(l)=0$ we get the same but with $\cos$ and $\sin$ in the both lines with opposite signs
$$\left| \begin{matrix} \cosh (cl) +\cos (cl) & \sinh (cl) + \sin (cl)\\ \sinh (cl) - \sin (cl) & \cosh (cl) + \cos (cl) \end{matrix}\right|=0$$
which rewrites as
$$[\cosh (cl) + \cos (cl)]^2 - \sinh^2 (cl) + \sin^2 (cl) = 2+2\cosh (cl) \cos (cl)=0.$$
We used $\cosh^2-\sinh^2=\cos^2+\sin^2=1$.

Now solve these equations graphically. To do so we plot $1/\cosh(x)$ in blue, $-1/\cosh(x)$ in cyan and $\cos(x)$ in red and compare intersections; solutions of $\cos(x)=1/\cosh(x)$ we denote by $\alpha_n$ and
solutions of $\cos(x)=-1/\cosh(x)$ we denote by $\beta_n$. Then one can see easily that $\alpha_n$  and  $\beta_{n+1}$ are very close to $(n+\frac{3}{2})\pi$ and there is $\beta_1$ between $\pi/2$ and $\pi$.

Wolframalpha gives
$\beta_1 \approx 1,875$,
$\beta_2 \approx 4,6941$,
$\beta_3 \approx 7,8548$

$\alpha_1\approx 4.730$,
$\alpha_2\approx 7,8532$.

« Last Edit: October 28, 2012, 09:08:42 AM by Victor Ivrii »

#### Victor Ivrii

Comparing $\alpha_1$ and $\beta_1$ (and recalling that frequencies are proportional to $c^2$ we conclude that the main frequency of beam clamped at both ends is $\approx 6$ times greater than of beam free on one end.
On the other hand, if $\gamma_n$ denotes $\pi n$ (beam supported but not clamped on both ends) we see that $\alpha_1\approx 3\gamma_1/2$ and therefore corresponding frequencies relate as $\approx 9/4$.