Author Topic: HA4 problem 2  (Read 3988 times)

Victor Ivrii

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HA4 problem 2
« on: February 12, 2015, 05:22:39 PM »
Oscillations of the beam which with the left end having fixed positions
and fixed directions (bricked into the walls: it is called ''clamped'')
and the free right end are described by an equation

u_{tt} + K u_{xxxx}=0, \qquad 0<x< l,
\label{eq-4.8}

with $K\>0$ and the boundary conditions

u(0,t)=u_{x}(0,t)=u_{xx}(l,t)=u_{xxx}(l,t)=0.
\label{eq-4.9}

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 (see (\ref{eq-4.7})).

d.  Bonus Prove that eigenvalues are simple, i.e. all eigenfunctions corresponding to the same eigenvalue are proportional.

Chaojie Li

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Re: HA4 problem 2
« Reply #1 on: March 26, 2015, 08:51:16 PM »
First separate variables. Let: $\Big {[}u(x,t) = X(x)T(t)\Big {]}$ in $\Big {[}u_{tt}(x,t) + K u_{xxxx} = 0, K > 0\Big {]}$

Then:$$\Big {[}u_{tt}(x,t) = X(x)T(t), u_{xxxx}(x,t) = X(x)T(t)\Big {]}$$
$$\Big {[}u_{tt}(x,t) + K u_{xxxx} = X(x)T(t) + K X(x)T(t) = 0\Big {]}$$
$$\Big {[}\frac{X(x)}{X(x)} = \frac{-T(t)}{K T(t)} = \lambda\Big {]}$$

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

So we are left with two ODEs in the form: $$\Big {[}X(x) = c^4 X(x), T(t) = -c^4 k^2 T(t)\Big {]}$$

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

Using the two boundary conditions: $\Big {[}u(0,t) = 0 = u_x(0,t)\Big {]}$

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

As we disregard the case where: $$\Big {[}X(0) \neq 0, X(x) \neq 0 \implies T(t)=0\Big {]}$$

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

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

$$X(l) = A c^2 \cosh(c l) + B c^2 \sinh(c l) - C c^2 \cos(c l) - D c^2 \sin(c l)$$
$$=A c^2 \cosh(c l) + B c^2 \sinh(c l) + A c^2 \cos(c l) + B c^2 \sin(c l)$$$$= A c^2 (\cosh(c l) + \cos(c l)) + B c^2 (\sinh(c l) + \sin(c l)) = 0$$

$$\Big {[}X(l) = A c^3 (\sinh(c l) - \sin(c l)) + B c^3 (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
Yielding:
$$\Big {[}A (\cosh(c l) + \cos(c l)) + B (\sinh(c l) + \sin(c l)) = 0\Big {]}$$
$$\Big {[}A (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
$$\Big {[}A = -B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))}\Big {]}$$
$$\Big {[}-B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
$$\Big {[}-\frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l) + (\cosh(c l) + \cos(c l)))\Big {]}$$
$$\Big {[}-\sinh(c l)^2 + \sin(c l)^2 + \cosh(c l)^2 + 2 \cosh(c l) \cos(c l) + \cos(c l)^2 =0\Big {]}$$

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

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

So our eigenvalues are those c which satisfy$\Big {[}\cosh(c l) \cos(c l) = -1 \blacksquare$

Next consider: $$\Big {[}\int_0^l X_n(x)X_m(x) dx, n \neq m, \lambda_n \neq \lambda_m\Big {]}$$

Then $$\Big {[}(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = \int_0^l \lambda_n X_n(x)X_m(x) - \lambda_m X_n(x)X_m(x) dx\Big {]}$$
$$\Big {[}\int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx = (X_n(x)X_m(x) - X_n(x)X_m(x))_{x=(0,l)} - \int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx\Big {]}$$
$$\Big {[}= 0 - (X_n(x)X_m(x) - X_n(x)X_m(x))_{x=(0,l)} + \int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx = 0 + 0 = 0\Big {]}$$

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

$$\Big {[}X(0) = 0, X(0) = 0, X(l) = 0, X(l) = 0\Big {]}$$

Then we have $$\Big {[}(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = 0, \lambda_n \neq \lambda_m \neq 0 \implies \int_0^l X_n(x)X_m(x) dx = 0\Big {]}$$

And our eigenfunctions are orthogonal.

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

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

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

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.
« Last Edit: March 30, 2015, 10:41:47 PM by Chaojie Li »

Victor Ivrii

You definitely miss some derivatives at $T$ and $X$. Please correct. Also no need to put formulae in the large square brackets, it looks funny. Should be like
$$u_{tt}(x,t) = X(x)T''(t),\qquad u_{xxxx}(x,t) = X^{(4)}(x)T(t)$$