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**Home Assignment 4 / Re: Web Bonus Problem--3**

« **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–

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–