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) ≡ 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–
