Ytukyg

I have Partial Differential equation in the form:

$$ frac{partial^2 y}{partial t^2} + frac{partial^4 y}{partial x^4} =0, quad 0<x<1 $$

Vibrating Beam

Boundary Conditions:
$$ y(0,t) = frac{partial y}{partial x}(0,t) = 0 \ y(1,t) = frac{partial^2 y}{partial x^2}(1,t) = 0 $$

Initial conditions:

$$y(x,0) = 0.1sin(pi x), frac{partial y}{partial t}(x,0) = 0$$

I have come up to the point where I have solved the boundary conditions and got the four equations:

$A+C=0$

$B+D=0$

$Asinh(beta) + Bcosh(beta) + Ccos(beta) + Dsin(beta) = 0 \$

$Asinh(beta)+Bcosh(beta)-Csin(beta)-Dcos(beta) = 0$

Not sure what to do next. I am supposed to find the Eigenvalues and Eigenfunctions.

We have $C=-A$ and $D=-B$, so

$$ A(sinh beta - cos beta) + B(cosh beta - sin beta) = 0 $$

$$ A(sinh beta + sin beta) + B(cosh beta + cos beta) = 0 $$

This guarantees a solution if the determinant of the coefficient matrix is zero

$$left|begin{matrix} sinhbeta - cosbeta && coshbeta -sinbeta \
sinhbeta + sinbeta && coshbeta + cosbeta end{matrix}right| = 0 $$

which, after some manipulation, gives
$$ e^{-beta} = sinbeta - cos beta = sqrt{2}sinleft(beta - frac{pi}{4}right) $$

You can solve this graphically as the intersection of the two functions. The first eigenvalue is $beta_0 approx 1.038415$. Since $e^{-beta}$ decays rapidly, the larger solutions are approximately $beta_n sim (n+1/4)pi $

EDIT: I'm not sure if your equations are actually correct. I tried solving for the eigenfunction and got
$$ X(x) = Acosh beta x + Bsinh beta x + Ccosbeta x + Dsinbeta x $$

And the four boundary conditions give
$$ X(0) = A + C = 0 $$
$$ X'(0) = B + D = 0 $$
$$ X(1) = Acoshbeta + Bsinhbeta + Ccosbeta + Dsinbeta = 0 $$
$$ X''(1) = Acoshbeta + Bsinhbeta - Ccosbeta - Dsinbeta = 0 $$

In this case, the eigenvalues satisfy
$$left|begin{matrix} coshbeta - cosbeta && sinhbeta -sinbeta \
coshbeta + cosbeta && sinhbeta + sinbeta end{matrix}right| = 0 $$
or
$$ tanh beta = tanbeta $$

Since $tanhbeta$ very quickly goes to $1$, $beta_n sim (n + 1/4)pi, n ge 1$. This is a very good approximation; the first zero is $beta_1 approx 3.9266 = 5pi/4 - 0.0004 $

EDIT 2: For the eigenfunctions, we have
$$ X(x) = A(coshbeta x - cosbeta x) + B(sinhbeta x - sinbeta x) $$

where $(A,B)$ satisfy
$$ A(coshbeta-cosbeta) + B(sinhbeta-sinbeta) = 0 $$
$$ A(coshbeta+cosbeta) + B(sinhbeta+sinbeta) = 0 $$

Note that these are actually the same equation (since the determinant is zero), so we can just pick any arbitrary pair that satisfies one of the above. For example, let $A = sinhbeta-sinbeta$ and $-B = coshbeta-cosbeta$, then the eigenfunction is (up to a constant)
$$ X_n(x) = (sinhbeta_n-sinbeta_n)(coshbeta_n x - cosbeta_n x) - (coshbeta_n-cosbeta_n)(sinhbeta_n x - sinbeta_n x) $$

To show these functions are mutually orthogonal, we use the definition of orthogonality
$$ int_0^1 X_n(x) X_m(x) dx = 0, m ne n $$

The integral is a bit messy, but it should reduce down to some expression in terms of the determinant.

EDIT 3: The B.C.s are $X(0) = X(1) = X'(0) = X''(1) = 0$. Using integration by parts, we can show that

begin{align}
int_0^1 {X_m}^{(4)} X_n &= {X_m}'''X_nBigg|_0^1 - int_0^1 {X_m}'''{X_n}', && X_n(0)=X_n(1)=0 \
&= -{X_m}''{X_n}'Bigg|_0^1 + int_0^1 {X_m}''{X_n}'', && {X_m}''(1) = {X_n}'(0) = 0 \
&= {X_m}'{X_n}''Bigg|_0^1 - int_0^1 {X_m}'{X_n}''', && {X_m}'(0) = {X_n}''(1) = 0 \
&= -X_m{X_n}^{(4)}Bigg|_0^1 + int_0^1 X_m{X_n}^{(4)}, && X_m(0) = X_n(1) = 0
end{align}

$$ implies int_0^1 {X_m}^{(4)}X_n - int_0^1 X_m{X_n}^{(4)} = 0 implies (beta_m^4 - beta_n^4) int_0^1 X_mX_n = 0 $$

since $beta_m ne beta_n$, the integral must be $0$