I have separated the variables of the following fourth order PDE
$$ \rho A \cdot \frac {\partial^2 y} {\partial t^2} + EI \cdot \frac {\partial^4 y} {\partial x^4} + P \cdot\frac {\partial^2 y} {\partial x^2} = 0 $$
with boundary conditions
$$ y(0,t) = \frac{\partial y}{\partial x}(0,t) = \frac{\partial^2 y}{\partial x^2}(L,t) = \frac{\partial^3 y}{\partial x^3}(L,t) = 0$$
and got
$$y(x,t) = \sum_{n=1}^{\infty} X_n(x) \cdot T_t(n)$$
where
$$X_n(x) = C_1 \cos(\delta x) + C_2 \sin(\delta x) + C_3 \cosh(\epsilon x) + C_4\sinh (\epsilon x)$$
and
$$T_n(t) = D_1 \cos(wt) + D_2 \sin(wt)$$
Using the first two boundary conditions, I got $C_1 = – C_3$ and $\frac{\delta}{\epsilon}C_2 = – C_4$. How can I use the remaining two boundary conditions to simplify the equation further?
$\delta = \pm i \sqrt{s_1+s_2}$, $\epsilon = \pm \sqrt{s_1-s_2}$
$\delta, \epsilon, \omega$ are constants.
Best Answer
For convenience, I'll divide through by the highest-order coefficient, and redefine the constants to the form:
$$ \frac{\partial^4 y}{\partial x^2} + 2\beta\frac{\partial ^2y}{\partial x^2} + \gamma^2 \frac{\partial ^2y}{\partial t^2} = 0 $$
Separating variables gives
$$ \frac{X^{(4)}}{X} + 2\beta\frac{X''}{X} + \gamma^2\frac{T''}{T} = 0 $$
As usual, both $X$ part and the $T$ part have to be equal to constants. Suppose $T''/T = -\omega^2$, then we can separate
\begin{align} T'' + \omega^2T &= 0 \\ X^{(4)} + 2\beta X'' - \gamma^2\omega^2 X &= 0 \end{align}
The $T$ equation has a sinusoidal solution as you described. For the $X$ equation, we have the following characteristic polynomial:
$$ r^4 + 2\beta r^2 - \gamma\omega^2 = 0 $$
which has a solution of the form
$$ r^2 = -\beta \pm \sqrt{\beta^2 + \gamma^2\omega^2} $$
where $\omega$ is the unknown eigenvalue. So the eigenfunction does indeed have the form
$$ X(x) = c_1\cos(\delta x) + c_2\sin(\delta x) + c_3\cosh(\epsilon x) + c_4\sinh(\epsilon x) $$
where
\begin{align} \epsilon &= \sqrt{-\beta + \sqrt{\beta^2+\gamma^2\omega^2}} \\ \delta &= \sqrt{\beta + \sqrt{\beta^2+\gamma^2\omega^2}} \end{align}
Apply the 2 B.C. at $x=0$, we find
$$ X(x) = A\big[\cosh(\epsilon x) - \cos(\delta x)\big] + B\big[\delta\sinh(\epsilon x) - \epsilon\sin(\delta x)\big] $$
The remaining B.C.'s are a bit messy, but we get:
$$\begin{align} A\big[\epsilon^2\cosh(\epsilon L) + \delta^2 \cos(\delta L) \big] + B\big[\delta\epsilon^2\sinh(\epsilon L) + \epsilon\delta^2\sin(\delta L) \big] &= 0 \\ A\big[\epsilon^3\sinh(\epsilon L) - \delta^3\sin(\delta L)\big] + B\big[\delta\epsilon^3\cosh(\epsilon L) + \epsilon\delta^3\cos(\delta L) \big]&= 0 \end{align} \tag{*} $$
Since this is a homogeneous system, it has non-zero solutions only if the determinant of the coefficient matrix is $0$, i.e.
$$ \begin{vmatrix} \epsilon^2\cosh(\epsilon L) + \delta^2 \cos(\delta L) & \delta\epsilon^2\sinh(\epsilon L) + \epsilon\delta^2\sin(\delta L) \\ \epsilon^3\sinh(\epsilon L) - \delta^3\sin(\delta L) & \delta\epsilon^3\cosh(\epsilon L) + \epsilon\delta^3\cos(\delta L) \end{vmatrix} = 0 $$
which I believe simplifies to
$$ \epsilon\delta(\epsilon^4 + \delta^4) + 2\epsilon^3\delta^3\cosh(\epsilon L)\cos(\delta L) + 2\delta^2\epsilon^2(\delta^2-\epsilon^2)\sinh(\epsilon L)\sin(\delta L) = 0 $$
You'll need to solve the above equation numerically to find $\omega$ in terms of $\beta$ and $\gamma$.
Once values of $\omega$ are found, the last step is to pick the arbitrary constants such that one of the two equations in $(*)$ is satisfied. Something like
$$ X(x) = \big[\delta\epsilon^2\sinh(\epsilon L) + \epsilon\delta^2\sin(\delta L) \big]\big[\cosh(\epsilon x) - \cos(\delta x)\big] - \big[\epsilon^2\cosh(\epsilon L) + \delta^2 \cos(\delta L) \big]\big[\delta\sinh(\epsilon x) - \epsilon\sin(\delta x)\big] $$
up to a constant.