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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.

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  • $\begingroup$ what is $y(x, t)$? $\endgroup$
    – tortue
    Commented Aug 9, 2018 at 10:17
  • $\begingroup$ @pointguard0 sorry, I fixed it. $\endgroup$
    – user572780
    Commented Aug 9, 2018 at 10:22
  • $\begingroup$ the solution will obviously depend on the value of $L$. Just differentiate two more times and plug-in the value of $L$ expression constants $C_1, C_2, C_3, C_4$ through $L$. $\endgroup$
    – tortue
    Commented Aug 9, 2018 at 10:25
  • $\begingroup$ @pointguard0 I have tried that but it does not cancel any terms or simplify the equation. $\endgroup$
    – user572780
    Commented Aug 9, 2018 at 11:22
  • $\begingroup$ The homogeneous system will force constraints on $\epsilon$ and $\delta$, depending on which variable is the eigenvalue. What was the original problem? $\endgroup$
    – Dylan
    Commented Aug 9, 2018 at 11:52

1 Answer 1

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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.

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  • $\begingroup$ $A$ and $B$ are non-zero only if the determinant of the coefficient matrix is $0$? Also, can you explain the last step? $\endgroup$
    – user572780
    Commented Aug 9, 2018 at 16:25
  • $\begingroup$ In terms of linear algebra, if there exists $\vec{u}\ne \vec{0}$ such that $\textbf{M}\vec{u} = \vec{0}$ then it must be true that $\det(M)=0$. Here $\vec{u}=(A,B)$ and $\textbf{M}$ is the coefficient matrix comprised of the terms dependent on $\omega$ and $L$ above. You can also think of it in terms of Kramer's Rule if you're familiar with that. $\endgroup$
    – Dylan
    Commented Aug 9, 2018 at 17:10
  • $\begingroup$ As for the last step, we pick the two constants $A$ and $B$ arbitrarily such that they satisfy one of the two equations above (I picked the first one). The constraint on $\omega$ will ensure that the second equation is automatically satisfied. $\endgroup$
    – Dylan
    Commented Aug 9, 2018 at 17:13
  • $\begingroup$ I was told it is possible to plot $$ \epsilon\delta(\epsilon^4 + \delta^4) + 2\epsilon^3\delta^3\cosh(\epsilon L)\cos(\delta L) + 2\delta^2\epsilon^2(\delta^2-\epsilon^2)\sinh(\epsilon L)\sin(\delta L) = 0 $$ as a function of $\omega$ and find the zeros. How would you do that? $\endgroup$
    – user572780
    Commented Aug 28, 2018 at 10:16
  • $\begingroup$ @user572780 $\epsilon$ and $\delta$ are already functions of $\omega$. You'll need a graphing calculator or a program (MATLAB, Mathematica, etc) to plot it and solve numerically. $\endgroup$
    – Dylan
    Commented Sep 16, 2018 at 6:30

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