0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ what is $y(x, t)$? $\endgroup$ – pointguard0 Aug 9 '18 at 10:17
  • $\begingroup$ @pointguard0 sorry, I fixed it. $\endgroup$ – user572780 Aug 9 '18 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 $L$. $\endgroup$ – pointguard0 Aug 9 '18 at 10:25
  • $\begingroup$ @pointguard0 I have tried that but it does not cancel any terms or simplify the equation. $\endgroup$ – user572780 Aug 9 '18 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 Aug 9 '18 at 11:52
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $A$ and $B$ are non-zero only if the determinant of the coefficient matrix is $0$? Also, can you explain the last step? $\endgroup$ – user572780 Aug 9 '18 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 Aug 9 '18 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 Aug 9 '18 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 Aug 28 '18 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 Sep 16 '18 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.