Help understanding modal projection in PDE with assumed solution form

[Like Tony Stark]

Homework Statement

Obtain mass and stiffness matrices given equation of motion

Relevant Equations

$$A_{11e} \frac{d^2 u_1}{dx^2} + (A_{12e} + A_{66e}) \frac{d^2 v_1}{dx,dy} + A_{66e} \frac{d^2 u_1}{dy^2} + \frac{G_2}{h_2} \left( u_3 - u_1 - d \frac{dw}{dx} \right) = (\rho_p h_p + \rho_s h_s) \frac{d^2 u_1}{dt^2}$$

Hello,

This is not homework but I am trying to replicate some results I found in a paper. In short, the situation is as follows. The following equation is given:

A_{11e} \frac{d^2 u_1}{dx^2} + (A_{12e} + A_{66e}) \frac{d^2 v_1}{dxdy} + A_{66e} \frac{d^2 u_1}{dy^2} + \frac{G_2}{h_2} \left( u_3 - u_1 - d \frac{dw}{dx} \right) = (\rho_p h_p + \rho_s h_s) \frac{d^2 u_1}{dt^2}

where u_1(x,y), v_1(x,y), u_3(x,y), and w(x,y) are given by:

u_1(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} U_{1mn}(t) \frac{dX(x)}{dx} Y(y)

u_3(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} U_{3mn}(t) \frac{dX(x)}{dx} Y(y)

v_1(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} V_{1mn}(t) X(x) \frac{dY(y)}{dy}

w(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} W_{mn}(t) X(x) Y(y)

with X(x) = \sin\left( \frac{m \pi x}{a} \right) and Y(y) = \sin\left( \frac{n \pi y}{b} \right).
All other quantities are constants.

As far as I understand, the author substitutes the proposed solutions and groups terms according to the temporal variable that accompanies them.
He then rewrites the equation as:

K_{11} U_{1mn} + K_{12} V_{1mn} + K_{13} U_{3mn} + K_{15} W_{mn} + M_{11} \ddot{U}_{1mn} = 0

where (according to him):

K_{11} = -A_{11e} \left( \frac{m \pi}{a} \right)^2 - A_{66e} \left( \frac{n \pi}{b} \right)^2 - \frac{G_2}{h_2}

K_{12} = -(A_{12e} + A_{66e}) \left( \frac{m \pi}{a} \right) \left( \frac{n \pi}{b} \right)

K_{13} = \frac{G_2}{h_2}

K_{15} = -\frac{G_2 d}{h_2} \left( \frac{m \pi}{a} \right)

M_{11} = -(\rho_p h_p + \rho_s h_s)

The problem is that he doesn’t explain how these elements K_{ij} and M_{ij} are obtained.
I assume he multiplied both sides by \frac{dX(x)}{dx} Y(y) and integrated over x \in [0,a] and y \in [0,b], as if applying orthogonality conditions.
However, when I do that, I only recover some of the terms:

K_{11} = -A_{11e} \left( \frac{m \pi}{a} \right)^2 - A_{66e} \left( \frac{n \pi}{b} \right)^2 - \frac{G_2}{h_2}

K_{12} = -(A_{12e} + A_{66e}) \left( \frac{n \pi}{b} \right)^2

K_{13} = \frac{G_2}{h_2}

K_{15} = -\frac{G_2 d}{h_2}

M_{11} = -(\rho_p h_p + \rho_s h_s)

I cross-checked my results with a Mathematica script, and they seem consistent.

Does anyone have an idea of what the author might be doing differently?

Thanks in advance.

 

[pasmith]

Could you identifty exactly which of your results disagree with the author's, and show you arrived at them?

The idea seems to be to arrive at a coefficient of \cos(m\pi x/a)\sin(n\pi y/b). I would suggest using that in your calculation instead of X'(x)Y(y), etc.

 

[Like Tony Stark]

Could you identifty exactly which of your results disagree with the author's, and show you arrived at them?

The idea seems to be to arrive at a coefficient of \cos(m\pi x/a)\sin(n\pi y/b). I would suggest using that in your calculation instead of X'(x)Y(y), etc.

Hi. I'm not sure if I understood your comment correctly, but in my post I wrote both the author's results and mine. We differ in K_{12} and K_{15}.

On the other hand, what you said about using \cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) instead of X'(x)Y(y) is the same, since X(x) = \sin\left(\frac{m\pi x}{a}\right).

 

[TSny]

Does anyone have an idea of what the author might be doing differently?

I get the same results as you. I don’t see how the author could get their result.

Is the paper available online?

 

[Like Tony Stark]

I get the same results as you. I don’t see how the author could get their result.

Is the paper available online?

"Newtonian and Variational Formulations of the Vibrations of Plates With Active Constrained Layer Damping" by Chul H. Park and Amr Baz. See eqs. 21, 35, 36 and appendix.