Solution of unsteady linearized potential flow PDE

[MarkoA]

Hi,

I have a problem following the solution of a linearized potential flow equation in a publication by Fung.

The problem describes potential flow over an oscillating plate. A boundary layer is approximated by defining a subsonic layer over the panel and supersonic flow above the subsonic flow. From the equation of motion (1) and (2) in combination with a standing wave condition of the wall (8) and traveling wave of the perturbations (9) and (10) it seems to be easy to get the solutions (12) and (13).

https://dl.dropboxusercontent.com/u/20358584/fung1.png
https://dl.dropboxusercontent.com/u/20358584/fung2.png

Can anyobody give me hint of how to get to this solution? The paper is the following:
[PLAIN]http://arc.aiaa.org/doi/abs/10.2514/3.1661[/PLAIN]
http://arc.aiaa.org/doi/abs/10.2514/3.1661

Many thanks in advance!

https://dl.dropboxusercontent.com/u/20358584/fung3.png

 

[Andy Resnick]

When I am confronted with "it's easy to see that..." I usually first try substituting the answer into the expression and seeing what happens- sometimes there's an oddball change of variables or trig identity involved.

 

[MarkoA]

I don't know. This doesn't help. What could he have done? I've heard that Duhamel's principle could be an approach of solving non-homogenious PDEs like the wave equation. Could the solution have something to do with this approach?

Substituting (13) in (2) gives:
$\begin{equation} [-\frac{1}{a_{\delta}^2} \omega^2 - \frac{2M_{\delta}}{a_{\delta}} \alpha_{\nu}\omega + \beta_{\delta}^2\alpha_{\nu}^2 + \zeta_{\delta}^2] \cdot e^{i(\omega t + \alpha_{\nu} x)} \cdot [C_{\nu} sin(\zeta_{\nu}y ) + D_{\nu} cos(\zeta_{\nu}y) ] = 0 \end{equation}$

 

[MarkoA]

Oh... the substitution was absolutely wrong. I need to find the correlation between the potential and z..

 

[MarkoA]

I made some progress to get equation (14) and (15). Not sure if I can already corellate the fourer constant alpha with the wave number.

$$The equation of motion:<br /> \begin{equation}<br /> \frac{1}{a^2}\frac{\partial^2\phi}{\partial t^2} + \frac{2M}{a} \frac{\partial^2 \phi}{\partial x \partial t} + \overline{\beta}^2 \frac{\partial^2 \phi}{\partial x^2} = \frac{\partial^2 \phi}{\partial y^2}<br /> \label{eq:01}<br /> \end{equation}<br /> The potential must oscillate harmonically:<br /> \begin{equation}<br /> \phi = \Psi(x,y)e^{i\omega t}<br /> \label{eq:02}<br /> \end{equation}<br /> This yields:<br /> \begin{equation}<br /> \Big(\frac{i\omega}{a}\Big)^2\Psi + 2i\frac{\omega M}{a} \frac{\partial \Psi}{\partial x} + \overline{\beta}^2 \frac{\partial^2 \Psi}{\partial x^2} = \frac{\partial^2 \Psi}{\partial y^2}<br /> \label{eq:03}<br /> \end{equation}<br /> A double Fourier transformation in x and y:<br /> \begin{equation}<br /> \Psi^* = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-i(\gamma y + \alpha x)} \Psi(x,y) dx dy<br /> \label{eq:04}<br /> \end{equation}<br /> If this Fourier transformation is applied to all terms of Eq.~(\ref{eq:03}) then \Phi^* cancels out and (\ref{eq:03}) can be written as:<br /> \begin{equation}<br /> \frac{\omega^2}{a^2} + 2 \frac{\omega M}{a} \alpha + \overline{\beta}^2 \alpha^2 = \gamma^2<br /> \end{equation}<br /> This is equation (14) in the publication from Fung. The same approach for Fungs equation (2) yields (15). Can I already assume that \gamma is \gamma_{\nu} and \alpha is \alpha_{\nu}?$$