# Solution of a wave equation

1. Nov 23, 2015

### MarkoA

Hi,

following the attached paper I try to find the general solution of the following wave equation:

$\frac{1}{a^2} \frac{\partial^2 \phi}{\partial t^2} + \frac{2M}{a}\frac{\partial^2 \phi}{\partial x \partial t} + \overline{\beta}^2 \frac{\partial^2 \phi}{\partial x^2} = \frac{\partial^2\phi}{\partial y^2}$ (1)

where $\overline{\beta}^2=M^2-1>0$. Furthermore $M$ is the Mach number and $a$ the sonic speed. For $x$ and $t$ the author states that $-\infty < t < \infty$ and $-\infty < x< \infty$.

The author of the paper finds the general solution to (1) being:
$\phi = \sum_{\nu = -\infty}^{\infty} E_{\nu} e^{i(\omega t + \alpha_{\nu}x - \gamma_{\nu}y)}$, (2)
with
$\gamma_{\nu}^2 = \frac{\omega^2}{a^2} + \frac{2 M \omega \alpha_{\nu}}{a} + (M^2-1)\alpha_{\nu}^2$. (3)

_________________________________________
This is my approach:
In order reduce the equation to an ODE I apply a Fourier transformation in x and t:
$\phi^* = \mathcal{F}(\phi) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi e^{-i(\omega t + \alpha_{\nu}x)}dtdx$ (4)

This yields:
$\frac{\omega^2}{a^2} \phi^* + \frac{2M\omega \alpha_{\nu}}{a}\phi^* + \overline{\beta}^2 \alpha^2_{\nu}\phi^* = \frac{d^2\phi^*}{dy^2}$ (5)

Thus, the characteristic polynomial of (5) is:
$\lambda^2 - \Big(\frac{\omega^2}{a^2} + \frac{2M\omega \alpha_{\nu}}{a} + \overline{\beta}^2 \alpha^2_{\nu}\Big) = 0$ (6).

So the eigenvalue should be:
$\lambda = \gamma_{\nu} = \pm \sqrt{\frac{\omega^2}{a^2} + \frac{2M\omega \alpha_{\nu}}{a} + \overline{\beta}^2 \alpha^2_{\nu}}$ (7)

If the eigenvalues are real and distinct the general solution of the differential equation is:
$\phi^* = C_1 e^{\lambda_1 y} + C_2 e^{\lambda_2 y}$ (8)
If the eigenvalues are complex the general solution should look like:
$\phi^* = C_1 \cos{(\Im{(\lambda)}y)}e^{\Re{(\lambda)} y } + C_2 \sin{(\Im{(\lambda)}y)}e^{\Re{(\lambda)} y }$ (9)

If my approach is not wrong I would apply equation (8). But in the given solution (2) of the paper there is an $i$ I am missing. And do you think the exponentail comes from the inverse Fourier transformation:
$\phi = \mathcal{F}^{-1}(\phi*) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi^* e^{i(\omega t + \alpha_{\nu}x)}d\omega d\alpha$ ?

That's where I get stuck. Do I need some of the boundary conditions the author gives to deduct the general solution?

Any help, ideas and discussions are very appreciated!

I have posted a similar thread before in PhysicsForums, but since this became more or less a list of my own thoughts I decided to start from scratch and to present my latest solution approach.

#### Attached Files:

• ###### Fung1963.pdf
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Last edited: Nov 23, 2015
2. Nov 25, 2015

### Geofleur

Here is an approach that amounts to using Fourier analysis, but is quicker and cleaner than actually carrying out the Fourier integrals. In a great variety of cases, we can gain insight into a PDE by using the ansatz $e^{i(kx - \omega t)}$ or, in two dimensions, $e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)} = e^{i(k_x x + k_y y -\omega t)}$. Let us then, substitute $\phi = e^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)}$ into the PDE.

This trick will turn the PDE into an algebraic equation because, when we differentiate an exponential, we get the exponential back again, times something. In other words, the exponential function is an eigenfunction of the differential operator and, indeed, of all powers of the differential operator. For example, using our ansatz, $\frac{\partial }{\partial t} \phi = (-i \omega) \phi$ and, indeed, $\left( \frac{\partial }{\partial t} \right)^n \phi = (-i\omega)^n \phi$. That means we can simply make the replacement $\left( \frac{\partial }{\partial t} \right)^n \rightarrow (-i\omega)^n$ in the differential equation, as well as similar replacements $\left( \frac{\partial }{\partial x} \right)^n \rightarrow (i k_x)^n$ and $\left( \frac{\partial }{\partial y} \right)^n \rightarrow (i k_y)^n$. Carrying out these replacements and dividing out $\phi$ in all terms gives

$\frac{-\omega^2}{a^2} + \frac{2Mk_x\omega}{a}-\frac{k_x^2}{\beta^2}+k_y^2 = 0$.

This relation tells us how $\omega$ must be related to $\mathbf{k}$. Because the group velocity of a wave is given by $\frac{d \omega}{dk}$ and, because a propagating disturbance disperses (i.e., spreads apart) when $\omega$ is not proportional to $k$, the relationship between $\omega$ and $\mathbf{k}$ is called a dispersion relationship. If we set $k_x = -\alpha_\nu$ and $k_y = \gamma_\nu$, we get the relation (3) from your post. Now the general solution will be a superposition of these basis solutions, each weighted by a different amplitude. This gives rise to equation (2) from your post (though I think, in terms of that equation, our ansatz is really the slightly different one, $e^{i(\omega t - \mathbf{k}\cdot \mathbf{r})}$).

Last edited: Nov 25, 2015
3. Dec 21, 2015

### MarkoA

Due to your help I finally managed to follow this publication. Thanks!!