- #1
MarkoA
- 13
- 1
Hi,
following the attached paper I try to find the general solution of the following wave equation:
[itex]
\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}
[/itex] (1)
where [itex]\overline{\beta}^2=M^2-1>0[/itex]. Furthermore [itex]M[/itex] is the Mach number and [itex]a[/itex] the sonic speed. For [itex]x[/itex] and [itex]t[/itex] the author states that [itex] -\infty < t < \infty[/itex] and [itex] -\infty < x< \infty[/itex].
The author of the paper finds the general solution to (1) being:
[itex]\phi = \sum_{\nu = -\infty}^{\infty} E_{\nu} e^{i(\omega t + \alpha_{\nu}x - \gamma_{\nu}y)}[/itex], (2)
with
[itex]\gamma_{\nu}^2 = \frac{\omega^2}{a^2} + \frac{2 M \omega \alpha_{\nu}}{a} + (M^2-1)\alpha_{\nu}^2[/itex]. (3)
_________________________________________
This is my approach:
In order reduce the equation to an ODE I apply a Fourier transformation in x and t:
[itex] \phi^* = \mathcal{F}(\phi) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi e^{-i(\omega t + \alpha_{\nu}x)}dtdx [/itex] (4)
This yields:
[itex] \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} [/itex] (5)
Thus, the characteristic polynomial of (5) is:
[itex] \lambda^2 - \Big(\frac{\omega^2}{a^2} + \frac{2M\omega \alpha_{\nu}}{a} + \overline{\beta}^2 \alpha^2_{\nu}\Big) = 0 [/itex] (6).
So the eigenvalue should be:
[itex]\lambda = \gamma_{\nu} = \pm \sqrt{\frac{\omega^2}{a^2} + \frac{2M\omega \alpha_{\nu}}{a} + \overline{\beta}^2 \alpha^2_{\nu}}[/itex] (7)
If the eigenvalues are real and distinct the general solution of the differential equation is:
[itex]\phi^* = C_1 e^{\lambda_1 y} + C_2 e^{\lambda_2 y}[/itex] (8)
If the eigenvalues are complex the general solution should look like:
[itex]\phi^* = C_1 \cos{(\Im{(\lambda)}y)}e^{\Re{(\lambda)} y } + C_2 \sin{(\Im{(\lambda)}y)}e^{\Re{(\lambda)} y }[/itex] (9)
If my approach is not wrong I would apply equation (8). But in the given solution (2) of the paper there is an [itex]i[/itex] I am missing. And do you think the exponentail comes from the inverse Fourier transformation:
[itex] \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 [/itex] ?
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.
following the attached paper I try to find the general solution of the following wave equation:
[itex]
\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}
[/itex] (1)
where [itex]\overline{\beta}^2=M^2-1>0[/itex]. Furthermore [itex]M[/itex] is the Mach number and [itex]a[/itex] the sonic speed. For [itex]x[/itex] and [itex]t[/itex] the author states that [itex] -\infty < t < \infty[/itex] and [itex] -\infty < x< \infty[/itex].
The author of the paper finds the general solution to (1) being:
[itex]\phi = \sum_{\nu = -\infty}^{\infty} E_{\nu} e^{i(\omega t + \alpha_{\nu}x - \gamma_{\nu}y)}[/itex], (2)
with
[itex]\gamma_{\nu}^2 = \frac{\omega^2}{a^2} + \frac{2 M \omega \alpha_{\nu}}{a} + (M^2-1)\alpha_{\nu}^2[/itex]. (3)
_________________________________________
This is my approach:
In order reduce the equation to an ODE I apply a Fourier transformation in x and t:
[itex] \phi^* = \mathcal{F}(\phi) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi e^{-i(\omega t + \alpha_{\nu}x)}dtdx [/itex] (4)
This yields:
[itex] \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} [/itex] (5)
Thus, the characteristic polynomial of (5) is:
[itex] \lambda^2 - \Big(\frac{\omega^2}{a^2} + \frac{2M\omega \alpha_{\nu}}{a} + \overline{\beta}^2 \alpha^2_{\nu}\Big) = 0 [/itex] (6).
So the eigenvalue should be:
[itex]\lambda = \gamma_{\nu} = \pm \sqrt{\frac{\omega^2}{a^2} + \frac{2M\omega \alpha_{\nu}}{a} + \overline{\beta}^2 \alpha^2_{\nu}}[/itex] (7)
If the eigenvalues are real and distinct the general solution of the differential equation is:
[itex]\phi^* = C_1 e^{\lambda_1 y} + C_2 e^{\lambda_2 y}[/itex] (8)
If the eigenvalues are complex the general solution should look like:
[itex]\phi^* = C_1 \cos{(\Im{(\lambda)}y)}e^{\Re{(\lambda)} y } + C_2 \sin{(\Im{(\lambda)}y)}e^{\Re{(\lambda)} y }[/itex] (9)
If my approach is not wrong I would apply equation (8). But in the given solution (2) of the paper there is an [itex]i[/itex] I am missing. And do you think the exponentail comes from the inverse Fourier transformation:
[itex] \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 [/itex] ?
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.
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