Differential ecuation of a parabolic antenna It's known that the paralel beams that beat in a parabolic antenna are reflected to a point. Moreover, the angle of incidence is the same as the angle of reflection. I'd like to know if someone can give me the differential ecuation of a parabolic antenna
 A: The equation is the wave equation
$$\partial_{ct}^{2}\vec{\Psi}+\Delta\vec{\Psi}=0$$
The solution is
$$\vec{\Psi}(t, \vec{r})=\int\vec{\Psi}_{i}(\omega, \vec{q})e^{i\omega{t}-\vec{q}\cdot{\vec{r}}}d\omega{d}^{3}q+\int\vec{\Psi}_{s}(\omega, \vec{q})e^{i\omega{t}+\vec{q}\cdot{\vec{r}}}d\omega{d}^{3}q$$
Where the indicies $i, s$ stand for the incident wave and scattered wave respectively. Assume that your paraboloid is given by $\Sigma=\{z=ax^{2}+ay^{2}\}$. Then you take the solution $\Psi$ and apply the boundary conditions at $\Sigma$. Assuming that the paraboloid is a perfect reflector, you would get something like
$$\vec{\Psi}|_{\vec{r}\in\Sigma}=0$$
$$\nabla\wedge\vec{\Psi}|_{\vec{r}\in\Sigma}=0$$
Which will give you the relation between incident and scattered signals. The intensity maxima of the scattered signal will lie at some point on the $z$ axis (no surprize))) the location of this point will be dependent on $a$ (i guess), so you want to adjust $a$ depending on the position of your receiver.
