# How to calculate desired light path in continuous medium with gradient refraction index

See the Figure below.

$O:(0,0)$ is the disk center of light source $\odot{O}$ with radius $3$.

Then the profile light rays of disk $O$ from the view point $B:(-14,0)$ is defined by segments $DB$ and $EB$ (also the tangent lines of $\odot{O}$ through $B$) when the refraction index is a constant value everywhere.

Now if the refraction index is defined as:

$$n(x,y)=\dfrac{e^{\tfrac{(x+15)^2+y^2+12}{(x+15)^2+y^2+11}}}{e}$$

How to determine the two profile light curves of disk $O$ from the viewpoint $B$?

I tried to establish:

$$\delta\int{n(x,y)}\rm{d}s=0$$ and the second order nonlinear ODE via Euler-Lagrange equation: $$y''(x)=\dfrac{2\left((x+15)y'(x)-y(x)\right)\left(y'(x)^2+1\right)}{\left(y(x)^2+x(x+30)+236\right)^2}$$ but don't know how to establish initial/boundary values of the ordinary differential equation and obtain a symbolic or numerical solution.

Update

Since $y(-14)=0$ is easily available, actually my question is only `how to determine another' $y'(-14)=?$ such that the ODE can be easily solved numerically?

Update

I tried some calculation, and it seems, for any numerical solution $y(x)$, it will be difficult to determine whether it is tangent to the circle $O$ or not:

• It appears to me that you want to know the path along which the light travels from point D to point B, where D is defined as to be on the edge of the red circle and perpendicular to the path, correct? Mar 13, 2015 at 14:19
• Yes exactly. Even when $D$ is a fixed point as defined in the figure, the ODE will become a boundary value problem and is more difficult than the ODE with both initial and boundary value conditions. Mar 13, 2015 at 14:22
• Would it be easier to change to a radial coordinate system centered at (-15,0)? Then $$n(r) = e^{\frac{r^2+2}{r^2+1}-1}$$ Mar 13, 2015 at 14:32
• It seems coordinate transformation does not make it easier; the key issue lies in solving the 2nd order nonlinear ODE obtained (with boundary value conditions) Mar 13, 2015 at 14:37
• Why votes for close? The OP didn't ask to solve his problem, he/she asked for some guidance. Those who voted for close, know how to solve? Then give a hand of help! Mar 13, 2015 at 14:59