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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 $B$?

enter image description here

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:

enter image description here enter image description here

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    $\begingroup$ 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? $\endgroup$
    – Oebele
    Mar 13 '15 at 14:19
  • $\begingroup$ 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. $\endgroup$ Mar 13 '15 at 14:22
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    $\begingroup$ Would it be easier to change to a radial coordinate system centered at (-15,0)? Then $$n(r) = e^{\frac{r^2+2}{r^2+1}-1}$$ $\endgroup$
    – Floris
    Mar 13 '15 at 14:32
  • $\begingroup$ 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) $\endgroup$ Mar 13 '15 at 14:37
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    $\begingroup$ 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! $\endgroup$
    – Sofia
    Mar 13 '15 at 14:59
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Yes, change to a radial coordinate system. You basically have an index profile of n(r) = e^(r2-A)/(r2-B). I don't know where you get this type of gradient profile. It is the first time I have seen it. Typically we consider gradient geometries of axial, radial, spherical, parabolic and such. But the path of the light from the center of your right sphere will curve through the sphere and when it exits, it will go in a straight line. I find your figure confusing. You seem to have a point source of light at the center of the left hand circle yet your text says that the light source is at O (center of the right side circle). I do not understand exactly what you are trying to do. Are you saying that your gradient index is inside the circle on the right and there is a point source at the center? Can you please be more specific about the set up. Also, light ALWAYS goes from left to right. It's a Law dude! So when you draw it, your light source should be on the left. It's just convention. Looking forward to more info. I've been working with gradient index glass for 25 years. I developed GRADIUM while at LightPath Technologies. Large scale axial gradient index glass.

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  • $\begingroup$ The red disk on the right hand side is actually the light source. I was interested in the intersection angle of upper and lower light rays (not a point light source, but a disk like the sun)when they met each other on the surface of the left disk. $\endgroup$ Oct 31 '16 at 6:13

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