I need to find a differential equation for a light ray in moving through a region with a non-discrete index of refraction. It's a problem about mirages and how the light ray curves downward towards the hot asphalt to minimize its flight time and then curves back up.
Given $n=n_0(1+\alpha y)$ and the speed of light in a medium as $v=c/n$, find the differential equation for the path $y(x)$ that the ray of light will take between the points $(x_1,y_1)$ and $(x_2,y_2)$.
My attempt:
$$T_{min}=\int_A^B \frac{ds}{v}=\int_{y_1}^{y_2}\frac{\sqrt{dx^2+dy^2}}{c/n_0(1+\alpha y)}$$
$$=\int_{y_1}^{y_2}\frac{n_0(1+\alpha y)\sqrt{(dx/dy)^2+1}}{c}dy$$
Therefore, $f=f(x,x',y)=\frac{n_0(1+\alpha y)\sqrt{(dx/dy)^2+1}}{c}=\frac{n_0(1+\alpha y)\sqrt{(x')^2+1}}{c}$
Now we use the Euler-Lagrange equation: $\frac{\partial f}{\partial x}=\frac{d}{dy}\frac{\partial f}{\partial x'}$
$$\frac{\partial f}{\partial x}=0$$
$$\frac{\partial f}{\partial x'}=\frac{x'n_0(1+\alpha y)}{(x'^2+1)^{1/2}}$$
Because $\frac{\partial f}{\partial x}=0$, $\frac{\partial f}{\partial x'}$ must be a constant, so:
$$\frac{x'n_0(1+\alpha y)}{(x'^2+1)^{1/2}}=const.=b$$
Solving for $x'$, we get:
$$x'=\sqrt{\frac{b}{n_0^2(1+\alpha y)^2-b}}$$
Now I'm pretty confused. I was following an example in the book (it was a time-minimization problem, but it wasn't about light/indexes of refraction) but it has led me to find $x'$. If I integrate this, I get this horrible mess: https://www.wolframalpha.com/input/?i=integral+of+sqrt%5Bb%2F(t%5E2(1%2Bay)%5E2-b)%5D+dy. Furthermore, I need to show that the curve is a catenary. Does anyone know what I might have done wrong?
The equation you derive is correct. Sometimes integrates are a bit opaque and you have to do work by hand. Try a $\cosh (t) = \frac{1+ay}{b}$ substitution.
For catenary relationship, compare your solution to it or relate the diff eqs .