# Fermat's Principle

1. Jun 6, 2009

### cscott

1. The problem statement, all variables and given/known data

Need to minimize $$\int_{(x_1,y_1)}^{(x_2,y_2)} n(x,y)~ds$$ where $$n(x,y)=e^y$$ and $$(x_1,y_1)=(-1,1)$$, $$(x_2,y_2)=(1,1)$$.

2. Relevant equations

Euler-Lagrange equation

3. The attempt at a solution

$$\frac{d}{dx}\frac{dF}{dy'} - \frac{dF}{dy}=0$$

$$0 - e^y y' = 0$$

y' = 0 so y = constant or y = 1 considering the initial conditions?

Last edited: Jun 6, 2009
2. Jun 6, 2009

### Dick

The independent variable here is time, s is arclength and ds=sqrt(x'^2+y'^2)*dt. So your F is not independent of y'=dy/dt.

3. Jun 7, 2009

### cscott

Oh ok, thanks. So,

$$\frac{dF}{dy} = e^y \dot{y} (\dot{x}^2+\dot{y}^2)^{1/2}$$

$$\frac{dF}{d\dot{y}} = e^y \dot{y} (\dot{x}^2+\dot{y}^2)^{-1/2}$$

But then is $$\frac{d}{dx}\frac{dF}{d\dot{y}} = 0$$?

4. Jun 7, 2009

### Dick

Thinking about it, I think what they actually want you to do is take x to be the parameter instead of t. ds=sqrt((dx/dt)^2+(dy/dt)^2)*dt=sqrt(1+(dy/dx)^2)*(dx/dt)*dt=sqrt(1+(dy/dx)^2)*dx. Doing it that way is much easier. So F=e^y*sqrt(1+y'^2) where y'=dy/dx. Try finding dF/dy, dF/dy' from that. And why would you think d/dx(dF/dy')=0? It's not a partial derivative.

5. Jun 7, 2009

### cscott

$$\frac{\partial F}{\partial y'}=e^y y'(1+y'^2)^{-1/2}$$

$$\frac{\partial F}{\partial y}=e^y (1+y'^2)^{1/2}$$

$$\frac{d}{dx}\frac{\partial F}{\partial y'} = e^y y'^2 (1+y'^2)^{-1/2} + e^y y'' (1+y'^2)^{-1/2} - e^y y'^2 y''(1+y'^2)^{-3/2}$$

$$\frac{d}{dx}\frac{\partial F}{\partial y'} - \frac{\partial F}{\partial y} = y'^2 (1+y'^2)^{-1/2} + y'' (1+y'^2)^{-1/2} - y'^2 y''(1+y'^2)^{-3/2} - (1+y'^2)^{1/2} = 0$$

Last edited: Jun 7, 2009
6. Jun 7, 2009

### Dick

dF/dy and dF/dy' look fine. d/dx(dF/dy') takes a bit of work...

7. Jun 7, 2009

### cscott

I'm wondering how I'm going to solve this ODE

8. Jun 7, 2009

### Dick

I got the same thing you did. Luckily, it simplifies. A lot. Set up dF/dy=d/dx(dF/dy') and multiply both sides by (1+y'^2)^(3/2), divide by e^y, etc. etc.

9. Jun 7, 2009

### cscott

$$y''-y'^2-1=0$$?

10. Jun 7, 2009

### Dick

That's it.

11. Jun 7, 2009

### cscott

Thanks for your help.

12. Jun 9, 2009

### cscott

Is there a way to not explicitly compute the derivative $$\frac{d}{dx}\frac{\partial F}{\partial dy'}$$ to make the ODE easier to solve? What's the method for this non-linear second-order?

13. Jun 9, 2009

### Dick

That's not so hard so solve. First substitute y'=u and solve for u.

14. Jun 9, 2009

### cscott

For some reason I was thinking reduction of order could only be done on linear ODEs. This is nonlinear, true?

15. Jun 9, 2009

### Dick

It's nonlinear, true. But there's no reason you can't use a substitution on it! Solve for u and then integrate u to get y.

Last edited: Jun 9, 2009
16. Jun 9, 2009

### cscott

Thanks. I have the full answer now.

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