Fermat's Principle: Minimizing Int. w/ Euler-Lagrange

[cscott]

Homework Statement

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).

Homework Equations

Euler-Lagrange equation

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?

 

[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.

 

[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?

 

[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.

 

[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

 

[Dick]

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

 

[cscott]

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

 

[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.

 

[cscott]

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

 

[Dick]

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

That's it.

 

[cscott]

Thanks for your help.

 

[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?

 

[Dick]

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

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

 

[cscott]

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

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

 

[Dick]

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

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.

 

[cscott]

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

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.

Thanks. I have the full answer now.