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

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cscott
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Homework Statement



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

Homework Equations



Euler-Lagrange equation

The Attempt at a Solution



[tex]\frac{d}{dx}\frac{dF}{dy'} - \frac{dF}{dy}=0[/tex]

[tex]0 - e^y y' = 0[/tex]

y' = 0 so y = constant or y = 1 considering the initial conditions?
 
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Oh ok, thanks. So,

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

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

But then is [tex]\frac{d}{dx}\frac{dF}{d\dot{y}} = 0[/tex]?
 
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.
 
[tex]\frac{\partial F}{\partial y'}=e^y y'(1+y'^2)^{-1/2}[/tex]

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

[tex]\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}[/tex]

[tex]\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[/tex]
 
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I'm wondering how I'm going to solve this ODE
 
[tex]y''-y'^2-1=0[/tex]?
 
Thanks for your help.
 
Is there a way to not explicitly compute the derivative [tex]\frac{d}{dx}\frac{\partial F}{\partial dy'}[/tex] to make the ODE easier to solve? What's the method for this non-linear second-order?
 
Dick said:
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?
 
cscott said:
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.
 
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Dick said:
That's not so hard so solve. First substitute y'=u and solve for u.

Dick said:
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.