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

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Homework Help Overview

The discussion revolves around minimizing the integral \(\int_{(x_1,y_1)}^{(x_2,y_2)} n(x,y)~ds\) where \(n(x,y)=e^y\) with specified endpoints. The context involves applying the Euler-Lagrange equation in the realm of calculus of variations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the formulation of the problem, questioning the independence of variables and the correct application of the Euler-Lagrange equation. There are discussions about the implications of parameterizing the problem differently and the resulting expressions for derivatives.

Discussion Status

Several participants have provided insights into the formulation of the problem and the derivation of necessary derivatives. There is an ongoing exploration of methods to simplify the resulting ordinary differential equation (ODE), with some participants suggesting substitutions to facilitate solving it.

Contextual Notes

Participants express uncertainty about the nature of the ODE, discussing its linearity and the applicability of certain solution methods. There is also mention of homework constraints that may influence the approach taken.

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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?
 
Last edited:
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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.
 
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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dF/dy and dF/dy' look fine. d/dx(dF/dy') takes a bit of work...
 
I'm wondering how I'm going to solve this ODE
 
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.
 
[tex]y''-y'^2-1=0[/tex]?
 
  • #10
cscott said:
[tex]y''-y'^2-1=0[/tex]?

That's it.
 
  • #11
Thanks for your help.
 
  • #12
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?
 
  • #13
cscott said:
[tex]y''-y'^2-1=0[/tex]?

That's not so hard so solve. First substitute y'=u and solve for u.
 
  • #14
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?
 
  • #15
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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  • #16
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.
 

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