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

In summary, the conversation discusses a homework problem involving minimizing an integral with a given function and initial conditions. The Euler-Lagrange equation is mentioned and the attempt at a solution involves taking x as the parameter instead of t. The conversation also covers finding the partial derivatives of the function and solving the resulting ODE. The method of substitution is used to solve the nonlinear second-order ODE, resulting in the final solution.
  • #1
cscott
782
1

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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  • #2
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
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]?
 
  • #4
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
[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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  • #6
dF/dy and dF/dy' look fine. d/dx(dF/dy') takes a bit of work...
 
  • #7
I'm wondering how I'm going to solve this ODE
 
  • #8
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
[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.
 

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

What is Fermat's Principle?

Fermat's Principle, also known as the Principle of Least Time, states that light (or any other form of electromagnetic radiation) will take the path that requires the shortest amount of time to travel between two points.

How does Fermat's Principle relate to the Euler-Lagrange equation?

Fermat's Principle can be mathematically described using the Euler-Lagrange equation, which is a fundamental equation in the calculus of variations. This equation allows us to find the path that minimizes the time taken by light to travel between two points.

What is the importance of minimizing the action in Fermat's Principle?

The action in Fermat's Principle represents the total time taken by light to travel between two points. By minimizing this action, we can find the path that will take the least amount of time for light to travel, which is the path that light will actually take in reality.

What are some real-world applications of Fermat's Principle?

Fermat's Principle has many applications in optics and engineering. It is used to design optical systems and predict the behavior of light in various mediums, such as lenses and mirrors. It is also used in the development of technologies like fiber-optic communication and laser technology.

How does Fermat's Principle relate to the concept of refraction?

Refraction is the bending of light as it passes through different mediums. Fermat's Principle explains why this bending occurs - because light takes the path that minimizes the time it takes to travel between two points, it will bend when it encounters a medium with a different optical density in order to reach its destination faster.

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