Calculus of variations for known derivative on both extremes

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Discussion Overview

The discussion revolves around minimizing a functional in the calculus of variations, specifically when boundary conditions involve known derivatives at both extremes. Participants explore different scenarios, including cases where only derivatives are known and where both function values and derivatives are specified at the boundaries.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that minimizing a functional with known boundary values leads to the Euler-Lagrange equation, but questions how to approach the problem when only derivatives at the boundaries are known.
  • Another participant suggests that if both function values and derivatives are known at the boundaries, the Lagrangian must involve higher order derivatives to avoid an overdetermined system.
  • A participant presents a specific example of finding the shortest path between two points with given derivatives, asserting that there is a solution but questioning how to find it.
  • Another participant counters that there are infinitely many solutions to the problem posed, illustrating this with examples of shifted solutions.
  • One participant expresses a belief that the solution is unique, prompting further inquiry into how to find the family of solutions when only derivatives are imposed.
  • A later reply discusses the scenario where all boundary conditions are known, suggesting that in the limit, the shortest path approaches a straight line, but admits to not fully understanding the implications.

Areas of Agreement / Disagreement

Participants express differing views on the uniqueness of solutions when only derivatives are specified, with some arguing for a family of solutions and others claiming uniqueness. The discussion remains unresolved regarding the methods to find solutions under the various boundary conditions presented.

Contextual Notes

Participants highlight the complexity of the problem, noting that the requirements for a well-posed problem depend on the order of derivatives in the Lagrangian and the number of boundary conditions imposed. There is also uncertainty regarding the implications of limiting cases and the behavior of solutions.

Gux
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Minimizing a functional:

When you know the values of the function y(x) on the boundary points y(x1) and y(x2), minimizing the functional ∫{L(x,y,y')} yields the Euler-Lagrange equation.

How can you minimize the functional if, instead, you know the derivatives y'(x1) and y'(x2)?

What if you know y(x1), y(x2), y'(x1) and y'(x2)?

Thank you for your help.
 
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Gux said:
Minimizing a functional:

When you know the values of the function y(x) on the boundary points y(x1) and y(x2), minimizing the functional ∫{L(x,y,y')} yields the Euler-Lagrange equation.

How can you minimize the functional if, instead, you know the derivatives y'(x1) and y'(x2)?

What if you know y(x1), y(x2), y'(x1) and y'(x2)?

Thank you for your help.

I don't have time to work out the first case right now, but in the second case where you know the functions and its derivatives at both boundaries, your Lagrangian will need higher order derivatives for the problem to be well-posed. If your Lagrangian only has a first derivative, you will get a second order differential equation. You only need two initial conditions or boundary values for a second order DE. If you had four, the system would be overdetermined. If you have four boundary conditions, I think your Lagrangian would need to depend on up to the 4th derivative of y to have a well-posed problem.
 
But...

Thank you for your reply, but:

Imagine you want to find the shortest way y(x) between two points (x1,y1) and (x2,y2), given that y'(x1)=0 and y'(x2)=1. Obviously there is a solution. How to find it?
 
Imagine you want to find the shortest way y(x) between two points (x1,y1) and (x2,y2), given that y'(x1)=0 and y'(x2)=1. Obviously there is a solution. How to find it?

Be careful. Here there is not a unique solution. There are an infinity of solutions. To see why imagine that y1(x) is a solution of the problem you just posed. Then y2(x)=y1(x)+2 is also a solution and so is y3(x)= y1(x)+7382.2119.
 
Yes, the solution is unique

Thank you:

(1) Ok, so if we impose only the derivatives, we have a family of solutions. How to find them?

(2) How to find the solution in case we impose y(x1), y(2), y'(x1) and y'(x2)?
 
I think I am strating to understand...

In case you know (x1), y(2), y'(x1) and y'(x2): in the limit, the shortest way is a straight line: the part of the curve around the point x1 should change to obtain the given derivative y1(x1), but "in the limit" this part of the curve disappears. So the Euler-Lagrange works. Anyway I don't understand this completely...
 

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