# Calculus of variations for known derivative on both extremes

1. May 6, 2013

### Gux

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)?

2. May 6, 2013

### Mute

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.

3. May 6, 2013

### Gux

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?

4. May 6, 2013

### the_wolfman

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.

5. May 6, 2013

### Gux

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)?

6. May 7, 2013

### Gux

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...