Find the extremals of with and subject to the constraint

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- Aug 25th 2010, 09:27 PMMaccamanExtremal trouble
Find the extremals of with and subject to the constraint

- Aug 26th 2010, 03:18 AMAckbeet
What ideas have you had so far?

- Aug 26th 2010, 05:51 AMMaccaman
Okay, first of all, could you please what extremals are in a physical sense? I don't quite understand that.

Anyway, I haven't got very far, but here is what I have.

Since is independent of , it follows from the Euler-Lagrange equation that

so

Therefore

I don't really know where to go from here. - Aug 26th 2010, 06:14 AMAckbeet
You're doing calculus of variations here. Calculus of variations is all about finding functions that maximize or minimize a function

. A functional, in this case, maps functions to numbers. Your functional in this problem is**al**

So you have to find an extremum, or extremal, (doesn't matter if it's a min or a max) of this functional over the collection of functions such that

and

I don't think your application of the Euler-Lagrange equations is correct. Moreover, in order to satisfy the integral constraint of

I think you're going to have to use the method of Lagrange multipliers. You should incorporate the Lagrange multiplier before you apply the Euler-Lagrange equation.

Does any of this ring a bell? - Sep 12th 2010, 12:52 AMMaccaman
Yes, when I originally posted this we hadn't done it in class yet and I was trying to get ahead. Now I have something......

Define

and

The Lagrangian is

We can apply the Euler-Lagrange equation to obtain

and therefore

So

(where c is a constant and k is a constant)

Then from we have

using

Using the constraint ,

and here is where I am having trouble. I don't think the next line is correct.

- Sep 13th 2010, 02:55 AMAckbeet
I don't think you're applying the B.C. x(2) = 1 correctly. Everything looks good up until that point. What I would do is this: once you've applied the B.C. x(0) = 0, you find out that C = k. So re-write x(t) using the value for k. Then apply the second boundary condition. That'll give you an equation relating lambda to C (assuming you've eliminated k - you could just as easily eliminate C. It will make no difference in the final answer.) Then apply the constraint equation. And no, I don't think you're applying the constraint equation correctly, either. Check that again, but with using the method I've outlined.

- Sep 14th 2010, 11:23 AMMaccaman
Okay. So we have determined that .

Then becomes

using

so

Using the constraint ,

Solving for Lambda....

Substituting back into k we have.....

which can be subbed back into x for the extremal - Sep 14th 2010, 11:59 AMAckbeet
You've still got a sign error in applying x(2) = 1. So here's how you get good at algebra: pretend someone is holding a gun to your head, and if you make a mistake, that someone is going to pull the trigger! Or, you can just work in a high-energy physics lab. With those 10,000 volt wires running around, if you make a mistake,

*you're dead.* - Sep 14th 2010, 06:12 PMMaccaman
Imaginary gun to my head.....check.

Working in a high-energy physics lab with 10,000 volt wires running around.....check.

Standing over a tank full of ill-tempered mutated see Bass and sharks with frickin laser-beams attached to their head for good measure....check.

Working through carefully this time I find the extremal to be

which at and at

I'm alive!(Rofl) - Sep 15th 2010, 02:59 AMAckbeet
Nope, you're dead. But you're only

*mostly dead*. Mostly dead is slightly alive. Your final answer still has one sign error in it.