Solving the Euler-Lagrange Equation for Calculus of Variations

insynC
Messages
66
Reaction score
0
Just did this in class today and was doing a problem to see if I understood it and I'm not sure I did. Thanks for any help

Homework Statement



Solve the Euler equation to make the following integral:

(integral from x1->x2)

∫ [(y')² + y²] dx

Homework Equations



Euler-Lagrange equation

∂F/∂y - d/dx (∂F/∂y') = 0

The Attempt at a Solution



Clearly F = (y')² + y²

In class we had been rearranging the integral so that ∂F/∂y = 0, which made the problem much simpler. I was unable to do this here as there was no apparent way to introduce ds to swap y' for x'. This was my concern as, although I didn't think all equations would be this simple, I thought most would reduce.

So instead I proceeded as follows:

∂F/∂y - d/dx (∂F/∂y') = 2y - d/dx (2y') = 2y - 2y'' = 0 => y'' = y

And this is easy enough to solve. But I'm concerned I've made a mistake getting there.

Have I made a mistake up to here or is there a better way to proceed?

Thanks
 
Physics news on Phys.org
AFAIK this is the best you can do. You can't eliminate the dF/dy term in general.
 
Cheers. I'm a bit rusty on my calculus, but is there anything wrong with the step:

d/dx(2y') = 2y''

My lecturer made note to be careful of the d/dx part, but I don't think that applies here, just want to check.
 
That's just fine.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top