# Calculus of variations

1. Jul 13, 2009

### daudaudaudau

1. The problem statement, all variables and given/known data

Consider the functional defined by
$$J(y)=\int_{-1}^1 x^4(y'(x))^2 dx$$

Without resorting to the Euler-Lagrange equation, prove that J cannot have a local minimum in the set
$$S=\{y\in C^2[-1,1]:\ y(-1)=-1,\ y(1)=1\}.$$

3. The attempt at a solution
I have thought about this one, but I have no clue how to do it without the Euler-Lagrange equation. Using the EL equation, I can find a solution which satisfies the boundary conditions, but is not continuous at $x=0$: $y=1/x^3$.

2. Jul 13, 2009

### Dick

Your functional J is nonnegative. And for every element y of S, J(y)>0, right? If you can show for every e>0, there is an element y_e of S that satisfies J(y_e)<e, you would be done.

3. Jul 13, 2009

### daudaudaudau

So you're saying that J(y) has no lower boundary, and I agree, but how does this exclude that J(y) could have a local minimum somewhere? It has no global minimum for sure.

4. Jul 13, 2009

### Dick

Good point. I was thinking of a global minimum. To show there is no local minimum you'll have to think of a way to create a variation of an element y in S which decreases J(y). Any ideas?

5. Jul 13, 2009

### daudaudaudau

\begin{align*} J(y+a\delta(x))&=\int_{-1}^1 x^4(y'(x)+a\delta'(x))^2dx\\ &=\int_{-1}^1 x^4(y'(x))^2dx+2a\int_{-1}^1x^4y'(x)\delta'(x)dx+a^2\int_{-1}^1x^4(\delta'(x))^2dx \end{align*}

The second-order term is always positive, so if the first-order term is zero, we have a minimum. But since $\delta(x)$ is arbitrary, and because $y(x)$ is differentiable, this means that $y'(x)=0$, and this solution cannot satisfy the boundary conditions.

6. Jul 13, 2009

### daudaudaudau

But I might as well perform an integration by parts on the first-order term and obtain the E-L equation which leads to the solution $y=1/x^3$, but this solution is certainly not a stationary point, or continuous.

I should mention that the first part of the question in my book is

a) Show that no extremals in $C^2[-1,1]$ exist which satisfy the boundary conditions $y(-1)=-1,\ y(1)=1$.

7. Jul 13, 2009

### Dick

Well, you want to ignore the terms that are quadratic in the variation. But, yes, that's the idea. So, yes, then if you integrate by parts, you get the EL equation times the variation. Since you know that nothing in S satisfies the EL equation, it must be nonzero someplace for every element of S. Then choose a variation which is nonzero near that point. This is all fine, but is this really 'without resorting to the Euler-Lagrange equation'?? Offhand, I'm not thinking of any other clever tricks right now.

8. Jul 13, 2009

### daudaudaudau

So could we just choose the derivative of the variation $\delta'(x)$ to be positive whenever $y'(x)$ is, and zero otherwise?

9. Jul 13, 2009

### Dick

Sort of, I guess. Though you would need to make sure the variation has two continuous derivatives. But all of this is really just using the Euler Lagrange procedure in a messy way, and isn't really a new way of showing there is no local minimum. I really wonder if they meant to say 'local minimum' in the question and not 'global minimum'.