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I had the following solution for this problem. I hope I'm not missing something important.

Find extremals for the functional $J[y]=\displaystyle\int_0^1(xy+y^2-2y^2y')dx$ with $y(0)=1$ and $y(1)=2$. For the Euler-Lagrange equation we have $$0=F_y-\dfrac{d}{dx}(F_y')=x+2y-4yy'-(-4yy')=x+2y\Rightarrow y=-\dfrac{x}{2}$$ which is inconsistent with the boundary conditions so there is no extremal for $J$.

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  • $\begingroup$ Now that I think about it, the above implies that there is no continuous solution. $\endgroup$ – Cristian Baeza Sep 2 '20 at 0:00
  • $\begingroup$ While the post is true, the no continuous solution is not the right conclusion. What you have is equivalent to the case where the critical point is outside of your domain of consideration (functions with $y(0) = 1$ and $y(1) = 2 $) That doesn't mean an extrema can't happen on the "boundary" of this domain but generally this is a tough problem to do analytically (you would need some tool equivalent to Lagrange multipliers, but for functionals of functions, not functions of points). And unlike the function case, the existence of an extrema is not guaranteed because the domain is not compact. $\endgroup$ – Ninad Munshi Sep 2 '20 at 0:19
  • $\begingroup$ oh boy!, thanks for that. $\endgroup$ – Cristian Baeza Sep 2 '20 at 14:23
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Yes, that's right. In particular, you're implicitly performing the extremization over some function space for which $J[y]$ is well-defined, e.g. $H^1([0,1])$. Suppose for contradiction that you have an extremum $y(x)$ that satisfies your boundary conditions. Then for any $\delta y(x) \in H^1([0,1])$ with $\delta y(0) = \delta y(1) = 0$, you've computed that $$\lim_{t\to 0} \frac{d}{dt} J[y + t\delta y] = \int_0^1 (x+2y)\delta y,$$ and since $y$ is continuous and not equal to $-x/2$ everywhere, you can construct a $\delta y$ (using a bump function e.g.) which contradicts the extremality of $y$.

If you imagine actually performing extremization (for instance, by repeatedly perturbing $y$ in a direction $\delta y$ that decreases $J$) you'll see that $y$ "looks more and more" like $y(x)=-x/2$, with derivatives that become larger and larger near the boundaries. As Ninad Munshi points out, since $H^1([0,1])$ is not compact, this process won't converge to a minimum in $H^1([0,1])$ (it should be easy to visualize that the sequence is "converging" to $y(x) = -x/2$ with two discontinuities at the endpoints).

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