No extremals satisfying the Euler equation - what does it mean? Consider the functional $J[y] = \int_{0}^{1}xyy^{'}dx$.
If I want to find extremals (a function $y=y(x)$ that makes the functional stationary) with boundary condition $y(0)=0$ , $y(1)=1$ for this functional, I can use the Euler equation. The solution to the Euler equation turns out to be $y=0$. This obviously doesn't satisfy the boundary conditions. The conclusion then is that no such extremals exist.
I'm now trying to understand what this tells us about the functional itself. Is it safe to assume that the functional is somewhat analogous to a monotonic function of n variables? In particular, for a function $\hat{y}$, $J[y] - J[\hat{y}]$ is never always non-negative or always non-positive for all $y$ in the neighborhood of $\hat{y}$ satisfying $||y-\hat{y}||<\epsilon$ for small $\epsilon>0$. How can I visualize the functional as in the case of the monotonic function?
Thanks
 A: In this case, you can use integration by parts to show that this functional is not bounded below:
$$ J[y] = \int_0^1 x(yy') \, dx = \frac{1}{2}\left[xy^2\right]_0^1- \frac{1}{2}\int_0^1 y^2 \, dx = 1-\int_0^1 y^2 \, dx, $$
so you can essentially make $J$ as negative as you like by choosing $y$ large enough away from $1$. This essentially shows that $J$ acts like a negative-definite quadratic form.
You can say a bit more: obviously you're also wondering about why there's no maximiser. To answer that, consider the function $y=x^{n}$, which satisfies the boundary conditions. However,
$$ J[y] = \frac{1}{2}\left( 1-\int_0^1 x^{2n} \, dx \right) = \frac{1}{2}\left( 1-\frac{1}{2n+1} \right), $$
so we can make the functional as close to $1/2$ as we like, and so if there is a maximum value, it is $1$. On the other hand, since $y$ has to be continuous to be differentiable, the integral $\int_0^1 y^2 \, dx$ has to be positive, so $1/2$ cannot be attained by continuous functions.
If you allow non-continuous functions and write $J$ in the form I have, the function
$$ y(x) = \begin{cases} 1 & x=1 \\ 0 & \text{else} \end{cases} $$
is a maximiser that satisfies the Euler-Lagrange equation almost everywhere, and in particular, in the interior of the interval $[0,1]$.

This basically means you need to be careful about what sorts of function you allow in your Euler-Lagrange problems: you can only minimise over a space of functions, which we normally take to be continuous with some sort of first derivatives, enough to get the Lagrangian to make sense, at least. For an example of a function of the same type, consider
$$ f(x) = 8x-x^4. $$
Now, $f$ has a maximum at $\sqrt[3]{2}$ and no minimum, but if we consider $f:\mathbb{Q} \to \mathbb{R}$, then $f$ has no maximum value.
