Is this a coincidece for harmonic equations.

Question

Let $ \Omega $ be a smooth bounded domain in $ \mathbb{R}^d $. Consider the functional problem $$ \min_{v\in H_0^1(\Omega)}\int_{\Omega}|\nabla v|^2dx. $$ It is easy to calculate the E-L equation of the problem above and obtain that the minimizer of such functional problem satisfies $ -\Delta v=0 $ weakly in $ \Omega $. For any $ u\in H_0^1(\Omega) $, since $ v $ is the minimizer, it follows that $$ \int_{\Omega}|\nabla v|^2dx\leq\int_{\Omega}|\nabla u|^2dx.\quad(*) $$ On the other hand, by testing the equation $ -\Delta v=0 $ with the function $ v-u $, we can deduce that $$ \int_{\Omega}\nabla v\cdot\nabla(v-u)dx=0. $$ In view of the inequality $ a\cdot (a-b)\geq \frac{1}{2}(|a|^2-|b|^2) $ for any $ a,b\in\mathbb{R}^n $, we can also obtain $(*)$. Here we have used two methods to obtain $(*)$. I want to ask is it a coincidence? Precisely speaking, for any inequality obtained from the E-L equation, if we can establish the same one by simply choose test function bu compare the value of the functional or not?

Answer 1

To some small degree this is a rule, but obtaining some inequality which would do the step can be impossible or hard.

Instead of functionals, look at the three functions $x^2$, $x^4$ and $x^3- \cos x$. All of the them have (local) minimum at $0$. The first two are convex. For the first one we are able to do the trick. For the second it will be possible, but harder.

For the third one, the claim is not true (i.e., the minimum is only local), so the trick per se fails. But if one works harder, there still be something possible locally by inequalities. However it is easier to work with the second derivative.

Fourth function will be $x^3$: For this function, $x=0$ satisfies the E-L condition $3x^2=0$, but the trick fails because $0$ is not a minimum.

--

The original problem: You have $u$ and $v$, you have the line $L=u \{ v + t(u-v) \}$ containing the two points. Take the restriction of the functional to the line. Is not it a quadratic function? Should not it be easy to find an inequality showing that the minimum is a minimum?

Our functional is very quadratic. It equals $B(u,u)$ for a nice bilinear form (scalar product, actually).