I am trying to follow this resource for learning the basics of Calculus of Variations: https://ocw.mit.edu/courses/mathematics/18-086-mathematical-methods-for-engineers-ii-spring-2006/readings/am72.pdf
I am reading page 5, and I see that it is talking about Calculus of Variations in 2 dimensions. Of course, we get a PDE as the integrand (equation 11):
$a\frac{\delta^2u}{\delta x^2}+2b\frac{\delta^2u}{\delta x \delta y}+c\frac{\delta^2u}{\delta y^2}=0$
Now, when I see $P(u)$, which comes immediately after that, I also see an explanation below it that says "If we minimize P we expect to reach (11) as its Euler equation. But there is more to it than that. To minimize P it should be positive definite".
I know that if P(u) is positive definite, then that means that the differential equation must be elliptical, and hence must have $ac>b^2$.
So, by this explanation I am wondering, can we only have elliptic PDEs in order to minimize P this way? It is saying "to minimize P it should be positive definite", so that is why I am wondering.
Thanks.
This is a general phenomenon; minimizers of variational problems satisfy an elliptic PDE. If $u$ minimises the functional $I[w],$ for any other function $v$ we get $i(t) = I[u+tv]$ satisfies $i''(t) \geq 0.$ Roughly speaking, this shows the associated 'Hessian' of $I[w]$ at $u$ is positive definite. Intuitively you can see why we get some kind of elliptic behavior as a result.
More precisely, can show that if $u : \Omega \rightarrow \mathbb R$ minimses the functional,
$$ I[w] = \int_{\Omega} F(x,w,Dw)\,\mathrm{d}x, $$
then for each $x \in \Omega$ and $\xi \in \mathbb R^n,$
$$ \sum_{i,j=1}^n \frac{\partial^2 F}{\partial p_i \partial p_j}(x,w,Dw) \xi_i\xi_j \geq 0. $$
Where $F = F(x,z,p).$ In the case the Euler-Lagrange equation is linear ($D^2F$ only depends on $x$) and $n=2,$ we recover the condition you seek.
The idea of the proof is to use the fact that $i''(t) \geq 0$ for a clever choice of $\phi.$ The details are a bit involved, but it is done in Evans's PDE book (chapter 8).
Typing this answer I realise what I said above may not be suited for someone who is learning about calculus of a variations from an applied viewpoint. I'll stick to the example in the linked PDF, namely the functional,
$$ P(u) = \frac12\iint_S \left[ a \left(\frac{\partial u}{\partial x}\right)^2 + b\left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial u}{\partial y}\right) + c\left(\frac{\partial u}{\partial y}\right)^2\right]\, \mathrm{d}x\,\mathrm{d}y.$$
Suppose a minimiser $u$ exists. Then the associated Euler-Lagrange equation is given by,
$$ a\frac{\partial^2 u}{\partial x^2} + 2b \frac{\partial^2 u}{\partial x\partial y} + c \frac{\partial^2 u}{\partial y^2} = 0.$$
Moreover, the second variation is given by,
$$i''(0) = \iint_S \left[ a \left(\frac{\partial v}{\partial x}\right)^2 + b\left(\frac{\partial v}{\partial x}\right)\left(\frac{\partial v}{\partial y}\right) + c\left(\frac{\partial v}{\partial y}\right)^2\right]\, \mathrm{d}x\,\mathrm{d}y \geq 0,$$
for all smooth functions $v : S \rightarrow \mathbb R^n$ vanishing on the boundary. I'll leave this as an exercise, but the derivation is similar to that of the first variation. It is not difficult to see that this forces the matrix $$ \begin{pmatrix} a & b \\ b & c \end{pmatrix},$$ to be non-negative definite.