# Is this book implying that elliptic PDE is the only one that can be minimized by calculus of variations?

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.

• I'm not sure what $P(u)$ refers to in your problem. Although you wrote a PDE involving $u$, the nature of $P(u)$ as something that can be "minimized" is unclear. – hardmath Jan 2 '18 at 17:26
• @hardmath The resource that I cited shows it. and I am trying not to overcrowd the question. – nundo Jan 2 '18 at 18:47
• You should use a link for additional (supplementary) material, not to convey the essence of your problem. Readers should be given a synopsis of the linked content, perhaps a verbatim quote of the most relevant statements there, to make an informed decision about whether to follow the link. It could also help with reconstruction if the link stops working. – hardmath Jan 2 '18 at 18:53
• The point is that when $P$ is convex, the Euler-Lagrange equation is elliptic. Convex problems have unique minimizers, which is why it is natural to assume convexity. There are many applications of the calculus of variations where the functional is not convex, so the Euler-Lagrange equation may not be elliptic. – Jeff Jan 3 '18 at 4:03
• I see what you're saying, thank you @Jeff. Now if the Euler-Lagrange equation is not elliptic (and hence $P$ is non-convex), then doesn't that imply that there can be multiple minimizers? And hence, it would mean that solving the Euler-Lagrange equation is necessary, but not sufficient to claim $u$ is a minimizer? – nundo Jan 6 '18 at 17:14

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.

• this is a good answer, thanks. – nundo Jan 6 '18 at 17:02
• “non-negative definite” = positive semi-definite, right? This isn’t totally clear to me. I do not see how $v$ relates to $u$, but maybe it is because I’m a beginner in this area. But thank you @ktoi – nundo Jan 6 '18 at 17:29
• @nundo Yup, that's correct. It depends on what definition you choose, but the idea is to show that $ax^2+bxy+cy^2\geq0$ for any real $x,y,$ by choosing v appropriately (this requires some care, since v needs to vanish on the boundary). – ktoi Jan 6 '18 at 17:32