What is the connection between calculus of variations and PDEs?

Question

I have heard of calculus of variations and I would like to know how it is related with partial differential equations. What type of equations (hyperbolic, parabolic, or elliptic) can be solved? Does it make it easier to solve some of these equations? i am interested in applications of this topic.

Answer 1

The surface level answer is that if $I[w]$ is a functional acting on an appropriate function space, the stationary points of the functional (defined appropriately) satisfies a PDE, namely the Euler-Lagrange equations. But you probably know that already.

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A lot of modern PDE theory is concerned with the existence, uniqueness and regularity of solutions. That is, given a particular PDE with initial/boundary conditions, we ask (a) whether a solution exists, (b) whether it is unique and (c) how 'regular' it is, for example how many times we can differentiate it. The important point is that we aren't interested in explicitly writing down a solution, provided we can prove it exists.

To do this, we usually use the following method:

1. Define a sufficiently general space of functions $X$ and an appropriate notion for a function $u \in X$ to be a 'weak solution' to the equation. We then (often exploiting compactness properties) prove a weak solution exists.
2. Prove that $u$ is sufficiently regular and is a solution to the PDE in the usual sense.

I've omitted the question of uniqueness here, as that tends to vary in approach and is often somewhat separate from the existence part.

The calculus of variations is useful for two reasons:

• It provides a way of doing step 1 of the above method.

• Moreover the weak solutions we assert exist often satisfy additional properties, which can be useful for step 2.

In the following, we will assume we are solving the equation $Mu = 0$ where $M$ is a (often nonlinear) differential operator in some domain in $\mathbb R^n,$ which coincides with the Euler-Lagrange equation of the functional $I[w].$

Existence: Under appropriate convexity and growth conditions, we can prove the existence of a weak solution by showing a stationary point of $I[w]$ exists. The idea is that if $I[w]$ is continuous on a compact subspace of functions, then it will attain a global maximum / minimum there. Then we can show this max/min satisfies the Euler-Lagrange equation in a weak sense.

In practice there are technicalities (we often can't show $I[w]$ is continuous with respect to the topology we impose on $X$), but the general idea is to show the existence of a minimum and to prove it also solves the PDE.

Additional information: If we have a stationary point $u$ is actually a minimum, then we have $I''[u] \geq 0,$ in the sense that for all nice functions $v,$ $$ \frac{d^2}{dt^2} I[u + tv] \geq 0. $$ By differentiating under the integral sign (which we can justify with appropriate hypotheses) we obtain the second variation. Hence our solution $u$ not only satisfies the PDE $Mu = 0$ in a weak sense, but it also satisfies another differential inequality. This is useful in proving that $u$ is actually more regular (smooth) than we initially assumed, which serves as a starting point for step 2.

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Finally, I will mention that the equations obtained are often elliptic. One can show that if $I[w]$ is convex and $u$ is a local minimum, then the associated equation has an elliptic structure. This is based on analysing the second variation and is explained in detail in Evan's PDE book (section 8.1.3).

Answer 2

Calculus of variations is about minimising/maximising real-valued functionals on function spaces. PDE that happen to be Euler-Lagrange equations can be reformulated as minimising such a functional.

As an example, suppose $U\subset \mathbb R^n$ is a domain (bounded, open, connected), and $f\in L^2(U)$, and consider the functional $E:H_0^1(U)\to \mathbb R$ $$ E(u):=\int_U\left(\frac{1}{2}|\nabla u|^2-fu\right)dx.$$ A minimiser of $E$ will then necessarily be a weak solution of the Poisson equation $$-\Delta u\vert_{U} = f,\qquad u\vert_{\partial U}= 0.$$