PDES FROM

LECTURE NOTES FOR MARCH 23

1. Dirichlet principle revisited

In the previous lecture we considered the Dirichlet problem for the Laplace equation,  ∆u = 0 in D, (1.1) u = h on ∂D, where, D ⊂ R3 is the given domain, and h : ∂D → R is a given . We learned that the solution minimizes the Dirichlet energy 1 Z (1.2) E[u] = |∇u|2. 2 D Namely, we proved: Theorem 1.3 (Dirichlet principle). Let u be the unique solution of the Dirichlet problem (1.1). If w is any other function in D satisfying w = h on ∂D, then (1.4) E[w] ≥ E[u]. In other words, our u minimizes the energy among all functions (with the same boundary condition).

Here is another way to think about this: Since u minimizes the energy functional E, it is in particular a critical point of E. Thus, for any test function v : D → R that vanishes on ∂D we have d (1.5) dε |ε=0E[u + εv] = 0. To use this, let us explicitly compute the first variation of the energy functional:  Z  d d 1 2 (1.6) dε |ε=0E[u + εv] = dε |ε=0 |∇(u + εv)| 2 D Z (1.7) = h∇v, ∇ui D Z (1.8) = − v∆u, D 2 LECTURE NOTES FOR MARCH 23 where we integrated by parts in the last step. Hence, we obtain Z (1.9) v∆u = 0. D Since this holds for all test functions v as above, we conclude that (1.10) ∆u = 0 in D. To recap, we have shown that whenever u is a critical point of E, then u solves the PDE (1.10). Finally, let us observe since the Dirichlet energy E is strictly convex, there is a unique critical point, and this critical point is a minimum.

2. Calculus of variations and nonlinear PDEs

Many PDEs in mathematics, physics, engineering, economics, etc, arise as critical points of some energy functional. There is a big area in mathematics, called the calculus of variations, which systematically studies such problems. Let us consider some examples:

Example 2.1 (Nonlinear Laplace equation). Consider the energy func- tional Z 1 1  (2.2) E[u] = |∇u|2 − up+1 . D 2 p + 1 Let us compute the first variation: Z d  p  (2.3) dε |ε=0E[u + εv] = h∇u, ∇vi − u v D Z (2.4) = − v∆u + up). D Hence, the critical points of E solve the nonlinear Laplace equation (2.5) ∆u + up = 0.

Example 2.6 (Nonlinear wave equation). Consider the action func- tional1 Z   1 2 1 2 1 p+1 (2.7) A[u] = (∂tu) − |∇u| + u , D 2 2 p + 1

1In physics, the action is the kinetic energy minus the potential energy. PDES FROM CALCULUS OF VARIATIONS 3 where now u also depends on time. The first variation is: Z d  p  (2.8) dε |ε=0A[u + εv] = ∂tu∂tv − h∇u, ∇vi + u v D Z 2 p (2.9) = − ∂t u + ∆u + u )v. D Hence, the critical points of A solve the nonlinear wave equation 2 p (2.10) ∂t u = ∆u + u .

Example 2.11 (Minimal surfaces). Another famous problem is to find the shape of a soap film that spans a given wire ring. When we dip the wire ring into the soap water, a ”” will form, i.e. a surface that minimizes area subject to the condition that its boundary is given by the wire. The geometry is simplest when the surface can be written as graph of a function z = u(x, y), where (x, y) ∈ D, and the wire ring is the graph of a function h : ∂D → R. We thus must minimize the area functional Z q 2 2 (2.12) A[u] = 1 + ux + uy D among all functions u : D → R satisfying the boundary condition u|∂D = h. We compute Z 1 (2.13) d | A[u + εv] = (u v + u v ). dε ε=0 p 2 2 x x y y D 1 + ux + uy Hence, after integration by parts, we see that critical points of the area functional satisfy the minimal surface equation ! ! u u (2.14) x + y = 0. p1 + u2 + u2 p1 + u2 + u2 x y x x y y The minimal surface equation shares some similarities with the Laplace equation uxx + uyy = 0, but is quite a bit more complicated due to its nonlinear nature. It is one of the most classical and well studied geometric PDEs.2 You can google “minimal surface” and look at some beautiful pictures. And if you have solved the bonus question from the last problem set, then you have already seen the catenoid.

2In case you are wondering what your professor is doing most of the time: I study the corresponding evolution equation, called the mean curvature flow.