Regularity of Minimisers and Constraints

6A.11 Calculus of Variations and Gradient Flows proved that convex, coercive energies have minimisers in $H^1$, and 6B.9 De Giorgi–Nash–Moser explained how De Giorgi–Nash–Moser makes such minimisers smooth, solving Hilbert's nineteenth problem, once one knows that their derivatives are weak solutions of a linearised equation. This chapter supplies that missing step and then adds constraints, which is where most variational problems of geometry and applications live.

Constraints come in two kinds. An equality constraint, such as $\int u^2 = 1$ or $|u| = 1$ pointwise, turns the Euler–Lagrange equation into an equation with a Lagrange multiplier: the eigenvalue equation $-\Delta u = \lambda u$ is the Euler–Lagrange equation of the Dirichlet energy on the unit sphere of $L^2$, and the harmonic map equation $-\Delta u = |\nabla u|^2u$ is that of the energy of maps into a sphere. An inequality constraint, such as $u \geq h$, gives a variational inequality: the solution satisfies the equation where the constraint is inactive and sits on the constraint elsewhere, and the boundary between the two regions, the free boundary, is an unknown of the problem. The model is the obstacle problem, a membrane stretched over an obstacle. In finance it is exactly the problem of pricing an American option, where the free boundary is the exercise boundary (6D.7 American Options and Free Boundaries).

By the end of this chapter you will be able to:

• prove interior $H^2$ regularity for minimisers of uniformly convex integrals, and explain the bootstrap to $C^\infty$;
• prove the Lagrange multiplier rule for an integral constraint in $H^1$, and derive nonlinear eigenvalue equations from it;
• derive the harmonic map equation for maps into a sphere;
• prove existence and uniqueness for the obstacle problem as a projection, derive the variational inequality and the complementarity conditions, and compute an explicit solution with smooth fit;
• explain the regularity of solutions of the obstacle problem and what the free boundary is.

1.1Contact problems

1.2Regularity of minimisers

Let $U \subset \mathbb{R}^n$ be bounded, and consider

over $w \in H^1_0(U)$, with $f \in L^2(U)$ and $L : \mathbb{R}^n \to \mathbb{R}$ of class $C^2$ and uniformly convex:

By 6A.11 Calculus of Variations and Gradient Flows a minimiser $u$ exists and is unique, and it is a weak solution of the Euler–Lagrange equation:

(The bound $|L_p(p)| \leq \Lambda|p| + |L_p(0)|$, from the upper bound on $L_{pp}$, justifies differentiating $I[u + sv]$ under the integral.)

Notice what was not needed: any continuity of the coefficients $a^{ij}_h$. The convexity of $L$ alone makes the difference-quotient equation uniformly elliptic. That is why the $H^2$ step works for every uniformly convex integrand.

From $H^2$ to $C^\infty$. Now $w = \partial_ku \in H^1_{\mathrm{loc}}$, and letting $h \to 0$ in the difference-quotient equation (testing with smooth $v$) shows that $w$ is a weak solution of

a divergence-form equation with bounded measurable, uniformly elliptic coefficients, exactly as in 6B.9 De Giorgi–Nash–Moser's discussion of Hilbert's nineteenth problem. If $f$ is smooth (or at least in $L^q$ for $q > n$), De Giorgi–Nash–Moser makes each $\partial_ku$ Hölder continuous: $u \in C^{1,\alpha}_{\mathrm{loc}}$. Then $a^{ij} = L_{pp}(\nabla u)$ is Hölder, Schauder theory (6B.7 Schauder Theory and the Continuity Method) gives $u \in C^{2,\alpha}$, then $a^{ij} \in C^{1,\alpha}$, and so on: if $L$ and $f$ are smooth, so is $u$ (Evans, section 8.3.2, carries out the bootstrap). Existence by minimisation, $H^2$ by difference quotients, $C^{1,\alpha}$ by De Giorgi–Nash–Moser, $C^\infty$ by Schauder: the complete regularity theory of scalar variational problems.

1.3Equality constraints and Lagrange multipliers

In finite dimensions, a minimum of $E(x)$ on the level set $\{G(x) = 0\}$, at a point where $\nabla G \neq 0$, satisfies $\nabla E = \lambda\nabla G$ for some number $\lambda$ (2B.9 The Inverse and Implicit Function Theorems). The same holds in function spaces, with the same proof via the implicit function theorem.

Let $G : \mathbb{R}\to\mathbb{R}$ be smooth with $g = G'$, and suppose $|g(z)| \leq C(1 + |z|)$ and $|g'(z)| \leq C$. Consider the constraint set

and the energy $I[w] = \frac12\int|\nabla w|^2$.

Existence of a constrained minimiser is the direct method again: a minimising sequence is bounded in $H^1_0$, a subsequence converges weakly in $H^1_0$ and strongly in $L^2$ (Rellich), and the constraint $\int G(w) = 0$ passes to the strong limit because $G$ grows at most quadratically. So the constrained minimum is attained whenever $\mathcal A$ is non-empty (Exercise 1.7).

1.3.1Maps into a sphere

A pointwise constraint behaves differently: at each point $x$ the value $u(x) \in \mathbb{R}^m$ must lie on the unit sphere $S^{m-1}$. Minimise the energy $\frac12\int_U|\nabla u|^2 = \frac12\int\sum_{i,a}(\partial_iu^a)^2$ over maps $u : U \to S^{m-1}$ with given boundary values. Variations must stay on the sphere, and the simplest way is to project: for a smooth $v$ with compact support, $u_\tau = \frac{u + \tau v}{|u + \tau v|}$ is admissible for small $\tau$.

The right side is the Lagrange multiplier: the force $-\Delta u$ that the energy exerts is balanced by a normal force, proportional to $u$, that keeps the map on the sphere. Taking the dot product with $u$ and using $u\cdot\Delta u = -|\nabla u|^2$ (from differentiating $u\cdot\partial_iu = 0$ again) confirms the multiplier is exactly $|\nabla u|^2$. Harmonic maps between Riemannian manifolds generalise this (6C.7 Harmonic Maps and Their Heat Flow); their regularity is the case of systems, where De Giorgi–Nash–Moser fails and minimisers can be singular: $u(x) = \frac{x}{|x|}$ is an energy-minimising map from the unit ball of $\mathbb{R}^3$ to $S^2$ with a singular point at the origin (Brezis, Coron and Lieb, 1986).

1.4The obstacle problem

Now an inequality constraint. Let $h$ be a smooth function on $\overline U$ (the obstacle) with $h < 0$ on $\partial U$, let $f \in L^2(U)$, and consider

$K$ is non-empty (it contains $\max(h, 0)\zeta$-type functions; for instance $h^+$ after a cut-off near the boundary, since $h < 0$ there), convex (if $w_1, w_2 \geq h$ then so is any convex combination) and closed in $H^1_0$ (an $H^1_0$-limit converges in $L^2$, and a subsequence almost everywhere, preserving $w \geq h$).

Choosing particular $w$ extracts the structure.

• Supersolution everywhere. For $\varphi \geq 0$ in $C_c^\infty(U)$, $w = u + \varphi \in K$, so $\int\nabla u\cdot\nabla\varphi \geq \int f\varphi$: $-\Delta u \geq f$ in the sense of distributions. The membrane is pushed up by the obstacle, never pulled down.
• Equation off the obstacle. If $u$ is continuous (which regularity theory gives), the set $\{u > h\}$ is open. For $\varphi \in C_c^\infty(\{u > h\})$, $u \pm\varepsilon\varphi \in K$ for small $\varepsilon$, so the inequality holds in both directions: $-\Delta u = f$ where $u > h$.
• Complementarity. Together:

At each point either the constraint is active ($u = h$, contact) or the equation holds; the "reaction force" $-\Delta u - f$ is a non-negative measure supported on the contact set. This is the linear complementarity structure, and discretised it is the problem solved for every American option on every trading day (6D.7 American Options and Free Boundaries).

The free boundary is $\partial\{u > h\}$, where the membrane leaves the obstacle. It is not given; it is part of the solution.

The condition that $u'$ be continuous across the free boundary is called smooth fit (or smooth pasting) in optimal stopping, and it determines the free boundary there exactly as here: the optimal exercise boundary of an American put is where the option value leaves the payoff tangentially (6D.7 American Options and Free Boundaries).

Regularity. For a smooth obstacle and smooth $f$, the solution is $C^{1,1}$: its second derivatives are bounded (Hans Lewy and Stampacchia 1969 in two dimensions; Haim Brezis and David Kinderlehrer 1974; Jens Frehse 1972). It cannot in general be $C^2$, since $\Delta u$ jumps across the free boundary, as in the example. The regularity of the free boundary itself is much harder: Luis Caffarelli proved in 1977 that it is smooth near "regular" points and classified the possible singular points, the start of a large modern theory.

1.5History

Hilbert's nineteenth problem (1900) was solved by De Giorgi and Nash in 1957–58; the difference-quotient method for second derivatives of minimisers goes back to Louis Nirenberg (1955) and to Charles Morrey's work on multiple integrals. The Lagrange multiplier rule is Lagrange's (1788) and its infinite-dimensional form was made rigorous in the twentieth century with the implicit function theorem in Banach spaces. Signorini posed his contact problem in 1933 and Fichera solved it in 1963; Stampacchia (1964) and Lions and Stampacchia (1967) developed the theory of variational inequalities. The regularity of the obstacle problem was developed by Lewy and Stampacchia (1969), Frehse (1972), Brezis and Kinderlehrer (1974), and the free boundary by Caffarelli (1977). The singular minimising map $\frac{x}{|x|}$ is due to Brezis, Coron and Lieb (1986).

1.6Exercises

Solution

The energies are bounded, so the sequence is bounded in $H^1_0$; by 4A.7 Weak Convergence and the Direct Method and Rellich (4A.11 Sobolev Embeddings and Critical Exponents) a subsequence converges weakly in $H^1_0$ and strongly in $L^2$. Strong $L^2$ convergence preserves $\int w^2 = \pi$, and weak lower semicontinuity of the norm gives $\frac12\int|u'|^2 \leq \liminf$. By Theorem 1.2, $-u'' = \lambda u$, so $u = \pm\sqrt2\sin x$ (normalised: $\int_0^\pi2\sin^2x = \pi$), the first eigenfunction, with energy $\frac\pi2$.

Solution

$\nabla E = \lambda\nabla G$ with $G = x_1^2 + x_2^2 - 1$: $(2x_1, 4x_2) = \lambda(2x_1, 2x_2)$, so either $x_2 = 0$, $\lambda = 1$ (minimum $1$ at $(\pm1, 0)$), or $x_1 = 0$, $\lambda = 2$ (the maximum). The multipliers are the eigenvalues of the matrix of the quadratic form, and the minimum on the sphere is the smallest eigenvalue: the Rayleigh quotient, which in infinite dimensions is $\lambda_1$ of the Laplacian.

Solution

(a) $\partial_iu = \partial_i\theta(-\sin\theta, \cos\theta)$, so $|\partial_iu|^2 = (\partial_i\theta)^2$. (b) $\Delta u = \Delta\theta(-\sin\theta, \cos\theta) - |\nabla\theta|^2(\cos\theta, \sin\theta)$. So $-\Delta u - |\nabla u|^2u = -\Delta\theta(-\sin\theta, \cos\theta)$, which vanishes iff $\Delta\theta = 0$.

Solution

(a) The parabola is concave, so the line through $(c, h(c))$ lies above it on $[c, 1]$ iff its slope $-\frac{h(c)}{1 - c}$ is at least the tangent slope $-2c$, i.e. $h(c) \leq 2c(1 - c)$, i.e. $\frac12 - c^2 \leq 2c - 2c^2$, i.e. $c^2 - 2c + \frac12 \leq 0$, i.e. $c \geq b$ (for $c < 1$). (b) On $[0, c]$, $u_c' = -2x$, contributing $\frac12\int_0^c4x^2 = \frac23c^3$; on $[c, 1]$ the slope is $-\frac{h(c)}{1-c}$, contributing $\frac12\frac{h(c)^2}{1 - c}$; double for both sides. (c) $E'(c) = 4c^2 + \frac{2h(c)h'(c)(1 - c) + h(c)^2}{(1 - c)^2}$. At $c = b$, $h(b) = 2b(1 - b)$ and $h'(b) = -2b$, so the fraction is $\frac{-8b^2(1 - b)^2 + 4b^2(1 - b)^2}{(1 - b)^2} = -4b^2$, and $E'(b) = 0$. At $c = \frac12$: $h = \frac14$, $h' = -1$, the fraction is $\frac{-\frac14 + \frac{1}{16}}{\frac14} = -\frac34$, and $E' = 1 - \frac34 > 0$.

Solution

$-u'' = 2$ on $(-b, b)$ and $0$ on $b < |x| < 1$, with no point masses since $u'$ is continuous. So $-u'' - f = -u'' \geq 0$; $u - h = 0$ on $[-b, b]$ and $> 0$ outside (the line lies above the concave parabola away from the tangency point); and the product $(u - h)(-u'')$ vanishes because on $[-b, b]$ the first factor is $0$ and outside the second is $0$.

Solution

(a) The variation gives $\int(4\nabla w\cdot\nabla v + Rwv) = \mu\int2wv$ for all $v$, i.e. $-4\Delta w + Rw = 2\mu w$; with $v = w$, $\lambda(g) = 2\mu\int w^2 = 2\mu$. (b) $w = e^{-f/2}$: $\nabla w = -\frac12w\nabla f$ and $\Delta w = w\big(\frac14|\nabla f|^2 - \frac12\Delta f\big)$. Then $-4\Delta w + Rw = w(-|\nabla f|^2 + 2\Delta f + R)$, and dividing by $w > 0$ gives $2\Delta f - |\nabla f|^2 + R = \lambda$.