Course 6 Book 6A: The Heat Equation and Its RelativesChapter 9

Calculus of Variations and Gradient Flows

Euler–Lagrange equations, minimal surfaces, soap films and gradient flows.

25 min read · Updated Oct 3, 2026

Read with Evans, Partial Differential Equations, sections 8.1 (introduction: Euler–Lagrange equations, second variation, systems) and 8.2 (existence of minimisers: coercivity, lower semicontinuity, convexity), and section 9.6 on gradient flows if your edition has it. Colding and Minicozzi, A Course in Minimal Surfaces, is a reference for the geometry.

In this chapter · 6 sections

9.1A soap film between two rings
9.2The Euler–Lagrange equation
9.3Existence of minimisers: the direct method
9.4Gradient flows
9.5History
9.6Exercises

Many of the equations in this book are the equations satisfied by something that minimises an energy. Laplace's equation minimises the Dirichlet energy (6A.2 Harmonic Functions); the minimal surface equation minimises area; a hanging chain minimises its potential energy. The calculus of variations turns " $u$ minimises $E$ " into a PDE, the Euler–Lagrange equation, and its second-order version says when a critical point is stable.

Turned around, an energy also produces an evolution: move downhill as fast as possible. That is a gradient flow, and three flows of this book are gradient flows: the heat equation decreases Dirichlet energy, curve shortening decreases length (6A.8 Curve Shortening and the First Geometric Flows), and the harmonic map heat flow decreases the energy of a map. The Ricci flow is a gradient flow too, but only after a change of point of view that Perelman discovered (12A.2 Ricci Flow as a Gradient Flow), and this chapter is the preparation for it.

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

derive the Euler–Lagrange equation of an integral functional, and the second variation;
prove the existence of minimisers by the direct method, and make Dirichlet's principle rigorous;
write the minimal surface equation, compute the catenoid between two rings and its critical separation;
recognise the heat equation, curve shortening and the harmonic map heat flow as gradient flows, and compute their energy identities;
explain why gradient descent in machine learning is a discretised gradient flow.

A soap film between two rings

In the world Data The catenoid and its critical separation

Dip two parallel coaxial wire rings of radius $R$ into soapy water and pull them apart slowly. A soap film spans them, shaped like a waisted tube. Surface tension makes the film minimise its area, and the area-minimising surfaces of revolution are catenoids, $r(z) = a\cosh(z/a)$ , obtained by rotating a catenary (Exercise 9.2). With the rings at $z = \pm d$ , the condition $a\cosh(d/a) = R$ becomes, with $x = d/a$ ,

\frac dR = \frac{x}{\cosh x}.

The function $\frac{x}{\cosh x}$ rises to a maximum of $0.6627$ at $x \approx 1.1997$ , where $x\tanh x = 1$ , and then falls (Figure 9.1). So for ring separations $h = 2d$ below $1.3255R$ there are two catenoids, a stable one with a wide waist and an unstable one with a narrow waist, which merge at the critical separation; beyond it there is no catenoid at all. Pull the rings further apart and the film's waist thins rapidly and the film pinches off, leaving a flat film across each ring. Even before the critical separation the two flat discs, of total area $2\pi R^2$ , have less area than the stable catenoid once $h$ exceeds about $1.055R$ ; between $1.055R$ and $1.3255R$ the catenoid is a local but not a global minimum, and a disturbance can still burst it.

This is a tabletop experiment and an exact computation, and it shows three things that recur in geometric analysis: a family of solutions can cease to exist when a parameter crosses a critical value, critical points come in stable and unstable pairs that merge there, and the loss of a solution is seen physically as a neck pinching. It is the existence question that 2B.3 Compactness previewed.

Figure 9.1. Left:

\frac{x}{\cosh x}

, whose maximum

0.6627

x \approx 1.1997

gives the critical half-separation

d/R

. A level

d/R

below the maximum has two solutions

x

: two catenoids. Right: profiles

r(z) = a\cosh(z/a)

between rings of radius

1

z = \pm0.5

: the stable wide catenoid (solid) and the unstable narrow one (dashed), computed from the two roots. As the rings move apart to

z = \pm0.6627

the two merge, and beyond that neither exists.

The Euler–Lagrange equation

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

E[u] = \int_UL(\nabla u(x), u(x), x)\,dx

over functions $u$ with given boundary values, with $L = L(p, z, x)$ smooth. If $u$ is a smooth minimiser, then for every $v \in C_c^\infty(U)$ the function $s \mapsto E[u + sv]$ has a minimum at $s = 0$ , so its derivative vanishes there (2A.10 Derivatives):

0 = \frac{d}{ds}E[u + sv]\Big|_{s=0} = \int_U\Big(\sum_iL_{p_i}\partial_iv + L_zv\Big)dx = \int_U\Big(-\sum_i\partial_i\big(L_{p_i}\big) + L_z\Big)v\,dx,

after integrating by parts (1A.10 Divergence, Curl and the Integral Theorems). This holds for all $v$ , so (4A.8 Distributions and Weak Derivatives)

-\sum_{i=1}^n\partial_i\big(L_{p_i}(\nabla u, u, x)\big) + L_z(\nabla u, u, x) = 0,

the Euler–Lagrange equation of $E$ . Any solution, minimiser or not, is a critical point of $E$ .

energy	$L(p, z)$	Euler–Lagrange equation
Dirichlet energy	$\frac12\lvert p\rvert^2$	$\Delta u = 0$
Dirichlet energy with source	$\frac12\lvert p\rvert^2 - fz$	$-\Delta u = f$
area of a graph	$\sqrt{1 + \lvert p\rvert^2}$	$\operatorname{div}\Big(\frac{\nabla u}{\sqrt{1 + \lvert\nabla u\rvert^2}}\Big) = 0$
$\frac12\lvert p\rvert^2 + F(z)$		$\Delta u = F'(u)$

The second line is Poisson's equation, and the third is the minimal surface equation: a graph is a critical point of area exactly when its mean curvature vanishes. In one dimension, with $t$ in place of $x$ , the same computation gives Lagrange's equations of mechanics: with $L = \frac12m\dot x^2 - V(x)$ the Euler–Lagrange equation is $m\ddot x = -V'(x)$ , Newton's law (Exercise 9.5 treats a classical example).

The second variation. At a minimiser the second derivative is also non-negative:

\frac{d^2}{ds^2}E[u + sv]\Big|_{s=0} = \int_U\Big(\sum_{i,j}L_{p_ip_j}\partial_iv\,\partial_jv + 2\sum_iL_{p_iz}v\,\partial_iv + L_{zz}v^2\Big)dx \geq 0.

Testing with rapidly oscillating $v$ shows that the first term must dominate, which forces Legendre's condition: the matrix $L_{p_ip_j}$ is positive semidefinite at a minimiser. When it is uniformly positive definite, the Euler–Lagrange equation is elliptic (6A.1 What a PDE Is). This is why equations coming from convex energies are elliptic, and their gradient flows parabolic. A critical point whose second variation is positive is stable; the narrow catenoid is unstable because the second variation is negative for a variation that pinches its waist further.

Existence of minimisers: the direct method

The Euler–Lagrange equation is a necessary condition. Existence of a minimiser is a separate question, and the answer is the direct method of 4A.6 Weak Convergence and the Direct Method: take a minimising sequence, extract a convergent subsequence by compactness, and show the limit is a minimiser by lower semicontinuity.

Theorem 9.1 Existence of minimisers

Suppose $L(p, z, x)$ is convex in $p$ , and coercive: $L(p, z, x) \geq \alpha|p|^2 - \beta$ with $\alpha > 0$ . Then for every $g \in H^1(U)$ , the energy $E$ has a minimiser among functions $u \in H^1(U)$ with $u - g \in H^1_0(U)$ .

Proof. (Outline; Evans, section 8.2.) Let $m = \inf E$ over the admissible class, and $u_k$ a minimising sequence. By coercivity, $\int|\nabla u_k|^2$ is bounded, and by the Poincaré inequality (4A.9 Sobolev Spaces) so is $\|u_k\|_{H^1}$ . A subsequence converges weakly in $H^1$ (4A.6 Weak Convergence and the Direct Method) and strongly in $L^2$ (Rellich, 4A.10 Sobolev Embeddings and Critical Exponents) to some $u$ in the admissible class. Convexity in $p$ makes $E$ weakly lower semicontinuous: $E[u] \leq \liminf E[u_k] = m$ . So $u$ is a minimiser.

For the Dirichlet energy this is Dirichlet's principle made rigorous: Riemann's 1851 argument assumed a minimiser exists, Weierstrass pointed out that it need not in general, and Hilbert's direct method, completed with Sobolev spaces, shows that here it does. The minimiser is a weak solution of the Euler–Lagrange equation, and the regularity theory of 6A.5 Weak Solutions and Elliptic Regularity shows it is smooth. Existence by minimisation, then regularity by bootstrapping, is the pattern for nonlinear problems too, including the minimisers of Perelman's $\mathcal F$ - and $\mathcal W$ -functionals (12A.2 Ricci Flow as a Gradient Flow, 12A.3 The 𝓦-Entropy).

Area is not coercive in this sense (it grows like $|p|$ , not $|p|^2$ ), and Plateau's problem, to find a surface of least area spanning a given wire loop in space, needed new ideas. Jesse Douglas and Tibor Radó solved it independently in 1930–31, minimising a different energy over parametrised discs; Douglas received one of the first Fields Medals in 1936 for this work.

In the world In use Form-finding in architecture

A tensile roof, a fabric or cable net held in tension between supports, works best when its surface is in equilibrium under uniform tension, which is the condition that it be a minimal surface or close to one. Frei Otto, at his Institute for Lightweight Structures in Stuttgart, found such forms experimentally with soap films stretched over wire frames, among other physical models; the film solves the minimal surface equation for the given boundary. The cable-net roofs of the Munich Olympic Park (1972), by Günter Behnisch with Frei Otto and Fritz Leonhardt, were developed from precise physical scale models before computer form-finding was available. Today the same problems are solved numerically, by minimising a discretised area or energy.

Gradient flows

Let $E$ be a smooth function on a Hilbert space $H$ (4A.4 Hilbert Spaces and Lax–Milgram). Its gradient $\nabla E(u)$ is the vector representing the derivative: $\frac{d}{ds}E[u + sv]|_{s=0} = \langle\nabla E(u), v\rangle$ for all $v$ (Riesz representation). The gradient flow is the ODE

\frac{du}{dt} = -\nabla E(u).

Along it,

\frac{d}{dt}E[u(t)] = \langle\nabla E(u), u_t\rangle = -\|\nabla E(u)\|^2 \leq 0:

the energy decreases, at the fastest possible rate for a given speed $\|u_t\|$ . It is constant exactly at critical points. Generic trajectories descend to local minima; special ones end at saddle points (Figure 9.2).

Figure 9.2. Gradient flow of

E(x, y) = (x^2 - 1)^2 + y^2

, with minima at

(\pm1, 0)

and a saddle at

(0, 0)

(computed). Trajectories cross the level curves at right angles and run downhill into a minimum; only the trajectories starting on the line

x = 0

end at the saddle. In infinite dimensions the same picture describes the heat equation and the other gradient flows of this chapter.

Which gradient you get depends on the inner product. For the Dirichlet energy on $H^1_0(U)$ , with the $L^2$ inner product,

\frac{d}{ds}\frac12\int|\nabla(u + sv)|^2\Big|_{s=0} = \int\nabla u\cdot\nabla v = \langle-\Delta u, v\rangle_{L^2},

so $\nabla E(u) = -\Delta u$ and the gradient flow is $u_t = \Delta u$ . The heat equation is the $L^2$ gradient flow of the Dirichlet energy, and

\frac{d}{dt}\frac12\int|\nabla u|^2 = -\int(\Delta u)^2 \leq 0

(Exercise 9.4). The same energy with a different inner product gives a different flow: with the $H^{-1}$ inner product it gives $u_t = -\Delta^2u$ , a fourth-order equation of the Cahn–Hilliard type.

Curve shortening is the gradient flow of length: 6A.8 Curve Shortening and the First Geometric Flows showed $\frac{dL}{dt} = -\int\kappa V\,ds$ for a normal velocity $V$ , so with respect to the $L^2(ds)$ inner product on normal velocities the gradient of $L$ is $\kappa$ , and $V = \kappa$ is steepest descent. Mean curvature flow is the gradient flow of area in the same way.

The harmonic map heat flow. For a map $u : M \to N$ between Riemannian manifolds, the energy $\frac12\int_M|du|^2$ has as its critical points the harmonic maps, and its gradient flow, introduced by James Eells and Joseph Sampson in 1964, deforms any map towards a harmonic one when $N$ has non-positive curvature. It is used directly in the Ricci flow: DeTurck's diffeomorphisms are solutions of a harmonic map heat flow (11A.3 Short-Time Existence and Uniqueness).

In the world In use Gradient descent is a discretised gradient flow

Training a machine-learning model means minimising a loss function $E(\theta)$ over millions or billions of parameters $\theta$ . The basic algorithm, gradient descent, is

\theta_{k+1} = \theta_k - \eta\nabla E(\theta_k),

which is Euler's method for the gradient flow $\theta' = -\nabla E(\theta)$ with time step $\eta$ , the learning rate. Too large a step and the iteration overshoots and diverges, exactly as an explicit scheme for the heat equation does when $\Delta t > \frac12(\Delta x)^2$ ; too small and progress is slow. Practical methods use random subsets of the data for each gradient (stochastic gradient descent) and add momentum and adaptive step sizes, but the continuous-time picture, energy decreasing along a flow, is how much of the theory of these methods is analysed.

Where this goes Ricci flow as a gradient flow, and min–max

The Ricci flow is not the gradient flow of any function of the metric alone; Hamilton asked whether it was, and Perelman answered: it is, up to diffeomorphisms, the gradient flow of his functional $\mathcal F(g, f) = \int(R + |\nabla f|^2)e^{-f}dV$ , with respect to an $L^2$ inner product weighted by $e^{-f}$ (12A.2 Ricci Flow as a Gradient Flow). One consequence is immediate from the picture above: a gradient flow cannot return to where it started, so the Ricci flow has no periodic solutions other than fixed points ("no breathers"). The other half of the variational picture, finding unstable critical points such as the narrow catenoid, is min–max theory: Birkhoff's sweepouts and the width of a three-manifold, which forces the Ricci flow on certain manifolds to become extinct (10A.8 Min–Max and Width, 12C.2 Finite Extinction).

History

The calculus of variations began with Johann Bernoulli's brachistochrone problem of 1696, and Euler and Lagrange derived its equations in the 1740s–1760s. Joseph Plateau's experiments with soap films were published in 1873, and Plateau's problem was solved by Douglas and Radó in 1930–31. Weierstrass's criticism of the Dirichlet principle dates from 1870 and Hilbert's rescue from 1900; the modern direct method with Sobolev spaces was developed through the twentieth century. Eells and Sampson introduced the harmonic map heat flow in 1964. The Munich Olympic roofs were completed in 1972. The Gateway Arch, analysed in Exercise 9.6, was completed in 1965.

Recall Where we stand

Minimisers of $\int L(\nabla u, u, x)\,dx$ satisfy the Euler–Lagrange equation $-\partial_i(L_{p_i}) + L_z = 0$ , and their second variation is non-negative, which forces ellipticity. Convex, coercive energies have minimisers in $H^1$ by the direct method, smooth by regularity. Area gives the minimal surface equation; between two rings the catenoid exists only up to separation $1.3255R$ , where a stable and an unstable catenoid merge. A gradient flow $u' = -\nabla E(u)$ decreases $E$ at the rate $\|\nabla E\|^2$ ; the heat equation, curve shortening and the harmonic map heat flow are gradient flows of energy, length and map energy. 6A.10 Entropy, Information and Diffusion studies a different monotone quantity along the heat equation: entropy.

Exercises

Exercise 9.2 Surfaces of revolution of least area

A surface obtained by rotating the graph $r = r(z) > 0$ , $-d \leq z \leq d$ , about the $z$ -axis has area $A = 2\pi\int_{-d}^dr\sqrt{1 + r'^2}\,dz$ . (a) Since the integrand does not depend on $z$ , the quantity $r'\frac{\partial L}{\partial r'} - L$ is constant along solutions of the Euler–Lagrange equation (the Beltrami identity); use it to show $\frac{r}{\sqrt{1 + r'^2}} = a$ for a constant $a$ . (b) Solve to get $r = a\cosh\frac{z - z_0}{a}$ .

Solution

(a) With $L = r\sqrt{1 + r'^2}$ , $r'L_{r'} - L = \frac{rr'^2}{\sqrt{1 + r'^2}} - r\sqrt{1 + r'^2} = -\frac{r}{\sqrt{1 + r'^2}}$ . (b) $r'^2 = \frac{r^2}{a^2} - 1$ , so $\frac{dr}{\sqrt{r^2 - a^2}} = \frac{dz}{a}$ and $\operatorname{arccosh}(r/a) = (z - z_0)/a$ .

Exercise 9.3 The critical separation

(a) Show that $\frac{x}{\cosh x}$ has its maximum where $x\tanh x = 1$ , and find $x \approx 1.1997$ by a few steps of Newton's method (2A.10 Derivatives). (b) Deduce $h_{\text{crit}} = 2R\cdot\frac{x}{\cosh x} \approx 1.3255R$ . (c) The same equation $x\tanh x = 1$ appeared for the Frank-Kamenetskii critical parameter in 6A.7 Nonlinear Parabolic Equations. Explain why: in both problems a one-parameter family of solutions of an ODE with $\cosh$ is matched to a boundary condition, and the critical value is where the matching has a double root.

Exercise 9.4 The energy identity for the heat equation

For a smooth solution of $u_t = \Delta u$ on a bounded $U$ with $u = 0$ on $\partial U$ , show that $\frac{d}{dt}\frac12\int|\nabla u|^2 = -\int(\Delta u)^2$ and $\frac{d}{dt}\frac12\int u^2 = -\int|\nabla u|^2$ . Deduce that $\int|\nabla u|^2$ is non-increasing and $\int u^2$ decreasing, and that the ratio $\frac{\int|\nabla u|^2}{\int u^2}$ (the Rayleigh quotient, 6A.5 Weak Solutions and Elliptic Regularity) is also non-increasing. (For the last, use Cauchy–Schwarz: $(\int|\nabla u|^2)^2 = (\int u\Delta u)^2 \leq \int u^2\int(\Delta u)^2$ .)

Exercise 9.5 The brachistochrone

A bead slides without friction from rest at the origin down a wire $y = y(x)$ (with $y$ measured downward) to a point $(x_1, y_1)$ . Its speed is $\sqrt{2gy}$ , so the time taken is $\int_0^{x_1}\frac{\sqrt{1 + y'^2}}{\sqrt{2gy}}\,dx$ . Use the Beltrami identity to show that the fastest curve satisfies $y(1 + y'^2) = c$ , and check that the cycloid $x = a(\theta - \sin\theta)$ , $y = a(1 - \cos\theta)$ satisfies it with $c = 2a$ .

Solution

$L = \frac{\sqrt{1 + y'^2}}{\sqrt y}$ (dropping $\sqrt{2g}$ ). $y'L_{y'} - L = -\frac{1}{\sqrt y\sqrt{1 + y'^2}}$ , constant, so $y(1 + y'^2) = c$ . For the cycloid, $y' = \frac{\sin\theta}{1 - \cos\theta}$ , so $1 + y'^2 = \frac{2}{1 - \cos\theta}$ and $y(1 + y'^2) = 2a$ .

Exercise 9.6 The hanging chain and the Gateway Arch

(a) A chain of uniform density hangs between two points; it minimises its potential energy $\int y\,ds$ subject to fixed length $\int ds$ . With a Lagrange multiplier $\lambda$ , minimise $\int(y - \lambda)\sqrt{1 + y'^2}\,dx$ , and show by the Beltrami identity that $y - \lambda = a\cosh\frac{x - x_0}{a}$ : a catenary. (b) An arch of constant thickness, standing in pure compression, should follow an inverted catenary. The Gateway Arch in St. Louis is thicker at its base than at its top, and its centroid curve is a weighted catenary $y = A\cosh(Bx) + C$ with $AB \neq 1$ (Osserman, "Mathematics of the Gateway Arch", Notices of the AMS, 2010). Explain why a chain with heavier links near its ends would hang in such a curve rather than a catenary.

Exercise 9.7 Step sizes

For $E(\theta) = \frac12\lambda\theta^2$ with $\lambda > 0$ , show that gradient descent $\theta_{k+1} = \theta_k - \eta\lambda\theta_k$ converges to $0$ if and only if $0 < \eta < \frac2\lambda$ . For $E(\theta) = \frac12\sum\lambda_i\theta_i^2$ , show the condition is $\eta < \frac{2}{\lambda_{\max}}$ , and that convergence is then slowest in the direction of $\lambda_{\min}$ . Relate this to the stability condition $\Delta t \leq \frac12(\Delta x)^2$ for the explicit scheme for the heat equation, whose largest eigenvalue is about $\frac{4}{(\Delta x)^2}$ .

Solution

$\theta_k = (1 - \eta\lambda)^k\theta_0$ , which tends to $0$ iff $|1 - \eta\lambda| < 1$ . With several directions each needs $\eta\lambda_i < 2$ ; the slowest factor is $1 - \eta\lambda_{\min}$ . The discrete Laplacian on a grid of spacing $\Delta x$ has eigenvalues up to about $\frac{4}{(\Delta x)^2}$ , so explicit Euler needs $\Delta t < \frac{2}{4/(\Delta x)^2} = \frac12(\Delta x)^2$ .

Exercise 9.8 Rehearsal: gradient flows have no non-trivial periodic orbits

Let $u(t)$ solve $u' = -\nabla E(u)$ , and suppose $u(T) = u(0)$ for some $T > 0$ . (a) Show that $\int_0^T\|\nabla E(u(t))\|^2dt = E(u(0)) - E(u(T)) = 0$ , so $u$ is constant. (b) Perelman shows that $\lambda(g) = \inf_f\mathcal F(g, f)$ is non-decreasing along the Ricci flow and constant only on Ricci-flat (steady soliton) metrics (12A.2 Ricci Flow as a Gradient Flow). A breather is a solution that returns to its initial metric up to diffeomorphism and scaling. Explain, as in (a), why a steady breather (no scaling) must be a steady soliton. Perelman's $\mathcal W$ -entropy does the same for shrinking breathers (12A.3 The 𝓦-Entropy).

Solution

(a) $\frac{d}{dt}E(u) = -\|\nabla E(u)\|^2$ ; integrating over $[0, T]$ gives the identity, and the integrand is non-negative and continuous, so $\nabla E(u(t)) = 0$ for all $t$ , and then $u' = 0$ . (b) $\lambda$ is invariant under diffeomorphisms, so a breather has $\lambda(g(T)) = \lambda(g(0))$ ; since $\lambda$ is non-decreasing, it is constant on $[0, T]$ , which by Perelman's monotonicity formula happens only for steady solitons.

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9.1A soap film between two rings

9.2The Euler–Lagrange equation

9.3Existence of minimisers: the direct method

9.4Gradient flows

9.5History

9.6Exercises

A soap film between two rings

The Euler–Lagrange equation

Existence of minimisers: the direct method

Gradient flows

History

Exercises