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Symmetry and Uniqueness via a Variational Approach

Speaker: Yao Yao
Date of Talk: June 23, 2025
Upstream link: UCI PDE Summer School

We begin with the Keller-Segel equation: \[\rho _ t = \Delta \rho + \nabla \cdot \left( \rho \nabla \left( \mathcal{N} \star \rho \right) \right),\] where \(\mathcal{N} = \frac{1}{2 \pi } \log \left\lvert x \right\rvert\). \(\rho \) represents the population density of a population of cells moving by Brownian motions, hence the \(\Delta \rho \) contribution. But additionally, these cells may emit some chemicals they’re attracted to (or repelled by), hence they’ll follow the concentration gradient of this chemical \(c\). This concentration gradient itself diffuses too, so \(c_t = \Delta c + \rho \). Writing \(\rho _t = \Delta \rho - \nabla\cdot \left( \rho \nabla c \right)\) yields a system of 2 equations called the parabolic-parabolic Keller-Segel equation.

But the diffusion of chemicals is often much faster than the movement of cells, and modelers decided to assume that \(c_t = 0\) (it reaches equilibrium too quickly to affect the evolution of \(\rho \)). Solving the now elliptic \(0 = \Delta c + \rho \) yields \(c = \mathcal{N} \star \rho \), recovering the parabolic-ellipt Keller-Segel equation we started with.

This is an example of an aggregation-diffusion equation: the Laplacian diffuses the population while the second term pulls everyone together. Which of the two forces wins out?

Mass \(M = \int \rho _0 dx\) is conserved by the equation, and there’s a critical mass that characterises which of the two forces “wins”. In fact, it’s \(M_c = 8 \pi \).

One can show, after some computations and rewriting things as a transport equation, that \(\rho _t = \nabla\cdot \left( \rho \cdot \frac{\delta E}{\delta \rho } \right)\), where \[E\left[ \rho \right] = \int \left( \rho \log \rho + \frac{1}{2}\rho \left( \rho \star \mathcal{N} \right) \right) dx.\] Note \(E\) is a Lyapunov function. Rescaling by \(\rho_ \lambda (x) = \lambda ^2 \rho \left( \lambda x \right)\) for some \(\lambda > 0\), we get \[E\left[ \rho _ \lambda \right] = E \left[ \rho \right] + 2 M \log\lambda - \frac{1}{4 \pi }M^2\log \lambda .\] Morally speaking, when \(M > 8 \pi \), it’s favourable to take \(\lambda \to \infty\), i.e. we expect aggregation to happen. Likewise, when \(M < 8 \pi \), taking \(\lambda \to 0\) is favourable, and we expect things to diffuse. This isn’t a formal proof, but one can prove that if \(M > 8 \pi \) and \(\rho _0\) has bounded second moment, the population density blows up in finite time. All you have to do is differentiate the second moment!

This kind of equation generalises: consider the evolution

\[\rho _t = \Delta \rho ^m + \nabla \cdot \left( \rho \nabla \left( W\star \rho \right) \right),\] where \(m \geq 1\), \(W\) is radially symmetric and increasing in \(r\), and everything is defined on \(\mathbb{R}^d\). The first term models local repulsion while the latter models nonlocal interactions between our cells.

When \(m > 2-\frac{2}{d}\), the diffusion term is so strong that global existence is guaranteed, though the asymptotic behaviour of solutions is shrouded in mystery.

Once again, we can create the free energy \[E\left[ \rho \right] = \underbrace{\frac{1}{m-1}\int \rho ^m dx }_{S\left[ \rho \right] \textrm{ (entropy)}} + \underbrace{\frac{1}{2}\int \rho \left( \rho \star W \right) dx }_{ I \left[ \rho \right] \textrm{ (interaction)}}.\] This is still (formally) a Lyapunov function, and better yet, solutions are gradient flows of \(E\) in certain Wasserstein spaces (though this requires additional assumptions on \(W\)). A similar analysis as before can identify if there is a critical mass (leveraging the Hardy-Littlewood-Sobolev inequality) depending on \(m\), though this is not our focus.

Our first question is whether or not there’s a stationary solution, and this is usually the first step to understanding long-term behaviour of solutions. This has been answered, as the energy admits global minimisers. This was shown with the concentation-compactness principle, and applying the Riesz rearrangement shows that such stationary solutions must be radially decreasing. Do we expect radial symmetry?

Suppose instead that \(\rho _s\) is a stationary non-radial solution. Then there must be some hyperplane cutting the mass of \(\rho \) in half asymmetrically. Then do a Steiner symmetrisation to decrease the energy. The idea is to use a layer-cake decomposition and translate each layer a little towards the dividing hyperplane.

Note that this only directly produces a contradiction when the problem has a gradient flow formulation. Otherwise, one is only exposed to the Euler-Lagrange equation, and the perturbations in the Steiner symmetrisation aren’t compatible with the derivation of the Euler-Lagrange equation! These perturbations go “the wrong way” and produce very large vertical changes near irregular points on \(\rho _s\). The fix is to manually adjust the Steiner symmetrisation to avoid large vertical changes when shifting our layers.

The main point is that a delicately chosen perturbation demonstrates symmetry!

Uniqueness is also a question, and it was only known for Newtonian, Riesz, and convex potentials. Another exceptional case \(m = 2\) with a \(C^2\) attractive potential \(W\) is known. In 2022, this was generalised: if \(m \geq 2\), \(W’\) is \(C^1\) away from the origin, and \(W’(r) \lesssim r ^{-d-1+ \delta }\) on \((0, 1)\), then the steady states for a given mass are unique up to translation.

The idea is to construct an interpolating curve \(\rho _t\) from \(\rho _ 0\) to \(\rho _ 1\) where the energy is convex. This forbids \(\rho _ 0 \) and \(\rho _1\) from both being critical points of the energy (here we need the gradient flow structure of the problem). These \(\rho _t\)’s are hard to find — linear interpolations and following the Wasserstein geodesic doesn’t work. Instead, one begins with \(\rho _0, \rho _1\) both radially decreasing step functions, linearly interpolating their heights (so the intermediates are superpositions of two such step functions). This can be extended to radially decreasing simple functions, then to any measurable functions. This just happens to work!

The condition \(m \geq 2\) is to ensure the convexity of the entropy, and the speaker et. al showed that for \(1 < m < 2\), in fact there are at least countably many stationary solutions. There still remain some open questions, though: uniqueness (either in the positive or negative) is unknown for \(m = 1\). When \(m > 2\), long-term behaviour is unknown (it’s hard to prevent mass from escaping to \(\infty\)).

Fluid Equations

We’ll begin with a fun toy problem:

There was a great description of the incompressible Euler equation, which states that for a velocity field \(u\) describing a fluid’s motion, one has \[u_t + u \cdot \nabla u = -\nabla p,\] where \(p\) is some implicitly defined pressure that makes \(\nabla \cdot u = 0\). Here, \(u_t\) is the acceleration at a point, but the \(u\cdot \nabla u \) corrects this measurement as if the observer was moving along with the fluid. Thus, the left side is the acceleration “relative to the fluid”.

One can rewrite this equation in a pressure-independent way via vorticity. In 2D, the vorticity is a scalar given by \[\omega = \nabla^\perp \cdot u,\] where \(\nabla^\perp\) is the usual gradient “vector” rotated \(90^\circ\). One now has the equation \[\omega _t + u \cdot \nabla \omega = 0\] with the inverse relation \(u = \nabla^\perp \left( \omega \star \mathcal{N} \right)\) given by the Biot-Savart law.

When one imposes the initial condition \(\omega _0 = \mathbf 1_D\) for some bounded domain \(D\subseteq \mathbb{R}^2\), in fact \(\omega (t) = \mathbf 1 _{D_t}\); how does \(D\) evolve over time? What are the stationary and/or uniformly rotating solutions? One can show (using similar computations as in the derivation of the vorticity form) that a “patch” solution is stationary if and only if (!) \(\mathbf 1_D * \mathcal{N}\) is locally constant on \(\partial D\).

For rotating solutions, one instead gets the condition that \(\mathbf 1_D \star \mathcal{N} - \frac{1}{2} \Omega \left\lvert x \right\rvert^2\) is locally constant on \(\partial D\) for some real parameter \(\Omega \) (representing the angular velocity of the rotation). Thus, we should ask what kinds of domains satisfy this equation.

Setting \(\psi = \mathbf 1_D \star \mathcal{N}\), one obtains the elliptic equation \(\Delta \psi = \mathbf 1 _{\left\lbrace \psi < c \right\rbrace}\) for some \(c\) depending on \(\Omega \)! This bears a strong resemblance to the obstacle problem.

In 2000, Fraenkel showed that if \(D\) is simply connected and \(\Omega = 0\), then \(D\) is a disc; this was based on the method of moving planes. Later, in 2014, Hmidi showed that when \(D\) is simply connected and \(\Omega = \frac{1}{2}\), then \(D\) is a disc, likewise when \(D\) is convex and \(\Omega < 0\).

On the other hand, any ellipse of radii \(a\neq b\) can be a rotating patch if \(\Omega = \frac{ab}{(a+b)^2} < \frac{1}{2}\) (Kirchhoff, 1876). Additionally, there are bifurcating rotating patches with \(m\)-fold symmetry with \(\Omega = \frac{m-1}{2m} < \frac{1}{2}\) (Burbea, 1982).

The speaker and coauthors showed that unconditionally rotating patches \(D\) are radially symmetric when \(\Omega \leq 0\) or \(\Omega \geq \frac{1}{2}\)! The proof is variational: when \(D\) is simply connected, find an energy functional for which rotating patches are critical points, then perform careful deformations of \(D\) when \(D\) is not radial.

A key ingredient of the proof is

This says to symmetrise the equation in two different ways: you can either solve the PDE and then symmetrise, or you can symmetrise first and solve. (Talenti additionally proved a generalised elliptic version and even a nonlinear version for generalised Laplace operators.)