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The Alt Phillips Free Boundary Problem

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

The Alt Phillips free boundary problem is the study of functions \(u\) that minimise the energy \[J(u) = \int _{\Omega } \left( \frac{1}{2}\left\lvert \nabla u \right\rvert^2 + W(u) \right) dx,\] where \(\Omega \) is some fixed domain and we have aconstraint \(u = \varphi > 0\) on \(\partial \Omega \). \(W\) is the “potential”, and we’ll impose that \(W(t) = 0\) for all \(t \leq 0\), \(W(t) \geq 0\) everywhere. This is the one-phase version of the problem, where we essentially rule out any possibility of \(u\) dipping below \(0\) when \(\varphi > 0\). Of course, when \(\varphi < 0\), the negative values of the potential become important, but this is outside the scope of this talk.

The Euler-Lagrange equation is simply \(\Delta u = W^\prime (u)\), and when \(W^\prime\) is Lipschitz, standard elliptic theory and the maximum principle imposes that \(u > 0\) everywhere. We are interested in the setting where \(u\) may have “patches” of zeroes, and \(\Gamma = \partial\left\lbrace u > 0 \right\rbrace\) is called the free boundary of the solution.

Here, we’ll focus on \(W(t) = \left( t^+ \right)^ \gamma \) for some \(\gamma < 2\). When \(\gamma =0\), this is called the Bernoulli one-phase free boundary (or the Alt-Caffarelli free boundary problem); when \(\gamma = 1\), this is the familiar obstacle problem.

We’ll focus on the case \(\gamma = 1\), and we’re interested in the geometry of the free boundary. This is sensitive only to the behaviour of \(W\) near \(0\)! Surprisingly, one can even consider \(\gamma < 0\) (though very negative values create infinite energy), and taking \(\gamma \to -2\) with some rescaling makes the energy tend to the perimeter of \(\Gamma \) (which ends up being a minimal surface).

General Results for \(\gamma \in [0, 2)\)

As a simple test case, let’s analyse the 1-D case. When \(u > 0\), we have \(u ^{\prime\prime} = W^\prime(u)\). Multiplying by \(u^\prime\) and integrating recovers \(\frac{1}{2}\left( u^\prime \right)^2 = W(u) + \mu \) for some constant \(\mu \).

Minimisers of \(J\) in this case must have \(\mu = 0\), and this can be proven with an “inner deformation”. Assuming \(u(x) = 0\) for \(x \leq 0\), \(u >0\) for \(x > 0\), and the boundary of \(\Omega \) ends at \(\delta \), we may write \[u_ \lambda (x) = u\left( \delta + \lambda (x - \delta ) \right),\] i.e. we’re deforming the region on which \(u\) is positive. Then \[J\left( u_ \lambda \right) = \int\left( \frac{\lambda }{2} \left\lvert \nabla u \right\rvert^2 + \frac{W(u)}{\lambda }\right) dx, \] and minimality at \(\lambda = 1\) forces \(\mu = 0\). Chasing the remaining computations, we expect the solutions to look like \(u_0(x) = c_0 x^ \alpha \), \(\alpha = \frac{2}{2 - \gamma }\). Note that \(\alpha \geq 1\) when \(\gamma \in [0, 2)\), and in particular, the \(\gamma = 0\) case shows that \(u\) should vary linearly between the zero patch and the boundary. We expect that one-dimensional slices near higher-dimensional free boundaries look like this profile.

So we want to do a scaling argument, and in fact the scaling \[\widetilde u(x) = \frac{u(rx)}{r^ \alpha }\] does not change our energy at all.

Regularity

The existence of a minimiser with prescribed boundary data \(\varphi \) is pretty standard, and one gets some \(u\in H^1(\Omega ) \cap L^ \gamma (\Omega )\) as the minimiser. We get the Euler-Lagrange equation \(\Delta u = \gamma u ^{\gamma - 1}\), but only when \(\gamma \geq 1\)! Otherwise, a negative power of \(u\) is possibly not integrable if \(u\) is ever \(0\), etc.

Instead of trying to be finnicky with this Euler-Lagrange equation from the getgo, one can observe that \(u\) is subharmonic, i.e. \(\Delta u \geq 0\). By taking a “downward” perturbation \(J(u - \epsilon \psi \) \geq J(u)\) and using the monotonicity of \(W\), we get \[\int \left\lvert \nabla (u - \epsilon \psi \right\rvert^2 dx \geq \int \left\lvert \nabla u \right\rvert^2\] for all \(\epsilon > 0\). This implies \(\Delta u \geq 0\) in a weak sense, but this lets us define \(u\) as an upper semicontinuous function! A big step up from \(H^1\).

Proof

When \(\gamma > 1\), we get \(\Delta u = W’(u)\), so \(\left\lvert \Delta u \right\rvert \leq C \left( 1 + u \right)\). The regular Harnack inequality now applies.

When \(\gamma \leq 1\), let’s define \[a = \frac{1}{\left\lvert \partial B_1 \right\rvert} \int _{\partial B_1} u\ dr \geq u(0) \geq C_0.\] The first inequality is by the maximum principle. Let \(\widetilde u = \frac{1}{a} u \), and let \(h\) be the harmonic function with \(h = \widetilde u \) on \(\partial B_1\) and \(h(0) = 1\). Now use the rescaled energy inequality with \(\widetilde u \): \[\begin{align*} \int \left( \frac{1}{2} \left\lvert \nabla \widetilde u \right\rvert^2 + a ^{\gamma - 2} \widetilde u ^{\gamma } \right) dx & \leq \int \left( \frac{1}{2} \left\lvert \nabla h \right\rvert^2 + a ^{\gamma - 2 }h ^ \gamma \right) dx \\ \implies \int \frac{1}{2} \left\lvert \nabla \left( \widetilde u - h \right) \right\rvert^2 dx & \leq a ^{ \gamma - 2 }\int h ^{\gamma } dx \\ & \leq a ^{\gamma - 2 }\int _{B_1} C \left( 1 + h \right)dx \\ & \leq C a ^{\gamma - 2}. \end{align*}\]

This second-to-last inequality is the usual Harnack inequality again, and the last inequality is by using the mean value property of \(h\). We may thus obtain that \[\left\lVert \widetilde u - h \right\rVert _{H^1\left( B_1 \right)} \leq C a ^{\frac{\gamma - 2}{2}},\] and eventually (after some Poincaré-like inequality) this trickles down to the average value of \(\widetilde u \) on \(\partial B _{\frac{1}{2}}\) being bounded from below by \(1 - C a ^{\frac{\gamma - 2}{2}}\). In particular, we get that if the average of \(u\) on \(\partial B_1\) is \(a\), then the average of \(u\) on \(\partial B _{\frac{1}{2}}\) is at least \(a \left( 1 - C a ^{\frac{\gamma - 2}{2}} \right)\), and we can iterate when \(a\) is large enough! Eventually you’ll get \(u(0) \geq \frac{a}{2}\), which gives half of the Harnack inequality.

Proof left as an exercise…the idea is to rescale and apply some interior estimates (not too sure what that means). Here, \(C ^{\alpha }\) means “\(\alpha - 1 \) Lipschitz derivatives” when \(\alpha \in \mathbb{N}\).

In addition to this, this comes with a nondegeneracy estimate: if \(0\in \Gamma \), then \(\max _{\partial B_r} u \geq c r ^{\alpha }\) for some constant \(c\) and for \(r\) sufficiently small. This is done by constructing an energy barrier and energy competitors. This will allow us to do some blowup procedures and establish a profile of the free boundary. A critical tool is

All you really have to do is compute the derivative and see what falls out. It’s only really important that \(u\) is a critical point of the energy; it doesn’t have to be a minimiser on the nose. Another annoying aspect is the inner product structure of \(\left\lvert \nabla u \right\rvert^2\); with other powers of the gradient, it’s not easy to do this at all.

We would very much like to obtain regularity of the free boundary by studying these blowup profiles. This entails (a) classifying \(\overline{u}\) (called “cones”) and (b) attaining some uniqueness of blowup limits. This second point is so that we can go backwards from the blowup profile to say something about the free boundary.

The Case \(\gamma = 1\).

Our Euler-Lagrange equation is \[\Delta \overline{u} = \chi _{\left\lbrace \overline{u} > 0 \right\rbrace}.\] We already know \(\left\lVert \overline{u} \right\rVert _{C ^{1, 1}} \leq C\), and of course \(\overline{ u }\) is homogenous of degree \(\alpha = 2\). Understanding these profiles comes in two steps:

1. In any unit direction, we show \(\Delta \left( \overline{u} _{ee}\right) = 0\) on \(\left\lbrace \overline{u} > 0\right\rbrace\). Taking a minimising sequence of points for \(\overline{u} _{ee}\) and performing a homogenous rescaling, we obtain a globally defined harmonic function with an interior minimum, hence \(\overline{u} _{ee}\) is a negative constant by harmonicity. This contradicts nonnegativity of \(\overline{u}\) everywhere! Thus \(\overline{u}\) is convex.
2. By convexity, \(\left\lbrace \overline{u} = 0 \right\rbrace\) is a convex cone. We split into two cases:
   • \(\left\lbrace \overline{u} = 0 \right\rbrace\) has nonempty interior. Without loss of generality, we may assume that every line parallel to the \(x_n\)-axis intersects the zero set. From one-dimensional calculus and convexity, \(\overline{u} _{x_n}> 0\) whenever \(\overline{u} > 0\). Every directional derivative solves the same spherical Laplacian eigenvalue problem, so every directional derivative is a scalar multiple of \(\overline{u} _{x_n}\). It follows that \(\overline{u}\) only depends on \(x_n\), so \(\overline{u} = \frac{1}{2}\left( x_n^+ \right)^2\).
   • \(\left\lbrace \overline{u} = 0 \right\rbrace\) has empty interior. Then it’s contained in a hyperplane, which has measure zero, hence \(\Delta \overline{u} = 1\) everywhere since we have continuity of \(u\). By Liouville’s theorem, \(\overline{u}\) is a quadratic polynomial: \(\overline{u} = \frac{1}{2} x^\top A x\) for some positive-semidefinite matrix \(A\) with \(\operatorname{tr}(A) = 1\).

Some remarks: suppose one has boundary data where the incidence set of the solution \(\left\lbrace u = 0 \right\rbrace\) has many components. By changing the boundary data, we can imagine two or more components colliding and fusing; what does that meeting ponit look like? It can’t be the first case in step 2 above, hence there must be a very “thin cusp” connecting the two. Second, changing \(\gamma \) doesn’t affect the first case, but now the second case becomes a lot harder to handle.

In joint work with Hui Yu, the speaker proved that when looking at blowup profiles as \(\gamma \to 1\), they can only converge to axisymmetric quaratic polynomials! This was done by analysing the critical points of the Weiss energy. This indicates that for \(\gamma \approx 1\), the problems are very rigid compared to the obstacle problem.

It’s not hard to show that if \(\frac{1}{2}\left( x_n^+ \right)^2\) is a blowup profile of \(u\), then it’s unique.

This is because of a “improveness of flatness” argument: if one has \[\left\lVert u - \frac{1}{2}\left( x_n^+ \right)^2 \right\rVert _{L^\infty \left( B_1 \right)} \leq \epsilon ,\] then for some unit vector \(\nu \) close to \(e_n\) and some \(\rho > 0\) one has \[\left\lVert u - \frac{1}{2}\left( (x\cdot \nu )^+ \right)^2 \right\rVert _{L^\infty \left( B_ \rho \right) } \leq \frac{1}{2}\epsilon \rho ^2.\] Now iterate!

One can show that all blowup profiles are unique, but the irregular profiles are not stable under perturbations of the boundary data. In fact, the irregular profiles are solutions to the thin obstacle problem! (This is the linearisation of the original PDE.) See Figalli-Serra for more.

The Case \(\gamma = 0\)

Recall that our differential equation is \(\Delta u = \gamma u ^{\gamma - 1}\) on \(\left\lbrace u > 0 \right\rbrace\). When \(\gamma = 1\) as above, the nonlinearity disappears; likewise, when \(\gamma = 0\), we instead arrive at the similarly linear \[\begin{cases} \Delta u = 0 & \textrm{on }\left\lbrace u > 0 \right\rbrace, \\ \left\lvert \nabla u \right\rvert = \sqrt 2 & \textrm{on }\Gamma . \end{cases}\]

Recall that this is the one-phase problem, wherein we are minimising (or finding critical points of) the energy \[J(u) = \int \left( \left\lvert \nabla u \right\rvert^2 + \chi _{\left\lbrace u > 0 \right\rbrace} \right) dx.\] This has some significant differences compared to the \(\gamma = 1\) case.

1. When \(u\) is a minimiser of \(J\), \(\left\lbrace u = 0 \right\rbrace\) has positive density. This is philosophically because of the optimal regularity. If \(u\) has a very skinny zero set, \(u\) is “almost harmonic”, yet it must also grow linearly in some direction. These together force the gradient to become large at some point, a contradiction of minimality.
2. One has \(u(x) \gtrsim \operatorname{dist}(x, \Gamma )\). This is the “reverse” optimal growth, which in general is not true (including when \(\gamma = 1\)).

In dimensions \(n \leq 4\), the only minimising cones are \(x_n^+\) up to rotation. That is, you can only encounter a singular cone in dimension \(5\). Contrast this with the obstactle problem: every dimension admits singular cones. (See Alt-Caffarelli for \(n = 2\); Caffarelli-Jerrison-Kenig for \(n = 3\); Jerrison-Savin for \(n = 4\).)

For higher dimensions, one can analyse the values of axisymmetric cones \(\overline{u}\) (possibly not minimal) on the unit sphere. Rewriting our PDE actually produces an ODE on the sphere, which can be solved to deduce the free boundary \(\Gamma \). In dimensions \(n \leq 6\), cones obtained with this process are unstable — there are many other solutions nearby that “cross over” the free boundary of the cone — hence cannot be minimal.

But in dimensions \(n \geq 7\), these cones are stable, and the nearby solutions foliate an annular neighbourhood of \(\Gamma \). This requires some computer assistance. Stability analysis only observes the behaviour of the cone and neighbouring solutions at \(\infty\), but to get minimality, one needs to additionally undestand the behaviour near the orgin. De Silva and Jerrison followed this technique and proved that axisymmetric cones are minimising in dimensions \(n \geq 7\).

The Case \(\gamma \in (1 - \delta , 1)\)

When \(\delta > 0\) is sufficiently small, one can find radial (hence singular) cones in dimension \(n = 3\), but not \(n = 2\). In dimension \(n = 4\), one just extends the radial cone axially to produce an axisymmetric minimising cone. But in addition, there is a very close minimising cone with a zero set of positive density! This shares properties of both the \(\gamma = 1\) and \(\gamma = 0\) scenarios, and it’s suggestive of a bifurcation occurring just below \(\gamma = 1\).

The Case \(\gamma \in (-2, 0)\)

Recall the modelling ODE in one dimension: \[\frac{1}{2}\left( u^\prime \right)^2 = u ^{\gamma } + \mu .\] When \(\gamma > 0\), the \(\mu \) dominates for \(u\approx 0\), and minimality dictates that \(\mu = 0\) (after a Taylor expansion, for example). When \(\gamma = 0\), the two terms are comparable, and this gives us the condition on the gradient we saw earlier.

When \(\gamma < 0\), the same analysis gives \[u = c_0 \left( x^+ \right)^ \alpha + c(\mu ) x ^{2- \alpha } + \cdots,\] and minimality yields \(c(\mu ) = 0\).

Traditional methods of understanding the condition \(c(\mu ) = 0\) don’t work very well because \(\nabla u \) is very ill-behaved. But by using the theory of viscosity solutions, one can begin to recover some familiar results from the \(\gamma \geq 0\) cases (optimal regularity, optimal growth, etc.).