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# REMARKS ON THE CONVEXITY OF FREE BOUNDARIES (SCALAR AND SYSTEM CASES)

Authors:
Algebra i analiz St. Petersburg Math. J.
Tom. 32 (2020), }4 S 1061-0022(XX)0000-0
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES
(SCALAR AND SYSTEM CASES)
L. EL HAJJ AND H. SHAHGHOLIAN
Dedicated to Nina Nikolaevna Ural tseva on the occasion of her 85th birthday
Abstract. Convexity is discussed for several free boundary value problems in exte-
rior domains that are generally formulated as
u=f(u) in \D, |∇u|=gon , u 0 in Rn,
where uis assumed to be continuous in Rn, Ω = {u > 0}(is unknown), u= 1 on
∂D, and Dis a bounded domain in Rn(n2). Here g=g(x) is a given smooth
function that is either strictly positive (Bernoulli-type) or identically zero (obstacle
type). Properties for fwill be spelled out in exact terms in the text.
The interest is in the particular case where Dis star-shaped or convex. The focus
is on the case where f(u) lacks monotonicity, so that the recently developed tool of
quasiconvex rearrangement is not applicable directly. Nevertheless, such quasicon-
vexity is used in a slightly diﬀerent manner, and in combination with scaling and
asymptotic expansion of solutions at regular points. The latter heavily relies on the
regularity theory of free boundaries.
Also, convexity for several systems of equations in a general framework is dis-
cussed, and some ideas along with several open problems are presented.
§1. Introduction
We discuss several free boundary value problems and their geometric properties such
as that of being star-shaped or convex for (continuous) solutions in the exterior of a given
domain. Starting with a bounded domain DRn(n2) and a given function fon R
(with properties to be speciﬁed) we consider the following semilinear partial diﬀerential
equation
(1.1)
u=f(u) in Rn\D,
u= 1 on ∂D,
u0 in Rn,
and ask whether the super-level sets of positive (continuous) solutions to this equation
inherit certain geometric properties of D. We also discuss the particular case of the so-
called Bernoulli problem, where the bulk equation in (1.1) is satisﬁed in Ω := {u > 0},
and an extra boundary condition |∇u|=gis imposed on Ω. We will be more exact on
these points later.
In order to give a fair chance to the problem for positive results, we need, to a certain
extent, restrict the problem to a reasonable class of right-hand sides f(u). The class we
shall consider is still very general and contains many classical free boundary problems as
well as some new ones.
2010 Mathematics Subject Classiﬁcation. Primary 35R35.
Key words and phrases. Convexity, starshapedness, uniqueness, system of equations.
H. Shahgholian was supported by Swedish Research Council.
c
2004 American Mathematical Society
1
2 L. EL HAJJ AND H. SHAHGHOLIAN
For the system case we consider equations of the type
(1.2)
u=f(x, u) in Dc,
u=kon D,
u0in Rn,
where u= (u1, u2,·, um) is continuous,
k= (k1,...,km) and D= (D1, D2,·, Dm),
with DjRn. We use the componentwise notation throughout the paper; in particular
u0means that ui0 for i= 1,···m.
When f=uFwith F“reasonably” smooth, one has a variational formulation of the
problem. The reason for introducing the system of equations is to initiate and promote a
possible development of qualitative theory for free boundary value problems for systems,
which scarcely exists now. Our discussions for systems are at a heuristic level and we
present some ideas and an example with hand-waving arguments. We also state and give
a heuristic proof of the existence of starshaped solutions whenever the given domains
and ingredients have this property. We hope that mathematical tools will come in the
future, and that the current discussion will tease the appetite of many enough for such
a development.
What concerns the scalar case, our approach combines tools from quasiconvexiﬁcation
[7, 13] and those of free boundary regularity of both Bernoulli type and obstacle type
problems. For the system case we also resort to recent results of regularity theory for
the corresponding problems. In simple terms our approach can be explained as follows.
Given a solution to the above problem, we take the quasiconvex rearrangement uthat is
larger than u. It is known that usatisﬁes the inequality ∆uf(u), provided f(u)0
and |∇u|>0 inside the set {u > 0}. Now if fis also monotone, one can conclude by the
comparison principle that u=u, and consequently the super-level sets of uare convex.
Since we lack this monotonicity, we can circumvent the diﬃculty by rescaling the solution
ut0(x) = u(t0x) such that ut0(x)u(x) and t0<1 is the largest such value. From
this, it must follow that the surfaces of uand ut0touch at some point(s) x0. If the
touch is in the set where u(x0)>0, then we use a standard argument combined with
the comparison principle between the two functions and reach a contradiction. If it is on
the boundary of the support, the matter is subtler, and we have to resort to regularity
theory and asymptotic expansion of the solutions, and ﬁnally comparison between them
The method in this paper can be generalized to other operators (such as fully nonlinear
operators) and a general right-hand side (depending on xas well as on u) whenever
the corresponding quasiconvexiﬁcation and regularity theory of the free boundaries are
available.
§2. Scalar equations
In this section, we make the following general assumptions on f(see [10] for similar
type of problems):
(2.1)
f(t)>0 for t > 0,
f(t) = 0 for t0,
f(t) = btα+o(tα) for t < t1,some t1>0 and 1< α < 1,
where b0. We also assume that
fis either left- or right-continuous and its discontinuities are isolated.(2.2)
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES 3
We can relax the third condition to allow bbe a function of x, with certain assumptions
to ﬁt into the context of quasiconvex rearrangement, see [7]. For clarity, we leave this
generalization to readers to ﬁgure out.
Property (2.2) includes examples such as the singular perturbation problem ∆uǫ=
1
ǫχ{0<uǫ}, with the above boundary value on D. It is well known that solutions to this
problem converge to the Bernoulli free boundary.
Theorem 1. Let Dbe a bounded convex domain in Rn(n2), and let fhave properties
(2.1)(2.2). Let g(x)be a Cβ-function in the entire space satisfying max(b, g(x)) b0
for some b0, and let g(x)be either identically zero or strictly positive. When g > 0,we
assume 1/g to be a concave function.
Then there exists a nonnegative function uwith Ω := {u > 0}solving1the free bound-
ary problem
(2.3)
u=f(u)in \D,
u= 1 on D,
|∇u|=gon ,
with convex super-level sets,i.e.,{u > l}is convex for all 0l1.
If in addition there is a point zDsuch that
(2.4) tg(t(xz) + z)is monotone nondecreasing in tfor all xRn,
then uis a unique solution.
Remark 1.The theorem can be generalized in two directions.
i) One may allow f=f(x, u, u), with f(x, s, t)/t2convex in both xand tfor all s;
see [7].
ii) One may also allow f(t) to take zero values for t > 0 as well as being in L1, through
approximation of fby smooth functions, and then passing to the limit, because the
convexity of the super-level sets will be preserved. A technical issue is that one needs to
make sure that the supports of the approximating solutions will stay uniformly compact.
In this case the uniqueness proof of the theorem will however not work anymore.
Proof. The existence of solutions follows in a standard way as in the literature, by using
the corresponding functional: we vaguely approximate the function fwith c0uαfrom
below, replace Dwith a larger ball containing D, and then use standard comparisons
with the corresponding symmetric functional; see for example [12] for similar arguments2.
Therefore we focus on the convexity of solutions only.
We let ube the quasiconcave envelope of the function u(see [7, 13]). By deﬁnition,
uis the function whose super-level sets are the closed convex hulls of the corresponding
super-level sets of u. If we take the support of uto be Ω = {u > 0}and denote by Ω
the convex hull of Ω, then we have (see [7])
={x:u>0}.
From [7, 13] we have
(2.5) ∆uf(u) in Ω,
1The notion of a solution here can be taken in the weak sense for the PDE, and in the continuous
sense for all boundary conditions including the supplementary gradient condition. We also remark that
for linear PDEs, weak and viscosity solutions are the same. See [14, Page 103].
2One can actually use the so-called nondegeneracy argument to show that any (local/global) mini-
mizer ushould satisfy supBR(z)uc0cR2/(1α)+u(z), as long as BR(z)Dcand u(z)>0. Since
u < 1 we have a uniform bound on Rand hence on the support of u. See, e.g., [2].
4 L. EL HAJJ AND H. SHAHGHOLIAN
in the sense of viscosity, and hence in the weak sense (since the Laplacian is linear), see
[14, page 103] for the deﬁnition of viscosity solutions and the equivalence of weak and
viscosity. We remark further that (see [4, Proposition 3.7])
(2.6) uC1because uC1.
Our aim is to prove that uu, using the so-called Lavrentiev method of scaling and
comparing. Hence we deﬁne
ut(x) = u(tx),t:= {x:tx }={x:ut(x)>0},
and set
t0:= sup{t:ut(x)u(x), x }.
Clearly t0<1, otherwise u=uand we are done. Therefore there is x0with u(x0) =
ut0(x0) such that either
A: x0(Ωt0)\D, or
B: x0t0.
In case A, we set s0:= u(x0) = ut0(x0)>0, and by the hypotheses (2.1)–(2.2) we
may consider two subcases:3
A1: fis continuous in a neighborhood of s0;
A2: fis continuous in a punctured neighborhood of s0.
For A1 we use the continuity of fto show that t2
0f(ut0(x)) f(u(x)), in a small
neighborhood of x0. Indeed, for any xclose to x0we have
(2.7) f(ut0(x)) f(ut0(x0)) + ω(r1),
where r1:= |ut0(x)ut0(x0)|and ωis the modulus of continuity for f. Once again using
the continuity of fand (2.7) we have (with r2:= |ut0(x)u(x0)|)
(2.8) f(ut0(x)) f(u(x0)) + ω(r2)f(u(x)) + ω(r1) + ω(r2).
Now by (2.8) and (2.5) it follows that
(2.9) ∆ut0=t2
0f(ut0(x)) t2
0f(u(x)) + t2
0(ω(r1) + ω(r2)) f(u(x)) u,
provided r1, r2are chosen suﬃciently small such that
ω(r1) + ω(r2)<1t2
0
t2
0
f(u(x)),
for xclose to x0. Hence ∆ut0uin Br(x0). Since also ut0uin a neighborhood
of x0and ut0(x0) = u(x0), we reach a contradiction to the strong comparison principle.
For the second subcase A2, we may assume for deﬁniteness that fis left-continuous.
The right-continuous case can be treated in exactly the same way as the left-continuous
case, and hence is left to the reader. Now take xBr(x0) with ut0(x)< ut0(x0) (since
we look at the left side of s0), and deﬁne K:= Br(x0)∩ {ut0(x)< s0}. Recall also that
s0:= u(x0) = ut0(x0)>0. Using a continuity argument as above, we easily arrive at
(2.8), and hence we deduce (2.9) for such x. From here we conclude, as argued above,
that
(2.10) ut0u, ut0(x0) = u(x0),ut0uin K.
Next, we observe that ∆ut0=t2
0f(ut0) is bounded and hence ut0is a C1function
locally near x0(for all 0 < β < 1). Recall (2.6) that uC1.
3We remark that case A2 covers also A1, but for clarity of the exposition we have considered the
cases separately.
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES 5
Along with (2.10), this implies
(2.11) |∇ut0(x0)|=|∇u(x0)|.
To this end we want to invoke Hopf’s boundary point lemma (in K) to contradict (2.11),
and conclude that u=u. To see that Hopf’s lemma can be invoked we only need (2.10),
the C1-regularity of both uand u, along with the interior C1,Dini-condition for Kat x0.
To show the last property, we use the fact that for each point zthe super-level
set Lz:= {u(x)> u(z)}has a support plane and hence the standard Hopf lemma for u
applies at zfrom the interior of the set Ω \Lz, and we can conclude that |∇u(z)|>0.
This together with the C1-regularity of umentioned above implies that Lzis C1in
a neighborhood of z. By the construction of u, the boundary of the super-level sets of
umust be C1, and hence Hopf’s lemma can be applied. This concludes the proof for
this case.
Now, we turn our attention to Case B, i.e., x0t0. In this case we know,
by the regularity theory, that near the special points xithat span \Ω, the free
boundary is smooth.4Without loss of generality, assume that x0= 0 and the inward
normal to at the origin is e1= (1,0,...,0). In particular, this means that the
inward normal to Ω at each extremal point yithat span the origin5is e1, and hence
(by the regularity theory for free boundaries) we have a local asymptotics for uat these
extremal points:
(2.12) u(x) = ax+
1+b(x+
1)κ+o(|x|κ),where a=g(x0), κ =2
1α.
Now, we consider two subcases, which appear in the assumptions (2.1).
B1: g0, i.e., u(x) = b(x+
1)κ+o(|x|κ).
B2: g > 0, i.e., u(x) = ax+
1+o(|x|), with a > 0.
For Case B1 a straightforward calculation in the 1-dimensional case gives
b=1
α(α1)11
.
Similarly, we have
(2.13) ut0(x) = b0(x+
1)κ+o(|x|κ),where b0=t2
0
α(α1)11
< b,
because t0<1,and 1
1α>0. For small δ > 0,for some xi(i= 1,...,k with kn) we
have
(2.14)
k
X
i=1
λixi=δe1and u(δe1) = u(xi),
where Pk
i=1 λi= 1, and all xidepend on δ. Observe that the ﬁrst equation in (2.14)
implies
(2.15)
k
X
i=1
λixi
1=δ.
4Observe that for any point xthe free boundary has a support plane and hence the free
boundary is C1at such points where there is some a priori ﬂatness from the exterior of the domain.
Indeed, also by property (2.1) our problem falls into the class of free boundaries that are well studied and
the regularity of the free boundaries for these cases is well known; see [1, 2, 10], and also generalization
to equations with right-hand side in some subsequent work.
5This means that Pk
i=1 λiyi=0, with kn
6 L. EL HAJJ AND H. SHAHGHOLIAN
Using (2.12)–(2.14), and the fact that ut0u, we have
b0δκ+o(δκ) = ut0(δe1)u(δe1) = u(xi) = b|xi
1|κ+o(|xi
1|κ).
Using the inequality b0< b, we can rephrase this as
|xi
1| ≤ (1 2ρ)δ+o(δ)(1 ρ)δ
for some ρ=ρ(b, b0, κ)>0, and provided δis small such that o(δ)ρδ. From this and
(2.15) we have
δ
k
X
i=1
λi|xi
1| ≤ (1 ρ)δ
k
X
i=1
λi= (1 ρ)δ,
For Case B2 we have the Bernoulli boundary condition, with g > 0. Once again using
the C1-regularity of Ω (see the comment in footnote 4), we conclude that
is C1regular. Similarly, all super-level sets of uwill have the same regularity. In
particular, we can conclude as in [16] that
h(x) := 1
|∇u|1
g
is convex on any line segment of \Ω, with end-points on Ω; we call such
line segments maximal. Since uu, we conclude that |∇u| ≤ |∇u|on \Ω,
and in particular at the end-points of any maximal line segments and hence h(x)0 at
such end-points. Now hbeing convex implies h0 on such line segments and therefore
|∇u| ≥ gon these line segments. This in turn implies that uis a subsolution to the
Bernoulli problem.
Next we use Lavrentiev’s scaling and comparison. For this, suppose without loss of
generality that zin the statement of the theorem is the origin, and deﬁne ut0(x) = u(t0x),
as was done above. Comparing ut0and u, we conclude that
g(x0)≤ |∇u|(x0)≤ |∇ut0|(x0) = t0|∇u|(t0x0) = t0g(t0x0),
which contradicts the assumption (2.4) on g.
Uniqueness for both cases can be proved in a similar fashion, by using the function u
and scaling the next solution vt(x) = v(tx), after which the above Lavrentiev argument
is applied.
This theorem can be extended to an “exterior-like” ring-shaped domain, as done in
[8]. A consequence of this is that one can state (with a straightforward proof as in the
above case) a general uniqueness and convexity theory for elliptic PDEs in ring-shaped
domains, for f(t)>0 and left- or right- continuous with isolated discontinuities. At this
point, one can use approximation (of the function f) to strengthen the results to the
existence of solutions to elliptic PDEs in convex rings with convex super-level sets. Since
the positivity of fis a crucial part in our approach, we ﬁnd it not worthwhile to enter
into any technical details here, and we hope to be back to this in the near future using
more elaborated combinations of convexity theory for justifying the existence theory for
solutions with quasiconvex super-level sets.
§3. Systems (A discussion)
Problems involving weakly coupled systems have been on rise recently, in the free
boundary community. Currently, there are two main directions that are pursued by the
community, one being cooperative, and the other competitive systems. We shall discuss a
few examples of cooperative systems here below. To do so, we start introducing vectorial
notation for vector-functions and vector-domains as u= (u1, u2,··· , um), and D=
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES 7
(D1, D2,··· , Dm), with DjRn. The term cooperative refers to the speciﬁc behavior
of the system, which forces the supports of all components of the vector-function coincide.
For the simplicity of presentation, we shall denote Ω = {|u|>0}, and k= (k1,··· , km),
with ki>0.
In general, one may consider equations of the type
(3.1)
u=f(x, u) in \D,
u=kon D,
G(u) = g(x) on .
In case |∇u|= 0 on Ω, the last equation is dropped out and the ﬁrst equation is replaced
by ∆u=f(x, u) in Dc, i.e.,
(3.2) (u=f(x, u) in Dc,
u=kon D.
When f=uFwith F“reasonably” smooth, one may ﬁnd solutions to (3.2) using
minimizers of the functional
(3.3) J(v) = ZRn
|∇v|2+F(x, v),
over {vW1,2
0(Rn) : v=kon D}. When F(x, u) = g2(x)χ{|u|>0}, one obtains (3.1)
with G(u) = |∇u|, which is a Bernoulli type free boundary problem for systems that
were studied in [6]:
(3.4) (u= 0 in \D,
|∇u|=g(x) on .
The particular case of F(u) = |u|, recently studied in [3], gives rise to obstacle type
problems for the systems
(3.5) ∆u=u
|u|χ{|u|>0}.
The next goal is to see what kind of domains Dcan be of interest to consider. This
has several possible scenarios; here are a few:
(i) Di=Dj,i, j = 1,2,...,m,
(ii) D1D2 · · · Dm,
(iii) D1D2 · · · Dm, and they are homothetic,
(iv) TDi6=.
We should also remark that for minimizers uof the functional Jwe always have
{ui>0}={uj>0}for all i, j, provided DiDj6=. This follows from the fact that in
such cases we can make variations in both directions (upward and downward), uiǫφi
and ujǫφj. Hence we have the Euler–Lagrange equation for both uiand ujwhenever
one of them is nonzero.
The question we want to raise is under what conditions on D, and equations above
we may expect geometric inheritance for solutions. For instance, will the star-shapness
of Dwith respect to a point zDimply the same for Ω? Will the convexity of all
components of Dimply the convexity of Ω?
We ﬁrst observe that in the most general case (iv), the convexity of the support of ui
is not trivial without further conditions on Di. Indeed, consider the following example
with m= 2.
8 L. EL HAJJ AND H. SHAHGHOLIAN
Example 1 (Negative result).Let
D1={x:k < x1< k, 1< x2<1}
and
D2={x:1< x1<1,k < x2< k},
where kis very large. By nondegeneracy in both problems (see [3, 6]), we see that the
free boundary has to be in a uniform neighborhood of D1D2and hence it cannot be
convex, for klarge.
Therefore, in what follows we consider cases (i) and (ii).
It is also noteworthy that from Theorem 1 it follows easily that for small perturbation
of Di(i= 1,...,m) the solution-domains Ω remain convex, provided the free boundary
for the perturbed problem is uniformly C2. This can be seen in an obvious way from the
continuity of the curvature in the C2-case.
When the Disatisfy D1D2 · · · Dm,and each Diis star-shaped with respect to
zD1, then one may obtain certain results, but still incomplete, because the regularity
of the free boundary is essential for our technique. Indeed, one may show through
a minimizing functional (whenever it applies) that the super-level sets of any global
minimizer are star-shaped with respect to z. If a solution is unique (this is true for the
cases when there is a minimizing functional that is strictly convex), then obviously it is
also star-shaped. This follows by standard star-shaped rearrangement arguments.
Uniqueness for the star-shaped case is subtler and many times requires smoothness of
the free boundary. However, we can still state and prove the following weaker version of
uniqueness.
Theorem 2. In equation (3.1), let DiDj(for i < j), and suppose all domains Di
are star-shaped with respect to the origin. Let further
f(x, u) = uF(x, u),and G(u) = |∇u|,
and let each component of f= (f1,...,fm)satisfy the assumptions (2.1)(2.2) in the u
variable;next,let gbe continuous and satisfy 6(for some a < 2)
tg(x)g(tx), taf(t|x|,p)f(|x|,p),t1,and p= (p1,...,pm).
We also assume that for some b0, b10 (with max(b0, b1)>0) either of the following
holds7
gb0,|f(x, t)| ≥ b1|t|αfor some 1< α < 1,and 0< t < t1,
and that f(t) = 0if min(t1,··· , tm)<0,and f(t)0for t0. Then there is a
solution of (3.1), with compact support,satisfying
u(tx)tu(x)t1.
Moreover,among the solutions that satisfy the free boundary condition |∇u|=g > 0on
,or those that have C1free boundary in the case where g= 0,there exists only one
solution to (3.1).
A proof of a general version of this theorem, and related results for nonlinear equations,
will appear later in a separate paper. Here we present a narrative sketch of the proof for
6The reason of taking a < 2 is of technical nature, but we believe that a= 2 should also work. This
appears in comparison when we use the scaling of utas was done in the proof of Theorem 1.
7Compare the assumptions in the scalar case.
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES 9
Proof. Starting with the corresponding functional
J(v) = ZDc
|∇v|2+F(x, v) + g2(x)χ{|v|>0},
we ﬁrst need to show that the conditions of the theorem reinforce the statement that the
support of a global minimizer should be compact. This can be done by using comparison
arguments for the functional and another one that gives larger solutions, and where
Diis replaced by a ball BR(0) Difor all i. See, e.g., [12] for such a comparison
argument. In other words, one can prove that replacing the functional with symmetric
and “competitive” integrands as well as replacing all Diwith a ball BR(along with
Schwarz symmetrization) will give a solution with larger support. From here one can
prove that the assumptions of the theorem will imply that we may take a minimizing
sequence of functions that all have uniformly bounded supports.
In a similar fashion as above, one can consider a decreasing star-shaped rearrange-
ment uof any global minimizier uthat reduces the functional value, unless u=u.
This would then imply that global minimizers are star-shaped. The assumptions in the
theorem are designed to allow the star-shaped rearrangement to work. This will show
the existence of star-shaped solutions as stated in the theorem.
The uniqueness follows with similar techniques as that in Theorem 1, with more care
as the system has several components to be treated. Indeed, having a solution proved
to be star-shaped as in the above paragraph, we may use scaling and comparison of
Lavrentiev in the case of g > 0, and use Hopf’s boundary principle for the case of g= 0.
For the latter, since the regularity of the free boundary for systems is so far at a foster
stage, we have imposed the extra condition of C1-regularity to obtain uniqueness. The
drill is standard.
Corollary 1. Retain the hypotheses of Theorem 2, and assume that
Di=Bri(0) = {|x|< ri},
f=f(|x|,u),and g:= g(|x|). Then there is a unique spherically symmetric solution
of (3.1) with compact support.
Proof. By Theorem 2, there exists a solution uto our problem. In what follows all
operations are componentwise.
If this solution is now spherically symmetric, we can deﬁne
v= inf
σu(σx),
where σis the class of all possible spherical rotations. In particular, ∆vf(|x|,v).
Similarly, we take w= supσu(σx), which is a subsolution, ∆wf(|x|,w).
If we have two diﬀerent solutions, we may take the minimum of the two solutions
u3:= min(u1,u2)
and then a second inﬁmum of all possible rotations of this minimum
u4= inf
σu3(σx).
A similar argument applies in taking the maximum of the two and then the supremum
of all possible rotations of this maximum.
In this way we have created one supersolution vand one subsolution wsatisfying
vw. Now, we can apply a scaling and comparison argument as we have done earlier,
i.e., for some largest possible t0<1 with vt0w, satisfying
vt0t2f(t0|x|,v)t2a
0f(|x|,v).
10 L. EL HAJJ AND H. SHAHGHOLIAN
From this an argument precisely as that in Theorem 1 works to deduce that ∆vt0w
in the vicinity of the touching point zas in Theorem 1. We leave out the obvious details.
Observe that the regularity of the boundary that was used in Theorem 1 is not needed
here, because the spheres are smooth and Hopf’s lemma applies.
3.1. Bernoulli type problems for systems. Convexity results for the Bernoulli type
problems, represented in the system of equations (3.4), seem for now out of reach, at least
with the methods we know of. It is also not straightforward what conditions we should
impose on the domains D(besides being convex and maybe Di’s being homothetic).
For F(x, v) = χ{|v|>0}in case (i), i.e., when D=D1=D2=..., one may reduce the
problem to the scalar case by letting vi=ui
λiand Ω = {|v|>0}, to arrive at
(3.6)
v= 0 in \D,
v= 0 on ,
v= 1 on D,
Pλ2
i|∇vi|2=g2on ,
where the equations and boundary values are componentwise. The ﬁrst three equations
imply that vi=vjfor all i, j, and hence the last condition turns into (Pλ2
i)|∇v1|2=g2.
Therefore this case reduces to the scalar case.
The main question concerning the Bernoulli problem is the following. Let D=
(D1,··· , Dm) with DiDi+1 and each Diconvex. Is there a unique solution u, with
Ω = {|u|>0}, to the Bernoulli free boundary problem
(3.7)
u= 0 in \D,
u= 0 on ,
u= 1 on D,
|∇u|= 1 on ,
with the set {|u|>0}convex? Uniqueness for this problem follows as in the scalar case,
by using Lavrentiev’s principle.8
3.2. Obstacle type problems for systems. The system case of the free boundary for
general F(x, p), even if Fis independent of xand convex in the p-variables, is of course
a nontrivial problem. Let us consider the particular case as in (3.5), i.e., F(x, v) = |v|, in
the minimization problem (3.3). As in the Bernoulli problem, case (i) reduces to scalar
cases. In fact since Di=Dj, for vi=ui
λiwe have ∆v=v
|u|, and if we let hij =vivj,
then (hij =hij
|u|in \D,
hij = 0 on (Ω \D).
Hence by the maximum principle we have hij 0, i.e., vi=vjfor all i, j, and it follows
that ui=λi
λjuj.
The harder case of DiDi+1 ,i= 1,··· , m 1, does not seem to admit an obvious
approach. For the scalar case, we know of two approaches, where the ﬁrst one is the
classical approach of using a quasiconvexity function (see [15, 17]) and the second and
more straightforward is the quasiconvex rearrangement as in [7, 13]. Both of these
approaches seems to be nonelementary to implement in this case.
8Observe that the boundary condition |∇u|= 1 tacitly assumes that the gradient is continuous up
to the boundary.
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES 11
There are other models of the system case, which we mention here. These are
a) ∆ui=1 + X
j6=i
ujχ{|u|>0},b) ∆ui=ui
Pm
j=1 ujχ{|u|>0},
where both of them can be reduced to the scalar case by deﬁning U=Pm
i=1 ui, which
gives us the following equations
c) ∆U= (m+mU)χ{U >0},d) ∆U=χ{U >0},
respectively, where we have used the relation χ{|u|>0}=χ{U>0}. Even such a reformula-
tion does not seem to work by earlier methods, unless Di=Djfor all i, j. Indeed in this
case the equation becomes scalar with a ﬁxed D=Di, with boundary values k:= Pki
on D. Since the right-hand side is monotone in both cases, we can invoke earlier results
of [13] to conclude that Uhas convex level sets. It is however not obvious whether each
component will be quasiconvex. This is left to readers to explore.
As a ﬁnal simple example we mention the following system:
ui=1 + (m1)uiX
j6=i
ujχ{|u|>0},
and with U=Puiwe have ∆U={U>0}, with Di=Djfor all i, j , and the
corresponding boundary value. This is an obstacle problem and it is well known that the
only solution to this problem is quasiconvex.
For the above problems, with DiDi+1 , the question of quasiconvexity for each
component, or any combination of the components, or even the convexity of the support
Ω = {|u|>0}remains tantalizing.
We know of no quasiconvexity results (or even problems) for systems in existing liter-
ature. Even the simplest equation such as
u= (1 + 2v)χ{u>0}in Dc,
v= (1 + u)χ{v>0}in Dc,
u= 1 on ∂D,
v= 2 on D,
does not seem to be easy to handle with existing methods for convexity. Here we have
chosen equations such that the system cannot be reduced to a scalar case. A ma jor
problem is that solutions in exterior domains become quasiconvex and not convex. This
means that any existing method would imply that the right-hand side of each equation
above is a quasiconvex function and this does not suﬃce to force through the existing
methods, as they require the right-hand sides be convex functions. This may also be an
indication that what we ask for is not true, but we have neither a counterexample.
Example 2. (A positive result) We shall now consider an ad hoc system, whose solution
can be shown to have convex level sets. Consider the equation
u= (10 ev)χ{u>0}in Dc
v= (10 eu)χ{v>0}in Dc,
u= 1 on ∂D,
v= 2 on D,
where euand evare the concave envelops of u, respectively, v.9This system is specially
designed to create quasiconvex solutions, by using standard techniques.
9The reason we take the concave envelop is that it is nonzero. On the other hand, the convex envelop
of a positive function can be zero.
12 L. EL HAJJ AND H. SHAHGHOLIAN
(u0= 10χ{u0>0}in Dc,
u0= 1 on D,
which has a (unique) nonnegative solution u0with convex level sets, that is u0is quasi-
convex (see [7]). We also remark that u01, by the maximum principle and hence the
concave envelop eu0does not exceed 1, because the constant linear function h1 is a
linear function above u0.
Next for this u0we ﬁnd a nonnegative solution v0to
(v0= (10 eu0)χ{v0>0}in Dc,
v0= 2 on ∂D.
Let v
0v0be the quasiconvex solution of v0, see [7]. The right-hand side g(x, v0) :=
(10 eu0)χ{v0>0}satisﬁes the hypothesis for the quasiconvexity theory (it is nonnegative,
monotone increasing in v0, and gis convex in the x-variables). Hence we can apply
Theorems 4.2 and 4.4 in [7] (see also [13]) to arrive at
v
0g(x, v
0)g(x, v0) = ∆v0.
Since both functions have the same boundary values on the set {v
0>0}\D, we conclude
by the comparison principle that v0=v
0.
We also have u0v0by the comparison principle for the obstacle-type problems (this
is straightforward and left to the readers) which gives eu0ev0, and hence solutions of
their corresponding obstacle problems must satisfy the reverse order v0u1.
Next we solve (u1= (10 ev0)χ{u1>0}in Dc,
u1= 1 on ∂D,
and observe that due to the same convexity argument and the comparison principle as
mentioned above, we see that u1is quasiconvex and u1u0.
This u1is used once again as before to ﬁnd v1that solves
(v1= (10 eu1)χ{v1>0}in Dc,
v1= 2 on ∂D.
Since the left-hand side of the equation is monotone increasing in v1and u1u0, and
hence eu1eu0, we deduce (by the comparison principle) that v1v0. Similarly, we
obtain v1u1.
Continuing in this vein we will have a sequence of pairs (ui, vi) solving the problem
ui= (10 evi1)χ{ui>0}in Dc,
vi= (10 eui)χ{vi>0}in Dc,
u= 1 on ∂D,
v= 2 on D,
for i= 1,2,..., and with (u0, v0) as obtained above.
We denote by
(u0, v0)(u1, v1)
the ordering
u0u1,and v0v1,
and this is repeated in the correct order (using the iteration above indeﬁnitely),
(ui, vi)(ui+1, vi+1 ).
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES 13
In other words uiis an increasing sequence and viis a decreasing sequence.
From here, with some footwork, one can obtain a convergence result and hence a
solution to our system above.
3.3. Singular perturbation for systems. In the scalar case, the Bernoulli problem
discussed earlier can also be obtained as a limit problem of the singular perturbation; the
literature is vast for this problem, so none mentioned none forgotten. A similar theory,
yet to be developed, seems plausible for the system case, which seems to have much more
possibilities and variations than its scalar counterpart.
Here we shall give a few examples of such models that might be interesting for the
readers. The simplest equation is of the form
(3.8) ∆ui=1
ǫχ{0<|u|}
in Dc, for a given (convex/star-shaped) D, and with u=kon D. This equation can be
reduced to the scalar case when k= (k,...,k), but stays a system with no possibility
of reduction when k= (k1,...,km), with k1< k2<··· < km. The more complicated
version would be to replace Dwith D= (D1,...,Dm) as earlier. This problem does not
seem to have a variational formulation, but one may study this through supersolution
techniques.
The following model equation
(3.9) ∆u=1
ǫ
u
|u|χ{0<|u|}
has a variational formulation and is a direct generalization of the singular perturbation
problem to the system. This is also related to the obstacle problem for the systems as
we mentioned above.
3.4. Serrin type problem for systems. Contrary to convexity problems, the sym-
metry methods, such as the moving plane technique, work very well for free boundary
value problems for systems, when there is a symmetry in the given equation, the bulk
domain, and boundary values. Here, we present the simplest example, leaving several
obvious generalizations towards other problems to the reader. The original Saint Venant
problem, with an overdetermination of the boundary gradient condition (here expressed
as a system) is to show that whenever there is a solution vector uto the following problem
(3.10)
u=kin Ω,
u=0on ,
|∇u|= 1 on ,
the domain Ω has to be a ball. For the scalar case, James Serrin [21] gave a very nice
proof of this based on the moving plane technique (of A. D. Alexandrov); see also [23]
for a diﬀerent proof.
For the system case, one may think of solutions being
u=k1|x|2
2n,with Ω = n|x|<n
|k|o.
Same techniques carry over to the system case above, mutatis mutandis, by reﬂection
principle as long as the boundary is C2, to make Serrin’s argument work. Indeed, using
the moving plane technique one makes comparison componentwise along with Hopf’s
exercise, we leave out the details to the readers.
14 L. EL HAJJ AND H. SHAHGHOLIAN
It is however more interesting to consider the semilinear case of Serrin’s problem for
systems, that is to prove spherical symmetry for the same system as in (3.10) where one
replaces kwith f(u), with mild assumptions on fsuch as Lipschitz regularity.
Another related problem is the discrete Bernoulli problem discussed in [22] (for the
scalar case), see also [11]. Let Ω be a domain (in Rn,n2) with C1boundary, and
r, l > 0 given constants, with l < maxu. Suppose the solution u= (u1,...,um) to the
problem
u=k,in Ω,u= 0 on
has the property that, for all xΩ,
(3.11) dist(x, Γl) = r, Γl={|u|=l}.
Then, Ω is necessarily a ball.
See [18, 19, 20] for related problems in the scalar case.
§4. A few problems to ponder!
Several natural questions that arise from our analysis are the following.
Problems for Subsection 3.1–3.2. Both Bernoulli and obstacle type problems for
systems seem to be very hard to treat with existing methods and tools. All problems
discussed for both these are open and new methods are needed to treat them. The
quasiconvexiﬁcation as used in the above does not apply, because each right-hand side
of the equation contains other components of the solution and the requirement is that
these have to be convex functions in order to apply the method of quasiconvex envelope.
As is seen from Example (2), we force this by an ad hoc assumption in order to obtain a
positive result.
We strongly believe that these problems, as well as their interior counterparts will be
of interest for future study. All we need is some starting ideas and new tools to treat the
systems.
Problems for Subsection 3.3. In the scalar case, the ﬁrst author of this paper gave a
proof of convexity for the singular perturbation problem [8], with a slightly diﬀerent argu-
ment. Our proof in §2 provides a more general result and is probably simpler. However,
none of these methods seem to work for the system case for the singular perturbation
problem. It would be nice to see some new ideas for how the system case can work.
Problems for Subsection 3.4.
(1) How far can one stretch the conditions on f? Can we allow each component of f
to change sign, or be in L1? This question is yet not answered in the scalar case.
(2) Can we derive a stability theory for the system case, as this was done for the
scalar case [5]?
(3) Try to prove symmetry results for the discrete Bernoulli problem in the system
case.
Problems in unbounded domains. Most of the above questions and open problems
can naturally be stated for unbounded domains and half-spaces. Several authors have
considered the scalar case of such problems and the literature is vast. The reader may
consult the paper [9], and the references therein for these problems.
It would be interesting to see if methods can be invented to treat not only problems
in bounded domains, but also in unbounded ones as developed in the literature.
REMARKS ON THE CONVEXITY OF FREE BOUNDARIES 15
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American University in Dubai, Dubai, UAE
KTH Royal institute of Technology, Stockholm, Sweden
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