EXISTENCE OF GROUND STATES AND FREE BOUNDARY PROBLEMS
FOR QUASILINEAR ELLIPTIC OPERATORS
Filippo Gazzola - James Serrin - Moxun Tang
Abstract. We prove the existence of non-negative non-trivial radial solutions of the
quasilinear equation
m
u + f(u) = 0 in R
n
and its associated free boundary problem,
where
m
denotes the m-Laplace operator. The nonlinearity f is subject to very weak
restrictions for u near zero. We allow general subcritical assumptions on f when n > m,
and also obtain existence for exponentially growing functions f(u) when m = n.
Our proofs use only standard shooting techniques from the theory of ordinary dif-
ferential equations; unlike well known earlier demonstrations of the existence of ground
states, we rely neither on critical point theory [BL] nor on the Emden-Fowler inversion
technique [AP]. Finally, the proof of the main result yields as a byproduct an a priori
estimate for the supremum of a ground state in terms of n, m and elementary parameters
of the nonlinearity f .
1. INTRODUCTION
Let
m
u = div(jDuj
m2
Du), m > 1, denote the degenerate m-Laplace operator. We
study the existence of radial ground states of the quasilinear elliptic equation
(1.1)
m
u+ f(u) = 0 in R
n
; n > 1;
and the existence of positive radial solutions of the homogeneous Dirichlet-Neumann free
boundary problem
(1.2)
m
u+ f(u) = 0; u > 0 in B
R
;
u =
@u
@n
= 0 on @B
R
;
where B
R
is an open ball in R
n
with radius R > 0. Here, a ground state is a non-negative,
non-trivial continuously dierentiable distribution solution u = u(x) of (1.1) which tends
to zero as jxj approaches innity.
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1
2 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
We shall assume throughout the paper the following general condition
(H1) f is locally Lipschitz continuous on (0;1), and
R
0
jf(s)jds 0 such that F (s) < 0 for 0 < s < , F () = 0 and f() > 0.
The behavior of f near zero is of crucial importance to our results. In the following
considerations we shall identify two mutually exclusive situations:
1. Regular case. f is continuous on [0; 1); clearly (H2) implies that f(0) 0
2. Singular case. f cannot be extended as a continuous function to [0; 1); we leave
the value of f at 0 undened.
When m = 2 we recall that (1.1) reduces to the classical nonlinear Euclidean scalar
eld equation
(1.3) u+ f(u) = 0 in R
n
:
Under the assumption that f is regular and f(0) = 0, together with conditions on the
behavior of f for large u, there are in the literature a number of well-known existence
theorems for radial ground states of (1.3); see in particular [CGM, BL, AP1, AP2].
Much less is known about ground states for the degenerate equation (1.1). Citti [Ci]
has proved existence when 1 < m < n, f(0) = 0, and f is bounded in [0;1), while
Franchi, Lanconelli and Serrin [FLS] have considered the case when f(s) is \sublinear"
for large s, in the sense that either f() = 0 for some nite > or
lim inf
s!1
jF (s)j
1=m
s
. Moreover, it was conjectured by Pucci and
Serrin (see [PuS2, Section 6.1]) that if f(0) < 0, or even if f is singular, then solutions
of the free boundary problem can be found for all m > 1. We shall address this question
in detail in Theorem 1.
GROUND STATES AND FREE BOUNDARY PROBLEMS 3
The purpose of this paper is to extend and unify the above results, by considering
throughout the general conditions (H1)-(H2) on f(u). Thus we allow f to be dierent
from zero, or even singular, at 0, and additionally we generalize earlier growth restrictions
on f(u) as u!1. Our results are thus new even for the classical eld equation (1.3).
In what follows, the two conditions
(1.4) jF (s)j (s) for 0 < s and
Z
0
j(s)j
1=m
ds =1;
where > 0 and : [0; )! R is a non-decreasing function with (0) = 0, and
(1.5)
Z
0
jF (s)j
1=m
ds ; f(s) = 0g
and put = 1 if f(s) > 0 for all s . Finally we introduce a function related to the
Pohozaev identity, namely
Q(s) = nmF (s) (nm)sf(s):
Our main existence result is then
Theorem 1. Let hypotheses (H1) and (H2) hold and suppose that one of the following
three conditions is satised:
(C1) 0 such that
(1.6) lim sup
s"
f(s)
( s)
m1
< k
0
:
(C2) =1 and n < m.
4 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
(C3) =1 and n m. Q(s) is locally bounded below near s = 0, and there exists b >
and k 2 (0; 1) such that Q(s) 0 for all s b and
(1.7) lim sup
s!1
Q(s
2
)
s
m1
f(s
1
)
n=m
=1;
where s
1
, s
2
is an arbitrary pair of numbers in [ks; s].
Then we have the following results:
(i) When f is regular and f(0) = 0 there exists a radial ground state u for equation
(1.1). If (1.4) is satised, then u is positive and the free boundary problem (1.2) has no
radial solution; if (1.5) is satised, then u has compact support and accordingly is also a
radial solution of (1.2) for some R > 0.
(ii) When lim sup
s#0
f(s) < 0, there are no ground states of (1.1), and (1.2) has a
radial solution for some R > 0.
(iii) When f is singular and lim sup
s#0
f(s) 0, then either (1.1) has a positive radial
ground state or (1.2) has a radial solution for some R > 0. If (1.4) is satised, then the
rst case occurs; if (1.5) is satised, the second case occurs.
Finally, if r = jxj, the function u = u(r) obeys u
0
(r) < 0 for all r > 0 such that
u(r) > 0.
Note that when m 2, (1.6) is automatically satised in view of (H1). In fact, (1.6)
could be omitted even for m > 2 by arguing as in Section 2.2 of [FLS]; we shall not
pursue this here since our main interest is in conditions (C2) and (C3). We underline
the fact that in case (C2) there are no conditions on f apart from (H1)-(H2). The local
boundedness assumption on Q in condition (C3) is automatic if n = m or f is regular,
or even if f is singular and bounded above by Const. s
1
.
Condition (1.7) is similar to one rst introduced by Castro and Kurepa [CK] for the
classical Laplace operator in a ball and later used in [GMS] for quasilinear operators.
Note nally that, in the important case when Q and f are monotone increasing for large
u, condition (1.7) can be written more simply
(1:7
0
) lim sup
s!1
Q(ks)
s
m1
f(s)
n=m
=1:
Note in particular the assertion of part (ii) of the theorem, that under the condition
lim sup
s#0
f(s) < 0 there are no ground states of (1.1), even for the nonradial case. On the
GROUND STATES AND FREE BOUNDARY PROBLEMS 5
other hand, if f is redened to be 0 when u = 0 (thus making it certainly discontinuous),
a radial distributional compact support ground state on R
n
is obtained. See the remark
preceding Section 3.1.
In part (iii) of Theorem 1, the two cases are distinguished exactly by the convergence
or divergence of the integrals in (1.4){(1.5) provided that jF (s)j (s), or in particular
if f(s) 0 for 0 < s . When neither (1.4) nor (1.5) holds, then we cannot tell which
case occurs. It is interesting to note however that there can exist positive ground states
of (1.1) even if f is singular near zero.
Our proofs use only standard techniques from the theory of ordinary dierential e-
quations, the shooting method and variational identities. In particular, unlike earlier
well-known demonstrations of the existence of ground states, we rely neither on critical
point theory [BL] nor on the Emden-Fowler inversion technique [AP]. It is additionally
worth emphasizing that the proof of Theorem 1 yields as a byproduct an a priori estimate
for the supremum norm of ground states in terms of n, m and the nonlinearity f ; see
Theorem 2 in Section 4.
A case of particular interest is the polynomial function
(1.8) f(s) = s
p
+ s
q
; p < q:
Clearly (1.8) satises (H1)-(H2) when p > 1, while = 1. First, for the case when
n > m the principal condition (1:7) (1:7
0
) is satised when q < , where is the critical
Sobolev exponent
=
(m 1)n+m
nm
:
This shows that (1.7) is essentially a subcritical assumption on f for large u. It is also
worth remarking that when q , there are no ground states of (1.1) for the function
(1.8), see [NS1, Theorem 3.2 and pages 180-181]. In the case n m, by (C2), (C3) and
(1.7
0
), we see that any powers 1 < p < q are allowed in (1.8); therefore, from Theorem
1 we get the following
Corollary 1. Let f be as in (1.8). Then
(i) there exists a radial ground state u of (1.1) provided either
n m; p > 0 or n > m; 0 < p < q < ;
moreover, u is positive for all r > 0 if and only if p m 1;
6 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
(ii) there exists a positive radial solution of (1.2) for some R > 0 provided either
n m; 1 < p < m 1 or n > m; 1 < p < m 1; q < :
The results of Corollary 1 can be applied to obtain existence of solutions for several
problems in physics. For example, when p = 0 and q = 1=2 in R
2
, (1.3) has been suggested
as a model in plasma physics for Tokamak equilibria with magnetic islands, see [MFW]
and [KK]. Also, as pointed out in [C], when 1 < p < 0 and q = 1, equation (1.3) is
related to the blow-up of self-similar solutions of a singular nonlinear parabolic problem,
a model proposed in [Lo] for studying force-free magnetic elds in a passive medium.
Theorem 1 shows a striking dierence between the cases n m and n < m. There
is also an important dierence between the cases n = m and n > m. When n > m we
have already noted the subcritical requirement q < for the nonlinearity (1.8).. When
n = m, however, there is no critical Sobolev exponent , and no optimal embedding
W
1;m
(R
n
)! L
+1
(R
n
). Instead, the appropriate embedding is into an Orlicz space [S],
and critical growth means exponential growth. For the classical eld equation (1.3) such
exponential behavior for f was treated, for example, in [BGK] and [AP2].
On the other hand, Theorem 1 is not directly satisfactory for exponential growth since
condition (1.7) then fails. To include such behavior, especially for the general equation
(1.1) when n = m, we need a renement of condition (1.7).
Theorem 1
0
. Let hypotheses (H1) and (H2) hold, and suppose that =1 and n = m.
Assume moreover that there exist constants b > , > 0 and > b+ such that
(1.9)
n1
F ( )
f()
> ;
where is an arbitrary number in [; ] and is a constant depending only on n and
on the behavior of f on the bounded interval [0; b], see (5.3).
Then the conclusions of Theorem 1 hold.
The constant has an extremely simple form in the natural case when f(s) has only
a single positive zero, say at s = a. Then clearly a < and F = F (a), so we can take
=
2
n
(n 1)b
2
F (a)
F (b)
n
;
or in the important subcase n = m = 2 (see footnote 1)
= b
2
3
2
F (a)
F (b)
2
:
GROUND STATES AND FREE BOUNDARY PROBLEMS 7
In Section 5, as a second major goal of the paper, we describe a class of exponentially
growing functions f satisfying (1.9). In particular, behaviors at innity like f(s)
s
p
exp(s
q
) are allowed for 1 < p < 1, 0 q < 1 and all > 0, and for q 1
provided lies in some appropriate range. When m = n = 2 Atkinson and Peletier [AP2]
and Berestycki, Gallouet and Kavian [BGK] have obtained existence results for all > 0
and for the range 1 q < 2, and also for various further cases when q 2. In Example
3 of Section 5 we comment further on the relation between our work and theirs.
The paper is organized as follows: in the next section we present some preliminary
results on the behavior of radial solutions; in Section 3 we prove Theorem 1 and in Section
4 we obtain an important a priori estimate for the supremum norm of ground states. We
prove Theorem 1
0
in Section 5, and nally discuss the symmetry and uniqueness of ground
states of (1.1) and (1.3) in Section 6.
2. PRELIMINARY RESULTS
We maintain the hypotheses (H1)-(H2) without further comment. Observe that a
nonnegative, nontrivial radial solution u = u(r) of (1.1) is in fact a solution of the
ordinary dierential initial value problem
(2.1)
(ju
0
j
m2
u
0
)
0
+
n 1
r
ju
0
j
m2
u
0
+ f(u) = 0;
u(0) = > 0; u
0
(0) = 0
for some initial value , where for our purposes the dimension n may be considered as
any real number greater than 1. The equation (2.1) can be rewritten as
(2.2) (r
n1
ju
0
j
m2
u
0
)
0
+ r
n1
f(u) = 0;
or equivalently, with w = w(r) = ju
0
(r)j
m2
u
0
(r),
(2.3) (r
n1
w)
0
= r
n1
f(u)
where of course w(0) = 0.
Lemma 1.1.1 and Corollary 1.2.5 in [FLS] show that any non-trivial radial solution of
(1.1) or (1.2) has initial value > ; therefore we take
(2.4) u(0) = 2 [; );
where the case u(0) = is included for convenience in later work.
For deniteness in what follows, we understand that a (classical) solution of (2.1) is a
function u which, together with w = ju
0
j
m2
u
0
, is of class C
1
on its domain of denition
and satises (2.1) there.
8 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
Lemma 2.1. Let (2.4) be valid. Then (2.1) has a unique (classical) solution u in a
neighborhood of the origin, which satises u
0
(r) < 0 for small r > 0.
Proof. Local existence and uniqueness of solutions of the Cauchy problem (2.1) is well-
known, see for example [NS1], [NS2] and Propositions A1, A4 of [FLS].
As proved in Lemma 1.1.1 of [FLS], also
w
0
(0) =
1
n
f(u(0)) =
1
n
f():
By (2.4) and assumption (H2) we have f() > 0. Hence w
0
(0) < 0 so that u
0
(r) < 0 for
small r > 0.
Continuation of the solution given by Lemma 2.1 is standard. We denote by J = (0; R),
R 1, the maximal open interval of continuation under the restriction
u > 0; 1 < u
0
< 0 in J:
Since clearly 0 < u < in J , it is also standard that the continuation and the corre-
sponding interval J is uniquely determined.
In the sequel we understand that every solution u of (2.1) is continued exactly to the
corresponding maximal domain J .
Since u is decreasing and positive on J it is obvious that lim
r"R
u(r) exists and is
non-negative. We denote this important limit by l.
Now dene the energy function
(2.5) E(r) =
m 1
m
ju
0
(r)j
m
+ F (u(r)); r 2 J:
By a straightforward calculation one nds that E is continuously dierentiable on J and
(2.6)
dE(r)
dr
=
n 1
r
ju
0
(r)j
m
:
Hence E(r) is decreasing and moreover bounded below since F (u) is bounded below, see
(H2).. This shows in particular that ju
0
(r)j is bounded on J .
More detailed characterizations of solutions are given in the following three lemmas.
We assume throughout that (2.4) holds.
Lemma 2.2. If R =1 then lim
r!1
u
0
(r) = 0.
Proof. Since E(r) is decreasing and bounded below, it is convergent as r ! 1. Then,
by (2.5) and the fact that F (u(r)) ! F (l), we see that u
0
(r) also approaches a limit,
necessarily zero, as r !1.
GROUND STATES AND FREE BOUNDARY PROBLEMS 9
Lemma 2.3. If R is nite then either u(r) ! 0 or u
0
(r) ! 0 as r ! R. In the rst
case, u
0
(R) = lim
r"R
u
0
(r) exists and u
0
(R) 0.
Proof. Since u
0
is negative and bounded on J , and R is assumed nite, the only obstacle
to continuation on J is for either u or u
0
to approach zero as r " R. This proves the rst
part of the lemma.
Next, since E approaches a limit as r tends to R it is clear that also u
0
approaches a
limit. But u
0
< 0 in J and the conclusion follows.
Lemma 2.4. l 2 [0; ).
Proof. Suppose for contradiction that l . Then obviously u > in J , so by (2.3) and
(H2)
(r
n1
w(r))
0
< 0;
that is, r
n1
w(r) is decreasing on J .
Now, if R is nite, then by the rst part of Lemma 2.3 we have u
0
(r) ! 0 as r " R.
In turn, r
n1
w(r) approaches 0 as r " R, while also r
n1
w(r) takes the value 0 at r = 0.
But this is absurd since r
n1
w(r) is decreasing on J .
If R = 1, then by (2.1) and Lemma 2.2 we get lim
r!1
w
0
(r) = f(l), where the
right hand side is negative by the assumption l . This is of course impossible since
w(r)! 0 as r!1. The proof is complete.
In the proof of Theorem 1 we will also need the following results, the rst of which is
an obvious consequence of Lemma 2.4.
Lemma 2.5. Suppose b 2 (; ) and 2 (b; ). Then there exists a unique value
R = R() 2 J such that u(R) = b.
Proposition 2.6. (Continuous dependence on initial data). Let u be a solution of (2.1)
with maximal domain J . Then for any r
0
2 J and > 0, there exists > 0 such that if
v is a solution of (2.1) with ju(0) v(0)j < , then v(r) is dened on [0; r
0
] and
sup
r2[0;r
0
]
(ju(r) v(r)j+ ju
0
(r) v
0
(r)j) < :
This can be obtained by combining the ideas of Propositions A3 and A4 in [FLS].
To conclude the section, we state a Pohozaev-type identity due to Ni, Pucci and Serrin,
see [NS1], [NS2], [PuS1], and also [ET].
10 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
Proposition 2.7. (Ni-Pucci-Serrin) Let u be a solution of (2.1) with domain J . Let
Q(s) = nmF (s) (nm)sf(s) and
P (r) = (nm)r
n1
u(r)u
0
(r)ju
0
(r)j
m2
+ (m 1)r
n
ju
0
(r)j
m
+mr
n
F (u(r))
= (nm)r
n1
u(r)u
0
(r)ju
0
(r)j
m2
+mr
n
E(r):
Then
P (r) =
Z
r
0
Q(u(t))t
n1
dt 8r 2 J:
3. PROOF OF THEOREM 1
Let < and let u
be a corresponding solution of (2.1) with maximal domain
J
= (0; R
), R
1. Also set l
= lim
r"R
u
(r); of course l
2 [0; ) by Lemma 2.4.
Dene the pair of sets,
I
= f 2 [; ) : R
0g:
Clearly I
+
and I
are disjoint; we shall show
Claim 1. 2 I
+
,
Claim 2. I
+
is open in [; ),
Claim 3. I
is non-empty,
Claim 4. I
is open.
Deferring the proof of these claims until subsections 3.1 - 3.3 below, we turn to the
demonstration of Theorem 1. First, in view of Claims 1-4 there must be some
2 (; )
which is neither in I
+
nor in I
. We denote the corresponding solution by u
, with
domain J
= (0; R
). Then l
= 0 (notation obvious) since
is not in I
+
. Moreover,
since
is also not in I
, either R
=1 or R
is nite and u
0
(R
) = 0, see Lemma 2.3.
In the rst case u
is a positive ground state of (1.1), in the second a solution of (1.2)
with R = R
.
When f is regular and f(0) = 0, the solution in the second case, when it is extended
to all r > R
by the value 0, becomes a compactly supported ground state of (1.1). The
rst statement of (i) is proved.
To obtain the remaining part of (i), observe that if (1.4) holds then Proposition 1.3.2
of [FLS] applies and the ground state u is necessarily positive, while if (1.5) is satised
then u has compact support in view of Proposition 1.3.1 of [FLS].
GROUND STATES AND FREE BOUNDARY PROBLEMS 11
In case (ii), there are no ground states of (1.1), radial or not. This is an immediate
consequence of Theorem 2 of [PuSZ] together with Remark 2 at the end of the proof of
that theorem. Indeed from this result any possible ground state in case (ii) would have
compact support, which would violate the dierential equation for large values of jxj. [If f
is redened to be 0 when u = 0 (thus making it certainly discontinuous), a distributional
compact support ground state on R
n
is however obtained. See the following remark.]
Case (iii) is obtained in almost exactly the same way as case (i).
Remark Under the assumptions of Theorem 1, if (1.1) has no ground states, then (1.2)
has a radial solution for some R > 0. In this case, let u be such a solution and extend it to
R
n
by u(r) 0 for r R. The resulting extension (still called u) is obviously in C
1
(R
n
).
Moreover, if we redene f to have the value 0 when u = 0, then the extension becomes
a distributional compactly support ground state of (1.1) provided lim sup
u#0
f(u)
12 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
E is decreasing, E(r) < 0 on J . Pick r
0
2 J ; then E(r
0
) < 0 and E(r) E(r
0
) for all
r 2 (r
0
; R). It follows that F (u(r)) E(r
0
) < 0 on this interval and thus l > 0. Hence
2 I
+
.
Proof of Claim 2. Let 2 I
+
. We denote the corresponding solution by u and its domain
by J = (0; R). Of course l = l 2 (0; ).
>From Lemmas 2.2 and 2.3 we have u
0
(r)! 0 as r " R. Hence
lim
r"R
E(r) = F (l) < 0:
Choose r
0
in J so that E(r
0
) < 0. If is taken suciently close to and u denotes
the corresponding solution of (2.1) with u(0) = , then applying Proposition 2.6 we can
arrange that (0; r
0
] J , where J is the domain of u, while also E(r
0
)
1
2
E(r
0
) < 0. As
in the proof of Claim 1, this implies that 2 I
+
.
3.2. Proof of Claim 3
Since F (0) = 0 and F (s) > 0 for < s < , we can dene
F = min
0
GROUND STATES AND FREE BOUNDARY PROBLEMS 13
Proof of Lemma 3.1. Let 2 (b; ), and let u be the solution of (2.1) with u(0) = ,
dened on the maximal domain J = (0; R), R 1. Suppose for contradiction that
=2 I
; then one sees easily from Lemmas 2.2{2.4 that l 2 [0; ) and u
0
(r)! 0 as r " R.
Let M = sup
[R;R)
ju
0
(r)j = ju
0
(R
1
)j, where R
1
2 [R;R); therefore formula (2.5) at r = R
together with (2.6) integrated on (R;R) gives
(3.2)
F (b) < E(R) = E(R) + (n 1)
Z
R
R
ju
0
(r)j
m
r
dr
(n 1)
M
m1
R
Z
R
R
ju
0
(r)jdr
= (n 1)
M
m1
R
u(R) l
(n 1)
M
m1
R
b;
where E(R) = lim
r"R
E(r) = F (l) 0 by (H2).
Similarly, (2.5) at r = R
1
together with (2.6) integrated on (R
1
; R) yields
(3.2
0
)
m 1
m
M
m
= E(R
1
) F (u(R
1
))
(n 1)
Z
R
R
1
ju
0
(r)j
m
r
dr F (u(R
1
))
(n 1)
M
m1
R
b+ F :
By (3.1) and (3.2) we get
M
m1
>
R
n 1
F (b)
b
n 1
R
b
F (b)
m
m 1
(F + F (b))
m1
;
so also
m 1
m
M >
n 1
R
b
F (b)
(F + F (b)):
Now from (3.2
0
) and these estimates for M , we nd that
F M
m1
m 1
m
M (n 1)
b
R
>
R
n 1
F (b)
b
(n 1)b
R
F + F (b)
F (b)
1
= F ;
a contradiction.
14 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
We shall apply this lemma to prove Claim 3 under the three dierent conditions (C1),
(C2) and (C3).
Condition (C1). Assume 0 for all r 2 (0; R]:
Using these inequalities and equation (2.3) there holds, for r 2 [0; R],
ju
0
(r)j
m1
= jw(r)j =
1
r
n1
Z
r
0
t
n1
f(u(t))dt
r
n
sup
[0;r]
f(u(t))
r
n
(k
0
+ 1)( u(r))
m1
:
In turn,
ju
0
(r)j cr
1=(m1)
( u(r)) cR
1=(m1)
+
Z
r
0
ju
0
(t)jdt
8r 2 [0; R]
where c = ((k
0
+ 1)=n)
1=(m1)
. Applying Gronwall's inequality, we obtain
ju
0
(r)j cR
1=(m1)
( ) exp(cR
m=(m1)
) 8r 2 [0; R]
which shows that R() ! 1 as " . Indeed, if R() remains bounded, then the
previous inequality implies that max
r2[0;R]
ju
0
(r)j ! 0, contradicting
b = u(0) u(R) R max
r2[0;R]
ju
0
(r)j:
Thus (3.1) is satised if is suciently close to , and so I
6= ;.
Condition (C2). Assume =1 and n < m.
Fix b > , and put C = C(b), the constant dened in (3.1). Suppose for contradiction
that I
is empty. Then by Lemma 3.1, for any > b there holds
(3.3) R = R() < C;
GROUND STATES AND FREE BOUNDARY PROBLEMS 15
where we recall that C is independent of . Dene
v(r) = r
n1
m1
ju
0
(r)j; r 2 J ;
then by equation (2.2) one has (v
m1
)
0
= r
n1
f(u), and therefore v is increasing on
[0; R]. Let V = v(R); then v(r) V , or equivalently,
ju
0
(r)j V r
n1
m1
on [0; R]:
Integrating this inequality over [0; R] leads to
b
m 1
m n
R
mn
m1
V:
Combining this with (3.3), there results
(3.4) V ( b)
m n
m 1
C
nm
m1
:
Now we introduce the function
(3.5) D(r) = r
em
E(r) =
m 1
m
(v(r))
m
+ r
em
F (u(r));
where em = m(n 1)=(m 1). Using (2.6), it follows that
(3.6) D
0
(r) = emr
em1
F (u(r)):
Let r 2 (R;R) and integrate (3.6) on [R; r] to obtain
m 1
m
(v(r))
m
=
m 1
m
V
m
+R
em
F (b) r
em
F (u(r)) + em
Z
r
R
t
em1
F (u(t))dt:
Dene for any constant a > 0
R
a
= minfC + a;Rg:
We assert that R
a
2 (R;R): This is obvious if R
a
= R; otherwise, if R
a
= C + a < R
then R
a
> R+ a (> R) by (3.3). Clearly 0 < u(r) < b for R < r < R
a
and we nd with
the help of (3.3) that
(v(r))
m
V
m
m
m 1
(C + a)
em
(F (b) + F ); R < r < R
a
:
16 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
It now follows from (3.4) that if R < r < R
a
, and is suciently large, then
(3.7) (v(r))
m
>
C + a
em
b
a
m
;
in turn, u
0
(r) < b=a. Now, if R
a
= R, then u
0
(R) = 0 since =2 I
, contradicting
u
0
(R) = lim
r"R
u
0
(r) b=a. Thus R
a
= C + a, leading to u
0
(r) < b=a for R < r 0 for 0 r < R. This shows that I
is not empty.
Condition (C3). Assume =1, n m and (1.7).
We may nd b > such that Q(s) 0 if s b. Choose > b=k, where k is the
constant specied in condition (C3), and let u be a corresponding solution of (2.1), with
domain J = (0; R). Dene R
k
to be the rst (unique by Lemma 2.5) point where u
reaches k: clearly R
k
< R. Also dene = (k; ) by f() = max
s2[k;]
f(s). In fact,
we can take = k
1
for some k
1
2 [k; 1]. For any r 2 (0; R
k
), integration of the identity
(2.3) over [0; r] gives
r
n1
ju
0
(r)j
m2
u
0
(r) =
Z
r
0
t
n1
f(u(t))dt
f()
n
r
n
;
thus
ju
0
(r)j
m2
u
0
(r)
f()
n
r;
or equivalently,
u
0
(r)
f()
n
1=(m1)
r
1=(m1)
:
Integrating this over [0; R
k
] leads to
(1 k)
m 1
m
f()
n
1=(m1)
R
m=(m1)
k
;
and therefore
(3.8) R
k
dn
m1
f()
1=m
; where d =
(1 k)
m
m 1
m1
:
GROUND STATES AND FREE BOUNDARY PROBLEMS 17
Since Q(s) 0 for s b, and Q(s) is locally bounded below near s = 0 by hypothesis
(C3), we can dene Q = inf
s>0
Q(s) 0.
Now assume for contradiction that I
is empty. Applying Proposition 2.7, we have
for R < r < R,
mr
n
E(r)
Z
r
0
Q(u(t))t
n1
dt =
Z
R
k
0
+
Z
R
R
k
+
Z
r
R
!
Q(u(t))t
n1
dt
Z
R
k
0
+
Z
r
R
!
Q(u(t))t
n1
dt since Q(u(t)) 0 for 0 < t < R
Z
R
k
0
Q(u(t))t
n1
dtQ
Z
r
R
t
n1
dt
Q(k
2
)
R
n
k
n
Q
r
n
n
; where Q(k
2
) = min
s2[k;]
Q(s); k
2
2 [k; 1]
Q(k
2
)
n
dn
m1
f(k
1
)
n=m
Q
n
r
n
by (3.8):
Let R
a
= minfC + a;Rg be as in the previous case. Recall that R
a
2 (R;R): Now
using (1.7) and noting that R
a
C + a, we may x > b=k so large that
(3.9) E(r) F (b) +
m 1
m
b
a
m
; R < r < R
a
:
It follows that
(3.10) R
a
= C + a < R
for this ; indeed otherwise R
a
= R From the fact that F (s) < F (b) when s < b, we get
(3.12) F (u(r)) < F (b) for R < r < R
a
:
Taking r = R
a
and using (3.11) gives E(R
a
) < F (b), which contradicts (3.9).
Finally, by (2.5), (3.9) and (3.12), we obtain u
0
(r) < b=a for R < r R
a
. As in the
previous case, this leads to u(R
a
) < 0, contradicting the fact that u(r) > 0 for 0 r < R.
18 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
Hence I
must be non-empty, and must in fact contain the value > b=k for which (3.9)
holds, completing the proof of Claim 3.
3.3. Proof of Claim 4
Let 2 I
and f
i
g be a sequence approaching as i ! 1. Let u be the solution
of (2.1) corresponding to u(0) = with maximal domain J = (0; R), R 0 since 2 I
. By Proposition
2.6 we can choose r
0
2 (R=2; R) such that 2d < E(r
0
) < 3d and
(3.13) R
i
> r
0
; d < E
i
(r
0
) < 4d; u
i
(r
0
) < 2u(r
0
) < ;
for i suciently large. Integrating (2.6) over [r
0
; R
i
] yields
(3.14)
E
i
(R
i
)E
i
(r
0
)
=
Z
R
i
r
0
n 1
r
ju
i
0
(r)j
m
dr
=
Z
u
i
(R
i
)
u
i
(r
0
)
n 1
r
ju
i
0
j
m1
du
(if R
i
=1; then u
i
(R
i
) = lim
r!1
u
i
(r))
n 1
r
0
sup
r
0
r
GROUND STATES AND FREE BOUNDARY PROBLEMS 19
Note that this remains valid if we replace r
0
by any r 2 (r
0
; R); in particular if r " R
we get u(r) ! 0 and thus E
i
(R
i
) d because of (3.13), which shows that
i
2 I
for
suciently large i. Thus I
is open.
4. A PRIORI ESTIMATES
The proof of Claim 3 in Section 3 gives, as a by product, an a priori estimate for the
supremum of a ground state u in terms of n, m and the nonlinearity f . First, in the
simple case (C1) it is clear that u(r) < for all r 0.
To treat case (C2), let b > and C = C(b) be given by (3.1). Suppose > b is not
in I
; then (3.3) and (3.4) hold and we can proceed as in the proof of Claim 3 to obtain
(3.7), provided that
m n
m 1
m
C
m(nm)
m1
( b)
m
>
m
m 1
(F + F (b)) +
b
a
m
(C + a)
em
;
or equivalently
(4.1) > b b+
m 1
m n
C
mn
m1
m
m 1
(F + F (b)) +
b
a
m
1=m
(C + a)
n1
m1
;
where a > 0 is an arbitrary number. By the nal argument in the proof of Claim 3 one
then obtains a contradiction. Consequently, when (4.1) is valid, then 2 I
and in turn
a radial ground state u of (1.1) (or a radial solution of (1.2)) has the upper bound b.
Next, we consider case (C3). Let b be such that Q(s) 0 for s b, and k 2 (0; 1].
Suppose > 0 is not in I
. Then we can proceed as in the proof of Claim 3 to obtain
(3.9), provided that
(4.2)
Q(k
2
)
dn
m1
f(k
1
)
n=m
Q+ nmF (b) + n(m 1)
b
a
m
(C + a)
n
;
> b=k; k
1
; k
2
2 [k; 1]; d = [(1 k)m=(m 1)]
m1
;
where we recall that C = C(b) is given by (3.1) and a > 0 is an arbitrary number. Then
by the remaining argument of Claim 3 one obtains a contradiction. Consequently, when
(4.2) is valid, then 2 I
and in turn the corresponding solution of (2.1) cannot be a
ground state or a solution of (1.2). This proves the following a priori estimates.
20 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
Theorem 2. Let hypotheses (H1) and (H2) hold. Let u be a radial ground state of (1.1)
or a radial solution of (1.2).
(i) Let n < m and =1. Then
(4.3) u(r) < b for all r 0;
where b is the constant given in (4.1), and b is any number larger than .
(ii) Let n m, =1. Suppose Q(s) is locally bounded below near s = 0 and Q(s) 0
for s b (> ). If (4.2) holds for all b, then
(4.4) u(r) < b for all r 0:
In connection with case (i), consider the example
(4.5) f(u) = u+ u
3
m = 4; n = 3:
Then
F (u) =
1
2
u
2
+
1
4
u
4
; =
p
2; F =
1
4
:
Choose b = 2; then
F (b) = 2; C = 2 3
3=4
;
see (3.1). Moreover,
b = 2 + 3
3 + 16=a
4
1=4
C
1=3
(C + a)
2=3
;
taking a = 2, which approximately minimizes the right hand side, we get b 26:6501.
Thus for example (4.5) any radial ground state u for (1.1) satises
ju(r)j < 27 for all r 0:
Note nally for this example that condition (1.5) is satised, so that from Theorem 1(i)
any ground state necessarily has compact support.
It is more complicated to nd explicit a priori estimates for the case (ii). From a
practical point of view, even for simple cases, the estimate (4.4) can give an extremely
large value for the supremum. Consider the example
(4.6) f(u) = u+ u
3
; m = 2; n = 3
GROUND STATES AND FREE BOUNDARY PROBLEMS 21
for which the corresponding ground state is known to be symmetric, positive and unique.
Then F (u), , F are as above. We have also
Q(u) = 2u
2
+
1
2
u
4
; Q = 2; d = 2(1 k):
Choosing b = 2 then gives F (b) = 2 and C = 3
p
2. Now, taking k
2
= k and k
1
= 1, the
left side of (4.2) is
2k
2
2
+
1
2
k
4
4
6(1 k)
1 +
2
3=2
;
and so inequality (4.2) becomes
(4.7) 3
p
6(1 k)
3=2
k
4
1 4=(k
2
2
)
(1 1=
2
)
3=2
14 +
12
a
2
3
p
2 + a
3
:
Finally choosing a = 4=3, which approximately minimizes the right hand side of (4.7),
and taking k = 8=11, condition (4.7) is satised for all 12; 500. Thus an a priori
estimate for the supremum of the ground state for (4.6) is
ju(r)j < 12; 500 for all r 0:
The case n = m = 2, f(u) = u + u
3
is also instructive. Here , F , d are the same
as for (4.6), while Q = 4F and Q = 1. Again let b = 2, then F (b) = 2, C = 3
p
2=2, and
(4.2) becomes
(4.8) 4(1 k)k
4
2
1 2=(k
2
2
)
1 1=
2
9 +
8
a
2
3
p
2
2
+ a
!
2
:
Choosing a = 5=4 and k = 4=5, condition (4.8) is satised for all > 22:5, a much more
practical value. Thus an a priori estimate for the supremum of the ground state in this
case is
ju(r)j < 22:5 for all r 0:
5. THE CASE n=m: EXPONENTIAL AND SUPERCRITICAL GROWTH
First we give the proof of Theorem 1
0
. We need to show Claims 1-4 of Section 3;
clearly only Claim 3 needs further argument.
Proof of Theorem 1
0
. As in Case (C3) of Section 4, in order to show that a value is
in I
it is enough to verify (4.2). In the present case, we have m = n, Q = n
2
F ; then
22 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
setting (1 k) = (note that the condition > b=k is thus equivalent to > b+ ) it
is easy to check that (4.2) reduces to
(5.1)
n1
F ( )
f()
>
(n 1)
n1
n
n
h
F + F (b) +
n 1
n
b
a
n
i
(C + a)
n
;
where a > 0 is an arbitrary number and (see (3.1))
C = C(b) = (n 1)
n
n 1
n1
n
b
F (b)
F + F (b)
n1
n
:
We take
a
n+1
=
n 1
n
C
F + F (b)
b
n
in (5.1), which minimizes the right hand side; after a fairly long calculation, (5.1) then
becomes
(5.2)
n1
F ( )
f()
>
(n 1)
n
b
n
n
n+1
1 +
nF
F (b)
+ n
n
n+1
!
n+1
:
Applying the elementary inequality
(x+ y)
p
2
p1
(x
p
+ y
p
); x; y; p 0;
we get
1 +
nF
F (b)
+ n
n
n+1
!
n+1
n
2
1
n
1 + n+
nF
F (b)
:
Thus (5.2), and consequently (4.2), follows if
n1
F ( )
f()
>
2
n
(n 1)b
F
F (b)
+ 1 +
1
n
n
;
which is just (1.9) if we dene
(5.3) =
2
n
(n 1)b
F
F (b)
+ 2
n
; F = min
0
GROUND STATES AND FREE BOUNDARY PROBLEMS 23
It is worth remarking that the left side of (1.9) depends only on the behavior of f for
large , and the right side only on the behavior of f on (0; b].
We shall now apply Theorem 1
0
to the case of nonlinearities f with exponential growth
as u!1. Suppose that 0, q 0, and
(5.4) f(s) = !(s) exp(s
q
) for s > ;
where !(s) is a Lipschitz continuous function for s > , with
(5.5) !
1
s
p
1
!(s) !
2
s
p
2
; !
1
> 0; !
2
> 0; p
1
p
2
:
Then the following result holds.
Theorem 3.1. Assume that (H1)-(H2) and (5.4)-(5.5) are satised and that n = m > 1,
=1. Then the conclusions of Theorem 1 hold if 0 q < 1 (p
2
p
1
)=n.
Proof. Without loss of generality we may assume . For any s > and q > 0, let
= (s)= s
1q
. Then for any s > + and s 2 [s ; s],
n1
(s)
F (s )
f(s)
!
1
s
(n1)(1q)
!
2
s
p
2
exp(s
q
)
Z
ss
1q
t
p
1
exp(t
q
)dt:
By L'Hopital's rule,
lim
s!1
n1
(s)
F (s )
f(s)
lim
s!1
!
1
!
2
s
(n1)(1q)
(s s
1q
)
p
1
(p
2
(n 1)(1 q))s
p
2
1
+ qs
p
2
+q1
exp
(s s
1q
)
q
s
q
=
!
1
exp(q)
q!
2
lim
s!1
(1 s
q
)
p
1
s
nq+p
2
p
1
n
=1;
since nq + p
2
p
1
n < 0. When q = 0, we take = s=2 and repeat the above
computation to get the same result. Now if is chosen suciently large, then (1.9) is
obviously satised and the proof is completed.
If q = 1 (p
2
p
1
)=n, the limit above is nite and the verication of (1.9) becomes
more involved. We give one theorem and one example for this case; for clarity, we shall
assume p
1
= p
2
, though more general cases can be discussed similarly.
24 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
Theorem 3.2. Assume that (H1)-(H2) and (5.4)-(5.5) are satised, and that n = m > 1,
= 1. If p
1
= p
2
and q = 1, then the conclusions of Theorem 1 hold if there exists
b > such that
(5.6) b
F
F (b)
+ 2
0 and > 0. It follows that
F (s) =
1
e
s
1 s
2
2
s
2
2
s
2
2
6
s
3
2
s
2
:
The last quantity takes a minimum value at s = 2=
2
, and thus
F jF (2=
2
)j =
2
3
3
4
:
Choose b = 4=
2
. We nd
F (b)
8
3
3
4
;
which leads to F=F (b) 1=4, a constant independent of , . Thus
b
F
F (b)
+ 2
9
:
To apply Theorem 3.2, we write !(s) = 1 [1 + (+ )s] =e
s
, and observe that corre-
spondingly p
1
= p
2
, !
1
!(s) !
2
= 1 and !
1
=!
2
! 1 as s ! 1. Obviously, (5.6)
follows if
(5.8)
> 9e
2
e
n 1
n
1=n
=
1
(n):
GROUND STATES AND FREE BOUNDARY PROBLEMS 25
For instance, if = 1 then (5.8) is satised if >
1
(n), or if = 1 then <
1
1
(n).
We mention that
1
(2) = 9
p
e and
1
(n)! 9e as n!1. It is easy to show that
1
(n)
is an increasing function of n, so that in fact we can use
1
= 9e for all n > 1.
The following result avoids the restriction q 1 (p
2
p
1
)=n, but at the expense of
more subtle considerations concerning the behavior of f(s).
Theorem 3.3. Assume that (H1)-(H2) and (5.4)-(5.5) are satised and that n = m > 1,
=1. Then the statement of Theorem 1 holds if there is a b > such that
(5.9) b
F
F (b)
+ 2
;
where is a constant depending only on n and the parameters in (5.4)-(5.5).
Proof. We choose = + 1, = 1=2 in (1.9). Then
n1
F ( )
f()
1
2
n1
!
1
!
2
R
+1=2
t
p
1
exp(t
q
)dt
( + 1)
p
2
exp(( + 1)
q
)
>
1
2
n
!
1
!
2
p
1
( + 1)
p
2
exp([( + 1)
q
q
]) = :
Thus (1.9) is satised if we choose
(5.10) =
1
n 1
n
2
1=n
:
This completes the proof.
It should be noted that the solutions in Theorem 1 which are obtained in this way
have the a priori bound ju(r)j < + 1.
Example 2. Consider the nonlinearity
(5.11) f(s) = e
s
q
1 s; > 0; > 0; q > 1:
By rescaling we can rst reduce to the case = 1. Then clearly f(s) s
q
s; thus
F (s)
1
2
s
2
1
2
(q + 1)
s
q1
:
26 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
The last quantity takes a minimum value at s =
1=(q1)
, yielding
F
q 1
2(q + 1)
s
2
:
Choose b = (q)
1=(q1)
= q
1=(q1)
s. We nd
F (b)
q 1
2(q + 1)
b
2
=
q 1
2(q + 1)
s
2
q
2=(q1)
:
Combining the last two inequalities gives
b
F
F (b)
+ 2
2q
1
q1
+ q
1
q1
1=(q1)
:
Using the previous elementary inequality, the right hand side can, for simplicity, be
replaced by the larger quantity 4(q)
1=(q1)
.
To apply Theorem 3.3, we write (5.11) in the form of (5.4) with = 1, p
1
= p
2
, and
observe that
1
2
!(s) = 1 (1 + s)=e
s
q
1;
provided 1 and s 2. Hence we can take !
1
= 1=2, !
2
= 1, p
1
= p
2
= 0 and = 2
in (5.5). For = 1=2 we then nd (see the proof of Theorem 3.3)
=
1
2
n+1
exp(2
q
3
q
):
In view of (5.10), condition (5.9) is now satised if
(q)
1=(q1)
1
8(n 1)
n
4
exp(2
q
3
q
)
1=n
:
Returning to the unscaled nonlinearity (5.11), we see that the conclusions of Theorem
1 now hold if
1=q
min
(
1;
1
q
1
8(n 1)
q1
n
4
exp(2
q
3
q
)
(q1)=n
)
:
Since the function (5.11) is logarithmically convex for large s, this example is covered
(when n = m = 2) by the results of [AP2]; also for 1 < q < 2 they require no restrictions
on and beyond ; > 0. See, however, the following Example 3.
GROUND STATES AND FREE BOUNDARY PROBLEMS 27
Example 3. Using the procedures illustrated in Examples 1 and 2 we can easily treat
other more complicated cases, e.g.,
(5.12) f(s) = s
p
(e
s
1) (+ )s
t
when p+ 1 = t > 1, and
(5.13) f(s) = s
p
(e
s
q
1) s
t
when p+ q > t > 1.
The work of [BGK] applies to (5.12) when p = 0, t = 1, with only the restriction
; > 0; and to (5.13) when t = 1, 0 q < 2, p > 1q also for all ; > 0. The results
of [AP2] do not apply to (5.12) because f is not logarithmically convex; they apply to
(5.13) when t 1, q > 1, p > t q, moreover for all ; > 0 when 1 < q < 2. Thus the
cases covered in [AP2] and [BGK] which are not included in our work are n = m = 2,
t = 1 [BGK], t > 1 [AP2], and 1 q < 2 for arbitrary ; > 0. Of course, for restricted
; , we include all these cases.
Finally, let the term e
s
q
in (5.13) be replaced by
e
s
q
+ sin(e
2s
q
)
when s (
1
log k)
1=q
and k is a large integer. Then, exactly as in Example 2, the
conclusions of Theorem 1 continue to hold whenever =
1=q
is suitably small.
On the other hand, the function e
s
q
+sin(e
2s
q
) is not logarithmically convex for any
q, so that the results of [AP2] no longer apply.
6. COMMENTS
We add some comments here concerning whether ground states of (1.1) are necessarily
radial, and whether radial ground states are themselves unique.
The classical paper of Gidas, Ni, Nirenberg [GNN] (see also [L] and [LN]) showed that
positive ground states of (1.3) are necessarily radially symmetric about some origin O,
provided that f is of class C
1+
in [0; ), with f(0) = 0; f
0
(0) < 0. Moreover, by virtue
of the strong maximum principle the a priori condition of positivity in this result is in
fact automatic.
Symmetry without recourse to the assumption f
0
(0) < 0 is more delicate. For simplic-
ity, we shall assume that f(0) = 0 and that f is a decreasing function on some interval
(0; ). Then, if
(6:1)
Z
0
ds
jF (s)j
1=2
=1;
28 FILIPPO GAZZOLA - JAMES SERRIN - MOXUN TANG
ground states of (1.3) indeed remain positive and symmetric, see [SZ]. On the other hand,
if (6.1) fails, then ground states necessarily have compact support, as shown in [PuSZ].
For this case, it was observed by [KS] and [FLS] that two or more compact support
ground states, with translated origins and disjoint supports, will still constitute a ground
state, which is clearly not radial: thus, in such cases equation (1.3) (and even (1.1)) can
have denumerable many non-radial ground states. To restore symmetry one can assume
additionally that the open support of the ground state is connected, see [SZ].
Turning to equation (1.1), the results are more subtle. First, if (1.4) holds and 1 2,
or if (1.4) fails, symmetry is known only under the fairly strong condition that the ground
state in question has only a single critical point, see [SZ], [BN].
For the study of uniqueness of radial ground states of (1.1) for the model nonlinearity
(1.8) and also for other types of nonlinearities, the reader is referred to [FLS], [PS2],
[PuS2] and references therein. In the last reference, uniqueness is established both for
problems (1.1) and (1.2) and whether or not f(0) = 0. The results here cover in particular
the examples discussed at the end of Section 4. Finally, for the casem = n > 1 uniqueness
holds for the model exponential nonlinearity (5.7), see [PuS3].
Throughout the paper we have considered only ground states for the entire space R
n
,
and solutions of the Dirichlet-Neumann problem (1.2). The related Dirichlet problem for
(1.1) or (1.3) in a bounded domain is of course important in itself, both when f(0) = 0
and f(0) 6= 0. For these problems the papers [AG, CK, CRT, CS, GOW, LM, M] are
particular relevant to the results here.
Acknowledgment. Part of this work was done while the rst author was a visitor at
the School of Mathematics at the University of Minnesota; he is grateful for the kind
hospitality there. F.G. was partially supported by the Italian CNR.
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Authors addresses:
Filippo Gazzola:
Dipartimento di Scienze T.A., Universita A. Avogadro, 15100 Alessandria, Italy
E-Mail: [email protected]
James Serrin, Moxun Tang:
School of Mathematics, University of Minnesota, Minneapolis, MN 55455
E-Mail: [email protected]
E-Mail: [email protected]
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