Nonlinear Analysis 57 (2004) 341 – 348
www.elsevier.com/locate/na
Existence and uniqueness of positive radial
solutions for the Lane–Emden system
Robert Dalmasso∗
Laboratoire LMC-IMAG, Equipe EDP, Tour IRMA, BP 53, F-38041 Grenoble Cedex 9, France
Received 22 April 2003; accepted 24 February 2004
Abstract
In this article, we consider (component-wise) positive radial solutions of a weakly
coupled system of elliptic equations in a ball with homogeneous nonlinearities. The existence is
well-known in general: We give a result for the remaining cases. The uniqueness is less studied:
We complement the known results.
? 2004 Elsevier Ltd. All rights reserved.
Keywords: Semilinear elliptic system; Positive radial solution
1. Introduction
In this paper, we study the existence and the uniqueness of (component-wise)
positive radial solutions of the semilinear elliptic system with homogeneous Dirichlet data

6u + vq = 0 in BR ;



6v + up = 0 in BR ;
(1.1)



u = v = 0 on @BR ;
where BR denotes the open ball of radius R centered at the origin in Rn (n ¿ 1) and p,
q ¿ 0. Eq. (1.1) is a natural extension of the well-known Lane–Emden equation and
thus referred to as the Lane–Emden system.
When q = 1 and p ∈ (0; 1) ∪ (1; +∞) the uniqueness of a positive radial solution
of problem (1.1) was established in [2] (see also the references therein). When p,
q ¿ 0 and pq ¡ 1 the uniqueness of a positive radial solution of problem (1.1) is just
∗
Tel.: +33-4-76-63-57-38; fax: +33-4-76-63-12-63.
E-mail address: [email protected] (R. Dalmasso).
0362-546X/$ - see front matter ? 2004 Elsevier Ltd. All rights reserved.
doi:10.1016/j.na.2004.02.018
342
R. Dalmasso / Nonlinear Analysis 57 (2004) 341 – 348
a particular case of a result obtained in [5]. Finally, when n = 1, q ¿ 1 and p ¿ 1, the
uniqueness of a positive solution follows from a general result given in [3]. Therefore,
it only remains to study the case where p, q ∈ (0; 1) ∪ (1; +∞) with pq ¿ 1 and the
case where p, q ¿ 0 with pq = 1.
We have the following theorem.
Theorem 1.1. (i) Let p, q ¿ 0 with pq = 1. Then the Lane–Emden system (1.1) has
at most one positive radial solution.
(ii) Let p, q ¿ 0 with pq = 1. Assume that the Lane–Emden system (1.1) has a
positive radial solution (u; v). Then all positive radial solutions are given by ( q u; v),
where ¿ 0 is an arbitrary constant.
When p, q ¿ 1, pq ¿ 1 and
1
1
n−2
+
¿
p+1 q+1
n
if n ¿ 3;
(1.2)
the existence of a positive solution was established in [1,6,8] for more general nonlinearities (see also [3] when n = 1). Moreover, it is well-known (see [7,11]) that when
n ¿ 3, p, q ¿ 0 and
1
1
n−2
+
;
6
n
p+1 q+1
problem (1.1) has no positive solution. If p, q ¿ 0 and pq ¡ 1, the existence of a
positive solution is a particular case of a result proved in [5]. Therefore, it remains
to examine the existence of positive radial solutions of problem (1.1) in the cases
described in the following theorem.
Theorem 1.2. (i) If p ¿ 1, q ∈ (0; 1) satisfy (1.2) and pq ¿ 1, then the Lane–Emden
system (1.1) has a positive radial solution.
(ii) Let p, q ¿ 0 with pq = 1. Then there exists R ¿ 0 such that the Lane–Emden
system (1.1) has a positive radial solution.
2. Proof of Theorem 1.1. (i) Let (u; v) and (w; z) be two positive radial solutions of
problem (1.1). Let s and t be deAned by
s=2
q+1
pq − 1
and
t=2
p+1
:
pq − 1
(2.1)
For ¿ 0 we set
w̃(r) = s w(r)
and
z̃(r) = t z(r);
We have

6w̃(r) + z̃(r)q = 0; 0 6 r 6 R=;



6z̃(r) + w̃(r)p = 0; 0 6 r 6 R=;



w̃(R=) = z̃(R=) = 0;
0 6 r 6 R=:
R. Dalmasso / Nonlinear Analysis 57 (2004) 341 – 348
343
where 6 denotes the polar form of the Laplacian
d
d
r n−1
:
6 = r 1−n
dr
dr
Choose such that s w(0) = u(0). Then we have
w̃(0) = u(0):
(2.2)
We want to show that
z̃(0) = v(0):
(2.3)
Suppose that z̃(0) ¡ v(0). If there exists a ∈ (0; min(R; R=)] such that z̃ − v ¡ 0 on
[0; a) and (z̃ − v)(a) = 0, then w̃ − u ¿ 0 on (0; a]. Indeed assume the contrary. Then
there exists b ∈ (0; a] such that (w̃ − u)(b) 6 0. Since 6(w̃ − u) = vq − z̃ q ¿ 0 on [0; b),
the maximum principle implies that w̃ − u ¡ 0 on [0; b), a contradiction with (2.2).
Now we have 6(z̃ − v) = up − w̃p ¡ 0 on (0; a] and the maximum principle implies
that z̃ − v ¿ (z̃ − v)(a) = 0 on [0; a), a contradiction. Thus z̃ − v ¡ 0 on [0; min(R; R=)].
Since

−v(R=) if ¿ 1;



if = 1;
(z̃ − v)(min(R; R=)) = 0



z̃(R)
if ¡ 1;
we deduce that ¿ 1. As before we show

−u(R=)



(w̃ − u)(min(R; R=)) = 0



w̃(R)
that w̃ − u ¿ 0 on (0; min(R; R=)]. We have
if ¿ 1;
if = 1;
if ¡ 1;
hence ¡ 1 and we have a contradiction. The case z̃(0) ¿ v(0) can be handled in the
same way. Thus (2.3) is proved.
Now we deAne the functions U , W , F and Gn by
U (r) = (u(r); v(r));
W (r) = (w̃(r); z̃(r));
0 6 r 6 R;
0 6 r 6 R=;
F(x; y) = (yq ; xp ); x; y ¿ 0
and

r−s
if n = 1;




r 

if n = 2;
s ln
Gn (r; s) =
s


s n−2 
s


1−
if n ¿ 3;

n−2
r
(2.4)
for 0 6 s 6 r. Using (2.2) and (2.3) and the fact that u (0) = w̃ (0) = v (0) = z̃ (0) = 0
we easily obtain
r
U (r) − W (r) =
Gn (r; s)(F(W (s)) − F(U (s))) ds
0
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R. Dalmasso / Nonlinear Analysis 57 (2004) 341 – 348
for r ∈ [0; min(R; R=)]. When p, q ¿ 1, F is locally Lipschitz continuous, and using
Gronwall’s lemma we obtain U = W on [0; min(R; R=)]. When p or/and q ∈ (0; 1), let
a ∈ (0; min(R; R=)) be Axed. Then u(0) ¿ u(r) ¿ u(a) ¿ 0, w̃(0)=u(0) ¿ w̃(r) ¿ w̃(a)
¿ 0, v(0) ¿ v(r) ¿ v(a) ¿ 0 and z̃(0) = v(0) ¿ z̃(r) ¿ z̃(a) ¿ 0 for r ∈ [0; a]. Since F
is locally Lipschitz continuous on (0; +∞) × (0; +∞), as before we obtain U = W on
[0; a]. By continuity we get U = W on [0; min(R; R=)]. Now we deduce that = 1 and
thus (u; v) = (w; z) on [0; R].
(ii) Let (u; v) be a positive radial solution of problem (1.1). Then, for any ¿ 0,
(w; z) = ( q u; v) is a positive radial solution of problem (1.1). Now let (w; z) be a
positive radial solution of (1.1). Choose ¿ 0 such that q u(0) = w(0) and deAne
w̃ = q u, z̃ =v. Then (w̃; z̃) is a positive radial solution of (1.1) such that w̃(0)=w(0).
Arguing as in part (i) we show that z̃(0) = z(0) and that (w̃; z̃) = (w; z).
Remark 2.1. Notice that our technique also applies when there is a homogeneous dependence on the radius |x|. More precisely, for p, q ¿ 0 and pq = 1, the following
system:

6u + |x| vq = 0 in BR ;



6v + |x| up = 0 in BR ;
(2.5)



u=v=0
on @BR ;
where , ¿ 0, has at most one positive radial solution (u; v). Indeed, the arguments
are the same with s and t in (2.1) replaced by
s=
2(q + 1) + + q
pq − 1
and
t=
2(p + 1) + + p
:
pq − 1
Now let p, q ¿ 0 with pq = 1. Assume that problem (2.5) has a positive radial
solution (u; v). Then all positive radial solutions are given by ( q u; v), where ¿ 0
is an arbitrary constant.
3. Proof of Theorem 1.2. We shall use a two-dimensional shooting argument for the
ordinary diIerential system associated to radial solutions of (1.1) [3,4,9,10]. We only
assume for the moment that p, q ¿ 0. Since we are only interested in positive solutions
we may extend v → vq and u → up to R by setting
q
p
v
if v ¿ 0;
u
if u ¿ 0;
g(v) =
and f(u) =
0
if v 6 0;
0
if u 6 0:
We introduce the one-dimensional (singular if n ¿ 2) initial value problem

6u(r) + g(v(r)) = 0; r ¿ 0;



6v(r) + f(u(r)) = 0; r ¿ 0;



u(0) = !; v(0) = "; u (0) = v (0) = 0;
where ! ¿ 0, " ¿ 0.
(3.1)
R. Dalmasso / Nonlinear Analysis 57 (2004) 341 – 348
345
We shall need a series of lemmas. We begin with a local existence and uniqueness
result. The proof is standard (see [3] if n = 1 and [9,10] when n ¿ 3).
Lemma 3.1. For any ! ¿ 0, " ¿ 0 there exists R ¿ 0 such that problem (3.1) on
[0; R] has a unique solution (u; v) ∈ (C 2 [0; R])2 .
In view of Lemma 3.1, for any !, " ¿ 0 problem (3.1) has a unique local solution:
Let [0; R!; " ) denote the maximum interval of existence of that solution (R!; " = +∞
possibly). If 0 ¡ q ¡ 1 or 0 ¡ p ¡ 1 the uniqueness of the solution could fail at any
point r where v(r) = 0 or u(r) = 0. In this case R!; " could also depend on the particular
solution itself. DeAne
P!; " = {s ∈ (0; R!; " ); u(!; "; r)v(!; "; r) ¿ 0
∀r ∈ [0; s]};
where (u(!; "; ·); v(!; "; ·)) is a solution of (3.1) in [0; R!; " ). Set
r!; " = sup P!; " :
Notice that the solution is unique on [0; r!; " ], so r!; " depends only on !, ".
The next lemma can be proved by direct integration of system (3.1).
Lemma 3.2. We have u (!; "; r) ¡ 0 and v (!; "; r) ¡ 0 for r ∈ (0; R!; " ).
Lemma 3.3. For any !, " ¿ 0 we have r!; " ¡ R!; " .
Proof. If not, there exist ! ¿ 0 and " ¿ 0 such that r!; " = R!; " . Suppose Arst that
R!; " ¡ ∞. Noting u = u(!; "; ·) and v = v(!; "; ·) we have
06u6!
and
Since
u (r) = −r 1−n
06v6"
r
0
on
sn−1 v(s)q ds
and
[0; R!; " ):
v (r) = −r 1−n
0
r
sn−1 u(s)p ds
for r ∈ (0; R!; " ), we conclude that u, v, u and v are bounded on [0; R!; " ) and we get
a contradiction with the deAnition of R!; " . Now assume that R!; " = +∞. When n ¿ 3
the result follows from Corollary 1.2 in [9]. When n = 1 we have u ¡ 0 on [0; +∞).
We deduce that
u (r) 6 u (1) ¡ 0
for all r ¿ 1;
from which we get
u(r) 6 u(1) + u (1)(r − 1)
for all r ¿ 1:
Thus, we can And r ¿ 1 such that u(r) ¡ 0 and we obtain a contradiction. Now
if n = 2, we have (ru (r)) ¡ 0 on (0; +∞). We deduce that
ru (r) 6 u (1) ¡ 0;
for all r ¿ 1;
from which we get
u(r) 6 u(1) + u (1) ln r;
for all r ¿ 1:
Thus we can And r ¿ 1 such that u(r) ¡ 0 and we obtain a contradiction.
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R. Dalmasso / Nonlinear Analysis 57 (2004) 341 – 348
Proposition 3.1. For any ! ¿ 0 there exists a unique " ¿ 0 such that u(!; "; r!; " ) =
v(!; "; r!; " ) = 0.
Proof. We Arst prove the uniqueness. Let ! ¿ 0 be Axed. Suppose that there exist
" ¿ % ¿ 0 such that u(!; "; r!; " ) = v(!; "; r!; " ) = u(!; %; r!; % ) = v(!; %; r!; % ) = 0. Using the
same arguments as in the proof of (2.3) we obtain a contradiction.
Now we prove the existence. Suppose that there exists ! ¿ 0 such that for any " ¿ 0
u(!; "; r!; " ) ¿ 0 or v(!; "; r!; " ) ¿ 0. DeAne the sets
B = {" ¿ 0; u(!; "; r!; " ) = 0
and
v(!; "; r!; " ) ¿ 0}
C = {" ¿ 0; u(!; "; r!; " ) ¿ 0
and
v(!; "; r!; " ) = 0}:
and
The proof of the proposition is completed by using the next two lemmas which
contradict the fact that (0; +∞) = B ∪ C.
Lemma 3.4. (i) Suppose B = ∅. Then there exists m ¿ 0 such that m 6 inf B.
(ii) Suppose C = ∅. Then there exists M ¿ 0 such that M ¿ sup C.
Lemma 3.5. B and C are open.
Proof of Lemma 3.4. We have
r
u(!; "; r) = ! − Gn (r; s)g(v(!; "; s)) ds;
0
and
v(!; "; r) = " −
0
r
Gn (r; s)f(u(!; "; s)) ds;
where Gn is deAned in (2.4).
(i) Let " ∈ B. Lemma 3.2 and (3.2) imply
1=2
2n!
r!; " ¿
;
"q
and from (3.3) we get
r!;"
"¿
Gn (r!; " ; s)u(!; "; s)p ds:
0
0 6 r ¡ R!; " ;
0 6 r ¡ R!; " ;
(3.2)
(3.3)
(3.4)
(3.5)
Suppose that inf B = 0 and let ("j ) be a sequence in B decreasing to zero. Then
r!; "j → +∞ by (3.4). From (3.5) we deduce that
1
"j ¿
Gn (r!; "j ; s)u(!; "j ; s)p ds
(3.6)
0
for j large. Using Lemma 3.2 and (3.2) we have
"jq
!
¿
(3.7)
u(!; "j ; r) ¿ ! −
2n
2
for r ∈ [0; 1] and j large. From (3.6) and (3.7), we get "j ¿ c for j large where c ¿ 0
is independent of j. This gives a contradiction.
R. Dalmasso / Nonlinear Analysis 57 (2004) 341 – 348
347
(ii) Suppose that sup C = +∞ and let ("j ) be a sequence in C increasing to +∞.
By virtue of Lemma 3.2 we have
0 ¡ u(!; "j ; r) 6 !
for r ∈ [0; r!; "j ]:
(3.8)
(3.3) and (3.8) imply that r!; "j → +∞ as j → +∞. Then we can assume that r!; "j ¿ 1
for all j and that
!p 6 "j
for all j:
(3.9)
Lemma 3.2, (3.3), (3.8) and (3.9) imply
2n − 1
"j 6 v(!; "j ; r) 6 "j for r ∈ [0; 1]
2n
and using (3.2) we deduce that u(!; "j ; 1) 6 ! − (2n − 1)q "jq =(2n)q+1 . But then
u(!; "j ; 1) ¡ 0 for j large contradicting (3.8).
Remark 3.1. When n = 1 and p, q ¿ 1 Lemma 3.5 is proved in [3]. It is easily seen
that the proof also holds when n ¿ 2. Now if 0 ¡ q ¡ 1 and p ¿ 1, the arguments
given in [3] do not apply since g is no longer Lipschitz continuous at 0; thus one
loses continuous dependence on the initial data at r = r!; " , a fact which was essential
in the proof. We shall use a result [10, Lemma 2.4] which applies in the Lipschitz and
nonLipschitz cases. In fact n ¿ 3 in [10], but the result used also holds when n = 1 or
2. Finally notice that f and g in [10] (see also [9]) are
f(x) = |x|p
and
g(x) = |x|q
for x ∈ R;
but the proof is the same with our deAnition of f and g.
Proof of Lemma 3.5. As explained in Remark 3.1 we shall use a result established
in [10]. DeAne
X = {(%; ") ∈ (0; +∞) × (0; +∞); v(%; "; r%; " ) ¿ u(%; "; r%; " ) = 0}
and
Y = {(%; ") ∈ (0; +∞) × (0; +∞); u(%; "; r%; " ) ¿ v(%; "; r%; " ) = 0}:
X and Y are open (see [9,10]). Now let h : R2 → R denote the second projection on
R. Since B = h({!} × (0; +∞) ∩ X ) and C = h({!} × (0; +∞) ∩ Y ), we deduce that B
and C are open and the lemma is proved.
Now we can complete the proof of Theorem 1.2.
(i) Let ! ¿ 0 be Axed. By Proposition 3.1 there exists a unique " ¿ 0 such that
u(!; "; r!; " ) = v(!; "; r!; " ) = 0. With s and t deAned in (2.1) we set
r s r t r!; " r!; " !; "
!; "
u !; ";
v !; ";
w(r) =
r
and z(r) =
r ; 0 6 r 6 R:
R
R
R
R
Then (w; z) is a positive radial solution of the Lane–Emden system (1.1).
(ii) follows from Proposition 3.1.
Remark 3.2. Notice that, by Lemma 3.3, X and Y are precisely A and B in [10] (B
and A in [9]).
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R. Dalmasso / Nonlinear Analysis 57 (2004) 341 – 348
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