Uniqueness of positive radial solutions for
semilinear ellipticequations on annular domains
Chun-Chieh Fu
a;∗, Song-Sun Lin
baDepartment of Mathematics, National Tsing-Hua University, Hsin-Chu, Taiwan bDepartment of Applied Mathematics, National Chiao-Tung University, Hsin-Chu, Taiwan
Received 1 February 1999; accepted 4 May 1999
Keywords: Uniqueness; Dirichlet boundary condition; Annular domain
1. Introduction
In this paper we shall study the uniqueness problem of positive radial solutions for semilinear ellipticequations on annular domains. Indeed, we consider the following equations:
2u + f(u) = 0 in 5; (1.1)
u = 0 on @5; (1.2)
where 5 = 5a;b= {x ∈ Rn; 0 ¡ a ¡ |x| ¡ b} is an annular domain in Rn; n ≥ 3.
It is well known that the uniqueness problems of solutions of (1.1) and (1.2) are both fundamental and often di8cult. During the last decade, there has been tremendous progress in studying these problems when 5 is a ball in Rn or entire Rn n ≥ 3, see,
e.g., [1,2,5,10,11]. However, very little was known concerning the annular domain until the work of Ni and Nussbaum [12]. Until very recently, Co<man [3] showed that when n = 3; f(u) = −u + up; 1 ¡ p ≤ 3, (1.1) and (1.2) have unique positive radial solution
for any annular domain, see also [13]. Then, it is of interest to extend Co<man’s result to a more general situation. For example, we may ask the following question. Question. Do (1.1) and (1.2) have unique positive radial solution when f(u) = −u + up; 1 ¡ p ≤ n=(n − 2), and n ≥ 4?
Work partially supported by the National Science Council of the Republic of China.
∗Corresponding author.
0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved. PII: S0362-546X(99)00303-X
In this paper, after carefully investigating the auxiliary functions which have been used in [3,11] and some techniques developed in dealing with the problem on ball [1], we can prove the uniqueness result in the case n = 4 and partial result when n = 5.
Indeed, we can prove the following general result.
Theorem 1. Assume f ∈ C1([0; ∞)) and satis1es the following conditions:
(A-1) f(u)(u − 1) ¿ 0 for u ¿ 0 and u = 1, (A-2) f(u)u ¿ f(u) for u ¿ 0,
(A-3) f(u)(u − 1) ¿ f(u) for u ¿ 1, (A-4) I(u) ¡ 0 for u ¿ 1,
where
I(u) = (n − 2)f(u)(u − 1) − nf(u): (1.3)
Then (1:1) and (1:2) have a unique positive radial solution on any annular domain in Rn; n ≥ 3.
One consequence of Theorem 1 is the following result.
Theorem 2. For 3 ≤ n ≤ 5; f(u) = −u + up; and 1 ¡ p ≤ min(4=(n − 2); n=(n − 2)).
Then (1:1) and (1:2) have a unique positive radial solution on any annular domain in Rn.
Higher dimensional cases are a subject of further study in future.
This paper is organized as follows: In Section 2, we present some preliminary results concerning the positive radial solutions on annular domains. In Section 3, we prove the main results.
2. Preliminaries
In this section we shall recall and prove some useful properties of positive radial solution of (1.1). We always assume (A-1) holds.
The radial symmetricsolution of (1.1) satisKes the following equation: u(r) +n − 1
r u(r) + f(u(r)) = 0; r ¿ a ¿ 0: (2.1)
Denote by u(·; ) the unique solution of Eq. (2.1) and initial conditions
u(a; ) = 0 and u(a; ) = ¿ 0: (2.2)
If u(·; ) has a zero, then denote its Krst zero by R()(¡ ∞); otherwise, R() = ∞. It is known that the solutions can be classiKed into the following three types:
N = { ¿ 0: u(r; ) ¿ 0 in (a; R()) and u(R(); ) = 0; R() ¡ ∞}; G = ¿ 0: u(r; ) ¿ 0 in (a; ∞) and lim
r→∞u(r; ) = 0
and
P = (0; ∞) \ (N ∪ G):
It is also proved that N = ∅ when (A-1) and (A-2) holds, see, e.g., [6,8,9]. Indeed, N ⊃(∗; ∞) for some ∗¿ 0. In this case, there is a unique R
0() ∈ (; R()) such that u(r; ) ¿ 0 in (a; R 0()) and u(r; ) ¡ 0 in (R 0(); R()); see, e.g., [4,9].
To study the uniqueness problem, it is useful to study the variation ’ of u with respect to , i.e.,
’(r; ) =@u(r; )@ : It is clear that ’ satisKes
’(r) +n − 1
r ’(r) + f(u(r))’(r) = 0; r ¿ a; (2.3)
’(a; ) = 0 and ’(a; ) = 1: (2.4)
It is also known that ’ has a zero in (a; R()) when (A-1) and (A-2) hold, see e.g., [7]. The Krst zero of ’ in (a; R()] is denoted by r1(). Denote the maximum ’()
of ’ in (a; r1); then it is easy to verify
’(r; ) ¿ 0 in (a; ()) and ’(r; ) ¿ 0 in ((); r
1()); (2.5)
see, e.g., [4,9].
The following lemma due to Co<man [3] is critical. Lemma 2.1. (Co<man [2,3]). If (A-1)–(A-3) hold; then
u(; ) ¿ 1: (2.6)
Proof. We Krst claim ’¡1
u in (a; ): (2.7)
Indeed, let V = (1=)u − ’. Then V satisKes V+n − 1
r V+ f(u)V = 1
(f(u)u − f(u)): (2.8)
From (2.3) and (2.8), we have
rn−1{V(r)’(r) − V (r)’(r)} =1
r
a s
By (A-2), the right-hand side is positive in (a; r1), which implies V(r)’(r) − V (r)’(r) ¿ 0 (2.9) in (a; r1). Hence, d dr V (r) ’(r) =’21(r){V(r)’(r) − V (r)’(r)} ¿ 0 in (a; r 1): Now, V (a) ’(a) = limr→a+
1 u(r; ) ’(r; )− 1 = 0: Therefore, we have V (r) ¿ 0 in (a; r1):
By (2.9) and the last inequality, we have V(r) ¿ 0 in (a; );
and thus (2.7) follows.
To prove (2.6), we need to introduce more auxiliary functions as follows: Denote F(u) = u 0 f(v) dv; (2.10) M(r) = u2(r) + 2F(u(r)); (2.11) and N(r) = u(r)’(r) + f(u(r))’(r): (2.12) It is easy to verify M(r) = −2(n − 1) r u2(r) (2.13) and N(r) = −2(n − 1) r u(r)’(r): (2.14)
Therefore, by (2.11) and (2.13), we have M(r) = 2− 2(n − 1) r
a
u2(s)
s ds: (2.15)
Since u(R(); ) = 0, we have M(R()) = u2(R()) ¿ 0: Hence (2.13) implies
Eq. (2.15) implies 2(n − 1) r a u2(s) s ds ¡ 2: (2.16)
Similarly, by (2.12) and (2.14) we have N(r) = − 2(n − 1)
r
a
1
s’(s)u(s) ds: (2.17)
By Schwartz inequality, (2.7) and (2.16), we have 2(n − 1) r a 1 s’s(s)u(s) ds ≤ 2(n − 1) r a 1 s’2(s) ds 1=2 r a 1 su2(s) ds 1=2 ¡ 2(n − 1) ·1 · r a 1 su2(s) ds ¡ :
By (2.17) and the last inequality, we have N() ¿ 0:
Since ’() ¿ 0 and ’() = 0, by (2.12) we have f(u(; )) ¿ 0:
By (A-1), we have u(; ) ¿ 1: The proof is complete.
Next, we shall prove the following lemma. Lemma 2.2. If (A-1)–(A-3) hold; then
u(r1(); ) ¿ 1: (2.18)
Proof. Let W = u − 1, then W satisKes W+n − 1
r W+ f(u)W = f(u)(u − 1) − f(u): (2.19)
By (2.3) and (2.19), we obtain rn−1(W’ − W’)|r = r s n−1{f(u(s))(u(s) − 1) −f(u(s))}’(s) ds ¿ 0 (2.20) for any r ∈ (; r1].
We now claim W (r1) ¿ 0. Otherwise, if there were ˆr ∈ (; r1] such that W ( ˆr) = 0,
then, by (2.20) and(2.7) we have ˆrn−1u( ˆr)’( ˆr) ¿ n−1u()’() ¿ 0 which implies
u( ˆr) ¿ 0:
Now (2.6) and u( ˆr) = 1 imply u( ˆr) ¡ 0, a contradiction. This implies W (r1) ¿ 0:
The proof is complete. 3. Proofs of main results
In this section, we shall prove Theorems 1 and 2. For any c ∈ R1, denote
vc(r) = ru(r) + c(u(r) − 1): (3.1) Then, v c(r) = (c + 2 − n)u(r) − rf(u(r)) (3.2) and vc satisKes v c(r) +n − 1r vc(r) + f(u(r))vc(r) = Ic(u(r)); (3.3) where
Ic(u) = cf(u)(u − 1) − (c + 2)f(u): (3.4)
For notational simplicity, when c = n − 2, we denote v = vn−2= ru(r) + (n − 2)(u(r) − 1)
and
I = In−2= (n − 2)f(u)(u − 1) − nf(u):
It is easy to verify
I(u) = (n − 2)f(u)(u − 1) − 2f(u) and
I(u) = (n − 2)f(u)(u − 1) + (n − 4)f(u):
The following lemma is crucial in studying the uniqueness problem. Lemma 3.1. If (A-1)–(A-4) hold; then
Proof. By (2.3) and (3.3), we have rn−1(v’ − v’)|r1 = r1 r n−1I(u(r))’(r) dr:
Now, (A-4) implies I(u(r)) ¡ 0 in (; r1). Therefore, we have
−rn−1
1 v(r1)’(r1) − n−1v()’() ¡ 0:
By (3.2), we have
−r1n−1v(r1)’(r1) + nf(u())’() ¡ 0:
Now (2.6) implies f(u()) ¿ 0, therefore v(r1) ¡ 0:
The proof is complete.
To prove the main result, we need to introduce the following auxiliary function Z which also has been used in [3]. Denote
Z(r) = v(r) · {r’(r) + (n − 2)’(r)} + r2f(u(r))’(r): (3.6)
Then it is easy to verify that
Z(r) = r’(r)J(u(r)); (3.7)
where
J (u) = (4 − n)f(u) − (n − 2)f(u)(u − 1): (3.8)
It is easy to verify that
J (u) ≤ 0 for u ¿ 1 (3.9)
when (A-3) holds.
After these preparations, we can now prove the following lemma which leads to the main result.
Lemma 3.2. If (A-1)–(A-4) hold; then ’ has exactly one zero in (a; R()].
Proof. Suppose that ’(r; ) has a second zero r2 ∈ (a; R()]. Then there exists ∈
(r1; r2) such that
’( ) + (n − 2)’( ) = 0: (3.10)
If (3.10) holds, then we claim that
u( ) ¡ 1: (3.11) Indeed, by (3.5), we have Z(r1) = r1v(r1)’(r1) ¿ 0: (3.12) By (3.7) and (3.9) we have Z(r) ≥ 0 in (r 1; ˆr); (3.13)
whenever u( ˆr) ≥ 1. Otherwise, if (3.11) were false, i.e., u( ) ≥ 1. Then (3.12) and(3.13) imply
Z( ) ¿ 0: (3.14)
On the other hand, by (3.6), (3.10), and (3.14), we have
2f(u( ))’( ) ¿ 0
which implies f(u( )) ¡ 0 or u( ) ¡ 1, a contradiction. Hence (3.11) follows. Now, let v0= ru(r) as in (3.1). Then I0= −2f(u), i.e., v0 satisKes
v 0(r) +n − 1r v0(r) + f(u(r))v0= −2f(u(r)): (3.15) By (2.3) and (3.15), we have rn−1(v 0’ − v0’)|r2= −2 r2 rn−1f(r(r))’(r) dr ¡ 0:
However, by (3.11), the left-hand side of the last equality is
−rn
2u(r2)’(r2) + nf(u( ))’( ) ¿ 0;
a contradiction. Therefore, ’ has no second zero in (a; R()). The proof is com-plete.
Remark 3.3. In the proof of Lemma 3.2, it is critical to establish (3.11) which is immediate when (3.5) holds. Therefore, to prove (3.5), more careful study is needed for v(r; ) when (A-4) does not hold.
Proof of Theorem 1. Since
u(R(); ) = 0 (3.16)
after di<erentiating (3.16) with respect to ∈ N, we have u(r(); )dR()
d + ’(R(); ) = 0: (3.17)
Therefore, Lemma 3.2 implies dR()
d ¡ 0 (3.18)
for any ∈ N. Now, it is easy to see N = (∗; ∞), for some ∗¿ 0. Otherwise, if there were (1; 2) ⊂ N, 0 ¡ 1¡ 2¡ ∞ and 1 ∈ N; 2 ∈ N, then 1 and 2 will be
in G, a contradiction to (3.18). The proof is complete. Now, we can apply Theorem 1 to prove Theorem 2.
Proof of Theorem 2. We will verify that f(u) satisKes all assumptions of Theorem 1. It is easy to see that f(u) satisKes (A-1)–(A-3). To prove (A-4), by (1.3) we have
I(u) = (n − 2)(pup−1− 1)(u − 1) − n(up− u), then
I(u) = (n − 2)(p(p − 1)up−2− 1)(u − 1) − n(pup−1− 1);
I(u) = p(p − 1)[(n − 4)up−2+ (u − 2)(p − 2)up−3(u − 1)]:
For 3 ≤ n ≤ 4; p − 2 ≤ n=(n − 2) − 2 = (4 − n)=(n − 2). Since I(1) = 0; I(1) =
−2(p − 1) ¡ 0, and
I(u) ≤ p(p − 1)[(n − 4)up−2+ (4 − n)up−3(u − 1)]
= p(p − 1)(n − 4)up−3≤ 0;
we have I(u) negative for u ¿ 1.
For 4 ¡ n ¡ 6, if I(u) is nonnegative somewhere in u ¿ 1, then there is a u0¿ 1
such that I(u0) = 0 and I(u0) ≥ 0. I(u0) = 0 is equivalent to 2(pup−10 − 1) = (n − 2)p(p − 1)up−20 (u0− 1): Hence, 0 ≤ u0I(u0) = p(p − 1)[(n − 4)up−10 + (n − 2)(p − 2)up−20 (u0− 1)] = p(p − 1)(n − 4)up−10 + 2(p − 2)(pup−10 − 1) = p[p(n − 2) − n]up−10 − 2(p − 2) ¡ p[p(n − 2) − n] − 2(p − 2) ≤ 0 for 1 ¡ p ≤ n − 24 ; a contradiction. The proof is complete. References
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