Solutions of semilinear elliptic equations with
asymptotic linear nonlinearity
Cheng-Hsiung Hsu
a; ∗, Yi-Wen Shih
baDepartment of Mathematics, National Central University, Chung-Li 320, Taiwan, ROC bDepartment of Applied Mathematics, National Chiao-Tung University, Hsin-Chu 300, Taiwan, ROC
Received 5 January 1999; accepted 15 September 2000
Keywords: Hopf lemma; Perturbation lemma; Variation method of Nehari-type; Minimal solution
1. Introduction
In this paper, we study the existence, uniqueness, and asymptotic behavior of positive solutions of the semilinear elliptic equations with asymptotic linear nonlinearity as follows:
6u + (u − b)2+ = 0 in ;
u = 0 on @; (1)
where is a bounded C2; domain in Rn, and ; and b are positive real numbers.
The solutions structure of (1) is quite di:erent when is equal to zero or not. When is equal to zero or a nonzero real number but small enough, there are some results of solutions structure of (1) in [1,9]. Here we extend these results to the case when is an arbitrary positive real number.
For convenience, we denote f=(u − b)2+ and the main results are as follows:
Theorem 1. Suppose ¿ 0; then there exists exactly one solution u(·; ) of (1) for
all ∈ (0; 1). Here 1 is the )rst eigenvalue of − in with Dirichlet boundary
condition.
Work partially supported by the National Science Council of the Republic of China.
∗Corresponding author. Tel.: +886-3-422-7151, ext: 5123; fax: +886-3-425-7379.
E-mail address: [email protected] (C.-H. Hsu).
0362-546X/02/$ - see front matter c 2002 Elsevier Science Ltd. All rights reserved.
The solutions obtained in Theorem 1 are the minimal solutions of (1) and can be proved by monotone iteration method easily. It is quite natural to ask the existence of solutions of (1) when is larger than 1. However, since the nonlinearity is asymptotic
linear, it is diGcult to apply Mountain–Pass Lemma to show the existence of solutions of (1). Therefore, due to the linear growth of f, we need to carefully estimate the energy levels in order to use the variation method of Nehari-type to prove the existence of solutions, see [8]. The result is as follows:
Theorem 2. For any ¿ 0; there exists ∗() ¿ 1 such that if ∈ (1; ∗()); then
(1) has at least two solutions.
The result in Theorem 2 can be obtained by using the blow up method to study the asymptotic behavior of the solutions. However, the global structure of the solutions of (1) is still open.
The paper is organized as follows. In Section 2, we show the existence of non-minimal solution of (1) when is small enough. In Section 3, we study the uniqueness and asymptotic behavior of solutions of (1). Applying the results of Sections 2 and 3, we prove the main theorems in Section 4.
2. Existence of non-minimal solutions
In this section, we will Hrstly use the perturbation method to show the existence of minimal solution of (1). The key lemma is as follows:
Lemma 1 (Perturbation Lemma). Let u be a minimal solution of (1). Consider
6u + f(u) + h(u) = 0 in ;
u = 0 on @; (2)
where h ∈ C1(R1; R1) and f
(·) + h(·) ¿ 0 in [0; ∞). Then there exists ≡ (u) ¿ 0
such that for ||h||∞¡ ; (2) has a minimal solution.
Proof. Let w and v be the solutions of 6w = − 1 in ; w = 0 on @ (3) and 6v + f (u)v = − v in ; v = 0 on @; (4)
respectively, where ¿ 0 is the Hrst eigenvalue and ||v||∞= 1.
DeHne Ju = u + tv + sw, where t and s are positive real numbers. By using
such that 6u + f(u) + h(u) = − f(u) − tf(u)v − tv − s + f(u+ tv + sw) + h(u) = f (u)(tv + sw) − tf(u)v − tv − s + h(u) = tv −1 2 + (f(u) − f(u)) + −1 2 tv + swf(u) − s + h(u) = tv −1 2 + f(u)(u− u) + −1 2 tv + swf(u) − s + h(u): (5) Since f
is bounded and ||u−u||∞6 ||tv+sw||∞, the Hrst term in (5) is negative
when s and t are small enough. By the Boundary Hopf Lemma, see [4], the second term in (5) is also negative if s is smaller than t and small enough. Therefore, take = s ¿ 0 and let ||h( Ju)||∞¡ , the right-hand side in (5) is negative. Hence Ju is a
supersolution of (2). It is obvious that 0 is a subsolution of (2). By monotone iteration method, we obtain a minimal solution of (2). The proof is complete.
In Lemma 1, we had assumed the existence of minimal solution of (1). It is easy to check that (1) has a minimal solution when = 0, i.e. f0(u) = |u − b|, since 0 and
b are subsolution and supersolution, respectively. In addition, by Lemma 1, we have the following theorem.
Theorem 3. There exists 0¿ 0; such that for any ∈ (0; 0) there exists ∗() ¿ 1
such that if ∈ (0; ∗()); (1) has a minimal solution.
Proof. We may rewrite the Eq. (1) as 6u + f0+ h(u) = 0;
where h(u) = (f− f0). Since (f− f0) 6 √, and by Lemma 1, there exists 0¿ 0
such that when 0 ¡ ¡ 0, the equations
6u + (f0(u) +√) = 0 in ;
u = 0 on @; (6)
have a minimal solution which is also a supersolution of (1). Since 0 is a subsolution of (1), the result follows.
From Theorem 3, we show the existence of minimal solution of (1) when is smaller than ∗. It is natural to study the existence of non-minimal solution of (1)
when ¡ ∗. The result is as follows:
Theorem 4. If is small enough and if 1¡ ¡ ∗(); then (1) has a non-minimal
Proof. We shall prove the theorem by variation method of Nehari-type. By Theorem 3, let u(·; ) be a minimal solution of (1). If (1) has a non-minimal solution u, say
u = w + u and w ¿ 0, then w satisHes
6w + (f(w + u) − f(u)) = 0 in ;
w = 0 on @: (7)
For convenience, let
g(x; w) = f(w + u) − f(u) and G(x; w) = w 0 g(x; v) dv; (8) then w satisHes 6w + g(x; w) = 0 in ; w = 0 on @: DeHne J(w) = 1 2|∇w|2− G(x; w); I(w) = |∇w| 2− wg(x; w); and M= {w ∈ H01: I(w) = 0}:
Since f is convex and w ¿ 0, the function g(x; w) in (8) is convex in w such that
g(x; w) = g(x; w) − g(x; 0) 6 g(x; w)w: (9)
By integrating (9) with respect to w, we have 2G(x; w) 6 g(x; w)w. Therefore, on M,
J(w) =2
wg(x; w) − 2G(x; w)
¿ 0;
i.e. J(w) is bounded below. Now, if we can show that M= ∅ for ¿ 1, then by the
Nehari method, see [6], we can obtain a non-minimal solution of (1).
To prove that M= ∅, let "1 be the Hrst eigenfunction of − in with Dirichlet
boundary condition and "2
1= 1. If 1¡ , then I(t"1) = 1t2− t"1g(x; t"1) = 1t2− t2 " 2 1f t"1+ 2u− 2b (t"1+ u) + f(u) ¡ 0;
when t tend to inHnity. On the other hand, let w1 be the eigenfunction with w12= 1
of the Hrst eigenvalue 1 of the following equation:
6w + f
(u)w = − 1w in ;
By the result of [2], we know 1¿ 0. Therefore, I(sw1) = s2 |∇w1|2− s w1g(x; sw1) = s2 |∇w 1|2− s [f(u )sw21+ O(s2)] = s2( 1+ O(s)); when s is near 0.
Hence, I(sw1) ¿ 0 when s is small enough. Therefore, M is non-empty and we
obtain a positive solution of (1) by using the Nehari method. The proof is complete.
3. Uniqueness and asymptotic behavior of solutions
In this section, we prove the uniqueness of solutions of (1) when is smaller than 1. In addition, we also use the blow up method to study the asymptotic behavior of
solutions as varies and which gives us the multiplicity of solutions. Theorem 5. For any ¿ 0; if 0 ¡ 6 1; then the solution of (1) is unique.
Proof. Suppose that u and v are solutions of (1). Let w = u − v, then w satisHes 6w + (f(u) − f(v)) = 0 in ;
w = 0 on @: (11)
By mean-value theorem, w satisHes 6w + f
( Jw)w = 0;
where Jw lies between u and v and − |∇w| 2+ f ( Jw)w2= 0: Since |f ( Jw)| ¡ 1, we have f( Jw)w2¿ 1w2. Hence, if 0 6 6 1, then w = 0
and the result follows. The proof is complete.
Corollary 1. The solution of (1) is unique in the set Bb= {u ∈ C0( J): ||u||∞6 b}.
Proof. In the proof of Theorem 5, we Hnd f
( Jw) is negative under the assumption
|w| 6 b. Hence, the uniqueness of solution follows in Bb.
From Theorem 5, we know that (1) can have non-minimal solutions only if ¿ 1.
To study the existence of non-minimal solutions, we need the following lemma: Lemma 2. Fix ¿ 0; if {u} is a family of solutions of (1) such that ||u||∞ tends
Proof. By the uniqueness result of Theorem 5, we know that is bounded away from zero, and it is easy to check that ∗() ¡ ∞. Hence {} is bounded and has
a convergent subsequence. We still use the same and assume that → J, where J ∈ (0; ∗()) and u satisHes 6u+ f(u) = 0 in ; u= 0 on @: (12) Let M= ||u||∞ and v=Mu ; then 6v+ f(vMM) v v= 0 in ; v= 0 on @: (13)
Since f(vM)=Mare bounded, we have f(vM)=(Mv) → 1 a.e. in and uniformly
in any compact subset D of . By standard elliptic estimates, we obtain v → v in
C2; as → J. Hence,
6v + Jv = 0 in ;
v = 0 on @: (14)
Since ||v||∞= 1 and v ¿ 0, we have J = 1, the Hrst eigenvalue of − and the result
follows.
If we denote the solutions set of (1) by S, then by the global bifurcation theorem of
Crandall and Rabinowitz in [2], Scontains a branch C which is unbounded. Therefore,
by Lemma 2, the following result is immediate.
Theorem 6. The solutions set S of (1) contains an unbounded component C. If
u∈ C and ||u||∞→ ∞; then → 1.
Remark 1. If = B1, the unit ball in R2. The uniqueness problem of non-minimal
solution is still open. 4. Proof of main theorems
In this section, we will use the results of previous sections to prove the main theorems.
Proof of Theorem 1.
If = 0, then (1) has a minimal solution for all ¿ 0. Now rewrite Eq. (1) as 6u + f0+ (f− f0) = 0:
Fig. 1.
Fig. 2.
Since (f− f0) 6 √, and by Lemma 1, (1) has a minimal solution for any Hxed
¿ 0 if is small enough. Let S be the solution set of (1), then by the global
bifurcation theorem in [2,7] and Theorem 5, we have the bifurcation diagrams of solutions of (1) as Fig. 1 or 2. In either case, there exists exactly one solution for ∈ (0; 1). The proof is complete.
Proof of Theorem 2.
First, by Theorems 4 and 6, we know that Theorem 2 is true if is small enough. Hence, the bifurcation diagram of solutions of (1) is roughly as in Fig. 3.
Let
& = {J∈ R+: Theorem 2 is false; when = J}:
If the result in Theorem 2 is false, then the bifurcation diagram of solutions must be Fig. 1. We will prove that the set & is an empty set. Suppose that & is a non-empty
Fig. 3.
set and denote by 0= inf &, then it is clear that 0= 0. Now, we can consider the
following two cases:
Case 1: 0∈ &. In this case, there exists 0¿ 1 such that Eq. (1) has a minimal
solution for ∈ [1; 0] when = 0. Hence, by Lemma 1, (1) must have a
mini-mal solution for some ¿ 1, when is suGciently close to 0, but this contradicts
0= inf &.
Case 2: 0∈ &. In this case, let ' be a positive Hxed real number. We will claim
that the choice of as in the proof of Lemma 1 is independent of for ∈ [0− '; 0).
The proof is similar as in proving Lemma 1. Let Ju be as the Ju in Lemma 1 with v = v; = and M ; K be positive real numbers such that
|f
(x)| 6 M and |f(x)| 6 K; for ∈ [0− '; 0) and x ¿ 0:
By (5), we have 6u + f(u) + h(u) 6 tv −1 2 + M||tv+ sw||∞ + −1 2 tv+ Ksw − s + h(u):
Now, we claim that there exist Jv 6 v for all ∈ [0− '; 0) with Jv ¿ 0 in and
@ Jv=@n ¡ 0 on @, where n is the unit outer normal of . Suppose that there exists no such Jv. Let {vn} be a convergence subsequence of {v} which converge to Jv ¿ 0
and Jv = 0 at some x0∈ or @ Jv=@n = 0 on some x0∈ @. If we consider the following
equations:
6vn+ f(un)vn= − nvn in ;
vn= 0 on @ (15)
and using the same method as in the proof of Lemma 2, we have 6 Jv + Jf Jv = − ˜ Jv in ;
for some ˜ ¿ 0. But (16) is a linear eigenvalue problem which gives us a contradiction. Hence, if s ¡ t and small enough, we have
6u + f(u) + h(u) ¡ − s + ||h(u)||∞6 0;
by choosing = s. Since the choice of is independent of when is suGciently close to 0 which contradicts 0= inf &, we have & = ∅ and complete the proof.
Acknowledgements
The authors would like to express their thanks to Prof. S.-S. Lin for some helpful suggestions and the referee for helpful comments.
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