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Existence of many positive nonradial solutions for a superlinear Dirichlet problem on thin annuli ∗
Alfonso Castro & Marcel B. Finan
Abstract
We study the existence of many positive nonradial solutions of a su- perlinear Dirichlet problem in an annulus inRN. Our strategy consists of finding the minimizer of the energy functional restricted to the Nehrai manifold of a subspace of functions with symmetries. The minimizer is a global critical point and therefore is a desired solution. Then we show that the minimal energy solutions in different symmetric classes have mu- tually different energies. The same approach has been used to prove the existence of many sign-changing nonradial solutions (see [5]).
1 Introduction
In this article we study the existence of many positive nonradial solutions of the equation
∆u+f(u) = 0 in Ω
u >0 in Ω (1)
u= 0 on∂Ω, where
Ω={x∈R3: 1− <|x|<1}.
The non-linearityf is of classC1(R) and satisfies the following conditions:
(C1)f(0) = 0 andf0(0)< λ1, whereλ1is the smallest eigenvalue of−∆ with zero Dirichlet boundary condition in Ω.
(C2)f0(u)>f(u)u for allu6= 0.
(C3)(Superlinearity) lim|u|→∞f(u)u =∞.
∗Mathematics Subject Classifications: 35J20, 35J25, 35J60
Key words: Superlinear Dirichlet problem, positive nonradial solutions, variational methods
c2000 Southwest Texas State University.
Published October 24, 2000.
21
(C4) (Subcritical growth) There exist constants C > 0 and p ∈ (1,5) such that
|f0(u)| ≤C(|u|p−1+ 1), ∀u∈R.
(C5) There exist constantsm∈(0,1) andρsuch that for|u|> ρ, uf(u)≥ 2
mF(u)>0, where F(u) =Ru
0 f(s)ds.
This paper is motivated by the work of Coffman [6], Li [7] and Lin [8]. In 1984, Coffman showed that for f(t) =−t+tp, wherep= 2N+ 1, N = 2, the above problem has many positive nonradial solutions. His result was extended by Li [7] to the N-dimensional case withN ≥4 orN = 2 and forp∈(1,NN−2+2), when the nonlinear term is λt+tp for λ ≤ 0. Our main result (see Theorem 1.1 below) concerns the case N = 3. We will show that for N = 3 our problem has many distinct nonradial solutions. We follow the strategy used in [6, 7] and [8]. That is, we look for the minimizer of the energy functional restricted to the Nehari manifold of a subspace of functions with symmetries. The minimizer is a global critical point and therefore is a solution to (1). Then we show that the minimal energy solutions in different symmetric classes has mutually different energies.
We would like to point out here that during the preparation of this article we were unaware of the papers by Byeon [1] and Catrina and Wang [2], where the above problem has been solved. However, our approach is different from theirs.
Our main result is the following
Theorem 1.1 Let conditions (C1) - (C5) be satisfied. Then, for each positive integerkthere exists1(k)>0such that if0< < 1(k)then (1) has kdistinct positive nonradial solutions.
We note that our argument works for N ≥ 2 and not only for the three dimensional case.
The approach used in proving Theorem 1.1 is similar to the one used in [7], i.e., we consider the functionals
J(u) = Z
Ω
1
2|∇u|2−F(u) dx and
γ(u) = Z
Ω
|∇u|2−uf(u) dx
onH01(Ω), and fork≥1, we consider the space of invariant functions H(,k) := Fix(Gk× {idR,−idR})
=
v∈H01(Ω) :v(g(x1, x2), T x3) =v(x1, x2, x3),
∀ (g, T)∈Gk× {idR,−idR}
=
v∈H01(Ω) :v(x1, x2, x3) =u(x1, x2,|x3|) for someu satisfyingu(g(x1, x2),|x3|) =u(x1, x2,|x3|)∀g∈Gk
where Gk = {gi : 0 ≤ i ≤ k−1} and gz = e2πi/kz, z ∈ C ' R2. Also, we consider the Nehari manifold
S(,k) ={v∈H(,k)\{0}:γ(v) = 0}.
Similarly, we define the space of functions H(,∞) := O(2)× {idR,−idR}
=
v∈H01(Ω) :v(g(x1, x2), T x3) =v(x1, x2, x3),
∀(g, T)∈O(2)× {idR,−idR}
=
v∈H01(Ω) :v(x1, x2, x3) =u(
q
x21+x22,|x3|), for some u , and the manifold
S(,∞) ={v∈H(,∞)\{0}:γ(v) = 0},
where O(2) denotes the space of 2×2 orthogonal matrices. Note that if u∈ H(,∞) thenuisθ−independent.
Associated with the above sets, we consider the numbers jk= inf
v∈S(,k)J(v) and j∞ = inf
v∈S(,∞)J(v).
We prove Theorem 1.1 by establishing the following lemmas:
Lemma 1.2 Fork= 1,2, ...,∞,jk is achieved by someu,k∈S(,k)andu,k is a critical point of J onH(,k).
Lemma 1.3 Letu,k be as in Lemma 1.2. Thenu,k is a critical point ofJ on H01(Ω).
Lemma 1.4 Given a positive integerk, there exists1(k)such that for0< <
1(k)we have
jk < j∞ .
Lemma 1.5 Forn= 2,3,4, ...,k= 1,2,3, ...,jkn < j∞ implies jk < jkn . This article is organized as follows: In Section 2, we prove Lemmas 1.2–1.5.
In Section 3, we prove Theorem 1.1.
2 Proof of Lemmas 1.2–1.5
Proof of Lemma 1.2: The proof follows from [3] combined with the facts that the embeddingsH(, k)⊂Lp(Ω) andH(,∞)⊂L∞(Ω) are compact (See [9])
♦
Lemma 1.3 is a result of the symmetric criticality principle: ifu,kis a critical point ofJ onH(,k), thenu,k is a critical point ofJ onH01(Ω) (See [9]).
To prove Lemma 1.4 we need the following results.
Lemma 2.1 (Poincar´e’s inequality) Let Ω⊂RN be a smooth bounded do- main with diameter d.Then for u∈H01(Ω) we have
Z
Ωu2dx≤ d2 N
Z
Ω|∇u|2dx .
Lemma 2.2 ([3]) 0 is a local minimum ofJ. Ifu∈H(, k)− {0}, then there exists a uniqueα=α(u)∈(0,∞)such that αu∈S(, k). Moreover, J(αu) = maxλ>0J(λu)>0.
Lemma 2.3 ([4]) For|v|> ρand s >2 we have svf(sv)≥Cs2/mvf(v) for some constant C >0.
Proof of Lemma 1.4: Let k ≥1 be an integer and > 0 to be chosen below.
According to Lemma 1.2, there existu,k∈S(, k) andu,∞∈S(,∞) such that jk =J(u,k) andj∞ =J(u,∞). By Lemma 1.3,u,kand u,∞ are solutions to Problem 1. Let Ωk be the set of points x= (x1, x2, x3)≡(r, θ, x3)∈ Ω such thatθ∈[0,2π/k].
Letj be a positive integer to be chosen independent ofkand. Define ω(r, θ, x3) =
u,∞(r,|x3|) sin (jkθ) forθ∈[0, π/(jk)]
0 forθ∈[π/(jk),2π/k],
where r= (x21+x22)1/2. Extend ω periodically to all of Ω in the θ direction.
Letz∈H(, k) be the resulting extension. Since u,∞>0 then z >0. By the chain rule we have
|∇ω(r, θ, x3)|2 = (u,∞)2rsin2(jkθ) + 1
r2(u,∞)2(jk)2cos2(jkθ) +|∇x3u,∞|2sin2(jkθ)
ifθ∈[0, π/(jk)], otherwise∇ω= 0.Thus,
|∇ω(r, θ, x3)|2≤(u,∞)2r+ 1
r2(jk)2(u,∞)2+|∇x3u,∞|2.
By Lemma 2.1 we have Z
Ω2jk
u2,∞dx1x2dx3≤ (diam(Ω2jk ))2 3
Z
Ω2jk
|∇u,∞|2dx1x2dx3. Thus,
Z
Ω
|∇z|2dx1x2dx3
= k
Z
Ω2jk
|∇ω|2dx1x2dx3
≤ k Z
Ω2jk
[(1 + 16(jk)2
27 2)(u,∞)2r+|∇x3u,∞|2]dx1x2dx3
≤ 2k Z
Ω2jk
((u,∞)2r+|∇x3u,∞|2)dx1x2dx3 (2)
= 2k Z
Ω2jk
|∇u,∞|2dx1x2dx3 provided that < 3√
3/(4jk). By Lemma 2.2 we can find α > 0 such that γ(αz) = 0. Let D={(r, θ, x3) :u,∞(r,|x3|)> ρ, θ∈[4jkπ ,2jkπ ]}. Suppose that α > √
2. Then for (r, θ, x3)∈ D we haveαsin (jkθ)>2. This, the fact that tf(t) is bounded from below, say byE, and Lemma 2.3 imply
Z
Ω
αzf(αz)dx1x2dx3
= k
Z
Ω2jk
αsin (jkθ)u,∞f(αsin (jkθ)u,∞)dx1x2dx3
≥ k
E|Ω2jk |+Cα2/m Z
D
u,∞f(u,∞)dx1x2dx3
(3)
= k E|Ω2jk |+Cα2/m( Z
u,∞>ρu,∞f(u,∞)r dr dx3)(
Z π/(2jk)
π/(4jk) dθ)
!
= kE|Ω2jk |+kC 4α2/m(
Z
u,∞>ρu,∞f(u,∞)r dr dx3)(
Z π/(jk)
0 dθ),
where |Ω|denotes the volume of Ω.
Now, let <1/4. ThenK = infu∈S(1/4,k)J(u)≤infu∈S(,k)J(u). Letu,∞
also denote the zero extension of u,∞ to all of Ω1/4. Then u,∞ ∈ S(1/4, k).
Thus,J(u,∞)≥K and consequently Z
Ω
|∇u,∞|2dx1x2dx3 ≥ 2K+ 2M|Ω| (4) where M = inf{F(t) :t∈R}.
If we chooseso that|Ω|< K/(−2M) then (4) implies Z
Ω
|∇u,∞|2dx1dx2dx3≥K (5)
and this leads to Z
u,∞>ρu,∞f(u,∞)r dr dx3
= Z
u,∞≥0u,∞f(u,∞)r dr dx3− Z
u,∞≤ρu,∞f(u,∞)r dr dx3
=
1− R
u,∞≤ρu,∞f(u,∞)r dr dx3 R
u,∞≥0u,∞f(u,∞)r dr dx3 Z
u,∞≥0
u,∞f(u,∞)r dr dx3
=
1−2πR
u,∞≤ρu,∞f(u,∞)r dr dx3 R
Ω|∇u,∞|2dx1x2dx3 Z
u,∞≥0u,∞f(u,∞)r dr dx3
≥ (1−C) Z
u,∞≥0u,∞f(u,∞)r dr dx3
where the constant C is independent of (, j, k). By choosing in such a way that 1−C >1/2 we conclude
Z
u,∞>ρu,∞f(u,∞)r dr dx3> 1 2 Z
Ω
u,∞f(u,∞)r dr dx3. This reduces (3) to
Z
Ω
αzf(αz)dx1dx2dx3≥kE|Ω2jk |+kCα2/m Z
Ω2jk
|∇u,∞|2dx1dx2dx3. (6) On the other hand, using (2) we have
Z
Ω
αzf(αz)dx1x2dx3 = α2 Z
Ω
|∇z|2dx1x2dx3
≤ 2kα2 Z
Ω2jk
|∇u,∞|2dx1x2dx3. (7) Combining (6) and (7) to obtain
(Cα2/m−2α2) Z
Ω2jk
|∇u,∞|2dx1x2dx3≤ C jk Hence,
Cα2/m−2α2 ≤ R C/(jk)
Ω2jk |∇u,∞|2dx1x2dx3
≤ R C
Ω|∇u,∞|2dx1x2dx3 ≤M1,
where we have used (5). But the functiong(α) =Cα2/m−2α2satisfiesg(0) = 0 and limα→∞g(α) = +∞. Thus, ifg(α)≤M1 then there is a constantK0 such thatα≤K0. By letting
α≤max√
2, K0 ≡K00 (8)
we conclude that α is bounded. Letw=αz ∈S(, k). Since F(t) is bounded from below, say by M, then (8) and (2) imply
J(w) = Z
Ω
|∇w|2
2 −F(w)
dx1x2dx3
= Z
Ω
α2|∇z|2
2 dx1x2dx3− Z
Ω
F(w)dx1x2dx3
≤ 2(K00)2k Z
Ω2jk
|∇u,∞|2
2 dx1x2dx3−kM|Ω2jk |
≤ 2(K00)2 j
Z
Ω
|u,∞|2
2 dx1x2dx3+kC jk
≤ (K00)2 j
Z
Ω
|∇u,∞|2
2 dx1x2dx3+1 j
Z
Ω
|∇u,∞|2
2 dx1x2dx3
≤ C j
Z
Ω
|∇u,∞|2
2 dx1x2dx3 (9)
where we have used (4) with chosen in such a way thatC < K and|Ω|<
K/(−2M). We claim that there exists a constantCsuch that Z
Ω
|∇u,∞|2
2 dx1x2dx3≤CJ(u,∞).
Indeed, since γ(u,∞) = 0 and by (C5) we have Z
Ω
u,∞f(u,∞)dx1x2dx3
= Z
Ω
|∇u,∞|2dx1x2dx3
= Z
Ω
(|∇u,∞|2−2F(u,∞))dx1x2dx3+ 2 Z
Ω
F(u,∞)dx1x2dx3
≤ 2(J(u,∞) +m 2
Z
Ω
u,∞f(u,∞)dx1x2dx3+C|Ω|).
It follows that (1−m)
Z
Ω
u,∞f(u,∞)dx1x2dx3≤2(J(u,∞) +C|Ω|). (10) On the other hand, using (4) and (C5) we have
J(u,∞) = Z
Ω
(1
2|∇u,∞|2−F(u,∞))dx1x2dx3
≥ (1−m 2 )
Z
Ω
u,∞f(u,∞)dx1x2dx3−C0|Ω|
≥ (1−m
2 )K−C0|Ω|.
By choosingin such a way that C0|Ω|< 14(1−m)K, we see thatJ(u,∞)>
(1−m4 )K. Now, we choose such that C|Ω| <(1−m4 )K. Using this in (10) we
obtain Z
Ω
u,∞f(u,∞)dx1x2dx3≤CJ(u,∞).
Also, using this in (9) we obtain J(w)≤C
jJ(u,∞).
Choosingj such thatj >2Cwe obtain J(w)< J(u,∞) and this concludes the
present proof. ♦
To prove Lemma 1.5 we need
Lemma 2.4 Let v∈H01(Ω). Then the function Pv(λ) =λvf(λv)
2 −F(λv) is increasing on (0,∞).
Proof. DifferentiatingP−v with respect toλwe find Pv0(λ) =λv2
2
f0(λv)−f(λv) λv
>0,
where we have used (C2). This completes the proof ♦ Proof of Lemma 1.5: Fixkand n. Let 0< < 1(kn). Lemma 1.2 guarantees the existence of a minimizer u,kn of J on S(, kn). From Lemma 1.3 we see that u,kn is a solution to (1). Furthermore, from Lemma 1.4 we know that u,knis nonradial. Now, by the regularity theory of elliptic equations we know that u,knis a C2 function. Let x= (r, θ) be the polar coordinates of x∈R2 and writeu=u,kn(r, θ,|x3|). Then
Z
Ω
|∇u|2dx1x2dx3= Z
(r,|x3|)
Z 2π
0 (u2r+ 1
r2u2θ+|∇x3u|2)r dr dθ dx3
and Z
Ω
F(u)dx1x2dx3= Z
(r,|x3|)
Z 2π
0 F(u)r dr dθ dx3. Define the function
v(r, θ,|x3|) =u(r,θ
n,|x3|), 0≤θ≤2π.
Then
v(r, θ+2π
k ,|x3|) = u(r,θ n+ 2π
kn,|x3|)
= u(r,θ n,|x3|)
= v(r, θ,|x3|).
It follows that v ∈ H(, k). On the other hand, it is easy to check that the following equalities hold
vr(r, θ,|x3|) =ur(r, θ n,|x3|), vθ(r, θ,|x3|) = 1
nuθ(r,θ n,|x3|),
∇x3v(r, θ,|x3|) =∇x3u(r, θ n,|x3|).
Therefore, Z
Ω
|∇v|2dx1x2dx3 = k Z
(r,|x3|)
Z 2π/k
0 (u2r(r, θ
n,|x3|) + 1
r2n2u2θ(r, θ n,|x3|) +|∇x3u(r, θ
n,|x3|)|2)r dr dθ dx3
= kn Z 2π
nk
0 (u2r(r, θ,|x3|) + 1
r2n4u2θ(r, θ,|x3|) +|∇x3u(r, θ,|x3|)|2)r dr dθ dx3
= Z
(r,|x3|)
Z 2π
0 (u2r(r, θ,|x3|) + 1
r2n4u2θ(r, θ,|x3|) +|∇x3u(r, θ,|x3|)|2)r dr dθ dx3.
Also Z
Ω
F(v)dx1x2dx3 = k Z
(r,|x3|)
Z 2π/k
0 F(u(r,θ
n,|x3|))r dr dθ dx3
= Z
(r,|x3|)
Z 2π
0 F(u(r, θ,|x3|))r dr dθ dx3. Sinceudoes not belong toS(,∞) we have
Z
(r,|x3|)
Z 2π
0 u2θ(r, θ,|x3|)r dr dθdx3>0. It follows that
γ(v) = Z
Ω
(|∇v|2−vf(v))dx1x2dx3
= Z
(r,|x3|)
Z 2π
0 (u2r(r, θ,|x3|) + 1
r2n4u2θ(r, θ,|x3|) +|∇x3u(r, θ,|x3|)|2−uf(u))r dr dθ dx3
<
Z
(r,|x3|)
Z 2π
0 (u2r(r, θ,|x3|) + 1
r2u2θ(r, θ,|x3|) +|∇x3u(r, θ,|x3|)|2−uf(u))r dr dθ dx3
= Z
Ω
(|∇u|2−uf(u))dx1x2dx3= 0.
This yields Z
Ω
|∇v|2dx1x2dx3<
Z
Ω
vf(v)dx1x2dx3. (11) Now, by Lemma 2.2 we can find 0 < α < 1 such that αv ∈ S(, k). Let w=αv∈S(, k). Using Lemma 2.4 and the definition ofjk we have
jk ≤ J(w) =J(αv) =Pv(α)
< Pv(1)
= Z
Ω
(1
2vf(v)−F(v))dx1x2dx3
= Z
Ω
(1
2uf(u)−F(u))dx1x2dx3
= J(u) =jkn .
Putting together all the arguments above we conclude a proof of the lemma ♦
3 Proof of Main Theorem
For any integerk≥1, according to Lemma 1.4 there exists 1(2k) such that if 0< < 1(2k) then
j2k< j∞ . It follows from Lemma 1.5 that
j2< j22< ... < j2k< j∞ . (12) According to Lemma 1.2, there existsu,i∈S(, i), i= 1, ..., k, such that
j2i =J(u,i).
Moreover, by Lemma 1.3 u,i is a solution of Problem 1, i = 1, . . . , k. By (12), {u,i}ki=1, are nonrotationally equivalent and nonradial. This completes
the proof of the theorem. ♦
References
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[5] A. Castro and M.B. Finan, Existence of many sign-changing nonradial solutions for a superlinear Dirichlet problem on thin annuli, Topological Methods in Nonlinear Analysis Vol. 13, No.2 (1999), 273–279
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Alfonso Castro
Division of Mathematics and Statistics University of Texas at San Antonio San Antonio, TX 78249 USA.
e-mail: [email protected] Marcel B. Finan
Department of Mathematics The University of Texas at Austin Austin, TX 78712 USA.
e-mail: [email protected]
