Singular limit of radial solutions in an annulus ∗

Francesca Gladiali † Massimo Grossi‡

Abstract

In this paper we study the radial solutions of the problem−∆u = λeu in Ωu = 0 on ∂Ω

(0.1)

where Ω is an annulus of IRN , N ≥ 2 and λ is close to zero.Among the other results we show the existence of a singular limit and

some qualitative properties of the solution.

1 Introduction

In this paper we study the problem−∆u = λeu in Ωu = 0 on ∂Ω, (1.1)

where Ω = x ∈ IRN : a < |x| < b is an annulus in IRN , N ≥ 2 and λ > 0 is areal parameter.

By the strong maximum principle any solution of (1.1) is strictly positiveinside Ω.

We restrict ourselves to radial solutions of (1.1). To this end we introducethe space H1

0,r(Ω) = v ∈ H10 (Ω) s.t. v(x) = v(|x|).

A classical result says that there exists λ∗ = λ∗(Ω) > 0 such that for λ ∈(0, λ∗), problem (1.1) has always a minimal solution that we call uλ and nosolution for λ > λ∗. The minimal solution uλ can be found applying the ImplicitFunction Theorem and moreover this is the unique stable solution of (1.1),namely

λ1(−∆− λeuλ) > 0. (1.2)

It is easy to see that if Ω = B(0, R) or Ω is an annulus the minimal solution uλis radial.

∗Supported by M.I.U.R., project “Variational methods and nonlinear differential equa-tions”.†Struttura dipartimentale di Matematica e Fisica, Universita di Sassari,via Vienna 2 -07100

Sassari, e-mail fgladiali@uniss.it.‡Dipartimento di Matematica, Universita di Roma “La Sapienza”, P.le A. Moro 2 - 00185

Roma, e-mail grossi@mat.uniroma1.it.

1

When Ω = B(0, 1) Gel’fand in [5] and Joseph and Lundgren in [6] haveclassified all the regular solutions of (1.1)(let us recall that by the result of Gidas,Ni and Niremberg ([4]) all positive solutions of (1.1) are radially symmetric).Note that the range of values λ for which a solution exists strongly depends onthe dimension N of the space.

If Ω is an annulus, the situation is different. The result of [4] is not stillapplicable and the nonradial bifurcation can occur (we cite for example ([2], [7],[10], [11], [12] among others).

Again in the case of the annulus another known result concerns the existenceof a second radial solution of (1.1) for λ ∈ (0, λ∗). This can be done, for example,by considering the functional

Jλ(u) :=12

∫ b

a

rN−1(u′)2dr − λ∫ b

a

rN−1euλ(eu − 1− u)dr

defined on H10,r(Ω) where uλ is the minimal solution. Using the well known

Ambrosetti-Rabinowitz Mountain Pass Theorem one gets the existence of acritical point of Jλ which gives rise to another radial solution ∀λ ∈ (0, λ∗).

In [11] Nagasaki and Suzuki showed the existence of other solutions andproved the total blow-up of the solution uλ using the Sturm comparison The-orem. Moreover they proved that the blow-up of a radial solution of (1.1) canonly occur at λ = 0.

Note that if λ is sufficiently close to zero we have additional information.Indeed, in this case, problem (1.1) has exactly two radial solutions (see S.S. Linin [7] and Nagasaki and Suzuki in [11]). In [7] S.S. Lin exhibits explicitly thesesolutions for N = 2 by using their invariance through a transformation group(this was previously observed by Gel’fand in [3]).

Starting from these results we study the asymptotic behavior of the radialblowing up solution uλ and we show the existence of a singular limit as λ tendsto 0. This will be done by considering the limit problem associated to (1.1). Tothis end the estimates proved in Section 4 are crucial.

In what follows we denote by uλ = uλ(r) any radial solution of (1.1) differentfrom uλ (the minimal solution).

Since the minimal solution is uniformly bounded we have

limλ→0

λ

∫ b

a

euλ(r)rN−1dr = 0

while, as we see in Proposition 2.1, for any other radial solution uλ it holds

limλ→0

λ

∫ b

a

euλ(r)rN−1dr = +∞.

Note that this follows easily for N = 2 (see [10]) using that no radial solutionscan have a finite number of blowing up points.Set

γλ = λ

∫ b

a

euλ(r)rN−1dr,

2

andδ2λ =

1λe‖uλ‖∞

. (1.3)

Since γλ → +∞ it is easy to see that δλ → 0 as λ → 0. Now we can state ourmain Theorem

Theorem 1.1. Let uλ be a solution of (1.1) such that λ∫ baeuλ(r)rN−1dr →∞.

Then, as λ→ 0δλuλ(r)→ 2

√2G(r, r0) in C0(Ω), (1.4)

where r0 is given by

r0 =

(b2−N + a2−N

2

) 12−N

if N ≥ 3√ab if N = 2.

(1.5)

Here G(r, s) denotes the Green function of the operator −u′′ − N−1r u′ for

r ∈ (a, b) with Dirichlet boundary conditions (see Section 5 for the explicit formof the Green function).

The previous theorem states the existence of a singular limit for the functionδλuλ. Note that the value of r0 also appears in another problem involvingpolynomial nonlinearities ([5]).

A consequence of the Theorem 1.1 is an estimate on the L∞ norm of uλ.

Proposition 1.2. We have

δλ‖uλ‖∞ =12

√2(1 + o(1))

rN−10N − 2(a2−N − b2−N ) if N ≥ 3

r0 log ba if N = 2,(1.6)

where o(1) converge to zero as λ goes to zero.

Our last result concerns the existence of a “limit problem” associated to(1.1).

Theorem 1.3. Let rλ ∈ (a, b) such that ‖uλ‖∞ = uλ(rλ) and uλ(r) = uλ(δλr+rλ)− ‖uλ‖∞. Then

uλ(r)→ U(r) in C1loc(IR). (1.7)

Here U(r) = log 4e√

2r

(1 + e√

2r)2is the unique solution of

−U ′′ = eU in IRU(0) = 0 U ′(0) = 0. (1.8)

The paper is organized as follows: in Section 2 we prove Proposition 2.1 andsome technical lemmas while in Section 3 we prove Theorem 1.3. Section 4 isdevoted to a crucial estimate on the function uλ. Finally Section 5 contains theproof of Theorem 1.1.

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2 Some preliminary estimates

Let uλ be any radial solution of (1.1) different from uλ. Note that since ∆uλ < 0and uλ > 0 inside (a, b), uλ(r) has exactly one maximum point rλ ∈ (a, b). Setuλ(rλ) = ‖uλ‖∞.Let us rewrite equation (1.1) in radial coordinates:

−(rN−1u′λ(r))′ = rN−1λeuλ(r) for r ∈ (a, b). (2.1)

Integrating from r to rλ we have

u′λ(r) =λ

rN−1

∫ rλ

r

euλ(s)sN−1ds. (2.2)

Finally let us define

γλ = λ

∫ b

a

euλ(s)sN−1ds. (2.3)

Proposition 2.1. Let uλ be any solution of (2.1) different from uλ. Thenγλ →∞ as λ→ 0.

Proof Let us suppose by contradiction that γλ is bounded for λ close to zero.From (2.2) we have

|u′λ(r)| ≤ Cγλ (2.4)

If γλ is bounded we get that |u′λ(r)| ≤ C for each r ∈ [a, b]. By the Ascoli-ArzelaTheorem uλ(r)→ u0 uniformly in [a, b], where u0 is a bounded function. Henceγλ → 0 as λ → 0. This implies that u′λ(r) → 0 uniformly in [a, b] and thenu0 = 0. This cannot happen since the minimal solution is isolated (see [9] formore details). ut

Corollary 2.2. We have that δλ → 0 as λ→ 0.

Proof It follows by the previous proposition and γλ ≤ λCe‖uλ‖∞ . ut

Lemma 2.3. Let uλ be a solution to (2.1) and γλ as in (2.3). We have

γλ = aN−1u′λ(a)− bN−1u′λ(b) (2.5)(u′λ(a))2

2− (N − 1)

∫ rλ

a

(u′λ(s))2

sds = λe‖uλ‖∞ − λ, (2.6)

(u′λ(a))2

2− (u′λ(b))2

2− (N − 1)

∫ b

a

(u′λ(s))2

sds = 0. (2.7)

Proof Integrating (2.1) we get

−∫ b

a

(rN−1u′λ(r))′dr = λ

∫ b

a

euλ(r)rN−1dr = γλ

4

and then (2.5) follows. To prove (2.6) and (2.7) we rewrite (2.1) as:

−u′′λ(r)− N − 1r

u′λ(r) = λeuλ(r) for r ∈ (a, b).

Multiplying by u′λ and integrating on (r1, r2), we get

(u′λ(r1))2

2− (u′λ(r2))2

2− (N − 1)

∫ r2

r1

(u′λ(s))2

sds = λ

∫ r2

r1

euλ(s)u′λ(s)ds.

Letting r1 = a and r2 = rλ we get (2.6), while letting r1 = a and r2 = b we get(2.7). ut

Remark 2.4. From (2.7) we deduce the following

(u′λ(a))2

2> (N − 1)

∫ b

a

(u′λ(s))2

sds; (2.8)

(u′λ(a))2 > (u′λ(b))2. (2.9)

Lemma 2.5. We have

lim infλ→0

u′λ(a)γλ

≥ c > 0. (2.10)

Proof Using (2.9) and that u′λ(b) < 0 while u′λ(a) > 0 for each λ, by (2.5) weget

1 = lim infλ→0

(aN−1u

′λ(a)γλ

− bN−1u′λ(b)γλ

)≤ lim inf

λ→0

(aN−1u

′λ(a)γλ

+ bN−1u′λ(a)γλ

)= (aN−1 + bN−1) lim inf

λ→0

u′λ(a)γλ

.

ut

Proposition 2.6. Let uλ be a solution of (2.1) and let γλ and δλ be as definedin (2.3) and (1.3). Then there exist constants c > 0 and C > 0 such that

c ≤ δλγλ ≤ C. (2.11)

Proof Multiplying (2.6) by 1γ2λ

we have

(u′λ(a))2

2γ2λ

− (N − 1)∫ rλ

a

(u′λ(s))2

sγ2λ

ds =λe‖uλ‖∞

γ2λ

− λ

γ2λ

. (2.12)

¿From (2.4) and the definition of δλ (see (1.3)), we get

λe‖uλ‖∞

γ2λ

≤ C ⇒ δλγλ ≥ c. (2.13)

5

The reverse inequality (2.11) is harder. Let us suppose by contradiction that

lim infλ→0

λe‖uλ‖∞

γ2λ

= 0. Let uλkγλk= vk. From (2.4) we have |v′k| ≤ C. Hence up to

a subsequence vk → v0 uniformly in [a, b].Inserting (2.8) in (2.6) we find

λe‖uλ‖∞ > λ+ (N − 1)∫ b

rλ

(u′λ(s))2

sds.

Dividing by γ2λ and passing to a subsequence we get

limk→∞

1γ2λk

∫ b

rλk

(u′λk(s))2

sds = 0. (2.14)

¿From (2.14) we have that v′k → 0 almost everywhere in (r0, b). Since rλk isalso the maximum point of vk we have

vk(rλk) = vk(rλk)− vk(b) =∫ b

rλk

v′k(s)ds→ 0 (2.15)

and then vk → 0 uniformly in [a, b].Dividing (2.1) by γλk and using vk as a test function we get∫ b

a

(v′k(r))2 rN−1dr =λ

γλk

∫ b

a

euλk (r)vk(r)rN−1dr ≤ supr∈[a,b]

|vk(r)| → 0.

This shows that v′k(r) → 0 almost everywhere in [a, b] and since |v′k| ≤ C wecan pass to the limit in (2.12) getting

limk→∞

(u′λk(a))2

2γ2λk

≤ limk→∞

λe‖uλk‖∞

γ2λk

= 0

and this gives a contradiction with (2.10). This proves that lim infλ→0

λe‖uλ‖∞γ2λ

> 0,

and the claim follows. ut

Lemma 2.7. Let uλ be a solution of (2.1), then

lim supλ→0

u′λ(b)γλ

< 0. (2.16)

Proof Let us argue by contradiction. Since u′λ(b)γλ

< 0 for any λ, let us supposethat

lim supλ→0

u′λ(b)γλ

= 0.

Integrating (2.1) in (rλ, b), we have

−bN−1u′λ(b) = λ

∫ b

rλ

sN−1euλ(s)ds.

6

Dividing by γλ, and passing to a subsequence we get

limk→∞

λk∫ brλk

sN−1euλk (s)ds

γλk= 0. (2.17)

Setting vk = uλkγλk

as in the proof of Proposition 2.6 we get that up to a subse-quence vk → v0 uniformly in [a, b] and from (2.17) that

v′k(r) =1

rN−1

λk∫ rλkr

sN−1euλk (s)ds

γλk→ 0 pointwise in (r0, b].

Arguing as in the last part of Proposition 2.6 we get a contradiction. ut

3 The rescaling argument

Let us define the function uλ : Iλ =[a− rλδλ

, b− rλδλ

]→ IR,

uλ(r) = uλ(δλr + rλ)− ‖uλ‖∞ (3.1)

where δλ and rλ are as stated before. Set aλ = a− rλδλ

and bλ = b− rλδλ

. Let usshow that rλ is far away from a and b.

The monotonicity result in [4] states that ∂uλ∂r

< 0 for a+ b2 < r ≤ b.

In order to show the reverse inequality near b let us perform the substitutiont = r2−N for N ≥ 3 and t = − log r for N = 2. Then uλ(t) = uλ(r) satisfies

∂2uλ∂t2

= −λρ(t)euλ in (t0, t1)uλ > 0 in (t0, t1)uλ(t0) = uλ(t1) = 0

(3.2)

where

ρ(t) =

(N − 2)2t

2(N−1)N−2 if N ≥ 3

e−2r if N = 2(3.3)

and (t0, t1) = (b2−N , a2−N ) for N ≥ 3 or (t0, t1) = (− log b,− log a) for N = 2.Applying Theorem 1’ in [4] we get ∂uλ

∂r(t) < 0 for t0 + t1

2 < t ≤ t1. This implies

that ∂uλ∂r

(r) > 0 in a ≤ r < r0, where r0 is as defined in Theorem 1.1.Hence

r0 ≤ rλ ≤a+ b

2.

Then we deduce limλ→0

bλ = +∞ and limλ→0

aλ = −∞.

Proof of Theorem 1.3 We have that uλ(r) satisfies the equation

−u′′λ(r)− (N − 1)δλδλr + rλ

u′λ(r) = euλ(r) in Iλ. (3.4)

7

By the definition of uλ(r), (2.4) and (2.11), we get

|u′λ(r)| = δλ|u′λ(δλr + rλ)| ≤ Cδλγλ ≤ C. (3.5)

Then from Ascoli-Arzela’s Theorem we have that uλk(r) → U uniformly oncompact sets of IR = limk→∞ Iλk . Then also u′′λk(r) is bounded in Iλk and sou′λk(r)→ U ′(r) uniformly on compact sets of IR. Here U is a solution of

−U ′′(r) = eU(r) in (−∞,+∞) ; U(0) = 0 , U ′(0) = 0. (3.6)

Hence uλ converge to U , and U is a solution to (3.6) in IR and then (see [5])

U(r) = log4e√

2r

(1 + e√

2r)2. (3.7)

ut

4 A useful estimate

Proposition 4.1. Let uλ be a solution of (2.1). We have, for λ small enough,that

uλ(t) ≤ w(t) = C1U(t) + C2 (4.1)

in Iλ for some positive constant C1 and C2, where U(t) is defined in (3.7) andIλ is as in Section 3.

Proof Let R > 0. Since uλ → U in C1loc(IR), we deduce that uλ(t) ≤ U(t) + 1

in [−R,R], if λ is small enough.

Rescaling (2.2) we have

u′λ(t) = −(δλt+ rλ)1−N∫ t

0

(δλz + rλ)N−1euλ(z)dz.

Since uλ → U in C1([−R,R]) we deduce that∫ R

0

(δλz + rλ)N−1euλ(z)dz ≥ 12aN−1

∫ R

0

eU(z)dz = aN−1CR > 0

if λ is small enough. Then, using that δλt+ rλ ≤ b, we get

u′λ(t) ≤ −C ′Rfor t > R and λ sufficiently small. This leads to

uλ(t) = uλ(R) +∫ t

R

u′λ(s)ds ≤ −D1t+D2

for some positive constant D1 and D2, for t > R and λ small enough.

This proves the claim for t > 0. The case t < 0 is analogous. ut

8

5 The main results

We start this section by recalling the explicit form of the Green function G(r, s)of the operator −u′′ − N−1

r u′ for r ∈ (a, b) with zero Dirichlet boundary condi-tions.

G(r, s) =sN−1

(N − 2)(a2−N − b2−N )

(b2−N − s2−N )(r2−N − a2−N ) for a ≤ r ≤ s

(s2−N − a2−N )(b2−N − r2−N ) for s < r ≤ b

for N ≥ 3, while for N = 2

G(r, s) =s

log b− log a

(log r − log a)(log b− log s) for a ≤ r ≤ s

(log s− log a)(log b− log r) for s < r ≤ b.

We just consider N ≥ 3 (the case N = 2 can be handled in the same way).

Proof of Theorem 1.1 Up to a subsequence we have r0 = limk→∞

rλk . Using the

Green representation formula, we can write

uλk(r) = λk

∫ b

a

G(r, s)euλk (s)ds =1δλk

∫Iλk

G(r, δλks+ rλk)euλk (s)ds.

Using estimate (4.1) and ew(r) ∈ L1(IR) we can pass to the limit in (5.1) getting

uλk(r) =1δλk

(G(r, r0)

∫ +∞

−∞eU(s)ds+ o(1)

).

¿From (2.4) and (2.11) we have |δλu′λ(r)| ≤ C. Hence δλkuλk(r)→ 2√

2G(r, r0)uniformly. Since we already know that a < r0 < b, we have to show thatlimλ→0 rλ = r0 and that it is given by (1.5).

Using the explicit expression of the Green function we derive

u′λk(r) =1

a2−N − b2−Nr1−N

(∫ r

a

sN−1(s2−N − a2−N )λkeuλk (s)ds

−∫ b

r

sN−1(b2−N − s2−N )λkeuλk (s)ds

).

Since u′λk(rλk) = 0 we get∫ rλk

a

sN−1(s2−N − a2−N )λkeuλk (s)ds =∫ b

rλk

sN−1(b2−N − s2−N )λkeuλk (s)ds

9

and rescaling∫ 0

aλk

(δλks+ rλk)N−1((δλks+ rλk)2−N − a2−N) euλk (s)ds =

(5.1)∫ bλk

0

(δλks+ rλk)N−1(b2−N − (δλks+ rλk)2−N

)euλk (s)ds

where aλk = a− rλkδλk

and bλk = b− rλkδλk

as in the proof of Theorem 1.3. Since

aλ → −∞ and bλ → +∞ as λ→ 0, passing to the limit in (5.1) we get

rN−10 (r2−N0 − a2−N )

∫ 0

−∞eU(s)ds = rN−1

0 (b2−N − r2−N0 )∫ +∞

0

eU(s)ds.

Since this can be done for each subsequence we can infer that limλ→0

rλ = r0 and

that (1.5) holds. utWe observe that using the boundness of ∂G

∂r(r, s) in (a, b) we have that

δλu′λ(r)→ 2

√2∂G∂r

(r, r0) pointwise in [a, r0) ∪ (r0, b].¿From the previous theorem we derive the proof of Proposition 1.2.

Proof of Proposition 1.2 ¿From (1.4) we have that

δλuλ(rλ) = 2√

2G(r0, r0)(1 + o(1)).

Using the definition of G and (1.5) we get (1.6). ut

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