A Neumann problem in exterior domain

manuscripta math. 106, 63 – 74 (2001)

© Springer-Verlag 2001

Daomin Cao · Marcello Lucia · Huan-Song Zhou

A Neumann problem in exterior domain Received: 26 May 1999 / Revised version: 2 April 2001 Abstract. We are concerned with the existence of radial solutions for the following Neumann problem  −u(x) = g(u(x)), x ∈ ,    u(x) > 0, in , (P) u(x) → 0, as |x| → +∞,    ∂u σ > 0, ∂n = −σ, ∂ denotes the normal interior derivative on ∂ and where  is an exterior domain in IR N , ∂n g satisfies certain assumptions.

1. Introduction We study the existence of radial solutions for the following Neumann problem (P) −u(x) = g(u(x)), x ∈ ; u(x) > 0, u(x) → 0, ∂u = −σ, ∂n where • = •

∂ ∂n



[1, ∞) IR N \ B, B

(1.1)

for all x ∈ ; as |x| → +∞;

(1.2) (1.3)

σ > 0;

(1.4)

if N = 1; is the unit ball in IR N if N ≥ 2;

denotes the normal exterior derivative on ∂B;

and g satisfies: (H1) g : R → R, and g(s) = 0 for s ≤ 0; D. Cao: Institute of Applied Mathematics, AMMS, Chinese Academy of Sciences, Beijing 100080, P.R. China M. Lucia: Università degli Studi di Roma“Tor Vergata", Dipartimento di Matematica, Via della Ricerca Scientifica, 00133 Roma, Italy H.-S. Zhou:Young Scientist Laboratory of Mathematical Physics, Wuhan Institute of Physics and Mathematics, Chinese Academy of Sciences, P.O.Box 71010, Wuhan 430071, P.R. China

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(H2) g is locally Lipschitz continuous; (H3) For some δ > 0, g(s) < 0 on (0, δ). Equation (1.1) arises in many contexts. As explained in [8], problem (P) with N = 2 is related to the study of the electrostatic potential outside a charged cylinder, in such case, u denotes the electrostatic potential and g(u) = −shu. Moreover, in the same paper, an existence result to problem (P) with N = 2 is also given for general g which is negative in (0, +∞) and strictly decreasing. In this paper, we discuss problem (P) without requiring the monotonicity of g and by assuming only g to be negative in some interval (0, δ). It is easy to see that the radial solutions of problem (P) can be obtained from the following initial value problem: 

 −u (r) − N−1 r u (r) = g(u(r)), r ∈ [1, +∞); u(1) = ξ > 0, u (1) = −σ < 0;

(1.5)

where u is now a function of r = |x| alone and ξ > 0 has to be determined in order to have: u is globally defined on [1, ∞),

u(r) > 0,

lim u(r) = 0.

r→+∞

(1.6)

We set,  α :=

+∞, if g(s) < 0 for all s > 0, inf{s > 0, g(s) ≥ 0}, otherwise.

(1.7)

Let us remark that hypothesis (H3) always implies α > 0 and that g(s) < 0 for all s ∈ (0, α). Our main result is the following: Theorem 1.1. Let g be a function satisfying conditions (H1) to (H3), α be defined  t √ by (1.7) and set G(t) = g(s)ds. Then, for every 0 < σ < −2G(α) there 0

exists ξ ∈ (0, α) such that the unique solution u satisfying (1.5) has the properties: u satisfies (1.6) and u (r) < 0 on [1, +∞). In particular, u is a radial solution to problem (P). Let us mention that some crucial differences occur between considering the problem (P) on exterior domain and the similar problem of solving (1.1) to (1.3) on the whole IR N . In fact, for the equation (1.1) with  = IR N , the existence of radial positive solutions vanishing at +∞ has been studied widely, see, e.g., [5], [4]. In these two papers, the case in which the nonlinearity g is negative on some interval (0, δ) is considered. However, it follows immediately from the maximum principle that g must change sign if  = IR N , this implies that α can not be +∞.

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But, α = +∞ is allowed for the problem that we consider in this paper. Moreover, as explained in [5], g is necessary to satisfy the following stronger conditions:  ζ g(s)ds > 0, (1.8) There exists ζ > 0 such that G(ζ ) : = 0

g(s) lim > 0, if α < +∞, s→α + s − α

(1.9)

which may not be required in our paper. If  = IR N , seeking for a radial solution of the equation (1.1) with properties (1.2)-(1.3) is equivalent to looking for a solution of the ordinary differential equation given in (1.5) on the whole interval (0, +∞) with the boundary conditions: u(0) = ξ, u (0) = 0. This problem is essentially solved in [4] by using the combination of a shooting method and a constrained variational method under conditions (1.8) and (1.9), in which a rescaling argument is used. However, the rescaling argument seems not applicable in the domain like [1, +∞). The main ideas of our paper come in fact from the paper of Berestycki, Lions and Peletier [4], but we use only the shooting method and choose the initial value in an interval near the origin instead of an interval with positive distance to the origin as in [4]. In this way we need only considerably weaker conditions on the nonlinearity g. Remarks. 1) Let g be a function satisfying (H1) to (H3) and negative on (0, ∞). Then, an application of the maximum principle and of Hopf Lemma (see Thm. 3.1 and Lemma 3.4 in [6]) show that problem (P) has no solutions if σ ≤ 0. For example, when N = 1 and g(s) = −s, a simple calculation shows that the solution to (1.1), (1.3), (1.4) is given by u(r) = σ e−r and that (1.2) is satisfied iff σ > 0. Thus, in the class of nonlinearities g we are considering, the assumption that σ > 0 is necessary to have existence of solution to Problem (P). 2) If σ = 0, finding solutions to (1.5) on a finite interval (1, R) has been intensively studied in recent years, such as Bandle, Coffman and Marcus [2], Bandle and Kwong [3], Lin [9], Rother [10], Simader and Sohr [11], etc.. The authors of papers [2], [3], [9] used a shooting method combined with the Sturm comparison theorem, and a variational method is used in [10]. In this paper, we will use a method similar to [3] to prove our main result.

2. The shooting method Setting X(r) = (u(r), u (r)), we see that problem (1.5) is equivalent to the following first-order initial value problem:  X  (r) = F (r, X(r)), r ∈ [1, ∞); (2.1) X(1) = (ξ, −σ );

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where, F : (0, ∞) × R → R 2

  N −1 x2 − g(x1 ) . (r, x1 , x2 )  → x2 , − r

2

Since g is locally Lipschitz continuous, the mapping F is locally Lipschitz continuous with respect to (x1 , x2 ) ∈ R2 . Thus, from [1, Prop. 7.6, p. 100] we deduce that problem (2.1) has a unique C 1 solution Xξ defined on a maximal interval [1, Rξ ) for all ξ ∈ R. In particular, the initial value problem (1.5) has a unique C 2 solution uξ defined on a maximal interval [1, Rξ ): uξ : [1, Rξ ) −→ R . For α given in (1.7), we define  Iα =

(0, α], if α < +∞; (0, +∞), if α = +∞.

Motivated by [4], we split Iα into the following three parts: I0 := {ξ ∈ Iα : uξ (r) > 0, uξ (r) < 0, ∀ r ∈ [1, Rξ )};

I+ := {ξ ∈ Iα : ∃r0 ∈ (1, Rξ ) such that uξ (r0 ) = 0, uξ (r) > 0 ∀ r ∈ [1, r0 ]}; I− := {ξ ∈ Iα : ∃r0 ∈ (1, Rξ ) such that uξ (r0 ) = 0, uξ (r) < 0 ∀ r ∈ [1, r0 ]}.

At this point of the paper, those sets could be empty. Actually, from the results of the next sections, we will see that this cannot happen. Let us note that all α appearing in the following refer to (1.7) and that we will always consider initial data ξ ∈ Iα , i.e. 0 < ξ ≤ α (if α < +∞) and 0 < ξ < +∞ (otherwise). We first prove the following simple fact : Proposition 2.1. Assume (H1)-(H2). Then, Iα = I0 I+ I− . Proof. Let ξ ∈ Iα . Since ξ > 0 and uξ (1) = −σ < 0, the solution uξ is positive and decreasing in a neighborhood of r = 1. We set, E1 := {r ∈ [1, Rξ ) : uξ (r) = 0}

E2 := {r ∈ [1, Rξ ) : uξ (r) = 0}.

There are four mutually exclusive cases. Case 1. inf E1 = inf E2 = +∞ (i.e. E1 = E2 = ∅). This means that uξ is always positive and decreasing on the interval [1, Rξ ). So ξ ∈ I0 . Case 2. inf E1 < inf E2 (in particular E1  = ∅). Set r0 := inf E1 . We have uξ (r0 ) = 0 (since uξ is of class C 2 ), and moreover r0  = 1 (since uξ (1) < 0). Thus, from the fact that inf E1 < inf E2 and ξ > 0, we get uξ (r) > 0 for all r ∈ [1, r0 ). So, we have ξ ∈ I+ . Case 3. inf E1 > inf E2 . Set r0 := inf E2 . Arguing as in case 2, we get ξ ∈ I− . Case 4. inf E1 = inf E2 < ∞. Set r0 := inf E1 . We have uξ (r0 ) = uξ (r0 ) = 0, with uξ solution of the initial value problem (1.5) which has a unique solution (using (H2)). From (H1), we deduce uξ ≡ 0, and this is impossible (ξ > 0).  

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We end this section by pointing out that if I0 is non empty, then this set consists of initial data ξ giving rise to a solution satisfying (1.5) and (1.6) and thus to a solution of problem (P). This will mainly be a consequence of the following result. Proposition 2.2. Let u be a solution of (1.5) which is globally defined on [1, ∞) and satisfies lim u(r) = L ≥ 0. Then, r→+∞

(i)



lim u (r) = lim u (r) = 0;

r→+∞

(ii) g(L) = 0.

r→+∞

Proof. Multiplying the differential equation in (1.5) by u and integrating from (1, R), we can have 1 1 G(ξ ) + σ 2 = G(u(R)) + [u (R)]2 + (N − 1) 2 2



R 1

[u (s)]2 ds. s

(2.2)

By this formula, it is easy to see that both of the following limits: 1  [u (R)]2 and R→∞ 2 lim



R

lim

R→+∞ 1

[u (s)]2 ds s

exist. Then, we may assume that for some L1 , lim u (R) = L1 . If L1 > 0, then there exists R > 1 such that u (r) >  u(r) − u(s) =

r s

L1 2

R→+∞

for all r ∈ (R, ∞). This would imply,

u (ξ )dξ ≥

L1 (r − s) 2

∀ r, s ∈ (R, ∞),

in contradiction with the fact that lim u(r) = L.A similar contradiction is obtained r→∞ if L1 < 0. Thus, L1 = 0. Hence, using again the equation in (1.5) we have u (R) = −g(L). R→+∞ R

lim u (R) = − lim g(u(R)) − (N − 1) lim

R→+∞

R→+∞

Arguing as we did for L1 , we derive that g(L) = 0 and thus  

lim u (R) = 0.

R→+∞

Remark. If the additional condition (1.9) holds, using the same argument as in [4] (Lemma I.2), it is possible to prove that L  = α (but we will not use this fact in our paper). Proposition 2.3. Let g be a function satisfying (H1) to (H3) and assume I0  = ∅. Then, for all ξ ∈ I0 , we have : (i) uξ is globally defined on [1, ∞), (ii) uξ is a solution of (1.5) and (1.6).

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Proof. (i) As explained at the beginning of this section, the initial value problem (1.5) is equivalent to the first order equation (2.1). Let ξ ∈ I0 and consider the set  

 γξ := Xξ (r) : r ∈ [1, Rξ ) = uξ (r), uξ (r) : r ∈ [1, Rξ ) .

(2.3)

From [1] (Corollary (7.7), p. 103), if the set γξ is bounded in R2 , we necessarily have Rξ = +∞. Since ξ ∈ I0 , the solution uξ is positive and strictly decreasing. Thus, 0 < uξ (r) ≤ ξ 0 < uξ (r) < α

∀ r ∈ [1, Rξ ). ∀ r ∈ (1, Rξ ).

(2.4) (2.5)

Moreover, uξ satisfied uξ (r)



 N −1  =− uξ (r) + g(uξ (r)) , r

r ∈ [1, Rξ ).

(2.6)

Using uξ < 0, (2.5) and the definition of α in (1.7), we see that the right-handside of (2.6) is positive. So uξ is a decreasing convex function. Thus, −σ < uξ (r) ≤ 0

∀ r ∈ [1, Rξ ).

(2.7)

Relations (2.4) and (2.7) imply that the set defined by (2.3) is bounded, which concludes the proof. (ii) Let ξ ∈ I0 . Since uξ is positive, decreasing and globally defined on [1, ∞) (by part (i)), we have lim uξ (r) = L,

r→+∞

with

0 ≤ L < ξ.

On the one hand, 0 ≤ L < α (since ξ ∈ Iα ). On the other hand, we have g(s) = 0 for all s ∈ (0, α) (by (H3) and the definition of α in (1.7)). Proposition 2.2 implies then that L = 0.   Thus, using previous proposition, Theorem 1.1 will be a consequence of the fact that I0  = ∅. To prove this, the strategy will be the following. 1) In Sect. 3, we prove that I+ and I− are non empty (see Prop. 3.1 and Prop. 3.5); 2) In Sect. 4, we prove that the sets I+ and I− are open in Iα (see Prop. 4.1); 3) The conclusion that I0  = ∅ will follow from Prop. 2.1 and the fact that Iα is connected.

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3. Existence of elements in I+ and I− Proposition 3.1. Assume (H1) to (H3) and that σ 2 < −2G(α). Then if α  = +∞, then α ∈ I+ ; if α = +∞, then there exists α0 ∈ (0, +∞) such that [α0 , +∞) ⊂ I+ . In particular, I+  = ∅. Proof. We consider first the case α < +∞. From Prop. 2.1, we know that (0, α] = I0 I+ I− . Assume that α ∈ I− . Taking R = r0 in (2.2) (where r0 is given by the definition of I− ), we have  r0  1 [uα (s)]2 1 G(α) + σ 2 = [uα (r0 )]2 + (N − 1) ds, 2 2 s 1 which is impossible since σ 2 < −2G(α). Hence α  ∈ I− . Assume that α ∈ I0 . By Propositions 2.2 and 2.3, we deduce that limr→∞ uα (r) = 0. Thus, letting now r → +∞ in (2.2), we have then  +∞  [uα (s)]2 1 1 ds + lim [u (r)]2 . G(α) + σ 2 = (N − 1) 2 s 2 r→+∞ α 1 The left-hand side being negative, we have a contradiction. Thus α  ∈ I0 . This means α ∈ I+ . On the other hand, if α = +∞, choose α0 ∈ (0, +∞) such that σ 2 < −2G(α0 ). Then just replacing α by α0 in the first case and following exactly the same way as above, we know that [α0 , +∞) ⊂ I+ .   We end this section by proving that the set I− is nonempty. We need first three lemmas. Lemma 3.2. Let a < b and the functions q1 , q2 : [a, b] −→ R be continuous with q1 (r) ≥ q2 (r) for each r ∈ [a, b]. Assume that u1 , u2 > 0 are two solutions of the following ordinary differential equations  u1 (r) + q1 (r)u1 (r) = 0, for r ∈ [a, b] u2 (r) + q2 (r)u2 (r) = 0, satisfying

u1 (a) = u2 (a), u1 (a) = u2 (a)

then u1 (r) ≤ u2 (r) for all r ∈ [a, b]. Proof. This result is the classical Sturm comparison theorem, see e.g., [7], pp. 229. We omit its proof here.  

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Lemma 3.3. Let ξ, σ˜ , K > 0 and w be a function defined by   2  w (r) − K w(r) = 0, r ∈ [a, +∞] w(a) = ξ    w (a) = −σ˜

(3.1)

Then, we have σ˜ . (i) w has a zero r0 in (a, +∞) if and only if ξ < K σ˜  (ii) w (r) < 0 for all r ∈ (a, r0 ] as soon as ξ < K .

Proof. (i) It is not difficult to see that the general solution for problem (3.1) is as follows w(r) = AeK(r−a) + Be−K(r−a) ,

(3.2)

where A, B are constants which satisfy  ξ =A+B −σ˜ = K(A − B) that is, 2A = ξ − Then, it follows from w(r) = 0 that

σ˜ σ˜ , 2B = ξ + . K K

r =a+

1 −B ln( ). 2K A

Therefore, B 1 B > 0 and ln(− ) > 0 A 2K A B A+B ⇐⇒ − > 1 ⇐⇒ < 0, A < 0 A A Kξ σ˜ ⇐⇒ < 0 ⇐⇒ ξ < . Kξ − σ˜ K

∃r0 ∈ (a, +∞) s.t. w(r0 ) = 0 ⇐⇒ −

σ˜ (ii). Let ξ < K , we now turn to proving w  (r) < 0 on (a, r0 ]. By noticing (3.2), we see that

w (r) = K(AeK(r−a) − Be−K(r−a) ). To show that w (r) < 0, it is enough to prove that Ae2K(r−a) − B < 0. Since ξ <

σ˜ K,

(3.3)

A < 0 , we know that (3.3) is equivalent to e2K(r−a) >

B . A

But, B A < −1, the above inequality is obviously true and so (3.3) holds, which completes the proof.  

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Lemma 3.4. Let g be a function satisfying (H1) to (H3), and uξ be the unique solution to (1.5). Assume that there exists R ∈ (1, Rξ ) such that 0 < uξ (r) < α,

∀ r ∈ (1, R).

(3.4)

Then, (i) If there exists r˜ ∈ (1, R) with uξ (˜r ) = 0, then uξ (˜r ) > 0 and uξ (˜r ) > 0. (ii) If uξ (R) = 0, then uξ (r) < 0 for all r ∈ (1, R). Proof. (i) By the uniqueness of the solution to (1.5), we have uξ (˜r )  = 0. Thus, using (1.5), (3.4) and definition 1.7 of α we deduce, uξ (˜r ) = −g(uξ (˜r )) > 0. (ii) Assume the existence of r1 ∈ (1, R) such that uξ (r1 ) = 0. From part (i), we deduce that uξ (˜r ) is a positive local minimum. On the other hand, by assumption uξ (R) = 0, we deduce that there is some r2 ∈ (r1 , R) such that uξ has a local maximum at r2 , that is, 0 < uξ (r2 ) < α,

uξ (r2 ) = 0

and

uξ (r2 ) ≤ 0,

which contradicts the conclusion of part (i).   Proposition 3.5. Let g be a function satisfying (H1) to (H3). Then, there exists ξ0 ∈ (0, α) such that (0, ξ0 ) ⊂ I− . Proof. By setting V (r) = r (N−1)/2 u(r), we see easily that u is a solution to the initial value problem (1.5) iff V satisfies,    V  (r) + (1 − N )(N − 3) + g(u) V (r) = 0, 4r 2 u (3.5)   ξ − σ. V (1) = ξ, V (1) = N−1 2 Let N (u) := sup{r ∈ [1, Rξ ) : u(r) > 0}. Then the ODE in (3.5) may be regarded for r ∈ [1, N (u)). Since g is locally Lipschitz continuous, for any α0 < α, there exists some constant k := k(α0 ) > 0 such that (1 − N )(N − 3) g(s) + > −k 2 , ∀ s ∈ (0, α0 ), ∀ r ∈ [1, +∞). 4r 2 s We compare problem (3.5) with the following problem:  W  (r) − k 2 W (r) = 0, r ∈ [1, +∞), W (1) = ξ, W  (1) = N−1 2 ξ − σ. Setting −σ˜ =

N−1 2 ξ

(3.6)

(3.7)

− σ , we choose ξ0 such that ξ0 < α0 and ξ0 <

2 σ. 2k + N − 1

(3.8)

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By Lemma 3.3, for all ξ ≤ ξ0 , there exists r0 ∈ (1, +∞) such that the unique solution W of (3.7) satisfies W (r0 ) = 0 and

0 < W (r) < ξ < α ∀ r ∈ (1, r0 ).

Now since V (1) = ξ < α0 , we have V (r) < α0 near r = 1. Thus, since we work on [1, +∞), we still have u(r) = r (1−N)/2 V (r) < α0 for r near 1. Thus, referring to (3.6), we deduce that (1 − N )(N − 3) g(u) + > −k 2 for r near 1. 4r 2 u It follows now from Lemma 3.2 that for r > 1 and near r = 1 V (r) ≤ W (r) < α. By continuation, we deduce that the above inequality still holds at least before V has a zero. We deduce then the existence of T0 ≤ r0 such that V (T0 ) = 0. It remains to show that u (r) < 0 for r ∈ (1, T0 ). Since, u(T0 ) = 0 and

u(r) ≤ V (r) < α, ∀ r ∈ (1, T0 ),

so, our result is concluded by applying Lemma 3.4.   4. Topological property of I+ and I− The next proposition is borrowed from paper [4]. Proposition 4.1. Under the assumptions (H1) to (H3), the sets I+ and I− are open subsets of Iα . Proof. We prove this proposition by using the fact that uξ (r) and uξ (r) depend continuously on ξ (see [1], Thm. 8.3, p.110). First, we prove the conclusion for I+ . Let us note that if ξ ∈ I+ , then, r0 = inf{ r > 1 : uξ (r) = 0, uξ (r) > 0} > 1, and

  0 < uξ (r) < α, uξ (r0 ) = 0,    uξ (r) < 0,

for all

r ∈ (1, r0 ],

for all

r ∈ [1, r0 ).

From Lemma 3.4 (part (i)), we deduce that uξ (r0 ) > 0. This implies that there exists some r1 > r0 such that uξ (r) > uξ (r0 )

for all

r ∈ (r0 , r1 ].

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Now by a continuity argument we find that for η near ξ we have uη (r1 ) > uη (r0 ), uη (r) > 0

for all

(4.1) r ∈ [1, r1 ].

(4.2)

From (4.1) and the fact that uη (1) < 0, we deduce the existence of r2 ∈ (1, r1 ) such that uη (r2 ) = 0. From (4.2) we also have uη (r) > 0 for r ∈ [1, r2 ]. This implies that η ∈ I+ and that I+ is open. Now we prove that I− is open. Let us first remark that if ξ ∈ I− then r0 = inf{r > 1 : uξ (r) = 0,

uξ (r) < 0} > 1,

  uξ (r) > 0, for all r ∈ [1, r0 ), uξ (r0 ) = 0,    uξ (r) < 0, for all r ∈ [1, r0 ].

and

Since uξ (r0 ) < 0, by the continuity of uξ (r) with respect to r, there exists a number r1 > r0 such that uξ (r) < 0

and uξ (r) < 0

for all

r ∈ (r0 , r1 ].

Then by a continuity argument one can find that for η near ξ ,  uη (r1 ) < 0, uη (r) < 0 for all r ∈ (1, r1 ]. Thus there exists r2 ∈ (1, r1 ) such that uη (r2 ) = 0. This implies that η ∈ I− and that I− is open.   With the above conclusions, we can give now the proof of Theorem 1.1. 5. Existence result Proof of Theorem 1.1. By Proposition 3.1 and Proposition 3.5 we know that I+ and I− are non empty. From propositions 2.1 and 4.1 they are open in Iα and disjoint. Since, Iα is connected, we have  I+ I−  = Iα . From Proposition 2.1, we deduce the existence of ξ ∈ I0 , and moreover from Proposition 3.1 we have ξ  = α. Thus, ξ ∈ (0, α) and Proposition 2.3 implies then that uξ is a strictly decreasing solution to (1.5) and (1.6).   Acknowledgements. This work was supported by NSFC and innovation funds of CAS. The second author would like to thank Institute of Applied Mathematics of CAS in Beijing and Wuhan Institute of Physics and Mathematics of CAS for supporting his visit in their institutes where this work was finished. The authors thank also the referee for his constructive suggestions.

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