Positive solutions of quasilinear elliptic equations in two-dimensional exterior domains

Nonlinear Analysis 42 (2000) 243–250www.elsevier.nl/locate/na

Positive solutions of quasilinear elliptic equationsin two-dimensional exterior domains

Adrian Constantina; ∗, Gabriele Villari b

aMathematisches Institut, Universit�at Z�urich, Winterthurerstrasse 190, Z�urich CH-8057, SwitzerlandbDipartimento di Matematica “U. Dini”, Universit�a degli Studi di Firenze, Viale Morgagni 67/A,

I-50134 Firenze, Italy

Received 1 April 1998; accepted 28 May 1998

Keywords: Quasilinear elliptic equation; Li�enard equation

1. Let us consider the quasilinear second-order elliptic equation

�u+ f(x; u) + g(u)x · ∇u|x|2 = 0; x∈GA; (1)

in an exterior domain GA= {x∈R2: |x|¿A}; here A¿0. Problems related to Eq. (1)were considered by several authors – we refer to [5–9, 11–13].We will prove under quite general assumptions on the functions f and g the exis-

tence of a positive solution to Eq. (1): using some results from the qualitative theoryof autonomous ordinary di�erential equations we will construct a positive radial super-solution v1 and a positive radial subsolution v2 to Eq. (1) such that v2(x)≤ v1(x) forx∈GA, and we establish the existence of a positive solution of Eq. (1) that is squeezedbetween v1 and v2.Among the equations of form (1) we have the semilinear Schr�odinger equation

�u+ f(x; u)= 0; x∈GA: (2)

For the particular case (2) our results cover cases where other investigations[1, 3, 5, 11, 12] on Eq. (2) are powerless.

∗ Corresponding author.E-mail address: [email protected] (A. Constantin)

0362-546X/00/$ - see front matter ? 2000 Elsevier Science Ltd. All rights reserved.PII: S0362 -546X(98)00343 -5

244 A. Constantin, G. Villari / Nonlinear Analysis 42 (2000) 243–250

2. Let F ∈C�(M × J ×N;R) (H�older continuous) for every bounded domainM ⊂GA, every bounded interval J ⊂R and every bounded domain N ⊂R2; here �∈(0; 1). Assume also that for every bounded domain M ⊂GA there exists a nonnegativecontinuous function lM such that

|F(x; t; y)| ≤ lM (|t|) (1 + |y|2); x∈M; t ∈R; y∈R2:A solution of the elliptic equation

�u+ F(x; u;∇u)= 0; x∈GA⊂R2 (3)

is de�ned to be a function u∈C2+�(M) for every bounded domain M ⊂GA, functionthat satis�es Eq. (3) at every x∈GA. Supersolutions of Eq. (3), that is, functions usatisfying �u + F(x; u;∇u)≤ 0, and subsolutions, i.e., functions u satisfying �u +F(x; u;∇u)≥ 0, are de�ned similarly.Let us denote SA= {x∈R2: |x|=A}.The following will be useful in the sequel:

Lemma [8]. Under the above assumptions on F, if there exists a positive superso-lution v1 and a nonnegative subsolution v2 to Eq. (3) in GA such that v2(x)≤ v1(x)for all x∈GA ∪ SA, then Eq. (3) has a solution u in GA such that v2(x)≤ u(x)≤ v1(x)throughout GA and u(x)= v1(x) for x∈ SA.

3. We will now prove the main result.

Theorem. Assume that there is a number �∈ (0; 1) such that f∈C�(M × J ;R)(H�older continuous) for every bounded domain M ⊂GA and every bounded intervalJ ⊂R and that g∈C�loc(R+;R+). If

0≤f(x; t)≤ a(|x|) b(t); t ∈R+; x∈R2;where a; b∈C(R+;R+) are such that b is bounded, b(0)= 0, and

lim supt→∞

[t2a(t) +

∫ t

0b(s) ds

]¡∞

then there is a bounded positive solution to Eq. (1) on GA if g is bounded and

lim supt→∞

∫ t

0|g(s)| ds¡∞:

Proof. Let K¿0 be such that

t2a(t)≤K; t ∈R+;and let us consider the autonomous di�erential equation of Li�enard type

h′′ + h′g(h) + b0(h)= 0; (4)

A. Constantin, G. Villari / Nonlinear Analysis 42 (2000) 243–250 245

where b0 ∈C1(R;R) is such that tb0(t)¿0 for t 6=0, b0(t)= −b0(−t) for t¿0, b0(t)≥Kb(t) for t≥ 0 with b0 bounded, and lim supt→∞

∫ t0 b0(s) ds¡∞: the construction of

such a function b0 is easy.The investigation of the qualitative behaviour of the solutions of Eq. (4) will be

crucial in our approach, providing the way to construct a supersolution to Eq. (1).For this reason, we present some results related to the Li�enard equation (4) which

has been widely investigated in the literature of nonlinear oscillations (see the papers[2, 4, 10, 14, 15]).Write Eq. (4) as the Li�enard system

h′= k − G(h); k ′=− b0(h); (5)

where G(h)=∫ h0 g(s) ds. Since b0 ∈C1(R;R) we have that uniqueness is guaranteed

for the solutions of Eq. (5). We now claim that all solutions of Eq. (5) are de�nedin the future. Indeed, b0 being bounded on R, we see that in �nite time k(t) cannotblow-up. Using this and the boundedness of G we observe that h(t) cannot blow-upeither in �nite time and therefore all solutions of Eq. (5) are global solutions.The curve k =G(h) is called the vertical isocline for Eq. (5) as the �rst component

of the vector �eld given by the di�erential system is zero on such a curve.We are looking for a solution of Eq. (5) that remains eventually positive. An in-

spection of system (5) shows that each trajectory crossing the vertical isocline at someh¿0 eventually becomes negative (see for instance [14]). Hence we are interested ina condition ensuring that the trajectory does not intersect the vertical isocline. This isprecisely what is studied in [2, 4, 15]. For instance, in [15], necessary and su�cientconditions are presented.Actually, the hypotheses we assumed guarantee that (see [15]) for every initial

data (h0; k0) with k0¿G(h0) and h0¿0, the positive semitrajectory of Eq. (5) passingthrough (h0; k0) at time t0 = ln(A) does not cross the vertical isocline; therefore, wehave that h(s) is positive for s≥ ln(A).Let us prove that we may choose h not only positive but also bounded on [ln(A);∞).If g 6≡ 0 we can choose k0; h0¿0 such that∞¿ lim

t→∞G(t)¿k0¿h0:

By the positivity of h and the second equation in Eq. (5), we �nd that k is non-increasing on [ln(A);∞). But k(t)¿G(h(t)) on [ln(A);∞), as we saw, so that h(t)is bounded from above in view of the relation G(h(t))≤ k0¡ lims→∞G(s) valid fort ∈ [ln(A);∞).Assume now that g≡ 0. De�ne B(t)= ∫ t

0 b0(s) ds for t≥ 0. Since b0 6≡ 0 we canchoose k0; h0¿0 such that

k202+ B(h0)¡ lim

s→∞B(s)¡∞:

In the present case, Eq. (5) is a Hamiltonian system with Hamiltonian

H (h; k)=k2

2+ B(h):


This quantity is conserved along the solution to Eq. (5) for time t ∈ [ln(A);∞), theinterval on which h(t)¿0. We infer the inequality

B(h(t))≤ k202+ B(h0)¡ lim

s→∞B(s)¡∞

from where we deduce that the increasing function h (recall k(t)¿0 by the abovediscussion of the dynamics) must be bounded on [ln(A);∞).We proved therefore that h can be chosen positive and bounded on [ln(A);∞).De�ne v1(x)=y(r)= h(s) for r= |x| ≥A and s= ln(r). We have

r[�v1(x) + f(x; v1) + g(v1)

x · ∇v1|x|2

]=ddr

{r@y@r

}+ rf(x; y) + g(y)

@y@r

= e−sh′′ + esf(x; h) + e−sg(h)h′

≤ e−s[h′′ + e2sa(es)b(h) + g(h)h′]

≤ e−s[h′′ + h′g(h) + b0(h)]= 0; r¿A;

so that v1 is a supersolution to Eq. (1). Clearly v2≡ 0 on SA ∪GA is a subsolution toEq. (1).By the Lemma we deduce that Eq. (1) has a solution u(x) in GA with 0≤ u(x)≤ v1(x)

for |x|¿A and u(x)= v1(x) for x∈ SA.Since g is bounded, there is a p¿0 such that

pA2≥ 2 + sups∈R

{|g(s)|}:

De�ne for �¿0,

u�(x) = infz∈SA

{u(z)}+ �e−p|x|2

= y(A) + �e−p|x|2; x∈ SA ∪GA;

where u(x) is the solution of Eq. (1) found above, and let us consider the operator

Lw=�w + g(u)x · ∇w|x|2 ; w∈C2(GA)∪C(GA):

Observe that

Lu� = �(4p2|x|2 − 4p)e−p|x|2 − 2�pg(u)e−p|x|2

≥ 2�p2|x|2e−p|x|2 − 2�pg(u)e−p|x|2

≥ 2�pe−p|x|2(pA2 − sup

s∈R{g(s)}

)¿0

≥−f(x; u(x))=�u(x) + g(u)x · ∇u|x|2 =L(u+ �e−pA2); x∈GA;


that is,

0¿L([u+ �e−pA

2]− u�

); x∈GA:

On the other hand,

[u(x) + �e−pA2]− u�(x)≥ 0; x∈ SA:

Since u� is bounded on GA and u(x)≥ 0 on GA, we have that the function w�(x) :=u(x) + �e−pA

2 − u�(x); x∈GA ∪ SA, has a �nite in�mum in GA ∪ SA. The existence ofan x0 ∈GA with

w�(x0)= infx∈GA∪SA

{w�(x)}

would imply �w�(x0)≥ 0 and ∇w�(x0)= 0, so that Lw�(x0)≥ 0, which is false as wesaw above. Therefore, the in�mum is either attained on SA or “at in�nity”, meaningthat there is a sequence {xn}n≥1 in GA such that

w�(xn)→ infx∈GA∪SA

{w�(x)} and |xn|→∞:

Let us �rst assume that for all �¿0 the in�mum is attained on SA, that is,

0≤ infx∈SA

{w�(x)}= infx∈GA∪SA

{w�(x)}:

We obtain

u�(x)≤ u(x) + �e−pA2 ; x∈GA ∪ SA:Letting �→ 0 in this relation, we get

0¡y(A)= infx∈SA

{u(x)}≤ u(x); x∈GA;

so that u is positive in GA.If for some �¿0 the in�mum is attained “at in�nity”, choose {xn}n≥1 in GA such

that w�(xn)→ inf x∈GA∪SA{w�(x)} and |xn|→∞. We then have that

w�(x) = u(x) + �e−pA2 − y(A)− �e−p|x|2

≥ limn→∞(u(xn) + �e

−pA2 − y(A)− �e−p|xn|2 ); x∈GA ∪ SA:

The existence of the limit on the right-hand side of the above inequality proves thatlimn→∞ u(xn) exists (and is certainly nonnegative). But then the inequality becomes

u(x)− �e−p|x|2 ≥ limn→∞ u(xn)≥ 0; x∈GA ∪ SA:

This implies that

u(x)≥ �e−p|x|2 ; x∈GA ∪ SA;and therefore u is positive in GA in this case too.


We ruled out all possibilities so that the positivity of u in GA is proved.The boundedness of the constructed supersolution (that dominates u in GA) shows

that u is also bounded in GA.

Example 1. Not all positive solutions in GA must be bounded. Indeed, the function

u(x; y) := ln(1 + ln√x2 + y2)

solves

�u+e−2u

2|x|2 +12e−u

x · ∇u|x|2 = 0 in G2

and is clearly unbounded.

Corollary. Assume that there is a number �∈ (0; 1) such that f∈C�(M × J ;R) forevery bounded domain M ⊂GA and every bounded interval J ⊂R. If

0≤f(x; t)≤ a(|x|) b(t); t ∈R+; x∈R2;where a; b∈C(R+;R+) are such that b(0)= 0; b is bounded; and

lim supt→∞

[t2a(t) +

∫ t

0b(s) ds

]¡∞

then there is a positive solution to Eq. (2) on GA.

4. The following example shows the applicability of our result.

Example 2. Consider the elliptic equation

�u+1

(1 + |x|2) (1 + u2) = 0; x∈G1: (6)

By the Corollary, there is a positive solution to Eq. (6) in G1. It is easy to see thatthe results from [3, 11] are not applicable.

In order to clarify the importance of the Corollary, consider Eq. (2) under theassumption

0≤f(x; t)≤ a(|x|) b(t); t ∈R+; x∈R2;where a; b∈C(R+;R+). If∫ ∞

A+1sa(s) ln(s) ds¡∞

we know by the result in [3] that there is a positive solution to Eq. (2) in GB forsome B≥A (to apply the result obtained in [3], replace b by the nondecreasing func-tion b1(t) := max0≤s≤t{b(s)}). Let us also point out that (see [12]) in order that the


equation

�u+ a(|x|) b(u)= 0; x∈GA⊂R2;for a∈C(R+;R+), b∈C1(R+; (0;∞)) increasing and sublinear, has a positive solutionin GB for some B≥A, it is necessary that∫ ∞

Asa(s) ds¡∞:

The natural question arises: are there positive solutions if sa(s) 6∈L1[A;∞) and, say,the function b∈C(R+;R+) is not increasing?The results in [3, 5, 11, 12] give no answer whereas the Corollary shows that it is

possible to allow limt→∞{t2a(t)}=1 if b is “small enough”.

Remark. Our approach is typically two-dimensional: seeking for a radial supersolutionto Eq. (1) in dimension n¿2 leads us to the study of a non-autonomous ordinarydi�erential equation so that one cannot exploit the qualitative study of the Li�enardequation.

Acknowledgements

The authors are grateful for the constructive suggestions made by the referee.

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Positive solutions of quasilinear elliptic equations in two-dimensional exterior domains

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