0 in Ω, (1.1) u= 0 on∂Ω, (1.2) u→0 as |x

Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 112, pp. 1–11.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

EXISTENCE OF SOLUTIONS FOR SEMILINEAR PROBLEMS WITH PRESCRIBED NUMBER OF ZEROS ON EXTERIOR

DOMAINS

JANAK JOSHI, JOSEPH IAIA

Abstract. In this article we prove the existence of an infinite number of radial solutions of ∆(u) +f(u) = 0 with prescribed number of zeros on the exterior of the ball of radiusR >0 centered at the origin inR^N wheref is odd with f <0 on (0, β),f >0 on (β,∞) whereβ >0.

1. Introduction In this article we study radial solutions of

∆(u) +f(u) = 0 in Ω, (1.1)

u= 0 on∂Ω, (1.2)

u→0 as |x| → ∞ (1.3)

where x∈Ω =R^N\BR(0) is the complement of the ball of radiusR >0 centered at the origin.

The functionf is odd, locally Lipschitz and is defined by f(u) =|u|^p−1u+g(u) withp >1,f⁰(0)<0 and lim

u→∞

g(u)

u^p = 0. (1.4) We assume that there existsβ >0 such thatf(0) =f(β) = 0 andF(u) =Ru

0 f(s)ds where

f <0 on (0, β),f >0 on (β,∞) (1.5) Asf is odd, it follows thatF(u) =Ru

0 f(s)dsis even. AlsoF has a unique positive zero,γ, withβ < γ <∞andF is bounded below by some−F0<0 so that

F <0 on (0, γ),F >0 on (γ,∞), andF ≥ −F0 on (0,∞). (1.6) Since we are interested in radial solutions of (1.1)–(1.3) we assume that u(x) = u(|x|) =u(r), where r=|x|=p

x²₁+x²₂+· · ·+x²_N so thatusolves u⁰⁰(r) +N−1

r u⁰(r) +f(u(r)) = 0 on (R,∞) whereR >0, (1.7) u(R) = 0, u⁰(R) =a >0. (1.8)

2010Mathematics Subject Classification. 34B40, 35B05.

Key words and phrases. Exterior domains; semilinear; superlinear; radial.

c

2016 Texas State University.

Submitted December 31, 2015. Published May 3, 2016.

1

We will occasionally denote the solution of the above by ua(r), to emphasize the dependence on the initial parametera.

Theorem 1.1. For each nonnegative integer n, there exists a solution u(r) of (1.7)–(1.8) on [R,∞) such that lim_r→∞u(r) = 0 and u(r) has exactly n zeros on (R,∞).

The radial solutions of (1.1), (1.3) have been well-studied when Ω =R^N. These include [1, 2, 6, 8, 10]. Recently there has been an interest in studying these problems onR^N\BR(0). These include [4, 5, 7, 9]. Here we use a scaling argument as in [8] to prove existence of solutions.

2. Preliminaries

ForR >0 existence and uniqueness of solutions of (1.7)-(1.8) on [R, R+) for some >0 and continuous dependence of solutions with respect toafollows from the standard existence-uniqueness theorem for ordinary differential equations [3].

For existence on [R,∞) we consider Ea(r) = 1

2u⁰²_a +F(ua). (2.1)

Using (1.7) we see that

E_a⁰(r) =−N−1

r u⁰²_a ≤0 (2.2)

soEa is non-increasing on [R,∞). Therefore 1

2u⁰²_a +F(ua) =Ea(r)≤Ea(R) =1

2a² forr≥R. (2.3) Therefore by (1.6),

1 2u⁰²_a ≤1

2a²+F0.

So for a fixedawe see thatu⁰_a is uniformly bounded and hence existence on all of [R,∞) follows.

Lemma 2.1. Let ua(r)be the solution of (1.7)-(1.8). Ifais sufficiently large then there existsr > R such thatua(r)> β. In particular, there existsra> Rsuch that u_a(r_a) =β.

Proof. Sinceu⁰_a(R) =a >0 we see thatua(r) is increasing on [R, R+δ) for some δ >0. Ifua(r) has a first critical pointMa> Rwithu⁰_a(r)>0 on [R, Ma) then we must haveu⁰_a(Ma) = 0, u⁰⁰_a(Ma)≤0. In factu⁰⁰_a(Ma)<0 (by uniqueness of solutions of initial value problems). Therefore from (1.7) it follows thatf(ua(Ma))>0 and using (5) we see thatua(Ma)> β.

On the other hand, ifua(r) has no critical point thenu⁰_a(r)>0 for eachr≥R.

Suppose now by the way of contradiction that ua(r) ≤ β for each r ≥R. Since ua(r) is increasing and bounded above then limr→∞ua(r) exists. Thus there exists L >0,L≤β such that

r→∞lim u_a(r) =L. (2.4)

SinceEa(r) is non-increasing and bounded below, it follows that limr→∞Ea(r) ex- ists. This implies limr→∞u⁰_a(r) exists and in fact limr→∞u⁰_a(r) = 0 since otherwise uawould become unbounded contradicting (2.4). Hence by (1.7), limr→∞u⁰⁰_a(r) ex- ists and as withu⁰_a(r) we see that lim_r→∞u⁰⁰_a(r) = 0. Taking limits in (1.7) we see thatf(L) = 0. SinceL >0 it follows thatL=β.

Suppose now that this is true for all values ofa >0. We then letya(r) = ^ua_a^(r) and we see that:

y_a⁰⁰+N−1

r y_a⁰ +f(aya)

a = 0. (2.5)

y_a(R) = 0, y_a⁰(R) = 1. (2.6)

Since

y⁰²_a

2 +F(ay_a) a²

0

=y_a⁰y⁰⁰_a+f(ay_a)

a y_a⁰ =−(N−1)

r y⁰²_a ≤0, it follows that

y_a⁰²

2 +F(aya) a² ≤ 1

2 ∀r≥R.

In addition, from (1.6) it follows that y⁰²_a

2 −F0

a² ≤ 1 2. Hence

y_a⁰² 2 ≤1

2 +F0

a² ≤1

ifais sufficiently large. Therefore|y⁰_a|is uniformly bounded ifais sufficiently large.

Also 0≤ua ≤β implies 0≤ya ≤ ^β_a ≤1 ifa is large soya is uniformly bounded.

And since ay_a is bounded it follows that ^f(ay_a^a) → 0 as a → ∞. Thus it follows from (2.5) that |y_a⁰⁰| is uniformly bounded for sufficiently large a. Hence by the Arzela-Ascoli theoremya →y and y⁰_a → y⁰ uniformly on the compact subsets of [R,∞) as a→ ∞ for some subsequence still denoted byya. Moreover from (2.6) we seey(R) = 0 and y⁰(R) = 1.

On the other hand, 0≤ya ≤ ^β_a so it follows that ya →0 as a→ ∞. Soy ≡0 and therefore y⁰ ≡ 0 which is a contradiction to y⁰(R) = 1. Hence there exists ra> Rsuch thatua(ra) =β and 0< ua< β on (R, ra).

Ifu⁰_a(ra) = 0 thenua ≡β by uniqueness of solutions of initial value problems.

But this contradicts the fact that u⁰_a(R) =a >0. Thusu⁰_a(ra)>0. Henceua(r) must get larger thanβ. Thus there existsra> Rsuch thatua(ra) =β, u⁰_a(ra)>0 andua< β on [R, ra). This completes the proof.

Lemma 2.2. If ais sufficiently large then ua(r)has a maximum at Ma> ra. In addition,|ua|has a global maximum at Ma andua(Ma)→ ∞asa→ ∞.

Proof. Suppose by the way of contradiction that u⁰_a(r)>0 for eachr > R. Then u_a(r)> β forr > r_a as we saw in the proof of the Lemma 2.1. Also as in Lemma 2.1, u⁰_a(r_a)>0 thus∃r_a1 > r_a such that u(r_a1)> β+ for some >0 and since u⁰_a>0, forr > ra₁ we have f(ua)≥f(β+)>0. Therefore,

u⁰⁰_a+N−1

r u⁰_a+f(β+)≤u⁰⁰_a+N−1

r u⁰_a+f(u_a) = 0 forr > r_a1. This implies

r^N−1u⁰_a(r)⁰

≤ −f(β+)r^N−1 forr > ra₁. Hence forr > r_a1 we have

r^N−1u⁰_a(r)< r^N_a1⁻¹u⁰_a(ra₁)−f(β+)r^N−1−r^N_a1⁻¹ N−1

→ −∞

as r→ ∞. This contradicts the assumption thatu⁰_a>0 forr > R. So ∃Ma > ra

such that u⁰_a(Ma) = 0 and u⁰⁰_a(Ma) ≤ 0. By uniqueness of solutions of initial value problems it follows that u⁰⁰_a(Ma) < 0 so Ma is a local maximum. Thus f(u_a(M_a)) >0 and therefore u_a(M_a) > β. To see this is a global maximum for

|ua|suppose there exists M_a2 > M_a with|ua(M_a2)|> u_a(M_a)> β. Then sinceF is even and increasing foru > β it follows that

F(ua(Ma2)) =F(|ua(Ma2)|)< F(ua(Ma)).

On the other hand,Ea is nonincreasing so

F(ua(Ma₂)) =Ea(Ma₂)≤Ea(Ma) =F(ua(Ma)), a contradiction. HenceM_a is the global maximum for |u_a|.

We now show that ua(Ma) → ∞as a → ∞. Suppose not. Then |ua(r)| ≤ C whereCis a constant independent of a. As in Lemma 2.1, let y_a(r) = ^ua_a^(r). Then as in Lemma 2.1, y_a → y with y ≡ 0 and y⁰(R) = 1, a contradiction. Hence u_a(M_a)→ ∞as a→ ∞. This proves the lemma.

Next we proceed to show thatu_a(r) has zeros on (R,∞) and the number of zeros increases asa→ ∞. First we letva(r) =ua(Ma+r). It follows thatva satisfies

v⁰⁰_a(r) + N−1

M_a+rv_a⁰(r) +f(va(r)) = 0 on [R,∞), (2.7) v_a(0) =u_a(M_a)≡λ

2

ap−1 and v⁰_a(0) = 0. (2.8) By Lemma 2.2, lima→∞ua(Ma) =∞and thusλa→ ∞as a→ ∞.

Next we letwλ_a(r) = λ⁻

2 p−1

a va(_λ^r

a) as in [8]. Then using (1.4) and (2.7)–(2.8) we see that

w_λ⁰⁰

a(r) + N−1 λ_aM_a+rw⁰_λ

a(r) +|w_λa|^p−1w_λa+g(λ

2 p−1

a wλ_a) λ

2p

ap−1

= 0, (2.9) wλ_a(0) = 1, w_λ⁰

a(0) = 0. (2.10)

Lemma 2.3. wλ_a →w uniformly on compact subsets of [0,∞) asa→ ∞and w satisfiesw⁰⁰+|w|^p−1w= 0.

Proof. From (1.4) we know thatf(u) =|u|^p−1u+g(u) withp >1 where ^g(u)_up →0 as u → ∞. Letting G(u) = Ru

0 g(s)ds then it follows that ^G(u)_up+1 →0 as u→ ∞.

Let w_λa(r) be the solution of the system (2.9)–(2.10) and E_λa(r) be the energy associated withw_λa(r) defined by

Eλ_a =w⁰²_λ

a

2 +|wλa|^p+1 p+ 1 + 1

λ

2(p+1) p−1

a

G(λ

2

ap−1wλ_a). (2.11) ThenE_λ⁰

a(r) = _λ^−(N−1)

aMa+rw_λ⁰²

a ≤0 which impliesEλ_a(r) is a non-increasing function ofr. Therefore,

Eλa(r)≤Eλa(0) = 1

p+ 1+ 1 λ

2(p+1) p−1

a

G(λ

2

ap−1).

Since ^G(u)_up+1 →0 asu→ ∞it follows forasufficiently large that E_λa(r)≤E_λa(0)≤ 1

p+ 1+ 1<2.

Also it follows that |G(u)| ≤ _2(p+1)¹ |u|^p+1 if|u| ≥T1 for someT1 >0. And since Gis continuous on the compact set|u| ≤T1, there exists a constantCG >0 such that|G(u)| ≤CG if|u| ≤T1. Thus

|G(u)| ≤CG+ 1

2(p+ 1)|u|^p+1 for allu.

Therefore if a is sufficiently large we see from this upper bound for Gand (2.11) that

w⁰²_λ

a

2 +|wλ_a|^p+1

p+ 1 ≤2− 1 λ

2(p+1) p−1

a

G(λ

2 p−1

a wλ_a)≤2 + CG

λ

2(p+1) p−1

a

+|wλ_a|^p+1 2(p+ 1). Thus ifais sufficiently large we have

w⁰²_λ

a

2 +|wλ_a|^p+1

2(p+ 1) ≤2 + CG

λ

2(p+1)

ap−1

≤3. (2.12)

Thereforewλ_a andw_λ⁰

a are uniformly bounded for largea. So by the Arzela-Ascoli theorem wλ_a → w uniformly on compact subsets of [0,∞) for some subsequence still labeledwλ_a.

Now using the definition off from (1.4) we have:

w⁰⁰_λ

a+ N−1 λ_aM_a+rw⁰_λ

a+|w_λa|^p−1w_λa+λ

−2p p−1

a g(λ

2 p−1

a w_λa) = 0, wλ_a(0) = 1, w⁰_λa(0) = 0.

Since lim_u→∞^g(u)_up = 0, it follows that for all >0 there exists aT2>0 such that

|g(u)| ≤|u|^pif|u|> T2and the continuity ofgon the compact set|u| ≤T2implies

|g(u)| ≤Cg for someCg>0 if|u| ≤T2. Thus,

|g(u)| ≤C_g+|u|^p for allu and hence

|g(λ

2 p−1

a wλ_a)| ≤Cg+λ

2p p−1

a |wλ_a|^p. Recall from (2.12) that|wλa| ≤[6(p+ 1)]^p+11 <4 for p >1. So:

|g(λ

2 p−1

a wλ_a)|

λ

2p

ap−1

≤ Cg+λ

2p p−1

a 4^p λ

2p

ap−1

= Cg

λ

2p

ap−1

+4^p. This implies

0≤lim sup

a→∞

|g(λ

2 p−1

a wλ_a)|

λ

2p p−1

a

≤lim sup

a→∞

Cg

λ

2p p−1

a

+4^p=4^p. This is true for each >0. Hence

a→∞lim

|g(λ

2 p−1

a wλ_a)|

λ

2p

ap−1

= 0. (2.13)

In addition, recall thatMa≥Rand so forr≥Rwe have 1

λaMa+r ≤ 1 (λa+ 1)R

and since |w⁰_λ

a| is uniformly bounded (by (2.12)) we see that _λ^N−1

aMa+rw_λ⁰

a →0 as a → ∞. From this and (2.13) we see that the second and fourth terms on the left-hand side of (2) go to 0 asa→ ∞. In addition,wλ_a is bounded by (2.12) and therefore it follows from (2) that|w_λ⁰⁰

a|is uniformly bounded.

Therefore by the Arzela-Ascoli theorem for some subsequence still labeled wλ_a

we have wλ_a → w and w⁰_λ

a → w⁰ uniformly on compact subsets of [0,∞) and from (2) we have lim_a→∞w⁰⁰_λ

a+|w|^p−1w= 0. Thus lim_a→∞w⁰⁰_λ

a exists and in fact lima→∞w⁰⁰_λa =w⁰⁰. Hence

w⁰⁰+|w|^p−1w= 0 (2.14)

w(0) = 1, w⁰(0) = 0. (2.15)

Therefore ¹₂w⁰²+_p+1¹ |w|^p+1= _p+1¹ .

It is straightforward to show that solutions of (2.14)–(2.15) are periodic with periodp

2(p+ 1)R1 0

√ dt

1−t^p+1 and they have an infinite number of zeros on [0,∞).

Sincewλa →wuniformly on compact subsets of [0,∞) asa→ ∞it follows that w_λa has zeros on (0,∞) and the number of zeros of w_λa gets arbitrarily large by takingasufficiently large. Recalling that

wλ_a(r) =λ^−p−12 ua(Ma+ r λ_a)

we see that ua(r) has zeros (R,∞) for large a and the number of zeros ofua(r)

increases asaincreases.

Next we examine (1.7)-(1.8) whena >0 is small.

Lemma 2.4. ra→ ∞ asa→0⁺ wherera is defined in Lemma 2.1.

Proof. From (2.3) we have ¹₂u⁰²_a +F(u_a)≤¹₂a²forr≥R, and from Lemma 2.2 we haveu⁰_a >0 on [R, r_a]. So rewriting this inequality and integrating on (R, r_a) gives

Z r_a

R

u⁰_a

pa²−2F(ua) ≤ Z r_a

R

1dr=ra−R.

Lettings=ua(r) we see that Z β

0

ds

pa²−2F(s) = Z r_a

R

u⁰_adr

pa²−2F(ua) ≤ra−R. (2.16) From (1.4) we havef⁰(0)<0, thusf(u)≥ −³₂|f⁰(0)|ufor smallu. Soa²−2F(u)≤

3

2|f⁰(0)|u²+a² for smalluand so pa²−2F(u)≤

r a²+3

2|f⁰(0)|u²≤a+ r3

2|f⁰(0)|ufor smallu.

Therefore,

1

pa²−2F(u) ≥ 1 a+

q3

2|f⁰(0)|u

for smallu.

So for somewith 0< < β we have Z

0

ds

pa²−2F(s) ≥ Z

0

ds a+q

3 2|f⁰(0)|s

= s 2

3|f⁰(0)|ln 1 +

r3

2|f⁰(0)|

a → ∞

asa→0⁺. Therefore from (2.16) and the above computation we see that ra−R≥

Z β

0

ds

pa²−2F(s) ≥ Z

0

ds

pa²−2F(s)→ ∞ as a→0⁺

thusra→ ∞asa→0⁺. Hence the lemma is proved.

Note that ifE(r₀)<0, then

u(r)>0 for eachr > r0. (2.17) Suppose not. Then there existsz > r0 such thatu(z) = 0 and soF(u(z)) = 0. By (2.2),E(r) is non-increasing soE(z)≤E(r0)<0. Therefore

0≤u⁰(z)²

2 = u⁰(z)²

2 +F(u(z)) =E(z)<0 which is impossible. Henceu(r)>0 for allr > r0.

Lemma 2.5. If a >0and ais sufficiently small thenu_a(r)>0for each r > R.

Proof. Assume by the way of contradiction thatua(za) = 0 for someza> R. Since ua(R) = 0 and u⁰_a(R) = a > 0 we see that ua(r) has a positive local maximum, Ma, withR < Ma < za and since the energy functionEa(r) is non-increasing then

0< Ea(za)≤Ea(Ma) =F(ua(Ma)).

Thus by (1.6) u_a(M_a) > γ and so in particular there exist p_a, q_a with R < p_a <

qa < Ma such that ua(pa) = ^β₂, ua(qa) =β and 0 < ua(r)< β for [R, qa). Then by (1.5) we see thatf(u_a)<0 on [R, q_a) sou⁰⁰_a+^N_r⁻¹u⁰_a >0 on [R, q_a) by (1.7).

ThereforeRr

R(r^N−1u⁰_a)⁰dr >0 from which it follows that r^N−1u⁰_a> R^N−1u⁰_a(R)>0 on [R, q_a).

Thusu_a(r) is increasing on [R, q_a). In addition, p_a → ∞as a→0⁺ for if thep_a were bounded then a subsequence would converge to say some finitep₀asa→0⁺. SinceE_a(r) is non-increasing this would implyu_a(r) andu⁰_a(r) would be uniformly bounded on [R, p₀+ 1] and so by the Arzela-Ascoli theorem for a subsequence u_a(r)→u₀(r) ≡0 asa→0⁺. On the other hand, ^β₂ =u_a(p_a)→u₀(p₀) = 0 as a→0⁺ which is a contradiction. Thus we see thatpa → ∞asa→0⁺.

Next we return to (2.3) and after rewriting we have u⁰_a

pa²−2F(u_a)≤1 for eachr≥R.

Integrating on [pa, qa] and setting ua(r) =t we obtain Z β

β 2

dt

pa²−2F(t)= Z qa

p_a

u⁰_a

pa²−2F(ua)dr≤ Z qa

p_a

1dr=q_a−p_a. (2.18) Now on [^β₂, β] we have 0< a²−2F(t)≤1 + 2|F(β)|if 0< a≤1. It follows that

Z β

β 2

dt

pa²−2F(ua) ≥ β 2p

1 + 2|F(β)| ≡c >0

for some constantc >0 and sufficiently smalla. Combining this with (2.18) we see that

q_a−p_a ≥c ifais sufficiently small. (2.19)

Now by the definition ofEa(r) it is straightforward to show that (r^2(N−1)Ea(r))⁰= (r^2(N−1))⁰F(ua).

Integrating on [p_a, q_a] gives

q^2(N−1)_a Ea(qa) =p^2(N−1)_a Ea(pa) + Z q_a

p_a

[r^2(N−1)]⁰F(ua)dr.

SinceF(u_a)≤F(^β₂)<0 on [p_a, q_a] we have p^2(N_a ⁻¹⁾Ea(pa) +

Z q_a

pa

(r^2(N−1))⁰F(ua)dr

≤p^2(N_a ⁻¹⁾Ea(pa)− |F(β

2)|[q^2(N−1)_a −p^2(N_a ⁻¹⁾].

But

p^2(N_a ⁻¹⁾Ea(pa) =R^2(N−1)Ea(R) + Z p_a

R

[r^2(N−1)]⁰F(ua)dr and

Z pa

R

[r^2(N−1)]⁰F(u_a)dr≤0 asF(u_a)≤0 on [R, p_a]. Thus

p^2(N_a ⁻¹⁾Ea(pa)≤R^2(N−1)Ea(R) =1

2a²R^2(N−1). Therefore,

q_a^2(N−1)E_a(q_a)≤ 1

2a²R^2(N−1)− |F(β 2)|h

q^2(N−1)_a −p^2(N−1)_a i . So

q^2(N_a ⁻¹⁾Ea(qa)≤ a²R^2(N−1)

2 − |F(β

2)|(q^2(N_a ⁻¹⁾−p^2(N−1)_a ) (2.20) Now by (2.19) we have

q_a^2(N−1)−p^2(N_a ⁻¹⁾≥(q_a−p_a)p^2N−3_a ≥c p^2N_a ⁻³,

and from earlier in the proof of this lemma we saw lim_a→0+p^2N−3_a = ∞. Thus q^2(N−1)a −p^2(N−1)a → ∞asa→0⁺.

It follows then from (2.20) thatqa^2(N−1)Ea(qa) is negative ifais sufficiently small.

Thus by (2.17) it follows that ua(r)> 0 for r ≥qa. Also, since we have u⁰_a > 0 on [R, qa] andua(R) = 0 we see thatua(r)>0 on (R,∞) ifais sufficiently small.

This completes the proof.

3. Proof of Theorem 1.1 Let

S0={a >0|ua(r)>0∀r > R}.

By Lemma 2.5 we know that fora >0 and asufficiently small that ua(r)>0 so S0 is nonempty. Also from Lemma 2.3 we know that ifais sufficiently large then ua(r) has zeros. HenceS0is bounded above and so the supremum ofS0exists. Let a0= sup(S0).

Lemma 3.1. ua₀(r)>0 on (R,∞).

Proof. Suppose by the way of contradiction that there existsz0such thatua₀(z0) = 0 andua(r)>0 on [R, z0). Thenu⁰_a0(z0)≤0 and by uniqueness in factu⁰_a0(z0)<0.

Thus ua0(r)<0 for z0 < r < z0+. Ifa < a0 and ais close enough to a0 then the continuity of solutions of boundary value problems with respect to the initial conditions implies thatu_a(r) also gets negative which contradicts the definition of a₀. Sou_a0(r)>0 on (R,∞). This completes the lemma.

Lemma 3.2. u_a0(r)has a local maximum,M_a0 > R.

Proof. Suppose not. Thenu⁰_a

0(r)>0 for all r≥R. Since E_a0(r)≤E_a0(R) for all r≥R, we have

u⁰²_a0(r)

2 +F(ua₀(r))≤ a²₀ 2 .

This impliesF(ua0(r))≤^a₂²⁰ and henceua0(r) is bounded. Since we are also assum- ingu⁰_a0(r)>0 it follows that limr→∞ua0(r) exists. Let us denote limr→∞ua0(r) = L. SinceE_a0(r) is a non-increasing function which is bounded below, it follows that lim_r→∞Ea₀(r) = lim_r→∞[^u

02 a0

2 +F(ua₀)] exists.

Since we also know that lim_r→∞ua₀(r) exists it follows that lim_r→∞u⁰_a0(r) ex- ists and in fact limr→∞u⁰_a0(r) = 0 (since otherwise ua₀(r) would be unbounded).

Therefore from (1.7) it follows that limr→∞u⁰⁰_a0(r) =−f(L) and in factf(L) = 0.

(Otherwise,u⁰_a0 would be unbounded but we knowu⁰_a0 →0). SoL=−β,0, or β.

Sinceua0(r)>0 andu⁰_a0(r)>0 thus L=β.

Now by the definition ofa₀ we knowu_a(r) has a zero ifa > a₀, sayu_a(z_a) = 0.

Next we show that

lim

a→a⁺₀

z_a =∞. (3.1)

Suppose not. Then|z_a| ≤K for some constantKand so there is a subsequence of z_a still denotedz_a such thatz_a →z₀ asa→a⁺₀. Butu_a(r)→u_a0(r) uniformly on the compact subset [R, z0+ 1] as a→a⁺₀ so 0 = lim_a→a+

0 ua(za) =ua₀(z0) which contradicts thatu_a0(r)>0 from Lemma 3.1. Thus lim_a→a+

0 z_a =∞. In addition, Ea(za) = ^u

02 a(z_a)

2 ≥0. Also:

r→∞lim Ea₀(r) = lim

r→∞[u⁰²_a0(r)

2 +F(ua₀(r))] =F(β)<0.

So there existsR₀> R such thatE_a0(R₀)<0.

Since lim_a→a0u_a(r) =u_a0(r) uniformly on the compact set [R, R₀+ 1], it follows that lim_a→a0E_a(R₀) = E_a0(R₀) < 0. Since E_a(R₀) < 0 < E_a(z_a) and E_a is non-increasing it follows thatza < R0ifais sufficiently close toa0.

However, by (3.1), we have za → ∞ as a →a⁺₀ which is a contradiction since R0<∞.

Hence ua₀(r) has a local maximum at r = Ma₀ for some Ma₀ > R. This

completes the proof.

Lemma 3.3. u⁰_a0(r)<0 if r > Ma₀.

Proof. Suppose u⁰_a0(ma₀) = 0 for some ma₀ > Ma₀. Then u⁰⁰_a0(ma₀) > 0 and so f(u(ma₀)) < 0. Since we also know that ua₀(r) > 0 (by Lemma 3.1) it follows that 0< ua0(ma0) < β. Therefore,Ea0(ma0) = F(ua0(ma0)) <0 and so by the continuity of the solution with respect to initial conditions we haveE_a(m_a0)<0 if ais sufficiently close toa₀.

Now by the definition ofa0 ifa > a0 thenua(r) has a zero,za, withEa(za)≥0 and by (31) we have seen that lima→a₀za = ∞. Since Ea is non-increasing we therefore have za < ma₀. Butza → ∞ as a→ a⁺₀ and ma₀ <∞so we obtain a

contradiction. This completes the proof.

So u⁰_a0(r) <0 for all r ≥ Ma₀. Also, ua₀(r) > 0 so lim_r→∞ua₀(r) = L with L ≥ 0. Since Ea₀(r) is non-increasing, we see as we did earlier that f(L) = 0.

ThusL= 0 orβ. We now showEa₀(r)≥0 for all r≥R. So suppose there is an r0> Rsuch thatEa₀(r0)<0. ThenEa(r0)<0 foraclose toa0and in particular if a > a0. But then we know that za exists and since Ea(za)≥0 it follows that za < r0 sinceEa is non-increasing. But this contradicts thatza → ∞from (3.1).

ThusEa₀(r)≥0 for allr≥R.

Let us suppose now that L = β. Since Ea(r) is non-increasing and bounded below:

r→∞lim Ea0(r, a0) exists.

This implies

r→∞lim u⁰²_a

0(r) exists

and as we have seen earlier this implies limr→∞u⁰_a0(r) = 0. Therefore, 0≤ lim

r→∞Ea₀(r) = lim

r→∞

u⁰²_a0(r)

2 +F(L) = 0 +F(β)<0.

which is a contradiction. Hence we must haveL= 0. i.e. lim_r→∞u_a0(r) = 0. Thus we have found a positive solutionu_a0(r) of (1.7)-(1.8) such that lim_r→∞u_a0(r) = 0.

Next we let

S₁={a >0|u_a(r) has one zero on (R,∞)}.

[8, Lemma 4] states that if u_ak(r) is a bounded solution of (1.7) on (0,∞) with k zeros and lim_r→∞u_ak(r) = 0 then if ais sufficiently close toa_k then u_a has at mostk+ 1 zeros on [0,∞). A nearly identical lemma holds for solutions of (1.7) on (R,∞). Applying this lemma witha₀we see thatu_a on (R,∞) has at most one zero ifais sufficiently close toa0.

On the other hand, for a > a0 we know that ua(r) has at least one zero on (R,∞) by the definition ofa0. Thus ifa > a0 andais sufficiently close toa0 then ua has exactly one zero and so we see thatS1 is nonempty. We also know S1 is bounded from above by Lemma 2.3 and so we let:

a₁= supS₁.

Using a similar argument as earlier we can show thatu_a1(r) has exactly one zero on (R,∞) and limr→∞u_a1(r) = 0. Continuing in this way we see that we can find an infinite number of solutions - one with exactlyn zeros on (R,∞) for each nonnegative integern- and with lim_r→∞u(r) = 0.

References

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Anal., Volume 82, 313-375, 1983.

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[4] A. Castro, L. Sankar, R. Shivaji; Uniqueness of nonnegative solutions for semipositone prob- lems on exterior domains,Journal of Mathematical Analysis and Applications, Volume 394, Issue 1, 432-437, 2012.

[5] J. Iaia; Loitering at the hilltop on exterior domains, Electronic Journal of the Qualitative Theory of Differential Equations, No. 82, 1-11, 2015.

[6] C. K. R. T. Jones, T. Kupper; On the infinitely many solutions of a semi-linear equation, SIAM J. Math. Anal., Volume 17, 803-835, 1986.

[7] E. Lee, L. Sankar, R. Shivaji; Positive solutions for infinite semipositone problems on exterior domains,Differential and Integral Equations, Volume 24, Number 9/10, 861-875, 2011.

[8] K. McLeod, W. C. Troy, F. B. Weissler; Radial solutions of ∆u+f(u) = 0 with prescribed numbers of zeros,Journal of Differential Equations, Volume 83, Issue 2, 368-373, 1990.

[9] L. Sankar, S. Sasi, R. Shivaji; Semipositone problems with falling zeros on exterior domains, Journal of Mathematical Analysis and Applications, Volume 401, Issue 1, 146-153, 2013.

[10] W. Strauss; Existence of solitary waves in higher dimensions,Comm. Math. Phys., Volume 55, 149-162, 1977.

Janak Joshi

Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX 76203-1430, USA

E-mail address:[email protected]

Joseph Iaia

Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX 76203-1430, USA

E-mail address:[email protected]