Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 117, pp. 1–16.

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

EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS

FOR SEMILINEAR EQUATIONS WITH BOUNDED

NONLINEARITIES ON EXTERIOR DOMAINS

JOSEPH IAIA

Abstract. In this article we study radial solutions of ∆u+K(r)f(u) = 0 on

the exterior of the ball of radius R > 0 centered at the origin in RN where
f is odd with f < 0 on (0, β), f > 0 on (β, δ), f ≡ 0 for u > δ, and where

the function K(r) is assumed to be positive and K(r) → 0 as r → ∞. The

primitive F (u) =
∫ u
0 f(t) dt has a “hilltop” at u = δ. With mild assumptions

on f we prove that if K(r) ∼ r−α with 2 < α < 2(N − 1) then there are n

solutions of ∆u + K(r)f(u) = 0 on the exterior of the ball of radius R such
that u→ 0 as r →∞ if R > 0 is sufficiently small. We also show there are no

solutions if R > 0 is sufficiently large.

1. Introduction

In this article we study radial solutions of

∆u+K(r)f(u) = 0 in Ω, (1.1)

u = 0 on ∂Ω, (1.2)

u→ 0 as |x| → ∞ (1.3)

where x ∈ Ω = RN\BR(0) is the complement of the ball of radius R > 0 centered
at the origin. We assume f : R → R is locally Lipschitz and there exist β, δ with
0 < β < δ such that f(0) = f(β) = f(δ) = 0 where:

(H1) f is odd, f ′(0) < 0, f < 0 on (0, β), f > 0 on (β, δ), f ′(δ−) < 0, f ≡ 0 on
(δ,∞).

It follows that F (u) =
∫ u

0
f(s) ds is even. We also assume that F has a unique

positive zero, γ, with β < γ < δ such that

(H2) F < 0 on (0, γ), F > 0 on (γ,∞).

Note from (H1) and (H2) it follows that F is bounded.
In an earlier paper [6] we studied (1.1), (1.3) when Ω = RN and K(r) ≡ 1.

Interest in the topic for this paper comes from recent papers [5, 12, 14] about
solutions of differential equations on exterior domains. In [7] we studied (1.1)-(1.3)
with K(r) ≡ 1 and Ω = RN\BR(0), in [8] we studied the case when K(r) ∼ r−α

with 0 < α < 2 and in [9] with α > 2(N − 1). In [7, 8, 9] we proved existence of an

2010 Mathematics Subject Classification. 34B40, 35B05.

Key words and phrases. Sublinear equation; radial solution; exterior domain.
c©2020 Texas State University.

Submitted January 6, 2020. Published December 1, 2020.

1



2 J. IAIA EJDE-2020/117

infinite number of solutions - one with exactly n zeros for each nonnegative integer
n such that u→ 0 as |x| → ∞.

When f grows superlinearly at infinity - i.e. limu→∞
f(u)
u = ∞, and Ω = RN .

problem (1.1), (1.3) has been extensively studied in [1, 2, 3, 11, 13, 15]. The type
of nonlinearity addressed here has not been studied as extensively [6, 7, 8].

When f grows sublinearly at infinity - i.e. limu→∞
f(u)
u = 0, but limu→∞ f(u) =

∞ and Ω = RN , problem (1.1), (1.3) has also been studied in [9, 10].
Since we are interested in radial solutions of (1.1)-(1.3) we assume that u(x) =

u(|x|) = u(r) where x ∈ RN and r = |x| =
√
x2

1 + · · ·+ x2
N so that u solves

u′′(r) +
N − 1

r
u′(r) +K(r)f(u(r)) = 0 on (R,∞) where R > 0, (1.4)

u(R) = 0, u′(R) = a > 0. (1.5)

We will assume that there exist constants k1 > 0, k2 > 0, and α > 0 such that

(H3) k1r
−α ≤ K(r) ≤ k2r

−α for 2 < α < 2(N − 1) on [R,∞).

In addition, we assume that

(H4) K is differentiable, limr→∞
rK′

K = −α and rK′

K + 2(N − 1) > 0 on [R,∞).

Note that (H4) implies r2(N−1)K(r) is increasing. Also since f ′(0) < 0 and f ′(δ−) <
0 then it follows from (H1) that there exist positive constants f0, f̄0, f1, f̄1 such that

f0 = inf
(0,β/2]

(
− f(u)

u

)
, f̄0 = sup

u6=0

(
− f(u)

u

)
, (1.6)

f1 = inf
[γ,δ)

( f(u)

δ − u

)
, f̄1 = sup

[β′,δ)

( f(u)

δ − u

)
(1.7)

where β < β′ < γ and F (β2 ) = F (β′).

Theorem 1.1. Let N > 2, R > 0, 2 < α < 2(N − 1) and (H1)–(H4) hold.

(a) There are n solutions of (1.1)-(1.3) on [R,∞) - one with exactly n zeros
for each nonnegative integer n if

γ
(

1 +
(h2f̄0

h1f1

)1/2)
< δ

and if R > 0 is sufficiently small.
(b) There are no solutions for any value of R > 0 of (1.1)-(1.3) if

β′ +
β

2

h1

h2

(f0

f̄1

)1/2

> δ.

(c) There are no solutions of (1.1)-(1.3) on [R,∞) if R > 0 is sufficiently
large.

We note that in Sankar, Sasi, and Shivaji [14] established existence of a positive
solution to a semipositone version of this problem using sub and super solutions.
We use different techniques here and are able to establish existence of multiple
solutions.



EJDE-2020/117 EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS 3

2. Preliminaries

We first suppose that U(r) solves (1.4) and then make the change of variables:

U(r) = u(r2−N ).

Then for 0 < t <∞ we see u satisfies

u′′ + h(t)f(u) = 0, (2.1)

where

h(t) =
t
2(N−1)
2−N K(t

1
2−N )

(N − 2)2
.

It follows from (H3) and (H4) that

h(t) > 0, h′(t) < 0, lim
t→0+

th′

h
= −q, h1t

−q < h(t) < h2t
−q

for t > 0, q =
2(N − 1)− α

N − 2
, hi =

ki
(N − 2)2

.

(2.2)

In addition, it follows from (H3), (H4) and (2.2) that

0 < q < 2. (2.3)

We also assume that

u(0) = 0, u′(0) = b > 0. (2.4)

We want to find b > 0 such that u(R2−N ) = 0 then U(r) = u(r2−N ) will satisfy
(1.1)-(1.3). Therefore for the rest of this paper we will study (2.1), (2.4) with
(H1)–(H4) and attempt to find solutions u such that u(R2−N ) = 0.

We first prove existence of a solution of (2.1), (2.4) assuming (H1)–(H4) on [0, ε]
for some ε > 0. Integrating (2.1) twice on (0, t) and using (2.4) gives

u(t) = bt−
∫ t

0

∫ s

0

h(x)f(u(x)) dx ds. (2.5)

Letting y(t) = u(t)
t and y(0) = b > 0 gives

y(t) = b− 1

t

∫ t

0

∫ s

0

h(x)f(xy(x)) dx ds. (2.6)

Now let S = {y ∈ C[0, ε] : y(0) = b > 0} with the supremum norm, ‖ · ‖, and define
T : S → C[0, ε] by

T (y) = b− 1

t

∫ t

0

∫ s

0

h(x)f(xy(x)) dx ds. (2.7)

We first observe that T : S → S. Next let K be the Lipschitz constant for f(u) in
a neighborhood of u = 0 and suppose 0 ≤ t ≤ ε. Then

|Ty1 − Ty2| ≤
1

t

∫ t

0

∫ s

0

h2K|xy1 − xy2|x−q dx ds

≤
∫ t

0

h2Kx
1−q|y1 − y2| dx

≤ h2K

2− q
ε2−q‖y1 − y2‖.



4 J. IAIA EJDE-2020/117

It follows from this and (2.3) that T is a contraction if ε > 0 is sufficiently small.
Thus by the contraction mapping principle [4] it follows that (2.7) has a fixed point
y in S and therefore u = ty is a solution of (2.5) on [0, ε] for some ε > 0.

Next let

E0(t) =
1

2
u′2 + h(t)F (u). (2.8)

By (2.1) we have E′0 = h′(t)F (u) and thus on ( ε2 , t) we obtain

1

2
u′2 + h(t)F (u) =

1

2
u′2(ε/2) + h(ε/2)F (u(ε/2)) +

∫ t

ε
2

h′(s)F (u(s)) ds.

Since F is bounded and since h, h′ are bounded on [ε/2,∞) it follows that u′ is
bounded on [ε/2,∞). It then follows that the solution of (2.1), (2.4) exists on [0, Q)
for all Q > 0 and thus we obtain a solution of (2.1), (2.4) on [0,∞).

Next let

E(t) =
1

2

u′2

h(t)
+ F (u). (2.9)

Using (2.1)-(2.2) and (2.4) we see that limt→0+ E(t) = 0 and

E′ = −u
′2h′(t)

h2(t)
≥ 0 for t > 0. (2.10)

Thus E is nondecreasing and E(t) > 0 for t > 0.

Lemma 2.1. Assume (H1)–(H4) and let u solve (2.1), (2.4). Then there exists
tγ,b > 0 such that u(tγ,b) = γ, u′(tγ,b) > 0, and 0 < u < γ on (0, tγ,b). In addition,
there exists t2,b with 0 < t2,b < tγ,b such that u(t2,b) = β/2.

Proof. We first observe from (2.4) that u is initially positive and increasing for
t > 0 small. If u has a local maximum M then F (u(M)) = E(M) > 0 thus
u(M) > γ by (H2) and so the existence of tγ,b follows. So now let us assume
u is positive, increasing, and 0 < u < γ for all t > 0. From (2.10) we have
1
2
u′2

h(t) + F (u) = E(t) ≥ E(ε) > 0 for t ≥ ε > 0. Since 0 < u < γ then F (u) ≤ 0 so
1
2
u′2

h(t) ≥ E(ε) for t ≥ ε. Thus

|u′| ≥
√

2E(ε)h(t) ≥
√

2E(ε)h1t
−q/2 > 0 for t ≥ ε. (2.11)

Therefore u′ > 0 for t ≥ ε. Integrating (2.11) on (ε, t) gives

γ ≥ u(t)− u(ε) ≥
√

2E(ε)h1

1− q
2

(t1−
q
2 − ε1−

q
2 ) for t ≥ ε. (2.12)

Recall 0 < q < 2 by (2.3) and so the left-hand side of (2.12) is bounded but the
right-hand side goes to infinity as t→∞. Therefore we obtain a contradiction and
so there exists tγ,b > 0 such that u(tγ,b) = γ and 0 < u < γ for 0 < t < tγ,b. In

addition, 1
2
u′2(tγ,b)
h(tγ,b)

= E(tγ,b) > 0 hence u′(tγ,b) > 0. Since u(0) = 0 it then follows

by the intermediate value theorem that there exists t2,b with 0 < t2,b < tγ,b such

that u(t2,b) = β
2 . This completes the proof. �

Lemma 2.2. Assume (H1)–(H4) and let u solve (2.1), (2.4). If limt→∞ u(t) = L ∈
R then f(L) = 0.



EJDE-2020/117 EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS 5

Proof. Since limt→∞ u(t) = L and u(0) = 0 then it follows that u is bounded for

all t ≥ 0. Also E′ ≥ 0 implies 1
2
u′2

h(t) + F (u) → A ≤ ∞ as t → ∞ and thus
1
2
u′2

h(t) → A−F (L). If A−F (L) > 0 then we obtain |u′| ≥ A1t
−q/2 for some A1 > 0

and for large t. Thus |u′| > 0 and so without loss of generality suppose that u′ > 0.

Integrating u′ ≥ A1t
−q/2 on (t0, t) gives u(t)− u(t0) ≥ A1

1− q2
(t1−

q
2 − t1−

q
2

0 )→∞ as

t→∞ but the left-hand side is bounded since limt→∞ u(t) = L. Thus we obtain a

contradiction and so we see that A − F (L) = 0. Therefore 1
2
u′2

h(t) + F (u) → F (L)

and since F (u) → F (L) it then follows that limt→∞
u′2

h(t) = 0. Therefore by (2.2)

we have

lim
t→∞

tq/2u′ = 0. (2.13)

Next note that (u
′

h )′ = u′′

h −
u′h′

h2 . Rewriting (2.1) we see limt→∞
u′′

h = −f(L). Also

by (2.2) and (2.13) for large t we have |u
′h′

h2 | ≤ 2q
h1
tq−1|u′| = 2q

h1
(tq/2u′) 1

t1−
q
2
→ 0 as

t→∞ since 0 < q < 2. Therefore limt→∞(u
′

h )′ = −f(L). Then by L’Hôpital’s rule

lim
t→∞

u′

th
= lim
t→∞

(u
′

h )

t
= lim
t→∞

(u
′

h )′

(t)′
= −f(L). (2.14)

Now suppose without loss of generality that f(L) > 0. Then from (2.2) and (2.14) it

follows −u′ ≥ |f(L)|h1

2 t1−q for large t and so integrating on (t0, t) gives u(t0)−u(t) ≥
|f(L)|h1

2(2−q) (t2−q − t2−q0 ) → ∞ as t → ∞ so u(t) → −∞ which contradicts that u is

bounded. Thus f(L) ≤ 0. A similar argument shows f(L) ≥ 0 hence f(L) = 0.
This completes the proof. �

Lemma 2.3. Assume (H1)–(H4) and let u solve (2.1), (2.4). Then limb→0+ t2,b =
limb→0+ tγ,b =∞ and

lim inf
b→0+

t
q/2
2,b u

′(t2,b) ≥
β

2

√
h1f0, (2.15)

lim sup
b→0+

t
q/2
γ,b u

′(tγ,b) ≤ γ
√
h2f̄0. (2.16)

Proof. We rewrite (2.1) as

u′′ = h(t)
(
− f(u)

u

)
u. (2.17)

Thus by (1.6), (2.2), and (2.17) we see that

u′′ ≤ h2f̄0u

tq
when u > 0.

Now let v2 solve

v′′2 =
h2f̄0

tq
v2, (2.18)

v2(0) = 0, v′2(0) = b > 0. (2.19)

Then v2 is positive and increasing for t > 0. Also by (1.6) and (2.2) we see that

(u′v2 − uv′2)′ =
(
h(t)

(
− f(u)

u

)
− h2f̄0

tq

)
uv2 ≤ 0 while u > 0.



6 J. IAIA EJDE-2020/117

Since u(0) = v2(0) = 0 we see then that u′v2 − uv′2 ≤ 0 while u > 0 and thus
(u/v2)′ ≤ 0. Since u′(0) = v′2(0) = b we see then that

0 < u ≤ v2. (2.20)

Also u′v2 − uv′2 ≤ 0 and 0 < u ≤ v2 imply that

u′

u
≤ v′2
v2

for u > 0. (2.21)

Next (2.18)-(2.19) can be solved explicitly and we obtain

v2 = bC
√
tI 1

2−q

(2
√
h2f̄0

2− q
t
2−q
2

)
(2.22)

where I 1
2−q

is the modified Bessel function of order 1
2−q with limt→0+ I 1

2−q
(t) = 0.

A well-known fact is that limt→0+
Iν(t)
tν = 1

2νΓ(ν+1) where Iν is the modified Bessel

function of order ν with limt→0+ Iν(t) = 0 and thus from this and (2.22) we see

C = Γ( 3−q
2−q )(

√
h2f̄0

2−q )−
1

2−q > 0. (Here Γ(x) is the Gamma function). It is also known

that Iν > 0, I ′ν > 0, and limt→∞
I′ν(t)
Iν(t) = 1. (Some other general facts about the

modified Bessel functions are included in the appendix).
Now using (2.20) we see that

β

2
= u(t2,b) ≤ v2(t2,b) = bC

√
t2,bI 1

2−q

(2
√
h2f̄0

2− q
t
2−q
2

2,b

)
. (2.23)

If the t2,b are bounded as b → 0+ then the right-hand side of (2.23) goes to
zero which contradicts that β > 0. Thus it must be that limb→0+ t2,b = ∞. Since
tγ,b > t2,b then also limb→0+ tγ,b =∞. This completes the first part of the lemma.

Denoting

s =
2
√
h2f̄0

2− q
t1−

q
2 and sγ,b =

2
√
h2f̄0

2− q
t
1− q2
γ,b (2.24)

It follows from (2.22) that

v′2(t) =
v2(t)

2t
+

√
h2f̄0 t

−q/2v2(t)
I ′ 1

2−q
(s)

I 1
2−q

(s)
.

Therefore

tq/2v′2(t)

v2(t)
=

1

2t1−
q
2

+

√
h2f̄0

I ′ 1
2−q

(s)

I 1
2−q

(s)
. (2.25)

Evaluating at tγ,b it follows from (2.21) and (2.25) that

t
q/2
γ,b u

′(tγ,b)

u(tγ,b)
≤ 1

2t
1− q2
γ,b

+

√
h2f̄0

I ′ 1
2−q

(sγ,b)

I 1
2−q

(sγ,b)
. (2.26)

As mentioned earlier it is well-known that lims→∞
I′ν(s)
Iν(s) = 1. Recalling that 0 <

q < 2 and that tγ,b →∞ as b→ 0+ then we see from (2.26) that

lim sup
b→0+

t
q/2
γ,b u

′(tγ,b) ≤ γ
√
h2f̄0.



EJDE-2020/117 EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS 7

In a similar way let v1 solve

v′′1 =
h1f0

tq
v1, (2.27)

v1(0) = 0, v′1(0) = b > 0. (2.28)

We note that v1 > 0 and v′1 > 0 for t > 0. Then we can similarly show that

v′1
v1
≤ u′

u
for 0 < u <

β

2
. (2.29)

Solving for v1 explicitly we have

v1 = bC1

√
tI 1

2−q

(2
√
h1f0

2− q
t
2−q
2

)
where C1 = Γ(

3− q
2− q

)(

√
h1f0

2− q
)−

1
2−q > 0. (2.30)

It follows from (2.29) and (2.30) that

t
q/2
2,b u

′(t2,b)

u(t2,b)
≥
t
q/2
2,b v

′
1(t2,b)

v1(t2,b)
=

1

2t
1− q2
2,b

+
√
h1f0

I ′ 1
2−q

(p2,b)

I 1
2−q

(p2,b)
(2.31)

where p2,b = 2
√
h1f0

2−q t
1− q2
2,b .

It is shown in the appendix that

I ′ν
Iν

+
ν

t
> 1 for t > 0 and ν > 1/2

from which it follows using (2.31) that

lim inf
b→0+

t
q/2
2,b u

′(t2,b) ≥
β

2

√
h1f0.

This completes the proof. �

Next we rewrite (2.1) as

u′′ + h(t)

(
f(u)

δ − u

)
(δ − u) = 0. (2.32)

From (1.7) and (2.2) we have

h(t)
( f(u)

δ − u

)
≥ h1f1

tq
on [γ, δ), (2.33)

h2f̄1

tq
≥ h(t)

( f(u)

δ − u

)
for u ∈ [β′, δ). (2.34)

So now we compare (2.32) to

w′′2 +
h1f1

tq
(δ − w2) = 0 (2.35)

w2(tγ,b) = u(tγ,b) = γ,w′2(tγ , b) = u′(tγ,b). (2.36)

and

w′′1 +
h2f̄1

tq
(δ − w1) = 0 (2.37)

w1(tb′) = u(tb′) = β′, w′1(tb′) = u′(tb′). (2.38)

Lemma 2.4. Assume (H1)–(H4) and let u solve (2.1), (2.4). Then w1 ≤ u when
u,w1 ∈ [β′, δ) where w1 is the solution of (2.37), (2.38). Also u ≤ w2 when
u,w2 ∈ [γ, δ) where w2 is the solution of (2.35)-(2.36).



8 J. IAIA EJDE-2020/117

Proof. It follows from (2.32) and (2.35) that(
(δ − w2)u′ − (δ − u)w′2

)′
+
(
h(t)

( f(u)

δ − u
)
− h1f1

tq

)
(δ − u)(δ − w2) = 0. (2.39)

By (2.33) it follows that the second term in (2.39) is ≥ 0 when u,w2 ∈ [γ, δ).
Therefore integrating (2.39) on (tγ,b, t) gives

(δ − w2)u′ − (δ − u)w′2 ≤ 0. (2.40)

Thus (δ − w2

δ − u

)′
≤ 0.

Integrating on (tγ,b, t) gives
δ − w2

δ − u
− 1 ≤ 0

which implies u ≤ w2 when u,w2 ∈ [γ, δ).
A nearly identical argument proves that

w1 ≤ u when u,w1 ∈ [β′, δ)

and

(δ − w1)u′ − (δ − u)w′1 ≥ 0. (2.41)

This completes the proof. �

Now (2.35) can be solved explicitly and we obtain

w2 = δ +
√
t
(
c1I 1

2−q

(2
√
h1f1

2− q
t
2−q
2

)
+ c2K 1

2−q

(2
√
h1f1

2− q
t
2−q
2

))
(2.42)

where I 1
2−q

and K 1
2−q

are the modified Bessel functions of order 1
2−q and c1, c2 are

constants. It is well-known for t > 0 that: Iν > 0, I ′ν > 0, Kν > 0 and K ′ν < 0.
We rewrite (2.42) as

w2 − δ = c1y1 + c2y2

where

y1(t) =
√
tI 1

2−q

(2
√
h1f1

2− q
t
2−q
2

)
, y2(t) =

√
tK 1

2−q

(2
√
h1f1

2− q
t
2−q
2

)
. (2.43)

A straightforward computation shows

c1 =
y′2(tγ,b)(w2(tγ,b)− δ)− y2(tγ,b)w

′
2(tγ,b)

y1(tγ,b)y′2(tγ,b)− y′1(tγ,b)y2(tγ,b)
, (2.44)

c2 =
−y′1(tγ,b)(w2(tγ,b)− δ) + y1(tγ,b)w

′
2(tγ,b)

y1(tγ,b)y′2(tγ,b)− y′1(tγ,b)y2(tγ,b)
. (2.45)

Another well-known fact about the modified Bessel functions Iν and Kν is that

Iν(t)K ′ν(t)− I ′ν(t)Kν(t) = −1

t
for t > 0. (2.46)

Next a straightforward computation using (2.43) and (2.46) shows

y1(t)y′2(t)− y′1(t)y2(t) = −(1− q

2
).

And so we see from (2.36), (2.44)-(2.45) that

c1 =
y′2(tγ,b)(δ − γ) + y2(tγ,b)u

′(tγ,b)

1− q
2

, (2.47)



EJDE-2020/117 EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS 9

c2 =
−y′1(tγ,b)(δ − γ)− y1(tγ,b)u

′(tγ,b)

1− q
2

. (2.48)

Note that y1(t) > 0 and y′1(t) > 0. In addition, u′(tγ,b) > 0 and δ − γ > 0 so it
follows from (2.48) that

c2 < 0. (2.49)

Lemma 2.5. Assume (H1)–(H4) and let u solve (2.1), (2.4). If b > 0 is sufficiently
small and if

γ
(

1 +
(h2f̄0

h1f1

)1/2)
< δ (2.50)

then c1 < 0.

Proof. We let

r =
2
√
h1f1

2− q
t1−

q
2 , rγ,b =

2
√
h1f1

2− q
t
1− q2
γ,b . (2.51)

It follows from (2.43) and (2.24) that

c1 =
1

1− q
2

[
(δ − γ)

( 1

2
√
tγ,b

K 1
2−q

(rγ,b) +
√
h1f1t

1−q
2

γ,b K
′
1

2−q
(rγ,b)

)
+
√
tγ,bK 1

2−q
(rγ,b)u

′(tγ,b)
]
.

Therefore

c1 =
1

1− q
2

t
1−q
2

γ,b K 1
2−q

(rγ,b)
[
(δ − γ)

( 1

2t
1− q2
γ,b

+
√
h1f1

K ′ 1
2−q

(rγ,b)

K 1
2−q

(rγ,b)

)
+ t

q/2
γ,b u

′(tγ,b)
]
.

(2.52)

Another well-known fact about the modified Bessel function is that limt→∞
K′ν(t)
Kν(t) =

−1. We also know that tγ,b → ∞ as b → 0+ by Lemma 2.3 and thus by (2.51)
we see rγ,b →∞ as b→ 0+. Thus from Lemma 2.3, (2.16), (2.50), and taking the
limit superior of the bracketed term in (2.52) gives

lim sup
b→0+

[
(δ − γ)

( 1

2t
1− q2
γ,b

+
√
h1f1

K ′ 1
2−q

(rγ,b)

K 1
2−q

(rγ,b)

)
+ t

q/2
γ,b u

′(tγ,b)
]

≤ (δ − γ)(−
√
h1f1) + γ

√
h2f̄0 =

√
h1f1

[
γ
(

1 +

√
h2f̄0

h1f1

)
− δ
]
< 0.

It follows from this and (2.52) that c1 < 0. This completes the proof. �

Lemma 2.6. Assume (H1)–(H4) and let u solve (2.1), (2.4). Let n be a positive

integer. If γ
(
1 +

√
h2f̄0
h1f1

)
< δ and b > 0 is sufficiently small then u has n zeros on

(0,∞).

Proof. From Lemma 2.5 it follows that c1 < 0 if b > 0 is sufficiently small and
(2.50) holds. In addition, c2 < 0 by (2.49). Since Iν → ∞ as t → ∞ and Kν > 0
then we see from (2.42) that w2 < δ for all t > 0. Since c1 < 0 and Iν → ∞
as t → ∞ it follows from (2.42) that w2 → −∞ as t → ∞ so w2 must have
a local maximum, Mw2 , and that w2(Mw2) < δ. Since u ≤ w2 by Lemma2.4
it follows that u(t) ≤ w2(t) ≤ w2(Mw2

) < δ. This implies that u also has a



10 J. IAIA EJDE-2020/117

local maximum for otherwise u would be increasing and have a limit, L, with
γ < L < δ which is impossible by Lemma 2.2. Thus u has a local max, Mb, and
since F (u(Mb)) = E(Mb) > 0 we have β < γ < u(Mb) ≤ w2(Mb) ≤ w2(Mw2) < δ.
Then from (2.1) we see u is concave down while β < u < δ and so there exists
xb > Mb such that u(xb) = β and u′(xb) < 0. Next recall from (2.10) that
E(t) ≥ E(Mb) for t > Mb and so

1

2

u′2

h(t)
+ F (u) ≥ F (u(Mb)) for t > Mb. (2.53)

Now for t > xb we have F (u) ≤ 0 and so from (2.53) we have

1

2

u′2

h(t)
≥ F (u(Mb)) for t > xb.

Thus by (2.2),

−u′ ≥
√

2F (u(Mb))h(t) ≥
√

2h1F (u(Mb)) t
−q/2 for t > xb.

Integrating this on (xb, t) gives

−u(t) + β ≥
√

2h1F (u(Mb))

1− q
2

(
t1−

q
2 − x1− q2

b

)
→∞ as t→∞

and so u must be negative. Thus there exists z1,b > xb such that u(z1,b) = 0. In
addition, 1

2u
′2(z1,b) = E(z1,b) > 0 so u′(z1,b) < 0.

Further, u′(z1,b) → 0 as b → 0+. To see this, recall from (2.8) that E′0 =
h′(t)F (u) and so integrating this on (tγ,b, z1,b) gives

1

2
u′2(z1,b) =

1

2
u′2(tγ,b) +

∫ z1,b

tγ,b

h′(x)F (u(x)) dx

≤ 1

2
u′2(tγ,b) + F1[h(tγ,b)− h(z1,b)]

(2.54)

where |F (u)| ≤ F1 for some constant F1. (Recall from (H1) and (H2) that F is
bounded). Since tγ,b and z1,b go to infinity as b → 0+ by Lemma 2.3 we see by
(2.2) that the second term in (2.54) goes to 0 as b → 0+. Also from (2.16) we see
that u′(tγ,b)→ 0 as b→ 0+. Thus from (2.54) we see u′(z1,b)→ 0 as b→ 0+.

Next, let u1(t) = −u(t). Then since f(u) is odd we see that u1 also solves (2.1).
Further u1(z1,b) = 0, u′1(z1,b) = −u′(z1,b) > 0, and u′1(z1,b)→ 0 as b→ 0+.

Now we can define v̄2 with v̄2 solving (2.18) with v̄2(z1,b) = 0, v̄′2(z1,b) = u′1(z1,b) >
0 and as in Lemma 2.1 there exists t̄γ,b > z1,b such that v̄2(t̄γ,b) = γ. As in Lemma
2.3 we can show that

u′1
u1
≤ v̄′2
v̄2
. (2.55)

We again can solve for v̄2 explicitly and see that

v̄2 = c̄1ȳ1 + c̄2ȳ2 (2.56)

where ȳ1 =
√
tI 1

2−q
(s) and ȳ2 =

√
tK 1

2−q
(s) and:

s =
2
√
h2f̄0

2− q
t
2−q
2 with sγ,b =

2
√
h2f̄0

2− q
t
2−q
2

γ,b .

Then

tq/2v̄′2 = c̄1t
q/2ȳ′1 + c̄2t

q/2ȳ′2.



EJDE-2020/117 EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS 11

As in Lemma 2.3 and with the facts that
I′ν
Iν
→ 1 and

K′ν
Kν
→ −1 as t→∞ then

lim
b→0+

t̄
q/2
γ,b ȳ

′
1(t̄γ,b)

ȳ1(t̄γ,b)
=

√
h2f̄0, (2.57)

lim
b→0+

t̄
q/2
γ,b ȳ

′
2(t̄γ,b)

ȳ2(t̄γ,b)
= −

√
h2f̄0. (2.58)

Thus from (2.56),

t̄
q/2
γ,b v̄

′
2(t̄γ,b)

v̄2(t̄γ,b)
=
c̄1t̄

q/2
γ,b ȳ

′
1(t̄γ,b) + c̄2t̄

q/2
γ,b ȳ

′
2(t̄γ,b)

c̄1ȳ1(t̄γ,b) + c̄2ȳ2(t̄γ,b)

=
c̄1
t̄
q/2
γ,b ȳ

′
1(t̄γ,b)

ȳ1(t̄γ,b)
+ c̄2

t̄
q/2
γ,b ȳ

′
2(t̄γ,b)

ȳ1(t̄γ,b)

c̄1 + c̄2
ȳ2(t̄γ,b)
ȳ1(t̄γ,b)

.

(2.59)

We note that c̄1 6= 0 for sufficiently small b > 0 for if so then

t̄
q/2
γ,b v̄

′
2(t̄γ,b)

v̄2(t̄γ,b)
=
t̄
q/2
γ,b ȳ

′
2(t̄γ,b)

ȳ2(t̄γ,b)

for sufficiently small b > 0 but the right-hand side goes to −
√
h2f̄0 < 0 while the

left-hand side is positive.
Since ȳ2 → 0, ȳ′2 → 0 and ȳ1 → ∞ as t → ∞ it follows from (2.57)-(2.59) that

t̄
q/2
γ,b v̄

′
2(t̄γ,b)

v̄2(t̄γ,b)
goes to

√
h2f̄0 as b→ 0+ and so by (2.55) we see that

lim sup
b→0

t̄
q/2
γ,b u

′
1(t̄γ,b) ≤ γ

√
h2f̄0.

As in Lemmas 2.4 and 2.6 it is then possible to show if b is sufficiently small and

γ
(
1+
√

h2f̄0
h1f1

)
< δ then u1 will have a zero and hence u will have a second zero, z2,b.

Continuing in this way we see that if b > 0 is sufficiently small and γ
(
1+
√

h2f̄0
h1f1

)
< δ

then u will have n zeros for any given integer n. This completes the proof. �

Lemma 2.7. Assume (H1)–(H4) and let u solve (2.1), (2.4). If

β′ +
β

2

h1

h2

(f0

f̄1

)1/2

> δ (2.60)

then u(t) > 0 for t > 0.

Proof. Since E is nondecreasing,

1

2

u′2(tb′)

h(tb′)
+ F (β/2) = E(tb′) ≥ E(t2,b) =

1

2

u′2(t2,b)

h(t2,b)
+ F (β/2)

thus by (2.2) and (2.15),

lim inf
b→0+

t
q/2
b′ u

′(tb′) ≥ lim inf
b→0+

√
h1

h2
t
q/2
2,b u

′(t2,b) ≥
√
h1

h2

√
h1f0

β

2
= h1

β

2

√
f0

h2
. (2.61)

Now (2.37) can be solved explicitly and we obtain

w1 = δ +
√
t
(
ĉ1I 1

2−q

(2
√
h2f̄1

2− q
t
2−q
2

)
+ ĉ2K 1

2−q

(2
√
h2f̄1

2− q
t
2−q
2

))
(2.62)



12 J. IAIA EJDE-2020/117

where I 1
2−q

and K 1
2−q

are the modified Bessel functions of order 1
2−q and ĉ1, ĉ2 are

constants. We rewrite this as

w1 − δ = ĉ1ŷ1 + ĉ2ŷ2 (2.63)

where

ŷ1(t) =
√
tI 1

2−q

(2
√
h2f̄1

2− q
t
2−q
2

)
, ŷ2(t) =

√
tK 1

2−q

(2
√
h2f̄1

2− q
t
2−q
2

)
. (2.64)

Again we see as in (2.44)-(2.45),

ĉ1 =
ŷ′2(tb′)(w1(tb′)− δ)− ŷ2(tb′)w

′
1(tb′)

ŷ1(tb′)ŷ′2(tb′)− ŷ′1(tb′)ŷ2(tb′)
, (2.65)

ĉ2 =
−ŷ′1(tb′)(w1(tb′)− δ) + ŷ1(tb′)w

′
1(tb′)

ŷ1(tb′)ŷ′2(tb′)− ŷ′1(tb′)ŷ2(tb′)
. (2.66)

So we see from (2.46) and (2.64) that

ŷ1(t)ŷ′2(t)− ŷ′1(t)ŷ2(t) = −(1− q

2
).

Then we see from (2.65)-(2.66) that

ĉ1 =
ŷ′2(tb′)(δ − β′) + ŷ2(tb′)u

′(tb′)

1− q
2

, (2.67)

ĉ2 =
−ŷ′1(tb′)(δ − β′)− ŷ1(tb′)u

′(tb′)

1− q
2

. (2.68)

Note that ŷ1(t) > 0 and ŷ′1(t) > 0. In addition, u′(tb′) > 0 and δ − β′ > 0 so it
follows that

ĉ2 < 0. (2.69)

Also

ĉ1 =
1

1− q
2

t
1−q
2

b′ K 1
2−q

(rb′)
[
(δ − β′)

( 1

2t
1− q2
b′

+
√
h2f1

K ′ 1
2−q

(rb′)

K 1
2−q

(rb′)

)
+ t

q/2
b′ u

′(tb′)
]
,

(2.70)

with rb′ =
2

2− q

√
h2f̄1 tb′

1− q2 . (2.71)

We show in the appendix that(K ′ν
Kν

+
ν

t

)
> −1 for t > 0 and ν >

1

2
. (2.72)

Now here we have ν = 1
2−q >

1
2 since q > 0 thus using (2.60) and (2.61) we obtain

in the bracketed term in (2.70),

(δ − β′)
( 1

2t
1− q2
b′

+
√
h1f1

K ′ 1
2−q

(rb′)

K 1
2−q

(rb′)

)
+ t

q/2
b′ u

′(tb′)

≥ (δ − β′)(−
√
h2f̄1) + h1

β

2

( f0

h2

)1/2

=

√
h2f̄1

[
− (δ − β′) +

β

2

h1

h2

(f0

f̄1

)1/2]
> 0.

(2.73)

It follows from this that ĉ1 > 0.



EJDE-2020/117 EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS 13

Now recall from (2.63) that w1 = δ+ĉ1ŷ1+ĉ2ŷ2 and w1(tb′) = β′ < δ, w′1(tb′) > 0.
It follows from (2.37) that w1 is concave up when w1 > δ1 and w1 is concave down
when w1 < δ1. Since ĉ1 > 0, ĉ2 < 0, ŷ1 → ∞ as t → ∞, and ŷ2 → 0 as t → ∞ it
follows therefore that it must be the case that w1 → ∞ as t → ∞ and thus there
exists td > tb′ with w1(td) = δ and w1 ≥ δ for t ≥ td. By Lemma 2.4 it follows that
there exists tδ < td such that u(tδ) = δ and u ≥ δ for t > tδ. It also follows from
Lemma 2.4 that u ≥ w1 > 0 for tb′ ≤ t ≤ tδ. From Lemma 2.1 we know u > 0 on
(0, tγ,b) and since tb′ < tγ,b it follows that u(t) > 0 for t > 0. This completes the
proof. �

3. Proof of Theorem 1.1

Proof. For the proof of part (a), from Lemma 2.6 we see that if R > 0 is sufficiently
small then R2−N is very large and so z1,b < R2−N . We also know that tγ,b → ∞
as b → 0+ and since z1,b > tγ,b it follows that u(t) > 0 on (0, R2−N ) if b > 0 is
sufficiently small. Thus by continuity with respect to initial conditions it follows
that there is b0 > 0 such that u(R2−N ) = 0. Thus we obtain a positive solution,

u0, of (2.1), (2.4) if R > 0 is sufficiently small and if γ
(
1 +

√
h2f̄0
h1f1

)
< δ. Similarly

if R > 0 is sufficiently small then z2,b < R2−N and if b > 0 is sufficiently small
then z2,b > R2−N . Then by continuity there exists a b1 such that u1(R2−N ) = 0.
Thus u1 is a solution with exactly one zero on (0, R2−N ). Continuing in this way
we see that if R is sufficiently small then there exists u0, u1, . . . , un such that uk
has k zeros on (0, R2−N ) and uk(R2−N ) = 0. This completes the proof part (a).

The proof of part (b) follows immediately from Lemma 2.7.
A proof of part(c) c can be found in [10] but we include it here for com-

pleteness. Suppose there is a solution of (1.4)-(1.5) such that limr→∞ u = 0.

Then a straightforward computation shows if E2(r) = 1
2
u′2

K + F (u) then E′2 =

−u′2

2K

(
2(N−1)+ rK′

K

)
≤ 0 for r ≥ R. Now if limr→∞ u = 0 it follows that E2(r) > 0

for r ≥ R. Now u cannot have an infinite number of extrema, Mk, with Mk → ∞
because if so F (u(Mk)) = E2(Mk) > 0 so |u(Mk)| > γ contradicting that u(r)→ 0
as r →∞. Also there could not be an infinite number of extrema with Mk ≤ L <∞
for if so then for some subsequence Mk → M and there would exist sk → M such

that |u′(sk)| → ∞ contradicting that 1
2
u′2

K − F0 ≤ E(r) ≤ E(R) = 1
2

a2

K(R) which

implies u′ is bounded on [R,M ]. Thus we see that u must have a largest extremum,
M , and without loss of generality let us suppose that M > R is a local maximum
and u′ < 0 for r > M . Then

1

2

u′2

K(r)
+ F (u) ≤ F (u(M)) for r > M.

Rewriting and integrating on (M,∞) using that α > 2 (from (H3)) gives∫ u(M)

0

dt√
2
√
F (u(M))− F (t)

=

∫ ∞
M

−u′(r) dr√
2
√
F (u(M))− F (u(r))

≤
∫ ∞
M

√
K dr

≤
√
k2M

1−α2
α
2 − 1

≤
√
k2R

1−α2
α
2 − 1

.

(3.1)



14 J. IAIA EJDE-2020/117

From (H2) we see that F is bounded below so there exists F0 > 0 such that
F (u) ≥ −F0 for all u. Also, u(M) > γ and F (u(M)) < F (δ) therefore we see that∫ u(M)

0

dt√
2
√
F (u(M))− F (t)

≥ γ√
2
√
F (δ) + F0

. (3.2)

Combining (3.1) and (3.2) gives

γ√
2
√
F (δ) + F0

≤
√
k2R

1−α2
α
2 − 1

. (3.3)

The right-hand side of (3.3) goes to zero as R→∞ which contradicts (3.3) if R > 0
is too large. Thus there are no solutions of (1.1)-(1.3) if R > 0 is sufficiently large.
This completes the proof of part (c). �

4. Appendix - Facts about modified Bessel functions

In this section we collect some facts about modified Bessel functions. There are
numerous texts which contain these results such as [4].

The modified Bessel functions Iν and Kν are linearly independent solutions of

y′′ +
1

t
y′ −

(
1 +

ν2

t2
)
y = 0 for t > 0, ν > 0 (4.1)

for which limt→0+ Iν(t) = 0 and limt→0+ Kν(t) =∞. They are normalized so that

lim
t→0+

Iν(t)

tν
=

1

2νΓ(ν + 1)
, lim

t→0+

Kν(t)

t−ν
= 2ν−1Γ(ν).

It can in fact be shown that

Iν(t) = tν
∞∑
n=0

ant
n, Kν(t) = t−ν

∞∑
n=0

bnt
n

for appropriate constants an, bn.
In addition it is known that Iν(t) > 0, Kν(t) > 0, I ′ν(t) > 0 and K ′ν(t) < 0 for

t > 0 and also Iν(t) ∼ et√
t
, Kν(t) ∼ e−t√

t
for large t.

It is also known that

lim
t→∞

I ′ν
Iν

= 1, lim
t→∞

K ′ν
Kν

= −1.

Another well-known fact is that

Iν(t)K ′ν(t)− I ′ν(t)Kν(t) = −1

t
for t > 0. (4.2)

In addition (K ′ν
Kν

+
ν

t

)
> −1 if ν >

1

2
, t > 0;(I ′ν

Iν
+
ν

t

)
> 1 if ν >

1

2
, t > 0.

We prove these last two facts.

Proof. First
( I′ν
Iν

+ ν
t

)
> 0 and limt→∞

( I′ν
Iν

+ ν
t

)
= 1. From (4.1) we see that

I ′′ν
Iν

+
1

t

(I ′ν
Iν

)
= 1 +

ν2

t2
.



EJDE-2020/117 EXISTENCE AND NONEXISTENCE OF RADIAL SOLUTIONS 15

Next, (I ′ν
Iν

+
ν

t

)′
=
I ′′ν
Iν
−
(I ′ν
Iν

)2

− ν

t2
.

Combining these gives(I ′ν
Iν

+
ν

t

)′
+
(I ′ν
Iν

)2

+
1

t

I ′ν
Iν

= 1 +
ν2 − ν
t2

.

Therefore, (I ′ν
Iν

+
ν

t

)′
+
(I ′ν
Iν

+
1

2t

)2

= 1 +
(ν − 1

2 )2

t2
.

And (I ′ν
Iν

+
ν

t

)′′
+ 2
(I ′ν
Iν

+
1

2t

)((I ′ν
Iν

)′
− 1

2t2

)
=
−2(ν − 1

2 )2

t3
. (4.3)

Now suppose
( I′ν
Iν

+ ν
t

)
has a local minimum for t > 0. Then

( I′ν
Iν

+ ν
t

)′
= 0 and( I′ν

Iν
+ ν

t

)′′ ≥ 0. Substituting into (4.3) gives( ν
t2

+
1

2t

) (ν − 1
2 )

t2
≤
−2(ν − 1

2 )2

t3

which is impossible since ν > 1
2 . Thus

( I′ν
Iν

+ ν
t

)
does not have a local minimum.

Since

lim
t→0+

(I ′ν
Iν

+
ν

t

)
=∞

it follows that
( I′ν
Iν

+ ν
t

)
is a decreasing function and since limt→∞

( I′ν
Iν

+ ν
t

)
= 1 it

follows that
( I′ν
Iν

+ ν
t

)
> 1 for t > 0.

Similarly,
(K′ν
Kν

+ ν
t

)
does not have a local minimum for ν > 1/2. We also know

lim
t→∞

(K ′ν
Kν

+
ν

t

)
= −1.

Thus
(K′ν
Kν

+ ν
t

)
> −1 for t > 0 and ν > 1/2. �

References

[1] H. Berestycki, P.L. Lions; Non-linear scalar field equations I, Arch. Rational Mech. Anal.,

Volume 82, 313-347, 1983.
[2] H. Berestycki, P.L. Lions; Non-linear scalar field equations II, Arch. Rational Mech. Anal.,

Volume 82, 347-375, 1983.

[3] M. Berger; Nonlinearity and functional analysis Academic Free Press, New York, 1977.
[4] G. Birkhoff, G. C. Rota; Ordinary Differential Equations, Ginn and Co.,. Boston, 1962.

[5] 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.

[6] J. Iaia, H. Warchall, F. B. Weissler; Localized solutions of sublinear elliptic equations: loiter-

ing at the hilltop, Rocky Mountain Journal of Mathematics, Volume 27, Number 4, 1131-1157,
1997.

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

[8] J. Iaia; Existence and nonexistence for semilinear equations on exterior domains, Journal of

Partial Differential Equations, Vol. 30, No. 4, 1-17, 2017.
[9] J. Iaia; ‘Existence of solutions for semilinear problems on exterior domains, Electronic Journal

of Differential Equations, Vol. 2020, No. 34, 1-10, 2020.



16 J. IAIA EJDE-2020/117

[10] J. Iaia; Existence and nonexistence of solutions for sublinear equations on exterior domains,

Electronic Journal of Differential Equations, Vol 2017, No. 214, 1-13, 2017.

[11] 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.

[12] 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.
[13] 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.

[14] 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.

[15] W. Strauss; Existence of solitary waves in higher dimensions, Comm. Math. Phys., Volume

55, 149-162, 1977.

Joseph A. Iaia

Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX

76203-5017, USA
Email address: iaia@unt.edu


	1. Introduction
	2. Preliminaries
	3. Proof of Theorem ??
	4. Appendix - Facts about modified Bessel functions
	References