Special Issue in honor of John W. Neuberger

Electronic Journal of Differential Equations, Special Issue 02 (2023), pp. 231–238.

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

EXISTENCE OF POSITIVE GLOBAL RADIAL SOLUTIONS TO

NONLINEAR ELLIPTIC SYSTEMS

SESHADEV PADHI, JULIO G. DIX

Dedicated to the memory of John W. Nueberger

Abstract. In this article we obtain global positive and radially symmetric
solutions to the system of nonlinear elliptic equations

div
(
φj(|∇u|)∇u

)
+ aj(x)φj(|∇u|)|∇u| = pj(x)fj(u1(x), . . . , uk(x)) ,

and in particular to Laplace’s equation

∆uj(x) = pj(x)fj(u1(x), . . . , uk(x)) ,

where j = 1, . . . , k, x ∈ RN , N ≥ 3, ∆ is the Laplacian operator, and ∇ is

the gradient. Also we state conditions for solutions to be bounded, and to be
unbounded. With an example we illustrate our results.

1. Introduction

In this article we study the existence and asymptotic behavior of positive radial
solutions to the system

div
(
φj(|∇u|)∇u

)
+ aj(x)φj(|∇u|)|∇u| = pj(x)fj(u1(x), . . . , uk(x)), (1.1)

and in particular to the system

∆uj(x) = pj(x)fj(u1(x), . . . , uk(x)) (1.2)

for x ∈ RN , N ≥ 3, and j = 1, . . . , k. Here ∆ is the Laplacian operator, ∇
is the gradient, aj , pj are radially symmetric functions, and φj is a continuously
differentiable function. Let r = |x| be the Euclidean norm of x in RN . We use the
same symbol to indicate a radial function in terms of x ∈ RN and in terms of r.

We prove that (1.1) has positive solutions which are global and radially sym-
metric. See Theorem 2.4 below. By doing this, we extend the existing results from
2 equations to k equations. The main difficulty is finding the proper function h
that allows us bounding the iterated solutions, see inequality (2.5) and hypothesis
(A5). Also we provide a small improvement in the solution estimates which allows
us stating conditions in a much simpler way than in the references; see Remarks
2.3 and 2.7 below.

2020 Mathematics Subject Classification. 35J25, 45F10, 35B08.

Key words and phrases. Radial solution; elliptic system.
©2023 This work is licensed under a CC BY 4.0 license.
Published March 27, 2023.

231



232 S. PADHI, J. G. DIX EJDE/SI/02

The motivation for this article comes the following references: Lair [8, 7] consid-
ered the system

∆u = p1(|x|)vα

∆v = p2(|x|)uβ ,

for x ∈ RN . Li, Zhang, and Zhang [9], and Zhang [14] studied the system

∆u = p1(|x|)f(v)

∆v = p2(|x|)g(u) ,

for x ∈ RN . Covei [2] considered the system

∆u = p1(|x|)f(u, v)

∆v = p2(|x|)g(u) ,
(1.3)

for x ∈ RN . Othman and Chemman [10] used a fixed point approach by assuming
the existence of subsolution and supersolution to study the existence of a large
solution to the system ∆pu = f1(x, u, v), ∆qv = f2(x, u, v). This is done in a smooth
bounded domain Ω of RN , N ≥ 2, u, v > 0 in Ω, u|∂Ω = +∞ and v|∂Ω = +∞, where
1 < p, q <∞ and ∆tw = div(|∇w|t−2∇w) for any 1 < t <∞. In an another work,
Alves and de Holanda [1] studied the system ∆u = Fu(x, u, v), ∆v = Fv(x, u, v)
from a variational point of view. However this approach does not apply to (1.2).
Zhou [15] considered the system

div
(
φ1(|∇u|)|∇u|

)
+ a1(|x|)φ1(|∇u|)|∇u| = p1(|x|)f1(u, v)

div
(
φ2(|∇v|)|∇v|

)
+ a2(|x|)φ2(|∇v|)|∇v| = p2(|x|)f2(u, v) ,

for x ∈ RN . By using a monotone iterative technique and the Arzela-Ascoli the-
orem, Yang et al. [13], studied the positive entire bounded radial solutions of the
Schrödinger elliptic system

div(G(|∇y|p−2)∇y) = b1(|x|)ψ(y) + h1(|x|)ϕ(z), x ∈ Rn (n ≥ 3),

div(G(|∇z|p−2)∇z) = b2(|x|)ψ(z) + h2(|x|)ϕ(y), x ∈ Rn,

where G is a nonlinear operator. Garćıa-Melián [4] studied the system

∆u = p1(|x|)uαvβ

∆v = p2(|x|)uγvη,

for x ∈ Ω ⊂ RN . For more results on elliptic boundary value problems see [3, 5, 6,
9, 12] and the references therein.

2. Results

To obtain a solution to (1.1), we build a sequence of functions, and then show that
the sequence converges to a solution. First using the integrating factor rN−1µj(r),
we obtain a radial version of (1.1),(

rN−1µj(r)φj
(
u′j(r)

)
u′j(r)

)′
= rN−1µj(r)pj(r)fj(u1(r), . . . , uk(r)),

uj(0) = αj ≥ 0, u′j(0) = 0, j = 1, . . . , k,
(2.1)

where

µj(r) = exp
(∫ r

0

aj(s) ds
)
.



EJDE-2023/SI/02 NONLINEAR ELLIPTIC SYSTEMS 233

To study this problem we use the following assumptions:

(A1) pj : [0,∞)→ [0,∞) are continuous and radially symmetric functions. The
initial values satisfy αj ≥ 0 with αj > 0 for at least one index j.

(A2) fj : [0,∞)k → [0,∞) are continuous and non-decreasing with respect to
each one of their arguments.

(A3) φj ∈ C1([0,∞), [0,∞)) for each j, and

ψj(r) := rφj(r) satisfies ψ′j(r) > 0 for r > 0 .

(A4) There exist positive constants B̃1, B̃2 and β2 ≥ β1 ≥ 1 such that B̃1t
β1 ≤

ψ(t) ≤ B̃2t
β2 for t > 0. By a contradiction argument we can show that

that there are positive constants B1 and B2 such that

B2y
1/β2 ≤ ψ−1

j (y) ≤ B1y
1/β1 for y > 0, j = 1, . . . , k .

(A5) There exists a continuous function h : [0,∞)→ [0,∞) such that

fj(u1, . . . , uk) ≤ h(u1 + · · ·+ uk), for 1 ≤ j ≤ k ;

for the β1 in (A4), ∫ ∞
1

1

h1/β1(t)
dt =∞ ;

and for each positive t0, there is positive constant h0, such that h0 ≤ h(t)
for all t ≥ t0.

Examples of functions satisfying (A3) include φj(t) = 1 which yields the Laplacian
operator, and φj(t) = tp−2 which yields the p-Laplacian operator. See more ex-
amples in [15]. We do not need to define φj for negative values because we show

later that u′j,n ≥ 0. Assumption (A4) was stated as β1 ≤ tψ′(t)
ψ(t) ≤ β2 in [15]. As an

example of a functions satisfying (A2) and (A5) we have fj(u1, . . . , uk) = uδ11 · · ·u
δk
k

with max{δi : 1 ≤ i ≤ k} ≤ β1, and h(t) = max
{

1,
(∑k

i ui
)max{δi:1≤i≤k}}

.
We define a sequence of functions converging to a solution of (2.1) as follows.

First we integrate (2.1) to obtain

u′j(r) = ψ−1
j

(r1−N

µj(r)

∫ r

0

sN−1µj(s)pj(s)fj
(
u1(s), . . . , uk(s)

)
ds
)
,

uj(r) = αj +

∫ r

0

ψ−1
j

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s)fj
(
u1(s), . . . , uk(s)

)
ds
)
dt .

(2.2)

For n = 0, we define u1,0 = α1, u2,0 = α2, . . . , uk,0 = αk. And for n ≥ 1, we define

uj,n(r) = αj+

∫ r

0

ψ−1
j

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s)fj
(
u1,n−1(s), . . . , uk,n−1(s)

)
ds
)
dt .

(2.3)
We use a simultaneous iteration. To compute uj,n we use uj,n−1 for j = 1 . . . , k
(all uj,n−1 are replaced by uj,n at the same time). However it is possible to use
successive iterations: the uj,n are used as they become available. Compute uj,n in
ascending order of j, and for computing uj+1,n use uj,n, instead of uj,n−1. This
technique called the Gauss-Seidel method when doing numerical approximations.

Lemma 2.1. Under assumptions (A1)–(A3), the sequence of functions {uj,n} is
non-decreasing with respect to n, and each function is non-decreasing.



234 S. PADHI, J. G. DIX EJDE/SI/02

Proof. To show that the sequence is non-decreasing we use induction on n. Since
fj , pj , µj , and ψj are non-negative, we have

uj,0(r) ≤ αj +

∫ r

0

ψ−1
j

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s)fj(α1, . . . , αk) ds
)
dt = uj,1(r)

for all r ≥ 0, which is the base step for induction. Now we assume that uj,n−1(s) ≤
uj,n(s) for s ≥ 0. As fj is non-decreasing in each argument, ψj is increasing, and
µj > 0, we have

uj,n(r) = αj +

∫ r

0

ψ−1
j

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s)fj
(
u1,n−1(s), . . . , uk,n−1(s)

)
ds
)
dt

≤ αj +

∫ r

0

ψ−1
j

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s)fj
(
u1,n(s), . . . , uk,n(s)

)
ds
)
dt ,

for all r ≥ 0, which completes the induction step.
To show that these functions are non-decreasing, we use

u′j,n(r) = ψ−1
j

(r1−N

µj(r)

∫ r

0

sN−1µj(s)pj(s)fj
(
u1,n(s), . . . , uk,n(s)

)
ds
)
≥ 0 ,

which indicates that uj,n(r) is non-decreasing with respect to r. �

Lemma 2.2. Under assumptions (A1)–(A5), the sequence {uj,n(r)} is uniformly
bounded on each interval 0 ≤ r ≤ r1.

Proof. Using (2.2), (A4), and that h(
∑
αi) ≥ fj(α1, . . . , αk), we have

u′j(r) ≤ ψ−1
j

(
h
(
u1,n(r) + · · ·+ uk,n(r)

)r1−N

µj(r)

∫ r

0

sN−1µj(s)pj(s) ds
)

≤ B1

(
h
(
u1,n(r) + · · ·+ uk,n(r)

)r1−N

µj(r)

∫ r

0

sN−1µj(s)pj(s) ds
)1/β1

.

(2.4)

Dividing by B1h
1/β1 , and summing over j, we have

1

B1h1/β1
(
u1,n(r) + · · ·+ uk,n(r)

) k∑
j=1

u′j,n(r)

≤
k∑
j=1

(r1−N

µj(r)

∫ r

0

sN−1µj(s)pj(s) ds
)1/β1

.

(2.5)

Integrating from r = 0 to r = r1 yields∫ ∑
ui,n(r1)

∑
αi

1

B1h1/β1(t)
dt ≤

k∑
j=1

∫ r

0

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s) ds
)1/β1

dt . (2.6)

Note that the right-hand side is independent of n, and finite as long as r1 < ∞.
Recall that by (A5), h(

∑
αi) is bounded below by a positive constant because

αi > 0 at least one index i. Then we can find a constant M(r1) such that the upper
limit of integration satisfies

k∑
j=1

uj,n(r1) ≤M(r1), ∀n ≥ 1 . (2.7)



EJDE-2023/SI/02 NONLINEAR ELLIPTIC SYSTEMS 235

Since 0 ≤ αj = uj,0(r) ≤ uj,n(r) ≤ uj,n(r1) for all r ∈ [0, r1], the sequence {uj,n(·)}
is uniformly bounded for 1 ≤ j ≤ k and 1 ≤ n on [0, r1]. �

Remark 2.3. In [2] and [11], the quantity ui,n(s) in the integrand in (2.2) is
substituted using the inequality

ui,n(r) ≤ αi +

∫ r

0

s1−N
∫ t

0

sN−1pi(s)fi
(
u1,n(s), . . . , uk,n(s)

)
ds dt

≤Mifi(u1,n(r), . . . , uk,n(r))
(

1 +

∫ r

0

s1−N
∫ t

0

sN−1pi(s)
)
,

where Mi = max{1, αi/fi(α1, . . . , αk)}. This inequality comes from Lemma 2.1
with ψj(r) = r, and µj = 1. This substitution does not improve estimate (2.2); on
the contrary it makes the estimate less accurate. By avoiding this substitution, our
estimates yield a small improvement and make our estimates simpler than theirs.
Also we use the upper bound h in (A5), while [2] and [11] used the estimate

fj
(
u1,n, . . . , uk,n

)
≤ fj

(
u1,n + · · ·+ uk,n, . . . , u1,n + · · ·+ uk,n

)
,

which can be larger than the h(u1,n + · · · + uk,n) used in (A5). Therefore, our
estimates provide another small improvement.

Theorem 2.4. Under assumptions (A1)–(A5) there is a positive solution to (2.1)
for all r ≥ 0, and hence a global radially symmetric solution to (1.1).

Proof. First we show that {uj,n} is equi-continuous, by finding a uniform bound for
{u′j,n(·)} on an interval [0, r1]. From the continuity of fj , pj , and ψj , we have that
the bound for {uj,n} in Lemma 2.2 provides a bound for the right-hand side of (2.4).
Therefore, {u′j,n(·)} is uniformly bounded on [0, r1]. By (1.1) and the continuity of

h, from (2.5), there exists a constant M̃(r1) such that 0 ≤ u′j,n(r) ≤ M̃(r1) for all
r in [0, r1]; i.e., u′j,n is uniformly bounded on [0, r1]. For x, y ∈ [0, r1], by the mean
value theorem,

|uj,n(y)− uj,n(x)| ≤ M̃(r1)|y − x| for 1 ≤ j ≤ k, 1 ≤ n .

Given ε > 0, we select δ ≤ ε/M̃(r1). If |x − y| < δ, then |uj,n(y) − uj,n(x)| < ε
which shows the equi-continuity of uj,n for 1 ≤ j ≤ k and 1 ≤ n.

Then by the Arzela-Ascoli theorem, there exists a subsequence of {uj,n} that
converges uniformly to a function uj . Since {uj,n} is non-decreasing in n, the
whole sequence converges uniformly to uj . Therefore,

uj(r) = αj +

∫ r

0

ψ−1
j

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s)fj
(
u1(s), . . . , uk(s)

)
ds
)
dt .

which provides a solution to (2.2) on [0, r1]. Noting that r1 can be arbitrarily large,
we complete the proof. �

Regarding the asymptotic behavior of solutions we have the following result.

Theorem 2.5. The solution obtained in Theorem 2.4 satisfies the following: (1) if

k∑
j=1

∫ ∞
a

( t1−N
µj(t)

∫ t

a

sN−1µj(s)pj(s) ds
)1/β1

dt <∞ for some a > 0 , (2.8)

then limr→∞ uj(r) <∞ for each j ∈ {1, . . . , k}.



236 S. PADHI, J. G. DIX EJDE/SI/02

(2) If for an index j,
∫∞

0

(
t1−N

µj(t)

∫ t
0
sN−1µj(s)pj(s) ds

)1/β1

dt < ∞, and fj(. . . )

is bounded, then limr→∞ uj(r) <∞.
(3) If for an index j,∫ ∞

a

( t1−N
µj(t)

∫ t

a

sN−1µj(s)pj(s) ds
)1/β2

dt =∞ , (2.9)

then limt→∞ uj(t) =∞.

Proof. (1) Under assumption (2.8), the right-hand side of (2.6) remains bounded
when r →∞, so the bound in (2.7) can be made independent of n and r1. That is
there exists a constant M such that

k∑
j=1

uj(r) ≤M, ∀r ≥ 0 .

Using that uj(·) is non-decreasing, we have the result in part (1).
(2) From the assumptions, we define two bounds: fj(. . . ) ≤ Dj and∫ r

0

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s) ds
)1/β1

dr ≤ D̃j , r ≥ 0 .

Then from (2.2), (A2), and (A4), we have

uj(r) ≤ αj +B1

(
Dj

)1/β1
D̃j for r ≥ 0 .

Then (2) follows from uj begin non-decreasing.
(3) From (2.2), (A2), and (A4), we have

uj(r) ≥ αj +B2

(
fj(α1, . . . , αk)

)1/β2

∫ r

0

( t1−N
µj(t)

∫ t

0

sN−1µj(s)pj(s) ds
)1/β2

dt .

Because at least one αi is positive, fj(α1, . . . , αk) > 0. Then the right-hand side
increases to infinity as r →∞, and conclusion (3) follows. �

Remark 2.6. The results from Theorems 2.4 and 2.5 apply to (1.2), by setting
φj(r) = 1, aj = 0, and B1 = B2 = β1 = β2 = 1.

Remark 2.7. Conditions P̄ (∞) <∞ and Q̄(∞) <∞ on [2, page 88] are equivalent
to (2.8) with j = 1, 2, ψ(r) = r, µj = 1 and β1 = 1. Also condition [11, ineq. (22)]
is equivalent to (2.8). However our assumption (2.8) is much easier to verify than
theirs.

Example 2.8. Consider the system(
rN−1r0.1

(
u′1(r)

)3)′
= rN−1r0.1 1

1 + r2.1
u1(r)u2(r), r ≥ 0,(

rN−1r0.2
(
u′2(r)

)3)′
= rN−1r0.2 1

1 + r2.2

(
u1(r)u2(r)

)1/2
, r ≥ 0,

u1(0) = 1, u2(0) = 0, u′1(0) = u′2(0) = 0 .

(2.10)

First we check that the assumptions in Theorem 2.5 are satisfied. Certainly func-
tions p1(r) = 1

1+r2.1 and p2(r) = 1
1+r2.2 satisfy (A1). Also f1(u1, u2) = u1u2 and

f1(u1, u2) = (u1u2)1/2 satisfy (A2).
To check (A3), we use φ1(r) = φ2(r) = r2 so ψ1(r) = ψ2(r) = r3, and ψ′1(r) =

3r2 > 0 for r > 0.



EJDE-2023/SI/02 NONLINEAR ELLIPTIC SYSTEMS 237

To check (A4), we have ψ−1
1 (y) = y1/3, thus y1/3 ≤ ψ−1

1 (y) ≤ y1/3 and B1 =
B2 = 1, β1 = β2 = 3.

Now we check (A5). Let us set h(t) = max{1, t2}. Then

f1(u1, u2) = u1u2 ≤ (u1 + u2)2 ≤ max{1, (u1 + u2)2} = h(u1 + u2)

and

f2(u1, u2) = (u1u2)1/2 ≤
(
(u1 + u2)2

)1/2
= u1 + u2

≤ max{1, (u1 + u2)2} = h(u1 + u2) .

Moreover, ∫ ∞
1

1

h1/β1(t)
dt =

∫ ∞
1

1

t2/3
dt =∞

and (A5) is satisfied.
Now we check (2.8). For j = 1, the inner integral is∫ r

a

sN−1s0.1 1

1 + s2.1
ds ≤

∫ r

0

sN−1s0.1 1

s2.1
ds =

1

N − 2
rN−2 .

While the outer integral in (2.8) is∫ ∞
a

r1−N 1

r0.1

1

N − 2
rN−2 dr =

1

N − 2

∫ ∞
a

1

r1.1
dr <∞.

For j = 2, the inner integral in (2.8) is∫ r

a

sN−1s0.2 1

1 + s2.2
ds ≤

∫ r

0

sN−1s0.2 1

s2.2
ds =

1

N − 2
rN−2 .

While the outer integral in (2.8) is∫ ∞
a

r1−N 1

r0.2

1

N − 2
rN−2 dr =

1

N − 2

∫ ∞
a

1

r1.2
dr <∞.

So (2.8) is satisfied. Therefore, by Theorems 2.4 and 2.5, there is a global solution
for which both u1 and u2 remain bounded as r →∞.

Acknowledgments. We want to thank the anonymous referee for checking the
original manuscript and for giving us valuable suggestions.

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and systems, Electronic Journal of Qualitative Theory of Differential Equations 2016 (2016),

no. 38, 1–10.

Seshadev Padhi
Department of Mathematics, Birla Institute of Technology, Mesra Ranchi, India

Email address: spadhi@bitmesra.ac.in

Julio G. Dix

Department of Mathematics, Texas State University, 601 University Drive, San Marcos,

TX 78666, USA
Email address: jd01@txstate.edu


	1. Introduction
	2. Results
	Acknowledgments

	References