CUBO A Mathematical Journal

Vol.10, N
o
¯ 04, (109–117). December 2008

Positive Solutions for Elliptic Boundary Value

Problems with a Harnack-Like Property

Toufik Moussaoui

Department of Mathematics, E.N.S.,

P.O. Box 92, 16050 Kouba, Algiers, Algeria

email: moussaoui@ens-kouba.dz

and

Radu Precup

Department of Applied Mathematics,

Babeş–Bolyai University, 400084 Cluj, Romania

email: r.precup@math.ubbcluj.ro

ABSTRACT

The aim of this paper is to present some existence results of positive solutions for

elliptic equations and systems on bounded domains of R
N

(N ≥ 1). The main tool is

Krasnosel’skii’s compression-expansion fixed point theorem.

RESUMEN

El objetivo de este art́ıculo es presentar algunos resultados de existencia de soluciones

positivas para ecuaciones elipticas y sistemas sobre dominios acotados de R
N

(N ≥ 1).

La principal herramienta es el teorema de punto fijo compresión-expansión de Kras-

nosel’skii.



110 Toufik Moussaoui and Radu Precup CUBO
10, 4 (2008)

Key words and phrases: Positive solution, elliptic boundary value problem, elliptic systems,

Harnack-like inequality, Krasnosel’skii’s compression-expansion fixed point theorem.

Math. Subj. Class.: 47H10, 35J65.

1 Introduction

In this paper, we are concerned with the existence of positive solutions for the elliptic boundary

value problem
{

−∆u = λ f (x, u) , in Ω,

u = 0, on ∂Ω,
(1.1)

and for the elliptic system















−∆u = α g (x, u, v) , in Ω,

−∆v = β h (x, u, v) , in Ω,

u = v = 0, on ∂Ω.

(1.2)

Here Ω is a bounded regular domain of R
N

(N ≥ 1), f : Ω×R+ −→ R+ and g, h : Ω×R
2
+ −→ R+

are continuous functions, and λ, α and β are real parameters. By a positive solution of problem

(1.1) we mean a function u ∈ C1
(

Ω, R
)

which satisfies (1.1) (with ∆u in the sense of distributions),

and with u (x) > 0 for all x ∈ Ω. A positive solution to problem (1.2) is a vector-valued function

(u, v) ∈ C1
(

Ω, R2
)

satisfying (1.2), with u, v ≥ 0 and u + v > 0 in Ω.

The main assumption will be a global weak Harnack inequality for nonnegative superharmonic

functions. By a superharmonic function in a domain Ω ⊂ RN we mean a function u ∈ C1(Ω, R)

with ∆u ≤ 0 in the sense of distributions, i.e.,

∫

Ω

∇u · ∇v ≥ 0 for every v ∈ C∞
0

(Ω, R) satisfying v(x) ≥ 0 on Ω.

We shall assume that the following global weak Harnack inequality holds:























There exists a compact set K ⊂ Ω and a number η > 0

such that u(x) ≥ η‖u‖0 for all x ∈ K

and every nonnegative superharmonic function

u ∈ C1(Ω, R) with u = 0 on ∂Ω.

(1.3)

Here by ‖u‖
0

we denote the sup norm in C
(

Ω, R
)

, i.e., ‖u‖0 = sup
x∈Ω

|u(x)|.

The connection between such type of inequalities and Krasnosel’skii’s compression-expansion

theorem when applied to boundary value problems was first explained in [4]. Also in [4] (see also

[1]), several comments on weak Harnack type inequalities can be found.



CUBO
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Positive Solutions for Elliptic Boundary Value ... 111

By a cone in a Banach space E we mean a closed convex subset C of E such that C 6= {0} ,

λC ⊂ C for all λ ∈ R+, and C ∩ (−C) = {0} .

Our main tool in proving the existence of positive solutions to problems (1.1) and (1.2) is

Krasnosel’skii’s compression-expansion theorem [3], [2]:

Theorem 1. Let E be a Banach space, C ⊂ E a cone in E, and assume that T : C −→ C is

a completely continuous map such that for some numbers r and R with 0 < r < R, one of the

following conditions is satisfied:

(i) ‖T u‖ ≤ ‖u‖ for ‖u‖ = r and ‖T u‖ ≥ ‖u‖ for ‖u‖ = R,

(ii) ‖T u‖ ≥ ‖u‖ for ‖u‖ = r and ‖T u‖ ≤ ‖u‖ for ‖u‖ = R.

Then T has a fixed point with r ≤ ‖u‖ ≤ R.

2 Existence results for Problem 1.1

In this section, E is the Banach space

C0(Ω, R) = {u ∈ C(Ω, R) : u = 0 on ∂Ω}

endowed with norm ‖.‖
0
, and C is the cone

C = {u ∈ C0(Ω, R+) : u(x) ≥ η‖u‖0 for all x ∈ K}. (2.1)

In order to state our results we introduce the notation

f0 = lim sup
y→0+

max
x∈Ω

f (x, y)

y
and f

∞
= lim inf

y→∞
min
x∈K

f (x, y)

y

f
0

= lim inf
y→0+

min
x∈K

f (x, y)

y
and f∞ = lim sup

y→∞
max
x∈Ω

f (x, y)

y
.

Also, for a function h : Ω → R, by h|
K

we mean the function h|
K

(x) = h (x) if x ∈ K and h|
K

(x) = 0 if x ∈ Ω \ K. For example, if 1 is the constant function 1 on Ω, then 1|
K

(x) = 1 if x ∈ K

and 1|
K

(x) = 0 for x ∈ Ω \ K.

Theorem 2. Suppose (1.3) holds. Then for each λ satisfying

1

f
∞

η ‖(−∆)−1 1|
K
‖
0

< λ <
1

f0‖(−∆)−11‖0
(2.2)

there exists at least one positive solution of problem (1.1).

Proof. Let λ be as in (2.2) and let ǫ > 0 be such that

1

(f
∞

− ǫ)η ‖(−∆)−1 1|
K
‖
0

≤ λ ≤
1

(f0 + ǫ)‖(−∆)−11‖0
. (2.3)



112 Toufik Moussaoui and Radu Precup CUBO
10, 4 (2008)

We know that u is a solution of problem (1.1) if and only if

u = λ (−∆)−1F u

where F : C(Ω, R) −→ C(Ω, R), F u(x) = f (x, u(x)) . Hence, a solution to problem (1.1) is a fixed

point of the operator T : C −→ C0(Ω, R) given by

T u = λ (−∆)−1F u.

We shall prove that the hypotheses of Theorem 1 are satisfied.

We have that the operator T satisfies
{

−∆(T u) = λ f (x, u) , in Ω,

T u = 0, on ∂Ω.

Then by the global weak Harnack inequality (1.3), one has T (C) ⊂ C. Moreover, T is completely

continuous by the Arzela-Ascoli Theorem.

Furthermore, by the definition of f0, there exists an r > 0 such that

f (x, u) ≤ (f0 + ǫ)u for 0 < u ≤ r and x ∈ Ω. (2.4)

Let u ∈ C with ‖u‖0 = r. Then using (2.4), the monotonicity of operator (−∆)
−1

and of norm

‖.‖
0
, and (2.3), we obtain

‖T u‖
0

= λ
∥

∥(−∆)−1F u
∥

∥

0

≤ λ(f0 + ǫ)‖u‖0
∥

∥(−∆)−11
∥

∥

0

≤ ‖u‖0.

Hence

‖T u‖0 ≤ ‖u‖0 for ‖u‖0 = r. (2.5)

By the definition of f
∞

, there is R > r such that

f (x, u) ≥ (f
∞

− ǫ)u for u ≥ ηR and x ∈ K.

Then, if u ∈ C with ‖u‖0 = R, we have

‖T u‖0 = λ
∥

∥ (−∆)−1F u
∥

∥

0

≥ λ
∥

∥ (−∆)−1 (F u)|
K

∥

∥

0

≥ λ(f
∞

− ǫ)η‖u‖0
∥

∥(−∆)−1 1|
K

∥

∥

0

≥ ‖u‖0.

Hence

‖T u‖0 ≥ ‖u‖0 for ‖u‖0 = R. (2.6)

Inequalities (2.5) and (2.6) show that the expansion condition (i) in Theorem 1 is satisfied. Now

Theorem 1 guarantees the existence of a fixed point u of T with r ≤ ‖u‖
0
≤ R.



CUBO
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Positive Solutions for Elliptic Boundary Value ... 113

Similarly, we have the following result:

Theorem 3. Suppose (1.3) holds. Then for each λ satisfying

1

f
0
η ‖(−∆)−1 1|

K
‖
0

< λ <
1

f∞‖(−∆)−11‖0
(2.7)

there exists at least one positive solution of problem (1.1).

Proof. Let λ be as in (2.7) and let ǫ > 0 be such that

1

(f
0
− ǫ)η ‖(−∆)−1 1|

K
‖
0

≤ λ ≤
1

(f∞ + ǫ)‖(−∆)−11‖0
. (2.8)

By the definition of f
0
, there exists an r > 0 such that

f (x, u) ≥ (f
0
− ǫ)u for 0 < u ≤ r and x ∈ K.

If u ∈ C and ‖u‖0 = r, then

‖T u‖0 = λ
∥

∥ (−∆)−1F u
∥

∥

0

≥ λ
∥

∥ (−∆)−1 (F u)|
K

∥

∥

0

≥ λ(f
0
− ǫ)η‖u‖0

∥

∥(−∆)−1 1|
K

∥

∥

0

≥ ‖u‖0.

Hence

‖T u‖0 ≥ ‖u‖0 for ‖u‖0 = r. (2.9)

By the definition of f∞, there is R0 > 0 such that

f (x, u) ≤ (f∞ + ǫ)u for u ≥ R0 and x ∈ Ω.

Let M be such that f (x, u) ≤ M for all u ∈ [0, R0] and x ∈ Ω, and let R be such that

R > r and M ≤ (f∞ + ǫ) R.

If u ∈ C with ‖u‖0 = R, then 0 ≤ u (x) ≤ (f∞ + ǫ) R for all x ∈ Ω. Consequently, also using (2.8),

we obtain

‖T u‖
0

= λ
∥

∥(−∆)−1F u
∥

∥

0

≤ λ(f∞ + ǫ)R
∥

∥(−∆)−11
∥

∥

0

≤ R

= ‖u‖0.

Hence

‖T u‖0 ≤ ‖u‖0 for ‖u‖0 = R. (2.10)

Inequalities (2.9) and (2.10) show that the compression condition (ii) in Theorem 1 is satisfied.

Now Theorem 1 guarantees the existence of a fixed point u of T with r ≤ ‖u‖
0
≤ R.



114 Toufik Moussaoui and Radu Precup CUBO
10, 4 (2008)

3 Existence results for Problem 1.2

In this section, we are concerned with the existence of positive solutions to the Dirichlet problem

(1.2) for elliptic systems.

Here E will be the Banach space C0(Ω, R
2
) := C0(Ω, R) × C0(Ω, R) endowed with the norm

‖(., .)‖0 given by

‖(u, v)‖0 = ‖u‖0 + ‖v‖0

and the cone in E will be C × C, where C is given by (2.1).

In order to state our results in this section we introduce the notation

g0 = lim sup
y+z→0+

max
x∈Ω

g (x, y, z)

y + z
and g

∞
= lim inf

y+z→∞
min
x∈K

g (x, y, z)

y + z

g
0

= lim inf
y+z→0+

min
x∈K

g (x, y, z)

y + z
and g∞ = lim sup

y+z→∞

max
x∈Ω

g (x, y, z)

y + z
.

The limits h0, h0, h∞ and h∞ are defined similarly.

Theorem 4. Suppose (1.3) holds. In addition assume that there are numbers p, q > 0 with
1

p
+

1

q
= 1 such that

1

g
∞

η ‖(−∆)−1 1|
K
‖
0

< α <
1

p g0‖(−∆)−11‖0
(3.1)

and
1

h
∞

η ‖(−∆)−1 1|
K
‖
0

< β <
1

q h0‖(−∆)−11‖0
. (3.2)

Then there exists at least one positive solution (u, v) of problem (1.2).

Proof. Let α, β be as in (3.1), (3.2) and let ǫ > 0 be such that

1

(g
∞

− ǫ)η ‖(−∆)−1 1|
K
‖
0

≤ α ≤
1

p (g0 + ǫ)‖(−∆)−11‖0

and
1

(h
∞

− ǫ)η ‖(−∆)−1 1|
K
‖
0

≤ β ≤
1

q (h0 + ǫ)‖(−∆)−11‖0
.

It is easily seen that a vector-valued function (u, v) is a solution of problem (1.2) if and only

if

u = α (−∆)−1G (u, v)

v = β (−∆)
−1

H (u, v)

where G, H : C(Ω, R2) −→ C(Ω, R),

G(u, v)(x) = g (x, u(x), v(x)) , H (u, v) (x) = h (x, u (x) , v (x)) .



CUBO
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Positive Solutions for Elliptic Boundary Value ... 115

Hence, (u, v) is a positive solution of (1.2) if it is a fixed point of the operator

T : C × C −→ C0(Ω, R
2
), T = (T1, T2)

where

T1 (u, v) = α (−∆)
−1G (u, v) , T2 (u, v) = β (−∆)

−1H (u, v) .

We shall prove that the hypotheses of Theorem 1 are satisfied.

Clearly the operator T = (T1, T2) satisfies















−∆(T1u) = α g (x, u, v) , in Ω,

−∆(T2v) = β h (x, u, v) , in Ω,

T1u = T2v = 0, on ∂Ω.

Then by the global weak Harnack inequality (1.3), we have T (C × C) ⊂ C × C. Moreover, T is

completely continuous by the Arzela-Ascoli Theorem.

By the definitions of g0 and h0, there exists an r > 0 with

g (x, u, v) ≤ (g0 + ǫ)(u + v) for u, v ≥ 0, 0 < u + v ≤ r and x ∈ Ω

and

h (x, u, v) ≤ (h0 + ǫ)(u + v) for u, v ≥ 0, 0 < u + v ≤ r and x ∈ Ω.

Let (u, v) ∈ C × C with ‖(u, v)‖
0

= r. We have

‖T1 (u, v)‖0 = α
∥

∥(−∆)−1G (u, v)
∥

∥

0

≤ α(g0 + ǫ)‖u + v‖0
∥

∥(−∆)−11
∥

∥

0

≤
1

p
‖u + v‖0

≤
1

p
(‖u‖0 + ‖v‖0)

=
1

p
‖(u, v)‖0.

Then ‖T1 (u, v) ‖0 ≤
1

p
‖(u, v)‖0. Similarly, we have

‖T2 (u, v)‖0 = β
∥

∥(−∆)−1H (u, v)
∥

∥

0

≤ β(h0 + ǫ)‖u + v‖0
∥

∥(−∆)−11
∥

∥

0

≤
1

q
‖u + v‖0

≤
1

q
(‖u‖0 + ‖v‖0)

=
1

q
‖(u, v)‖0.



116 Toufik Moussaoui and Radu Precup CUBO
10, 4 (2008)

Thus ‖T2 (u, v) ‖0 ≤
1

q
‖(u, v)‖0. Combining the above two inequalities, we obtain

‖T (u, v) ‖0 = ‖T1 (u, v) ‖0 + ‖T2 (u, v) ‖0 ≤ (
1

p
+

1

q
)‖(u, v)‖0 = ‖(u, v)‖0.

Next by the definitions of g
∞

and h
∞

, there is R > 0 such that

g (x, u, v) ≥ (g
∞

− ǫ)(u + v) for u, v ≥ 0, u + v ≥ ηR and x ∈ K

and

h (x, u, v) ≥ (h
∞

− ǫ)(u + v) for u, v ≥ 0, u + v ≥ ηR and x ∈ K.

Let (u, v) ∈ C ×C with ‖(u, v)‖
0

= R. Then for each x ∈ K, u (x) ≥ η ‖u‖
0

and v (x) ≥ η ‖v‖
0
.

Hence (u + v) (x) ≥ η (‖u‖
0

+ ‖v‖
0
) , that is (u + v) (x) ≥ ηR for all x ∈ K. Consequently,

G (u, v) (x) ≥ (g
∞

− ǫ) (u + v) (x) for all x ∈ K.

Furthermore

‖T1 (u, v) ‖0 = α
∥

∥(−∆)−1G (u, v)
∥

∥

0

≥ α
∥

∥(−∆)−1 G (u, v)|
K

∥

∥

0

≥ α(g
∞

− ǫ)
∥

∥(−∆)−1 (u + v)|
K

∥

∥

0

≥ α(g
∞

− ǫ)
∥

∥(−∆)−1 u|
K

∥

∥

0

≥ α(g
∞

− ǫ)η‖u‖0
∥

∥(−∆)−1 1|
K

∥

∥

0

≥ ‖u‖0.

Similarly, we have

‖T2 (u, v) ‖0 ≥ ‖v‖0.

The above two inequalities give

‖T (u, v) ‖0 ≥ ‖(u, v)‖0.

Thus condition (i) in Theorem 1 is satisfied. Now Theorem 1 guarantees the existence of a fixed

point (u, v) of T with r ≤ ‖(u, v)‖
0
≤ R.

In a similar way, one can prove:

Theorem 5. Suppose (1.3) holds. In addition assume that there are numbers p, q > 0 with
1

p
+

1

q
= 1 such that

1

g
0
η ‖(−∆)−1 1|

K
‖
0

< α <
1

p g∞‖(−∆)−11‖0

and
1

h
0
η ‖(−∆)−1 1|

K
‖
0

< β <
1

q h∞‖(−∆)−11‖0
.

Then there exists at least one positive solution (u, v) of problem (1.2).

Received: April 2008. Revised: April 2008.



CUBO
10, 4 (2008)

Positive Solutions for Elliptic Boundary Value ... 117

References

[1] D. Gilbarg and N. Trudinger, Elliptic Partial Differential Equations of Second Order,

Springer, Berlin, 1983.

[2] A. Granas and J. Dugundji, Fixed Point Theory, Springer, New York, 2003.

[3] M.A. Krasnoselskii, Topological Methods in the Theory of Nonlinear Integral Equations,

Cambridge University Press, New York, 1964.

[4] R. Precup, Positive solutions of semi-linear elliptic problems via Krasnoselskii type theorems

in cones and Harnack’s inequality, Mathematical Analysis and Applications, AIP Conf. Proc.,

835, Amer. Inst. Phys., Melville, NY, 2006, 125–132.


	N9-Moussaoui-Precup2