CUBO, A Mathematical Journal

Vol. 24, no. 03, pp. 393–412, December 2022

DOI: 10.56754/0719-0646.2403.0393

Two nonnegative solutions for two-dimensional
nonlinear wave equations

Svetlin Georgiev1, B

Mohamed Majdoub2,3

1 Department of Differential Equations,

Faculty of Mathematics and Informatics,

University of Sofia, Sofia, Bulgaria.

svetlingeorgiev1@gmail.com B

2 Department of Mathematics, College of

Science, Imam Abdulrahman Bin Faisal

University, P. O. Box 1982, Dammam,

Saudi Arabia.

3 Basic and Applied Scientific Research

Center, Imam Abdulrahman Bin Faisal

University, P.O. Box 1982, 31441,

Dammam, Saudi Arabia.

mmajdoub@iau.edu.sa

ABSTRACT

We study a class of initial value problems for two-dimensional

nonlinear wave equations. A new topological approach is

applied to prove the existence of at least two nonnegative

classical solutions. The arguments are based upon a recent

theoretical result.

RESUMEN

Estudiamos una clase de problemas de valor inicial para

ecuaciones de onda no lineales en dimensión dos. Se aplica

un nuevo enfoque topológico para probar la existencia de al

menos dos soluciones clásicas no negativas. Los argumentos

se basan en un resultado teórico reciente.

Keywords and Phrases: Hyperbolic Equations, positive solution, fixed point, cone, sum of operators.

2020 AMS Mathematics Subject Classification: 47H10, 58J20, 35L15.

Accepted: 26 September, 2022

Received: 21 September, 2021

©2022 S. Georgiev et al. This open access article is licensed under a Creative Commons

Attribution-NonCommercial 4.0 International License.

http://cubo.ufro.cl/
http://dx.doi.org/10.56754/0719-0646.2403.0393
https://orcid.org/0000-0001-8015-4226
https://orcid.org/0000-0001-6038-1069
mailto:svetlingeorgiev1@gmail.com
mailto:mmajdoub@iau.edu.sa


394 S. Georgiev & M. Majdoub CUBO
24, 3 (2022)

1 Introduction

Global existence for nonlinear wave equations is an important mathematical topic. Mathemati-

cians, including F. John, S. Kleinerman, L. Hörmander, etc., have made investigations to this

subject. The first non-trivial long-time existence result was established by F. John and S. Klein-

erman in [19], where it is proved the almost global existence for a class 3D quasilinear scalar wave

equations. Global existence for 3D quasilinear wave equations was established firstly by S. Klein-

erman in [20] and by D. Christodoulou, independently by S. Kleinerman, in [5]. The problem in

2D case is quite delicate. Introducing the ghost weight, in [1] was proved the global well-posedness

for a class 2D nonlinear wave equations. Using a class Hardy-type inequality depending on the

compact support of the initial data, in [21] was proved almost global existence for 2D case. Here

we propose a new approach for investigations for classical solutions of a class 2D nonlinear wave

equations. We investigate for existence of at least two positive solutions for the following IVP

utt − ∆u = f(t, x, u, ut, ux), t > 0, x = (x1, x2) ∈ R2,

u(0, x) = u0(x), x = (x1, x2) ∈ R2, (1.1)

ut(0, x) = u1(x), x = (x1, x2) ∈ R2,

where ∆u = ux1x1 + ux2x2, ux = (ux1, ux2).

The initial value problem (1.1) has attracted considerable attention in the mathematical community

and the well-posedness theory in the Sobolev spaces for polynomial type nonlinearities has been

extensively studied. The case of exponential nonlinearity was recently investigated (see [18] and

references therein). In particular, if the nonlinearity f and the initial data u0, u1 are smooth then

the Cauchy problem (1.1) has a classical local (in time) solution. This follows from Duhamel’s

formula via the usual fixed point argument in the space Hsloc × H
s−1
loc , s > 2. Such an s guarantee

that u, ut, ∇u are in L∞. Note that u ∈ Hsloc means that the H
s norm over a ball centered at x0

and with radius 1 is uniformly bounded by a constant independent of x0. We refer the reader to

[23] and references therein for more properties and information on nonlinear wave equations. In

[17] is proved existence and uniqueness of generalized solutions of the first initial boundary value

problem for strongly hyperbolic systems in bounded domains. In the case when

f(t, x, u, ut, ux) = f(u(x)), t > 0, x ∈ R2,

and

u0(x) = u1(x) = 0, x ∈ R2,

the problem (1.1) is investigated in [14] where the authors prove existence of at least one nontrivial

classical solution of the problem (1.1).



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Two nonnegative solutions for two-dimensional nonlinear... 395

We make the following assumptions on the non-linearity and initial data trough the paper.

(H1) u0, u1 ∈ C2(R2),

0 ≤ u0, |u0x1|, |u0x1x1|, |u0x2|, |u0x2x2| ≤ r,

0 ≤ u1, |u1x1|, |u1x1x1|, |u1x2|, |u1x2x2| ≤ r on R
2,

where r > 0 is a given constant.

(H2) f ∈ C([0, ∞) × R6),

0 ≤ f(t, x, w1, w2, w3, w4)

≤
l∑

j=1

(aj(t, x)|w1|pj + bj(t, x)|w2|pj + cj(t, x)|w3|pj + dj(t, x)|w4|pj ) ,

(t, x) ∈ [0, ∞) × R2, where aj, bj, cj, dj ∈ C([0, ∞) × R2),

0 ≤ aj, bj, cj, dj ≤ a, pj > 0, j ∈ {1, . . . , l},

where a > 0 and l ∈ N are given constants.

Our main result reads as follows.

Theorem 1.1. Suppose (H1) and (H2). Then the IVP (1.1) has at least two nonnegative classical

solutions.

To prove our main result we use a new topological approach. This approach can be used for

investigations for existence of at least one and at least two classical solutions for initial value

problems, boundary value problems and initial boundary value problems for some classes ordinary

differential equations, partial differential equations and fractional differential equations (see [2,

3, 4, 7, 10, 12, 13, 15, 16] and references therein). So far, for the authors they are not known

investigations for existence of multiple solutions for the IVP (1.1).

The paper is organized as follows. In the next section, we give some auxiliary results. In Section

3, we prove our main result. In Section 4, we give an example.



396 S. Georgiev & M. Majdoub CUBO
24, 3 (2022)

2 Auxiliary Results

Let X be a real Banach space.

Definition 2.1. A mapping K : X → X is said to be completely continuous if it is continuous
and maps bounded sets into relatively compact sets.

The concept for k-set contraction is related to that of the Kuratowski measure of noncompactness

which we recall for completeness.

Definition 2.2. Let ΩX be the class of all bounded sets of X. The Kuratowski measure of non-

compactness α : ΩX → [0, ∞) is defined by

α(Y ) = inf


δ > 0 : Y =

m⋃
j=1

Yj and diam(Yj) ≤ δ, j ∈ {1, . . . , m}


 ,

where diam(Yj) = sup{∥x − y∥X : x, y ∈ Yj} is the diameter of Yj, j ∈ {1, . . . , m}.

For the main properties of measure of noncompactness we refer the reader to [6].

Definition 2.3. For a given number k ≥ 0, a map K : X → X is said to be k-set contraction if
it is continuous, bounded and

α(K(Y )) ≤ kα(Y )

for any bounded set Y ⊂ X.

Obviously, if K : X → X is a completely continuous mapping, then K is 0-set contraction.

Definition 2.4. Let X and Y be real Banach spaces. A mapping K : X → Y is said to be
expansive if there exists a constant h > 1 such that

∥Kx − Ky∥Y ≥ h∥x − y∥X

for any x, y ∈ X.

Definition 2.5. A closed, convex set P in X is said to be a cone if

(1) αx ∈ P for any α ≥ 0 and for any x ∈ P,

(2) x, −x ∈ P implies x = 0.



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Two nonnegative solutions for two-dimensional nonlinear... 397

Let P ⊂ X be a cone and define

P∗ = P\{0},

Pr1 =
{
u ∈ P : ∥u∥ ≤ r1

}
,

Pr1,r2 =
{
u ∈ P : r1 ≤ ∥u∥ ≤ r2

}
for positive constants r1, r2 such that 0 < r1 ≤ r2. The following result will be used to prove
Theorem 1.1 . We refer the reader to [8] and [11] for more details.

Theorem 2.6. Let P be a cone of a Banach space E; Ω a subset of P and U1, U2 and U3 three
open bounded subsets of P such that U1 ⊂ U2 ⊂ U3 and 0 ∈ U1. Assume that T : Ω → P is an
expansive mapping with constant h > 1, S : U3 → E is a k-set contraction with 0 ⩽ k < h − 1 and
S(U3) ⊂ (I − T)(Ω). Suppose that (U2 \ U1) ∩ Ω ̸= ∅, (U3 \ U2) ∩ Ω ̸= ∅, and there exists u0 ∈ P∗

such that the following conditions hold:

(i) Sx ̸= (I − T)(x − λu0), for all λ > 0 and x ∈ ∂U1 ∩ (Ω + λu0),

(ii) there exists ϵ > 0 such that Sx ̸= (I − T)(λx), for all λ ≥ 1 + ϵ, x ∈ ∂U2 and λx ∈ Ω,

(iii) Sx ̸= (I − T)(x − λu0), for all λ > 0 and x ∈ ∂U3 ∩ (Ω + λu0).

Then T + S has at least two non-zero fixed points x1, x2 ∈ P such that

x1 ∈ ∂U2 ∩ Ω and x2 ∈ (U3 \ U2) ∩ Ω

or

x1 ∈ (U2 \ U1) ∩ Ω and x2 ∈ (U3 \ U2) ∩ Ω.

Note that (see [9]) the function

G(t, x, τ, ξ) = −
1

2π

H(t − τ − |x − ξ|)√
(t − τ)2 − |x − ξ|2

, t, τ > 0, x, ξ ∈ R2,

where |x − ξ| =
√
(x1 − ξ1)2 + (x2 − ξ2)2, is the Green function for the two-dimensional wave

equation

utt − ∆u = h(t, x), t > 0, x = (x1, x2) ∈ R2,

u(0, x) = ut(0, x) = 0, x = (x1, x2) ∈ R2,

where H(·) denotes the Heaviside function. Observe that

G(t, x, τ, ξ) ≤ 0, t, τ > 0, x, ξ ∈ R2.



398 S. Georgiev & M. Majdoub CUBO
24, 3 (2022)

A key lemma in our proof is the following.

Lemma 2.7. For h1, h2, p > 0, we have∣∣∣∣
∫
R2

∫ ∞
0

(h1 + h2τ)
pG(t, x, τ, ξ)dτdξ

∣∣∣∣ ≤ (h1 + h2t)pI(t), (t, x) ∈ (0, ∞) × R2, (2.1)
where I(t) = t3 + t2(1 + | log t|).

Proof. Let h1, h2, p > 0 and t > 0. One has

∣∣∣∣
∫
R2

∫ ∞
0

(h1 + h2τ)
pG(t, x, τ, ξ)dτdξ

∣∣∣∣ ≤ 12π
∫
|x−ξ|≤t

∫ t−|x−ξ|
0

(h1 + h2τ)
p√

(t − τ)2 − |x − ξ|2
dτdξ

≤
(h1 + h2t)

p

2π

∫
|x−ξ|≤t

(
log(t +

√
t2 − |x − ξ|2) − log |x − ξ|

)
dξ

=
(h1 + h2t)

p

2π

(∫
|x−ξ|≤t

log(t +
√
t2 − |x − ξ|2)dξ −

∫
|x−ξ|≤t

log |x − ξ|dξ

)

≤
(h1 + h2t)

p

2π

(
log(2t)

∫
|x−ξ|≤t

dξ − 2π
∫ t
0

r1 log r1dr1

)

=
(h1 + h2t)

p

2π

(
πt2 log(2t) − π

(
t2 log t −

t2

2

))
≤

(h1 + h2t)
p

2

(
t2 log(1 + 2t) + t2| log t| +

t2

2

)
≤

(h1 + h2t)
p

2

(
2t3 + t2| log t| +

t2

2

)
≤ (h1 + h2t)p

(
t3 + t2(1 + | log t|)

)
.

This gives (2.1) as desired.

We make the change u = v + u0 + tu1. Then, we get the IVP

vtt − ∆v = f(t, x, v + u0 + tu1, vt + u1, vx + u0x + tu1x) + ∆u0 + t∆u1

= f1(t, x, v, vt, vx), t > 0, x ∈ R2, (2.2)

v(0, x) = vt(0, x) = 0, x ∈ R2.

Lemma 2.8. Suppose (H2). If wk ∈ R, |wk| ≤ b, k ∈ {1, . . . , 4}, for some positive b, then

f(t, x, w1, w2, w3, w4) ≤ 4a
l∑

j=1

bpj .



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Two nonnegative solutions for two-dimensional nonlinear... 399

Proof. We have

0 ≤ f(t, x, w1, w2, w3, w4)

≤
l∑

j=1

(aj(t, x)|w1|pj + bj(t, x)|w2|pj + cj(t, x)|w3|pj + dj(t, x)|w4|pj )

≤
l∑

j=1

(abpj + abpj + abpj + abpj )

= 4a

l∑
j=1

bpj , (t, x, w1, w2, w3, w4) ∈ [0, ∞) × R6.

This completes the proof.

Let E = C2([0, ∞) × R2) and for any u ∈ E, denote

∥u∥ = max
{
∥u∥∞, ∥ut∥∞, ∥utt∥∞∥uxj ∥∞, ∥uxjxj ∥∞, j ∈ {1, 2}

}
,

provided that it is finite, where

∥v∥∞ = sup
(t,x)∈[0,∞)×R2

|v(t, x)|.

Lemma 2.9. Suppose (H1) and (H2). Let v ∈ E, ∥v∥ ≤ b, for some positive b. Then

f(t, x, v + u0 + tu1, vt + u1, vx + u0x + tu1x) ≤ 4a
l∑

j=1

(b + r(1 + t))pj , (t, x) ∈ [0, ∞) × R2.

Proof. Let

w1 = v + u0 + tu1,

w2 = vt + u0 + tu1,

w3 = ux1 + u0x1 + tu1x1,

w4 = vx2 + u0x2 + tu1x2.

Then

|wj| ≤ b + r(1 + t), j ∈ {1, . . . , 4}, t ≥ 0.

Hence and Lemma 2.8, we get the desired result. This completes the proof.



400 S. Georgiev & M. Majdoub CUBO
24, 3 (2022)

Lemma 2.10. Suppose (H1) and (H2). Let v ∈ E, ∥v∥ ≤ b, for some positive b. Then

|f1(t, x, v, vt, vx)| ≤ 4a
l∑

j=1

(b + r(1 + t))pj + 2r(1 + t), (t, x) ∈ [0, ∞) × R2.

Proof. By (H1), we get

|∆u0| ≤ 2r, |∆u1| ≤ 2r on R2.

Using Lemma 2.9, we obtain

|f1(t, x, v, vt, vx)| ≤ f(t, x, v + u0 + tu1, vt + u1, vx + u0x + tu1x) + |∆u0| + t|∆u1|

≤ 4a
l∑

j=1

(b + r(1 + t))pj + 2r(1 + t), (t, x) ∈ [0, ∞) × R2.

This completes the proof.

Now, applying Lemma 2.10 and (2.1), we obtain the following result.

Lemma 2.11. Suppose (H1) and (H2). Then

∣∣∣∣
∫
R2

∫ ∞
0

G(t, x, τ, ξ)f1(τ, ξ, v(τ, ξ), vt(τ, ξ), vx(τ, ξ))dτdξ

∣∣∣∣
≤


4a l∑

j=1

(b + r(1 + t))pj + 2r(1 + t)


I(t)

≤


4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj tpj + 2r(1 + t)


I(t), (t, x) ∈ [0, ∞) × R2.

Take a nonnegative function g ∈ C([0, ∞) × R2). Suppose that v ∈ E is a solution to the integral
equation.

0 =
1

8

∫ t
0

∫ x1
0

∫ x2
0

(x1 − s1)2(x2 − s2)2(t − t1)2g(t1, s1, s2)v(t1, s1, s2)ds2ds1dt1

−
1

16π

∫ t
0

∫ x1
0

∫ x2
0

(x1 − s1)2(x2 − s2)2(t − t1)2g(t1, s1, s2)
∫ ∞
−∞

∫ ∞
−∞

∫ ∞
0

G(t1, s1, s2, t2, ξ1, ξ2)

× f1(t2, ξ1, ξ2, v(t2, ξ1, ξ2), vt(t2, ξ1, ξ2), vx(t2, ξ1, ξ2))dt2dξ2dξ1ds2ds1dt1, (2.3)

t ≥ 0, (x1, x2) ∈ R2. We differentiate three times in t, three times in x1 and three times in x2 the
equation (2.3) and we obtain

0 = g(t, x)v(t, x) −
1

2π
g(t, x)

∫
R2

∫ ∞
0

G(t, x, τ, ξ)f1(τ, ξ, v(τ, ξ), vt(τ, ξ), vx(τ, ξ))dτdξ,



CUBO
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Two nonnegative solutions for two-dimensional nonlinear... 401

t ≥ 0, x ∈ R2, whereupon

0 = v(t, x) −
1

2π

∫
R2

∫ ∞
0

G(t, x, τ, ξ)f1(τ, ξ, v(τ, ξ), vt(τ, ξ), vx(τ, ξ))dτdξ,

t ≥ 0, x ∈ R2. Hence, using the Green function, we conclude that v is a solution of the IVP (2.2).
Thus, any solution v ∈ E of the integral equation (2.3) is a solution to the IVP (2.2).

(H3) Let m > 0 be large enough and A, r1, L1, R1 be positive constants that satisfy the following

conditions

r1 < L1 < R1, r1 < r, R1 >

(
2

5m
+ 1

)
L1,

A


R1 + 4a l∑

j=1

(2(R1 + r))
pj + 4a

l∑
j=1

(2r)pj + 2r


 < L1

20
.

(H4) There exists a nonnegative function g ∈ C([0, ∞) × R2) such that

q(t, x1, x2) =

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)g(t1, s1, s2)

×
(
1 + |x1 − s1| + (x1 − s1)2

)(
1 + |x2 − s2| + (x2 − s2)2

)
×
(
1 + (t − t1) + (t − t1)2

)1 +

1 + t1 + l∑

j=1

t
pj
1


I(t1)


ds2ds1dt1

≤ A, (t, x1, x2) ∈ [0, ∞) × R2.

In the last section we will give an example for the constants m, A, r, L1, R1 and R and for a

function g that satisfy (H3) and (H4). For v ∈ E, define the operator

Fv(t, x1, x2) =
1

8

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)2g(t1, s1, s2)

× v(t1, s1, s2)ds2ds1dt1

−
1

16π

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)2g(t1, s1, s2)

×
∫ ∞
−∞

∫ ∞
−∞

∫ ∞
0

G(t1, s1, s2, t2, ξ1, ξ2)

× f1(t2, ξ1, ξ2, v(t2, ξ1, ξ2), vt(t2, ξ1, ξ2), vx(t2, ξ1, ξ2))dt2dξ2dξ1ds2ds1dt1,

(t, x1, x2) ∈ [0, ∞) × R2.

Lemma 2.12. Suppose (H1)–(H3). Then, for v ∈ E, ∥v∥ ≤ b, for some positive b, we have

∥Fv∥ ≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 .



402 S. Georgiev & M. Majdoub CUBO
24, 3 (2022)

Proof. Using Lemma 2.11 and (H3), we get

|Fv(t, x1, x2)| ≤
1

8

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)2g(t1, s1, s2)

× |v(t1, s1, s2)|ds2ds1dt1

+
1

16π

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)2g(t1, s1, s2)

×
∣∣∣∣
∫ ∞
−∞

∫ ∞
−∞

∫ ∞
0

G(t1, s1, s2, t2, ξ1, ξ2)

× f1(t2, ξ1, ξ2, v(t2, ξ1, ξ2), vt(t2, ξ1, ξ2), vx(t2, ξ1, ξ2))dt2dξ2dξ1
∣∣∣∣ds2ds1dt1

≤ bA + 4a
l∑

j=1

(2(b + r))pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)2

× g(t1, s1, s2)I(t1)ds2ds1dt1

+ 4a

l∑
j=1

(2r)pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)2

× g(t1, s1, s2)t
pj
1 I(t1)ds2ds1dt1

+ 2r

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)2

× g(t1, s1, s2)(1 + t1)I(t1)ds2ds1dt1

≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 ,

(t, x1, x2) ∈ [0, ∞) × R2, and

∣∣∣∣ ∂∂tFv(t, x1, x2)
∣∣∣∣ ≤ 14

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)g(t1, s1, s2)

× |v(t1, s1, s2)|ds2ds1dt1

+
1

8π

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)g(t1, s1, s2)

×
∣∣∣∣
∫ ∞
−∞

∫ ∞
−∞

∫ ∞
0

G(t1, s1, s2, t2, ξ1, ξ2)

× f1(t2, ξ1, ξ2, v(t2, ξ1, ξ2), vt(t2, ξ1, ξ2), vx(t2, ξ1, ξ2))dt2dξ2dξ1
∣∣∣∣ds2ds1dt1

≤ bA

+ 4a

l∑
j=1

(2(b + r))pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)

× g(t1, s1, s2)I(t1)ds2ds1dt1



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Two nonnegative solutions for two-dimensional nonlinear... 403

+ 4a

l∑
j=1

(2r)pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)

× g(t1, s1, s2)t
pj
1 I(t1)ds2ds1dt1

+ 2r

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2(t − t1)

× g(t1, s1, s2)(1 + t1)I(t1)ds2ds1dt1

≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 ,

(t, x1, x2) ∈ [0, ∞) × R2, and

∣∣∣∣ ∂2∂t2 Fv(t, x1, x2)
∣∣∣∣ ≤ 14

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2g(t1, s1, s2)

× |v(t1, s1, s2)|ds2ds1dt1

+
1

8π

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2g(t1, s1, s2)

×
∣∣∣∣
∫ ∞
−∞

∫ ∞
−∞

∫ ∞
0

G(t1, s1, s2, t2, ξ1, ξ2)

× f1(t2, ξ1, ξ2, v(t2, ξ1, ξ2), vt(t2, ξ1, ξ2), vx(t2, ξ1, ξ2))dt2dξ2dξ1
∣∣∣∣ds2ds1dt1

≤ bA + 4a
l∑

j=1

(2(b + r))pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2

× g(t1, s1, s2)I(t1)ds2ds1dt1

+ 4a

l∑
j=1

(2r)pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2

× g(t1, s1, s2)t
pj
1 I(t1)ds2ds1dt1

+ 2r

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x1 − s1)2(x2 − s2)2

× g(t1, s1, s2)(1 + t1)I(t1)ds2ds1dt1

≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 ,

(t, x1, x2) ∈ [0, ∞) × R2, and

∣∣∣∣ ∂∂x1 Fv(t, x1, x2)
∣∣∣∣ ≤ 14

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)|x1 − s1|(x2 − s2)2(t − t1)2g(t1, s1, s2)

× |v(t1, s1, s2)|ds2ds1dt1

+
1

8π

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)|x1 − s1|(x2 − s2)2(t − t1)2g(t1, s1, s2)



404 S. Georgiev & M. Majdoub CUBO
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×
∣∣∣∣
∫ ∞
−∞

∫ ∞
−∞

∫ ∞
0

G(t1, s1, s2, t2, ξ1, ξ2)

× f1(t2, ξ1, ξ2, v(t2, ξ1, ξ2), vt(t2, ξ1, ξ2), vx(t2, ξ1, ξ2))dt2dξ2dξ1
∣∣∣∣ds2ds1dt1

≤ bA

+ 4a

l∑
j=1

(2(b + r))pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)|x1 − s1|(x2 − s2)2(t − t1)2

× g(t1, s1, s2)I(t1)ds2ds1dt1

+ 4a

l∑
j=1

(2r)pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)|x1 − s1|(x2 − s2)2(t − t1)2

× g(t1, s1, s2)t
pj
1 I(t1)ds2ds1dt1

+ 2r

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)|x1 − s1|(x2 − s2)2(t − t1)2

× g(t1, s1, s2)(1 + t1)I(t1)ds2ds1dt1

≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 ,

(t, x1, x2) ∈ [0, ∞) × R2, and

∣∣∣∣ ∂2∂x21 Fv(t, x1, x2)
∣∣∣∣ ≤ 14

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x2 − s2)2(t − t1)2g(t1, s1, s2)

× |v(t1, s1, s2)|ds2ds1dt1

+
1

8π

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x2 − s2)2(t − t1)2g(t1, s1, s2)

×
∣∣∣∣
∫ ∞
−∞

∫ ∞
−∞

∫ ∞
0

G(t1, s1, s2, t2, ξ1, ξ2)

× f1(t2, ξ1, ξ2, v(t2, ξ1, ξ2), vt(t2, ξ1, ξ2), vx(t2, ξ1, ξ2))dt2dξ2dξ1
∣∣∣∣ds2ds1dt1

≤ bA

+ 4a

l∑
j=1

(2(b + r))pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x2 − s2)2(t − t1)2

× g(t1, s1, s2)I(t1)ds2ds1dt1

+ 4a

l∑
j=1

(2r)pj
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x2 − s2)2(t − t1)2

× g(t1, s1, s2)t
pj
1 I(t1)ds2ds1dt1

+ 2r

∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)(x2 − s2)2(t − t1)2

× g(t1, s1, s2)(1 + t1)I(t1)ds2ds1dt1



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Two nonnegative solutions for two-dimensional nonlinear... 405

≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 ,

(t, x1, x2) ∈ [0, ∞) × R2. As above, one can obtain

∣∣∣∣ ∂∂x2 Fv(t, x1, x2)
∣∣∣∣ ≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 ,

(t, x1, x2) ∈ [0, ∞) × R2, and

∣∣∣∣ ∂2∂x22 Fv(t, x1, x2)| ≤ A

b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 ,

(t, x1, x2) ∈ [0, ∞) × R2. Consequently

∥Fv∥ ≤ A


b + 4a l∑

j=1

(2(b + r))pj + 4a

l∑
j=1

(2r)pj + 2r


 .

This completes the proof.

3 Proof of the Main Result

Let

P̃ = {u ∈ E : u ≥ 0 on [0, ∞) × R2}.

With P we will denote the set of all equi-continuous families in P̃. Note that Fv ≥ 0 for any
v ∈ P. Let ϵ > 0. For v ∈ E, define the operators

Tv(t, x) = (1 + mϵ)v(t, x) − ϵ
L1
10

,

Sv(t, x) = −ϵFv(t, x) − mϵv(t, x) − ϵ
L1
10

,

(t, x) ∈ [0, ∞) × R2. Note that any fixed point v ∈ E of the operator T + S is a solution to the
IVP (2.2). Define

U1 = Pr1 = {v ∈ P : ∥v∥ < r1},

U2 = PL1 = {v ∈ P : ∥v∥ < L1},

U3 = PR1 = {v ∈ P : ∥v∥ < R1},



406 S. Georgiev & M. Majdoub CUBO
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R2 = R1 +
A

m


R1 + 4a l∑

j=1

(2(R1 + r))
pj + 4a

l∑
j=1

(2r)pj + 2r


 + L1

5
,

Ω = PR2 = {v ∈ P : ∥v∥ ≤ R2}.

1. For v1, v2 ∈ Ω, we have

∥Tv1 − Tv2∥ = (1 + mϵ)∥v1 − v2∥,

whereupon T : Ω → E is an expansive operator with a constant 1 + mϵ > 1.

2. For v ∈ PR1, we get

∥Sv∥ ≤ ϵ∥Fv∥ + mϵ∥v∥ +
L1
10

≤ ϵ
(
A


R1 + 4a l∑

j=1

(2(R1 + r))
pj + 4a

l∑
j=1

(2r)pj + 2r


 + mR1 + L1

10

)
.

Therefore S(PR1) is uniformly bounded. Since S : PR1 → E is continuous, we have that
S(PR1) is equi-continuous. Consequently S : PR1 → E is a 0-set contraction.

3. Let v1 ∈ PR1. Set
v2 = v1 +

1

m
Fv1 +

L1
5m

.

Note that by the second inequality of (H3) and by Lemma 2.12, it follows that ϵFv+ϵL1
5

≥ 0
on [0, ∞) × R2. We have v2 ≥ 0 on [0, ∞) × R2 and

∥v2∥ ≤ ∥v1∥ +
1

m
∥Fv1∥ +

L1
5m

≤ R1 +
A

m

(
R1 + 4a

l∑
j=1

(2(R1 + r))
pj + 4a

l∑
j=1

(2r)pj + 2r

)
+

L1
5

= R2.

Therefore v2 ∈ Ω and
−ϵmv2 = −ϵmv1 − ϵFv1 − ϵ

L1
10

− ϵ
L1
10

or

(I − T)v2 = −ϵmv2 + ϵ
L1
10

= Sv1.

Consequently S(PR1) ⊂ (I − T)(Ω).

4. Suppose that there exists an v0 ∈ P∗ such that T(v−λv0) ∈ P, v ∈ ∂Pr1, v ∈ ∂Pr1
⋂
(Ω+λu0)



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Two nonnegative solutions for two-dimensional nonlinear... 407

and Sv = v − λv0 for some λ ≥ 0. Then

r1 = ∥v − λv0∥

= ∥Sv∥

≥ −Sv(t, x)

= ϵFv(t, x) + ϵmv(t, x) + ϵ
L1
10

≥ ϵ
L1
20

, (t, x) ∈ [0, ∞) × R2,

because by the second inequality of (H3) and by Lemma 2.12, it follows that ϵFv + ϵL1
20

≥ 0
on [0, ∞) × R2.

5. Suppose that for any ϵ1 > 0 small enough there exist a u ∈ ∂PL and λ1 ≥ 1 + ϵ1 such that
λ1u ∈ PR1 and

Su = (I − T)(λ1u). (3.1)

In particular, for ϵ1 >
2

5m
, we have u ∈ ∂PL, λ1u ∈ PR1, λ1 ≥ 1 + ϵ1 and (3.1) holds. Since

u ∈ ∂PL and λ1u ∈ PR1, it follows that(
2

5m
+ 1

)
L < λ1L = λ1∥u∥ ≤ R1.

Moreover,

−ϵFu − mϵu − ϵ
L

10
= −λ1mϵu + ϵ

L

10
,

or

Fu +
L

5
= (λ1 − 1)mu.

From here,

2
L

5
≥
∥∥∥∥Fu + L5

∥∥∥∥ = (λ1 − 1)m∥u∥ = (λ1 − 1)mL
and

2

5m
+ 1 ≥ λ1,

which is a contradiction.

Therefore all conditions of Theorem 2.6 hold. Hence, the IVP (2.2) has at least two solutions v1

and v2 so that

r1 < ∥v1∥ < L1 < ∥v2∥ < R1,

and

u = v1 + u0 + tu1, w = v2 + u0 + tu1



408 S. Georgiev & M. Majdoub CUBO
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are two different positive solutions of the IVP (1.1). This completes the proof.

4 An Example

Let

l = 1, p1 =
3

5
, R1 = r = 1, a = 200, L1 =

1

2
, r1 =

1

100
,

m = 1050, ϵ = 50, A =
1

1010
, R = 100.

Then

R1 >

(
2

5m
+ 1q

)
L1, r1 < L1 < R1, r1 <

L1
20

.

Also,

A


R1 + 4a l∑

j=1

(2(R1 + r))
pj + 4a

l∑
j=1

(2r)pj + 2r


 = 1

1010
(
1 + 800 · (4)2 + 800 · 4 + 2

)
<

1

40
=

L1
20

.

Consequently (H3) holds. Now, we will construction the function g in (H4). Let

h(x) = log
1 + s11

√
2 + s22

1 − s11
√
2 + s22

, l(s) = arctan
s11

√
2

1 − s22
, s ∈ R.

Then

h′(s) =
22

√
2s10(1 − s22)

(1 − s11
√
2 + s22)(1 + s11

√
2 + s22)

,

l′(s) =
11
√
2s10(1 + s20)

1 + s40
, s ∈ R.

Therefore

−∞ < lim
s→±∞

(1 + s + s2)h(s) < ∞,

−∞ < lim
s→±∞

(1 + s + s2)l(s) < ∞.

Hence, there exists a positive constant C1 so that

(1 + s + s2)

(
1

44
√
2
log

1 + s11
√
2 + s22

1 − s11
√
2 + s22

+
1

22
√
2
arctan

s11
√
2

1 − s22

)
≤ C1, s ∈ R.



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Two nonnegative solutions for two-dimensional nonlinear... 409

Note that by [22, p. 707, Integral 79], we have

∫
dz

1 + z4
=

1

4
√
2
log

1 + z
√
2 + z2

1 − z
√
2 + z2

+
1

2
√
2
arctan

z
√
2

1 − z2
.

Let

Q(s) =
s10

(1 + s44)(1 + s + s2)2(1 + ((1 + s + s2)I(s))2)
, s ∈ R,

and

g1(t, x1, x2) = Q(t)Q(x1)Q(x2), t ∈ [0, ∞), x1, x2 ∈ R.

Then there exists a constant C2 > 0 so that

C2 ≥
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)g1(t1, s1, s2)

×
(
1 + |x1 − s1| + (x1 − s1)2

)(
1 + |x2 − s2| + (x2 − s2)2

)
×
(
1 + (t − t1) + (t − t1)2

)(
1 + (1 + t1 + t

2
1)I(t1)

)
ds2ds1dt1, (t, x1, x2) ∈ [0, ∞) × R2.

Now, we take

g(t, x1, x2) =
1

1020C2
g1(t, x1, x2), (t, x1, x2) ∈ [0, ∞) × R2.

Then

A =
1

1010

≥
∫ t
0

∫ x1
0

∫ x2
0

sign(x1)sign(x2)g(t1, s1, s2)

×
(
1 + |x1 − s1| + (x1 − s1)2

)(
1 + |x2 − s2| + (x2 − s2)2

)
×
(
1 + (t − t1) + (t − t1)2

)(
1 + (1 + t1 + t

2
1)I(t1)

)
ds2ds1dt1, (t, x1, x2) ∈ [0, ∞) × R2.

Now, consider the IVP

utt − ux1x1 − ux2x2 = w(t)u
3
5 , (t, x1, x2) ∈ (0, ∞) × R2,

u(0, x) = ut(0, x) = 0, (x1, x2) ∈ R2, (4.1)

where

w(t) =




10(9t2 − 9t + 2) t ∈ [0, 1]

20 t > 1.



410 S. Georgiev & M. Majdoub CUBO
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Here l = 1,

a1(t, x1, x2) = |w(t)| ≤ a = 200,

b1(t, x1, x2) = c1(t, x1, x2) = d1(t, x1, x2) = 0,

(t, x1, x2) ∈ [0, ∞) × R2, and

u0(x) = u1(x) = 0 ≤ 1 = r, (x1, x2) ∈ R2.

We have that (H1) and (H2) hold. The IVP (4.1) has two nonnegative solutions u1(t, x) = 0,

(t, x) ∈ [0, ∞) × R2, and

u2(t, x) =




(t(1 − t))5 (t, x) ∈ [0, 1] × R2

0 (t, x) ∈ (1, ∞) × R2.

Acknowledgements

The authors thank the reviewers for the careful reading of the manuscript and helpful comments.



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Two nonnegative solutions for two-dimensional nonlinear... 411

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	Introduction
	Auxiliary Results
	Proof of the Main Result
	An Example