Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 131, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF SOLUTIONS FOR A COUPLED QUASILINEAR
SYSTEM ON TIME SCALES
MOULAY RCHID SIDI AMMI, AGNIESZKA B. MALINOWSKA
Abstract. In this work we study a quasilinear system of equations on time
scales. Using the Krasnoselskii fixed-point theorem, sufficient conditions are
given for the existence of positive solutions.
1. Introduction
In this paper we consider the existence of positive solutions for the system of
dynamic equations
∆
− u∆ (t)
= λf (v(t)), ∀t ∈ [0, T ]T ,
(1.1)
∆
− v ∆ (t)
= λg(u(t)), ∀t ∈ [0, T ]T ,
satisfying the boundary conditions
u∆ (0) − βu∆ (η) = 0,
∆
∆
v (0) − βv (η) = 0,
u(T ) − βu(η) = 0,
v(T ) − βv(η) = 0,
(1.2)
where η ∈ [0, T ]T , β ∈ R, such that 0 < β < 1. We are seeking for a pair (u, v) of
solutions for the system (1.1)-(1.2). Our general assumptions are:
(H1) the functions f, g belong to C([0, ∞), [0, ∞));
(H2) the following limits exist as real numbers:
f0 := lim+ f (x)/x,
g0 := lim+ g(x)/x,
x→0
f∞ := lim f (x)/x,
x→∞
x→0
g∞ := lim g(x)/x,
x→∞
f∞ g∞ 6= 0.
The theory of dynamic equations on time scales (more generally, on measure
chains) was introduced in 1988 by Stefan Hilger in his PhD thesis (see [10, 11]).
The theory presents a structure where, once a result is established for a general
time scale, then special cases can be obtained by taking the particular time scale.
If T = R, then we have the result for differential equations. Choosing T = Z we
immediately get the result for difference equations. A great deal of work has been
done since 1988, unifying and extending the theories of differential and difference
equations, and many results are now available in the general setting of time scales,
2000 Mathematics Subject Classification. 37C25, 34B18, 47H10.
Key words and phrases. Time scales; fixed point theorem; positive solution; cone.
c
2011
Texas State University - San Marcos.
Submitted March 1, 2011. Published October 11, 2011.
1
2
M. R. SIDI AMMI, A. B. MALINOWSKA
EJDE-2011/131
see [1, 2, 3, 4] and references therein. In this paper we prove existence of positive
solutions to the problem (1.1)-(1.2) on a general time scale T.
The outline of this paper is as follows. In section 2, we give some preliminary
results with respect to the calculus on time scales. For more details see for example
[7, 8]. Section 3 is devoted to the existence of positive solutions using fixed-point
theory. We are concerned with determining values of λ (eigenvalues) for which there
exist positive solutions of (1.1)–(1.2). In [12], a Green function plays a fundamental
role to define an appropriate operator on a suitable cone and to prove existence of
solutions for a system of dynamic equations. Differently, here we will use the
well known Guo-Krasnoselskii fixed point theorem without introducing the Green
function.
2. Preliminary results on time scales
Now, we introduce some basic concepts of time scales, preliminaries and lemmas
that will be needed later. For a deep details, the reader can see [1, 2, 3, 4, 6, 7] and
references therein. A time scale T is an arbitrary nonempty closed subset of real
numbers R. The operators σ and ρ from T to T which are defined in [10, 11],
σ(t) = inf{τ ∈ T : τ > t} ∈ T,
ρ(t) = sup{τ ∈ T : τ < t} ∈ T
are called the forward jump operator and the backward jump operator, respectively.
The point t ∈ T is left-dense, left-scattered, right-dense, right-scattered if ρ(t) =
t, ρ(t) < t, σ(t) = t, σ(t) > t, respectively. If T has a right scattered minimum m,
define Tk = T − {m}; otherwise set Tk = T. If T has a left scattered maximum M ,
define Tk = T − {M }; otherwise set Tk = T.
Let f : T → R and t ∈ Tk (assume t is not left-scattered if t = sup T), we
define f ∆ (t) to be the number (provided it exists) such that for all > 0 there is a
neighborhood U of t such that
|f (σ(t)) − f (s) − f ∆ (t)(σ(t) − s)| ≤ |σ(t) − s|,
for all s ∈ U.
We call f ∆ (t) the delta derivative of f at t. If T = R, then f ∆ coincides with the
usual derivative f 0 . If T = Z, then f ∆ = ∆f := f (t + 1) − f (t) is the forward
difference.
Similarly, for t ∈ T (assume t is not right-scattered if t = inf T), the nabla
derivative of f at the point t is defined in [5] to be the number f ∇ (t) (provided it
exists) with the property that for all > 0 there is a neighborhood U of t such that
|f (ρ(t)) − f (s) − f ∇ (t)(ρ(t) − s)| ≤ |ρ(t) − s|,
for all s ∈ U.
If T = R, then f ∆ (t) = f ∇ (t) = f 0 (t). If T = Z, then f ∇ (t) = ∇f := f (t) − f (t − 1)
is the backward difference operator.
We say that a function f is left-dense continuous (i.e., ld-continuous), provided
f is continuous at each left-dense point in T and its right-sided limit exists at
each right-dense point in T. It is well-known that if f is ld-continuous and if
F ∇ (t) = f (t), then we can define the nabla integral by
Z b
f (t)∇t = F (b) − F (a)
a
for all a, b ∈ T. A function f : T → R is called rd-continuous if it is continuous
at right-dense points and if its left-sided limit exists at left-dense points. If f is
EJDE-2011/131
EXISTENCE OF SOLUTIONS
3
rd-continuous and F ∆ (t) = f (t), then we define the delta integral by
Z b
f (t)∆t = F (b) − F (a)
a
for all a, b ∈ T.
From now on, T is a closed subset of R such that 0 ∈ Tk , T ∈ Tk . Let E =
Cld ([0, T ], R) which is a Banach space of ld-continuous functions with the maximum
norm kuk = max[0,T ] |u(t)|.
The main tool in this paper is an application of the Guo-Krasnoselskii fixed point
theorem for operators leaving a Banach space cone invariant.
Theorem 2.1 (Krasnosel’skii [13]). Let B be a Banach space, and let P ⊂ B be a
cone in B. Assume that Ω1 and Ω2 are open subsets of B with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 ,
and let
G : P ∩ (Ω2 \Ω1 ) → P
be a completely continuous operator such that either
(i) kGuk ≤ kuk, u ∈ P ∩ ∂Ω1 , and kGuk ≥ kuk, u ∈ P ∩ ∂Ω2 ; or
(ii) kGuk ≥ kuk, u ∈ P ∩ ∂Ω1 , and kGuk ≤ kuk, u ∈ P ∩ ∂Ω2 .
Then, G has a fixed point in P ∩ (Ω2 \Ω1 ).
3. Main Results
By a positive solution of the eigenvalue problem (1.1)-(1.2), we understand a
pair of function (u(t), v(t)) which is positive on [0, T ]T , and satisfies (1.1)-(1.2).
Lemma 3.1. Assume that hypotheses (H1)–(H2) are satisfied. Then a pair of
functions (u(t), v(t)) is a solution of (1.1)-(1.2) if and only if u(t), v(t) ∈ E and
(u(t), v(t)) satisfies the system integral equations:
Z t
Z s
u(t) = u(0) +
A1 − λ
f (v(r))∆r ∆s,
(3.1)
0
0
Z t
Z s
v(t) = v(0) +
A2 − λ
g(u(r))∆r ∆s,
(3.2)
0
0
where
Z η
Z T
1 β
h1 (s)∆s −
h1 (s)∆s ,
1−β
0
0
Z η
Z T
1 v(0) =
β
h2 (s)∆s −
h2 (s)∆s ,
1−β
0
Z0 s
−h1 (s) =
λf (v(r))∆r − A1 ,
0
Z η
λβ
A1 = u∆ (0) = −
f (v(r))∆r,
1−β 0
Z s
−h2 (s) =
λg(u(r))∆r − A2 ,
0
Z η
λβ
∆
g(u(r))∆r.
A2 = u (0) = −
1−β 0
u(0) =
4
M. R. SIDI AMMI, A. B. MALINOWSKA
Proof. Necessity. Integrating (1.1) we have
Z
∆
∆
u (s) = u (0) −
EJDE-2011/131
s
λf (v(r))∆r.
0
On the other hand, by the boundary condition (1.2) we have
Z η
∆
∆
∆
u (0) = βu (η) = β u (0) −
λf (v(r))∆r .
0
Then
Z η
−λβ
f (v(r))∆r.
1−β 0
Z s
∆
u (s) = −λ
f (v(r))∆r + A1 .
A1 = u∆ (0) =
It follows that
0
Integrating the above equation we obtain
Z t
Z
u(t) = u(0) +
A1 − λ
0
s
f (v(r))∆r ∆s.
(3.3)
0
Moreover, by (3.3) and the boundary condition (1.2), we have
Z T
Z s
u(0) = u(T ) −
A1 − λ
f (v(r))∆r ∆s
0
0
T
Z
= βu(η) −
A1 − λ
0
Z
= β u(0) +
η
Z0
A1 − λ
0
Z
T
−
s
Z
f (v(r))∆r ∆s
s
f (v(r))∆r ∆s
0
Z
s
A1 − λ
0
f (v(r))∆r ∆s.
0
Then
Z η
Z s
1 β
A1 − λ
f (v(r))∆r ∆s
u(0) =
1−β
0
0
Z s
Z T
−
A1 − λ
f (v(r))∆r ∆s .
0
0
Then we have (3.1). Similar arguments need to be done to prove (3.2).
Sufficiency. Simple calculations by taking the delta derivative of u(t) and v(t)
lead to the result.
Lemma 3.2. Suppose that (H1)–(H2) hold, then a solution (u, v) of (1.1)-(1.2)
satisfies u(t) ≥ 0 and v(t) ≥ 0, for t ∈ [0, T ]T .
Proof. Since 0 < β < 1, we have
u(0)
β
=
1−β
β
1−β
≥ 0,
≥
Z s
Z T
1
A1 − λ
f (v(r))∆r ∆s −
A1 − λ
f (v(r))∆r ∆s
1−β 0
0
0
0
Z s
Z T
Z s
nZ η o
A1 − λ
f (v(r))∆r ∆s
A1 − λ
f (v(r))∆r ∆s −
Z
η
0
Z
s
0
0
0
EJDE-2011/131
EXISTENCE OF SOLUTIONS
5
and
T
Z
u(T ) = u(0) +
h1 (s)∆s
0
η
T
β
1−β
Z
≥
β
1−β
≥
β
1−β
Z T
β
h1 (s)∆s
1−β 0
0
Z T
Z η
h1 (s)∆s −
h1 (s)∆s
=
h1 (s)∆s −
0
Z
Z
T
h1 (s)∆s +
0
h1 (s)∆s
0
η
Z
h1 (s)∆s −
−β
≥
1−β
Z
β
1−β
Z
≥
1
1−β
0
T
0
h1 (s)∆s
η
T
(−h1 (s))∆s ≥ 0.
η
If t ∈ [0, T ]T , then
u(t) = u(0) +
Z t
0
Z
s
Z
A1 − λ
f (v(r))∆r ∆s
0
T
≥ u(0) +
Z
A1 − λ
0
s
f (v(r))∆r ∆s
0
= u(T ) ≥ 0 .
Arguing exactly as above, we have v(t) ≥ 0. The proof is now complete.
Lemma 3.3. Suppose that (H1)-(H2) are satisfied, then
u(T ) = inf u(t) ≥ ρu(0) = ρkuk,
t∈T
where 1 ≥ ρ =
−η
β TT−βη
≥ 0.
Rs
Proof. Since u∆ (0) ≤ 0 we have u∆ (s) = u∆ (0) − λ 0 f (v(r))∆r ≤ 0, then u∆ ≤ 0,
which implies that u is a non-increasing function on [0, T ]T . Moreover, we have
kuk = u(0), inf t∈(0,T )T u(t) = u(T ). Hence, from the concavity of u(t) that each
point on chord between (0, u(0)) and (T, u(T )) is below the graph of u(t), we have
u(T ) ≥ u(0) + T
u(T ) − u(η)
.
T −η
On other terms
T u(η) − ηu(T ) ≥ (T − η)u(0).
Using the boundary condition (1.2), it follows
(
T
− η)u(T ) ≥ (T − η)u(0).
β
Then
u(T ) ≥ β
T −η
u(0).
T − βη
6
M. R. SIDI AMMI, A. B. MALINOWSKA
EJDE-2011/131
For our construction of an operator G, define a cone K ⊂ E by
K = u ∈ E : u(t) ≥ 0 on T and u(t) ≥ ρkuk, for t ∈ T .
(3.4)
Also define the positive numbers
−1 β
β(T − η)η p
−1 ρ(f∞ − ε)(g∞ − ε)
,
(g∞ − )η
,
(1 − β)
1−β
−1 2T 2
2√2T 2 p
−1 (f
+
ε)(g
+
ε)
,
λ2 = min
(g
+
)
.
0
0
0
(1 − β)2
(1 − β)2
λ1 = max
Theorem 3.4. Assume that conditions (h1)–(H2) are satisfied. Then, for each λ
satisfying λ1 < λ < λ2 , there exists a pair (u, v) satisfying (1.1)-(1.2) such that
u(t) ≥ 0 and v(t) ≥ 0 on [0, T ]T .
Proof. The main idea of the proof is to use Theorem 2.1. First observe that we
seek suitable fixed points in the cone K 3.2 of the integral operator
Gu(t) = u(0) +
Z t
Z
A1 − λ
0
s
f (v(r))∆r ∆s,
(3.5)
0
where v is a function of u given by (3.2).
Lemma 3.5. Let G defined by (3.5), then
(i) G(K) ⊆ K.
(ii) G : K → K is completely continuous.
Notice that by Lemma 3.2 and Lemma 3.3, (i) is satisfied. On the other hand
standard arguments show that G is completely continuous.
Since β < 1 we have
Z η
Z T
1 Gu(t) ≤ u(0) =
−β
h1 (s)∆s +
h1 (s)∆s
1−β
0
0
Z T
2
|h1 (s)|∆s.
≤
1−β 0
(3.6)
Now, from the definitions of f0 and g0 , there exists H1 > 0 such that
f (x) ≤ (f0 + )x, g(x) ≤ (g0 + )x,
for all x such 0 < x ≤ H1 .
(3.7)
Let u ∈ K with kuk = H1 . We have
|v(r)| = v(0) +
Z
r
h2 (s)∆s
Z0 r
|h2 (s)|∆s
≤ |v(0)| +
Z0 η
≤ |v(0)| +
|h2 (s)|∆s.
0
(3.8)
EJDE-2011/131
EXISTENCE OF SOLUTIONS
7
Furthermore,
|h2 (s)| = A2 − λ
Z
s
g(u(r))∆r
Z0 s
≤ |A2 | + λ
|g(u(r))|∆r
0
≤ |A2 | + λ(g0 + )|s|kuk
Z η
λβ
|g(u(r))|∆r + λ(g0 + )|s|kuk
≤
1−β 0
λβ
≤
(g0 + )ηkuk + λ(g0 + )|s|kuk
1−β
λT
(g0 + )kuk.
≤
1−β
(3.9)
Then
Z
T
|h2 (s)|∆s ≤
0
λT 2
(g0 + )kuk.
1−β
(3.10)
Then using (3.10), we have
β
|v(0)| = |
1−β
Z
β
1−β
Z
1+β
≤
1−β
Z
≤
≤ λT 2
η
0
η
T
1
h2 (s)∆s −
1−β
Z
1
1−β
Z
|h2 (s)|∆s +
0
h2 (s)∆s|
0
T
|h2 (s)|∆s
0
T
(3.11)
|h2 (s)|∆s
0
1+β
(g0 + )kuk.
(1 − β)2
Then from (3.8)-(3.11), we obtain
λT 2
1+β
(g
+
)kuk
+
(g0 + )kuk
0
(1 − β)2
(1 − β)
2λT 2
≤
(g0 + )kuk.
(1 − β)2
|v(r)| ≤ λT 2
(3.12)
The choice of λ2 it yields
kvk ≤
2λT 2
2λ2 T 2
(g
+
)kuk
≤
(g0 + )kuk ≤ kuk = H1 .
0
(1 − β)2
(1 − β)2
Then by using (3.7) and (3.13) we obtain
Z η
Z η
λβ
λβ
|A1 | ≤
|f (v(r))|∆r ≤
(f0 + )
|v(r)|∆r.
1−β 0
1−β
0
(3.13)
(3.14)
It follows from (3.12)-(3.14) that
λβ
2λT 2
(f0 + )η
(g0 + )kuk
1−β
(1 − β)2
2λ2 βT 2 η
=
(f0 + )(g0 + )kuk.
(1 − β)3
|A1 | ≤
(3.15)
8
M. R. SIDI AMMI, A. B. MALINOWSKA
Since h1 (s) = A1 − λ
Rs
EJDE-2011/131
f (v(r))∆r, it follows that
Z s
|h1 (s)| ≤ |A1 | + λ
|f (v(r))|∆r.
0
0
We obtain from (3.7) and (3.12), that
Z s
Z s
|f (v(r))|∆r ≤
(f0 + )kvk∆r ≤ T (f0 + )kvk
0
0
≤
2λT 3
(f0 + )(g0 + )kuk.
(1 − β)2
It follows from (3.15)-(3.16) that
2λ2 βT 2 η
2λ2 T 3
|h1 (s)| ≤
(f
+
)(g
+
)kuk
+
(f
+
)(g
+
)kuk
0
0
0
0
(1 − β)3
(1 − β)2
4λ2 T 3
(f0 + )(g0 + )kuk.
≤
(1 − β)3
(3.16)
(3.17)
Combining inequalities (3.12)-(3.16), we have, from (3.6) and the choice of λ2 ,
Z T
2
kGuk ≤
|h1 (s)|∆s
1−β 0
8λ2 T 4
≤
(f0 + )(g0 + )kuk
(1 − β)4
8T 4
(f0 + )(g0 + )kuk = kuk.
≤ λ22
(1 − β)4
So kGuk ≤ kuk. If we set Ω1 = {u ∈ E : kuk < H1 }, then
kGuk ≤ kuk
for u ∈ K ∩ ∂Ω1 .
Now, from the definitions of f∞ and g∞ , there exists H 2 > 0 such that
f (x) ≥ (f∞ − ε)x,
g(x) ≥ (g∞ − ε)x,
for x ≥ H 2 > 0.
(3.18)
Let H2 = max{2H1 , Hρ2 }. Let u ∈ K with kuk = H2 , min u(t) ≥ ρkuk ≥ H 2 .
On the other hand,
Z t
Gu(t) = u(0) +
h1 (s)∆s
β
=
1−β
≥
β
1−β
β
≥
1−β
Z
h1 (s)∆s −
0
1
1−β
Z
T
Z
h1 (s)∆s +
0
Z T
β
h1 (s)∆s
1−β 0
0
Z T
Z η
h1 (s)∆s −
h1 (s)∆s
Z
t
h1 (s)∆s
0
η
h1 (s)∆s −
−β
≥
1−β
Z
β
1−β
Z
≥
0
η
0
T
0
h1 (s)∆s
η
T
(−h1 (s))∆s.
η
(3.19)
EJDE-2011/131
EXISTENCE OF SOLUTIONS
9
Furthermore, from (3.2) and as in (3.19) we have
Z T
β
v(t) ≥
(−h2 (s))∆s.
1−β η
Since A2 ≤ 0, we have from (3.18)
s
Z
−h2 (s) ≥ λ
g(u(r))∆r − A2
0
s
Z
≥λ
g(u(r))∆r
Z s
≥ λ(g∞ − )
u(r)∆r
Z0 s
ρkuk∆r
≥ λ(g∞ − )
0
0
≥ λ(g∞ − )ρkuks.
Then by the choice of λ1 ,
Z T
λβ
(g∞ − )ρkuk
s∆s
v(t) ≥
1−β
η
β
(g∞ − )ρkukη
≥λ
1−β
β
≥ λ1
(g∞ − )ρηkuk
1−β
≥ ρkuk ≥ H2 .
Since A1 ≤ 0, then by (3.20) we have
Z s
Z
−h1 (s) = λ
f (v(r))∆r − A1 ≥ λ
0
s
Z
f (v(r))∆r ≥ λ(f∞ − )
0
s
Z
h2 (s)∆s ≥
0
β
1−β
Z
T
(−h2 (s))∆s.
η
Applying Lemma 3.3, we have
Z
s
g(u(r))∆r − A2
−h2 (s) = λ
0
Z
s
≥λ
g(u(r))∆r
Z0 η
≥λ
g(u(r))∆r
Z η
≥ λ(g∞ − )
u(r)∆r
0
0
≥ λ(g∞ − )ρηkuk.
Consequently,
v(r) ≥ λ
β
(T − η)(g∞ − )ρηkuk.
1−β
η
v(r)∆r.
0
We also have
v(r) = v(0) +
(3.20)
10
M. R. SIDI AMMI, A. B. MALINOWSKA
Then
−h1 (s) ≥ λ2 (T − η)
EJDE-2011/131
βρη 2
(f∞ − )(g∞ − )kuk.
1−β
Finally, for λ ≥ λ1 , we obtain
Gu(t) =
β
1−β
T
Z
(−h1 (s))∆s
η
β 2 η 2 (T − η)2
ρ(f∞ − )(g∞ − )kuk
(1 − β)2
β 2 η 2 (T − η)2
≥ λ21
ρ(f∞ − )(g∞ − )kuk
(1 − β)2
≥ kuk = H2 .
Hence, kGuk ≥ kuk. So, if we set Ω2 = u ∈ E : kuk < H2 , then
\
kGuk ≥ kuk, for u ∈ K ∂Ω2 .
≥ λ2
T
Applying Theorem 2.1, we obtain that G has a fixed point u ∈ K (Ω2 /Ω1 ). With
v being defined by
Z t
Z s
v(t) = v(0) +
A2 − λ
g(u(r))∆r ∆s,
0
0
the pair (u, v) is a solution of (1.1)-(1.2) for the given λ. The proof is now complete.
We point out that, under technical calculus, results of this paper can be extended
to the following p-Laplacian case with 2 ≤ p ≤ +∞:
∆
− φp (u∆ (t))
= λf (v(t)), ∀t ∈ [0, T ]T ,
(3.21)
∆
− φp (v ∆ (t))
= λg(u(t)), ∀t ∈ [0, T ]T ,
where φp (ξ) = |ξ|p−2 ξ. If p = 2 we obtain problem (1.1).
Acknowledgements. This work was supported by the project New Explorations
in Control Theory Through Advanced Research (NECTAR) cofinanced by Fundação
para a Ciência e a Tecnologia (FCT), Portugal, and the Centre National de la
Recherche Scientifique et Technique (CNRST), Morocco. A.B. Malinowska is a
senior researcher at the Center for Research and Development in Mathematics and
Applications (CIDMA) at the University of Aveiro, Portugal, under the support of
BUT grant S/WI/02/2011.
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Moulay Rchid Sidi Ammi
Faculty of Sciences and Techniques, Moulay Ismail University, AMNEA Group, Errachidia, Morocco
E-mail address: sidiammi@ua.pt
Agnieszka B. Malinowska
Faculty of Computer Science, Bialystok University of Technology, 15-351 Bialystok,
Poland
E-mail address: abmalinowska@ua.pt