Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 117, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLUTIONS TO SYSTEMS OF PARTIAL DIFFERENTIAL
EQUATIONS WITH WEIGHTED SELF-REFERENCE AND
HEREDITY
PHAM KY ANH, NGUYEN THI THANH LAN, NGUYEN MINH TUAN
Abstract. This article studies the existence of solutions to systems of nonlinear integro-differential self-referred and heredity equations. We show the
existence of a global solution and the uniqueness of a local solution to a system of integro-differential equations with given initial conditions.
1. Introduction
Self-referred and hereditary phenomena play an important role in applied sciences, especially that in studying evolution processes of biology. Mathematically,
these phenomena can be described by the following model: let A : X → R and
B : X → R be two functionals defined on a function space X. Consider the equation
Au(x, t) = u(Bu(x, t), t),
(1.1)
where u = u(x, t), (x, t) ∈ R×[0, +∞) is an unknown function satisfying some initial
data at t = 0, A and B are differential or/and integral operators. For example, if
Z t
u(x, τ )dτ,
(1.2)
Bu(x, t) =
0
then B is called a hereditary operator. As the unknown function u in the righthand side of the equation (1.1) depends on itself, equation (1.1) may be called a
self-reference equation.
Some special cases of (1.1) were originally studied by Volterra in the 20 century
(see [9, 10] and references therein). It is noticeable to say that some authors considered the variable t as the complex one. In the simple case when B is an identity
operator, Eder [4] obtained the existence, uniqueness, analyticity of solutions, and
the analytic dependence of solutions of the real-variable equation
u0 (t) = u(u(t)).
Si and Cheng [12] investigated a more general functional-differential equation
u0 (t) = u(at + bu(t)),
(1.3)
2000 Mathematics Subject Classification. 35F25, 45G15, 47J35, 47N60, 92D15.
Key words and phrases. Hereditary; self-referred; non-linear integro-differential equations;
recursive scheme.
c
2012
Texas State University - San Marcos.
Submitted June 18, 2012. Published July 14, 2012.
1
2
P. K. ANH, N. T. T. LAN, N. M. TUAN
EJDE-2012/117
where a 6= 1 and b 6= 0 are complex numbers, and u : C → C is the unknown
complex-variable function. In particular, by constructing a convergent power series
solution v(t) of a companion equation of the form
βv 0 (βt) = v 0 (t)[v(β 2 t) − av(βt) + a],
the authors [12] obtained the analytic solution of (1.3) which is of the form
v(βv −1 (t)) − at
.
b
As a development of (1.3), Cheng, Si, and Wang [23] studied the equation
αt + βu0 (t) = u at + bu0 (t) ,
where α and β are complex numbers. The main results of [23] are the existence
theorems for the analytic solutions, and an explicit solution via symmetric methods.
Equations of the form (1.1) attract attention of many authors. More investigations can be found in [13, 14, 15, 16, 17, 18, 7, 2, 5, 8, 3, 9, 10, 11, 6, 20, 21, 1], and
references therein.
In recent years, Pascali and Miranda obtained many results concerning the selfreferred functional-differential equations [8, 9, 10, 11]. For instance, the authors in
[9] studied the initial-value problem
Z
1 t
∂
u(x, t) = u(
u(x, s)ds, t), x ∈ R, t ∈ [0, T ],
∂t
t 0
u(x, 0) = u0 (x), x ∈ R.
(1.4)
The authors claimed that under some suitable conditions problem (1.4) has a unique
bounded and continuous solution. Observe that the unknown u in the right-hand
side of (1.4) contains a weighted hereditary operator
1
(Bu)(t) :=
t
Z
t
u(x, s)ds.
0
Motivated by the long list of works on self-referred functional-differential equations as mentioned above, we study the following system of two partial-differential
equations with self-reference and weighted heredity
Z
∂
1 t
u(x, t) = u f u(x, t) + v
u(x, s)ds + ϕ(u(x, t)), t , t
∂t
t 0
(1.5)
Z
∂
1 t
v(x, t) = v g v(x, t) + u
v(x, s)ds + ψ(v(x, t)), t , t ,
∂t
t 0
associated with the initial conditions
u(x, 0) = u0 (x),
v(x, 0) = v0 (x),
(1.6)
where f, g, ϕ, ψ, u0 and v0 are given functions satisfying some suitable conditions.
This work is devoted to the uniqueness of local solution, and the existence of a
global solution of this problem.
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SOLUTIONS TO SYSTEMS OF PDES
3
2. Preliminaries
To study problem (1.5)-(1.6), we reduce it to the following system of two integral
equations
Z t 1 Z s
u(x, t) = u0 (x) +
u f u(x, s) + v
u(x, τ )dτ + ϕ(u(x, s)), s , s ds,
s 0
0
Z t 1 Z s
v(x, t) = v0 (x) +
v(x, τ )dτ + ψ(v(x, s)), s , s ds.
v g v(x, s) + u
s 0
0
(2.1)
Proposition 2.1. If problem (2.1) has a solution (u, v), then the pair of functions
(u, v) solves problem (1.5)-(1.6).
We omit the proof of this proposition, as it is quite simple. Therefore, we shall
investigate problem (2.1) hereafter. We now define the sequences of real functions
{un }n≥1 , {vn }n≥1 as follows:
Z t u1 (x, t) = u0 (x) +
u0 f u0 (x) + v0 u0 (x) + ϕ u0 (x)
ds,
0
Z t v1 (x, t) = v0 (x) +
v0 g v0 (x) + u0 v0 (x) + ψ v0 (x)
ds,
0
Z t un+1 (x, t) = u0 (x) +
un f un (x, s)
0
(2.2)
1 Z s
un (x, τ )dτ + ϕ un (x, s) , s , s ds,
+ vn
s 0
Z t vn+1 (x, t) = v0 (x) +
vn g vn (x, s)
0
1 Z s
vn (x, τ )dτ + ψ vn (x, s) , s , s ds,
+ un
s 0
for x ∈ R and t > 0.
We should give the following additional conditions on the functions u0 , v0 and
f, g, ϕ, ψ:
(A1) u0 and v0 are bounded and Lipschitz continuous on R.
(A2) f, g, ϕ and ψ are Lipschitz continuous on R.
The functional inequalities in the next lemma are useful for proving the main results.
Lemma 2.2. Assume that the functions u0 , v0 and f, g, ϕ, and ψ satisfy conditions
as in (A1)–(A2). For any n ≥ 1 there exist two continuous, non-negative functions
defined on R+ , say Mn (t) and Nn (t), such that the following two inequalities hold:
|un+1 (x, t) − un+1 (y, t)| ≤ Mn+1 (t)|x − y|,
n ∈ N, x, y ∈ R
|vn+1 (x, t) − vn+1 (y, t)| ≤ Nn+1 (t)|x − y|,
n ∈ N, x, y ∈ R.
Moreover, there is a positive constant T1 such that the non-negative function sequences {Mn (t)}n≥1 , {Nn (t)}n≥1 are uniformly bounded on the interval (0, T1 ];
i.e., there exists a constant G0 > 0 such that 0 < Mn (t), Nn (t) ≤ G0 for every
t ∈ (0, T1 ], and for any n ≥ 1.
4
P. K. ANH, N. T. T. LAN, N. M. TUAN
EJDE-2012/117
Proof. For n = 0, we have
|u0 (x) − u0 (y)| ≤ M0 |x − y|,
|v0 (x) − v0 (y)| ≤ N0 |x − y|
for any x, y ∈ R and for some M0 > 0, N0 > 0. Let P, Q, $, σ > 0 such that
|f (α1 ) − f (α2 )| ≤ P |α1 − α2 |,
α1 , α2 ∈ R
|g(β1 ) − g(β2 )| ≤ Q|β1 − β2 |,
β1 , β2 ∈ R
|ϕ(γ1 ) − ϕ(γ2 )| ≤ $|γ1 − γ2 |,
γ1 , γ2 ∈ R
|ψ(η1 ) − ψ(η2 )| ≤ σ|η1 − η2 |,
η1 , η2 ∈ R.
(2.3)
For n = 1 we have
|u1 (x, t) − u1 (y, t)| ≤ M1 (t)|x − y|,
where
M1 (t) = M0 + t(M02 P + M02 N0 + M02 N0 $),
and
|v1 (x, t) − v1 (y, t)| ≤ N1 (t)|x − y|,
where
N1 (t) = N0 + t(N02 Q + M0 N02 + M0 N02 σ).
For n = 2, we derive
|u2 (x, t) − u2 (y, t)| ≤ M2 (t)|x − y|,
where
M2 (t) = M0 +
Z t
M12 (s)P + N1 (s)
0
1 d
2s ds
s
Z
M1 (τ )dτ
2
+ M12 (s)N1 (s)$ ds,
0
and
|v2 (x, t) − v2 (y, t)| ≤ N2 (t)|x − y|,
where
N2 (t) = N0 +
Z t
N12 (s)Q + M1 (s)
0
1 d
2s ds
s
Z
N1 (τ )dτ
2
+ M1 (s)N12 (s)σ ds.
0
We can inductively prove that
|un+1 (x, t) − un+1 (y, t)| ≤ Mn+1 (t)|x − y|,
(2.4)
where
Mn+1 (t) = M0 +
Z t
Mn2 (s)P + Nn (s)
0
1 d
2s ds
s
Z
Mn (τ )dτ
2
+ Mn2 (s)Nn (s)$ ds,
0
and
|vn+1 (x, t) − vn+1 (y, t)| ≤ Nn+1 (t)|x − y|,
(2.5)
where
Nn+1 (t) = N0 +
Z t
Nn2 (s)Q + Mn (s)
0
1 d
2s ds
s
Z
Nn (τ )dτ
2
+ Mn (s)Nn2 (s)σ ds.
0
Clearly, the functions Mn+1 (t) and Nn+1 (t) are non-negative and continuous on
R. We shall prove that each one of the function sequences {Mn+1 }n≥1 (t) and
{Nn+1 }n≥1 (t) is uniformly bounded on some (0, T1 ]. Indeed, by choosing constants
K0 , H0 , and I0 > 0 fulfilling the following conditions
N 0 + K0 ≤ H0
M0 + K0 ≤ I0
G0 = max{H0 , I0 },
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SOLUTIONS TO SYSTEMS OF PDES
5
there exists a number T1 > 0 such that
(M02 P + M02 N0 + M02 N0 $)t ≤ K0 ,
(N02 Q
Then
+
∀t ∈ (0, T1 ]
M0 N02 + M0 N02 σ)t ≤ K0 , ∀t
(G20 P + G30 + G30 $)t ≤ K0 ,
(G20 Q + G30 + G30 σ)t ≤ K0 .
∈ (0, T1 ]
(2.6)
0 ≤ M1 (t) − M0 = (M02 P + M02 N0 + M02 N0 $)t ≤ K0 ,
(2.7)
0 ≤ N1 (t) − N0 = (N02 Q + M0 N02 + M0 N02 σ)t ≤ K0 .
It follows that
0 ≤ M1 (t) ≤ K0 + M0 ≤ I0 ≤ G0 ,
(2.8)
0 ≤ N1 (t) ≤ K0 + N0 ≤ H0 ≤ G0 .
Similarly,
t
Z
(G20 P + G30 + G30 $)ds = (G20 P + G30 + G30 $)t ≤ K0 ,
0 ≤ M2 (t) − M0 ≤
0
Z
0 ≤ N2 (t) − N0 ≤
(2.9)
t
(G20 Q
+
G30
+
G30 σ)ds
=
(G20 Q
+
G30
+
G30 σ)t
≤ K0 .
0
From these inequalities, we have
0 ≤ M2 (t) ≤ M0 + K0 ≤ I0 ≤ G0 ,
0 ≤ N2 (t) ≤ N0 + K0 ≤ H0 ≤ G0 .
(2.10)
By induction on n we obtain
0 ≤ Mn+1 (t) ≤ M0 + K0 ≤ G0 ,
0 ≤ Nn+1 (t) ≤ N0 + K0 ≤ G0 ,
for every t ∈ (0, T1 ], T1 > 0. The lemma is proved.
(2.11)
We can see that Lemma 2.2 concerns the properties of the functions {un (x, t)}
and {vn (x, t)}, while Lemma 2.3 concerns the recursive sequences {un+1 (x, t) −
un (x, t)} and {vn+1 (x, t) − vn (x, t)}.
Lemma 2.3. Assume that the functions u0 , v0 and f, g, ϕ, and ψ satisfy conditions
as in (A1)–(A2). For any n ≥ 1 there exist two nonnegative, continuous functions,
say An (t) and Bn (t), satisfying the following two inequalities:
|un+1 (x, t) − un (x, t)| ≤ An+1 (t),
x ∈ R, t ∈ R+ ,
|vn+1 (x, t) − vn (x, t)| ≤ Bn+1 (t),
x ∈ R, t ∈ R+ .
Moreover, there is a positive constant T2 such that the both series with general terms
An (t), and Bn (t) are uniformly convergent on (0, T2 ].
Proof. We have
|u1 (x, t) − u0 (x)| ≤ tku0 kL∞ := A1 (t),
|v1 (x, t) − v0 (x)| ≤ tkv0 kL∞ := B1 (t).
Similarly,
|u2 (x, t) − u1 (x, t)| ≤
Z t
0
A1 (s) 1 + M0 P + M0 N0 $ + M0 B1 (s)
6
P. K. ANH, N. T. T. LAN, N. M. TUAN
+ M0 N 0
1
s
Z
s
EJDE-2012/117
A1 (τ )dτ ds := A2 (t),
0
and
|v2 (x, t) − v1 (x, t)| ≤
Z t
B1 (s) 1 + N0 Q + M0 N0 σ + N0 A1 (s)
0
Z
1 s
+ M0 N 0
B1 (τ )dτ ds := B2 (t).
s 0
By induction on n, we conclude that
|un+1 (x, t) − un (x, t)| ≤ An+1 (t),
(2.12)
where
An+1 (t) =
Z t
An (s) 1 + Mn−1 (s)P + Mn−1 (s)Nn−1 (s)$
0
Z
1 s
An (τ )dτ ds;
+ Bn (s)Mn−1 (s) + Mn−1 (s)Nn−1 (s)
s 0
and
|vn+1 (x, t) − vn (x, t)| ≤ Bn+1 (t),
(2.13)
where
Bn+1 (t) =
Z t
Bn (s) 1 + Nn−1 (s)Q + Mn−1 (s)Nn−1 (s)σ
0
Z
1 s
Bn (τ )dτ ds.
+ An (s)Nn−1 (s) + Mn−1 (s)Nn−1 (s)
s 0
For a number h ∈ (0, 1/2), we can choose T2 > 0 such that the following two
inequalities hold for any t ∈ (0, T2 ],
1
,
2
1
(1 + G0 Q + G0 + G20 σ + G20 )t ≤ h < ,
2
(1 + G0 P + G0 + G20 $ + G20 )t ≤ h <
(2.14)
By (2.14) and Lemma 2.2,
0 ≤ An+1 (t) ≤ (1 + G0 P + G20 $ + G20 )tkAn kL∞ + G0 tkBn kL∞
≤ h(kAn kL∞ + kBn kL∞ ),
(2.15)
and
0 ≤ Bn+1 (t) ≤ (1 + G0 Q + G20 σ + G20 )tkBn kL∞ + G0 tkAn kL∞
≤ h(kAn kL∞ + kBn kL∞ ).
(2.16)
By induction on n, we obtain
0 ≤ An+1 (t), Bn+1 (t) ≤ hn kA1 k∞ + kB1 k∞ ,
for t ∈ (0, T2 ]. Therefore, the series with general terms An (.) and Bn (.) uniformly
converge on the interval (0, T2 ]. Lemma 2.3 is proved.
EJDE-2012/117
SOLUTIONS TO SYSTEMS OF PDES
7
Remark 2.4. It is easy to prove inductively that
|un+1 (x, t)| ≤ et ku0 k∞ ,
|vn+1 (x, t)| ≤ et kv0 k∞ .
If we consider T such that 0 < T ≤ min{T1 , T2 }, the functions un (x, t), vn (x, t)
are bounded uniformly with respect to variable x ∈ R, for t ∈ (0, T ]. On the other
hand, due to (2.8) and Lemma 2.2, the functions un (x, t), vn (x, t) are uniformly
Lipschitz continuous with respect to each of the variables x ∈ R and t ∈ (0, T ].
For serving the existence of a global solution to problem (2.1), we propose some
assumptions on the functions u0 , v0 and f, g, ϕ, ψ, that are different from (A1)–(A2)
in Lemma 2.3. Namely, assume that
(B1) u0 and v0 are non-negative, non-decreasing, bounded and lower semi-continuous on R.
(B2) f, g, ϕ and ψ are non-decreasing and lower semi-continuous.
Lemma 2.5. Suppose that the functions u0 , v0 and f, g, ϕ and ψ fulfill the conditions as in (B1)–(B2). Then the functions {un (x, t)}n≥1 and {vn (x, t)}n≥1 possess
the following properties:
(C1) un and vn are non-negative.
(C2) un and vn are non-decreasing with respect to each one of variables x ∈
R, t ∈ (0, T ]; more precisely un+1 ≥ un , vn+1 ≥ vn .
(C3) un and vn are lower semi-continuous with respect to x, for every t ∈
(0, +∞).
(C4) un and vn are Lipschitz continuous with respect to t, uniformly bounded
with respect to x ∈ R.
Proof. We have
u1 (x, t) ≥ u0 (x) ≥ 0,
∀x ∈ R, t ∈ (0, +∞),
v1 (x, t) ≥ v0 (x) ≥ 0,
∀x ∈ R, t ∈ (0, +∞).
For t1 , t2 ∈ (0, +∞), t1 < t2 , and for x ∈ R, we have
Z t2 u1 (x, t2 ) = u0 (x) +
u0 f u0 (x) + v0 u0 (x) + ϕ u0 (x) ds
0
Z t1 ≥ u0 (x) +
u0 f u0 (x) + v0 u0 (x) + ϕ u0 (x)
ds
(2.17)
(2.18)
0
= u1 (x, t1 ),
Z t2 v0 g v0 (x) + u0 v0 (x) + ψ v0 (x)
v1 (x, t2 ) = v0 (x) +
ds
0
Z t1 ≥ v0 (x) +
v0 g v0 (x) + u0 v0 (x) + ψ v0 (x)
ds
(2.19)
0
= v1 (x, t1 ).
Similarly, for all x1 , x2 ∈ R, x1 < x2 , for all t ∈ (0, +∞), we derive
Z t u1 (x1 , t) = u0 (x1 ) +
u0 f u0 (x1 ) + v0 u0 (x1 ) + ϕ u0 (x1 )
ds
0
Z t ≤ u0 (x2 ) +
u0 f u0 (x2 ) + v0 u0 (x2 ) + ϕ u0 (x2 )
ds
0
= u1 (x2 , t),
(2.20)
8
P. K. ANH, N. T. T. LAN, N. M. TUAN
Z
t
v1 (x1 , t) = v0 (x1 ) +
EJDE-2012/117
v0 g v0 (x1 ) + u0 v0 (x1 ) + ψ v0 (x1 )
ds
0
Z
≤ v0 (x2 ) +
t
v0 g v0 (x2 ) + u0 v0 (x2 ) + ψ v0 (x2 )
ds
(2.21)
0
= v1 (x2 , t).
Using (2.18)–(2.21), we can prove inductively that
un (x, t2 ) ≥ un (x, t1 ),
un (x2 , t) ≥ un (x1 , t),
∀t ∈ (0, +∞), x2 > x1 ,
vn (x, t2 ) ≥ vn (x, t1 ),
vn (x2 , t) ≥ vn (x1 , t),
∀x ∈ R, t2 > t1 ,
∀x ∈ R, t2 > t1 ,
(2.22)
∀t ∈ (0, +∞), x2 > x1 .
Also, we can prove that (see also remark 2.4)
0 ≤ un (x, t) ≤ un+1 (x, t) ≤ eT ku0 kL∞ ,
0 ≤ vn (x, t) ≤ vn+1 (x, t) ≤ eT kv0 kL∞ ,
for all x ∈ R, t ∈ (0, T ] and n ∈ N. On the other hand,
Z
t2
|un+1 (x, t1 ) − un+1 (x, t2 ) ≤ ku0 kL∞ eT ds ≤ ku0 kL∞ eT |t2 − t1 |,
t1
Z
t2
|vn+1 (x, t1 ) − vn+1 (x, t2 )| ≤ kv0 kL∞ eT ds ≤ kv0 kL∞ eT |t2 − t1 |.
(2.23)
(2.24)
(2.25)
t1
Relations (2.24) and (2.25) ensure that un and vn satisfy (C4 ). Since the sequences
(un ) and (vn ) are non decreasing, above and upper bounded, there exist the limits
u∞ (x, t) = lim un (x, t),
n
v∞ (x, t) = lim vn (x, t).
n
(2.26)
Since u0 , v0 , f, g, ϕ and ψ are lower
semi-continuous and
non-decreasing, the functions f (u0 ), g(v0 ), v0 u0 + ϕ(u0 ) , and u0 v0 + ψ(v0 ) are lower semi-continuous
and non-decreasing (see[21, Lemma 3]). Hence, u0 f (u0 ) + v0 u0 + ϕ(u0 ) , and
v0 g(v0 )+u0 v0 +ψ(v0 ) are lower semi-continuous and non-decreasing, too. Thus,
the lower semi-continuity and the decrease of u1 (x, t) and v1 (x, t) are established.
By induction on n, we can conclude that un (x, t) and vn (x, t) are lower semicontinuous and non-decreasing. Lemma 2.5 is proved.
3. Main results
Theorem 3.1 (Uniqueness of local solutions). Assume that the functions f , g,
ϕ, ψ, u0 , and v0 satisfy (A1)–(A2). Then there exists a positive constant T? such
that (2.1) has a unique solution on R × (0, T∗ ] denoted by {u∗ , v∗ }. Moreover, the
functions u∞ , v∞ are Lipschitz continuous and bounded with respect to each of the
variables x ∈ R, and t ∈ (0, T? ].
Theorem 3.2 (Existence of global solutions). Assume that f , g, ϕ, ψ, u0 and v0
satisfy (B1)–(B2). There exist two functions u∞ , v∞ : R × (0, +∞) → R satisfying
(2.1) for t ∈ (0, +∞). Moreover, these solutions have the properties similar to those
of {un (x, t)}n≥1 and {vn (x, t)}n≥1 as in Lemma 2.5; namely, the functions u∞ , v∞
possess the properties (C1)–(C4).
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SOLUTIONS TO SYSTEMS OF PDES
9
Proof of Theorem 3.1. Write T∗ := min{T1 , T2 }. By Lemmas 2.2, and 2.3, the
limits u∞ (x, t), v∞ (x, t) of the sequences {un (x, t)}n≥1 , {vn (x, t)}n≥1 are bounded
on R × (0, T∗ ], Lipschitz continuous with respect to each of variables, and satisfy
problem (2.1).
Now, suppose that (u? , v? ) is another solution of (2.1) on R × (0, T∗ ] with the
same given data. We have
1 Z t
u? (x, s)ds + ϕ(u? (x, t)), t , t
u? f (u? (x, t)) + v?
t 0
1 Z t
− u∞ f (u∞ (x, t)) + v∞
u∞ (x, s)ds + ϕ(u∞ (x, t)), t , t t 0
1 Z t
u? (x, s)ds + ϕ(u? (x, t)), t , t
≤ ku? − u∞ kL∞ + u∞ f (u? (x, t)) + v?
t 0
1 Z t
u∞ (x, s)ds + ϕ(u∞ (x, t)), t , t − u∞ f (u∞ (x, t)) + v∞
t 0
≤ (1 + M∞ (t)P + M∞ (t)N∞ (t) + M∞ (t)N∞ (t)$)ku? − u∞ kL∞
+ M∞ (t)kv? − v∞ kL∞ .
From the above inequality and Lemma (2.2) we obtain
|u? (x, t) − u∞ (x, t)|
≤ 1 + G0 P + G20 + G20 $ tku? − u∞ kL∞ + G0 tkv? − v∞ kL∞ .
In addition, we have
1 Z t
g(u
,
v
)
+
u
v? (x, s)ds + ψ(v? (x, t)), t , t
v?
? ?
?
t 0
1 Z t
− v∞ g(u∞ , v∞ ) + u∞
v∞ (x, s)ds + ψ(v∞ (x, t)), t , t t 0
≤ 1 + N∞ (t)Q + M∞ (t)N∞ (t)σ kv? − v∞ kL∞ + N∞ (t)ku? − u∞ kL∞ .
(3.1)
(3.2)
By (3.2) and Lemma (2.2), we find
|v? (x, t) − v∞ (x, t)|
≤ 1 + G0 Q + G20 + G20 σ tkv? − v∞ kL∞ + G0 tku? − u∞ kL∞ .
(3.3)
Combining (3.1) and (3.3), we obtain
|u? (x, t) − u∞ (x, t)|
≤ 1 + G0 P + G0 + G20 + G20 $ t max{ku? − u∞ kL∞ , kv? − v∞ kL∞ },
(3.4)
|v? (x, t) − v∞ (x, t)|
≤ 1 + G0 Q + G0 + G20 + G20 σ t max{ku? − u∞ kL∞ , kv? − v∞ kL∞ }.
(3.5)
and
Taking account of (2.14), (3.4) and (3.5), we have
|u? (x, t) − u∞ (x, t)| ≤ h max{ku? − u∞ kL∞ , kv? − v∞ kL∞ },
|v? (x, t) − v∞ (x, t)| ≤ h max{ku? − u∞ kL∞ , kv? − v∞ kL∞ },
(3.6)
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P. K. ANH, N. T. T. LAN, N. M. TUAN
EJDE-2012/117
for all t ∈ (0, T0 ], x ∈ R. Finally, we conclude that
max{ku? − u∞ kL∞ , kv? − v∞ kL∞ } ≤ h max{ku? − u∞ kL∞ , kv? − v∞ kL∞ }.
The last inequality ensures the uniqueness of the solution. Theorem 3.1 is proved.
Proof of Theorem 3.2. Thanks to (2.22) and (2.23), the following two limits exit:
u∞ (x, t) = sup un (x, t),
n
v∞ (x, t) = sup vn (x, t).
(3.7)
n
We shall prove that u∞ (x, t), v∞ (x, t) satisfy (2.1). From (3.7) we have
un+1 (x, t) − u0 (x)
Z t 1 Z s
un (x, τ )dτ + ϕ un (x, s) , s , s ds
un f un (x, s) + vn
=
s 0
0
Z t
1 Z s
≤
u∞ f u∞ (x, s) + v∞
u∞ (x, τ )dτ + ϕ u∞ (x, s) , s , s ds,
s 0
0
(3.8)
and
vn+1 (x, t) − v0 (x)
Z t 1 Z s
vn (x, τ )dτ + ψ vn (x, s) , s , s ds
vn g vn (x, s) + vn
=
s 0
0
Z t
1 Z s
≤
v∞ g v∞ (x, s) + u∞
v∞ (x, τ )dτ + ψ v∞ (x, s) , s , s ds.
s 0
0
As un (x, t), and vn (x, t) are non-decreasing, we have
1 Z t
un+p (x, s)ds + ϕ un+p (x, t) , t , t
un+p f un+p (x, t) + vn+p
t 0
1 Z t
≥ un f un+p (x, t) + vn+p
un+p (x, s)ds + ϕ un+p (x, t) , t , t ,
t 0
(3.9)
(3.10)
(3.11)
and
1 Z t
vn+p g vn+p (x, t) + un+p
vn+p (x, s)ds + ψ vn+p (x, t) , t , t
t 0
1 Z t
≥ vn g vn+p (x, t) + un+p
vn+p (x, s)ds + ψ vn+p (x, t) , t , t .
t 0
(3.12)
From (3.10) and (3.12) we deduce
Z t
un+p f un+p (x, s)
lim
p→∞ 0
1 Z s
(3.13)
+ vn+p
un+p (x, τ )dτ + ϕ un+p (x, s) , s , s ds
s 0
Z t Z
1 s
≥
un f u∞ (x, s) + v∞
u∞ (x, τ )dτ + ϕ u∞ (x, s) , s , s ds,
s 0
0
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SOLUTIONS TO SYSTEMS OF PDES
11
and
Z
t
vn+p g vn+p (x, s)
p→∞ 0
1 Z s
vn+p (x, τ )dτ + ψ vn+p (x, s) , s , s ds
+ un+p
s 0
Z t 1 Z s
v∞ (x, τ )dτ + ψ v∞ (x, s) , s , s ds.
≥
vn g v∞ (x, s) + u∞
s 0
0
lim
(3.14)
Hence,
lim[un+p+1 (x, t) − u0 (x)]
p
Z t
= lim
un+p f un+p (x, s)
p
0
1 Z s
(3.15)
un+p (x, τ )dτ + ϕ un+p (x, s) , s , s ds
+ vn+p
s 0
Z t 1 Z s
≥
un f u∞ (x, s) + v∞
u∞ (x, τ )dτ + ϕ u∞ (x, s) , s , s ds,
s 0
0
and
lim[vn+p+1 (x, t) − v0 (x)]
p
Z t
= lim
vn+p g vn+p (x, s)
p
0
1 Z s
vn+p (x, τ )dτ + ψ vn+p (x, s) , s , s ds
+ un+p
s 0
Z t 1 Z s
≥
vn g v∞ (x, s) + u∞
v∞ (x, τ )dτ + ψ v∞ (x, s) , s , s ds.
s 0
0
(3.16)
By (3.15)-(3.16) we find that
Z t
u∞ (x, t) − u0 (x) ≥
u∞ f u∞ (x, s)
0
1 Z s
+ v∞
u∞ (x, τ )dτ + ϕ u∞ (x, s) , s , s ds,
s 0
(3.17)
Z t
v∞ (x, t) − v0 (x) ≥
v∞ g v∞ (x, s)
0
1 Z s
v∞ (x, τ )dτ + ψ v∞ (x, s) , s , s ds.
+ u∞
s 0
(3.18)
and
Combining (3.8)-(3.9) and (3.17)-(3.18) we obtain
Z t
u∞ (x, t) − u0 (x) =
u∞ f u∞ (x, s)
0
1 Z s
+ v∞
u∞ (x, τ )dτ + ϕ u∞ (x, s) , s , s ds,
s 0
(3.19)
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P. K. ANH, N. T. T. LAN, N. M. TUAN
EJDE-2012/117
and
Z
v∞ (x, t) − v0 (x) =
t
v∞ v∞ (x, s)
0
1 Z s
v∞ (x, τ )dτ + ψ v∞ (x, s) , s , s ds.
+ u∞
s 0
(3.20)
The above equalities imply that (u∞ , v∞ ) is a solution of (2.1).
On the other hand, it is easily seen that u∞ , v∞ are Lipschitz continuous in t on
(0, +∞). The proof is complete.
4. Illustrative Example
Consider the initial-value problem for a system of integro-differential equations
(1.5)-(1.6) with the following data:
(
1 − |x| if |x| ≤ 1
u0 (x) =
0
otherwise
v0 (x) = 1
f (u) = u,
for all x ∈ R,
g(v) = v,
ϕ(u) = ψ(v) = 0.
We compute the successive approximations as follows:
Z t u1 (x, t) = u0 (x) +
u0 f (u0 (x)) + v0 (u0 (x) + ϕ(u0 (x))) ds
0
Z t
Z t
= u0 (x) +
u0 (u0 (x) + 1)ds = u0 (x) +
0ds = u0 (x),
0
0
Z t v1 (x, t) = v0 (x) +
v0 g(v0 (x)) + u0 (v0 (x) + ψ(v0 (x))) ds
0
Z t
=1+
1ds = 1 + t.
(4.1)
0
Similarly, u2 (x, t) = u0 (x), v2 (x, t) = 1 + t + (t2 /2). Suppose that
un (x, t) = u0 (x),
vn (x, t) =
n
X
ti
i=0
i!
.
(4.2)
We can prove inductively that
un+1 (x, t) = u0 (x),
vn+1 (x, t) =
n+1
X i
i=0
t
.
i!
Tending n to infinity we obtain
u? (x, t) = u0 (x),
v? (x, t) = et .
(4.3)
In fact, we can choose u0 (x) as a nonnegative, Lipschitz continuous function having
a compact support, and v0 (x) = c as a constant function. Due to the symmetry of
the system, the functions u0 and v0 are interchangeable.
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SOLUTIONS TO SYSTEMS OF PDES
13
Concluding remarks. Mathematically, one can provide acceptable assumptions
on equations, and add suitable restrictions on initial data of problems so that the
solution exists uniquely. Therefore, both the existence and uniqueness of solutions
of self-referred and heredity problems, in general, remain considerable challenges to
attempts at generalization, namely (see also [4, 8, 9, 10, 11]),
(1) The uniqueness/non-uniqueness of global solutions, with also relaxed condition on data.
(2) Structure of the solution set.
(3) Numerical solution for the above mentioned system.
Acknowledgements. This work is partially supported by the Vietnam National
Foundation for Science and Technology Development (NAFOSTED).
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Pham Ky Anh
Department of Mathematics, School of Science, Vietnam National University, 334 Nguyen
Trai Str., Hanoi, Vietnam
E-mail address: anhpk@vnu.edu.vn, anhpk2009@gmail.com
Nguyen Thi Thanh Lan
Faculty of Mathematics and Applications, Saigon University, 273 An Duong Vuong str.,
w. 3, dist. 5, Ho Chi Minh City, Vietnam
E-mail address: nguyenttlan@gmail.com
Nguyen Minh Tuan
Department of Mathematics, University of Education, Vietnam National University,
G7 Build., 144 Xuan Thuy Rd., Cau Giay Dist., Hanoi, Vietnam
E-mail address: tuannm@hus.edu.vn