Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 4, Article 82, 2003
ON AN INTEGRAL INEQUALITY WITH A KERNEL SINGULAR IN TIME AND
SPACE
NASSER-EDDINE TATAR
K ING FAHD U NIVERSITY OF P ETROLEUM AND M INERALS
D EPARTMENT OF M ATHEMATICAL S CIENCES
D HAHRAN 31261, S AUDI A RABIA .
tatarn@kfupm.edu.sa
Received 07 March, 2003; accepted 01 August, 2003
Communicated by D. Bainov
A BSTRACT. In this paper we deal with a nonlinear singular integral inequality which arises in
the study of partial differential equations. The integral term is non local in time and space and
the kernel involved is also singular in both the time and the space variable. The estimates we
prove may be used to establish (global) existence and asymptotic behavior results for solutions
of the corresponding problems.
Key words and phrases: Nonlinear integral inequality, Singular kernel.
2000 Mathematics Subject Classification. 42B20, 26D07, 26D15.
1. I NTRODUCTION
We consider the following integral inequality
Z Z t
F (s)ϕm (s, y)
(1.1)
ϕ(t, x) ≤ k(t, x) + l(t, x)
dyds, x ∈ Ω, t > 0,
1−β |x − y|n−α
Ω 0 (t − s)
where Ω is a domain in Rn (n ≥ 1) (bounded or possibly equal to Rn ), the functions k(t, x),
l(t, x) and F (t) are given positive continuous functions in t. The constants 0 < α < n, 0 <
β < 1 and m > 1 will be specified below.
This inequality arises in the theory of partial differential equations, for example, when treating the heat equation with a source of polynomial type

 ut (t, x) = ∆u(t, x) + um (t, x), x ∈ Rn , t > 0, m > 1
 u(0, x) = u (x), x ∈ Rn .
0
If we write the (weak) solution using the fundamental solution G(t, x) of the heat equation
ut (t, x) = ∆u(t, x),
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
The author would like to thank King Fahd University of Petroleum and Minerals for its financial support.
030-03
2
NASSER - EDDINE TATAR
namely,
Z tZ
Z
G(t, x − y)u0 (y)dy +
u(t, x) =
Rn
0
G(t − s, x − y)um (s, y)dyds
Rn
and take into account the Solonnikov estimates of this fundamental solution (see [14] for instance), then one is led to an inequality of type (1.1).
The features of this inequality, which make it difficult to deal with, are the singularities of
the kernel in both the space and time variables. It is also non integrable with respect to the
time variable. The standard methods one can find in the literature (see the recent books by
Bainov and Simeonov [1] and Pachpatte [12] and the references therein) concerning regular
and/or summable kernels cannot be applied in our situation. Indeed, these methods are based
on estimates involving the value of the kernels at zero and/or some Lp −norms of the kernels.
In contrast, there are very few papers dealing explicitly with singular kernels similar to ours.
Let us point out, however, some works concerning integral equations with singularities in time.
In Henry [4, Lemmas 7.1.1 and 7.1.2], a similar inequality to (1.1) with only the integral with
respect to time, namely
Z t
ψ(t) ≤ a(t) + b (t − s)β−1 sγ−1 ψ(s)ds, β > 0, γ > 0
0
i.e. the linear case (m = 1) has been treated. The case m > 1 has been considered by Medved
in [9], [10]. More precisely, the following inequality
Z t
ψ(t) ≤ a(t) + b(t) (t − s)β−1 sγ−1 F (s)ψ m (s)ds, β > 0, γ > 0
0
was discussed. The result was used to prove a global existence and an exponential decay result
for a parabolic Cauchy problem with a source of power type and a time dependent coefficient,
namely

 ut + Au = f (t, u), u ∈ X,
 u(0) = u ∈ X
0
with kf (t, u)k ≤ tκ η(t) kukm
α , m > 1, κ ≥ 0, where A is a sectorial operator (see [4]) and
k·kα stands for the norm of the fractional space X α associated to the operator A (see also [11]).
This, in turn, has been improved and extended to integro-differential equations and functional
differential equations by M. Kirane and N.-E. Tatar in [6] (see also N.-E. Tatar [13] and S.
Mazouzi and N.-E. Tatar [7, 8] for more general abstract semilinear evolution problems).
Here, we shall combine the techniques in [9, 10], based on the application of Lemma 2.1
and the use of Lemma 2.3 below, with the Hardy-Littlewood-Sobolev inequality (see Lemma
2.2) to prove our result. We will give sufficient conditions yielding boundedness by continuous
functions, exponential decay and polynomial decay of solutions to the integral inequality (1.1).
The paper is organized as follows. In the next section we prepare some notation and lemmas
needed in the proofs of our results. Section 3 contains the statement and proof of our result and
two corollaries giving sufficient conditions for the exponential decay and the polynomial decay.
Finally we point out that our results hold (a fortiori) for weakly singular kernels in time.
2. P RELIMINARIES
In this section we introduce some material necessary for our results. We will use the usual
Lp -space with its norm k·kp .
J. Inequal. Pure and Appl. Math., 4(4) Art. 82, 2003
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O N AN I NTEGRAL I NEQUALITY
WITH A
Lemma 2.1. (Young inequality) We have, for
K ERNEL S INGULAR IN T IME AND S PACE
1
r
=
1
ρ
3
+ 1q , the inequality
kf gkr ≤ kf kρ · kgkq .
Lemma 2.2. Let α ∈ [0, 1) and β ∈ R. There exists a positive constant C = C(α, β) such that


Ceβt ,
if β > 0;



Z t

C(t + 1), if β = 0;
s−α eβs ds ≤

0



 C,
if β < 0.
Lemma 2.3. (Hardy-Littlewood-Sobolev inequality)
Let u ∈ Lp (Rn ) (p > 1), 0 < γ < n and nγ > 1 − p1 , then (1/ |x|γ ) ∗ u ∈ Lq (Rn ) with
1
= nγ + p1 − 1. Also the mapping from u ∈ Lp (Rn ) into (1/ |x|γ ) ∗ u ∈ Lq (Rn ) is continuous.
q
See [5, Theorem 4.5.3, p. 117].
Lemma 2.4. Let a(t), b(t), K(t), ψ(t) be nonnegative, continuous functions on the interval
I = (0, T ) (0 < T ≤ ∞), ω : (0, ∞) → R be a continuous, nonnegative and nondecreasing
function, ω(0) = 0, ω(u) > 0 for u > 0 and let A(t) = max0≤s≤t a(s), B(t) = max0≤s≤t b(s).
Assume that
Z
t
ψ(t) ≤ a(t) + b(t)
K(s)ω(ψ(s))ds, t ∈ I.
0
Then
ψ(t) ≤ W
−1
Z t
W (A(t)) + B(t)
K(s)ds , t ∈ (0, T1 ),
0
where
Z
v
W (v) =
v0
W
−1
dσ
, v ≥ v0 > 0,
ω(σ)
is the inverse of W and T1 > 0 is such that
Z t
W (A(t)) + B(t)
K(s)ds ∈ D(W −1 )
0
for all t ∈ (0, T1 ).
See [2] (or [5]) for the proof.
Lemma 2.5. If δ, ν, τ > 0 and z > 0, then
Z z
1−ν
z
(z − ζ)ν−1 ζ δ−1 e−τ ζ dζ ≤ M (ν, δ, τ ) ,
0
1−ν
where M (ν, δ, τ ) = max (1, 2
) Γ (δ) 1 +
δ
ν
τ −δ .
See [6] for the proof of this lemma.
3. E STIMATION
In this section we state and prove our result on boundedness and also present an exponential
and a polynomial decay result.
Theorem 3.1. Assume that the constants α, β and m are such that 0 < α < βn, 0 < β < 1
and m > 1.
J. Inequal. Pure and Appl. Math., 4(4) Art. 82, 2003
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4
NASSER - EDDINE TATAR
(i) If Ω = Rn , then for any r satisfying max
(m−1)n m
,β
α
<r<
mn
,
α
we have
kϕ(t, x)kr ≤ Up,r,ρ (t)
with
Up,r,ρ (t) = 2
m(p−1)
r
1
K(t) p
m
(1−m)r
Z t
p−1
p
× 1 − 2m(p−1) (m − 1)C1 C2 K(t)m−1 L(t)eεpt
,
e−εps F p (s)ds
0
max0≤s≤t kk(s, ·)kpr
(3.1)
where K(t) =
, L(t) = max0≤s≤t kl(s, ·)kpρ , p = mr and ρ =
nr
for some ε > 0. Here C1 and C2 are the best constants in Lemma 2.2 and
αr−(m−1)n
Lemma 2.3, respectively. The estimation is valid as long as
Z t
1
m−1
εpt
K(t)
L(t)e
e−εps F p (s)ds ≤ m(p−1) (m − 1)C1p−1 C2p .
2
0
(ii) If Ω is bounded, then
kϕ(t, x)kr̃ ≤ Up,r,ρ (t)
n
for any r̃ ≤ r where p, r and ρ are as in (i). If moreover, r < nβ−α
(but not necessarily
(m−1)n
r >
) that is m
< r < min mn
, n , then this estimation holds for any
α
β
α nβ−α
1
β
Proof.
(3.2)
(3.3)
<p≤
r
m
provided that ρ >
nr
.
n−(nβ−α)r
(i) Suppose that Ω = Rn . By the Minkowski inequality and the Young inequality
(Lemma 2.1), we have
Z Z t
m
F
(s)ϕ
(s,
y)
dsdy
kϕ(t, x)kr ≤ kk(t, x)kr + kl(t, x)kρ n−α
n
1−β |x − y|
R
0 (t − s)
q
for r, ρ and q such that 1r = ρ1 + 1q .
Let p be such that p1 = 1q + αn and p0 its conjugate i.e.
inequality, we see that
Z t
(t − s)β−1 F (s)ϕm (s, y)ds
1
p
+
1
p0
= 1. Using the Hölder
0
Z
≤
t
(β−1)p0
(t − s)
εp0 s
e
p10 Z
ds
0
t
p1
e−εps F p (s)ϕmp (s, y)ds ,
0
for any positive constant ε. We have multiplied by eεs · e−εs before applying Hölder
inequality.
nr
, we see that p = mr . By our assumption on r it is easy to see
Choosing q = (nm−αr)
that q > 1, p > 1 and 1 + (β − 1)p0 > 0. Therefore, we may apply Lemma 2.2 to get
Z t
0
0
0
(t − s)(β−1)p eεp s ds < C1 eεp t .
0
Hence, inequality (3.3) becomes
Z t
p1
Z t
1
0
(t − s)β−1 F (s)ϕm (s, y)ds ≤ C1p eεt
e−εps F p (s)ϕmp (s, y)ds .
0
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O N AN I NTEGRAL I NEQUALITY
(3.4)
WITH A
K ERNEL S INGULAR IN T IME AND S PACE
5
It follows that
Z Z t
m
F
(s)ϕ
(s,
y)
dsdy
n−α
n
1−β |x − y|
R
0 (t − s)
q
Z Z
p1
t
1
dy
p0 εt −εps p
mp
≤ C1 e e
F (s)ϕ (s, y)ds
n−α .
Rn
|x − y|
0
q
As r <
mn
,
α
we may apply the results in Lemma 2.3 to obtain
Z Z
p1
t
dy
−εps p
mp
e
F (s)ϕ (s, y)ds
n−α Rn
|x − y|
0
q
Z
p1 t
−εps p
mp
≤ C2 e
F (s)ϕ (s, ·)ds ,
0
(3.5)
p
with p as above and C2 is the best constant in the result of Lemma 2.3. From (3.2), (3.4)
and (3.5) it appears that
Z
p1 t
kϕ(t, x)kr ≤ kk(t, x)kr + C3 eεt kl(t, x)kρ e−εps F p (s)ϕmp (s, ·)ds 0
p
or
(3.6)
kϕ(t, x)kr ≤ kk(t, x)kr + C3 eεt kl(t, x)kρ
Z
t
Z
Rn
e−εps F p (s)ϕmp (s, x)dsdx
p1
,
0
1
0
where C3 = C1p C2 . Inequality (3.6) can also be written as
Z
εt
kϕ(t, x)kr ≤ kk(t, x)kr + C3 e kl(t, x)kρ
p1
t
−εps
e
F
p
(s) kϕ(s, x)kmp
mp
ds
.
0
Observe that by our choice of p we have r = mp. It follows that
(3.7)
kϕ(t, x)kr ≤ kk(t, x)kr + C3 eεt kl(t, x)kρ
t
Z
e−εps F p (s) kϕ(s, x)krr
p1
ds .
0
Applying the algebraic inequality
(a + b)p ≤ 2p−1 (ap + bp ), a, b ≥ 0, p > 1,
we deduce from (3.7) that
(3.8)
kϕ(t, x)kpr
p−1
≤2
kk(t, x)kpr
εpt
+ C4 e
kl(t, x)kpρ
Z
t
e−εps F p (s) kϕ(s, ·)krr ds,
0
where C4 = 2p−1 C3p . Let us put ψ(t) = kϕ(t, x)kr/m
, then (3.8) takes the form
r
Z t
p
p
p−1
εpt
ψ(t) ≤ 2 kk(t, ·)kr + C4 e kl(t, x)kρ
e−εps F p (s)ψ m (s)ds.
0
J. Inequal. Pure and Appl. Math., 4(4) Art. 82, 2003
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NASSER - EDDINE TATAR
1
By the Lemma 2.4 with ω(u) = um , W (v) = 1−m
(v 1−m − v01−m ) and W −1 (z) =
1
(1 − m)z + v01−m 1−m , we conclude that
Z t
−1
p−1
εpt
−εps p
ψ(t) ≤ W
W 2 K(t) + C4 e L(t)
e
F (s)ds
0
1
1−m
Z t
m−1 εpt
p−1
p−1
−εps p
e L(t)
e
F (s)ds
,
≤ 2 K(t) 1 − (m − 1)C4 2 K(t)
0
where K(t) and L(t) are as in the statement of the theorem.
(ii) If Ω is bounded, then the first part in the assertion (ii) of the theorem follows from
the argument in (i) by an extension procedure and the application of the embedding
n
n
Lr (Ω) ⊂ Lr̃ (Ω) for r̃ ≤ r. Now if r < nβ−α
we choose q such that r < q < nβ−α
and
np
nr
ρ > n−(nβ−α)r . The Young inequality is therefore applicable. Also as 1 < q < n−αp , the
Hardy-Littlewood-Sobolev inequality (Lemma 2.3) applies (see also [3, p. 660] when
Ω is bounded).
In what follows, in order to simplify the statement of our next results, we define v := r or r̃
according to the cases (i) or (ii) in Theorem 3.1, respectively.
Corollary 3.2. Suppose that the hypotheses of Theorem 3.1 hold. Assume further that k(t, x)
and l(t, x) decay exponentially in time, that is k(t, x) ≤ e−k̃t k̄(x) and l(t, x) ≤ e−l̃t ¯l(x) for
some positive constants k̃ and ˜l. Then ϕ(t, x) is also exponentially decaying to zero i.e.,
kϕ(t, x)kv ≤ C5 e−µt , t > 0
(3.9)
for some positive constants C5 and µ provided that
Z ∞
m−1 k̄ (x)
¯l(x)
F p (s)ds ≤
r
ρ
0
1
2m(p−1)
(m − 1)C6p−1 C2p ,
where C6 is the best constant in Lemma 2.2 (third estimation) and the other constants are as in
(i) and (ii) of Theorem 3.1.
Proof. From the inequality (1.1) we have
−k̃t
ϕ(t, x) ≤ e
(3.10)
−l̃t ¯
k̄(x) + e
t
Z Z
l(x)
Ω
0
F (s)ϕm (s, y)
dsdy.
(t − s)1−β |x − y|n−α
After multiplying by e−mµt · emµt , where µ = min{k̃, ˜l}, we use the Hölder inequality to get
Z t
(t − s)β−1 F (s)ϕm (s, y)ds
0
Z
≤
t
(β−1)p0 −mp0 µt
(t − s)
e
p10 Z
ds
p1
t
p
mpµt
F (s)e
ϕ
mp
(s, y)ds
.
0
0
As in the proof of Theorem 3.1, 0 < (1 − β)p0 < 1. If C6 is the best constant in Lemma 2.2,
then we may write
Z t
p1
Z t
(3.11)
(t − s)β−1 F (s)ϕm (s, y)ds ≤ C6
F p (s)ϕmp (s, y)ds .
0
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O N AN I NTEGRAL I NEQUALITY
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From (3.10) and (3.11) it appears that
p1
Z Z t
µt
p
mpµt
mp
e ϕ(t, x) ≤ k̄(x) + C6 ¯l(x)
F (s)e
ϕ (s, y)ds
Ω
0
7
dyds
.
|x − y|n−α
r
Taking the L −norm and applying the Minkowski inequality and then the Young inequality
(Lemma 2.1), we find
eµt kϕ(t, x)kr ≤ k̄ (x)r
Z Z
p1
t
dyds
p
mpµt mp
¯
F (s)e
ϕ (s, y)ds
+ C6 l(x) ρ n−α ,
Ω
|x − y|
0
q
with
1
r
+ 1q . Applying Lemma 2.3, we arrive at
Z
p1 t
µt
p
mpµt mp
k̄
¯
F (s)e
ϕ (s, y)ds e kϕ(t, x)kr ≤ (x) r + C2 C6 l(x) ρ 0
=
1
ρ
p
or
(3.12)
eµt kϕ(t, x)kr ≤ k̄ (x)r + C2 C6 ¯l(x)ρ
t
Z
F p (s)empµt kϕ(s, x)kmp
r
p1
ds .
0
Taking both sides of (3.12) to the power p, we obtain
(3.13)
µpt
e
kϕ(t, x)kpr
p−1
≤2
k̄ (x)p + C7 ¯l(x)p
r
t
Z
ρ
F p (s)empµt kϕ(s, x)kmp
r ds.
0
Next, putting
χ(t) := eµpt kϕ(t, x)kpr ,
the inequality (3.13) may be written as
p−1
χ(t) ≤ 2
k̄ (x)p + C7 ¯l(x)p
r
ρ
Z
t
F p (s)χm (s)ds.
0
The rest of the proof is essentially the same as that of Theorem 3.1.
In the following corollary we consider the somewhat more general inequality
Z Z t
sδ F (s)ϕm (s, y)
(3.14)
ϕ(t, x) ≤ k(t, x) + l(t, x)
dyds, x ∈ Ω, t > 0
1−β |x − y|n−α
Ω 0 (t − s)
for some specified δ.
Corollary 3.3. Suppose that the hypotheses of Theorem 3.1 hold. Assume further that k(t, x) ≤
t−k̂ k̄(x) and 1 + δp0 − mp0 min{k̂, 1 − β} > 0. Then any ϕ(t, x) satisfying (3.14) is also
polynomially decaying to zero
kϕ(t, x)kv ≤ C8 t−ω , C8 , ω > 0
provided that
k̄ (x)m−1 L(t)
Z
r
t
eεps F p (s)ds ≤
0
1
2m(p−1)
(m − 1)C9p−1 C2p
where C9 is the best constant in Lemma 2.5.
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8
NASSER - EDDINE TATAR
Proof. Let us consider the inequality
t
Z Z
−k̂
sδ F (s)ϕm (s, y)
dyds.
(t − s)1−β |x − y|n−α
ϕ(t, x) ≤ t k̄(x) + l(t, x)
(3.15)
Ω
0
Multiplying by s−m min{k̂,1−β} e−εs · sm min{k̂,1−β} eεs , we obtain
Z t
(3.16)
(t − s)β−1 sδ F (s)ϕm (s, y)ds
0
Z
≤
t
(β−1)p0 δp0 −mp0 min{k̂,1−β} −εp0 s
(t − s)
s
e
p10
ds
0
Z
×
t
smp min{k̂,1−β} epεs F p (s)ϕmp (s, y)ds
p1
.
0
As 1 + δp0 − mp0 min{k̂, 1 − β} > 0, we may apply Lemma 2.5 to the first term in the right
hand side of (3.16) to get
Z t
(3.17)
(t − s)β−1 sδ F (s)ϕm (s, y)ds
0
β−1
Z
≤ Mt
t
s
mp min{k̂,1−β} pεs
e
p
F (s)ϕ
mp
p1
(s, y)ds .
0
Using (3.17) we infer from inequality (3.15) that
tmin{k̂,1−β} ϕ(t, x) ≤ k̄(x)
p1
Z Z t
mp min{k̂,1−β} pεs p
mp
s
e F (s)ϕ (s, y)ds
+ M l(t, x)
Ω
0
dy
.
|x − y|n−α
Next, after using the Hardy-Littlewood-Sobolev inequality and defining
φ(t, x) := tp min{k̂,1−β} kϕ(t, x)kpr ,
we proceed as in Theorem 3.1 to find a (uniform) bound for φ(t, x).
Remark 3.4. The investigation of (1.1) with a weakly singular kernel in time, that is
Z Z t −γ(t−s)
e
F (s)ϕm (s, y)
ϕ(t, x) ≤ k(t, x) + l(t, x)
dyds, γ > 0
1−β |x − y|n−α
Ω 0 (t − s)
is simpler since this kernel is summable (in time).
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O N AN I NTEGRAL I NEQUALITY
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