Electronic Journal of Differential Equations, Vol. 2008(2008), No. 108, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
LARGE TIME BEHAVIOR OF SOLUTIONS TO SECOND-ORDER
DIFFERENTIAL EQUATIONS WITH p-LAPLACIAN
MILAN MEDVEĎ, EVA PEKÁRKOVÁ
Abstract. We study asymptotic properties of solutions for certain secondorder differential equation with p-Laplacian. The main purpose is to investigate when all global solutions behave at infinity like nontrivial linear functions.
Making use of Bihari’s inequality and its Dannan’s version, we obtain results
for differential equations with p-Laplacian analogous which extend those known
in the literature concerning ordinary second order differential equations.
1. Introduction
In this paper, we study asymptotic properties of the second-order differential
equation with p-Laplacian
(|u0 |p−1 u0 )0 + f (t, u, u0 ) = 0,
p ≥ 1.
(1.1)
In the sequel, it is assumed that all solutions of (1.1) are continuously extendable
throughout the entire real axis. We refer to such solutions as to global solutions.
We shall prove sufficient conditions under which all global solutions are asymptotic
to at + b, as t → +∞, where a, b are real numbers. The problem for ordinary
second order differential equations without p-Laplacian has been studied by many
authors, e. g. by Cohen [6], Constantin [7], Dannan [8], Kusano and Trench
[9, 10], Rogovchenko [13], Rogovchenko [14], Tong [15] and Trench [16]. Our results
are more close to those obtained in the papers [13, 14]. The main tool of the
proofs are the Bihari’s and Dannan’s integral inequalities. We remark that sufficient
conditions on the existence of global solutions for second order differential equations
and second order functional-differential equations with p-Laplacian are proved in
the papers [1, 2, 3, 4, 11]. Many references concerning differential equations with
p-Laplacian can be found in the paper by Rachunková, Staněk and Tvrdý [12],
where boundary value problems for such equations are treated.
Let
u(t0 ) = u0 , u0 (t0 ) = u1 ,
(1.2)
where u0 , u1 ∈ R be initial condition for solutions of (1.1).
We say that a solution u(t) of (1.1) possesses the property (L) if u(t) = at+b+o(t)
as t → ∞, where a, b are real constants.
2000 Mathematics Subject Classification. 34C11.
Key words and phrases. Second order differential equation; p-Laplacian; Bihari’s inequality;
asymptotic properties; Dannan’s inequality.
c
2008
Texas State University - San Marcos.
Submitted June 9, 2008. Published August 11, 2008.
1
2
M. MEDVEĎ, E. PEKÁRKOVÁ
EJDE-2008/108
2. Main results
Theorem 2.1. Let p ≥ 1, r > 0 and t0 > 0. Suppose that the following conditions
are satisfied:
(1) f (t, u, v) is a continuous function in D = {(t, u, v) : t ∈< t0 , ∞), u, v ∈ R},
where t0 > 0
(2) There exist continuous functions h, g : R+ =< 0, ∞) → R+ such that
|u| r
|v|r , (t, u, v) ∈ D,
|f (t, u, v)| ≤ h(t)g
t
where for s > 0 the function g(s) is positive and nondecreasing,
Z ∞
h(s)ds < ∞,
t0
and if we denote
Z
x
G(x) =
t0
ds
,
sr/p g(sr/p )
then
Z
∞
G(∞) =
t0
ds
p
=
r
sr/p g(sr/p )
∞
Z
a
p
τ r −1 dτ
= ∞,
τ g(τ )
where a = (t0 )r/p .
Then any global solution u(t) of the equation (1) possesses the property (L).
Proof. Without loss of generality we may assume t0 = 1. Let u(t) be a solution of
(1.1), (1.2). Then
Z t
0
p
0
p−1 0
(u (t)) ≤ |u (t)| u (t) ≤ c2 +
|f (s, u(s), u0 (s))|ds,
(2.1)
1
where c2 = |u1 |p . Let w(t) be the right-hand side of inequality (2.1). Then
u0 (t) ≤ w(t)1/p
and
Z
u(t) ≤ c1 +
t
w(s)1/p ds ≤ c1 + (t − 1)w(t)1/p ≤ t[c1 + w(t)1/p ],
(2.2)
1
where c1 = |u0 |, i.e.
u(t) ≤ t[c1 + w(t)1/p ], t ≥ 1.
Applying the inequality (A + B)p ≤ 2p−1 (Ap + B p ), A, B ≥ 0 and the assumption
(2) of Theorem 2.1 we obtain from (2.2):
|u(t)| p
≤ 2p−1 cp1 + 2p−1 w(t)
t
Z t
(2.3)
|u(s)| r
≤ 2p−1 cp1 + 2p−1 c2 +
h(s)g
|u0 (s)|r ds.
s
0
Let
d = 2p−1 (cp1 + c2 ),
Then
|u(t)| p
t
Z
≤d+
t
H(s)g
1
H(t) = 2p−1 h(t).
|u(s)| r
s
|u0 (s)|r ds := z(t);
(2.4)
(2.5)
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LARGE TIME BEHAVIOR OF SOLUTIONS
3
i.e.,
|u(t)| r
t
≤ z(t)r/p .
From the assumption (2) of Theorem 2.1 and the inequality (2.1) it follows
0
p
|u (t)| ≤
up1
t
Z
+
h(s)g
|u(s)| 1
r
s
|u0 (s)|r ds ≤ z(t);
i.e. we have
|u0 (t)|p ≤ z(t).
Since g(s) is nondecreasing, the inequality (2.3) yields
g
|u(t)| r
t
≤ g(z(t)r/p )
and so we conclude for t ≥ 1,
t
Z
H(s)g(z(t)r/p )z(t)r/p ds.
z(t) ≤ d +
1
From the assumption (2) of Theorem 2.1 it follows that the inverse G−1 of G is
defined on the interval (G(+0), ∞). Applying the Bihari theorem (see [5]) we obtain
Z ∞
z(t) ≤ G−1 G(d) + 2p−1
h(s)ds := K < ∞.
1
Therefore the inequality (2.4) yields
|u0 (t)| ≤ L := K 1/p
and from (2.3) we have
|u(t)|
≤ L.
t
Since
Z
t
|f (s, u(s), u0 (s))|ds ≤
1
t
Z
1
|u(s)| r
)|u0 (s)|r ds ≤ z(t) ≤ K
h(s)g(
s
R∞
for t ≥ 1, the integral 1 |f (s, u(s), u0 (s))|ds exists. From (2.5) it follows that
there exists a ∈ R such that
lim u0 (t) = a.
t→∞
By the l’Hospital rule, we can conclude that
Rt
u1 + 1 u0 (τ )dτ
u(t)
lim
=
= lim u0 (t) = a.
t→∞ t
t→∞
t
Therefore there exist b ∈ R such that u(t) = at + b + o(t).
4
M. MEDVEĎ, E. PEKÁRKOVÁ
EJDE-2008/108
Example 1. Let t0 = 1, p ≥ r > 0,
u p−r h
|u| r i
f (t, u, u0 ) = η(t)t1−α e−t
(u0 )r , t ≥ 1,
ln 2 +
t
t
(2.6)
where 0 < α < 1, η(t) is a continuous function on interval h1, ∞) with K =
supt≥1 |η(t)| < ∞.
The function f (t, u, u0 ) can be written in the form
u (2.7)
f (t, u, u0 ) = h(t)g [ ]r (u0 )r ,
t
p
where h(t) = η(t)t1−α e−t , g(u) = u r −1R ln(2 + |u|). Obviously g(u)
R ∞ is positive,
∞
continuous and nondecreasing function, 1 |h(s)|ds < KΓ(α) = K 0 s1−α e−s ds
and
Z ∞ p −1
Z ∞
Z ∞
τ r dτ
dτ
dτ
=
>
= ∞.
(2.8)
τ g(τ )
τ ln(2 + τ )
(2 + τ ) ln(2 + τ )
1
1
1
Thus we have proved that all conditions of Theorem 1 are satisfied. This means that
for every solution u(t) of the initial value problem (1.1), (1.2) there exist numbers
a, b such that u(t) = at + b + o(t) as t → ∞.
Theorem 2.2. Let p ≥ 1, r > 0 and t0 > 0. Suppose the following conditions are
satisfied:
(1) The function f (t, u, v) is continuous in D = {(t, u, v) : t ∈< t0 , ∞), u, v ∈
R},
(2) There exist continuous functions h1 , h2 , h3 , g1 , g2 : R+ → R+ such that
|u| r
|f (t, u, v)| ≤ h1 (t)g1
+ h2 (t)g2 (|v|r ) + h3 (t), (t, u, v) ∈ D,
t
R∞
where Hi := t0 hi (s)ds < ∞, i = 1, 2, 3, for s > 0 the functions g1 (s),
g2 (s) are nondecreasing and if
Z x
ds
G(x) =
r/p ) + g (sr/p )
g
(s
1
2
t0
then
Z
∞
G(∞) =
t0
p
ds
=
r
g1 (sr/p ) + g2 (sr/p )
Z
a
∞
p
τ r −1 dτ
= ∞,
g1 (τ ) + g2 (τ )
where a = (t0 )r/p .
Then any global solution u(t) of the equation (1) possesses the property (L).
Proof. Without loss of generality we may assume t0 = 1. By the standard existence
results, it follows from the continuity of the function f that equation (1.1) has
solution u(t) corresponding to the initial data u(1) = u0 , u0 (1) = u1 . Two times of
integration (1.1) from 1 to t, yields for t ≥ 1
Z t
p
0
p
0
p−1 0
f (s, u(s), u0 (s))ds,
(2.9)
(u (t)) ≤ |u (t)| u (t) = u1 −
1
Z
t
h
i1/p
u(t) ≤ u0 + (t − 1) up1 −
f (s, u(s), u0 (s))ds
.
(2.10)
1
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LARGE TIME BEHAVIOR OF SOLUTIONS
5
It follows from (2.9) and (2.10) that for t ≥ 1,
|u0 (t)| ≤ w(t)1/p ,
|u(t)| ≤ t c1 + w(t)1/p ,
Rt
where c1 = |u0 |, c2 = |u1 |p , w(t) = c2 + 1 |f (s, u(s), u0 (s))|ds. Using the assumption (2) we obtain for t ≥ 1
Z t
|u(s)| h
r
h1 (s)g1
|u0 (t)| ≤ c2 +
ds
s
1
Z t
Z t
i1/p
+
h2 (s)g2 (|u0 (s)|r )ds +
h3 (s)ds
,
1
1
t
|u(t)|
r
≤ c1 + c2 +
h1 (s)g1
ds
t
s
1
Z t
Z t
i1/p
+
h2 (s)g2 (|u0 (s)|r )ds +
h3 (s)ds
.
Z
h
|u(s)| 1
1
p
p−1
p
p
Applying the inequality (A + B) ≤ 2 (A + B ), where A, B ≥ 0, we obtain
Z t
|u(s)| |u(t)| p
r
≤d+
ds
H1 (s)g1
t
s
1
(2.11)
Z t
Z t
+
H2 (s)g2 (|u0 (s)|r )ds +
H3 (s)ds.
1
1
where d = 2p−1 (cp1 + c2 ), Hi (t) = 2p−1 hi (t), i = 1, 2, 3. Denote by z(t) the righthand side inequality (2.11)
|u(t)| r
≤ z(t)r/p .
(2.12)
|u0 (t)|r ≤ z(t)r/p ,
t
Since the function g1 (s) and g2 (s) are nondecreasing for s > 0, we obtain
h |u(t)| ir g1 |u0 (t)|r ≤ g1 z(t)r/p , g1
≤ g2 z(t)r/p .
t
Thus, for t ≥ 1,
Z t
Z t
Z t
z(t) ≤ d +
H1 (s)g1 (z(s)r/p )ds +
H2 (s)g2 (z(s)r/p )ds +
H3 (s)ds. (2.13)
1
1
1
Furthermore, due to evident inequality
H1 (s)g1 (z(s)r/p ) + H2 (s)g2 (z(s)r/p ) ≤ (H1 (s) + H2 (s))(g1 (z(s)r/p ) + g2 (z(s)r/p ))
(2.14)
By (2.14), we have
Z t
z(t) ≤ d + H̄3 +
(H1 (s) + H2 (s))(g1 (z(s)r/p ) + g2 (z(s)r/p ))ds;
1
i.e.,
z(t) ≤ d + 2p−1 h̄3 + 2p−1
Z
t
(h1 (s) + h2 (s))(g1 (z(s)r/p ) + g2 (z(s)r/p ))ds. (2.15)
1
Applying Bihari’s inequality (see [5]) to (2.15), we obtain, for t ≥ 1,
Z t
−1
p−1
p−1
z(t) ≤ G
G(d + 2 h̄3 ) + 2
(h1 (s) + h2 (s))ds ,
1
6
M. MEDVEĎ, E. PEKÁRKOVÁ
where
Z
EJDE-2008/108
x
ds
,
+ g2 (sr/p )
1 g1
and G−1 (x) is the inverse function for G(x) defined for x ∈ (G(+0), ∞). Note that
G(+0) < 0, and G−1 (x) is increasing.
Now, let
K = G(d + 2p−1 h̄3 ) + 2p−1 (h̄1 + h̄2 ) < ∞.
Since G−1 (x) is increasing, we have
G(x) =
(sr/p )
z(t) ≤ G−1 (K) < ∞;
so it yields
|u(t)|
≤ G−1 (K), |u0 (t)| ≤ G−1 (K).
t
Using assumption (2) of the Theorem 2.2, we have
Z t
|u| r
|f (s, u(s), u0 (s))|ds ≤ h1 (t)g1
+ h2 (t)g2 (|u0 (s)|r ) + h3 (t)
t
1
≤ z(t) ≤ G−1 (K),
Rt
where t ≥ 1, the integral 1 |f (s, u(s), u0 (s))|ds converges, and there exists an a ∈ R
such that
lim u0 (t) = a.
t→∞
Example 2. Let t0 = 1, p ≥ r > 0,
u r u p−r ln 2 +
t
t
+ η2 (t)t1−α2 e−t v p−r ln(3 + v r ) + η3 (t)t1−α3 e−t
f (t, u, v) = η1 (t)t1−α1 e−t
where 0 < αi < 1, ηi (t) are continuous functions on [1, ∞), Ki = supt≥1 |ηi (t)| < ∞,
i = 1, 2, 3. Then f (t, u, u0 ) can be written as
u f (t, u, v) = h1 (t)g1 [ ]r + h2 (t)g2 (v r ) + h3 (t),
t
p
p
where hi (t) = ηi (t)t1−αi e−t , i = 1, 2, 3, g1 (u) = u r ln(2 + u), g2 (u) = u r ln(2 + u).
Then
u |f (t, u, v)| ≤ |h1 (t)|g1 [ ]r + |h2 (t)|g2 (|v|r ) + |h3 (t)|,
t
where (t, u, v) ∈ D = {(t, u, v) : t ∈ h1, ∞), u, v ∈ R}, |hi (t)| ≤ Ki Γ(αi ), i = 1, 2, 3
and obviously we have
p
Z ∞
τ r −1 dτ
G(∞) =
g1 (τ ) + g2 (τ )
1
p
Z ∞
τ r −1 dτ
=
p
τ r [ln(2 + τ ) + ln(3 + τ )]
1
Z
1 ∞
dτ
≥
= ∞.
2 1 (3 + τ ) ln(3 + τ )
This means that all assumptions of Theorem 2.2 are satisfied and thus any global
solution u(t) of the equation (1) possesses the property (L).
Theorem 2.3. Let t0 > 0. Suppose that the following assumptions hold:
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LARGE TIME BEHAVIOR OF SOLUTIONS
7
(i) there exist nonnegative continuous function h1 , h2 , g1 , g2 : R+ → R+ such
that
|u| r
|f (t, u, v)| ≤ h1 (t)g1
+ h2 (t)g2 (|v|r );
t
(ii) for s > 0 the function g1 (s), g2 (s) are nondecreasing, and
g1 (αu) ≤ ψ1 (α)g1 (u),
g2 (αu) ≤ ψ2 (α)g2 (u)
for α ≥ 1, u ≥ 0, where the functions ψ1 (α), ψ2 (α) are continuous for
α
≥ 1;
R∞
(iii) t0 hi (s)ds = Hi < ∞, i = 1, 2.
Assume that there exists a constant K ≥ 1 such that
Z +∞
ds
K −1 (ψ1 (K) + ψ2 (K))2p−1 (H1 + H2 ) ≤
r/p ) + g (sr/p )
g
(s
1
2
t0
p
Z
p +∞
τ r −1 dτ
=
,
r a
g1 (τ ) + g2 (τ )
where a = (t0 )r/p . Then any global solution u(t) of the equation (1.1) with initial
data u(t0 ) = u0 , u0 (t0 ) = u1 such that (|u0 | + |u1 |)p ≤ K possesses the property
(L).
Proof. Without loss of generality we may assume t0 = 1. Arguing in the same way
as in Theorem 2.1, we obtain by assumption (i) of Theorem 2.3
Z t
Z t
h
u(s) i1/p
|u0 (t)| ≤ |u1 |p +
h1 (s)g1 [
]r ds +
h2 (s)g2 (|u0 (s)|r )ds
(2.16)
s
1
1
Z t
Z t
h
u(s) i1/p
|u(t)|
≤ |u0 | + |u1 |p +
]r ds +
h1 (s)g1 [
h2 (s)g2 (|u0 (s)|r )ds
(2.17)
t
s
1
1
where t ≥ 1.
Z t
Z t
|u(t)| p
u(s) ≤ K + 2p−1
h1 (s)g1 [
]r ds +
h2 (s)g2 (|u0 (s)|r )ds , (2.18)
t
s
1
1
where K = 2p−1 (|u0 |p + |u1 |p ) ≥ (|u0 | + |u1 |)p . Denoting by z(t) the right-hand
side of inequality (2.18) we have by (2.16) and (2.18)
|u(t)| r
|u0 (t)|r ≤ z(t)r/p ,
≤ z(t)r/p .
(2.19)
t
Since the function g1 (s), g2 (s) are nondecreasing for s > 0, for t ≥ 1, (2.19) yields
Z t
Z t
z(t) ≤ K + 2p−1
h1 (s)g1 z(s)r/p ds +
h2 (s)g2 (z(s)r/p ) ds.
(2.20)
1
1
By assumption (ii) of Theorem 2.3, the functions g1 (u), g2 (u) belong to the class H.
Furthermore, if g1 (u) and g2 (u) belong to the class H with corresponding multiplier
function ψ1 (α), ψ2 (α) respectively, then the sum g1 (u) + g2 (u). Applying Bihari’s
Theorem (see [5]) to (2.20), we have for t ≥ 1
Z t
z(t) ≤ KW −1 (K −1 (ψ1 (K) + ψ2 (K)))2p−1
(h1 (s) + h2 (s))ds,
(2.21)
1
where
Z
W (u) =
1
u
ds
,
g1 sr/p + g2 sr/p
8
M. MEDVEĎ, E. PEKÁRKOVÁ
EJDE-2008/108
and W −1 (u) is inverse function for W (u). Inequality (2.21) holds for all t ≥ 1
because
(K −1 (ψ1 (K) + ψ2 (K))2p−1 (H1 + H2 ) = L < ∞.
Since W −1 (u) is increasing, we get
z(t) ≤ KW −1 (L) < ∞,
so it follows from (2.19), (2.20) that
|u(t)|
≤ KW −1 (L), |u0 (t)| ≤ KW −1 (L).
t
The rest of the proof is similar to that of Theorem 2.2 and thus it is omitted.
Example 3. Let t0 > 0. Consider (1.1) with p ≥ 1,
p
q
= 2,
f (t, u, v) = h1 (t)u2 = h2 (t)v 2 ,
η1 (t) 1−α1 −t
e ,
t2 t
(2.22)
1−α2 −t
where h1 (t) =
h2 (t) = η2 (t)t
e , 0 < αi ≤ 1, ηi (t), i = 1, 2 are
continuous functions on the interval h0, ∞) with Ki = supt≥t0 |ηi (t)| < ∞. Then
we can write
u 2
f (t, u, v) = η1 (t)t1−α1 e−t
+ η2 (t)t1−α2 e−t v 2
(2.23)
t
and
|f (t, u, u0 )| ≤ K1 Γ(α1 )g1 (u) + K2 Γ(α2 )g2 (u0 ),
(2.24)
2
0
0 2
where g1 (u) = u , g2 (u ) = (u ) . The functions g1 , g2 satisfy the condition (ii) of
Theorem 2.3 with ψ1 (α) = ψ2 (α) = α2 and
p
Z ∞
Z ∞
dτ
τ r −1 dτ
=
= ∞.
(2.25)
g
(τ
)
+
g
(τ
)
τ
1
2
t0
t0
Thus all assumptions of Theorem 2.3 are satisfied and therefore any global solution
u(t) of the equation (1.1) (independently on the initial values u0 , u1 ) possesses the
property (L).
Theorem 2.4. Let t0 > 0. Suppose that the assumptions (i) and (iii) of Theorem
2.3 hold, while (ii) is replaced by
(ii’) for s > 0 the functions g1 (s), g2 (s) are nonnegative, continuous and nondecreasing, g1 (0) = g2 (0) = 0 and satisfy a Lipschitz condition
|g1 (u + v) − g1 (u)| ≤ λ1 v,
|g2 (u + v) − g2 (u)| ≤ λ2 v,
where λ1 , λ2 are positive constants.
Then any global solution u(t) of (1.1) with initial data u(t0 ) = u0 , u0 (t0 ) = u1 such
that |u0 |p + |u1 |p ≤ K possesses property (L).
Proof. Applying [8, Corollary 2] to (2.20), we have for t ≥ 1
Z t
p−1
z(t) ≤ K + 2
(h1 (s) + h2 (s))(g1 (K) + g2 (K))
t0
Z
t
(λ1 + λ2 )(h1 (τ ) + h2 (τ ))dτ ds
t0
≤ K + 2p−1 (H1 + H2 )(g1 (K) + g2 (K)) exp 2p−1 (λ1 + λ2 )(H1 + H2 )
× exp 2p−1
< +∞.
The proof can be completed with the same argument as in Theorem 2.2.
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LARGE TIME BEHAVIOR OF SOLUTIONS
9
Theorem 2.5. Let t0 > 0. Suppose that there exist continuous functions h, g1 , g2 :
R+ → R+ such that
|f (t, u, v)| ≤ h(t)g1
|u| r
t
g2 (|v|r ),
where for s > 0 the functions g1 (s), g2 (s) are nondecreasing;
Z ∞
h(s)ds < ∞,
t0
and if we denote
Z
x
G(x) =
t0
R
p ∞
ds
g1 (sr/p )g2 (sr/p )
p
−1
r
r
τ
p
then G(+∞) = r a g1 (τ
)g2 (τ ) dτ = +∞, where a = (t0 ) . Then any global solution u(t) of the equation (1.1) possesses the property (L).
Proof. Without loss of generality we may assume t0 = 1. Arguing as in the proof
of Theorem 2.2, we obtain for t ≥ 1
Z t
h
u(s) i1/p
0
p
|u (t)| ≤ |u1 | +
h(s)g1 [
]r g2 (|u0 (s)|r )ds
,
s
1
Z
t
h
u(s) i1/p
|u(t)|
(2.26)
≤ |u0 | + |u1 |p +
h(s)g1 [
]r g2 (|u0 (s)|r )ds
,
t
s
1
Z t
|u(t)| p
u(s) p−1
]r g2 (|u0 (s)|r )ds,
≤C +2
h(s)g1 [
t
s
1
where C = 2p−1 (|u0 |p + |u1 |p ) ≥ (|u0 | + |u1 |)p . Denoting by z(t) the right-hand
side of inequality (2.26) and using the assumptions of the Theorem 2.5, we have for
t≥1
Z t
z(t) ≤ 1 + C + 2p−1
h(s)g1 (z r/p )g2 (z r/p )ds.
(2.27)
1
Applying Bihari’s inequality (see [5]) to (2.27), for t ≥ 1, we obtain
Z t
−1
p−1
z(t) ≤ G
G(1 + C) + 2
h(s)ds ≤ G−1 (K),
1
where
Z
G(w) =
1
w
ds
,
g1 (sr/p )g2 (sr/p )
−1
and G (w) is the inverse function for G(w). The function G−1 (w) is defined for
w ∈ (G(+0), ∞), where G(+0) < 0, it is increasing, and
Z ∞
p−1
K = G(1 + C) + 2
h(s)ds < ∞.
1
The rest of proof is similar that of Theorem 2.2 and thus is omitted.
10
M. MEDVEĎ, E. PEKÁRKOVÁ
EJDE-2008/108
Example 4. Let t0 = 1, p ≥ r > 0,
h u p−r u r i 34 h
i 14
f (t, u, v) = η(t)t1−α e−t
ln 2 +
· v p−r ln(2 + v r ) ,
t
t
where η(t) is a continuous function on h1, ∞) with K = supt∈h1,∞) η(t) < ∞. Let
h p
i3/4
h p
i1/4
g1 (u) = u r −1 ln(2 + u)
, g2 (v) = v r −1 ln(2 + v)
, h(t) = η(t)t1−α e−t .
Then
u f (t, u, v) = h(t)g1 [ ]r g2 (v r )
t
and
p
Z
Z
p ∞ τ r −1
p ∞
dτ
dτ =
r 1 g1 (τ )g2 (τ )
r 1 τ ln(2 + τ )
Z
p ∞
dτ
>
= +∞.
r 1 (2 + τ ) ln(2 + τ )
G(+∞) =
Obviously |f (t, u, v)| can be estimated as in Theorem 2.5. Thus all assumptions of
Theorem 2.5 are satisfied and this means that any global solution of the equation
(1.1) possesses the property (L).
Theorem 2.6. Let t0 > 0. Suppose that the following conditions hold:
(i) there exist nonnegative continuous functions h, g1 , g2 : R+ → R+ such that
h |u(t)| ir |f (t, u, v)| ≤ h(t)g1
g2 (|v|r )
t
(ii) for s > 0 the functions g1 (s), g2 (s) are nondecreasing; and
g1 (αu) ≤ ψ1 (α)g1 (u),
g2 (αu) ≤ ψ2 (α)g2 (u)
for
R ∞ α ≥ 1, u ≥ 0, where the functions ψ1 (α), ψ2 (α) are continuous for α ≥ 1;
(iii) t0 h(s)ds = H < +∞.
Assume also that there exists a constant K ≥ 1 such that
p
Z ∞
Z
p ∞ τ r −1 dτ
ds
=
K −1 Hψ1 (K)ψ2 (K) ≤
,
r a g1 (τ )g2 (τ )
g1 (sr/p )g2 (sr/p )
1
(2.28)
r
where a = (t0 ) p . Then any global solution u(t) of the equation (1.1) with initial
data u(t0 ) = u0 , u0 (t0 ) = u1 such that 2p−1 (|u0 |p +|u1 |p ) ≤ K possesses the property
(L).
Proof. Without loss of generality we assume that t0 = 1. With the same argument
as in Theorem 2.2, for t ≥ 1, we have
Z t
i1/p
h
|u(s)| r
g2 (|u0 (s)|r )ds
,
|u0 (t)| ≤ |u1 |p +
h(s)g1
s
1
Z t
h
|u(s)| i1/p
|u(t)|
r
p
≤ |u0 | + |u1 | +
h(s)g1
g2 (|u0 (s)|r )ds
.
t
s
1
Applying the inequality (A + B)p ≤ 2p−1 (Ap + B p ), A, B ≥ 0 we obtain
|u(t)| p
h Z t |u(s)| i
r
p−1
p
p
p−1
≤ 2 (|u0 | + |u1 | ) + 2
g1
g2 (|u0 (s)|r )ds . (2.29)
t
s
1
EJDE-2008/108
LARGE TIME BEHAVIOR OF SOLUTIONS
11
Denoting by z(t) the right-hand side of inequality (2.29), for t ≥ 1, we obtain
Z t
z(t) ≤ K +
H(s)g1 (z(s)r/p )g2 (z(s)r/p )ds,
(2.30)
1
p−1
p
p
where K = 2 (|u0 | + |u1 | ) and H(t) = 2p−1 h(t). Assumption (ii) implies that
the functions g1 (u), g2 (u) belong to the class H. Furthermore, it follows from [6,
Lemma 1] that if g1 (u) and g2 (u) belong to the class H with the corresponding
multiplier functions ψ1 (α) and ψ2 (α) respectively, then the product g1 (u)g2 (u)
also belongs to H and the corresponding multiplier function is ψ1 (α)ψ2 (α). Thus,
applying [8, Theorem 1] to (2.30), for t ≥ 1, we have
Z t
−1
−1
z(t) ≤ KW
K ψ1 (K)ψ2 (K)
H(s)ds ,
(2.31)
1
where
Z
u
W (u) =
1
ds
,
g1 (sr/p )g2 (sr/p )
(2.32)
and W −1 (u) is the inverse function for W (u). Evidently, inequality (2.31) holds for
all t ≥ 1 since by (2.28)
Z t
K −1 ψ1 (K)ψ2 (K)
H(s)ds ∈ Dom(W −1 )
(2.33)
1
for all t ≥ 1. The rest of the proof is analogous to that of Theorem 2.2 and is
omitted.
Theorem 2.7. Let t0 > 0. Suppose that assumptions (i) and (iii) of Theorem 2.6
hold, while (ii) is replaced by
(ii’) for s > 0 the functions g1 (s), g2 (s) are continuous and nondecreasing,
g1 (0) = g2 (0) = 0, and satisfy a Lipschitz condition
|g1 (u + v) − g1 (u)| ≤ λ1 v,
|g2 (u + v) − g2 (u)| ≤ λ2 v,
where λ1 , λ2 are positive constants.
Then any global solution u(t) of the equation (1.1) with initial data u(t0 ) = u0 ,
u0 (t0 ) = u1 such that |u0 |p + |u1 |p ≤ K possesses the property (L).
Proof. Without loss of generality we may assume t0 = 1. Applying [8, Corollary 2]
to (2.30), we have for t ≥ 1
Z t
Z t
z(t) ≤ K + g1 (K)g2 (K)
H(s) exp λ1 λ2
H(τ )dτ ds
1
1
≤ K + H̄g1 (K)g2 (K) exp (λ1 λ2 H̄) < +∞.
The proof of the above theorem can be completed with the same argument as in
Theorem 2.2.
Acknowledgements. The first author was supported by Grant No. 1/0098/08
from the Slovak Grant Agency VEGA-SAV-M. The second author was supported
by Grant No. 201/08/0469 from Grant Agency of the Czech Republic.
12
M. MEDVEĎ, E. PEKÁRKOVÁ
EJDE-2008/108
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Milan Medveď
Department of Mathematical Analysis and Numerical Mathematics, Faculty of Mathematics, Physics and Informatics, Comenius University, Mlýnská dolina, 842 48 Bratislava,
Slovakia
E-mail address: medved@fmph.uniba.sk
Eva Pekárková
Department of Mathematics and Statistics, Faculty of Science, Masaryk University,
Janáčkovo nám. 2a, CZ-602 00 Brno, Czech Republic
E-mail address: pekarkov@math.muni.cz