Applied Mathematics Letters 22 (2009) 1394–1400
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Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
A global nonexistence theorem for viscoelastic equations with arbitrary positive initial energy Yanjin Wang Institute of Applied Physics and Computational Mathematics, P.O. Box 8009-15, Beijing, 100088, China
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Article history: Received 16 September 2008 Accepted 18 January 2009
abstract In this letter we consider the nonlinear viscoelastic equation utt − ∆u +
t
Z
g (t − τ )∆u(τ )dτ + ut = |u|p−1 u
in Ω × (0, ∞)
0
Keywords: Blow-up Arbitrary positive initial energy Viscoelastic equation
with Dirichlet boundary conditions. Under some appropriate assumptions on g and the initial data, a blow-up result with arbitrary positive initial energy is established for 1 < n p < n− (if n = 1, 2, then 1 < p < ∞). 2 © 2009 Elsevier Ltd. All rights reserved.
1. Introduction Consider the following wave equation with nonlinear viscoelastic term:
Z t utt − ∆u + g (t − τ )∆u(τ )dτ + ut |ut |m−1 = |u|p−1 u in Ω × (0, ∞), 0
u(t , x) = 0, x ∈ ∂ Ω , t ≥ 0, u(0, x) = u0 (x), ut (0, x) = u1 (x), x ∈ Ω .
(1.1)
where Ω is a bounded domain of Rn (n ≥ 1) with a smooth boundary ∂ Ω , p > 1,Rm ≥ 0 and g is a positive function. t Before going any further, we consider Eq. (1.1) without the viscoelastic term, 0 g (t − τ )∆u(τ )dτ . In this case, Eq. (1.1) is known as the wave equation. The equation has been extensively studied by many people. As regards nonexistence of a global solution, Levine [1] firstly showed that the solutions with negative initial energy are non-global for some abstract wave equation with linear damping. Later Levine and Serrin [2] studied blow-up of a class of more generalized abstract wave equations. And then Pucci and Serrin [3] claimed that the solution with positive initial energy, which is appropriately bounded, blows up in finite time. In 2001, Levine and Todorova [4] proved that there exist some initial data with arbitrary positive initial energy such that the corresponding solution of wave equations blows up in finite time. Then Todorova and Vitillaro [5] improved the blow-up result above. However, they did not give a sufficient condition for the initial data such that the corresponding solution blows up in finite time with arbitrary positive initial energy. Recently, for Eq. (1.1) with g ≡ 0 and m = 1, Gazzalo and Squassina [6] established the condition for initial data with arbitrary positive initial energy such that the corresponding solution blows up in finite time. Very recently, the author [7] studied blow-up of solutions of the Klein–Gordon equation with arbitrary positive initial energy. Now we return to Eq. (1.1) with g 6≡ 0. Cavalcanti et al. [8] firstly studied (1.1) with m = 1 and a localized damping a(x)ut . They obtained an exponential rate of decay with some assumption on g. And then Messaoudi [9] obtained the global existence of solutions for the viscoelastic equation; there he also obtained a blow-up result with negative initial energy. Furthermore, he improved his blow-up result in [10].
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Y. Wang / Applied Mathematics Letters 22 (2009) 1394–1400
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In the present letter, we consider Eq. (1.1) for m = 1. We will establish a blow-up result for Eq. (1.1) with arbitrary positive initial energy for m = 1. Throughout the rest of this letter, we always let m = 1. We next state some assumptions on g: Assumption 1.1. g ∈ C 1 ([0, ∞)) is a non-negative and non-increasing function satisfying 1 − κ := 1 −
∞
Z
g (τ )dτ > 0. 0 t
Assumption 1.2. The function e 2 g (t ) is of positive type in the following sense: t
Z
v(s)
Z
s
e
s−τ 2
g (s − τ )v(τ )dτ ds ≥ 0
0
0
∀v ∈ C 1 ([0, ∞)) and ∀t > 0. We note that Assumption 1.1 is also used in [9,10]. For the definition of a(t ) of positive type in detail, we refer the readers to [11]. And an example of a(t ) of positive type is positive, decreasing, convex a(t ) (see [12]). Thus, it is obvious that g (t ) = e−t with 0 < < 1 satisfies Assumptions 1.1 and 1.2. Now we are in a position to state our main blow-up result for Eq. (1.1) with m = 1 n Theorem 1.1. Suppose that g (s) satisfies Assumption 1.1, 1.2, and κ < p+1 . Let 1 < p < n− when n ≥ 3, 1 < p < ∞ when 2 1 2 n = 1, 2. If the initial data (u0 , u1 ) ∈ H0 (Ω ) × L (Ω ) satisfy the following conditions: p−1
E (0) > 0,
Z Ω
(1.2)
u0 (x)u1 (x)dx > 0,
(1.3)
I (u0 ) < 0,
ku0 k22 >
(1.4) 4(p + 1)
((p − 1) − (p + 1)κ) min{χ , 1}
E (0),
(1.5)
where χ is the constant of the Poincaré inequality on Ω , and I (v) is defined as p+1
I (v) = k∇vk22 − kvkp+1 ,
(1.6)
then the corresponding solution u(t ) of Eq. (1.1) blows up in a finite time T > 0. 2. Proof of Theorem 1.1 Before we start to prove Theorem 1.1, it is necessary to state the local existence theorem for Eq. (1.1), whose proof follows the arguments in [8,13]. n Theorem 2.1. Suppose that 1 < p < n− when n ≥ 3 and 1 < p < ∞ when n = 1, 2, and g satisfies Assumption 1.1. For 2 (u0 , u1 ) ∈ H01 (Ω ) × L2 (Ω ), the problem (1.1) has a unique local solution
u ∈ C ([0, T ); H01 (Ω )),
u1 ∈ C ([0, T ); L2 (Ω ))
for the maximum existence time T > 0. We here put the energy function as E (t ) =
1 2
kut (t , ·)k22 +
1
t
Z
2
g (s)dx k∇ u(t , ·)k22 +
1− 0
1 2
(g ◦ ∇ u)(t ) −
1 p+1
1 kukpp+ +1 ,
(2.1)
where
(g ◦ v)(t ) =
t
Z
g (t − τ )kv(t , ·) − v(τ , ·)k22 dτ . 0
As in [9,10], we see that d dt
E (t ) = −
Z
1 |ut (t , x)| dx + g 0 (t − τ ) 2 2
Ω
Z
1
Ω
|∇ u(τ ) − ∇ u(t )|2 dxdτ − g (t )k∇ u(t , ·)k22 ≤ 0. 2
(2.2)
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Lemma 2.1. Assume that g (t ) satisfies Assumptions 1.1 and 1.2, and H (t ) is a function that is twice continuously differentiable, satisfying
H 00 (t ) + H 0 (t ) >
t
Z
Z
g (t − τ )
Ω
0
H (0) > 0,
H 0 (0) > 0
∇ u(τ , x)∇ u(t , x)dxdτ
(2.3)
for every t ∈ [0, T0 ), where u(t ) is the corresponding solution of Eq. (1.1) with u0 and u1 . Then the function H (t ) is strictly increasing on [0, T0 ). Proof. We consider the following auxiliary ODE:
h00 (t ) + h0 (t ) =
t
Z
g (t − τ ) 0
h(0) = H (0),
Z Ω
h (0) = 0 0
∇ u(τ )∇ u(t )dxdτ
(2.4)
for every t ∈ [0, T0 ). It is easy to find a solution of the ODE (2.4) as h(t ) = h(0) +
t
Z
e−ξ − e−t e−ξ
0
ξ
Z
g (ξ − τ )
Z Ω
0
∇ u(ξ , x)∇ u(τ , x)dxdτ dξ
(2.5)
for every t ∈ [0, T0 ). By a direct computation we obtain that h0 (t ) =
t
Z
eξ e−t
0
= e− t
ξ
Z
g (ξ − τ )
Z Ω
0
Z Z t Ω
ξ
e 2 ∇ u(ξ , x)
∇ u(ξ , x)∇ u(τ , x)dxdτ dξ ξ
Z
0
e
ξ −τ 2
g (ξ − τ )
τ
e 2 ∇ u(τ , x) dτ dξ dx
0
≥0 for every t ∈ [0, T0 ), where the last inequality comes from Assumption 1.2, which implies that h(t ) ≥ h(0) = H (0). Moreover, we see that H 0 (0) > h0 (0). We next show that H 0 (t ) > h0 (t ) for t ≥ 0.
(2.6)
Assume that (2.6) is not valid, which means that there exists a time t0 satisfying that t0 = min{t ≥ 0 : H 0 (t ) = h0 (t )}.
(2.7)
By the continuity of the solutions for the ODEs (2.3) and (2.5) we see that t0 > 0 and H (t0 ) = h (t0 ). Then we have the following ODE: 0
0
(H 00 (t ) − h00 (t )) + (H 0 (t ) − h0 (t )) > 0 H (0) − h(0) = 0, (H 0 (0) − h0 (0)) > 0
for every t ∈ [0, T0 ). This ODE can be solved as H 0 (t0 ) − h0 (t0 ) > e−t0 (H 0 (0) − h0 (0)) > 0, which contradicts H 0 (t0 ) = h0 (t0 ). Thus, we see that H 0 (t ) > 0, which implies our desired result.
Lemma 2.2. Suppose that (u0 , u1 ) ∈ H01 (Ω ) × L2 (Ω ) satisfies
Z Ω
u0 (x)u1 (x)dx ≥ 0.
(2.8)
If the local solution u(t ) of Eq. (1.1) exists on [0, T ) and satisfies I (u(t )) < 0,
(2.9)
where I is defined as p+1
I (u(t )) := k∇ u(t , ·)k22 − ku(t , ·)kp+1 , then ku(t , ·)k22 is strictly increasing on [0, T ).
(2.10)
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Proof. Since u(t ) is the local solution of Eq. (1.1), by a simple computation we have 1 d2
Z
2 dt 2
Ω
|u(t , x)|2 dx =
Z
(|ut (t )|2 + uutt )dx Z Z Z Z |∇ u|2 dx = |ut (t , x)|2 dx − uut dx + |u|p+1 dx − Ω Ω Ω Ω Z Z t g (t − τ ) ∇ u(τ )∇ u(t )dxdτ + 0 Ω Z Z Z t g (t − τ ) ∇ u(τ )∇ u(t )dxdτ , >− uut dx + Ω
Ω
Ω
0
where the last inequality uses (2.9) Then we obtain d2
Z
dt 2
Ω
|u(t , x)| dx + 2
d dt
Z
|u(t , x)| dx > 2
Ω
t
Z
g (t − τ )
Z Ω
0
∇ u(τ )∇ u(t )dxdτ .
Therefore, this lemma comes from Lemma 2.1. Proof of Theorem 1.1. We next prove Theorem 1.1 in two steps. First, by a contradiction argument we claim that I (u(t )) < 0
(2.11)
and 4(p + 1)
ku(t , ·)k22 >
((p − 1) − (p + 1)κ) min{χ , 1}
E (0)
(2.12)
for every t ∈ [0, T ). Suppose that there exists a time t1 such that t1 = min{t ∈ (0, T ) : I (u(t ) = 0)} > 0.
(2.13)
By the continuity of the solution u(t , x) as a function of t, we see that I (u(t )) < 0 when t ∈ [0, t1 ) and I (u(t1 )) = 0. Thus by Lemma 2.2 we obtain that
ku(t , ·)k22 > ku0 k22 4(p + 1) E (0) ((p − 1) − (p + 1)κ) min{χ , 1}
>
for every t ∈ [0, t1 ) In addition, it is obvious that ku(t , ·)k22 is continuous on [0, t1 ]. Thus the following inequality is obtained:
ku(t1 , ·)k22 >
4(p + 1)
((p − 1) − (p + 1)κ) min{χ , 1}
E (0).
(2.14)
On the other hand, it follows from (2.1), (2.2) and (2.13) that 1 2
t1
Z 1−
g (s)ds k∇ u(t1 , ·)k22 +
0
1 2
(g ◦ ∇ u)(t1 ) −
1 p+1
1 ku(t1 , ·)kpp+ +1 ≤ E (0).
Noting the fact that I (u(t1 )) = 0, we then have
1−κ 2
−
1 p+1
k∇ uk22 ≤ E (0).
Thus, by the Poincaré inequality and κ < p+1 we see that p−1
ku(t1 , ·)k22 ≤
2(p + 1) E (0). ((p − 1) − (p + 1)κ) min{χ , 1}
(2.15)
Obviously, there is a contradiction between (2.14) and (2.15). Thus, we have proved that (2.11) is true for every t ∈ [0, T ). Furthermore, by Lemma 2.2 we see that (2.12) is also valid on [0, T ). Secondly, we prove that the solution of Eq. (1.1) blows up in a finite time. We now suppose that T is sufficiently large. And then we define the following auxiliary function:
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G(t ) = ku(t , ·)k22 +
t
Z
kuτ (τ , ·)k22 dτ + (t2 − t )ku0 k22 + a(t3 + t )2 ,
(2.16)
0
where t2 , t3 and a are positive constants, which will be determined in the sequel. By direct computation, we obtain that G0 (t ) = 2(u(t ), ut (t )) + 2
t
Z
(u(τ ), uτ (τ ))dτ + 2a(t3 + t )
(2.17)
0
and 1 00 G (t ) = 2
Z
|ut |2 dx +
ZΩ
|ut |2 dx +
= Ω
Z
|u|p+1 dx −
ZΩ
g (t − τ )
In order to estimate the term rs ≤
r2 2
+
t
Z
g (t − τ ) 0
Z Ω
∇ u(τ )∇ u(t )dxdτ + a
|∇ u|2 dx + a
Ω
Z Ω
0
|∇ u|2 dx +
ZΩ
Ω
t
Z +
|u|p+1 dx −
Z
Rt 0
∇ u(t )(∇ u(τ ) − ∇ u(t ))dxdτ +
t
Z
g (t − τ ) 0
g (t − τ )
R
Ω
Z Ω
|∇ u(t )|2 dx.
(2.18)
∇ u(t )(∇ u(τ ) − ∇ u(t ))dxdτ , we will use the following Young’s inequality:
b2 2
for any > 0, where r ≥ 0 and s ≥ 0. We then have t
Z
g (t − τ ) 0
Z Ω
|∇ u(t )|(|∇ u(τ ) − ∇ u(t )|)dxdτ ≤
1 2
t
Z
g (τ )dτ k∇ u(t , ·)k22 + 0
2
(g ◦ ∇ u)(t ),
(2.19)
where = (p−12k . And satisfies )(1−κ)
κ (p − 1)(1 − κ) = , 2
(p − 1)(1 − κ) −
(2.20)
p + 1 − > 0,
(2.21) p2 −1
where the second inequality comes from κ < p+1 < p2 +1 . Substituting (2.1) and (2.19) for the second and fifth terms of the right hand side of (2.18), respectively, we obtain p−1
Z t 1 G (t ) ≥ (p + 3) |ut | dx + (p − 1) − p − 1 + g (τ )dτ k∇ u(t )k22 Ω 0 + (p + 1 − )(g ◦ ∇ u)(t ) − 2(p + 1)E (t ) + 2a Z 1 ≥ (p + 3) |ut |2 dx + (p − 1) − p − 1 + κ k∇ u(t )k22 Ω Z t + (p + 1 − )(g ◦ ∇ u)(t ) − 2(p + 1)E (0) + 2(p + 1) kut k2 dx + 2a Z
00
2
0
≥ (p + 3)
Z Ω
|ut |2 dx + 2(p + 1)
t
Z
kut k2 dx + 0
(p − 1)(1 − κ) 2
χku0 k22 − 2(p + 1)E (0) + 2a,
(2.22)
where the last inequality follows from Lemma 2.2, the Poincaré inequality, (2.20) and (2.21), which means that G00 (t ) > 0 for every t ∈ (0, T ). Thus by G0 (0) ≥ 0 we see that G(t ) and G0 (t ) is strictly increasing on [0, T ). p−1 Noting the condition (1.5) and κ < p+1 , we see that
(p − 1)(1 − κ) 2
χku0 k22 − 2(p + 1)E (0) > 0.
Thus we can let a satisfy
(p + 1)a <
(p − 1)(1 − κ) 2
χku0 k22 − 2(p + 1)E (0).
Y. Wang / Applied Mathematics Letters 22 (2009) 1394–1400
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Moreover, we let t2 and t3 satisfy that t2 ≥
p − 1 G(0) G0 (0)
4
p−1
Z
2
Ω
u0 (x)u1 (x)dx + at3
≥ ku0 k22 .
Now set A := ku(t , ·)k22 +
t
Z
ku(τ , ·)k22 dτ + a(t3 + t )2 , 0
B :=
1 0 G (t ),
2
C := kut (t , ·)k + 2 2
t
Z
kuτ (τ , ·)k22 dτ + a. 0
Since we have assumed that the solution u(t , x) of Eq. (1.1) exists for every t ∈ [0, T ) where T is sufficiently large, we have G(t ) ≥ A, G00 (t ) ≥ (p + 3)C for every t ∈ [0, t2 ). Then it follows that G00 (t )G0 (t ) −
p+3 4
(G0 (t ))2 ≥ (p + 3)(AC − B2 ).
Furthermore, we have
Z
2
(su(t , x) + ut (t , x)) dx + 2
As − 2Bs + C = Ω
t
Z
ksu(τ , ·) + uτ (τ , ·)k22 dτ + a[s(t3 + t ) + 1]2 ≥ 0 0
for every s ∈ R, which implies that B2 − AC ≤ 0. Thus, we obtain G00 (t )G0 (t ) − As
p+3 4
p+3 4
(G0 (t ))2 ≥ 0.
> 1, letting α =
p−1 , 4
we have
(G−α )0 = −α G−α−1 G0 (t ) < 0, p+3 0 (G−α )00 = −α G−α−2 G00 G(t ) − (G (t ))2 4
(2.23) (2.24)
for every t ∈ (0, T ), which means that the function G−α is concave. Obviously G(0) > 0; then from (2.24) it follows that the (p−1)G(0) function G−α → 0 when t < T and t → T (T < 4G0 (0) ). Therefore, we see that there exists a finite time T > 0 such that lim ku(t , ·)k2 → ∞.
(2.25)
t →T −
Thus, the proof of Theorem 1.1 is completed.
Acknowledgement The author would like to thank Prof. Mark Levi for his suggestions for improving the manuscript. The study was partly supported by NSAF(10676005). References [1] [2] [3] [4]
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