Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 20, pp. 1–7.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
BLOW-UP SOLUTIONS FOR A NONLINEAR WAVE EQUATION
WITH POROUS ACOUSTIC BOUNDARY CONDITIONS
SHUN-TANG WU
Abstract. We study a nonlinear wave equation with porous acoustic boundary conditions in a bounded domain. We prove a finite time blow-up for certain
solutions with positive initial energy.
1. Introduction
We consider the following system of nonlinear wave equations with porous acoustic boundary conditions:
utt − ∆u + α(x)u + φ(ut ) = j1 (u)
u(x, t) = 0
in Ω × [0, T ),
on Γ0 × (0, T ),
ut (x, t) + f (x)zt + g(x)z = 0
on Γ1 × (0, T ),
∂u
− h(x)zt + ρ(ut ) = j2 (u) on Γ1 × (0, T ),
∂ν
u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ Ω,
z(x, 0) = z0 (x),
x ∈ Γ1 ,
(1.1)
(1.2)
(1.3)
(1.4)
(1.5)
(1.6)
where Ω is a bounded domain in Rn (n ≥ 1) with a smooth boundary Γ = Γ0 ∪ Γ1 .
Here, Γ0 and Γ1 are closed and disjoint. Let ν be the unit normal vector pointing
to the exterior of Ω and α : Ω → R, f, g, h : Γ1 → R and j1 , j2 : R → R are given
functions.
The system (1.1)–(1.6) is a model of nonlinear wave equations with acoustic
boundary conditions which are described by (1.3) and (1.4). These boundary conditions were introduced by Morse and Ingard [12] and developed by Beale and Rosencrans [2, 3, 4]. In recent years, questions related to wave equations with acoustic
boundary conditions have been treated by many authors [5, 10, 11, 13, 14, 15, 16].
For example, Frota and Larkin [11] studied (1.1)–(1.6) with φ = ρ = j1 = j2 = 0
and they established the exponential decay result for suitably defined solutions. Recently, as j1 = j2 = 0, Graber [6, 7] showed that the systems (1.1)-(1.6) generates a
well-posed dynamical system by using semigroup theory. When one considers the
presence of the double interaction between source and damping terms, both in the
interior of Ω and on the boundary Γ1 , the analysis becomes more difficult. Very
2000 Mathematics Subject Classification. 35L70, 35B40.
Key words and phrases. Blow-up; acoustic boundary conditions; exponential growth.
c
2013
Texas State University - San Marcos.
Submitted August 28, 2012. Published January 23, 2013.
1
2
S.-T. WU
EJDE-2013/20
recently, Graber and Said-Houari [8] studied this challenging problem and obtained
several results in local existence, global existence, the decay rate and blow-up results. Particularly, in the absence of boundary source, that is j2 = 0, for certain
initial data, the authors proved that the solution is unbounded and grows as an
exponential function. However, the possibility of the solution that blows up in finite time is not addressed in that paper. Therefore, the intention of this paper is
to investigate the blow-up phenomena of solutions for system (1.1)-(1.6) without
imposing the boundary source. In this way, we can extend this unbounded result
of [8] to a blow-up result with positive initial energy.
The content of this paper is organized as follows. In section 2, we state the local
existence result and the energy identity which is crucial in establishing the blow-up
result in finite time. In section 3, we study the blow-up problem for the initial
energy being positive.
2. Preliminaries
In this section, we present some material which will be used throughout this
work. First, we introduce the set
HΓ10 = {u ∈ H 1 (Ω) : u|Γ0 = 0},
and endow HΓ10 with the Hilbert structure induced by H 1 (Ω), we have that HΓ10 is
a Hilbert space. For simplicity,R we denote k · kp = k · kLp (Ω) ,R k · kp,Γ = k · kLp (Γ) ,
1 ≤ p ≤ ∞, kuk2α = k∇uk22 + Ω α(x)u2 (x)dx and kuk2gh = Γ1 g(x)h(x)u2 (x)dΓ.
The following assumptions for problem (1.1)-(1.6) were used in [8].
(A1) The functions j1 (s) = |s|p−1 s and j2 (s) = 0, where p ≥ 1 is such that
HΓ10 (Ω) ,→ Lp+1 (Ω).
(A2) φ, ρ : R → R are continuous and increasing functions with φ(0) = ρ(0) = 0.
In addition, there exist positive constants ai and bi i = r, q such that
ar |s|r+1 ≤ φ(s)s ≤ br |s|r+1 , r ≥ 1,
q+1
aq |s|
q+1
≤ ρ(s)s ≤ bq |s|
, q ≥ 1.
(2.1)
(2.2)
(A3) The functions α, f, g, h are essentially bounded such that f > 0, g > 0,
h > 0 and α ≥ 0. (If α = 0, Γ0 is assumed to have a non-empty interior
such that the Poincarè inequality is applicable.)
Next, the energy function associated with problem (1.1)–(1.6), with j2 = 0, is
defined as
1
1
1
kuk2α + kzk2gh −
kukp+1
(2.3)
E(t) = kut k22 +
p+1 .
2
2
p+1
Then, we are ready to state the following local existence result and energy identity.
Lemma 2.1 ([8]). Suppose that (A1)–(A3) hold, and that u0 ∈ HΓ10 (Ω), u1 ∈ L2 (Ω)
and z0 ∈ L2 (Γ1 ). Then the system (1.1)–(1.6) with j2 = 0 admits a unique solution
(u, z) such that, for T > 0,
u ∈ C([0, T ); HΓ10 (Ω)) ∩ C 1 ([0, T ); L2 (Ω)),
Moreover, the energy satisfies
Z TZ
Z
E(0) = E(T ) +
φ(ut )ut dxdt +
0
Ω
0
T
Z
Γ1
z ∈ C([0, T ); L2 (Γ1 )).
(ρ(ut )ut + f hzt2 )dΓdt.
(2.4)
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BLOW-UP SOLUTIONS
3
Note that (2.4) shows that the energy is a non-increasing function along trajectories.
3. Blow-up of Solutions
In this section, we state and prove our main result. First, we define a functional
G which helps in establishing desired results. Let
G(x) =
1 2 B1p+1 p+1
x −
x ,
2
p+1
x > 0,
where B1−1 = inf{k∇uk2 : u ∈ HΓ10 (Ω), kukp+1 = 1}. Then, G has a maximum at
p+1
− p−1
λ1 = B1
with the maximum value
1
1
E1 ≡ G(λ1 ) = ( −
)λ2 .
2 p+1 1
The next Lemma will play an important role in proving our result.
Lemma 3.1 ([8]). Suppose that (A1)–(A3) hold, and that u0 ∈ HΓ10 (Ω), u1 ∈ L2 (Ω)
and z0 ∈ L2 (Γ1 ). Let (u, z) be a solution of (1.1)–(1.6) with j2 = 0. Assume that
E(0) < E1 and k∇u0 k2 > λ1 . Then there exists λ2 > λ1 such that, for all t ≥ 0,
(kuk2α + kzk2gh )1/2 ≥ λ2 ,
(3.1)
kukp+1 ≥ B1 λ2 .
(3.2)
Now, we are ready to state and prove our main result. Our proof technique
follows the arguments of [8] and some estimates obtained in [9].
Theorem 3.2. Suppose that (A1)–(A3) hold, and that u0 ∈ HΓ10 (Ω), u1 ∈ L2 (Ω)
z0 ∈ L2 (Γ1 ). Assume further that p > max(r, 2q − 1). Then any solution of (1.1)(1.6) with j2 = 0and satisfying E(0) < E1 and k∇u0 k2 > λ1 blows up at a finite
time.
Proof. We suppose that the solution exists for all time and we reach to a contradiction. To achieve this, we set
H(t) = E1 − E(t),
t ≥ 0.
(3.3)
Then, by (2.4), we see that H 0 (t) ≥ 0. From (3.1), the definition of E(t) and
1
E1 = ( 21 − p+1
)λ21 , we deduce that, for all t ≥ 0,
1
1
1
0 < H(0) ≤ H(t) ≤ E1 − λ22 +
kukp+1
kukp+1
p+1 ≤
p+1 .
2
p+1
p+1
(3.4)
Let
A(t) = H
1−σ
Z
ε
(t) + ε
uut dx −
2
Ω
Z
2
Z
f hz dΓ − ε
Γ1
huzdΓ,
(3.5)
Γ1
where ε is a positive constant to be specified later and
0 < σ < min{
p−r
p−1
,
}.
r(p + 1) 2(p + 1)
(3.6)
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S.-T. WU
EJDE-2013/20
Then
A0 (t) = (1 − σ)H(t)−σ H 0 (t) + εkut k22 − εkuk2α + εkukp+1
p+1
Z
Z
+ε
hg|z(x, t)|2 dΓ − ε
uφ(ut )dx
Γ1
Ω
Z
−ε
uρ(ut )dΓ + 2εH(t) − 2εH(t)
Γ1
(3.7)
ε(p − 1)
2
kukp+1
= (1 − σ)H(t)−σ H 0 (t) + 2εkut k22 +
p+1 + 2εkzkgh
p+1
Z
Z
−ε
uφ(ut )dx − ε
uρ(ut )dΓ − 2εE1 + 2εH(t).
Ω
Γ1
We observe from (3.2) that
p+1 p+1 −1
E1 = E1 (B1p+1 λp+1
)(B1p+1 λp+1
)−1 ≤ E1 kukp+1
λ2 ) .
2
2
p+1 (B1
(3.8)
Inserting (3.8) into (3.7), we have
2
A0 (t) ≥ (1 − σ)H(t)−σ H 0 (t) + 2εkut k22 + εc1 kukp+1
p+1 + 2εkzkgh
Z
Z
−ε
uφ(ut )dx − ε
uρ(ut )dΓ + 2εH(t),
Ω
(3.9)
Γ1
where
p−1
p−1
− 2E1 (B1p+1 λp+1
− 2E1 (B1p+1 λp+1
)−1 >
)−1 = 0.
2
1
p+1
p+1
By (2.1), Hölder inequality and Young’s inequality, we see that, for δ1 > 0,
Z
− r+1
br δ1r+1
br rδ1 r
r+1
kukr+1 +
kut kr+1
(3.10)
uφ(ut )dx ≤
r+1 ,
r+1
r+1
Ω
c1 =
A substitution of (3.10) into (3.9) leads to
A0 (t) ≥ (1 − σ)H(t)−σ H 0 (t) + 2εkut k22 + εc1 kukp+1
p+1
Z
uρ(ut )dΓ
+ 2εkzk2gh − ε
(3.11)
Γ1
− r+1
b δ r+1
r
br rδ1
r 1
−ε
kukr+1
kut kr+1
r+1 +
r+1 + 2εH(t),
r+1
r+1
At this point, for a large positive constant M1 to be chosen later, picking δ1 such
− r+1
r
that δ1
= M1 H(t)−σ and using the fact
q+1
H 0 (t) ≥ ar kut kr+1
r+1 + aq kut kq+1,Γ +
Z
f hzt2 dΓ
(3.12)
Γ1
by (2.4) and (A1) we have
εrbr M1
A0 (t) ≥ (1 − σ −
)H(t)−σ H 0 (t) + 2εkut k22 + εc1 kukp+1
p+1
ar (r + 1)
Z
εbr M1−r
+ 2εkzk2gh − ε
uρ(ut )dΓ −
H(t)σr kukr+1
r+1 + 2εH(t).
r
+
1
Γ1
In addition, using (3.4) and the inequality
1
χγ ≤ χ + 1 ≤ (1 + )(χ + ω), ∀χ ≥ 0, 0 < γ ≤ 1, ω > 0,
ω
(3.13)
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BLOW-UP SOLUTIONS
5
1
with χ = p+1
kukp+1
p+1 and ω = H(0) and noting that p > r, and 0 < σr +
by (3.6), we have
r+1
p+1
σr
H(t)σr kukr+1
kukp+1
p+1
r+1 ≤ c2 H(t)
r+1
σr+ p+1
1
≤ c3
kukp+1
p+1
p+1
1
≤ c3 d
kukp+1
p+1 + H(t) ,
p+1
p−r
r+1
where c2 = vol(Ω) p+1 and c3 = (p + 1) p+1 · c2 and d = 1 +
with (3.13), we obtain
1
H(0) .
r+1
p+1
≤1
(3.14)
Combining (3.14)
εrbr M1
)H(t)−σ H 0 (t) + 2εkut k22 + ε(c1 − c4 )kukp+1
p+1
ar (r + 1)
Z
uρ(ut )dΓ.
+ 2εkzk2gh + ε(2 − (p + 1)c4 )H(t) − ε
A0 (t) ≥ (1 − σ −
(3.15)
Γ1
br c3 dM1−r
with c4 = (p+1)(r+1) . Next, we will follow the arguments as in [9] to estimate
the last term on the right hand side of (3.15). For this purpose, let us recall the
following trace and interpolation theorems [1, 17]
kukq+1,Γ ≤ CkukW s,q+1 ,
which holds for some positive constant C, q ≥ 0, 0 < s < 1 and s >
(3.16)
1
q+1 .
W 1−θ,τ (Ω) = [H 1 (Ω), Lp+1 (Ω)]θ ,
(3.17)
θ
where τ1 = 1−θ
2 + p+1 , θ ∈ [0, 1] and [·, ·]θ denotes the interpolation bracket. We
p−1
1
note from q ≥ 1 and p > 2q −1 that q+1
≤ 2(p−q)
< 1. Then, we choose β satisfying
p−1
≤β<1
2(p − q)
(3.18)
and select θ such that
1−θ =
which imply that 1 − θ >
inequality, we have
1
,
β(q + 1)
1
q+1
τ=
2(p + 1)
,
(1 − θ)(p + 1) + 2θ
and τ ≥ q + 1. From (3.16), (3.17) and Young’s
θ
kukq+1,Γ ≤ CkukW 1−θ,q+1 (Ω) ≤ CkukW 1−θ,τ (Ω) ≤ Ckuk1−θ
α kukp+1
1
1−
1
= Ckukαβ(q+1) kukp+1β(q+1)
2β 2 (q+1)−2β 2β
(2β 2 −1)(q+1)
q+1
≤ C kukα + kukp+1
,
where C is a generic positive constant. Further, as in [9], there exists β satisfying
2β 2 (q+1)−2β
(p+1)β
(3.18) such that (2β
2 −1)(q+1) =
q+1 . Thus,
(p+1)β 2β
q+1
kukq+1,Γ ≤ C kukαq+1 + kukp+1
.
(3.19)
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S.-T. WU
EJDE-2013/20
Besides, we observe from the definition of E(t) by (2.3) and E(0) < E1 that
p+1
1
1
2
2 kukα ≤ E1 + p+1 kukp+1 . Hence, by (2.2) and (3.19), we have
Z
(p+1)β
2β
q+1
|
uρ(ut )dΓ| ≤ bq kukq+1,Γ kut kqq+1,Γ ≤ c5 (kukαq+1 + kukp+1
)kut kqq+1,Γ
Γ1
≤ c6 (E1 +
β
1
q
q+1 ku k
kukp+1
t q+1,Γ ,
p+1 )
p+1
β
β
with c5 = bq C and c6 = c5 (2 q+1 + (p + 1) q+1 ). Thus, using Young’s inequality,
further requiring σ such that 0 < σ < 1−β
q+1 , and exploiting (3.4) and (3.12), we
obtain, for δ > 0,
Z
|
uρ(ut )dΓ|
Γ1
β−1
1
1
1
q+1
p+1 q+1
kukp+1
kuk
kut kqq+1,Γ
E
+
1
p+1
p+1
p+1
p+1
β−1
h i
1
1
q+1
q+1
≤ c6 E1 +
kukp+1
kukp+1
δ E1 +
p+1
p+1 + cδ kut kq+1,Γ
p+1
p+1
β−1
β−1
1
−1
q+1 +σ H(t)−σ H 0 (t).
kukp+1
≤ c6 δH(0) q+1 E1 +
p+1 + cδ c6 aq H(0)
p+1
≤ c6 E1 +
Thus, (3.15) becomes
A0 (t) ≥ (1 − σ − εc7 )H(t)−σ H 0 (t) + 2εkut k22
β−1
c6 δH(0) q+1 + ε c1 − c4 −
kukp+1
p+1
p+1
β−1
+ 2εkzk2gh + ε(2 − (p + 1)c4 )H(t) − c6 δεH(0) q+1 E1 ,
where c7 =
arrive at
rbr M1
ar (r+1)
β−1
q+1 +σ . Employing the estimate (3.8) again, we
+ cδ c6 a−1
q H(0)
A0 (t) ≥ (1 − σ − εc7 )H(t)−σ H 0 (t) + 2εkut k22 + ε(c1 − c4 − δc8 )kukp+1
p+1
+ 2εkzk2gh + ε(2 − (p + 1)c4 )H(t),
where
1
+ E1 (B1p+1 λp+1
)−1 ).
2
p+1
Now, we choose M1 large enough such that 2 − (p + 1)c4 > 0 and c1 − c4 > c21 .
Once M1 is fixed, we select δ small enough such that c21 − δc8 > 0. Then, pick ε
small enough such that 1 − σ − εc7 ≥ 0 and A(0) > 0. Thus, there exists K > 0
such that
2
A0 (t) ≥ εK(kut k22 + kukp+1
p+1 + H(t) + kzkgh ),
(3.20)
A(t) ≥ A(0) > 0, for t ≥ 0.
β−1
c8 = c6 H(0) q+1 (
On the other hand, from the result of Graber et al [8, Lemma 6.5], we have
1
2
A(t) 1−σ ≤ c9 kut k22 + kukp+1
+
H(t)
+
kzk
t ≥ 0.
(3.21)
gh ,
p+1
Combining (3.21) with (3.20), we obtain
1
A0 (t) ≥ c10 A(t) 1−σ , t ≥ 0,
(3.22)
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BLOW-UP SOLUTIONS
7
where ci , i = 9, 10, are positive constants. Thus, inequality (3.22) leads to a blow-up
result in a finite time T with
1−σ
0<T ≤
σ .
c10 σA(0) 1−σ
The proof is complete.
Acknowledgments. We would like to thank the anonymous referees for their
important comments which improve this article.
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Shun-Tang Wu
General Education Center, National Taipei University of Technology, Taipei, 106, Taiwan
E-mail address: stwu@ntut.edu.tw