Volume 2007, Article ID 56350, 13 pagesdoi:10.1155/2007/56350 Research Article Existence and Asymptotic Stability of Solutions for Hyperbolic Differential Inclusions with a Source Term J
Trang 1Volume 2007, Article ID 56350, 13 pages
doi:10.1155/2007/56350
Research Article
Existence and Asymptotic Stability of Solutions for Hyperbolic Differential Inclusions with a Source Term
Jong Yeoul Park and Sun Hye Park
Received 10 October 2006; Revised 26 December 2006; Accepted 16 January 2007 Recommended by Michel Chipot
We study the existence of global weak solutions for a hyperbolic differential inclusion with a source term, and then investigate the asymptotic stability of the solutions by using Nakao lemma
Copyright © 2007 J Y Park and S H Park This is an open access article distributed un-der the Creative Commons Attribution License, which permits unrestricted use, distri-bution, and reproduction in any medium, provided the original work is properly cited
1 Introduction
In this paper, we are concerned with the global existence and the asymptotic stability of weak solutions for a hyperbolic differential inclusion with nonlinear damping and source terms:
y tt − Δy t −div
|∇ y | p −2∇ y
+Ξ= λ | y | m −2y inΩ×(0,∞),
Ξ(x,t) ∈ ϕ
y t(x, t)
a.e (x, t) ∈Ω×(0,∞),
y =0 on∂Ω×(0,∞),
y(x, 0) = y0(x), y t(x, 0) = y1(x) inx ∈Ω,
(1.1)
whereΩ is a bounded domain inRNwith sufficiently smooth boundary ∂Ω, p≥2,λ > 0,
andϕ is a discontinuous and nonlinear set-valued mapping by filling in jumps of a locally
bounded functionb.
Recently, a class of differential inclusion problems is studied by many authors [2,6,7,
11,14–16,19] Most of them considered the existence of weak solutions for differential inclusions of various forms Miettinen [6] Miettinen and Panagiotopoulos [7] proved the existence of weak solutions for some parabolic differential inclusions J Y Park et al [14] showed the existence of a global weak solution to the hyperbolic differential inclusion
Trang 2(1.1) withλ =0 by making use of the Faedo-Galerkin approximation, and then consid-ered asymptotic stability of the solution by using Nakao lemma [8] The background of these variational problems are in physics, especially in solid mechanics, where noncon-vex, nonmonotone, and multivalued constitutive laws lead to differential inclusions We refer to [11,12] to see the applications of differential inclusions
On the other hand, it is interesting to mention the existence and nonexistence of global solutions for nonlinear wave equations with nonlinear damping and source terms [4,5,
10,13,18] in the past twenty years Thus, in this paper, we will deal with the existence and the asymptotic behavior of a global weak solution for the hyperbolic differential inclusion (1.1) involvingp-Laplacian, a nonlinear, discontinuous, and multivalued damping term
and a nonlinear source term The difficulties come from the interaction between the p-Laplacian and source terms As far as we are concerned, there is a little literature dealing with asymptotic behavior of solutions for differential inclusions with source terms The plan of this paper is as follows InSection 2, the main results besides notations and assumptions are stated InSection 3, the existence of global weak solutions to problem (1.1) is proved by using the potential-well method and the Faedo-Galerkin method In
Section 4, the asymptotic stability of the solutions is investigated by using Nakao lemma
2 Statement of main results
We first introduce the following abbreviations: Q T =Ω×(0,T), ΣT = ∂Ω×(0,T),
· p = · L p( Ω), · k,p = · W k,p( Ω) For simplicity, we denote · 2by · For every
q ∈(1,∞), we denote the dual ofW01,qbyW −1,q withq = q/(q −1) The notation (·,·) for theL2-inner product will also be used for the notation of duality pairing between dual spaces
Throughout this paper, we assume thatp and m are positive real numbers satisfying
2≤ p < m < N p
2(N − p)+ 1 (2≤ p < m < ∞ifp ≥ N). (2.1) Define the potential well
ᐃ=y ∈ W01,p(Ω)| I(y) = ∇ y p p − λ y m
m > 0
Thenᐃ is a neighborhood of 0 in W1,p
0 (Ω) Indeed, Sobolev imbedding (see [1])
and Poincare’s inequality yield
λ y m ≤ λc m ∗ ∇ y m
p ≤ λc ∗ m ∇ y m p − p ∇ y p p, ∀ y ∈ W01,p(Ω), (2.4)
wherec ∗is an imbedding constant fromW01,p(Ω) to Lm(Ω) From this, we deduce that
I(y) > 0 (i.e., y ∈ᐃ) as∇ y p < (λ −1c − m
∗ )1/(m − p)
Trang 3For later purpose, we introduce the functionalJ defined by
J(y) : = 1
p ∇ y p p − λ
Obviously, we have
J(y) = 1
m I(y) +
m − p
Define the operatorA : W01,p(Ω)→ W −1,p
(Ω) by
Ay = −div
|∇ y | p −2∇ y
thenA is bounded, monotone, hemicontinuous (see, e.g., [3]), and
(Ay, y) = ∇ y p p,
Ay, y t
= 1
p
d
dt ∇ y p p fory ∈ W01,p(Ω) (2.8) Now, we formulate the following assumptions
(H1) Letb : R → Rbe a locally bounded function satisfying
b(s)s ≥ μ1s2, b(s) ≤ μ2| s |, fors ∈ R, (2.9) whereμ1andμ2are some positive constants
(H2)y0∈ ᐃ, y1∈ L2(Ω), and
0< E(0) =1
2y1 2
+ 1
p ∇ y0 p
p − m λy0m
m < m − p
2mp
λc m
∗2(m −1)p
p/(m − p) (2.10)
The multivalued functionϕ : R →2Ris obtained by filling in jumps of a functionb :
R → Rby means of the functionsb ,b ,b, b : R → Ras follows:
b (t) =ess inf
| s − t |≤ b(s), b (t) =ess sup
| s − t |≤ b(s), b(t) =lim
→0 +b (t), b(t) =lim
→0 +b (t), ϕ(t) = b(t), b(t)
.
(2.11)
We will need a regularization ofb defined by
b n(t) = n
∞
whereρ ∈ C0∞((−1, 1)),ρ ≥0 and −11ρ(τ)dτ =1 It is easy to show thatb nis continuous for alln ∈ Nandb ,b ,b, b, b nsatisfy the same condition (H1) with a possibly different constant ifb satisfies (H1) So, in the sequel, we denote the different constants by the same symbol as the original constants
Trang 4Definition 2.1 A function y(x, t) is a weak solution to problem (1.1) if for everyT > 0,
y satisfies y ∈ L ∞(0,T; W01,p(Ω)), yt ∈ L2(0,T; W1,2(Ω))∩ L ∞(0,T; L2(Ω)), ytt ∈ L2(0,T;
W −1,p (Ω)), there exists Ξ∈ L ∞(0,T; L2(Ω)) and the following relations hold:
T
0
y tt(t), z
+
∇ y t(t), ∇ z
+∇ y(t)p −2
∇ y(t), ∇ z
+
Ξ(t),zdt
=
T
0
λy(t)m −2
y(t), z
dt, ∀ z ∈ W01,p(Ω),
Ξ(x,t) ∈ ϕ
y t(x, t)
a.e (x, t) ∈ Q T,
y(0) = y0, y t(0)= y1.
(2.13)
Theorem 2.2 Under the assumptions (H1) and (H2), problem ( 1.1 ) has a weak solution Theorem 2.3 Under the same conditions of Theorem 2.2 , the solutions of problem ( 1.1 ) satisfy the following decay rates.
If p = 2, then there exist positive constants C and γ such that
and if p > 2, then there exists a constant C > 0 such that
where E(t) =(1/2) y t(t) 2+ (1/ p) ∇ y(t) p p −(λ/m) y(t) m
In order to prove the decay rates ofTheorem 2.3, we need the following lemma by Nakao (see [8,9] for the proof)
Lemma 2.4 Let φ :R +→ R be a bounded nonincreasing and nonnegative function for which there exist constants α > 0 and β ≥ 0 such that
sup
t ≤ s ≤ t+1
φ(s) 1+β
≤ α
φ(t) − φ(t + 1)
Then the following hold.
(1) If β = 0, there exist positive constants C and γ such that
(2) If β > 0, there exists a positive constant C such that
3 Proof of Theorem 2.2
In this section, we are going to show the existence of solutions to problem (1.1) using the Faedo-Galerkin approximation and the potential method To this end let { w j } ∞
j =1
be a basis in W1,p(Ω) which are orthogonal in L2(Ω) Let Vn =Span{ w1,w2, , w n }
Trang 5We choosey n0andy n1inV nsuch that
y0n −→ y0 inW01,p, y n1−→ y1 inL2(Ω) (3.1) Lety n(t) =n
j =1g jn(t)w jbe the solution to the approximate equation
y n tt(t), w j
+
∇ y n t(t), ∇ w j
+
Ay n(t), w j
+
b n
y t n(t)
,w j
=λy n(t)m −2
y n(t), w j
,
y n(0)= y n
0, y n
t(0)= y n
1.
(3.2)
By standard methods of ordinary differential equations, we can prove the existence of a solution to (3.2) on some interval [0,t m) Then this solution can be extended to the closed interval [0,T] by using the a priori estimate below.
Step 1 (a priori estimate) Equation (3.1) and the conditiony0∈ᐃ imply that
I
y n
0
= ∇ y n
0 p
p − λy n
0 m
m −→ I
y0
Hence, without loss of generality, we assume thatI(y n0)> 0 (i.e., y0n ∈ ᐃ) for all n
Sub-stitutingw jin (3.2) byy n t(t), we obtain
d
dt E
n(t) + ∇ y n t(t) 2
+
b n
y t n(t)
, y t n(t)
where
E n(t) =1
2y n
t(t) 2
+ 1
p ∇ y n(t)p
p − λ
my n(t)m
m
=1
2y n
t(t) 2
+J
y n(t)
.
(3.5)
Integrating (3.4) over (0,t) and using assumption (H1), we have
1
2y n
t(t) 2
+J
y n(t)
+
t
0∇ y n
t(τ) 2
SinceE n(0)→ E(0) and E(0) > 0, without loss of generality, we assume that E n(0)< 2E(0)
for alln Now, we claim that
Assume that there exists a constantT > 0 such that y n(t) ∈ ᐃ for t ∈[0,T) and y n(T) ∈
∂ ᐃ, that is, I(y n(T)) =0 From (2.6), (3.4), and (3.5), we obtain
J
y n(T)
= m − p
pm ∇ y n(T)p
p ≤ E n(T) ≤ E n(0)< 2E(0), (3.8) and therefore
∇ y n(T)
p <
2pm
m − p E(0)
1/ p
Trang 6Combining this with (2.4) and using (2.10), we see that
λy n(T)m
m < λc m
∗
2pm
m − p E(0)
(m − p)/ p
∇ y n(T)p
p
< m − p
2(m −1)p ∇ y n(T)p
p < ∇ y n(T)p
p,
(3.10)
where we used the fact that (m − p)/2(m −1)p < 1 This gives I(y n(T)) > 0, which is a
contradiction Therefore (3.7) is valid From (2.6), (3.6), and (3.7),
1
2y n
t(t) 2
+m − p
pm ∇ y n(t)p
p+
t
0∇ y t n(s) 2
ds < 2E(0). (3.11)
By (H1) and (3.11), it follows that
b n
y n
t(t) 2
≤ μ2 y n
t(t) 2
here and in the sequel we denote byc a generic positive constant independent of n and t.
It follows from (3.11) and (3.12) that
y n
is bounded inL ∞
0,T; W01,p(Ω),
y t n
is bounded inL ∞
0,T; L2(Ω)∩ L2
0,T; W1,2(Ω),
b n
y n t
is bounded inL ∞
0,T; L2(Ω),
(3.13)
and sinceA : W01,p(Ω)→ W −1,p
(Ω) is a bounded operator, it follows from (3.13) that
Ay n
is bounded inL ∞
Finally, we will obtain an estimate fory n
tt Since the imbeddingW01,p(Ω) L m(Ω) is con-tinuous, we have
y n(t)m −2
y n(t), z ≤ y n(t)m −1
m z m ≤ cy n(t)m −1
1,p z 1,p (3.15) From (3.2), it follows that
T
0
y n
tt(t), z
dt
≤T
0 − Ay n(t), z
−∇ y n
t(t), ∇ z
−b n
y n
t(t)
,z
+λy n(t)m −2
y n(t), zdt, ∀ z ∈ V m,
(3.16) and hence we obtain from (3.13)–(3.15) that
T
y n
tt(t) 2
Trang 7Step 2 (passage to the limit) From (3.13), (3.14), and (3.17), we can extract a subse-quence from{ y n }, still denoted by{ y n }, such that
y n −→ y weakly star inL ∞
0,T; W01,p(Ω),
y n t −→ y t weakly inL2
0,T; W1,2(Ω),
y n
t −→ y t weakly star inL ∞
0,T; L2(Ω),
y tt n −→ y tt weakly inL2
0,T; W −1,p (Ω),
Ay n −→ ζ weakly star inL ∞
0,T; W −1,p (Ω),
b n
y n
−→ Ξ weakly star in L ∞
0,T; L2(Ω).
(3.18)
Considering that the imbeddingsW01,p(Ω) L2(Ω) and W1,2(Ω) L2(Ω) are compact and using the Aubin-Lions compactness lemma [3], it follows from (3.18) that
y n −→ y strongly inL2
Q T
y n
t −→ y t strongly inL2
Q T
Using the first convergence result in (3.18) and the fact that the imbeddingW01,p(Ω)
L2(m −1)(Ω) (p < m < N p/2(N − p) + 1 if N > p and p < m < ∞ifp ≥ N) is continuous,
we obtain
y nm −2
y n 2
L2 (Q T)=
T 0
Ω
y n(x, t) 2(m −1)
This implies that
y nm −2
y n −→ ξ weakly inL2
Q T
On the other hand, we have from (3.19) thaty n(x, t) → y(x, t) a.e in Q T, and thus| y n(x, t) | m −2y n(x, t) → | y(x, t) | m −2y(x, t) a.e in Q T Therefore, we conclude from (3.22) that
ξ(x, t) = | y(x, t) | m −2y(x, t) a.e in Q T
Lettingn → ∞in (3.2) and using the convergence results above, we have
T 0
y tt(t), z
+
∇ y t(t), ∇ z
+
ζ(t), z
+
Ξ(t),zdt
=
T
λy(t)m −2
y(t), z
dt, ∀ z ∈ W01,p(Ω).
(3.23)
Trang 8Step 3 ((y,Ξ) is a solution of (1.1)) Letφ ∈ C1[0,T] with φ(T) =0 By replacingw jby
φ(t)w jin (3.2) and integrating by parts the result over (0,T), we have
y n
t(0),φ(0)w j
+
T 0
y n
t(t), φ t(t)w j
dt =
T 0
∇ y n
t(t), φ(t) ∇ w j
dt
+
T 0
Ay n(t), φ(t)w j
dt +
T 0
b n
y n t(t)
,φ(t)w j
dt
−
T 0
λy n(t)m −2
y n(t), φ(t)w j
.
(3.24)
Similarly from (3.23), we get
y t(0),φ(0)w j
+
T 0
y t(t), φ t(t)w j
dt =
T 0
∇ y t(t), φ(t) ∇ w j
dt
+
T 0
ζ(t), φ(t)w j
dt +
T 0
Ξ(t),φ(t)w j
ds
−
T 0
λy(t)m −2
y(t), φ(t)w j
.
(3.25)
Comparing between (3.24) and (3.25), we infer that
lim
n →∞
y n t(0)− y t(0),w j
This implies thaty n
t(0)→ y t(0) weakly inW −1,p (Ω) By the uniqueness of limit, yt(0)=
y1 Analogously, takingφ ∈ C2[0,T] with φ(T) = φ (T) =0, we can obtain thaty(0) = y0 Now, we show thatΞ(x,t) ∈ ϕ(y t(x, t)) a.e in Q T Indeed, since y t n → y t strongly in
L2(Q T) (see (3.20)), y n t(x, t) → y t(x, t) a.e in Q T Letη > 0 Using the theorem of Lusin
and Egoroff, we can choose a subset ω ⊂ Q Tsuch that| ω | < η, y t ∈ L2(Q T \ ω), and y n
t →
y tuniformly onQ T \ ω Thus, for each > 0, there is an M > 2/ such that
y n
t(x, t) − y t(x, t)<
2 forn > M, (x, t) ∈ Q T \ ω. (3.27) Then, if| y n
t(x, t) − s | < 1/n, we have | y t(x, t) − s | < for all n > M and (x, t) ∈ Q T \ ω.
Therefore, we have
b
y t(x, t)
≤ b n
y n t(x, t)
≤ b
y t(x, t)
, ∀ n > M, (x, t) ∈ Q T \ ω. (3.28) Letφ ∈ L1(0,T; L2(Ω)), φ ≥0 Then
Q T \ ω b
y t(x, t)
φ(x, t)dx dt ≤
Q T \ ω b n
y n
t(x, t)
φ(x, t)dx dt
≤
Q \ ω b
y t(x, t)
φ(x, t)dx dt.
(3.29)
Trang 9Lettingn → ∞in this inequality and using the last convergence result in (3.18), we obtain
Q T \ ω b
y t(x, t)
φ(x, t)dx dt ≤
Q T \ ω Ξ(x,t)φ(x,t)dxdt
≤
Q T \ ω b
y t(x, t)
φ(x, t)dx dt.
(3.30)
Letting →0+in this inequality, we deduce that
Ξ(x,t) ∈ ϕ
y t(x, t)
and lettingη →0+, we get
Ξ(x,t) ∈ ϕ
y t(x, t)
It remains to show thatζ = Ay From the approximated problem and the convergence
results (3.18)–(3.22), we see that
lim sup
n →∞
T
0
Ay n(t), y n(t)
dt ≤y1,y0
−y t(T), y(T)
+
T 0
y t(t), y t(t)
dt −1
2∇ y(T) 2
+1
2∇ y0 2
−
T 0
Ξ(t), y(t)dt
+
T 0
λy(t)m −2
y(t), y(t)
dt.
(3.33)
On the other hand, it follows from (3.23) that
T
0
ζ(t), y(t)
dt =y1,y0
−y t(T), y(T)
+
T 0
y t(t), y t(t)
dt
−1
2∇ y(T) 2
+1
2∇ y0 2
−
T 0
Ξ(t), y(t)dt
+
T 0
λy(t)m −2
y(t), y(t)
dt.
(3.34)
Combining (3.33) and (3.34), we get
lim sup
n →∞
T 0
Ay n(t), y n(t)
dt ≤
T 0
ζ(t), y(t)
SinceA is a monotone operator, we have
0≤lim sup
n →∞
T 0
Ay n(t) − Az(t), y n(t) − z(t)
dt
≤
T
ζ(t) − Az(t), y(t) − z(t)
dt, ∀ z ∈ L2
0,T; W01,p(Ω).
(3.36)
Trang 10By Mintiy’s monotonicity argument (see, e.g., [17]),
ζ = Ay inL2
0,T; W −1,p
Therefore the proof ofTheorem 2.2is completed
4 Asymptotic behavior of solutions
In this section, we will prove the decay rates (2.14) and (2.15) inTheorem 2.3by apply-ingLemma 2.4 To prove the decay property, we first obtain uniform estimates for the approximated energy
E n(t) =1
2y n
t(t) 2
+ 1
p ∇ y n(t)p
p − m λy n(t)m
and then pass to the limit Note thatE n(t) is nonnegative and uniformly bounded Let us
fix an arbitaryt > 0 From the approximated problem (3.2) andw j = y t n(t), we get
d
dt E
n(t) + ∇ y t n(t) 2
= −b n
y n t(t)
,y t n(t)
≤ − μ1 y n
t(t) 2
This implies thatE n(t) is a nonincreasing function Setting F2
n(t) = E n(t) − E n(t + 1) and
integrating (4.2) over (t, t + 1), we have
F n2(t) ≥
t+1
t ∇ y n t(s) 2
+μ1 y n
t(s) 2
ds ≥λ1+μ1 t+1
t
y n
t(s) 2
whereλ1> 0 is the constant λ1 v 2≤ ∇ v 2,∀ v ∈ W01,2(Ω) By applying the mean value theorem, there existt1∈[t, t + 1/4] and t2∈[t + 3/4, t + 1] such that
y n t
t i ≤ 2
λ1+μ1
Now, replacingw jbyy n(t) in the approximated problem, we have
Ay n(t), y n(t)
− λy n(t)m −2
y n(t), y n(t)
= −y n
tt(t), y n(t)
−∇ y n
t(t), ∇ y n(t)
−b n
y n
t(t)
,y n(t)
...tt(t) 2
Trang 7Step (passage to the limit) From (3.13), (3.14), and (3.17), we can...
Trang 6Combining this with (2.4) and using (2.10), we see that
λy... same symbol as the original constants
Trang 4Definition 2.1 A function y(x, t) is a weak solution