entropy Article
Existence of Solutions to a Nonlinear Parabolic Equation of Fourth-Order in Variable Exponent Spaces Bo Liang 1, *, Xiting Peng 1 and Chengyuan Qu 2 1 2
*
School of Science, Dalian Jiaotong University, Dalian 116028, China;
[email protected] Department of Mathematics, Dalian Nationalities University, Dalian 116600, China;
[email protected] Correspondence:
[email protected]; Tel.: +86-411-8410-6838
Academic Editor: Angelo Plastino Received: 17 September 2016; Accepted: 15 November 2016; Published: 18 November 2016
Abstract: This paper is devoted to studying the existence and uniqueness of weak solutions for an initial boundary problem of a nonlinear fourth-order parabolic equation with variable exponent vt + div(|∇4v| p( x)−2 ∇4v) − |4v|q( x)−2 4v = g( x, v). By applying Leray-Schauder’s fixed point theorem, the existence of weak solutions of the elliptic problem is given. Furthermore, the semi-discrete method yields the existence of weak solutions of the corresponding parabolic problem by constructing two approximate solutions. Keywords: fourth-order parabolic equation; semi-discretization; variable exponent
1. Introduction We mainly study the following fourth-order parabolic equations with variable exponents: vt + div(|∇4v| p( x)−2 ∇4v) − |4v|q( x)−2 4v = g( x, v),
( x, t) ∈ Ω T ,
(1)
v( x, t) = 4v( x, t) = 0,
( x, t) ∈ Γ T ,
(2)
x ∈ Ω,
(3)
v( x, 0) = v0 ( x ),
where Ω is an open, bounded domain in R N , ∂Ω ∈ C1 . Define Ω T = Ω × (0, T ) and Γ T = ∂Ω × (0, T ). If p is a constant (especially p ≡ 2 and q ≡ 2), the Equation (1) has the structure of the classical Cahn–Hilliard problem, which is often used to describe the evolution of a conserved concentration field during phase separation in physics. It is also related to the thin-film equation if |∇4v| p( x)−2 becomes v p , which can analyze the motion of a very thin layer of viscous incompressible fluids along an include plane. There have been some results related to the existence, uniqueness and properties of solutions to the fourth-order degenerate parabolic equations (see [1,2]). The paper [3] has studied the existence of the Cahn–Hilliard equation and the reader may refer to [4] to obtain its physical background. For the constant exponent case of (1), the paper [5] has given the existence and uniqueness of weak solutions. For the problems in variable exponent spaces, the papers [6–8] have studied the existence of some fourth-order parabolic equations with a variable exponent, and [9] has given the Fujita type conditions for fast diffusion equation. For the research of the existence and long-time behavior of the fourth-order partial differential equations, the entropy functional method is often applied in order to obtain the necessary estimates and to show the entropy dissipation. The large time behavior of solutions of the thin film equation ut + (un u xxx ) x = 0 was addressed in [10,11] by the entropy function method. For 0 < n < 3, [12] proved the existence of (1) in the distributional sense and obtained the exponentially fast convergence in L∞ -norm via the entropy method of a regularized problem. We apply the idea of the entropy method to deal with the corresponding problems with variable exponents. Entropy 2016, 18, 413; doi:10.3390/e18110413
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In this paper, we apply the Leray-Schauder’s fixed point theorem to prove the existence of weak solutions of the corresponding elliptic problem of (1)–(3) in order to deal with the nonlinear source. Furthermore, the semi-discrete method yields the existence of weak solutions of the parabolic problem by constructing two approximate solutions. We will show the effect of the variable exponents and the second-order nonlinear diffusion to the degenerate parabolic Equation (1). 1.1. Preliminaries We introduce some elementary concepts and lemmas related to the variable exponent spaces in this part. Let p( x ) ≥ 1 be a continuous function in Ω and we define the variable exponent space as follows:
L p( x ) (Ω ) =
v( x ) : v is measureable and T p(.) (v) =
with the norm
Z Ω
|v| p(x) dx < ∞
v kvk p(x) = inf γ > 0 : T p(.) ≤1 . γ
It is easy to check that the variable exponent space L p( x) (Ω) becomes the classical Lebesgue space L p (Ω) when p( x ) is a positive constant. For convenience, we list some definitions and notations of the generalized Lebesgue–Sobolev space W k,p( x) (Ω): n o W k,p(x) (Ω) =: v( x ) ∈ L p(x) (Ω) : D α v ∈ L p(x) (Ω), |α| ≤ k ,
kvkW k,p(x) =:
∑
| D α v | p( x ) ,
|α|≤k 1,p( x )
E1 =: {v ∈ W0 E2 =: {v ∈
L
p0 ( x )
1,p( x ) H01 (Ω) ∩ W0 (Ω) ∩ W 2,p(x) (Ω) ∩ W 2,q(x) (Ω)|4v
(Ω)}, 1,p( x )
∈ W0
(Ω)},
1 1 + = 1. (Ω) denotes the dual space with 0 p (x) p( x )
k,p( x )
Moreover, W0
1,p( x )
dual space of W0
1,p( x )
(Ω) ∩ W 2,p(x) (Ω) ∩ W 2,q(x) (Ω)| 4v ∈ W0
−1,p0 ( x )
denotes the closure of C0∞ (Ω) in W k,p( x) (Ω)−norm, W0
(Ω) denotes the
(Ω). For any positive continuous function θ ( x ), we define θ − = inf θ ( x ), θ + = sup θ ( x ). x ∈Ω
x ∈Ω
Throughout the paper, C and Ci (i = 1, 2, 3, ...) denote the general positive constants independent of solutions and may change from line to line. In the following, we list some known results for the variable exponent spaces (see [13,14]). Lemma 1. Letting f ∈ L p( x) (Ω), one has
(1) k f k p(x) < 1(= 1; > 1) ⇐⇒ T p(.) ( f ) < 1(= 1; > 1); p+
p−
p−
p+
(2) k f k p(x) < 1 =⇒ k f k p(x) ≤ T p(.) ( f ) ≤ k f k p(x) ; k f k p(x) ≥ 1 =⇒ k f k p(x) ≤ T p(.) ( f ) ≤ k f k p(x) ; (3)k f k p(x) → 0 ⇐⇒ T p(.) ( f ) → 0; k f k p(x) → ∞ ⇐⇒ T p(.) ( f ) → ∞. 1,p( x )
Lemma 2. (Poincaré’s inequality) Letting f ∈ W0 k f k p(x) ≤ C k∇ f k p(x) .
(Ω), there exists a positive constant C such that
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R 0 Lemma 3. (Hölder’s inequality) Letting f ∈ L p( x) (Ω) and g ∈ L p ( x) (Ω), one has Ω f gdx ≤ 1 + p10− k f k p(x) k gk p0 (x) ≤ 2k f k p(x) k gk p0 (x) . p− 1.2. Results In (1), we require that p( x ) and q( x ) are two continuous functions in Ω and p− , q− > 1. Besides, the nonlinear source term g( x, v) ∈ C1 (Ω × R) satisfies the growth condition:
| g( x, v)| ≤ K |v|l (x) + s( x ),
v ∈ (−∞, +∞), x ∈ Ω,
(4)
0
where K is a positive constant, l ( x ) is a continuous function in Ω and s( x ) ∈ L p ( x) (Ω). Furthermore, by letting π ( x ) =: l ( x ) p0 ( x ), we require that π+ N p( x ) < 1, 0 ≤ l ( x ) ≤ . p− ( N − p( x )) p0 ( x )
(5)
The corresponding steady-state problem of (1)–(3) has the form: div(|∇4v| p( x)−2 ∇4v) − |4v|q( x)−2 4v = g( x, v) in Ω,
(6)
v = 4v = 0 on ∂Ω.
(7)
The weak solution is defined in the following sense. Definition 1. A function v ∈ E1 is said to be a weak solution of (6) and (7) provided that
−
Z Ω
1,p( x )
for each φ ∈ W0
|∇4v| p(x)−2 ∇4v∇φdx −
Z Ω
|4v|q(x)−2 4vφdx =
Z Ω
g( x, v)φdx
(8)
0
(Ω) ∩ Lq(x) (Ω) and g( x, v( x )) ∈ L p (x) (Ω).
The following theorem gives the existence of solutions. 1,p( x )
Theorem 1. Let v0 ∈ W0
(Ω). There exists at least a weak solution of (6) and (7) satisfying Definition 1.
For the evolution equation case, we define the weak solution of (1)–(3) as following. Definition 2. A function v is said to be a weak solution of (1)–(3) provided that −
1,p( x )
(i) v ∈ C ([0, T ]; L2 (Ω T )) ∩ L p (0, T; W0 0−
1,p( x )
−
(Ω)) ∩ Lq (0, T; W 2,q(x) (Ω)),
0−
p (0, T; W −1,p( x ) ( Ω )), v ( x, 0) = v ( x ) a.e. in Ω; 4v ∈ L p (0, T; W0 (Ω)), ∂v 0 ∂t ∈ L + + 1,p( x ) (ii) For any φ ∈ L p (0, T; W0 (Ω)) ∩ Lq (0, T; W 2,q(x) (Ω)), one has
Z TD ∂v
∂t
0
+
Z E , φ dt =
Z TZ 0
Ω
TZ 0
Ω
|∇4v| p(x)−2 ∇4v∇φdxdt
|4v|q(x)−2 4vφdxdt +
Z TZ 0
Ω
g( x, v)φdxdt.
The existence of solutions is the following theorem. 1,p( x )
1,p( x )
Theorem 2. Let p0− > 1, q0− > 1, v0 ∈ H01 (Ω) ∩ W0 (Ω) and 4v0 ∈ W0 (Ω). There exists at least a weak solution of (1)–(3). Moreover, the solution of (1)–(3) is unique when g( x, v) = µ(v( x, t) − b( x )) where µ is a constant and 0 b ( x ) ∈ L p ( x ) ( Ω ).
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This paper is organized as follows. In Section 2, we prove the existence and uniqueness of weak solution to the steady-state problem by using Leray-Schauder’s fixed point theorem. In Section 3, we prove the existence of the solution to an evolution equation by applying the semi-discrete method with necessary uniform estimates. 2. Steady-State Problem In order to apply the fixed point theorem, we consider a steady-state problem with the source g( x ): div(|∇4v| p( x)−2 ∇4v) − |4v|q( x)−2 4v = g( x ) in Ω, v = 4v = 0 on ∂Ω.
(9) (10)
By constructing an energy functional and obtaining its minimizer, we have the following existence of weak solutions. 0
Lemma 4. Let g ∈ L p ( x) (Ω). There exists a unique weak solution v ∈ E1 of (9) and (10) satisfying
− =
Z
Z Ω
Ω
|∇4v|
p( x )−2
∇4v∇4φdx −
Z Ω
|4v|q(x)−2 4v4φdx
g( x )4φdx
(11)
for any φ ∈ E1 . Proof. Introduce a functional
z(v) =
Z Ω
1 |∇4v| p(x) dx + p( x )
Z Ω
1 |4v|q(x) dx − q( x )
Z Ω
g4vdx.
(12)
For the last term, Hölder’s inequality, the Young inequality, the Sobolev embedding theorem (see [15]) and the L p -theory of the second-order elliptic equation (see [16]) gives Z g4vdx ≤ k gk p0 ( x) k4vk p( x)) Ω Z n o 0 0 1 p− p+ ≤C |4v| p(x) dx max k gk p0 ( x) , k gk p0 ( x) + ε ε Ω n o 1Z 0 0 1 p+ p− |∇4v| p(x) dx. ≤ C max k gk p0 (x) , k gk p0 (x) + 2 Ω p( x )
(13)
On the other hand, (12) implies
−C ≤ inf z(ν) ≤ z(0) = 0. ν∈ E1
Hence there exists a sequence {vk }∞ k =1 ∈ E1 such that
z(vk ) → inf z(ν)(k → ∞). ν∈ E1
Equations (13) and (14) give
k∇4vk k p(x) ≤ C, k4vk kq(x) ≤ C,
(14)
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which implies that z(vk ) is bounded and thus Lemmas 1–3 yield
kvk k2,p(x) ≤ k4vk k p(x) ≤ C k∇4vk k p(x) (Z 1 Z ≤ C1 max
Ω
|∇4vk |
p−
p( x )
,
Ω
|∇4vk |
p( x )
1 p+
)
≤ C2 ,
and
kvk k2,q(x) ≤ k4vk kq(x) (Z ≤ C3 max
Ω
|4vk |
q( x )
1 q−
1,p( x ) T
It shows that vk belongs to the space W0 a function v ∈ E1 such that
1,p( x )
vk * v weakly in W0
,
Z Ω
|4vk |
W 2,p(x)
T
q( x )
1 q+
)
≤ C4 .
W 2,q(x) uniformly, and then there exists
(Ω) ∩ W 2,p(x) (Ω) ∩ W 2,q(x) (Ω),
1,p( x )
4vk * 4v weakly in W0
( Ω ) ∩ L q ( x ) ( Ω ).
Furthermore, since z(v) is weakly lower semi-continuous on E1 , we have inf z(ν) ≤ z(v) ≤ lim inf z(vk ) = inf z(ν), k→∞
ν∈ E1
ν∈ E1
i.e., v is a minimizer of z(·) and z(v) = infν∈E1 z(ν). It guarantees that v is a weak solution of (9) and (10). The uniqueness is obvious and we omit the details. Now, we consider the problem (6) and (7) with the nonlinear source g( x, v). Lemma 5. Letting v( x ) ∈ E1 be a weak solution of (6) and (7), one has kvk E1 ≤ C. Proof. Multiplying (8) by v gives Z Ω
=−
|∇4v| p(x) dx +
Z Ω
Z Ω
|4v|q(x) dx
g( x, v)4vdx ≤ 2k gk p0 ( x) k4vk p( x)
≤C k gk p0 (x) k∇4vk p(x) Z 1 p0− p0+ ≤ |∇4v| p(x) dx + C max{k gk p0 (x) , k gk p0 (x) }. 4 Ω
(15)
By Lemmas 2 and 3 and L p -estimate (see [16]), we conclude that Z
0
Ω
| g( x, v)| p (x) dx ≤ C
Z
0
Ω
(K |v|l (x) + s( x )) p (x) dx
≤ CM p ≤ ≤
0+
Z
0
|v|l (x) p (x) dx + C
Z
|s( x )| p
Ω Ω l ( x ) p0 ( x ) l ( x ) p0 ( x ) C kvk 1,p(x) + C ≤ C k4vk p( x) W l ( x ) p0 ( x ) C k∇4vk p( x) + C,
0( x )
dx + C
+C (16)
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and thus p0−
p0+
−
+
max{k gk p0 ( x) , k gk p0 ( x) } ≤ C max{k∇4vkπp( x) , k∇4vkπp( x) } + C
≤
1 4
Z Ω
|∇4v| p(x) dx + C.
(17)
Equations (15)–(17) yield Z Ω
|∇4v| p(x) dx +
Z Ω
|4v|q(x) dx ≤ C.
(18)
It completes the proof of Lemma 5. ∗
Proof of Theorem 1. Letting ω ∈ L p ( x) (Ω) and δ ∈ [0, 1] where we choose p∗ ( x ) such that ∗ E1 ,→ L p ( x) (Ω) is compact, we consider the auxiliary problem div(|∇4v| p( x)−2 ∇4v) − |4v|q( x)−2 4v = δg( x, ω ) in Ω, v = 4v = 0 on ∂Ω. Lemma 4 ensures its existence and so we can define the fixed point operator T : [0, 1] × L p
∗ (x)
(Ω) −→ L p
∗ (x)
( Ω ),
(δ, ω ) 7−→ v and T (ω, 0) = 0. ∗ If ω ∈ L p ( x) (Ω) satisfies T (ω, δ) = ω, we can check that kω k E1 ≤ C where C > 0 is ∗ independence of ω and δ from the idea of Lemma 5. The compact embedding E1 ,→ L p ( x) (Ω) can ensure that T is a continuous and compact operator. Leray-Schauder’s fixed point theorem yields the existence of solutions of (6) and (7). 3. Evolution Equation In this section, we study the existence solutions of (1)–(3). For this purpose, we establish a semi-discrete problem at first: 1 (v − vk−1 ) + div(|∇4vk | p(x)−2 ∇4vk ) − |4vk |q(x)−2 4vk h k = g( x, vk−1 ), x ∈ Ω, vk ( x, t) = 4vk ( x, t) = 0, x ∈ ∂Ω, where vk = v( x, kh), h =
T n,k
(19) (20)
= 1, 2, · · · , n and n ∈ N.
Lemma 6. Assume v0 ∈ E2 . (19) and (20) admits a unique weak solution vk ∈ E2 satisfying i
∑
Z
i
Z
k =1 Ω
≤
∑
k =1 Ω
|∇vk |2 dx + h
i
∑
Z
h 2
i
k =1 Ω
|∇vk−1 |2 dx +
∑
|∇4vk | p(x) dx + h Z
k =1 Ω
i
∑
Z
k =1 Ω
|4vk |q(x) dx
|∇4vk−1 | p(x) dx + CT,
(21)
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and Z
|∇vi |2 dx +
Ω
i
+h ≤
∑
Z
k =1 Ω
Z
h 2
i
∑
|∇4vi | p(x) dx +
k =1 Ω
h 2
Z Ω
|∇4vi | p(x) dx
|4vk |q(x) dx h 2
|∇v0 |2 dx +
Ω
Z
Z Ω
|∇4v0 | p(x) dx + CT.
(22)
Proof. According to the argument of the Section 2, we conclude that the problem (19) and (20) has a unique weak solution vk ∈ E2 satisfying 1 h
Z
+
Ω
∇(vk − vk−1 ) · ∇φdx +
Z Ω
Z Ω
|∇4vk | p(x)−2 |∇4vk | · ∇4φdx
|4vk |q(x)−2 4vk 4φdx = −
Z Ω
g( x, vk−1 )4φdx
(23)
for anyφ ∈ C0∞ (Ω). Letting φ = vk in (23), we have 1 |∇vk |2 dx + |∇4vk | p(x) dx + |4vk |q(x) dx h Ω Ω Ω Z Z 1 ∇vk · ∇vk−1 dx − g( x, vk−1 )4vk dx = h Ω Ω Z Z Z 1 1 ≤ |∇vk−1 |2 dx + |∇vk |2 dx − g( x, vk−1 )4vk dx. 2h Ω 2h Ω Ω Z
Z
Z
(24)
Similar to the proof of (16) and (17), we get
−
Z Ω
g( x, vk−1 )4vk dx
0 1 |∇4vk | p(x) dx + C | g( x, vk−1 )| p (x) dx 2 Ω Ω Z Z 1 1 p( x ) ≤ |∇4vk | dx + |∇4vk−1 | p(x) dx + C. 2 Ω 4 Ω
≤
Z
Z
(25)
By (24) and (25), one has 1 1 |∇vk |2 dx + |∇4vk | p(x) dx + |4vk |q(x) dx 2h Ω 2 Ω Ω Z Z 1 1 ≤ |∇vk−1 |2 dx + |∇4vk−1 | p(x) dx + C. 2h Ω 4 Ω Z
Z
Z
(26)
Hence, for any 1 ≤ i ≤ n, we obtain i
∑
Z
i
Z
k =1 Ω
≤
∑
k =1 Ω
|∇vk |2 dx + h
i
∑
Z
h 2
i
k =1 Ω
|∇vk−1 |2 dx +
∑
|∇4vk | p(x) dx + h Z
k =1 Ω
i
∑
Z
k =1 Ω
|4vk |q(x) dx
|∇4vk−1 | p(x) dx + CT.
(27)
It completes the proof of (21) and (22) obtained from (21). Now, we are in the position to define the first approximate solution of (1)–(3) v (n) ( x, t) =
n
∑ χk (t)vk (t)
k =1
(28)
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where χk (t) is the characteristic function over the interval ((k − 1)h, kh] for k = 1, 2, ..., n. For this approximate solution, we have the following uniform estimates. Lemma 7. One has
kv (n) kL∞ (0,T;H1 (Ω)) + k4v (n) k 0
+ k|∇4v + k g(·, v
(n) p( x )−2
|
(n)
∇4v
)kL p0− (Ω
(n)
−
1,p( x )
L p (0,T;W0
(Ω))
kL p0− (ΩT ) + k|4v (n) |q(x)−2 4v (n) kLq0 (Ω
T)
≤ C.
T)
(29)
Proof. By Lemma 6 and
k∇v (n) k2L2 (Ω) = k∇vk ( x )k2L2 (Ω) ≤ C,
(30)
k∇v (n) kL∞ (0,T;L2 (Ω)) ≤ C.
(31)
we have the estimate On the other hand, we have
k4v (n) k
−
1,p( x )
L p (0,T;W0
n
=
h
=
)
T
Z 0
p−
∑ k4vk k1,p(x) dx
!
p−
k4v (n) k1,p(x) dx
1 p−
1 p−
k =1
n
≤
Ch
∑
k =1
p− k∇4vk k p(x) dx
!
1 p−
n
≤C h
∑(
k =1
!
Z Ω
|∇4vk | p(x) dx + 1)
1 p−
1
≤ C ( C + T ) p− .
(32)
Letting i = n in (22), we get Z TZ Ω
0
=h
|∇4v (n) | p(x) dxdt +
n Z
∑
k =1 Ω
|∇4vk | p(x) dx + h
Z TZ 0
n Z
∑
k =1 Ω
Ω
|4v (n) |q(x) dxdt
|4vk |q(x) dx ≤ C.
Another approximate solution is defined as follows: v(n) ( x, t) =
n
∑ χk (t)[ϑk (t)vk (x) + (1 − ϑk (t))vk−1 (x)],
(33)
k =1
where ( ϑk ( t ) =
t h
0,
− ( k − 1),
if t ∈ ((k − 1)h, kh], otherwise.
We also obtain some uniform estimates for this approximate solution. Lemma 8. One has
∂v(n)
+ v(n) L∞ (0,T;H1 (Ω)) ≤ C.
p0− 0 (x) − 1,p 0 ∂t L (0,T;W (Ω))
(34)
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Proof. By (33), we get ∂v(n) 1 = ∂t h
n
∑ χ k ( v k − v k −1 )
(35)
k =1
and then Z T (n) p0−
∂v 0
∂t
W −1,p
0 (x)
(Ω)
dt ≤
E p0− , ψ dt ∂t
Z T D (n) ∂v
0
≤C
Z T n
0
∑
Z Ω
k =1
Z T +C 0
+C
χk (t) n
p0− χk (t) g( x, vk−1 )ψdx dt Ω Z
∑
k =1
Z T n 0
∑ χk (t)
k =1
≤C k∇4v (n) k
p0− |∇4vk | p(x)−2 ∇4vk ∇ψdx dt
Z Ω
p0− 0− L p (Ω T )
|4vk |
q( x )−2
p0− 4vk ψdx dt
+ C k g( x, vk−1 )k
p0− 0− L p (Ω T )
≤C, 1,p( x )
for any ψ ∈ W0
(Ω) with kψk
1,p( x )
W0
(Ω)
≤ 1. It follows from (22) and (33) that
(n) m
v m
L (0,T;H01 (Ω))
≤C
m
=C
m
≤C
Ω
0
m
|∇v
(n) 2
Z T Z n
Ω
0
=C m ≤C
Z T Z
n Z (kh)
∑
∑
k =1
∑h
k =1 m+ m 2
Z
dt
2 m 2 dt χk (t)[ϑk (t)∇vk ( x ) + (1 − ϑk (t))∇vk−1 ( x )] dx
Z
k =1 ( k −1) h n
| dx
m2
Ω
[ϑk (t)∇vk ( x ) + (1 − ϑk (t))∇vk−1 ( x )] 2 dx 2
Ω
2
(|∇vk ( x )| + |∇vk−1 ( x )| )dx
m2 dt
m2
T
with C > 0 independent of m. Perform the limit m → ∞ to get
(n)
v
L∞ (0,T;H01 (Ω))
= lim v(n) Lm (0,T;H1 (Ω)) ≤ C. m→∞
0
Proof of Theorem 2. By (29), we can seek a subsequence of v (n) (still denoted by itself) and two 0 0 functions ν ∈ L p ( x) (Ω T ), ν0 ∈ Lq ( x) (Ω T ) such that v (n) * v weakly * in L∞ (0, T; H01 (Ω)), −
1,p
4v (n) * 4v weakly in L p (0, T; W0 (Ω)), 0
|∇4v (n) | p(x)−2 ∇4v (n) * ν weakly in L p (x) (Ω T ), 0
|4v (n) |q(x)−2 4v (n) * ν0 weakly in Lq (x) (Ω T ),
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as n → ∞. 0 It is easy to check that there exists a positive integer r such that W −1,p (Ω) ,→ H−r (Ω) and thus compact
the embedding H01 (Ω) ,→ L2 (Ω) ,→ H−r (Ω), the uniform estimate (34) and the Aubin lemma [17] yield the existence of a subsequence of v(n) and a function $ such that, as n → ∞, 0− ∂$ ∂v(n) * weakly in L p (0, T; W −1,p( x) (Ω)), ∂t ∂t v(n) * $ weakly ∗ in L∞ (0, T; H01 (Ω)),
v(n) → $ strongly in C ([0, T ]; L2 (Ω)), v(n) → $ a.e. in Ω T . Moreover, (23) gives, for any φ ∈ C0∞ (Ω T ), Z T Z (n) (n) 0 Ω (v − v )φdxdt Z T Z n χ ( t )( 1 − ϑ ( t ))( v − v ) φdxdt = ∑ k k k k −1 0 Ω k =1 Z T n Z h ∑ χk (1 − ϑk ) (|∇4vk | p( x)−2 ∇4vk ∇φ = 0
Ω
k =1
q( x )−2
4vk φ + g( x, vk−1 )φ)dxdt
+ |4vk | Z Z T Z (n) q( x )−2 (n) (n) p( x )−2 (n) + |4 v | 4 v φdx ≤h |∇4 v | ∇4 v ∇ φdx Ω 0 Ω Z Z g( x, v0 )φdx dt g( x, v (n) )φdx + + Ω
Ω
≤Ch → 0, as n → ∞. By the continuity of g and $ = v a.e. in Ω T , we have g( x, v (n) ) → g( x, v) a.e. in Ω T . Furthermore, the estimate (see Lemma 7) k g( x, v (n) )kL p0 (x) (Ω ) < ∞ gives T
0
g( x, v (n) ) * g( x, v) weakly in L p ( x) (Ω T ). Applying (23) and (35), we obtain, for any test function φ, Z T D (n) ∂v
∂t
0
=
Z TZ 0
+
|∇4v (n) | p(x)−2 ∇4v (n) ∇φdxdt
Ω Z TZ 0
+
0
−
Ω
Z TZ Z T
E , φ dt
Ω
|4v (n) |q(x)−2 4v (n) φdxdt g( x, v Z
( n −1) h Ω
By taking n → ∞, we have
(n)
)φdxdt + h
g( x, vn )φdxdt.
Z Ω
g( x, v0 )φ( x, 0)dx
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Z TD ∂v
∂t
0
Z E , φ dt =
TZ Ω
0
ν∇φdxdt +
Z TZ Ω
0
ν0 φdxdt +
Z TZ 0
Ω
g( x, v)φdxdt.
It remains to prove ν = |∇4v| p( x)−2 ∇4v and ν0 = |4v|q( x)−2 4v. By using v as a test function in (23) and integrating by part, we get 1 2
Z
=−
Ω
|∇v( x, T )|2 dx −
Z TZ Ω
0
1 2
Z Ω
|∇v0 |2 dx +
Z TZ Ω
0
ν∇4vdxdt +
Z TZ 0
Ω
ν0 4vdxdt
g( x, v)4vdxdt.
(36)
On the other hand, (23) implies 1 2
Z
|∇v (n) ( x, T )|2 dx −
Ω
+
Z TZ Ω
0
≤−
Z TZ Ω
0
+
Z T
1 2
Z Ω
|∇v0 |2 dx
|∇4v (n) | p(x) dxdt +
Z TZ 0
g( x, v (n) )4v (n) dxdt − h Z
( n −1) h Ω
Ω
|4v (n) |q(x) dxdt
Z Ω
g( x, v0 )v0 dx
g( x, vn )vn dxdt.
(37)
For any test functions φ, φ1 and constant ε > 0, we have Z TZ Ω
0
(|ρ(n) | p(x)−2 ρ(n) − |ηε | p(x)−2 ηε )(ρ(n) − ηε )dxdt ≥ 0,
(38)
and Z TZ
(n)
Ω
0
(n)
(n)
(|ρ1 |q(x)−2 ρ1 − |ηε |q(x)−2 η1ε )(ρ1 − η1ε )dxdt ≥ 0, (n)
where ρ(n) = ∇4v (n) , ηε = ∇4(v − εφ)) and ρ1 By (37)–(39), we arrive at 1 2
Z
+
Ω
|∇v( x, T )|2 dx −
Z TZ Ω
0
+
Z TZ Ω
0
+
Z TZ Ω
0
+
Z TZ
Z Ω
|∇v0 |2 dx −
Z TZ 0
Ω
|∇4(v − εφ)| p(x) dxdt
|∇4(v − εφ)| p(x)−2 ∇4(v − εφ)∇4vdxdt ν∇4(v − εφ)dxdt −
Z TZ Ω
0
|4(v − εφ1 )|q(x) dxdt
|4(v − εφ1 )|q(x)−2 4(v − εφ1 )4vdxdt 0
Ω
0
1 2
= 4v (n) , η1ε = 4(v − εφ1 ).
ν 4(v − εφ1 )dxdt ≤ −
Z TZ 0
Ω
g( x, v)4vdxdt.
Moreover, (36) gives Z TZ 0
+
[|∇4(v − εφ)| p(x)−2 ]∇4(v − εφ) − ν]∇4φdxdt
Ω Z TZ 0
Ω
[|4(v − εφ1 )|q(x)−2 ]4(v − εφ) − ν0 ]4φdxdt ≤ 0.
(39)
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By letting ε → 0, we obtain Z TZ Ω
0
[|∇4v| p(x)−2 ]∇4v − ν]∇4φdxdt +
Z TZ Ω
0
[|4v|q(x)−2 4v − ν0 ]4φ1 dxdt
≤ 0. By letting φ1 = 0, we have Z TZ 0
Ω
[|∇4v| p(x)−2 ]∇4v − ν]∇4φdxdt ≤ 0.
The arbitrariness of φ yields |∇4v| p( x)−2 ∇4v = ν, a.e. in Ω T . Similarly, we can obtain = ν0 , a.e. in Ω T .
|4v|q(x)−2 4v
Proof of Uniqueness. Let v1 ( x ) and v2 ( x ) be two weak solutions to (1)–(3) and φ = v1 − v2 . By taking 4φ as the test function, we get 1 2
Z tZ
+
0
Ω
Z tZ 0
+
Ω
Z tZ 0
Ω
d |∇φ|2 dxdt dt
[|∇4v1 | p(x)−2 ]∇4v1 − |∇4v2 | p(x)−2 ]∇4v2 ]∇4φdxdt [|4v1 |
q( x )−2
4v1 − |4v2 |
q( x )−2
4v2 ]4φdxdt = µ
Z tZ 0
Ω
|∇φ|2 dxdt.
It implies Z Ω
|∇φ|2 dx ≤ 2µ
Z tZ 0
Ω
|∇φ|2 dxdt,
where we have used the fact (| x |m−2 x − |y|m−2 y)( x − y) ≥ 0 for m > 1 and x, y ∈ R (or R N ). By Gronwall’s inequality, we obtain v1 ( x, t) = v2 ( x, t) a.e. in Ω T . Acknowledgments: This work was supported by NSFC (No. 11201045, 11401078) and the Education Department Science Foundation of Liaoning Province of China(No. JDL2016029). Author Contributions: Bo Liang completed the main study. Xiting Peng carried out the results of this article. Chengyuan Qu verified the calculation. All authors have read and approved the final manuscript. Conflicts of Interest: The authors declare no conflict of interest.
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