Keywords: quasilinear parabolic equation; L ∞ estimates; asymptotic behavior of solution... It is an interesting problem to prove the existence of global solution ut of 1.2 or 1.1 and to
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L^{\infty} estimates of solutions for the quasilinear parabolic equation with
nonlinear gradient term and L^1 data
Boundary Value Problems 2012, 2012:19 doi:10.1186/1687-2770-2012-19
Caisheng Chen (cshengchen@hhu.edu.cn) Fei Yang (yangfei3022@163.com) Zunfu Yang (qq348450449@163.com)
ISSN 1687-2770
Article type Research
Submission date 10 August 2011
Acceptance date 15 February 2012
Publication date 15 February 2012
Article URL http://www.boundaryvalueproblems.com/content/2012/1/19
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© 2012 Chen et al ; licensee Springer.
Trang 2L∞ estimates of solutions for the quasilinear parabolic
equation with nonlinear gradient term and L1 data
College of Science, Hohai University, Nanjing 210098, P R China
∗Corresponding author: cshengchen@hhu.edu.cn
Email addresses:
FY: yangfei3022@163.com ZY: qq348450449@163.com
Abstract
In this article, we study the quasilinear parabolic problem
(
u t − div(|∇u| m ∇u) + u|u| β−2 |∇u| q = u|u| α−2 |∇u| p + g(u), x ∈ Ω, t > 0,
u(x, 0) = u0(x), x ∈ Ω; u(x, t) = 0, x ∈ ∂Ω, t ≥ 0, (0.1)
where Ω is a bounded domain in RN , m > 0 and g(u) satisfies |g(u)| ≤ K1|u| 1+ν
with 0 ≤ ν < m By the Moser’s technique, we prove that if α, β > 1, 0 ≤ p <
q, 1 ≤ q < m + 2, p + α < q + β, there exists a weak solution u(t) ∈ L ∞ ([0, ∞), L1) ∩
L ∞loc((0, ∞), W01,m+2 ) for all u0 ∈ L1(Ω) Furthermore, if 2q ≤ m + 2, we derive the
L ∞ estimate for ∇u(t) The asymptotic behavior of global weak solution u(t) for small initial data u0 ∈ L2(Ω) also be established if p + α > max{m + 2, q + β}.
Keywords: quasilinear parabolic equation; L ∞ estimates; asymptotic behavior of solution
2000 Mathematics Subject Classification: 35K20; 35K59; 35K65
In this article, we are concerned with the initial boundary value problem of the quasilinear parabolic equation with nonlinear gradient term
(
u t − div(|∇u| m ∇u) + u|u| β−2 |∇u| q = u|u| α−2 |∇u| p + g(u), x ∈ Ω, t > 0,
u(x, 0) = u0(x), x ∈ Ω, u(x, t) = 0, x ∈ ∂Ω, t ≥ 0, (1.1)
where Ω is a bounded domain in RN with smooth boundary ∂Ω and m > 0, α, β > 1, 0 ≤ p <
q, 1 ≤ q < m + 2.
Trang 3Recently, Andreu et al in [1] considered the following quasilinear parabolic problem
(
u t − ∆u + u|u| β−2 |∇u| q = u|u| α−2 |∇u| p , x ∈ Ω, t > 0, u(x, 0) = u0(x), x ∈ Ω, u(x, t) = 0, x ∈ ∂Ω, t ≥ 0, (1.2)
where α, β > 1, 0 ≤ p < q ≤ 2, p + α < q + β and u0 ∈ L1(Ω) By the so-called stability theorem
with the initial data, they proved that there exists a generalized solution u(t) ∈ C([0, T ], L1) for
(1.2), in which u(t) satisfies A k (u) ∈ L2([0, T ], W01,2) and
Z
Ω
J k (u(t) − φ(t))dx +
t
Z
0
Z
Ω
(∇u · ∇A k (u − φ) + u|u| β−2 |∇u| q A k (u − φ))dxds
=
t
Z
0
Z
Ω
(u|u| α−2 |∇u| p A k (u − φ) − A k (u − φ)φ s )dxds +
Z
Ω
J k (u0− φ(0))dx
(1.3)
for ∀t ∈ [0, T ] and ∀φ ∈ L2([0, T ], W01,2 ) ∩ L ∞ (Q T ), where Q T = Ω × (0, T ], and for any k > 0,
A k (u) =
u − k ≤ u ≤ k,
(1.4)
J k (u) is the primitive of A k (u) such that J k(0) = 0 The problem similar to (1.2) has also been extensively considered, see [2–6] and the references therein It is an interesting problem to prove
the existence of global solution u(t) of (1.2) or (1.1) and to derive the L ∞ estimate for u(t) and
∇u(t).
Porzio in [7] also investigated the solution of Leray-Lions type problem
u t = div(a(x, t, u, ∇u)), (x, t) ∈ Ω × (0, +∞),
u(x, 0) = u0(x), x ∈ Ω, u(x, t) = 0, (x, t) ∈ ∂Ω × (0, +∞),
(1.5)
where a(x, t, s, ξ) is a Carath´ eodory function satisfying the following structure condition
a(x, t, s, ξ)ξ ≥ θ|ξ| m , f or ∀(x, t, s, ξ) ∈ Ω × R+× R1× R N (1.6)
with θ > 0 and u0 ∈ L q (Ω), q ≥ 1 By the integral inequalities method, Porzio derived the L ∞
decay estimate of the form
ku(t)k L ∞(Ω)≤ Cku0k α L q(Ω)t −λ , t > 0 (1.7)
with C = C(N, q, m, θ), α = mq(N (m − 2) + mq) −1 , λ = N (N (m − 2) + mq) −1
In this article, we will consider the global existence of solution u(t) of (1.1) with u0∈ L1(Ω)
and give the L ∞ estimates for u(t) under the similar condition in [1] More specially, we will study the behavior of solution u(t) as t → 0+ Obviously, if m = 0 and g ≡ 0, problem (1.1)
is reduced to (1.2) We remark that the methods used in our article are different from that of
[1] In L ∞ estimates, we use an improved Morser’s technique as in [8–10] Since the equation
Trang 4in (1.1) contains the nonlinear gradient term u|u| α−2 |∇u| p and u|u| β−2 |∇u| q, it is difficult to
derive L ∞ estimates for u(t) and ∇u(t).
This article is organized as follows In Section 2, we state the main results and present some
Lemmas which will be used later In Section 3, we use these Lemmas to derive L ∞ estimates of
u(t) Also the proof of the main results will be given in Section 3 The L ∞ estimates of ∇u(t) are considered in Section 4 The asymptotic behavior of solution for the small initial data u0(x)
is investigated in Section 5
Let Ω be a bounded domain in RN with smooth boundary ∂Ω and k · k r , k · k 1,r denote the
Sobolev space L r (Ω) and W 1,r (Ω) norms, respectively, 1 ≤ r ≤ ∞ We often drop the letter Ω
in these notations
Let us state our precise assumptions on the parameters p, q, α, β and the function g(u) (H1) the parameters α, β > 1, 0 ≤ p < q < m + 2 < N, p + α < q + β and q(α − 1) ≥ p(β − 1), (H2) the function g(u) ∈ C1 and ∃K1≥ 0 and 0 ≤ ν < max{q + β − 2, m}, such that
|g(u)| ≤ K1|u| 1+ν , ∀u ∈ R1,
(H3) the initial data u0∈ L1(Ω),
(H4) 2q ≤ 2 + m, α, β < 2 + m(1 + 1/N )/2,
(H5) the mean curvature of H(x) of ∂Ω at x is non-positive with respect to the outward
normal
Remark 2.1 The assumptions (H1) and (H3) are similar to as in [1]
Definition 2.2 A measurable function u(t) = u(x, t) on Ω × [0, ∞) is said to be a global weak solution of the problem (1.1) if u(t) is in the class
C([0, ∞), L1) ∩ L ∞loc((0, ∞), W01,m+2)
and u|u| β−2 |∇u| q , u|u| α−2 |∇u| p ∈ L1
loc([0, ∞) × Ω), and for any φ = φ(x, t) ∈ C1([0, ∞), C1
0(Ω)), the equality
T
Z
0
Z
Ω
n
−uφ t + |∇u| m ∇u∇φ + u|u| β−2 |∇u| q φ
o
dxdt
=
Z
Ω
(u0(x)φ(x, 0) − u(x, T )φ(x, T ))dx +
T
Z
0
Z
Ω
(u|u| α−2 |∇u| p + g(u))φdxdt
(2.1)
is valid for any T > 0.
Remark 2.3 In [1], the concept of generalized solution for (1.2) was introduced A similar concept can be found in [7, 11] By the definition, we know that weak solution is the generalized solution Conversely, a generalized solution is not necessarily weak solution
Our main results read as follows
Trang 5Theorem 2.4 Assume (H1)–(H3) Then the problem (1.1) admits a global weak solution
u(t) which satisfies
u(t) ∈ L ∞ ([0, ∞), L1) ∩ C([0, ∞), L1) ∩ L ∞loc((0, ∞), W01,m+2 ), u t ∈ L2loc((0, ∞), L2) (2.2) and the estimates
Furthermore, if (H4) is satisfied, the solution u(t) has the following estimates
T
Z
0
s 1+r ku t (s)k22ds ≤ C0, (2.4)
k∇u(t)k m+2 ≤ C0t −(1+λ)/(m+2) , 0 < t ≤ T, (2.5)
with r > λ = N (mN + m + 2) −1 and C0 = C0(T, ku0k1)
Theorem 2.5 Assume (H1)–(H5) Then the solution u(t) of (1.1) has the following L ∞
gradient estimate
with σ = (2 + 2λ + N )(mN + 2m + 4) −1 and C0= C0(T, ku0k1)
Remark 2.6 The estimates (2.3) and (2.6) give the behavior of ku(t)k ∞ and k∇u(t)k ∞ as
t → 0+
Theorem 2.7 Assume the parameters α, β > 1, γ ≥ 0, 0 ≤ q < m + 2 < N and p <
m + 2 < p + α, α ≤ (m + 2 − p)(1 + 2N −1)
Then, ∃d0> 0, such that u0 ∈ L2(Ω) with ku0k2< d0, the initial boundary value problem
(
u t − div(|∇u| m ∇u) + γu|u| β−2 |∇u| q = |u| α−2 u|∇u| p , x ∈ Ω, t > 0,
u(x, 0) = u0(x), x ∈ Ω, u(x, t) = 0, x ∈ ∂Ω, t ≥ 0, (2.7)
admits a solution u(t) ∈ L ∞ ([0, ∞), L2) ∩ W01,m+2, which satisfies
where C = C(ku0k2)
Theorem 2.8 Assume the parameters γ > 0, α, β > 1, 1 ≤ p < q < m + 2 < N and
τ = N (µ − q)(q + β) ≤ 2(q2+ N β) with µ = (qα − pβ)/(q − p) > q + β.
Then, ∃d0> 0, such that u0 ∈ L2 with ku0k2 < d0, the initial boundary value problem
(
u t − div(|∇u| m ∇u) + γu|u| β−2 |∇u| q = |u| α−2 u|∇u| p , x ∈ Ω, t > 0
u(x, 0) = u0(x), x ∈ Ω; u(x, t) = 0, x ∈ ∂Ω, t ≥ 0 (2.9)
admits a solution u(t) ∈ L ∞ ([0, ∞), L2) ∩ W01,m+2 which satisfies
ku(t)k2≤ C(1 + t) −1/(q+β−2) , t ≥ 0. (2.10)
Trang 6where C = C(ku0k2).
To obtain the above results, we will need the following Lemmas
Lemma 2.9 (Gagliardo–Nirenberg type inequality) Let β ≥ 0, N > p ≥ 1, q ≥ 1 + β and
1 ≤ r ≤ q ≤ pN (1 + β)/(N − p) Then for |u| β u ∈ W 1,p(Ω), we have
kuk q ≤ C01/(β+1) kuk 1−θ r k|u| β uk θ/(β+1) 1,p
with θ = (1 + β)(r −1 − q −1 )/(N −1 − p −1 + (1 + β)r −1 ), where the constant C0 depends only on
p, N
The Proof of Lemma 2.9 can be obtained from the well-known Gagliardo–Nirenberg–Sobolev inequality and the interpolation inequality and is omitted here
Lemma 2.10 [10] Let y(t) be a nonnegative differentiable function on (0, T ] satisfying
y 0 (t) + At λθ−1 y 1+θ (t) ≤ Bt −k y(t) + Ct −δ , 0 < t ≤ T with A, θ > 0, λθ ≥ 1, B, C ≥ 0, k ≤ 1 Then, we have
y(t) ≤ A −1/θ (2λ + 2BT 1−k)1/θ t −λ + 2C(λ + BT 1−k)−1 t 1−δ , 0 < t ≤ T.
In this section, we derive a priori estimates of the assumed solutions u(t) and give a proof of
Theorem 2.4 The solutions are in fact given as limits of smooth solutions of appropriate approx-imate equations and we may assume for our estapprox-imates that the solutions under consideration are sufficiently smooth
Let u 0,i ∈ C02(Ω) and u 0,i → u0 in L1(Ω) as i → ∞ For i = 1, 2, , we consider the
approximate problem of (1.1)
u t − div
³
(|∇u|2+ i −1)m2∇u
´
+ u|u| β−2 |∇u| q = u|u| α−2 |∇u| p + g(u), x ∈ Ω, t > 0,
u(x, 0) = u 0,i (x), x ∈ Ω, u(x, t) = 0, x ∈ ∂Ω, t ≥ 0.
(3.1)
The problem (3.1) is a standard quasilinear parabolic equation and admits a unique smooth
solution u i (t)(see Chapter 6 in [12]) We will derive estimates for u i (t) For the simplicity of notation, we write u instead of u i and u k for |u| k−1 u where k > 0 Also, let C, C j be generic
constants independent of k, i, n changeable from line to line.
Lemma 3.1 Let (H1)–(H3) hold Suppose that u(t) is the solution of (3.1), then u(t) ∈
L ∞ ([0, ∞), L1)
Proof Let n = 1, 2, , and
f n (s) =
1, 1n ≤ s ns(2 − ns), 0 ≤ s ≤ 1
n
−ns(2 + ns), − n1 ≤ s ≤ 0
−1, s < −1n
It is obvious that f n (s) is odd and continuously differentiable in R1 Furthermore, |f n (s)| ≤
1, f 0
n (s) ≥ 0 and f n (s) → sign(s) uniformly in R1
Trang 7Multiplying the equation in (3.1) by f n (u) and integrating on Ω, we get
Z
Ω
f n (u)u t dx +
Z
Ω
|∇u| m+2 f n 0 (u)dx +
Z
Ω
u|u| β−2 f n (u)|∇u| q dx
≤
Z
Ω
u|u| α−2 f n (u)|∇u| p dx +
Z
Ω
g(u)f n (u)dx
(3.2)
and the application of the Young inequality gives
Z
Ω
u|u| α−2 f n (u)|∇u| p dx ≤ 1
4 Z
Ω
u|u| β−2 f n (u)|∇u| q dx + C1
Z
Ω
|u| µ−1 dx, (3.3)
where µ = (qα − pβ)(q − p) −1 ≥ 1, i.e q(α − 1) ≥ p(β − 1).
In order to get the estimate for the third term of left-hand side in (3.2), we denote
F n (u) =
u
Z
0
(s|s| β−2 f n (s)) 1/q ds, u ∈ R1.
It is easy to verify that F n (u) is odd in R1 Then, we obtain from the Sobolev inequality that
1
4
Z
Ω
u|u| β−2 f n (u)|∇u| q dx =1
4 Z
Ω
|∇F n (u)| q dx
≥λ0
Z
Ω
|F n (u)| q dx = λ0
Z
Ωn
|F n (u)| q dx + λ0
Z
Ωc n
|F n (u)| q dx
(3.4)
with some λ0 > 0 and
Ωn = {x ∈ Ω||u(x, t)| ≥ n −1 }, Ω c n = Ω\Ω n , n = 1, 2,
We note that |F n (u)| q ≤ n −(q+β−1) in Ωc n and
Z
Ωc n
|F n (u)| q dx ≤ n −(q+β−1) |Ω|.
On the other hand, we have |u(x, t)| ≥ n −1 in Ωn and
|F n (u)| ≥
|u|
Z
n −1
(s|s| β−2 f n (s)) 1/q ds ≥ q
q + β − 1
³
|u| q+β−1 q − n − q+β−1 q
´
in Ωn
This implies that there exists λ1 > 0, such that
λ0
Z
Ωn
|F n (u)| q dx ≥ λ1
Z
Ωn
|u| q+β−1 dx − λ1|Ω|n −(q+β−1) (3.5)
Trang 8Then it follows from (3.4)–(3.5) that
1 4 Z
Ω
u|u| β−2 f n (u)|∇u| q dx ≥ λ1
Z
Ω
|u| q+β−1 dx − C2n −(q+β−1) (3.6)
with some C2 > 0.
Similarly, we have from the assumption (H2) and the Young inequality that
Z
Ω
|g(u)f n (u)|dx ≤K1
Z
Ω
|u| 1+ν |f n (u)|dx
≤K1
Z
Ω
|u| 1+ν dx ≤ λ1
2 Z
Ωn
|u| q+β−1 dx + C2(1 + n −1−ν ).
(3.7)
Furthermore, the assumption µ < q + β implies that
C1
Z
Ωn
|u| µ−1 dx ≤ λ1
2 Z
Ωn
|u| q+β−1 dx + C2. (3.8)
Then (3.2)–(3.3) and (3.6)–(3.8) give that
Z
Ω
f n (u)u t dx +1
2 Z
Ω
u|u| β−2 f n (u)|∇u| q dx ≤ C3
³
1 + n −1−ν + n −(q+β−1)
´
. (3.9)
Letting n → ∞ in (3.9) yields
d
dt ku(t)k1+
1 2 Z
Ω
|u| β−1 |∇u| q dx ≤ C3. (3.10)
Note that
Z
Ω
|u| β−1 |∇u| q dx =
µ
q
q + β − 1
¶qZ
Ω
|∇u1+β−1 q | q dx ≥ 2λ2kuk q+β−11
with some λ2 > 0 Then (3.10) becomes
d
dt ku(t)k1+ λ2ku(t)k
q+β−1
1 ≤ C3. (3.11)
This gives that u(t) ∈ L ∞ ([0, ∞), L1) if u0∈ L1 2.
Remark 3.2 The differential inequality (3.10) implies that the solution u i (t) of (3.1)
satisfies
T
Z
0
Z
Ω
|u i | β−1 |∇u i | q dxdt ≤ C0 for i = 1, 2, (3.12)
with C0 = C0(T, ku0k1)
Trang 9Lemma 3.3 Assume (H1)–(H4) Then, for any T > 0, the solution u(t) of (3.1) also
satisfies the following estimates:
where λ = N (mN + m + 2) −1 , C0 = C0(T, ku0k1)
Proof Multiplying the equation in (3.1) by u k−1 , k ≥ 2, we have
1
k
d
dt ku(t)k
k
k + (k − 1)
µ
m + 2
k + 2
¶m+2
k∇u k+m m+2 k m+2 m+2+
Z
Ω
|u| β+k−2 |∇u| q dx
≤
Z
Ω
|u| α+k−2 |∇u| p dx + K1
Z
Ω
|u| ν+k dx.
(3.14)
It follows from the H¨older and Sobolev inequalities that
K1
Z
Ω
|u| ν+k dx ≤ Ckuk θ1
k kuk θ2
1 kuk θ3
s ≤ Ckuk θ1
k k∇u k+m m+2 k (m+2)θ3 k+m
m+2
≤ k − 1
2
µ
m + 2
k + 2
¶m+2
k∇u k+m k m+2 m+2 + Ck σ kuk k k ,
in which θ1 = kλ(m − ν + (m + 2)N −1 ), θ2 = νλ(m + 2)N −1 , θ3 = νλ(k + m), σ = νλ, s =
N (k + m)(N − m − 2) −1
Note that
Z
Ω
|u| α+k−2 |∇u| p dx ≤ 1
4 Z
Ω
|u| β+k−2 |∇u| q dx + C
Z
Ω
|u| µ+k−2 dx
and
1 2 Z
Ω
|u| β+k−2 |∇u| q dx ≥ C1k −qR
Ω
|∇u q+β+k−2 q | q dx
with some C1 independent of k and µ = (qα − pβ)(q − p) −1 < q + β.
Without loss of generality, we assume k > 3 − µ Similarly, we derive
C
Z
Ω
|u| µ+k−2 dx ≤ Ckuk µ1
k−2 kuk µ2
1 kuk µ3
k ∗ ≤ Cξ µ2
1 kuk µ1
k kuk µ3
k ∗
≤ Ckuk µ1
k k∇u q k /q k qµ3/q k
q ≡ A k
with ξ1= supt≥0 ku(t)k1 and
µ1= λ0(k − 2)(q + β − µ + qN −1 ), µ2 = λ0µqN −1 , µ3= λ0µq k ,
λ0= (q + β + q/N ) −1 , k ∗ = q k N (N − q) −1 , q k = q + β + k − 2.
Then, for any η > 0,
A k ≤ Cηk∇u q k /q k q q + Cη −θ 0 /θ kuk µ1θ 0
Trang 10with µλ0θ = 1, (1 − µλ0)θ 0 = 1.
Note that µ1θ 0 < k Let η = C1
2C k −q Then it follows from (3.15) that
A k ≤ C1
2 k
−q k∇u q k /q k q q + Ck γ (kuk k k+ 1) (3.16)
with γ = qθ 0 θ −1 = qµλ0/(1 − µλ0) Then, (3.14) becomes
1
k
d
dt kuk
k
k+k − 1
2
µ
m + 2
k + 2
¶m+2
k∇u k+m k m+2 m+2+C1
2 k
−q k∇u q k /q k q q ≤ Ck σ0(kuk k k+ 1) or
d
dt kuk
k
k + C1k −m k∇u k+m m+2 k m+2 m+2 ≤ Ck 1+σ0(kuk k k+ 1) (3.17)
with σ0 = max{σ, γ} = max{νλ, γ}.
Now we employ an improved Moser’s technique as in [8, 9] Let {k n } be a sequence defined
by k1 = 1, k n = R n−2 (R − m − 1) + m(R − 1) −1 (n = 2, 3, ) with R > max{m + 1, m + 4 − µ} such that k n ≥ 3 − µ(n ≥ 2) Obviously, k n → ∞ as n → ∞.
By Lemma 2.9, we have
ku(t)k k n ≤ C
m+2 m+kn
0 ku(t)k 1−θ n
k n−1 k∇u m+kn m+2 k
θn(m+2) m+kn
with θ n = RN (1 − k n−1 k −1
n )(m + 2 + N (R − 1)) −1
Then, inserting (3.18) into (3.17) (k = k n), we find that
d
dt ku(t)k
k n
k n + C1C −
m+2 θn
0 k n −m ku(t)k (1−1/θ n )(m+k n)
k n−1 ku(t)k (m+k n )/θ n
k n
≤ Ck 1+σ0
n (ku(t)k k n
k n + 1), 0 < t ≤ T,
(3.19)
or
d
dt ku(t)k
k n
k n + C1C −
m+2 θn
0 k n −m ku(t)k m−β n
k n−1 ku(t)k k n +β n
k n ≤ Ck 1+σ0
n (ku(t)k k n
k n + 1), (3.20)
where β n = (m + k n )θ −1
n − k n , n = 2, 3, It is easy to see that
θ n → θ0 = N (R − 1)
m + 2 + N (R − 1) , β n k
−1
N (R − 1) , as n → ∞.
Denote
y n (t) = ku(t)k k n
k n , 0 < t ≤ T.
Then (3.20) can be rewritten as follows
y 0 n (t) + C1C − m+2 θn k −m n (y n (t)) 1+β n /k n ku(t)k m−β n
k n−1 ≤ Ck 1+σ0
n (y n (t) + 1). (3.21)
We claim that there exist a bounded sequence {ξ n } and a convergent sequence {λ n }, such
that
ku(t)k k n ≤ ξ n t −λ n , 0 < t ≤ T. (3.22)