Acta Mathematica Scientia 2010,30B(6):1865–1880 http://actams.wipm.ac.cn
COMPLEXITY OF ASYMPTOTIC BEHAVIOR OF SOLUTIONS FOR THE POROUS MEDIUM EQUATION WITH ABSORPTION∗
)
Yin Jingxue (
)
Wang Liangwei (
Department of Mathematics, Jilin University, Changchun 130012, China E-mail:
[email protected]
Huang Rui (
)
School of Mathematical Sciences, South China Normal University, Guangzhou 510031, China
Abstract In this paper we analyze the large time behavior of nonnegative solutions of the Cauchy problem of the porous medium equation with absorption ut − Δum + γup = 0, 2N where γ ≥ 0, m > 1 and p > m + N2 . We will show that if γ = 0 and 0 < μ < N(m−1)+2 , 1 2N or γ > 0 and p−1 < μ < N(m−1)+2 , then for any nonnegative function ϕ in a nonnegative countable subset F of the Schwartz space S(RN ), there exists an initial-value u0 ∈ C(RN ) μ with lim u0 (x) = 0 such that ϕ is an ω-limit point of the rescaled solutions t 2 u(tβ ·, t), x→∞
where β =
2−μ(m−1) . 4
Key words complexity; asymptotic behavior; porous medium equation 2000 MR Subject Classification
1
35B40; 35K65
Introduction
In this paper we investigate the asymptotic behavior of solutions of the Cauchy problem of the porous medium equation with absorption ∂u − Δum + γup = 0, ∂t u(x, 0) = u0 ,
in RN × (0, ∞),
(1.1)
in RN ,
(1.2)
where γ ≥ 0, m > 1, p > m+ N2 and N ≥ 1 are constants, u0 is a nonnegative and appropriately smooth function. The question of asymptotic behavior is an important research subject for evolution equations. In particular, it is well studied for the existence of simple attractor to the problem of (1.1)–(1.2), see for example [15] and [16]. For the corresponding results of other evolution ∗ Received
April 14, 2010. This work is partially supported by National Natural Science Foundation of China, partially supported by Specialized Research Fund for the Doctoral Program of Higher Education, and partially supported by Graduate Innovation Fund of Jilin University (20101045).
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equations, one can see also [1, 3, 13–19, 21]. But the investigation about complexity of asymptotic behavior of evolution equations, is a recent matter for mathematicians, where most of the works are devoted to the linear case m = 1, γ = 0, see [5–8, 20]. It were V´azquez and Zuazua 1 [20] who considered the rescaled solutions u(t 2 ·, t), and prove that the set of accumulation N points in L∞ loc (R ) as t → ∞ coincides with the set of {S(1)(ϕ)}, where ϕ ranges over the set of their accumulation points of the family {u0 (λ·)} in the weak-star topology of L∞ (RN ) as λ → ∞. Furthermore, Cazenave, Dickstein and Weissler considered more general rescaled soluμ tions t 2 u(tβ ·, t) with 0 < μ < N and β ≥ 12 , and showed that such solutions present multiscale asymptotic behavior as t → ∞ in L∞ (RN ), see [5, 6, 9]. Their approach is based on the explicit expression of the solutions for the Cauchy problem of the heat equation |x − y|2 N u0 (y)dy u(x, t) = (4πt)− 2 exp − 4π RN and the linearity of the heat equation, see also [8]. It is worthy of noticing that β = 12 is a critical value for the rescaled exponent β. In fact, if 0 < β < 12 , then the nonzero ω-limits set μ of the nonnegative rescaled solutions t 2 u(tβ ·, t) is an empty set, see [6]. In the same time, there have been some investigations devoted to the nonlinear case. As for the case m = 1, γ = 1 and p > N2 + 1, Cazenave, Dickstein and Weissler have discussed the complexity of large time behavior of the rescaled solutions tμ u(tβ ·, t) for non-critical case 12 < 2 β < 12 min { N2 , p − σ2 } with μ = 2βσ, where p−1 < σ < N , and for the critical case β = 12 with 2 μ = p−1 respectively, see [6, 10]. For quasilinear case m > 1, the corresponding problems have also attracted much attention. In 2002, V´azquez and Zuazua [20] investigated the complexity 1 of large time behavior of the rescaled solutions u(t 2 ·, t) for the problem (1.1)–(1.2) with γ = 0. In our pervious paper [23], we consider the supercritical case β > 2−μ(m−1) of the complexity 4 μ β 2 of asymptotic behavior of the rescaled solutions of t u(t x, t) of the problem (1.1)–(1.2) with 2N 0 < μ < N (m−1)+2 and γ = 0. Also, Carrillo and V´ azquez [4] show that if the nonnegative ∞ N 1 N initial-value u0 ∈ L (R ) L (R ) with finite second moment RN |x|2 u0 (x)dx = 1, then the complicated asymptotic behavior of solutions for some filtration equations ∂u ∂t − ΔΦ(u) = 0 may arise, and they proved that the simplest asymptotic behavior implies a homogeneous nonlinearity (i.e., Φ(u) = um ). Our interest here is to reveal an another important fact besides the one stated by [4]. Exactly speaking, we are much interested in the critical case β = 2−μ(m−1) , and we aim to show 4 that for any countable subset F of the Schwartz’s space S(RN ), there exists a nonnegative initial 2N value u0 ∈ C(RN ) \ L1 (RN ) with lim |u0 (x)| = 0 such that if γ = 0 and 0 < μ < N (m−1)+2 , |x|→∞
2 2N or if γ > 0, p > N2 + 1 and p−1 < μ < N (m−1)+2 , then the ω-limit set of the rescaled μ β solutions t 2 u(t ·, t) of problem (1.1)–(1.2) contains the set F . This is a natural continuation of our previous work [23], where for the supercritical case β > 2−μ(m−1) and for any countable 4 subset F of the Schwartz’s space satisfying ϕ(0) = 0 if ϕ ∈ F , we construct an initial value u0 ∈ C(RN )\L1 (RN ) with lim |u0 (x)| = 0 for the problem of (1.1)–(1.2) with γ = 0 such that |x|→∞
μ
2N ϕ is an ω-limit point of the rescaled solutions t 2 u(tβ ·, t) with 0 < μ < N (m−1)+2 in L∞ (RN ) provided that ϕ ∈ F . Just as did for the linear case, the treatments for super-critical case, sub-linear case and linear case are quite different. So, in dealing with our present problem for the critical case
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β = 2−μ(m−1) , some different approaches should be proposed. Lacking of the explicit expression 4 for the solutions of the problem (1.1)–(1.2) with γ = 0 is one of our main difficulties. We overcome this difficulty by using the commutative relations between the semigroup operators and the dilation operators, and the L1 -contraction principle. The next hurdle for our problem is caused by the nonlinearity of the porous medium equation. Owing to the finite propagation property and the regular property of the solutions to the problem (1.1)–(1.2) with γ = 0, we can transform the nonlinear problem into a linear-like problem to study. For the problem (1.1)–(1.2) with γ > 0, we first compare the long time behavior of the solutions of this problem with the problem of the porous medium equation without absorption to get that under some conditions μ the rescaled solutions t 2 u(tβ ·, t) of this problem can convergence to the rescaled solutions μ N t 2 v(tβ ·, t) of the problem of the porous medium equation without absorption in L∞ loc (R ) as t → ∞. Then we use the comparison principle and the result of the problem (1.1)–(1.2) with γ = 0 to get the desired result. This paper is organized as follows. In Section 2, we give some definitions and concepts. Section 3 is devote to dealing with the complexity asymptotic behavior of the medium porous equation without absorption, namely the problem (1.1)–(1.2) with γ = 0. In Section 4, we explore the complexity asymptotic behavior of the medium porous equation with absorption, namely the problem (1.1)–(1.2) with γ > 0.
2
The Preliminaries and Definitions
To study the complexity of long time behavior, we need adopt some concepts as in [21, 23]. For f ∈ L1loc (RN ) and r > 0, let N (m−1)+2 |f |r = sup R− m−1 |f (x)|dx, |x|≤R
R≥r
N
then define the space X = X(R ) as X ≡ {f ∈ L1loc (RN ); |f |1 < ∞}, and equip this space with the norm | · |1 . Hence it is a Banach space, and any norm | · |r , r > 0, is an equivalent norm. For f ∈ X, we define
(f ) = lim |f |r . r→∞
The space X0 = X0 (RN ) is then defined by X0 ≡ {f ∈ X; (f ) = 0}. Note that L1 (RN ) ⊂ X0 ⊂ X ⊂ L1loc (RN ) with continuous inclusions. Similarly, C0 (RN ) ⊂ X0 with continuous inclusion. We use the symbol C0+ (RN ) to denote a subset of the space C0 (RN ) = {f ∈ C(RN ); lim |f (x)| = 0} |x|→∞
C0+ (RN ) ≡ {f ∈ C0 (RN ); f (x) ≥ 0} and the notation S + (RN ) to represent a subset of the Schwartz’s space S(RN ) S + (RN ) ≡ {f ∈ S(RN ); f (x) ≥ 0}.
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From these definitions, we can see that S + (RN ) is dense in C0+ (RN ). Definition 2.1 A nonnegative measurable function u = u(x, t) defined in ST = RN × (0, T ], T > 0, is a weak solution of the problem (1.1)–(1.2) with γ = 0 if u ∈ L∞ (RN ) and for every test function φ ∈ C 2,1 (ST ) which vanishes for large |x|, the following identity holds (uφt + um Δφ)dxdt + u0 (x)φ(x, 0)dx = u(x, T )φ(x, T )dx. RN
ST
RN
For any u0 ∈ X0 , the existence and uniqueness of such a solution is well established in [19]. Moreover, when γ = 0, the problem (1.1)–(1.2) generates a bounded continuous semigroup in the space X0 given by S(t) : u0 → u(x, t),
(2.1)
that is, S(t)u0 ∈ C([0, ∞); X0 ), see [21]. We now introduce the definitions of scalings and present the commutative relations between the semigroup operators and the dilation operators as in [23]. For any λ, μ, β > 0 and for any u0 ∈ X0 , the space-time dilation Γμ,β is defined as λ follows: μ,β 2 μ 2β 2 Γμ,β λ [u0 ] ≡ Dλ [S(λ t)u0 ] = λ u(λ ·, λ t), where the dilation Dλμ,β is given by Dλμ,β w(x) ≡ λμ w(λ2β x), and S(t) is the semigroup given by (2.1). From the definitions of Dλμ,β and S(t), we can get the following commutative relations between the semigroup operators S(t) and the dilation operators Dλμ,β , see details in [23]: μ,β 2 2−4β−μ(m−1) t)[Dλμ,β u0 ]. Γμ,β λ u0 = Dλ [S(λ t)u0 )] = S(λ
(2.2)
The function set n→∞ μ,β [S(tn )u0 ] −→ f in L∞ (RN ) ω μ,β (u0 ) ≡ f ∈ C0+ (RN ); ∃tn → ∞ s.t. D√ tn is called Ω-limit set.
3
The Case Without Absorption
In this section, we consider the following porous medium equation without absorption, i.e., γ = 0 in the problem (1.1)–(1.2) ∂u − Δum = 0, ∂t
in RN × R+ ,
(3.1)
in RN .
(3.2)
u(x, 0) = u0 (x),
, we will obtain a stronger result about the problem (3.1)– For the critical case β = 2−μ(m−1) 4 (3.2) than that in our previous paper [23]. Using the regularity property of the solutions of the problem (3.1)–(3.2) at t ≥ τ > 0, in the following theorem we can omit the condition of ϕ(0) = 0 which is required in Theorem 2.2 of [23].
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2N . Assume F is a countable Theorem 3.1 Let E be a countable subset of 0, N (m−1)+2 subset of S + (RN ). Suppose μ ∈ E, (3.3) and
2 − μ(m − 1) . (3.4) 4 Then for all ϕ ∈ F and for all μ, β satisfying (3.3)–(3.4), there exists an initial-value u0 ∈ C0+ (RN ) with the following property: there exists a sequence tn → ∞ such that β=
n→∞
μ,β [S(tn )u0 ] −−−→ S(1)ϕ in L∞ (RN ). D√ t n
In other words, S(1)ϕ ∈ ω μ,β (u0 ) for all ϕ ∈ F and for all μ and β satisfying (3.3) and (3.4). To prove this theorem, we need a series of lemmas. The proof of the first lemma can be found in [23]. Lemma 3.1 For every u0 ∈ X, there exists a time T = T (u0 ) and a weak solution u(x, t) of the problem (3.1)–(3.2) in QT . Moreover, for 0 < t ≤ T (u0 ), where T (u0 ) = C (u0 )−(m−1) ,
(3.5)
we have |u(·, t)|r ≤ C|u0 |r , and N
2
2
|u(x, t)| ≤ Ct− N (m−1)+2 R m−1 |u0 |rN (m−1)+2
if
|x| ≤ R,
r ≤ R.
(3.6)
We need the following concepts as in [15] and [21] to give our main lemmas. Let d(x) ≡ sup{R; u0 (y) = 0 a.e. in BR (x)} be the distance from x to the support of u0 and let us introduce the following symbol to denote the positive set of u(x, t) at time t, 0 ≤ t < ∞, Ω(t) ≡ {x ∈ RN ; u(x, t) > 0}. We also define for the set Ω(t) the ρ-neighborhood as Ωρ (t) ≡ {x ∈ RN ; d(x, Ω(t)) ≤ ρ}, where d(x, Ω(t)) is the distance from x to Ω(t). The proof of the following lemma is similar to the evolution p-Laplace equation, see [15] and [24], which is based on Lemma 3.1. Lemma 3.2 Let u(x, t) be a nonnegative solution of the problem (3.1)–(3.2). Given x0 ∈ RN , if B(x0 ) = sup R− R>0
N (m−1)+2 m−1
BR (x0 )
u0 (y)dy < ∞,
then u(x0 , t) = 0 where BR (x0 ) = {y; |x0 − y| < R}.
for 0 < t ≤ C(p, N )B(x0 )−(m−1) ,
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Proof We can assume that x = 0 by translation. Note that B(0) = lim |u|r . r→0
Assuming R = r in (3.6), then letting r → 0, we can get the desired result by (3.5). Now we consider the finite propagation property of the porous medium equation and give the precise speed of this propagation. Lemma 3.3 Let u(x, t) be a nonnegative solution of the problem (3.1)–(3.2). And assume the initial data u0 satisfies 0 ≤ u0 ∈ L∞ (RN ). Then for any 0 ≤ t1 < t2 < ∞, we have Ω(t2 ) ⊂ Ωρ(t2 −t1 ) (t1 ), where Ωρ(t2 −t1 ) (t1 ) is the ρ neighborhood of Ω(t1 ) with m−1
1
2 ρ(t2 − t1 ) = C(t2 − t1 ) 2 u0 L∞ (RN ) .
Proof Without loss of generality, we can restrict our consideration to the case of t1 = 0. Assume x0 ∈ RN with d(x0 ) > 0. For any r ≥ 0, let R = d(x0 ) + r ≥ d(x0 ), then N (m−1)+2 − N (m−1)+2 m−1 u0 (y)dy ≤ Cu0 L∞ (RN ) R− m−1 RN R BR (x0 )
2
2
= Cu0 L∞ (RN ) R− m−1 ≤ Cu0 L∞ (RN ) d(x0 )− m−1 . If R ≤ d(x0 ), then
BR (x0 )
So B(x0 ) =
sup R−
N (m−1)+2 m−1
R≥d(x0 )
u0 (y)dy = 0.
BR (x0 )
2
u0 (y)dy ≤ Cu0 L∞ (RN ) d(x0 )− m−1 .
Then, Lemma 3.2 implies u(x0 , t) = 0,
−(m−1)
for all 0 ≤ t ≤ Cu0 L∞ (RN ) d(x0 )2 .
This means Ω(t) ⊂ Ωρ(t) (0), m−1 2 L∞ (RN )
1
where ρ(t) = Cu0 t 2 . So we complete the proof of this lemma. The following lemma can also be proved in the same way as above lemma, see [19]. We will give the proof in a different way. Lemma 3.4 Assume u(x, t) is a nonnegative solution of (3.1)–(3.2) with the initial value u0 such that 0 ≤ u0 ∈ L1 (RN ). Then for any 0 ≤ t1 < t2 < ∞, Ω(t2 ) ⊂ Ωρ(t2 −t1 ) (t1 ),
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where Ωρ(t2 −t1 ) (t1 ) is the ρ neighborhood of Ω(t1 ) with m−1
1
. ρ(t2 − t1 ) = C(t2 − t1 ) N (m−1)+2 u0 LN1(m−1)+2 (RN ) Proof Similar to Lemma 3.3, we can also restrict our consideration to the case of t1 = 0. For any 0 < t < ∞, we select a sequence of times tk = 2−k t → 0 as k → ∞. We then consider the evolution in the time intervals Ik = [tk , tk−1 ], that is, we will estimate the increase of the support in these time intervals. From the L1 –L∞ smoothing effect, at each initial time t = tk , we have 2
N − N (m−1)+2
u(tk )L∞ (RN ) ≤ C(p, N )u0 LN1(m−1)+2 tk (RN )
.
The result of Lemma 3.3 obviously implies Ω(tk−1 ) ⊂ Ωρ(tk−1 −tk ) (tk ), m−1
1
2 where ρ(tk−1 − tk ) = Cu(tk )L∞ (t − tk ) 2 . Iterating, we have (RN ) k−1
Ω(t) ⊂ Ωρ(t) (0), where ρ(t) = C ≤C
∞ k=1 ∞
m−1
1
2 u(tk )L∞ (t − tk ) 2 (RN ) k−1 m−1
1
u0 LN1(m−1)+2 tkN (m−1)+2 (RN )
k=1 m−1
1
= Cu0 LN1(m−1)+2 t N (m−1)+2 (RN )
∞
k
2− N (m−1)+2
k=1
≤ Cu0
m−1 N (m−1)+2 L1 (RN )
t
1 N (m−1)+2
.
So we complete the proof of Lemma 3.3. We also need the following lemma. Its proof can be found in [23]. Lemma 3.5 Let F be a countable subset of S + (RN ) and E be a countable subset of
4N 0, (N (m−1)+2)(2−μ(m−1)) . Then there exist a constant M > 0, a sequence {σj }j≥1 and a sequence {ϕj }j≥1 such that 1. Every element of E × F occurs infinitely often in the sequence {(σj , ϕj )}j≥1 . 2. The sequence {ϕj }j≥1 satisfies max(ϕj L1 , ϕj L∞ ) ≤ max(j, M ),
for all j ≥ 1.
(3.7)
Proof of Theorem 3.1 For given E and F , let {(σj , ϕj )}j≥1 ⊂ E × F be as that in Lemma 3.5. Then, we define the initial-value u0 as follows: u0 (x) =
∞ j=1
−σj
χj (x)aj
ϕj (x/aj ),
(3.8)
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where χj (x) is the cut-off function on set Bj = {x; 2−j < |x| < 2j } relative to set Bj = {x; 2−j+1 < |x| < 2j−1 }. The sequence {aj }j≥1 will be determined by the following method. Firstly, suppose a0 ≥ 1 is large enough to satisfy exp x ≥ (4x)δ where δ =
4N (N (m−1)+2)(2−μ(m−1)) .
for all x ≥ a0 ≥ 1,
Let
⎧ ⎪ ⎨ a1 = a0 ,
N −σj−1 N −σj a ⎪ ⎩ aj = max λj2βj , aj−1 , j > 1, , exp σj−1 j
(3.9)
where λj will be determined later, we thus have aj ≥ exp
aj−1 aj−1 ≥ 4aj−1 . > exp σj δ
Iterating, we, see that aj+k ≥ 4k aj
for all j, k ≥ 1,
which implies, via (3.7), (3.8) and (3.9), u0 L∞ ≤
∞
χj (x/aj )aj −σj ϕj (x/aj )L∞ ≤
j=1
∞
max (M, j)aj −σj < ∞.
j=1
Therefore, the sequence of (3.7) is convergent in C0 (RN ). So u0 ∈ C0 (RN ). Note also, from (3.7), that u0 ≥ 0. We thus have u0 ∈ C0+ (RN ) ⊂ X0+ . So S(t)u0 is well defined for all 0 ≤ t < ∞. The hypothesis of (3.4) clearly indicates that 4βj + μj (m − 1) − 2 = 0 Let
for all j ≥ 1.
1
λn = an2βn for all n ≥ 1. In light of (2.2), we then have Γμλnn ,βn [u0 ] = S(t)[Dλμnn ,βn (un−1 + vn + wn+1 )], where un−1 =
n−1
χj (x/aj )aj −σj ϕj (x/aj ),
(3.10)
(3.11)
j=1
vn = χn (x/an )an −σn ϕn (x/an )
(3.12)
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and wn+1 =
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χj (x/aj )aj −σj ϕj (x/aj ).
j=n+1
So, we have Dλμnn ,βn un−1 L1 ≤
n−1
N −σj
λμnn −2N βn aj
χj ϕj L1 ≤ Cn2
j=1
λμnn −2N βn , μn−1 −2N βn−1 λn−1
Dλμnn ,βn vn L1 ≤ ϕn L1 and Dλμnn ,βn wn+1 L∞ ≤
∞
∞
λμnn a−σ j ϕj L∞ ≤
j=n+1
−σj
λμnn aj
ϕj L∞ .
j=n+1
Therefore, from (3.9), we can get (Dλμnn ,βn wn+1 )L∞ ≤ aσn
∞
(j + 1)aj+1 −σj+1 ≤
j=n
∞
n→∞
(j + 1)aσn aj+1 −σj+1 −−−→ 0.
(3.13)
j=n
The definitions of un−1 , vn and wn+1 , Lemma 3.3 and Lemma 3.4 imply that suppS(t)[Dλμnn ,βn (un−1 )] 2βn−1 m−1 Cn2 λ2N βn−1 −μn−1 N (m−1)+2 1 2n−1 λn−1 n−1 N (m−1)+2 , + t ⊂ x ∈ RN ; |x| ≤ βn −μn n λ2β λ2N n n suppS(t)[Dλμnn ,βn (un−1 + vn )] m−1 N (m−1)+2 Cn2 λ2N βn−1 −μn−1 1 n−1 N n N (m−1)+2 + ϕn L1 t ⊂ x ∈ R ; |x| ≤ 2 + βn −μn λ2N n and suppS(t)[Dλμnn ,βn (wn+1 )] 2βn+1 m−1 1 2−n−1 λn+1 N N (m−1)+2 N (m−1)+2 ∞ . − (wn+1 L ) t ⊂ x ∈ R ; |x| ≥ n λ2β n Without loss of generality, we can assume 0 < T < n. First note, from (3.3) and (3.4), that for all n ≥ 1, 2N βn − μn > 0. Then for fixed a1 , a2 , · · · , an−1 and 0 < t < T < ∞, we can get that if letting λn large enough in (3.9), then suppS(t)[Dλμnn ,βn (un−1 )] ⊂ {x ∈ RN ; |x| ≤ 2}. Now letting λn+1 large enough in (3.9), we thus obtain that suppS(t)[Dλμnn ,βn (un−1 + vn )]
suppS(t)[Dλμnn ,βn (wn+1 )] = ∅.
(3.14)
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So, Γμλnn ,βn [u0 ] = S(t)(Dλμnn ,βn [un−1 + vn ]) + S(t)[Dλμnn ,βn wn+1 ],
(3.15)
i.e., superposition holds as long as the supports are disjoint. For any (σ, ϕ) ∈ E × F , Lemma 3.5 implies that there exists a sequence of integers {nk }k≥1 going to infinity such that σnk = σ,
ϕnk = ϕ,
for all k ≥ 1. The hypothesis of ϕ ∈ S(RN ) and the expressions of (3.11) and (3.12) clearly imply, via L1 -contraction principle, that for any τ > 0, [unk −1 + vnk ]) − S(τ /2)(ϕ)L1 ≤ Dλμ,β [unk −1 + vnk ] − ϕL1 → 0 S(τ /2)(Dλμ,β n n k
k
as nk → ∞. Note also that [unk −1 + vnk ]L∞ ≤ C(τ )Dλμ,β [unk −1 + vnk ]L1 ≤ C(τ, ϕL1 ). S(τ /2)Dλμ,β n n k
k
We thus obtain a subsequence, still denoted by S(τ /2)Dλμ,β [unk −1 + vnk ], which satisfies n k
S(τ /2)Dλμ,β [unk −1 nk
+ vnk ] S(τ /2)ϕ weak start in L∞ (RN ) as nk → ∞.
Therefore, the regularity of the solutions for all t ≥ τ >
τ 2
> 0 indicates that
nk →∞
N S(t)Dλμ,β [unk −1 + vnk ] −−−→ S(t)ϕ in L∞ loc (R ) n
(3.16)
k
uniformly for all 0 < τ ≤ t < T < ∞. We will prove the claim that for 0 < T < ∞, if 0 < τ ≤ t < T , then nk →∞
(unk −1 + vnk )] − S(t)ϕL∞ (RN ) −−−→ 0. S(t)[Dλμ,β n
(3.17)
k
Indeed, (unk −1 + vnk )] − S(t)ϕL∞ (RN ) S(t)[Dλμ,β n k
≤
S(t)[Dλμ,β (unk −1 nk
+ vnk )] − S(t)ϕL∞ (BR0 )
+S(t)[Dλμ,β (unk −1 + vnk )]L∞ (RN \BR0 ) + S(t)ϕL∞ (RN \BR0 ) , n k
(3.18)
where BR0 = {x ∈ RN ; |x| < R0 } is a ball in RN . From Lemma 3.2, we know that for 0 < t < T < ∞, the value of S(t)ϕ in RN \ BR0 depends only on the value of ϕ in 1
m−1
2 {x ∈ RN ; |x| ≥ R0 − CT 2 ϕL∞ }.
So, the hypothesis of ϕ ∈ S + (RN ) indicates that for any > 0, there exists a Rk1 > 0 such that for all R0 ≥ Rk1 , S(t)ϕL∞ (RN \BR0 ) ≤ ϕ
{x∈RN ;|x|≥R0 −CT
1 2
m−1 2 ϕ L∞
}
<
. 3
(3.19)
The expression (3.14) obviously means that for 0 < T < ∞, if Rk > 2, nk ≥ T and 0 ≤ t < T , then RN \ BR0 unk −1 ]) = ∅. supp(S(t)[Dλμ,β n k
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In other words, for 0 < T < ∞, if R0 ≥ Rk > 2, nk ≥ T and 0 ≤ t < T , then the value of S(t)[Dλμ,β (unk −1 + vnk )] n k
unk −1 . The obvious fact 0 ≤ Dλμ,β vnk ≤ ϕ in RN \ BR0 is independent of the initial-value Dλμ,β nk nk implies that for all t > 0, S(t)Dλμ,β vnk ≤ S(t)ϕ. n k
We then obtain from (3.19) that for any given > 0 and 0 < T < ∞, if R0 > max (2, Rk1 ), nk ≥ T and 0 ≤ t < T , then S(t)[Dλμ,β (unk −1 + vnk )]L∞ (RN \BR0 ) n k
=
S(t)[Dλμ,β vnk ]L∞ (RN \BR0 ) nk
≤ S(t)ϕL∞ (RN \BR0 ) <
. 3
(3.20)
For given , R0 > 0, we deduce from (3.16) that there exist nk1 such that for 0 < T < ∞, if nk ≥ nk1 ≥ T and 0 < τ ≤ t < T , then S(t)[Dλμ,β (unk −1 + vnk )] − S(t)ϕL∞ (BR0 ) < n k
. 3
(3.21)
Here we have used the fact that BR0 is a relative compact set in RN . So, from (3.18)–(3.21), we find that for 0 < T < ∞, if 0 < τ ≤ t < T , nk0 ≥ T and nk ≥ nk0 ≥ nk1 , then S(t)[Dλμ,β (unk −1 + vnk )] − S(t)ϕL∞ (RN ) n k
≤
S(t)[Dλμ,β (unk −1 nk
+ vnk )] − S(t)ϕL∞ (BR0 )
+S(t)[Dλμ,β (unk −1 nk
+ vnk )]L∞ (RN \BR0 ) + S(t)ϕL∞ (RN \BR0 ) < .
So, we complete the proof of (3.17). By (3.10), (3.13), (3.15) and (3.17), we get that for 0 < τ < T < ∞, Γμ,β λn [u0 ] − S(t)ϕL∞ (RN ) k
≤ S(t)[Dλμ,β wnk +1 ]L∞ (RN ) + S(t)[Dλμ,β (unk −1 + vnk )] − S(t)ϕL∞ (RN ) n n k
≤
k
Dλμ,β (wnk +1 )L∞ (RN ) nk
+
S(t)[Dλμ,β (unk −1 nk
nk →∞
+ vnk )] − S(t)ϕL∞ (RN ) −−−→ 0
uniformly for any t ∈ [τ, T ). That is, nk →∞
Γμ,β λn [u0 ] −−−→ S(t)ϕ, k
in
L∞ ([τ, T ); L∞ (RN )).
(3.22)
Setting t = 1 and tnk = λ2nk in (3.22), we can get n→∞
μ,β [S(tn )u0 ] −−−→ S(1)ϕ in L∞ (RN ). D√ tn
The proof of this theorem is complete. Remark 3.1 For the critical case β = 2−μ(m−1) , we cannot divide the expression (3.10) 4 into three terms which is the base of our previous work [23]. We overcome this difficulty by the regularity of the solutions of problem (3.1)–(3.2) and the L1 -contraction principle for the soltuions of problem (3.1)–(3.2).
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The Case with Absorption
In this section, we consider the Cauchy problem of the nonlinear porous medium equation with absorption, i.e., γ > 0 in the problem (1.1)–(1.2) ∂u − Δum + γup = 0, ∂t u(x, 0) = U0 (x),
in
R N × R+ ,
in RN ,
(4.1) (4.2)
where p > m + N2 and γ > 0 are constants. We write R+ = (0, ∞), S = RN × R+ , ST = RN × (0, T ] for T > 0 and STτ = RN × (τ, T ] for 0 < τ < T . we first give the definition of the weak solutions of the problem (4.1)–(4.2) as follows. Definition 4.1 By a solution of the problem (4.1)–(4.2), we shall mean a nonnegative function u ∈ L∞ (ST ) which satisfies the identity [uζt + um Δζ − γup ζ]dxdt + U0 (x)ζ(x, 0)dx = u(x, T )ζ(x, T )dx RN
ST
RN
for any ζ ∈ C 2,1 (ST ) which vanishes for large |x|. The existence and uniqueness of such a solution is well established in [2]. The continuity of this solution in ST is shown in [11]. In this section, we show the following analogue of Theorem 3.1. Theorem 4.1 Let 2 p>m+ , N and 2 < σ < N. p−m Suppose F is a countable subset of S + (RN ). For all j ≥ 1, fix μj = μ = and βj = β =
2σ , σ(m − 1) + 2
1 μ = . 2σ σ(m − 1) + 2
Then for all ϕ ∈ F , let u0 ∈ C0+ (RN ) be given by Theorem 3.1. If u is a solution of the problem (4.1)–(4.2) with the initial value U0 = u0 , then there exists a sequence tn → ∞ such that t→∞
Dμ,β −−→ S(1)ϕ in C0 (RN ), 1 u(·, tn ) − tn2
where S(t) is the semigroup operator as given in (2.1). To prove this theorem, we need the following lemma. Lemma 4.1 Suppose p > m + N2 . Let u be a solution of the problem (4.1)–(4.2). If 0 ≤ U0 (x) ∈ L∞ (RN ) has the property |x|σ U0 (x) ≤ C,
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where
2 p−m
1877
< σ < N , then σ
1
1
t 2+σ(m−1) |u(t 2+σ(m−1) x, t) − W (t 2+σ(m−1) x, t)| → 0 as
t→∞
uniformly on any compact set K of RN , where W is the solution of the problem ∂W = ΔW m , ∂t
in S,
W (x, 0) = U0 (x),
RN .
in
Proof We first consider the following problem ∂V − ΔV m = 0, ∂t V (x, 0) = C|x|−σ ,
in
RN × (0, T ),
in RN \ {0}.
Define the functions Wk (x, t) = k σ W (kx, k 2+σ(m−1) t), uk (x, t) = k σ u(kx, k 2+σ(m−1) t) and Vk (x, t) = k σ V (kx, k 2+σ(m−1) t). Using comparison principle, we know that for all k > 0, (x, t) ∈ ST , V (x, t) = Vk (x, t) ≥ Wk (x, t) ≥ uk (x, t). As in [16], we obtain that there exists a constant C > 0 such that for all k > 0, uk (x, t) ≤ Wk (x, t) ≤ CVk (x, t + From the definition of V (x, t), we have τ 0
1 ). k σ(m−1)+2
V (x, t)dxdt ≤ Cτ
(4.3)
(4.4)
B1
and ⎧ N −σq+γ ⎪ τ γ ⎪ ⎪ ⎪ ⎪ τ ⎨ τ log 1 τ V q (x, t)dxdt ≤ Cτ + C ⎪ B1 0 ⎪ log (1 + k γ τ ) ⎪ ⎪ ⎪ ⎩ −N +σq−γ k
if
N − σq + γ > 0, N = σq,
if
N = σq,
if
N − σq + γ = 0,
if
N − σq + γ < 0,
(4.5)
where q > 1, γ = σ(m − 1) + 2 and k 2 τ ≥ 1, see details in [16]. Then, for any T > t > 0, uk and Wk satisfy the integral identity −α [ξt (Wk − uk ) + Δξ(Wkm − um ξupk ]dxdt k ) − γk = RN
τ Sτ +ST
(Wk (x, T ) − uk (x, T ))ξ(x, T )dx,
(4.6)
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where α = σ(p − 2) − 2 > 0, ξ ∈ Cc1,2 (ST ) which vanishes for large x. For any > 0, by (4.3), (4.4) and (4.5), there exists τ1 > 0 such that if 0 < τ ≤ τ1 , then (4.7) [ξt (Wk − uk ) + Δξ(Wkm − um k )]dxdt < . 3 Sτ Thanks to the properties of uk , there exists k1 such that for all k ≥ k1 , γk −α ξupk dxdt < . 3 ST Note that 1
|U0 |1 = sup R−N + m−1
(4.8)
R≥1
U0 (x)dx ≤ C. BR
By comparison principle, we get that for any τ , R > 0, k > 0 and T1 > T , Wk (x, t), uk (x, t) ∈ L1 ((τ, T1 ) × BR ) and
Wk (x, t) (1 +
1
|x|2 ) m−1
,
L∞ ((τ, T1 ) × BR )
uk (x, t) 1
(1 + |x|2 ) m−1
C([τ, T1 ] × L1loc (RN ))
∈ L∞ ((τ, T1 ) × RN ).
For any R0 > 0, T > τ > 0 and any nonnegative function θ(x) ∈ Cc∞ (BR0 ), we solve the inverse-time problem in QRT = BR (0) × (τ, T ) ⎧ ⎪ ⎪ φt + ank Δφ = 0, ⎨ ⎪ ⎪ ⎩
in
BR (0) × (τ, t),
φ = 0,
on ∂BR (0) × (τ, T ),
φ(x, T ) = θ,
for
(4.9)
BR (0),
where R = R0 + 1 and ank is a smooth approximation of ⎧ m m ⎨ Wk − uk if Wk =
uk , Wk − uk ak = ⎩ if Wk = uk mum−1 k such that 0 < ank < n1 . For any > 0, taking test function ξ = ηφ in (4.6) and using the duality method as in [21] and [22], we can get that there exist R1 > R0 + 1 and n0 such that for any 0 < τ < τ1 and nk > n0 , (Wk − uk )η(ak − ank )Δφdxdt + (Wkm − um k )(2∇η∇φ + φΔη)dxdt < , (4.10) τ τ 3 ST ST where η is the smooth cut-off function on set BR1 − (0) relative to set BR1 −2 (0). Combining (4.6), (4.7), (4.8), (4.9) with (4.10), there exist constants τ , k0 > k1 > 0 such that for any k ≥ k0 and T > τ , (Wk (x, T ) − uk (x, T ))θ(x)dx < . Now letting zk = Wk − uk ≥ 0, we obtain that zk (x, T )θ(x)dx < .
(4.11)
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Recall that U0 (x) ∈ L∞ (RN ). Thus, the families of solutions {Wk } and {uk } are uniformly bounded in S \ {(0, 0)} whence they are equicontinuous on compact subsets of S \ {(0, 0)}, see [12]. These properties of Wk (x, t) and uk (x, t) imply that there exist a subsequence {zk (x, t)} and a function z(x, t) ∈ C(ST1 ) such that zk → z as k → ∞ uniformly on any compact subset of ST1 . So, (4.11) indicates that for T > τ , z(x, T ) = 0 uniformly on compact subset of RN . Therefore, using (4.11) again, we obtain that the entire sequence zk (·, T ) = Wk (·, T ) − uk (·, T ) (4.12) convergence to 0 uniformly on compact subset of RN if T > τ . Now letting T = 1 and t = k 2+σ(m−1) in (4.12), we have σ
1
1
t→∞
t 2+σ(m−1) |u(t 2+σ(m−1) ·, t) − W (t 2+σ(m−1) ·, t)| −→ 0 uniformly on any compact subset of RN . So we complete the proof of this lemma. Using comparison principle and the result of Theorem 3.1, we can prove Theorem 4.1. 2 2N Proof of Theorem 4.1 Hypotheses of this theorem imply that 0 < p−1 < μ < N (m−1)+2
μ and β = 2σ = 2−μ(m−1) . Now let βj = β and μj = μ for all j ≥ 1 in (3.8). So, from the 4 proof of Theorem 3.1, we can get that for any > 0, there exist R > 0 and tnk0 such that if tnk ≥ tnk0 , then
Γμ,β 1 [u0 ]L∞ (RN \BR ) tn2k
μ,β ≤ S(1)[Dμ,β 1 wnk +1 ]L∞ (RN ) + S(1)[D 1 (unk −1 + vnk )]L∞ (RN \BR ) < . tn2k
tn2k
(4.13)
Using the comparison principle, we find that μ
tn2k u(tβnk ·, tnk )L∞ (RN \BR ) μ,β ≤ Dμ,β 1 [S(tnk )u0 ]L∞ (RN \BR ) = Γ 1 [u0 ]L∞ (RN \BR ) < . tn2k
tn2k
(4.14)
The definition of u0 in (3.8) clearly implies that |x|σ u0 (x) ≤ C,
for all x ∈ RN .
Therefore, Theorem 3.1 and Lemma 4.1 indicate that there exist R, t0 > 0 satisfying if tnk > max (t0 , tnk0 ), then μ
tn2k u(tβnk ·, tnk ) − S(1)ϕL∞ (RN ) μ
μ,β β 2 ≤ Γμ,β 1 [u0 ] − tnk u(tnk ·, tnk )L∞ (RN ) + Γ 1 [u0 ] − S(1)ϕL∞ (RN ) tn2k
μ 2 nk
tn2k
≤ t u(tβnk ·, tnk )L∞ (RN \BR ) + Γμ,β 1 [u0 ]L∞ (RN \BR ) tn2k
μ
μ,β β 2 +Γμ,β 1 [u0 ] − S(1)ϕL∞ (RN ) + Γ 1 [u0 ] − tnk u(tnk ·, tnk )L∞ (RN \BR ) < . tn2k
tn2k
Here we have used the inequalities (4.13) and (4.14). So, we complete the proof of this theorem.
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