Time periodic solutions of porous medium equation

Research Article

Received 19 June 2009 Published online 6 April 2010 in Wiley Online Library

(wileyonlinelibrary.com) DOI: 10.1002/mma.1307MOS subject classification: 35 K 50; 35 K 55; 35 K 65; 35 B 33

Time periodic solutions of porous mediumequation

Jun Zhoua,b∗† and Chunlai Mub

Communicated by R. Showalter

In this article, we study the time periodic solutions to the following porous medium equation under the homogeneousDirichlet boundary condition:

ut −�um =ua(a(x, t)−b(x, t)ub) in X×(−∞,+∞).

The existence of nontrivial nonnegative solution is established provided that 0�a<m. The existence is also proved inthe case a=m but with an additional assumption minX×[0,T] a(x, t)>k1, where k1 is the first eigenvalue of the operator −Dunder the homogeneous Dirichlet boundary condition. We also show that the support of these solutions is independentof time by providing a priori estimates for their upper bounds using Moser iteration. Further, we establish the attractivityof maximal periodic solution using the monotonicity method. Copyright © 2010 John Wiley & Sons, Ltd.

Keywords: periodic solutions; Moser iteration; attractivity; monotonicity method

1. Introduction

In this article, we consider the degenerate parabolic problem

ut −�um = u�(a(x, t)−b(x, t)u�) in �×(−∞,+∞) (1)

u(x, t) = 0 on ��×(−∞,+∞) (2)

u(x, t+T) = u(x, t) in �×(−∞,+∞), (3)

where m>1, 0��m, �>0, the functions a(x, t) and b(x, t) are positive, smooth and T-periodic in t with 0<a−�a(x, t)�a+<+∞ and0<b−�b(x, t)�b+<+∞. � is a bounded domain of RN (N�2) with smooth boundary ��.

In population dynamics, the classical logistic growth model, or the Verhulst model, takes the form u′ =mu(1−bu). In this model, itis assumed that the dynamics is independent of spatial location (a more realistic interpretation is that only the average populationdensity is studied), and the growth and self-limitation (competition) of the species are constant (independent of time). In the lastfew decades, many modifications to the model have been introduced, and increasingly sophisticated models have been studied.For example, the diffusion model ut −�u=mu(1−bu) takes into account spatial dependence, and assumes that population diffusesto areas of lower concentration through random movement. Further, the heterogeneous environment model takes into accountspatial dependence of the growth rate and competition rate by taking m and b to be functions of x.

The problem (1)–(3) describes the evolution of the population density of a species living in a domain �, called the habitat. Theterm �um models a tendency to avoid crowding and the reaction term u�(a−bu�) represents the contribution of the populationsupply due to births and deaths(see [1, 2]). Reaction–diffusion equations with such reaction terms can be regarded as generalizationsof Fisher or Kolomogrov–Petrovsky–Piscunov equations. The homogeneous Dirichlet boundary conditions express the inhospitality

aSchool of Mathematics and Statistics, Southwest University, Chongqing, 400715, People’s Republic of ChinabCollege of Mathematics and Physics, Chongqing University, Chongqing, 400044, People’s Republic of China∗Correspondence to: Jun Zhou, School of Mathematics and Statistics, Southwest University, Chongqing, 400715, People’s Republic of China.†E-mail: [email protected]

Contract/grant sponsor: Fundamental Research Funds for the Central Universities; contract/grant number: XDJK2009C069Contract/grant sponsor: NNSF of China; contract/grant number: 10771226Contract/grant sponsor: Natural Science Foundation Project of CQ CSTC; contract/grant number: 2007BB0124

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Copyright © 2010 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2010, 33 1942–1954

J. ZHOU AND C. MU

of the boundary. The time dependence of the coefficients reflects the fact that the time periodic variations of the habitat are takeninto account (e.g. seasonal fluctuations or tidal effects).

When m=1, (1) becomes a semi-linear periodic-parabolic equation, and the problem has been studied by many researchers (see[3--13]). At the same time, the study of quasi-linear periodic-parabolic equations also attracted many authors (see [9--11, 14--16]). Inparticular, Hess, Pozio and Tesei [9] used the monotonicity methods to deal with the equation

ut =�um +a(x, t)u, (4)

where m>1 and a is a function periodic in t and Mizoguchi [10] applied the Leray–Schauder degree theory to investigate theequations with super-linear forcing term

ut =�um +h(t)f (u), (5)

where m>1, h is a positive function and f satisfies some structure conditions which are fuled automatically by f =up with p>m.Recently, in [17], Pang et al. considered the periodic solution of the following equation:

ut =div(|∇u|p−2∇u)+u�(a(x, t)−b(x, t)u�), (6)

with Dirichlet boundary condition in a bounded domain �⊂RN with smooth boundary ��, where p>2, 1��p−1, �>0. Thefunctions a and b are continuous and T-periodic. They proved the existence of periodic solutions and established the attractivity ofmaximal periodic solutions.

In this article, we consider problem (1)–(3) and we want to generalize the results of p-Laplace equation (6) to porous mediumEquation (1). Since m>1, Equation (1) is degenerate whenever u=0 and need not admit classical solutions. Hence we are led toweak solutions.

Definition ALet Q=�×(0, T). A function u is said to be a weak solution of problem (1)–(3) in [0, T] if u∈L∞((0, T), L∞(�))∩C([0, T], Lm(�)),um ∈L2((0, T), H1

0(�)) and u satisfies ∫ ∫Q

[u��

�t−∇um ·∇�+u�(a(x, t)−b(x, t)u�)�

]dx dt =0, (�)

for any �∈C1(Q) with �=0 for (x, t)∈��×(0, T).A sub-solution is defined by replacing the equality ′′ =′′ by the inequality ”�” in (�) for ��0 and a super-solution is defined by

replacing the equality ′′ =′′ by the inequality ”�” in (�) for ��0.

Let CT (Q) be the set of functions in C(�×R), which are T-periodic in t and C∗T (Q) be the set if all functions in C(�×R) satisfying

the Dirichlet boundary condition (2) and the periodicity condition (3).

Definition BA function u is called a T-periodic solution of (1) and (2) if it is a solution in [0, T] such that u∈CT (Q). A function u is a T-periodic sub-solution if it is a sub-solution in [0, T] such that u(·, 0)�u(·, T) in �. A function u is a T-periodic super-solution if it is a super-solutionin [0, T] such that u(·, T)�u(·, 0) in �.

The following comparison principle is essential to prove the asymptotic behavior of the solutions to (1) and (2), which was provedin [18]:

Comparison principle Let u and u be a pair of sub- and super-solutions of (1)–(2) in [0, T] with u(x, 0)�u(x, 0). Then u�u in Q.First, we study the existence of weak solution of problem (1)–(3) and our main results reads as follows:

Theorem 1.1Suppose that m>1 and 0��<m. Then the problem (1)–(3) admits at least one nontrivial nonnegative weak solution.

Theorem 1.2Suppose that �=m>1 and a−>�1, where �1 is the first eigenvalue of the operator −� under the homogeneous Dirichlet boundarycondition. Then the problem (1)–(3) admits at least one nontrivial nonnegative weak solution.

Next, we study the asymptotic behavior of the solution of problem (1)–(3). For simplicity, we consider the following problem:

ut −�um = u�(a(x, t)−b(x, t)u�) in �×(0,+∞), (8)

u(x, t) = 0 on ��×(0,+∞), (9)

u(x, 0) = u0(x) in �, (10)

where 1��<m, and the other conditions are the same as problem (1)–(3). Our main results reads as follows:

Theorem 1.3Let u(x, t) be a nonnegative periodic solution of (8) and (9). Then the support suppu(·, t) is independent of t.


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J. ZHOU AND C. MU

Theorem 1.4The problem (8) and (9) has a maximal periodic solution U(x, t) which is positive in �. Moreover, if u(x, t) is a solution of theinitial-boundary value problem (8)–(10) with u0(x)�0, then for any �>0, there exists t0 depending on u0(x) and �, such that

0�u(x, t)�U(x, t)+�

for x ∈� and t�t0.

This article is organized as follows: in the next section, we consider the existence of periodic solutions by using Leray–Schauderfixed point theorem and a priori estimate and prove Theorems 1.1 and 1.2. In Section 3, we provide a priori estimates for the upperbound of these periodic solutions by using Moser iteration and prove Theorems 1.3 and 1.4.

2. Existence of periodic solutions

In this section, we consider the existence of periodic solutions of problem (1)–(3) and prove Theorems 1.1 and 1.2. First, let usgive some preliminaries, which are essential to prove the existence of periodic solutions of problem (1)–(3). We need the followinglemma, whose proof can be found in [14, 19]:

Lemma 2.1Assume that a(r) is a continuous and positive function satisfying

��a(r)�� for all r ∈R,

with some positive constants � and �. Let K be a given positive constant. Then for h∈CT (Q), there exists an unique function

v ∈ (∩p>1W2,1p (Q))∩C∗

T (Q) satisfying

wt =a(w)(�w−Kw+h).

Moreover, the operator � defined by w =�(h) is continuous and compact from CT (Q) into itself, and

‖w‖W2,1

p (Q)�C‖h‖L∞(Q),

for each p>1 and some constant C depending only on �, � and p.

Proofs of Theorems 1.1 and 1.2.Step 1 (Transform of the regularized problem). First, let us consider the following regularized problem:

u�t −�um� = (um

� −�)�

m (a(x, t)−b(x, t)(um� −�)

�m ) in �×(−∞,+∞), (11)

u�(x, t) = �1m on ��×(−∞,+∞), (12)

u�(x, t+T) = u�(x, t) in �×(−∞,+∞), (13)

with small constant � satisfying 0<�<1 and then seek for a nontrivial nonnegative periodic solution as a limit point of approximatesolutions u� of problem (11)–(13). Set v� =um

� −�. Then v� ∈C∗T (Q) and

v�t =m(v�+�)1− 1m (�v�+v

�m� (a(x, t)−b(x, t)v

�m� )). (14)

If 0�v� ∈C∗T (Q) satisfies (14), then u� = (v�+�)1/m satisfies problem (11)–(13). In what follows, indeed we can prove that Equation

(14) has a solution v� ∈C∗T (Q) such that v��1, where � is a positive constant to be determined later and �1 is a fixed-positive

eigenfunction corresponding to the first eigenvalue �1 of the following eigenvalue problem:

−�� = �� in �,

� = 0 on ��.(15)

It is clear that �1>0 and �1>0 in �.Clearly, Equation (14) is equivalent to

��t =m(��+��1 +�)1− 1m (��+(��+��1)

�m (a(x, t)−b(x, t)(��+��1)

�m )−�1��1), (16)

where �� =v�−��1. To seek for its nonnegative solutions, we consider the equation

��t =m(��++��1 +�)1− 1m (��+(��++��1)

�m (a(x, t)−b(x, t)(��++��1)

�m )−�1��1), (17)

where ��+ =max{��, 0}. If ��0 satisfies (17), then �� also satisfies (16) and hence v� =��+��1 and u� = (��+��1 +�)1/m satisfy(14) and (11), respectively. Moreover, ��0 implies that u��(��1)1/m.

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J. ZHOU AND C. MU

For h∈CT (Q), we define �(h) by the unique solution of �� ∈C∗T (Q) of the equation

��t =m(��++��1 +�)1− 1m (��−K��++h),

where K>0 is a constant to be determined later. From Lemma 2.1, we know � is well-defined and compact operator from CT (Q)into itself.

Step 2 (Application of the Leray–Schauder fixed point theorem). Now, let us consider the following equation with parameter ∈ [0, 1] :

��t =m(��++��1 +�)1− 1m (��−K��++�(��)), (18)

where ⎧⎪⎪⎨⎪⎪⎩

�(��)=K��++(��1 +�(��))�

m (a(x, t)−b(x, t)(��1 +�(��))�m )−�1��1,

�(��)=min

{��+,

(a+b−

)m�

}.

In the following, we will choose the constants K and � to ensure �(��)�0.Case 1 0��<m. It is suffices to set

� = min

⎧⎪⎪⎪⎨⎪⎪⎪⎩

(a−2�1

) mm−�

max� �1,

a�m−

21+ 2m� ×max

{b

�m+ , b

�m−}

×max� �1

, 1

⎫⎪⎪⎪⎬⎪⎪⎪⎭

,

K = 22m+1

� + �m × b

1+ �m+

am�

−×(

max�

�1 +(

a+b−

)m�

) �+�m

.

In fact,

�(��) � K��++(��1 +�(��))�

m

[(a−4

−2�m b(��1)

�m

)+(a−

4−2

�m b(�(��))

�m

)]+ a−

2(��1)

�m −�1��1

� K��++(��1 +�(��))�

m

(a−4

−2�m b(�(��))

�m

).

Assume (a− / 4)−2�/mb(�(��))�/m�0, otherwise �(��)�0 holds. Then

��+�(

a−2

�m 4b

)m�

,

thus noticing ��1 and the choice of K , we get

�(��) � K��+−2�m b(��1 +�(��))

�m (�(��))

�m

� K

(a−

22+ �m b+

)m�

−2�m b+

(max

��1 +

(a+b−

)m�

) �+�m

� 0.

Case 2 �=m>1. It is suffices to set

�=min

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

(a−−�1

21+ �m ×b+

)m�

max� �1, 1

⎫⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎭

, K =21+ m� + �

m × b1+ m

�

+(a−−�1)

m�

×(

max�

�1 +(

a+b−

)m�

)m+�m

.

In fact,

�(��) = K��++(��1 +�(��))(a(x, t)−b(x, t)(��1 +�(��))�m )−�1��1

� K��++(��1 +�(��))(a−−�1 −b(x, t)(��1 +�(��))�m )


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45

J. ZHOU AND C. MU

� K��++(��1 +�(��))(a−−�1 −2�m b(��1)

�m −2

�m b(�(��))

�m )

= K��++(��1 +�(��))

[(a−−�1

2−2

�m b(��1)

�m

)+(

a−−�1

2−2

�m b(�(��))

�m

)]

� K��++(��1 +�(��))

(a−−�1

2−2

�m b(�(��))

�m

).

Assume that (a−−�1) / 2−2�/mb(�(��))�/m�0, otherwise �(��)�0 holds. Then

��+�(

a−−�1

21+ �m b

)m�

,

and thus

�(��) � K��+−2�m b(��1 +�(��))(�(��))

�m

� K

(a−−�1

21+ �m b+

)m�

−2�m b+

(max

��1 +

(a+b−

)m�

)m+�m

� 0.

Suppose that �� is a solution of (18) in C∗T (Q). Then form the standard regularity theory and maximum principle, it follows that ��

is a nonnegative classical solution. Hence, �� satisfies

��t =m(��+��1 +�)1− 1m (��−K��+�(��)), (19)

where ⎧⎪⎪⎨⎪⎪⎩

�(��)=K��+(��1 +�(��))�

m (a(x, t)−b(x, t)(��1 +�(��))�m )−�1��1,

�(��)=min

{��,

(a+b−

)m�

}.

It is easy to see that for any ∈ [0, 1],

maxQ

��(x, t)�(

a+b−

)m�

. (20)

If fact, if ��(x0, t0)=maxQ ��(x, t)≡�>(a+ / b−)m/�, then x0 ∈�, (� / �t)��|(x0 ,t0) =0 and at (x0, t0), we have

�(x0, t0) = �(�)

= K�+(��1(x0, t0)+�(�))(a(x0, t0)−b(x0, t0)(��1(x0, t0)+(�(�))�m ))−�1��1(x0, t0)

� K�+�

(�1(x0, t0)+

(a+b−

)m�

)(a(x0, t0)−b(x0, t0)

a+b−

)−�1��1(x0, t0)

� K�−�1��1(x0, t0).

Since ��|(x0 ,t0)�0, it follows that at (x0, t0),

��|(x0 ,t0) −K�+�(�)�(−1)K�−�1��1<0,

for ∈ [0, 1], and hence from (19), we get (� / �t)��|(x0 ,t0)�0, which contradicts (� / �t)��|(x0 ,t0) =0.According to the Leray–Schauder fixed point theorem, the compactness of the one-parameter operator �◦� with ∈ [0, 1] and

the estimate (20) on the solution �� ∈C∗T (Q) of (19) imply the existence of a fixed point �� ∈C∗

T (Q) of �◦�, namely, the existence if

a solution �� ∈C∗T (Q) of (19) for =1. From this and inequality (20), it follows that the problem (16) admits a nonnegative solution

�� ∈C∗T (Q).

Step 3 (Estimate of the approximate solutions). Since the inequality (20) and the relationship u� = (��+��1 +�)1/m from the step 1,we get the following estimate:

0��um� ��+�max

��1 +

(a+b−

)m�

�1+�max�

�1 +(

a+b−

)m�

. (21)

19

46


J. ZHOU AND C. MU

Fist, let us estimate ‖�um� / �t‖L2(Q). Multiplying Equation (11) by �um

� / �t and integrating over � yields

∫�

�u�

�t

�um�

�tdx =

∫�

�um�

�um�

�tdx+

∫�

(um� −�)

�m

�um�

�t(a−b(um

� −�)�m ) dx.

Integrating by part and using Young’s inequality with parameter �, we get

1

m

∫�

u1−m�

∣∣∣∣�um�

�t

∣∣∣∣2

dx+ 1

2

d

dt

∫�

|∇um� |2 dx��

2

∫�

∣∣∣∣�um�

�t

∣∣∣∣2

dx+ 1

2�

∫�

(um� −�)

2�m (a−b(um

� −�)�m )2 dx. (22)

Using (21), we get

u1−m� �

(1+�max

��1 +

(a+b−

)m�

)− m−1m

≡C1, (23)

and

(um� −�)

2�m

(a−b(um

� −�)�m

)2�(

�max�

�1 +(

a+b−

)m�

) 2�m⎛⎝a++b+

(�max

��1 +

(a+b−

)m�

) �m⎞⎠

2

≡ C2. (24)

Choosing �=C1 / m in (22) and combining (22)–(24), we get

C1

2m

∫�

∣∣∣∣�um�

�t

∣∣∣∣2

dx+ 1

2

d

dt

∫�

|∇um� |2 dx� mC2|�|

2C1. (25)

Integrating (25) with respect to t from 0 to T , and noticing that ∇um� (T)=∇um

� (0), we get

C1

2m

∫ ∫Q

∣∣∣∣�um�

�t

∣∣∣∣2

dx dt� mC2|�|T2C1

.

i.e. ∫ ∫Q

∣∣∣∣�um�

�t

∣∣∣∣2

dx dt� m2C2|�|TC2

1

. (26)

Second, let us estimate ‖∇um� ‖L2(Q). Multiplying Equation (11) by um

� and integrating over � yields

∫�

�u�

�tum

� dx =∫

��um

� um� dx+

∫�

(um� −�)

�m um

� (a−b(um� −�)

�m )dx. (27)

Since �um� / �−→� ∣∣�� 0 (−→� is the unit outward normal of ��), we get

∫�

�um� um

� dx =∫

��um

��um

�

�−→� d (x)−∫

�|∇um

� |2 dx�−∫

�|∇um

� |2dx, (28)

where d (x) denote the measure of ��. Similar to (24), we get

(um� −�)

�m um

�

(a−b(um

� −�)�m

)�(

�max�

�1 +(

a+b−

)m�

) �m(

1+�max�

�1 +(

a+b−

)m�

)m

×⎛⎝a++b+

(�max

��1 +

(a+b−

)m�

) �m⎞⎠

≡ C3. (29)

Combining (27)–(29), and using Höder’s inequality, we get∫�

|∇um� |2dx � C3 +

∫�

∣∣∣∣�u�

�tum

�

∣∣∣∣dx

� C3 +(∫

�

∣∣∣∣�u�

�tum−1

�

∣∣∣∣2

dx

) 12 (∫

�u2

�

) 12


19

47

J. ZHOU AND C. MU

� C3 +(

1

m

∫�

∣∣∣∣�um�

�t

∣∣∣∣2

dx

) 12⎛⎝∫

�

(1+�max

��1 +

(a+b−

)m�

) 2m

dx

⎞⎠

12

≡ C3 +C4

(∫�

∣∣∣∣�um�

�t

∣∣∣∣2

dx

) 12

. (30)

Integrating (30) with respect to t from t to T and using (26) and Höder’s inequality, we get

∫ ∫Q

|∇um� |2dx dt � C3T +C4

∫ T

0

(∫�

∣∣∣∣�um�

�t

∣∣∣∣2

dx

) 12

dt

� C3T +C4T12

(∫ ∫Q

∣∣∣∣�um�

�t

∣∣∣∣2

dx dt

) 12

� C3T + mTC4C122 |�| 1

2

C1. (31)

Step 4 (Completion of the proof). From (21), (26) and (31), we know that there exists a function u∈L∞(Q), such that u� →u, a.e.in Q, except for a subsequence. It is easy to see that um ∈L2((0, T), H1

0(�)), �um / �t ∈L2(Q) and u satisfies the integral equation inDefinition A.

∀t1, t2 ∈ [0, T], t1<t2, using (26) and Hölder’s inequality, we get

∣∣∣∣∫

�um

� (x, t2)dx−∫

�um

� (x, t1)dx

∣∣∣∣2 =∣∣∣∣∫ t2

t1

∫�

�um�

�tdx dt

∣∣∣∣2

�∫ t2

t1

∫�

(�um

�

�t

)2

dx dt|�|(t2 −t1)

�∫ ∫

Q

(�um

�

�t

)2

dx dt|�|(t2 −t1)

� m2|�|2C2T

C21

(t2 −t1).

From the above equality, we get the following estimate for u :∣∣∣∣∫

�um(x, t2)dx−

∫�

um(x, t1)dx

∣∣∣∣2 � m2|�|2C2T

C21

(t2 −t1).

The above inequality implies u∈C([0, T], Lm(�)).Finally, for any �∈C∞

0 (�), set f�(t)=∫� um� (x, t)�(x)dx. Using (26) and Hölder’s inequality, we get

∫ T

0|f ′� (t)|2dt =

∫ ∫Q

(�um

�

�t

)2

�2 dx dt�‖�2‖L∞(�)m2C2|�|T

C21

.

It follows that f�(t)→ f (t)=∫� um(x, t)�(x)dx uniformly on [0, T]. On the other hand, since f�(t+T)= f�(t), we see that f (t+T)= f (t),that is ∫

�um(x, t+T)�(x) dx =

∫�

um(x, t)�(x) dx,

which implies that um(x, t+T)=um(x, t) for all t ∈R and almost x ∈�, due to the arbitrariness of �(x). The proofs of Theorems 1.1and 1.2 are completed. �

3. Asymptotic behavior

In this section, we consider the asymptotic of problem (8)–(10) and our main methods come from [17]. First, we give a prioriestimates for the upper bound of periodic solutions of (8) and (9) by Moser iteration. These estimate will play a crucial role inproving the existence and attractivity of the maximal periodic solution of (8) and (9).

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J. ZHOU AND C. MU

Lemma 3.1If u(x, t) is a nonnegative continuous periodic function that satisfies

d

dt‖u(t)‖r+2

r+2 +C1

∥∥∥∇(ur+m+1

2 +1)∥∥∥2

2�C2(r+1)1+�‖u(t)‖r+2

r+2, (32)

for all r�0, where ‖u(t)‖r+2r+2 =∫� |u(x, t)|r+2dx, then there exists a constant M, which is independent of u and r such that

‖u(t)‖L∞(�)�M. (33)

Here, u(t)=u(·, t) and ��0 and Ci(i=1, 2) are constants independent of u and r

ProofAssume that ‖u‖L∞(�) �=0 and for k =1, 2,. . ., we set

uk(t)=urk+m+1

2 +1, �k = 2(rk +2)

rk +m+3, rk =2k +m−1.

Obviously, �k<2, rk =2k−1 +rk−1. By (32), we have

d

dt‖uk(t)‖�k

�k +C1‖∇uk(t)‖22�C2(rk +1)1+�‖uk(t)‖�k

�k . (34)

In the sequel, we use the generic symbol C to denote constants independent of k, which may take different values in differentcases.

Using Gagliardo–Nirenberg’s inequality, we have

‖uk(t)‖�k �C‖∇uk(t)‖�k2 ‖uk(t)‖1−�k

1 , (35)

where

�k = (rk −m+1)(N+2)

4N(rk +2)∈ (0, 1).

By (35) and the fact that ‖uk(t)‖1 =‖uk−1(t)‖�k−1�k−1 , we have

d

dt‖uk(t)‖�k

�k � −C‖uk(t)‖2�k�k ‖uk(t)‖

2(�k−1)�k

1 +C(rk +1)1+�‖uk(t)‖�k�k

� −C‖uk(t)‖2�k�k ‖uk−1(t)‖

2(�k−1)�k−1�k

�k−1 +C(rk +1)1+�‖uk(t)‖�k�k . (36)

Thus, setting �k =max{1, supt ‖uk(t)‖�k } and noticing that �k ∈ (0, 1), we obtain

d

dt‖uk(t)‖�k

�k �‖uk(t)‖�k (rk+1)

rk+2�k ×

⎡⎣−C‖uk(t)‖

2�k

− �k (rk+1)rk+2

�k ×�

2(�k−1)�k−1�k

k−1 +C(rk +1)1+�‖uk(t)‖�k

rk+2�k

⎤⎦ (37)

To estimate (rk +1)1+�‖uk(t)‖�k

rk+2�k , we use the following Young’s inequality

ab��ap′ +�− q′

p′ 1

q′(

1

p′) q′

p′bq′

,

where p′, q′>1 and 1 / p′+1 / q′ =1, let us set

a = ‖uk(t)‖�k

rk+2 , b= (rk +1)1+�, �= 1

2�

2(�k−1)�k−1�k

k−1 ,

p′ = �k = 2(rk +2)

�k�k−rk −1= 4N(rk +2)(rk +m+3)

(N+2)(rk −m+1)−rk −1,

and obtain

(rk +1)1+�‖uk(t)‖�k

rk+2�k � 1

2 ‖uk(t)‖2�k

− �k (rk+1)rk+2

�k ×�

2(�k−1)�k−1�k

k−1 +C(rk +1)�k (1+�)�k−1 ×�

2(1−�k )�k−1�k (�k−1)

k−1 . (38)

Here, we have used the fact p′ =�k>�>1 for some � independent of k. In fact, it is easy to verify that limk→+∞ �k =+∞.


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J. ZHOU AND C. MU

Denoting

ak = (1+�)�k

�k −1, bk = 2(1−�k)�k−1

�k(�k −1),

and combining (38) with (37), we have

d

dt‖uk(t)‖�k

�k �d

dt‖uk(t)‖

�k (rk+1)rk+2

�k

⎡⎣− C

2‖uk(t)‖

2�k

− �k (rk+1)rk+2

�k ×�

2(�k−1)�k−1�k

k−1 +C(rk +1)ak �bkk−1

⎤⎦ .

Hence,

(rk +2)d

dt‖uk(t)‖

�krk+2�k �− C

2‖uk(t)‖

2�k

− �k (rk+1)rk+2

�k ×�

2(�k−1)�k−1�k

k−1 +C(rk +1)ak �bkk−1. (39)

The periodicity of uk(x, t) implies that there exists time t0 at which ‖uk(t)‖�k reaches its maximum and thus the left-hand side of(39) vanished. It then follows that:

‖uk(t)‖�k �

⎡⎣C(rk +1)ak ×�

bk+ 2(1−�k )�k−1�k

k−1

⎤⎦

1ck

,

where

ck = 2

�k− �k(rk +1)

rk +2= �k�k

rk +2.

Therefore, we conclude that

‖uk(t)‖�k �[C(rk +1)�k ]rk+2�k�k ×�

2(1−�k )(rk+2)�k−1(�k−1)�k�k

k−1 .

Noticing that (rk +2) / [(�k −1)�k]=�k / (2−�k�k), by boundedness of (rk +2) / (�k�k) and ak , we get

‖uk(t)‖�k �C2�k ×�

2(1−�k )�k−12−�k�k

k−1 ,

where � is a positive constant independent of k.Since �k =c2(rk +2) / (rk +m+3<2) implies that

2(1−�k)�k−1

2−�k�k�2(1−�k)�k−1

2−2�k=�k−1<2

and �k−1�1, we get

‖uk(t)‖�k �CAk�2k−1,

or

ln‖uk(t)‖�k � ln�k� ln C+k ln A+2 ln�k−1,

where A=2�>1. Thus

ln‖uk(t)‖�k � ln C×k−2i=0 2i +2k−1 ln�1 + ln A(k−2

i=0 (k− i)2i)

� (2k−1 −1) ln C+2k−1 ln�1 +(3×2k−1 −k−2) ln A,

or

‖u(t)‖rk+2�[C2k−1−1 ×�2k−1

1 ×A3×2k−1−k−2]2

rk+m+3 . (40)

On the other hand, it follows from (34) with k =1 that

d

dt‖u1(t)‖�1

�1+C1‖∇u1(t)‖2

2�C2‖u1(t)‖�1�1

, (41)

where 1<�1 = (m+3) / (m+2)<2. By the Hölder’s inequality and the Sobolev’s inequality, we have

‖u1(t)‖�1�|�|1�1

− 12 ‖u1(t)‖2�|�|

1�1

− 12 ‖∇u1(t)‖2. (42)

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J. ZHOU AND C. MU

Combining (42) with (41) yields

d

dt‖u1(t)‖�1

�1+C1‖u1(t)‖2

�1�C2‖u1(t)‖�1

�1.

By the Young’s inequality, it follows that

d

dt‖u1(t)‖�1

�1+C1‖u1(t)‖2

�1�C2, (43)

where Ci (i=1, 2) are constants independent of u1(x, t).The periodicity of u1(x, t) implies that there exists time t0 at which ‖u1(t)‖�k reaches its maximum and thus the term (d / dt)‖u1(t)‖�1

�1

in (43) vanished. It then follows that

‖u1(t)‖�1�C. (44)

Combining (44) with (40), noticing the definition of �1 and rk , we get

‖u(t)‖rk+2 � [C2k−1−1 ×�2k−1

1 ×A3×2k−1−k−2]2

2k+2m+2

� [C2k−1−1 ×max{1, C}2k−1 ×A3×2k−1−k−2]2

2k+2m+2 . (45)

Let k →+∞ in (45), we obtain

‖u(t)‖L∞(�)�C×max{1, C}×A3.

Which implies (33). The proof of Lemma 3.1 is completed. �

Now we can use Lemma 3.1 to prove the following lemma.

Lemma 3.2Let u(x, t) be a nonnegative periodic solution of (8) and (9), then there exists a constant C independent of u(x, t), such that

‖u(·, t)‖L∞(�)�C.

ProofMultiplying Equation (8) by ur+1 (r�0) and integrating by parts over �, we obtain

1

r+2

d

dt‖u(t)‖r+2

r+2 + 4m(r+1)

(r+m+3)2

∥∥∥∇(ur+m+1

2 +1(t))∥∥∥2

2�‖a(x, t)‖L∞(�×(0,T))‖u(t)‖r+�+1

r+�+1,

i.e.

d

dt‖u(t)‖r+2

r+2 + 4m(r+1)(r+2)

(r+m+3)2‖∇(u

r+m+12 +1(t))‖2

2�(r+2)‖a(x, t)‖L∞(�×(0,T))‖u(t)‖r+�+1r+�+1. (46)

Set f (r)= (r+1)(r+2) / (r+m+3)2, it is easy to show f ′(r)>0. So

4m(r+1)(r+2)

(r+m+3)2� 8m

(m+3)2≡C1. (47)

On the other hand

(r+2)‖a(x, t)‖L∞(�×(0,T)) = ‖a(x, t)‖L∞(�×(0,T))r+2

r+1(r+1)

� 2‖a(x, t)‖L∞(�×(0,T))(r+1)

≡ C2(r+1). (48)

Combining (46)–(48), we get

d

dt‖u(t)‖r+2

r+2 +C1‖∇(ur+m+1

2 +1(t))‖22�C2(r+1)‖u(t)‖r+�+1

r+�+1. (49)

If N>2, using the Hölder’s inequality, we have

‖u‖r+�+1r+�+1 =

∫�

ur+�+1(x) dx�C(�)‖u(t)‖�r+2‖u(t)‖

q,

where

�= (r+2)(m−�+2)

m+1, = (�−1)(r+m+3)

m+1, q= N(r+m+3)

N−2,


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J. ZHOU AND C. MU

and C(�) is some constant depending only on �. Using the Sobolev’s imbedding theorem, we get

‖u(t)‖r+m+3q �C‖∇(u

r+m+12 +1(t))‖2

2.

Then, we obtain ∫�

|u(x, t)|r+�+1 dx � C‖u(t)‖�r+2‖∇(u

r+m+12 +1(t))‖

2 r+m+32

� C‖u(t)‖(r+2)(m−�+2)

m+1r+2 ‖∇(u

r+m+12 +1(t))‖

2(�−1)m+1

2 .

Further, the Young’s inequality with parameter � implies

C2(r+1)‖u(t)‖r+�+1r+�+1 � C2C(r+1)‖u(t)‖

(r+2)(m−�+2)m+1

r+2 ‖∇(ur+m+1

2 +1(t))‖2(�−1)m+1

2

� C′1�−

�−1m−�+2 (r+1)

m+1m−�+2 ‖u(t)‖r+2

r+2 +C′2�‖∇(u

r+m+12 +1(t))‖2

2,

where

C′1 = m−�+2

m+1 (C2C)m+1

m−�+2

and C′2 = (�−1) / (m+1). Take � small enough such that C′

2��C1 / 2 and denoting C3 =C′1�−(�−1)/ (m−�+2), then the above inequality

reads as

C2(r+1)‖u(t)‖r+�+1r+�+1�C3(r+1)

m+1m−�+2 ‖u(t)‖r+2

r+2 + C1

2‖∇(u

r+m+12 +1(t))‖2

2. (50)

Combining (50) with (49) gives

d

dt‖u(t)‖r+2

r+2 + C1

2‖∇(u

r+m+12 +1(t))‖2

2�C3(r+1)1+�‖u(t)‖r+2r+2, (51)

where �= (�−1) / m�0.If N=2, then we use Holder’s inequality to obatin

‖u(t)‖r+�+1r+�+1�‖u(t)‖�

r+2‖u(t)‖ r+m+3,

and use Sobolev’s inequality to estimate ‖u(t)‖ r+m+3. Similarly, we get (50) and (51). The proof of Lemma 3.2 is completed. �

Proof of Theorem 1.3To prove Theorem 1.3, it is sufficient to prove the following argument: If u(x0, 0)>0 for some x0 ∈�, then u(x0, t)>0 for any t>0.

Let d =dist(x0,��) and �=N(m−1)+2. From u(x0, 0)>0, it follows that

u(x, 0)�� for x ∈B�(x0)∈�,

for some small constant �, �∈ (0, 1).By Lemmas 3.1 and 3.2, we can see that there exists a constant c�1 such that the periodic solution u(x, t) satisfies

ut −�um +cu�0. (52)

Denote

�(x, t)=e−t��N�− N� (t)

⎡⎣1−

(|x−x0|�

1� (t)

)2⎤⎦

1m−1

+, (53)

where

�(t)=��+ 2m�

(m−1)c�m−1�N(m−1)(1−e−c(m−1)t),

then �(x, t) satisfies

�t −��m +c�=0 in Bd(x0)×(0,+∞).

It is easy to see that u(x, 0)��(x, 0) for x ∈Bd(x0) and �|�Bd(x0)×[0,+∞) =0 for r small enough. Therefore, by comparison principle, weget u(x, t)��(x, t) for x ∈Bd(x0), t ∈ (0,+∞) and thus u(x0, t)��(x0, t)>0 for any t>0. The proof of Theorem 1.3 is completed. �

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J. ZHOU AND C. MU

Proof of Theorem 1.4Let �1 and �1(x) be the first eigenvalue and its corresponding eigenvalue function of the following eigenvalue problem:

−��(x) = ��(x) in �′,

� = 0 on ��′,

where �′ ⊃⊃� is a bounded domain with smooth boundary. It is easy to see �1>0 and �1(x)>0 in � and we can make max� �1(x)=1(see [20]).

Let K be a positive constant with

�1Km−� min�

�1(x)�a+,

and write u=K�1/m1 . Then u is a super-solution of (8)–(9). Define the poincare map P : C(�)→C(�) by

P(u0(x))=u(x, T), un(x)=Pn(u(x)), n=1, 2,. . . .

Then un+1(x)�un(x)�u(x) by the comparison principle. Similar to the proof of Theorem 4.1 in [3], we can see that U(x, t), the solutionof (8)–(10) with initial value u∗(x)= limn→∞ un(x), is a periodic solution of (8) and (9). Moreover, by Lemma 3.2, any nonnegativesolution u(x, t) of (8) and (9) satisfies u(x, t)�C0 for (x, t)∈Q. Therefore, if we take K>C0 / min� �1/m

1 , then by the comparison principleu∗(x)�u(x, 0) and thus U(x, t)�u(x, t), which implies that U(x, t) is a maximal periodic solution of (8) and (9).

Now, for any giving value u0(x)�0, taking

K�max

⎧⎨⎩‖u0‖L∞(�)

min� �1m1

,

(a+

�1 min� �1

) 1m−�

⎫⎬⎭ ,

and taking v(x, t) to be the solution of (8)–(10) with initial value v(x, 0)=K�1/m1 , we have

u(x, t+pT)�v(x, t+pT),

for any (x, t)∈Q, p=0, 1, 2,. . ..As limp→∞ v(x, t+pT)=v∗(x, t), it follows that for any �>0, there exists p0 such that

u(x, t+pT)�v∗(x, t)+��U(x, t)+�,

for all p�p0 and (x, t)∈Q. The proof of Theorem 1.4 is complete. �

Acknowledgements

The first author is supported by Fundamental Research Funds for the Central Universities (XDJK2009C069), the second author issupported by NNSF of China (10771226) and in part by Natural Science Foundation Project of CQ CSTC (2007BB0124).

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