wenting cong jian-guo liu

A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL

Wenting Cong∗

School of Mathematics

Jilin UniversityChangchun 130012, China

and

Department of Physics and Department of MathematicsDuke University

Durham, NC 27708, USA

Jian-Guo Liu

Department of Physics and Department of Mathematics

Duke University

Durham, NC 27708, USA

(Communicated by Tao Luo)

Abstract. This paper investigates the existence of a uniform in time L∞

bounded weak solution for the p-Laplacian Keller-Segel system with the su-percritical diffusion exponent 1 < p < 3d

d+1in the multi-dimensional space Rd

under the condition that the Ld(3−p)p norm of initial data is smaller than a

universal constant. We also prove the local existence of weak solutions and a

blow-up criterion for general L1 ∩ L∞ initial data.

1. Introduction. In this paper, we study the following p-Laplacian Keller-Segelmodel in d ≥ 3:

∂tu = ∇ ·(|∇u|p−2∇u

)−∇ · (u∇v) , x ∈ Rd, t > 0,

−∆v = u, x ∈ Rd, t > 0,u(x, 0) = u0(x), x ∈ Rd,

(1)

where p > 1. 1 2 is calledthe slow p-Laplacian diffusion. Especially, the p-Laplacian Keller-Segel model turnsto the original model when p = 2.

The Keller-Segel model was firstly presented in 1970 to describe the chemotaxisof cellular slime molds [13, 14]. The original model was considered in 2D, ∂tu = ∆u−∇ · (u∇v), x ∈ R2, t > 0,

−∆v = u, x ∈ R2, t > 0,u(x, 0) = u0(x), x ∈ R2.

(2)

2010 Mathematics Subject Classification. Primary: 35K65, 35K92, 92C17.Key words and phrases. Chemotaxis, fast diffusion, critical space, global existence, monotone

operator, non-Newtonian filtration.The first author is supported by NSFC grant 11271154.∗ Corresponding author: Wenting Cong.

687

http://dx.doi.org/10.3934/krm.2016012

688 WENTING CONG AND JIAN-GUO LIU

u(x, t) represents the cell density, and v(x, t) represents the concentration of thechemical substance which is given by the fundamental solution

v(x, t) = Φ(x) ∗ u(x, t),

where

Φ(x) =

− 12π log |x|, d = 2,

1d(d−2)α(d)

1|x|d−2 , d ≥ 3,

α(d) is the volume of the d-dimensional unit ball. In this model, cells are attractedby the chemical substance and also able to emit it.

One natural extension of the original Keller-Segel model is the degenerate Keller-Segel model in the multi-dimension with m > 1, ∂tu = ∆um −∇ · (u∇v), x ∈ Rd, t > 0,

−∆v = u, x ∈ Rd, t > 0,u(x, 0) = u0(x), x ∈ Rd,

(3)

which has been widely studied [2, 4, 7, 8, 15, 22, 23, 24, 25]. Another natural exten-sion is the degenerate p-Laplacian Keller-Segel model in the multi-dimension sincethe porous medium equation and the p-Laplacian equation are all called nonlineardiffusion equations. Work in these two models has frequent overlaps both in phe-nomena to be described, results to be proved and techniques to be used. The porousmedium equation and the p-Laplacian equation are different territories with someimportant traits in common. The evolution p-Laplacian equation is also called thenon-Newtonian filtration equation which describes the diffusion with the diffusiv-ity depending on the gradient of the unknown. The comprehensive and systematicstudy for these two equations can be found in Vazquez [27], DiBenedetto [10] andWu, Zhao, Yin and Li [28].

In the p-Laplacian Keller-Segel model, the exponent p plays an important role.When p = 3d

d+1 , if (u, v) is a solution of (1), constructing the following mass invariantscaling for u and a corresponding scaling for v

uλ(x, t) = λu(λ

1dx, λt

),

vλ(x, t) = λ1− 2d v(λ

1dx, λt

),

(4)

then (uλ, vλ) is also a solution for (1) and hence p = 3dd+1 is referred to the critical

exponent. For the general exponent p, (uλ, vλ) satisfies the following equationut = λ(1+ 1

d )p−3∇ ·(|∇u|p−2∇u

)−∇ · (u∇v) ,

−∆v = u.(5)

If(1 + 1

d

)p−3 < 0 which is called the supercritical case, the aggregation dominates

the diffusion for high density(large λ) which leads to the finite-time blow-up, andthe diffusion dominates the aggregation for low density(small λ) which leads to theinfinite-time spreading. If

(1 + 1

d

)p − 3 > 0 which is called the subcritical case,

the aggregation dominates the diffusion for low density(small λ) which preventsspreading, while the diffusion dominates the aggregation for high density(large λ)which prevents blow-up. At the end of Section 5, we have the theorem of theexistence of a global weak solution for (1) in the subcritical case.

In the supercritical case, there is a Lq space, where q = d(3−p)p . The q is crucial

when studying the existence and blow-up results of the p-Laplacian Keller-Segel

A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 689

model and almost all the results are related to the initial data ‖u0(·)‖Lq(Rd). Also

considering model (1), if (u, v) is a solution, then

uλ(x, t) = λu(λ

3−pp x, λt

),

vλ(x, t) = λ3− 6p v(λ

3−pp x, λt

),

is also a solution of (1). Furthermore, the scaling of u(x, t) preserves the Lq norm‖uλ‖Lq = ‖u‖Lq . For 1 < p < 3d

d+1 , if ‖u0‖Lq(Rd) < Cd,p, where Cd,p is a universalconstant depending on d and p, then we will show that there exists a global weaksolution. Since the initial condition u0 ∈ L1

+ ∩ L∞(Rd), we can prove that weaksolutions are bounded uniformly in time by using the bootstrap iterative method(See[3], [19]). With no restriction of the Lq norm on initial data, we prove the localexistence of a weak solution. This result also provides a natural blow-up criterion for1 3dd+1 , there exists a global weak solution of (1) without

any restriction of the size of initial data.In the process of proving the existence of a global weak solution of (1), we

combine the Aubin-Lions Lemma with the monotone operator theory. The theoryof monotone operators was proposed by Minty [20, 21]. Then the theory was used toobtain the existence results for quasi-linear elliptic and parabolic partial differentialequations by Browder [5, 6], Leray and Lions [17], Hartman and Stampacchia [12],DiBenedetto and Herrero [11].

The paper is organized as follows. In Section 2, we define a weak solution, intro-duce a Sobolev inequality with the best constant and some lemmas. In Section 3, wegive the a priori estimates of our weak solution. In Section 4, we prove the theoremabout uniformly in time L∞ bound of weak solutions using a bootstrap iterativemethod. In Section 5, we construct a regularized problem to prove the existence ofa global weak solution. Finally, in Section 6, we discuss the local existence of weaksolutions and a blow-up criterion.

2. Preliminaries. The generic constant will be denoted by C, even if it is differentfrom line to line. At the beginning, we define a weak solution of (1) in this paper.

Definition 2.1. (Weak solution) Let u0 ∈ L1+ ∩ L∞(Rd) be initial data and T ∈

(0,∞). v(x, t) is given by the fundamental solution

v(x, t) =1

d(d− 2)α(d)

∫Rd

u(y, t)

|x− y|d−2dy.

Then (u, v) is a weak solution to (1) if u satisfies

(i) Regularity:

u ∈ L∞(0, T ;L1

+(Rd))∩ Lp

(0, T ;W 1,p(Rd)

)∩ L2

(0, T ;L

2dd+2 (Rd)

),

∂tu ∈ Lpp−1

(0, T ;W−2, p

p−1 (Rd)).

(ii) ∀ ψ(x, t) ∈ C∞c([0, T )× Rd

),∫ T

0

∫Rdu(x, t)ψt(x, t) dxdt =

∫ T

0

∫Rd|∇u(x, t)|p−2∇u(x, t) · ∇ψ(x, t) dxdt

− 1

2dα(d)

∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)

|x− y|2u(x, t)u(y, t)

|x− y|d−2dxdydt

−∫Rdu0(x)ψ(x, 0)dx. (6)

The following lemma is a Sobolev inequality with the best constant which wasidentified by Talenti [26] and Aubin [1].

Lemma 2.2. (Sobolev inequality) Let 1 < p < d. If the function u ∈ W 1,p(Rd),then

‖u‖Lp∗(Rd) ≤ K(d, p)‖∇u‖Lp(Rd), (7)

where p∗ = dpd−p and

K(d, p) = π−12 d−

1p

(p− 1

d− p

)1− 1p

[Γ(1 + d

2 )Γ(d)

Γ(dp )Γ(1 + d− dp )

] 1d

. (8)

Next two lemmas are proposed by Bian and Liu [2].

Lemma 2.3. Assume y(t) ≥ 0 is a C1 function for t > 0 satisfying y′(t) ≤ γ −βy(t)a for γ ≥ 0, β > 0 and a > 0. Then

(i) for a > 1, y(t) has the following hyper-contractive property:

y(t) ≤(γ

β

) 1a

+

[1

β(a− 1)t

] 1a−1

, t > 0,

(ii) for a = 1, y(t) decays as

y(t) ≤ γ

β+ y(0)e−βt,

(iii) for a < 1, γ = 0, y(t) has the finite time extinction, which means that there

exists a Text satisfying 0 < Text ≤ y1−a(0)β(1−a) such that y(t) = 0 for all t > Text.

Lemma 2.4. Assume f(t) ≥ 0 is a non-increasing function for t > 0, y(t) ≥ 0 isa C1 function for t > 0 and satisfies y′(t) ≤ f(t)−βy(t)

afor some constants a > 1

and β > 0, then for any t0 > 0 one has

y(t) ≤(f(t0)

β

) 1a

+

(β(a− 1)(t− t0)

)− 1a−1

, for t > t0.

With the additional condition that y(0) is bounded, we have Lemma 2.5 whichcan be proved by contradiction arguments.

Lemma 2.5. Assume y(t) ≥ 0 is a C1 function for t > 0 satisfying y′(t) ≤ γ −βy(t)a for γ > 0, β > 0 and a > 0. If y(0) is bounded, then

y(t) ≤ max

(y(0),

(γ

β

) 1a

), t > 0.

3. A priori estimates of weak solutions. In this section, we prove Theorem 3.1which is concerning a priori estimates of weak solutions of (1).

Theorem 3.1. Let d ≥ 3, 1 0, where Cp,d =[

qpp

Kp(d,p)(q−2+p)p

] 13−p

is a universal constant, let (u,v) be a non-negative weak solution of (1). Then

u ∈ L∞(R+;Lq(Rd)

), u ∈ Lq+1

(R+;Lq+1(Rd)

)and ∇u

q−2+pp ∈ Lp

(R+;Lp(Rd)

).

Furthermore, following a priori estimates hold true:

(i) For 1 < p < 2dd+1 , ‖u(·, t)‖Lq(Rd) has the finite time extinction. The extinct

time Text satisfies

0 < Text ≤ T0,

where T0 depends on d, p,A(d, p), ‖u0‖L1(Rd) and ‖u0‖Lq(Rd).

(ii) For p = 2dd+1 , ‖u(·, t)‖Lq(Rd) decays exponentially in time

‖u(·, t)‖Lq(Rd) ≤ ‖u0‖Lqe−Ct,

where C is a constant depending on d, p,A(d, p), ‖u0‖L1(Rd) and ‖u0‖Lq(Rd).

(iii) For 2dd+1 < p < 3d

d+1 , ‖u(·, t)‖Lq(Rd) decays in time

‖u(·, t)‖Lq(Rd) ≤‖u0‖Lq

(1 + Ct)

q−1

q(p−2+pd ),

where C depends on d, p,A(d, p), ‖u0‖L1(Rd) and ‖u0‖Lq(Rd).And for any 1 ≤ h ≤ q, ‖u(·, t)‖Lh(Rd) decays in time

‖u(·, t)‖Lh(Rd) ≤‖u0‖

q(h−1)h(q−1)

Lq ‖u0‖q−hh(q−1)

L1

(1 + Ct)

h−1

h(p−2+pd )

.

For any q < h <∞, u(x, t) has hyper-contractive property

‖u(·, t)‖hLh(Rd) ≤ C

(t− (q+ε−1)(h−q+1)(h−1)

ε(p−2+pd )(h+p−3+

pd ) + t

− h−1

p−2+pd

),

where C is a constant depending on h, d, p, A(d, p) and ‖u0‖L1 , ε > 0 satisfies(q+ε)pp

Kp(d,p)(q+ε−2+p)p − ‖u0‖3−pLq ≥A(d,p)

2 .

Proof. Step 1. (The Lq estimate for 1 < p < 3dd+1 ). Multiplying the first equation

in problem (1) by quq−1 and integrating it over Rd, we obtain

d

dt‖u(·, t)‖q

Lq(Rd)=

∫Rd∇ ·(|∇u|p−2∇u

)quq−1 dx−

∫Rd∇ · (u∇v)quq−1 dx

= −q(q − 1)

∫Rduq−2|∇u|p dx+ (q − 1)

∫Rd∇uq · ∇v dx

= − q(q − 1)pp

(q − 2 + p)p

∥∥∥∇u q−2+pp (t)

∥∥∥pLp(Rd)

+ (q − 1)‖u‖q+1Lq+1(Rd). (9)

Now we estimate the second term on the right hand side. Firstly, by using theinterpolation inequality, we obtain that

‖u‖q+1Lq+1(Rd) ≤ ‖u‖

d(q−2+p)pd+pq−2d

L(q−2+p)dd−p

‖u‖q(pd+pq+p−3d)pd+pq−2d

Lq

=∥∥∥u q−2+p

p

∥∥∥ dppd+pq−2d

Ldpd−p

‖u‖q(pd+pq+p−3d)pd+pq−2d

Lq

=∥∥∥u q−2+p

p

∥∥∥pL

dpd−p‖u‖3−pLq , (10)

where the last equality holds since dpd+pq−2d = 1 and q(pd+pq+p−3d)

pd+pq−2d = 3 − p from

q = d(3−p)p . Then using the Sobolev inequality (7), (10) turns to

‖u‖q+1Lq+1(Rd) ≤ K

p(d, p)∥∥∥∇u q−2+p

p

∥∥∥pLp‖u‖3−pLq , (11)

where K(d, p) is given by (8). Substituting (11) into (9), we have

d

dt‖u‖qLq + (q − 1)

(qpp

(q − 2 + p)p−Kp(d, p)‖u‖3−pLq

)∥∥∥∇u q−2+pp

∥∥∥pLp≤ 0. (12)

Since ‖u0(·)‖Lq(Rd) <[

qpp

Kp(d,p)(q−2+p)p

] 13−p

=: Cp,d, following two estimates hold

true

‖u(·, t)‖Lq(Rd) < ‖u0(·)‖Lq(Rd) < Cp,d, (13)

(q − 1)Kp(d, p)(C3−pp,d − ‖u0‖3−pLq

)∫ ∞0

∥∥∥∇u q−2+pp

∥∥∥pLp

ds ≤ Cp,d.

Combining (11) with two estimates above, we obtain

u(x, t) ∈ L∞(R+;Lq(Rd)

), (14)

u(x, t) ∈ Lq+1(R+;Lq+1(Rd)

), (15)

∇uq−2+pp (x, t) ∈ Lp

(R+;Lp(Rd)

). (16)

Step 2. (The Lq decay estimate). By using the interpolation inequality and (11),we have

‖u(·, t)‖qLq(Rd) ≤ ‖u‖(q+1)(q−1)

q

Lq+1 ‖u‖1q

L1

≤[Kp(d, p)

∥∥∥∇u q−2+pp

∥∥∥pLp‖u‖3−pLq

] q−1q

‖u‖1q

L1 , (17)

i.e. ∥∥∥∇u q−2+pp

∥∥∥pLp≥

‖u‖q2

q−1−3+p

Lq

Kp(d, p)‖u0‖1q−1

L1

=

(‖u‖qLq

)1+p−2+

pd

q−1

Kp(d, p)‖u0‖1q−1

L1

, (18)

since ‖u(·, t)‖L1 ≤ ‖u0‖L1 . Substituting (18) into (12) yields that

d

dt‖u(·, t)‖qLq +

(q − 1)A(d, p)

‖u0‖1q−1

L1

(‖u‖qLq

)1+p−2+

pd

q−1 ≤ 0, (19)

where we denote A(d, p) := C3−pp,d − ‖u0‖3−pLq .

Next we discuss the inequality (19) in three different situations.

(a) If 1+p−2+ p

d

q−1 > 1, i.e. 2dd+1 < p < 3d

d+1 , we can prove that ‖u(·, t)‖Lq(Rd) decaysin time

‖u(·, t)‖Lq(Rd) ≤‖u0‖Lq

(1 + Ct)

q−1

q(p−2+pd ), (20)

where C =A(d,p)(p−2+ p

d )(‖u0‖qLq )p−2+

pd

q−1

‖u0‖1q−1

L1

.

(b) If 1 +p−2+ p

d

q−1 = 1, i.e. p = 2dd+1 , ‖u(·, t)‖Lq(Rd) decays exponentially in time

‖u(·, t)‖Lq(Rd) ≤ ‖u0‖Lq(Rd)e−Ct,

where C = (q−1)A(d,p)

q‖u0‖1/(q−1)

L1(Rd)

.

(c) If 0 < 1 +p−2+ p

d

q−1 < 1, i.e. 1 < p < 2dd+1 , ‖u(·, t)‖Lq(Rd) has the finite

time extinction. The extinct time Text satisfies 0 < Text ≤ T0, where T0 =

‖u0‖−q(p−2+

pd )

q−1

Lq(Rd)‖u0‖1/(q−1)

L1(Rd)

−A(d,p)(p−2+ pd )

.

Step 3. (The Lh decay estimate for any 1 ≤ h ≤ q when 2dd+1 < p < 3d

d+1 ). Using

the interpolation inequality and (20), we have

‖u(·, t)‖Lh(Rd) ≤ ‖u(·, t)‖q(h−1)h(q−1)

Lq(Rd)‖u(·, t)‖

q−hh(q−1)

L1(Rd)≤‖u0‖

q(h−1)h(q−1)

Lq ‖u0‖q−hh(q−1)

L1

(1 + Ct)

h−1

h(p−2+pd )

. (21)

Step 4. (The hyper-contractive property for any q < h <∞ when 2dd+1 0 such that

(q + ε)pp

Kp(d, p)(q + ε− 2 + p)p− ‖u0‖3−pLq ≥

A(d, p)

2. (22)

In the same way of obtaining (9)-(11), we obtain

d

dt‖u(·, t)‖rLr(Rd) = − r(r − 1)pp

(r − 2 + p)p

∥∥∥∇u r−2+pp (t)

∥∥∥pLp

+ (r − 1)‖u‖r+1Lr+1 , (23)

and

‖u‖r+1Lr+1(Rd) ≤ ‖u‖

d(r−2+p)(r+1−q)rd+pd+pq−qd−2d

L(r−2+p)dd−p

‖u‖q(pd+pr+p−3d)rd+pd+pq−qd−2d

Lq

=∥∥∥u r−2+p

p

∥∥∥ dp(r+1−q)rd+pd+pq−qd−2d

Ldpd−p

‖u‖q(pd+pr+p−3d)rd+pd+pq−qd−2d

Lq

=∥∥∥u r−2+p

p

∥∥∥pL

dpd−p‖u‖3−pLq

≤ Kp(d, p)∥∥∥∇u r−2+p

p

∥∥∥pLp‖u‖3−pLq , (24)

where the third equality holds since d(r+1−q)rd+pd+pq−qd−2d = 1 and q(pd+pr+p−3d)

rd+pd+pq−qd−2d =

3 − p, and the last inequality holds from the Sobolev inequality. Then combining

(22), (23) and (24) together, we have

d

dt‖u(·, t)‖rLr +

(r − 1)Kp(d, p)A(d, p)

2

∥∥∥∇u r−2+pp

∥∥∥pLp≤ 0. (25)

By using the interpolation inequality and (24), we have

‖u(·, t)‖rLr(Rd) ≤ ‖u‖(r+1)(r−1)

r

Lr+1 ‖u‖1r

L1

≤[Kp(d, p)

∥∥∥∇u r−2+pp

∥∥∥pLp‖u‖3−pLq

] r−1r

‖u‖1r

L1

≤[Kp(d, p)

∥∥∥∇u r−2+pp

∥∥∥pLp‖u‖

r(3−p)(q−1)q(r−1)

Lr ‖u0‖(3−p)(r−q)q(r−1)

L1

] r−1r

‖u0‖1r

L1 ,

i.e. ∥∥∥∇u r−2+pp

∥∥∥pLp≥

(‖u‖rLr )1+

p−2+pd

r−1

Kp(d, p)‖u0‖1r−1

(1+

p(r−q)d

)L1

, (26)

since ‖u‖L1(Rd) ≤ ‖u0‖L1(Rd). Substituting (26) into (25) yields that

d

dt‖u(·, t)‖rLr + β1

(‖u‖rLr

)1+p−2+

pd

r−1 ≤ 0, β1 :=(r − 1)A(d, p)

2‖u0‖1r−1

(1+

p(r−q)d

)L1

. (27)

Solving this inequality by using Lemma 2.3, we have

‖u(·, t)‖rLr ≤ C(r)t− r−1

p−2+pd . (28)

Hyper-contractive estimates of Lh norm for h ≥ r. For h ≥ r > q, usingthe interpolation inequality, Sobolev inequality and Young’s inequality together, weobtain

‖u‖h+1Lh+1(Rd) ≤ ‖u‖

d(h−2+p)(h+1−r)hd+pd+pr−rd−2d

L(h−2+p)dd−p

‖u‖r(pd+ph+p−3d)hd+pd+pr−rd−2d

Lr

=∥∥∥uh−2+p

p

∥∥∥ dp(h+1−r)hd+pd+pr−rd−2d

Ldpd−p

‖u‖r(pd+ph+p−3d)hd+pd+pr−rd−2d

Lr

≤ Kdp(h+1−r)

hd+pd+pr−rd−2d (d, p)∥∥∥∇uh−2+p

p

∥∥∥ dp(h+1−r)hd+pd+pr−rd−2d

Lp‖u‖

r(pd+ph+p−3d)hd+pd+pr−rd−2d

Lr

≤ hpp

2(h− 2 + p)p

∥∥∥∇uh−2+pp

∥∥∥pLp

+ C(h, r)(‖u‖rLr

)1+h−r+1r−q , (29)

wheredp(h+ 1− r)

hd+ pd+ pr − rd− 2d=

pd(h+ 1− r)d(h+ 1− r) + p(r − q)

< p.

Considering (9) with h = q, we have

d

dt‖u(·, t)‖hLh(Rd) = − h(h− 1)pp

(h− 2 + p)p

∥∥∥∇uh−2+pp (t)

∥∥∥pLp(Rd)

+ (h− 1)‖u‖h+1Lh+1(Rd)

≤ − h(h− 1)pp

2(h− 2 + p)p

∥∥∥∇uh−2+pp (t)

∥∥∥pLp(Rd)

+ C(h, r)(‖u‖rLr

)1+h−r+1r−q . (30)

Substituting (28) into (30) yields that

d

dt‖u(·, t)‖hLh(Rd) ≤ −

h(h− 1)pp

2(h− 2 + p)p

∥∥∥∇uh−2+pp

∥∥∥pLp(Rd)

+C(h, r)t− (r−1)(h−q+1)

(p−2+pd )(r−q) . (31)

By the same way of obtaining (26), we obtain

∥∥∥∇uh−2+pp

∥∥∥pLp≥

(‖u‖hLh

)1+p−2+

pd

h−1

Kp(d, p)‖u0‖1

h−1 (1+p(h−q)d )

L1

. (32)

Then (31) turns to

d

dt‖u(·, t)‖hLh(Rd) ≤ −β2

(‖u‖hLh

)1+p−2+

pd

h−1

+ C(h, r)t− (r−1)(h−q+1)

(p−2+pd )(r−q) , (33)

where β2 = h(h−1)pp

2(h−2+p)pKp(d,p)‖u0‖1

h−1 (1+p(h−q)d )

L1

.

Using Lemma 2.4 with y(t) = ‖u(·, t)‖hLh(Rd), a = 1 +p−2+ p

d

h−1 > 1, β = β2 > 0

and f(t) = C(h, r)t− (r−1)(h−q+1)

(p−2+pd )(r−q) , for any t > t0 > 0, we have

‖u(·, t)‖hLh(Rd) ≤ C(h, r)t− (q+ε−1)(h−q+1)(h−1)

ε(p−2+pd )(h+p−3+

pd )

0 + C(h)(t− t0)− h−1

p−2+pd . (34)

By choosing t0 = t2 , we obtain that for any t > 0

‖u(·, t)‖hLh(Rd) ≤ C

(t− (q+ε−1)(h−q+1)(h−1)

ε(p−2+pd )(h+p−3+

pd ) + t

− h−1

p−2+pd

), (35)

where C is a constant depending on h, d, p, A(d, p) and ‖u0‖L1 , ε satisfies (22).

4. The uniformly in time L∞ estimate of weak solutions. In this section,we prove our theorem about uniformly in time L∞ boundness of weak solutions byusing a bootstrap iterative method. At the beginning of this section, we prove thefollowing proposition concerning Lh norm estimates of weak solutions for 1 < h <∞.

Proposition 1. Let d ≥ 3, 1 0, where Cp,d =[

qpp

Kp(d,p)(q−2+p)p

] 13−p

is a universal constant, let (u, v) be a non-negative weak solu-

tion of (1). Then u(x, t) satisfies for any t > 0

‖u(·, t)‖hLh(Rd) ≤ C‖u0‖q(h−1)q−1

Lq(Rd), 1 < h ≤ q, (36)

where C depends on h, q, and ‖u0‖L1 , and

‖u(·, t)‖hLh(Rd) ≤ Chu , q < h <∞, (37)

where Chu is a constant depending on d, p, h, ‖u0‖L1 and ‖u0‖Lh , ε > 0 satisfies(q+ε)pp


2 .

Actually, the proof of Proposition 1 is almost the same as the proof of Theorem3.1, except for the different initial condition u0 ∈ L1

+(Rd) ∩ Lh(Rd) for 1 < h <∞.

Proof. Using the same method in Step 1 of Theorem 3.1, we have for all t > 0

‖u(·, t)‖Lq(Rd) < ‖u0(·)‖Lq(Rd) < Cp,d.

Then we discuss in two different situations with respect to h.

For 1 < h ≤ q, using the interpolation inequality, we have

‖u(·, t)‖hLh(Rd) ≤ ‖u0(·)‖q−hq−1

L1(Rd)‖u0(·)‖

q(h−1)q−1

Lq(Rd). (38)

For q < h <∞, letting r := q+ ε ≤ h <∞, there exists ε > 0 small enough suchthat

(q + ε)pp

Kp(d, p)(q + ε− 2 + p)p− ‖u0‖3−pLq ≥

A(d, p)

2.

Then (25) also holds true, i.e.

d

dt‖u(·, t)‖rLr +

(r − 1)Kp(d, p)A(d, p)

2

∥∥∥∇u r−2+pp

∥∥∥pLp≤ 0.

Since q < r ≤ h, we have u0 ∈ Lr(Rd) and

‖u(·, t)‖Lr(Rd) ≤ ‖u0(·)‖Lr(Rd), (39)

for all t > 0. Combining (30), (32) and (39) together, we obtain

d

dt‖u(·, t)‖hLh(Rd) ≤ −β3

(‖u‖hLh

)1+p−2+

pd

h−1

+ C(h, r)(‖u0‖rLr

)1+h−r+1r−q , (40)

where β3 := h(h−1)pp

2(h−2+p)pKp(d,p)‖u0‖1

h−1 (1+p(h−q)d )

L1

> 0. Using Lemma 2.5 with y(t) =

‖u(·, t)‖hLh , a = 1 +p−2+ p

d

h−1 > 0, β = β3 > 0 and γ = C(h, r)(‖u0‖rLr

)1+h−r+1r−q > 0,

for any t > 0, we have

‖u(·, t)‖hLh ≤ max

‖u0‖hLh , C(h, r)

(‖u0‖rLr

) (h−q+1)(h−1)

ε(h+p−3+pd)

≤ max

‖u0‖hLh , C(h)

(‖u0‖hLh

) (h−q+1)(q+ε−1)

ε(h+p−3+pd)

=: Chu , (41)

where ε satisfies (q+ε)pp


2 .

Next, we prove the uniformly in time L∞ boundness of u(x, t) by using a boot-strap iterative technique [3, 19] with Proposition 1 and an additional initial condi-tion u0 ∈ L∞(Rd).


p . If u0 ∈ L1+(Rd) ∩ L∞(Rd)

and A(d, p) = C3−pp,d −‖u0‖3−pLq > 0, where Cp,d =

[qpp

Kp(d,p)(q−2+p)p

] 13−p

is a universal

constant, let (u, v) be a non-negative weak solution of (1). Then for any t > 0,

‖u(·, t)‖L∞(Rd) ≤ C(d, p,K0),

where K0 = max

1, ‖u0‖L1(Rd), ‖u0‖L∞(Rd)

.

Proof. We denote

hk = 3k +d(3− p)

p+ 1, for k ≥ 1.

Multiplying the first equation in (1) by hkuhk−1 and integrating, we have

d

dt‖u(·, t)‖hk

Lhk (Rd)= − hk(hk − 1)pp

(hk − 2 + p)p

∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ (hk − 1)‖u‖hk+1Lhk+1 . (42)

Step 1. (The Lhk estimate for 1 < p ≤ 2) Taking 0 < C1 ≤ hk(hk−1)pp

2(hk−2+p)p is a fixed

constant, then (42) turns to

d

dt‖u(·, t)‖hk

Lhk (Rd)≤ −2C1

∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ hk‖u‖hk+1Lhk+1 . (43)

Using the interpolation inequality and Sobolev inequality together, we obtain

‖u‖hk+1Lhk+1(Rd)

≤ ‖u‖(hk+1)θ

L(hk−2+p)d

d−p‖u‖(hk+1)(1−θ)

Lhk−1

=∥∥∥uhk−2+p

p

∥∥∥ p(hk+1)θ

hk−2+p

Ldpd−p

‖u‖(hk+1)(1−θ)Lhk−1

≤ Kp(hk+1)θ

hk−2+p (d, p)∥∥∥∇uhk−2+p

p

∥∥∥ p(hk+1)θ

hk−2+p

Lp‖u‖(hk+1)(1−θ)

Lhk−1, (44)

where

θ =d(hk − 2 + p)(hk − hk−1 + 1)

(hk + 1)((hk − 2 + p)d− hk−1(d− p)

) ,1− θ =

hk−1(hkp+ pd− 3d+ p)

(hk + 1)((hk − 2 + p)d− hk−1(d− p)

) .Since hk−1 = 3k−1 + d(3−p)

p + 1 > d(3−p)p , it is easy to see that p(hk+1)θ

hk−2+p < p. Then

using Young’s inequality and (44), we have

hk‖u‖hk+1Lhk+1(Rd)

≤ 1

aδa1

∥∥∥∇uhk−2+p

p

∥∥∥pLp

+1

bδ−b1 K

p(hk+1)θb

hk−2+p (d, p)hbk‖u‖(hk+1)(1−θ)bLhk−1

≤ C1

∥∥∥∇uhk−2+p

p

∥∥∥pLp

+ C2(hk)hbk‖u‖(hk+1)(1−θ)bLhk−1

, (45)

where

a =hk − 2 + p

(hk + 1)θ=d(hk − hk−1 + 1) + hk−1p+ pd− 3d

d(hk − hk−1 + 1)> 1,

b =hk − 2 + p

hk − 2 + p− (hk + 1)θ=d(hk − hk−1 + 1) + hk−1p+ pd− 3d

hk−1p+ pd− 3d> 1,

δ1 = (C1a)1a , C2(hk) =

1

b(C1a)−

baK

p(hk+1)θb

hk−2+p (d, p).

We can see that C2(hk) is uniformly bounded since a → 2d+p2d and b → 2d+p

p as

k →∞. Substituting (45) into (43) yields to

d

dt‖u(·, t)‖hk

Lhk (Rd)≤ −C1

∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ C2(hk)hbk

(‖u‖hk−1

Lhk−1

)γ1, (46)

where

γ1 =(hk + 1)(1− θ)b

hk−1=hkp+ pd− 3d+ p

hk−1p+ pd− 3d< 3.

Next, we estimate∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

. By using the interpolation inequality and

Sobolev inequality, we have

‖u‖hkLhk (Rd)

≤ ‖u‖hkβL

(hk−2+p)dd−p

‖u‖hk(1−β)

Lhk−1

=∥∥∥uhk−2+p

p

∥∥∥ phkβ

hk−2+p

Ldpd−p

‖u‖hk(1−β)

Lhk−1

≤ Kphkβ

hk−2+p (d, p)∥∥∥∇uhk−2+p

p

∥∥∥ phkβ

hk−2+p

Lp‖u‖hk(1−β)

Lhk−1, (47)

where

β =d(hk − 2 + p)(hk − hk−1)

hk((hk − 2 + p)d− hk−1(d− p)

) ,1− β =

hk−1(hkp+ pd− 2d)

hk((hk − 2 + p)d− hk−1(d− p)

) .Since it is easy to see that phkβ

hk−2+p < p, then using Young’s inequality, we have

‖u‖hkLhk (Rd)

≤ 1

a′δa′

2

∥∥∥∇uhk−2+p

p

∥∥∥pLp

+1

b′δ−b

′

2 Kphkβb

′hk−2+p (d, p)‖u‖hk(1−β)b′

Lhk−1

≤ C1

∥∥∥∇uhk−2+p

p

∥∥∥pLp

+ C3(hk)(‖u‖hk−1

Lhk−1

)γ2, (48)

where

a′ =hk − 2 + p

hkβ=d(hk − hk−1) + hk−1p+ pd− 2d

d(hk − hk−1)> 1,

b′ =hk − 2 + p

hk − 2 + p− hkβ=d(hk − hk−1) + hk−1p+ pd− 2d

hk−1p+ pd− 2d> 1,

δ2 = (C1a′)

1a′ , C3(hk) =

1

b′(C1a

′)−b′a′K

phkβb′

hk−2+p (d, p),

γ2 =hk(1− β)b′

hk−1=

hkp+ pd− 2d

hk−1p+ pd− 2d< 3.

We can also check that C3(hk) is uniformly bounded as k → ∞. Combining (46)and (48) together, we have

d

dt‖u‖hk

Lhk≤ −‖u‖hk

Lhk+ C2(hk)hbk

(‖u‖hk−1

Lhk−1

)γ1+ C3(hk)

(‖u‖hk−1

Lhk−1

)γ2. (49)

Since C2(hk) and C3(hk) are both uniformly bounded as k → ∞, we can choose aconstant C4 > 1 which is an upper bound of C2(hk) and C3(hk). Then by hk > 1and b > 1, we have for any t > 0,

d

dt‖u‖hk

Lhk≤ −‖u‖hk

Lhk+ C4h

bk

[(‖u‖hk−1

Lhk−1

)γ1+(‖u‖hk−1

Lhk−1

)γ2 ]. (50)

Step 2. (The Lhk estimate for 2 < p < 3dd+1 ) By changing form of (42), we have

d

dt

[(hk − 2 + p)p−2‖u‖hk

Lhk

]= − hk(hk − 1)pp

(hk − 2 + p)2

∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ (hk − 1)(hk − 2 + p)p−2‖u‖hk+1Lhk+1

≤ −2C ′1

∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ C5h2k‖u‖

hk+1Lhk+1 , (51)

where 0 < C ′1 ≤hk(hk−1)pp

2(hk−2+p)2 is a fixed constant and C5 is also a fixed constant

satisfying (hk − 1)(hk − 2 + p)p−2 ≤ C5h2k since hk > 1 and p < 3. Using Young’s

inequality and (44), we have

C5h2k‖u‖

hk+1Lhk+1 ≤

1

aδa3

∥∥∥∇uhk−2+p

p

∥∥∥pLp

+1

bδ−b3 (C5)bK

p(hk+1)θb

hk−2+p h2bk ‖u‖

(hk+1)(1−θ)bLhk−1

≤ C ′1∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ C ′2(hk)h2bk ‖u‖

(hk+1)(1−θ)bLhk−1

, (52)

where

a =hk − 2 + p

(hk + 1)θ=d(hk − hk−1 + 1) + hk−1p+ pd− 3d

d(hk − hk−1 + 1)> 1,

b =hk − 2 + p

hk − 2 + p− (hk + 1)θ=d(hk − hk−1 + 1) + hk−1p+ pd− 3d

hk−1p+ pd− 3d> 1,

δ3 = (C ′1a)1a , C ′2(hk) =

1

b(C ′1a)−

ba (C5)bK

p(hk+1)θb

hk−2+p (d, p).

We can see that C ′2(hk) is uniformly bounded as k → ∞. Since (hk−1 − 2 +

p)(p−2)(hk+1)(1−θ)b

hk−1 ≥ 1, (52) turns to

C5h2k‖u‖

hk+1Lhk+1 ≤ C ′1

∥∥∥∇uhk−2+p

p

∥∥∥pLp

+ C ′2(hk)h2bk

[(hk−1 − 2 + p)p−2‖u‖hk−1

Lhk−1

]γ1,

(53)where

γ1 =(hk + 1)(1− θ)b

hk−1=hkp+ pd− 3d+ p

hk−1p+ pd− 3d< 3.

Substituting (53) into (51), we obtain

d

dt

[(hk − 2 + p)p−2‖u‖hk

Lhk

]≤ −C ′1

∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ C ′2(hk)h2bk

[(hk−1 − 2 + p)p−2‖u‖hk−1

Lhk−1

]γ1. (54)

Next, by using Young’s inequality and (47), we have

(hk − 2 + p)p−2‖u‖hkLhk≤ 1

a′δa′

4

∥∥∥∇uhk−2+p

p

∥∥∥pLp

+1

b′δ−b

′

4 Kphkβb

′hk−2+p (d, p)(hk − 2 + p)(p−2)b′‖u‖hk(1−β)b′

Lhk−1

≤ C ′1∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ C ′3(hk)hb′

k ‖u‖hk(1−β)b′

Lhk−1, (55)

where

a′ =hk − 2 + p

hkβ=d(hk − hk−1) + hk−1p+ pd− 2d

d(hk − hk−1)> 1,

b′ =hk − 2 + p

hk − 2 + p− hkβ=d(hk − hk−1) + hk−1p+ pd− 2d

hk−1p+ pd− 2d> 1,

δ4 = (C ′1a′)

1a′ , C ′3(hk) =

C6

b′(C ′1a

′)−b′a′K

phkβb′

hk−2+p (d, p),

and C6 is a fixed constant such that (hk − 2 + p)(p−2)b′ ≤ C6hb′

k . We can see that

C ′3(hk) is uniformly bounded as k →∞. Also since (hk−1− 2 + p)(p−2)hk(1−β)b′

hk−1 ≥ 1,(55) turns to

(hk − 2 + p)p−2‖u‖hkLhk≤C ′1

∥∥∥∇uhk−2+p

p (t)∥∥∥pLp

+ C ′3(hk)hb′

k

[(hk−1 − 2 + p)p−2‖u‖hk−1

Lhk−1

]γ2, (56)

where

γ2 =hk(1− β)b′

hk−1=

hkp+ pd− 2d

hk−1p+ pd− 2d< 3.

Combining (54) and (56) together, we have

d

dt

[(hk − 2 + p)p−2‖u‖hk

Lhk

]≤ −(hk − 2 + p)p−2‖u‖hk

Lhk+ C ′2(hk)h2b

k

[(hk−1 − 2 + p)p−2‖u‖hk−1

Lhk−1

]γ1+ C ′3(hk)hb

′

k

[(hk−1 − 2 + p)p−2‖u‖hk−1

Lhk−1

]γ2. (57)

Since C ′2(hk) and C ′3(hk) are both uniformly bounded as k → ∞, we can choose aconstant C7 > 1 which is an upper bound of C ′2(hk) and C ′3(hk). Then by hk > 1and 2b > b′ > 1, we have for any t > 0,

d

dt

[(hk − 2 + p)p−2‖u‖hk

Lhk

]≤ −(hk − 2 + p)p−2‖u‖hk

Lhk

+ C7h2bk

[(hk−1 − 2 + p)p−2‖u‖hk−1

Lhk−1

]γ1+[(hk−1 − 2 + p)p−2‖u‖hk−1

Lhk−1

]γ2.

(58)

Step 3. (The uniform L∞ estimate for 1 1 is an upper bound of C4 and C7. Then (50) and (58) turn to

d

dtyk(t) ≤ −yk(t) + C8h

2bk

(yγ1k−1 + yγ2k−1

). (59)

Multiplying et to both sides of (59), we have

d

dt

(etyk(t)

)≤ C8hk

2b

(yγ1k−1 + yγ2k−1

)et ≤ 2C8hk

2b max

1, sup

t≥0y3k−1(t)

et. (60)

Solving this ODE, we obtain for t ≥ 0,

yk(t) ≤ e−tyk(0) + 2C8hk2b max

1, sup

t≥0y3k−1(t)

(1− e−t

)≤ 2C8hk

2b max

1, yk(0), sup

t≥0y3k−1(t)

. (61)

It is easy to see that

hk2b =

(3k +

d(3− p)p

+ 1

)2b

≤ C032bk

(d(3− p)

p+ 1

)2b

, (62)

where C0 is an appropriate positive constant. Combining (61) and (62) together,we can see

yk(t) ≤ C932bk

(d(3− p)

p+ 1

)2b

max

1, yk(0), sup

t≥0y3k−1(t)

,

for all 1 < p < 3dd+1 , where C9 = 2C0C8. Then after some iterative steps, we have

yk(t) ≤

(C9

(d(2−m)

2+ 1

)2b) 3k−1

2

32b( 3k+1

4 − k2−34 )

·max

1,

k−1∑i=0

supt≥0

y3i

k−i(0), supt≥0

y3k

0 (t)

. (63)

Denote K0 = max

1, ‖u0‖L1(Rd), ‖u0‖L∞(Rd)

, then

yk(0) =

‖u0‖hkLhk ≤ max

‖u0‖hkL1 , ‖u0‖hkL∞

, 1 < p ≤ 2,

(hk − 2 + p)p−2‖u0‖hkLhk , 2 < p < 3dd+1 ,

i.e.

yk(0) ≤Khk

0 , 1 < p ≤ 2,

(hk − 2 + p)p−2Khk0 , 2 < p < 3d

d+1 ,(64)

and for any 1 < p < 3dd+1 ,

limk→∞

y1hk

k (0) ≤ K0, (65)

since limk→∞

(hk − 2 + p)p−2hk = 1 for 2 < p < 3d

d+1 . Furthermore, we also have

max

1,

k−1∑i=0

supt≥0

y3i

k−i(0)

≤ kKqk

0 .

Taking the power 1hk

to both sides of (63) and letting k →∞, we obtain

‖u(·, t)‖L∞(Rd) ≤ C max

supt≥0

y0(t), K0

, (66)

where C = 33(2d+p)

2p C129

(d(2−m)

2 + 1)d+1

since b → 2d+pp as k → ∞. Recalling (37)

in Proposition 1, it shows that

y0(t) = ‖u(·, t)‖q+2Lq+2(Rd)

≤ Cq+2u . (67)

Then (66) turns to

‖u(·, t)‖L∞(Rd) ≤ C(d, p,K0).

5. Global existence of weak solutions. The following Lemma proved in [9,Lemma 2.1] is necessary for the existence of weak solutions of problem (1) in thesupercritical case.

Lemma 5.1. For any η, η′ ∈ Rd, there exists(|η|p−2

η − |η′|p−2η′)· (η − η′) ≥

C1 (|η|+ |η′|)p−2 |η − η′|2, p > 1,C2|η − η′|p, p ≥ 2,

where C1 and C2 are two positive constants only depending on p.

p . If u0 ∈ L1+(Rd) ∩ L∞(Rd)

and A(d, p) = C3−pp,d −‖u0‖3−pLq > 0, where Cp,d =

[qpp

Kp(d,p)(q−2+p)p

] 13−p

is a universal

constant. Then there exists a non-negative global weak solution (u, v) of (1), suchthat all a priori estimates in Theorem 3.1 and the uniform L∞ estimate in Theorem4.1 hold true.

Proof. We separate the proof of Theorem 5.2 into four steps. In Step 1, we constructthe regularized problem of (1) and show that all a priori estimates in Theorem 3.1and the uniform L∞ estimate in Theorem 4.1 hold true. Furthermore, we obtain theuniform estimate of ∇uε. In Step 2 and 3, by applying the Aubin-Lions Lemma, weprove that a non-negative weak solution of the regularized problem (68) convergesstrongly to a non-negative weak solution of (1) in a bounded domain. Finally, inStep 4, using the weak convergence and strong convergence estimates obtained inStep 1-3, we prove the existence of a global weak solution of (1) with monotoneoperators.

Step 1. (The regularized problem and a priori estimates) We consider the regu-larized problem of (1) for ε > 0,

∂tuε = ∇ ·(|∇uε|p−2∇uε

)+ ε∆uε −∇ · (uε∇vε) , x ∈ Rd, t > 0,

−∆vε = Jε ∗ uε, x ∈ Rd, t > 0,uε(x, 0) = u0ε(x), x ∈ Rd,

(68)

where d ≥ 3, 1 < p < 3dd+1 and Jε(x) = 1

εdJ(xε

), J(x) = 1

α(d)

(1 + |x|2

)− d+22

satisfying∫Rd Jε(x) dx = 1. A simple computation shows that vε can be expressed

by

vε(x, t) =1

d(d− 2)α(d)

∫Rd

uε(y, t)(|x− y|2 + ε2

) d−22

dy, (69)

where α(d) is the volume of the d-dimensional unit ball. The initial conditionu0ε(x) ∈ C∞(Rd) is a sequence of approximation for u0(x), which satisfies thatthere exists δ > 0 such that for all 0 < ε < δ,

u0ε(x) > 0, x ∈ Rd,

u0ε(x) ∈ Lr(Rd), for all r ≥ 1,

‖u0ε(·)‖L1(Rd) = ‖u0(·)‖L1(Rd),

‖u0ε(·)‖L∞(Rd) ≤ C.According to the classical theory for parabolic equations [16], the regularized prob-lem has a global smooth non-negative solution (uε, vε) with the regularity for allr ≥ 1,

uε ∈ L∞(R+;Lr(Rd)

)∩ Lr+1

(R+;Lr+1(Rd)

).

Then we want to show that all a priori estimates in Theorem 3.1 hold true for ourregularized problem. we take a cut-off function 0 ≤ ψ1(x) ≤ 1, satisfying

ψ1(x) =

1, if |x| ≤ 1,0, if |x| ≥ 2,

where ψ1(x) ∈ C∞c (Rd). Define ψR(x) := ψ1( xR ), then we know that limR→∞

ψR(x) =

1, |∇ψR(x)| ≤ C1

R and |∆ψR(x)| ≤ C2

R2 for x ∈ Rd, C1, C2 are constants. We can


also define ψ1p

R(x) := ψ1p

1 ( xR ) and choose a constant C3, such that

∣∣∣∣∇ψ 1p

R(x)

∣∣∣∣ ≤ C3

R .

Multiplying the first equation of (68) by quq−1ε ψR(x) and integrating over Rd, we

obtain

d

dt

∫RduqεψR(x) dx+

q(q − 1)pp

(q − 2 + p)p

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p ψR(x) dx

+4ε(q − 1)

q

∫Rd

∣∣∣∇u q2ε ∣∣∣2 ψR(x) dx

= −(q − 1)

∫Rduqε∆vεψR(x) dx− q

∫Rd|∇uε|p−2

uq−1ε ∇uε · ∇ψR(x) dx

+

∫Rduqε∇vε · ∇ψR(x) dx+ ε

∫Rduqε∆ψR(x) dx. (70)

Integrating (70) from 0 to t yields that∫Rduqε(x, t)ψR(x) dx+

q(q − 1)pp

(q − 2 + p)p

∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p ψR(x) dxds

+4ε(q − 1)

q

∫ t

0

∫Rd

∣∣∣∇u q2ε ∣∣∣2 ψR(x) dxds

=

∫Rduq0ε(x)ψR(x) dx− (q − 1)

∫ t

0

∫Rduqε∆vεψR(x) dxds

− q∫ t

0

∫Rd|∇uε|p−2

uq−1ε ∇uε · ∇ψR(x) dxds

+

∫ t

0

∫Rduqε∇vε · ∇ψR(x) dxds+ ε

∫ t

0

∫Rduqε∆ψR(x) dxds. (71)

For the second term on the right hand side of (71), by using Holder’s inequality, wehave

−(q − 1)

∫ t

0

∫Rduqε∆vεψR(x) dxds ≤ (q − 1)

∫ t

0

∫RduqεJε ∗ uε dxds

≤ (q − 1)

∫ t

0

‖uε‖qLq+1‖Jε ∗ uε‖Lq+1 ds

≤ (q − 1)

∫ t

0

‖uε‖q+1Lq+1 ds ≤ C(ε), (72)

since uε ∈ Lr+1(R+;Lr+1(Rd)

)for any r ≥ 1. Then we can use the dominated

convergence theorem as R →∞ for any small ε > 0 later. Next, we want to provethat last three terms on the RHS of (71) go to 0 as R→∞.

First, by using Holder’s inequality and Young’s inequality of convolution [18,pp.107], we obtain∫ t

0

∫Rduqε∇vε · ∇ψR(x) dxds ≤ C1

R

∫ t

0

∫Rduqε |∇vε| dxds

≤ C1

R

∫ t

0

‖uε‖qLd(q+1)d+1

‖∇vε‖Ld(q+1)d−q

ds

≤ C

R

∫ t

0

‖uε‖qLd(q+1)d+1

∥∥∥ x

|x|d∥∥∥L

dd−1w

‖uε‖Ld(q+1)d+1

ds


≤ C

R

∫ t

0

‖uε‖q+1

Ld(q+1)d+1

ds. (73)

Then using the interpolation inequality, (73) yields to∫ t

0

∫Rduqε∇vε · ∇ψR(x) dxds ≤ C

R

∫ t

0

‖uε‖q+1dq

L1 ‖uε‖(q+1)(dq−1)

dq

Lq+1 ds

≤C(‖u0‖L1

)R

(∫ t

0

‖uε‖q+1Lq+1 ds

) dq−1dq

≤ C(ε)

R, (74)

since uε ∈ Lq+1(R+;Lq+1(Rd)

)for q > 1.

Second, from uε ∈ L∞(R+;Lr(Rd)

)for all r ≥ 1, we have

ε

∫ t

0

∫Rduqε∆ψR(x) dxds ≤ C(t, ε)

R2. (75)

Third, by using Holder’s inequality, we have

− q∫ t

0

∫Rd|∇uε|p−2


≤ q∫ t

0

∫Rd|∇uε|p−1

uq−1ε |∇ψR(x)| dxds

= q

∫ t

0

∫Rd|∇uε|p−1

u(q−2)(p−1)

pε ψ

p−1p

R (x)uq−2+pp

ε ψ− p−1

p

R (x) |∇ψR(x)| dxds

= C

∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p−1

ψp−1p

R (x)uq−2+pp

ε

∣∣∣∣∇ψ 1p

R(x)

∣∣∣∣ dxds≤ C

R

(∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε


) p−1p(∫ t

0

‖uε‖q−2+pLq−2+p ds

) 1p

. (76)

Then we should prove∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p ψR(x) dxds is bounded in order to show

(76) goes to 0 as R→∞. Using Young’s inequality for (76), we obtain

− q∫ t

0

∫Rd|∇uε|p−2


≤ C

Rpp−1

∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p ψR(x) dxds+1

p

∫ t

0

‖uε‖q−2+pLq−2+p ds

≤ C

Rpp−1

∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p ψR(x) dxds+ C(t, ε), (77)

since q − 2 + p ≥ 1. Combining (71), (72), (74), (75) and (77) together, we have∫Rduqε(x, t)ψR(x) dx+

[q(q − 1)pp

(q − 2 + p)p− C

Rpp−1

] ∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε


+4ε(q − 1)

q

∫ t

0

∫Rd

∣∣∣∇u q2ε ∣∣∣2 ψR(x) dxds

≤∫Rduq0ε(x)ψR(x) dx+ C(t, ε) +

C(t, ε)

R+C(t, ε)

R2. (78)

Taking R large enough, we can see that q(q−1)pp

(q−2+p)p −C

Rpp−1

> 0. Then (78) shows

that ∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p ψR(x) dxds ≤ C(t, ε), (79)

since u0ε ∈ Lq(Rd) when R is large enough. Substituting (79) into (76), we have

− q∫ t

0

∫Rd|∇uε|p−2

uq−1ε ∇uε · ∇ψR(x) dxds ≤ C(t, ε)

R. (80)

Until now, we have proved that last three terms on the RHS of (71) go to 0 asR→∞. Using the dominated convergence theorem, when R→∞, (71) turns to∫

Rduqε(x, t) dx+

q(q − 1)pp

(q − 2 + p)p

∫ t

0

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p dxds

+4ε(q − 1)

q

∫ t

0

∫Rd

∣∣∣∇u q2ε ∣∣∣2 dxds

=

∫Rduq0ε(x)dx− (q − 1)

∫ t

0

∫Rduqε∆vε dxds, (81)

i.e., for any t > 0,

d

dt

∫Rduqε(x, t) dx+

q(q − 1)pp

(q − 2 + p)p

∫Rd

∣∣∣∣∇u q−2+pp

ε

∣∣∣∣p dx+4ε(q − 1)

q

∫Rd

∣∣∣∇u q2ε ∣∣∣2 dx

= (q − 1)

∫RduqεJε ∗ uε dx ≤ (q − 1)‖uε‖q+1

Lq+1

≤ (q − 1)Kp(d, p)

∥∥∥∥∇u q−2+pp

ε

∥∥∥∥pLp‖uε‖3−pLq , (82)

where last two inequalities can be obtained by the same method of (72) and (11).Then we have

d

dt‖uε‖qLq + (q − 1)

(qpp

(q − 2 + p)p−Kp(d, p)‖uε‖3−pLq

)∥∥∥∥∇u q−2+pp

ε

∥∥∥∥pLp≤ 0, (83)

which is same to (12), and all a priori estimates in Theorem 3.1 hold true for oursolution of the regularized problem. We also have following uniformly boundedestimates,

‖uε‖L∞(R+;L1+∩Lq(Rd)) ≤ C, (84)

‖uε‖Lq+1(R+;Lq+1(Rd)) ≤ C, (85)∥∥∥∥∇u r−2+pp

ε

∥∥∥∥Lp(R+;Lp(Rd))

≤ C, 1 < r ≤ q. (86)

Additionally, since u0 ∈ L∞(Rd), we let u0ε(x) also satisfy ‖u0ε‖L∞(Rd) ≤ C, where

C is a positive constant independent of ε. Then from the Theorem 4.1, we have theuniformly bounded estimate

‖uε(·, t)‖L∞(Rd) ≤ C. (87)

For q ≥ 2, i.e. 1 < p ≤ 3dd+2 , by taking r = 2 in (86), we have

‖∇uε‖Lp(R+;Lp(Rd)) ≤ C.

For 1 < r ≤ q < 2, i.e. 3dd+2 < p ≤ 3d

d+1 , by using (87), we obtain

C ≥∫R+

∫Rd

∣∣∣∣∇u r−2+pp

ε

∣∣∣∣p dxdt = C

∫R+

∫Rdur−2ε |∇uε|p dxdt

≥ C(‖uε‖L∞

) ∫R+

∫Rd|∇uε|p dxdt,

where C is a positive constant. From two estimates above, we have

‖∇uε‖Lp(R+;Lp(Rd)

) ≤ C, (88)

for all 1 < p < 3dd+1 .

Step 2. (The time regularity of uε) In this step, we want to estimate ∂tuε in anybounded domain in order to use the Aubin-Lions Lemma. For any test functionϕ(x) which satisfies ϕ ∈W 2,p(Ω) and ‖ϕ‖W 2,p(Ω) ≤ 1, we have

|〈∂tuε, ϕ〉| ≤ |〈|∇uε|p−2∇uε,∇ϕ〉|+ ε|〈uε,∆ϕ〉|+ |〈uε∇vε,∇ϕ〉|

≤ ‖∇uε‖p−1Lp(Ω) + ε‖uε‖

Lpp−1 (Ω)

+ ‖uε∇vε‖L

pp−1 (Ω)

. (89)

Then for any T > 0, since ‖uε(·, t)‖L∞(Rd) ≤ C and ‖∇uε‖Lp(R+;Lp(Rd)) ≤ C, using

Sobolev inequality, we obtain∫ T

0

‖∂tuε‖pp−1

W−2,

pp−1 (Ω)

dt ≤ C(∫ T

0

‖∇uε‖pLp(Ω) dt+ εpp−1

∫ T

0

‖uε‖pp−1

Lpp−1 (Ω)

dt

+

∫ T

0

‖uε∇vε‖pp−1

Lpp−1 (Ω)

dt

)≤ C(T )(1 + ε

pp−1 ) + C

∫ T

0

‖∇vε‖pp−1

Lpp−1 (Ω)

dt

≤ 2C(T ) + C

∫ T

0

‖∆vε‖pp−1

Ldp

dp−d+p (Ω)dt

≤ C(T ). (90)

Then we have ‖∂tuε‖L

pp−1

(0,T ;W

−2,pp−1 (Ω)

) ≤ C.

Step 3. (The application of the Aubin-Lions Lemma) It is easy to see that

‖uε‖Lp(0,T ;Lp(Ω)) ≤ C(Ω, T ),

from ‖uε(·, t)‖L∞(Rd) ≤ C, where Ω is any bounded domain. Then we obtain

that ‖uε‖Lp(

0,T ;W 1,p(Ω)) ≤ C. By the Sobolev Embedding Theorem, we have

W 1,p(Ω) →→ Lp(Ω) since p < d and p < dpd−p . Until now, we already have

‖uε‖Lp(

0,T ;W 1,p(Ω)) ≤ C,

‖∂tuε‖L

pp−1

(0,T ;W

−2,pp−1 (Ω)

) ≤ C,and W 1,p(Ω) →→ Lp(Ω) → W−2, p

p−1 (Ω). By the Aubin-Lions Lemma, thereexistes a subsequence of uε without relabeling such that

uε → u, in Lp(0, T ;Lp(Ω)

). (91)

Step 4. (The existence of a global weak solution) Next, we will show that (u, v) isa weak solution of the problem (1). The crucial idea in this step follows the proofof Theorem 2.2.1 in [28, p171]. The weak formulation for uε is that for any testfunction ψ(x) ∈ C∞c

([0, T )× Rd

)and any 0 < T <∞,∫ T

0

∫Rduε(x, t)ψt(x, t) dxdt+ ε

∫ T

0

∫Rduε(x, t)∆ψ(x, t) dxdt

=

∫ T

0

∫Rd|∇uε(x, t)|p−2∇uε(x, t) · ∇ψ(x, t) dxdt−

∫Rdu0ε(x)ψ(x, 0)dx

− 1

2dα(d)

∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)] · (x− y)

|x− y|2 + ε2uε(x, t)uε(y, t)(|x− y|2 + ε2

) d−22

dxdydt.

(92)

Next, we separate the proof of this step into three parts.

(i) Since uε → u in Lp (0, T ;Lp(Ω)), using Holder’s inequality, we have∫ T

0

∫Ω

|uε − u| dxdt ≤ C(Ω)

∫ T

0

‖uε − u‖Lp(Ω) dt

≤ C(Ω, T )‖uε − u‖Lp(

0,T ;Lp(Ω))

→ 0, as ε→ 0,

i.e.uε → u, in L1

(0, T ;L1(Ω)

). (93)

(ii) Next we have∣∣∣∣∣∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)

|x− y|2 + ε2uε(x, t)uε(y, t)(|x− y|2 + ε2

) d−22

dxdydt

−∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)

|x− y|2u(x, t)u(y, t)

|x− y|d−2dxdydt

∣∣∣∣∣≤

∣∣∣∣∣∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)uε(x, t)uε(y, t)

·

(1

|x− y|d− 1(|x− y|2 + ε2

) d2

)dxdydt

∣∣∣∣∣+

∣∣∣∣∣∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)] · (x− y)

|x− y|d

·(uε(x, t)uε(y, t)− u(x, t)u(y, t)

)dxdydt

∣∣∣∣∣=: I1 + I2. (94)

In order to estimate I1, we have

1

|x− y|d− 1(|x− y|2 + ε2

) d2

=dε2(

|x− y|2 + ε2) d+2

2

≤ dε

|x− y|d+1

ε

|x− y|≤ dε

|x− y|d+1, (95)

since ε is small enough. Then using Hardy-Littlewood-Sobolev inequality, I1satisfies

I1 ≤ dε

∣∣∣∣∣∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)uε(x, t)uε(y, t)

|x− y|d+1dxdydt

∣∣∣∣∣≤ Cε

∫ T

0

∫Ω

∫Ω

uε(x, t)uε(y, t)

|x− y|d−1dxdydt ≤ Cε

∫ T

0

‖uε‖2L

2dd+1 (Ω)

dt

≤ CεT → 0, as ε→ 0. (96)

For I2, also using Hardy-Littlewood-Sobolev inequality, we have

I2 ≤ C

∣∣∣∣∣∫ T

0

∫Ω

∫Ω

[uε(x, t)− u(x, t)

]uε(y, t)

|x− y|d−2dxdydt

∣∣∣∣∣+ C

∣∣∣∣∣∫ T

0

∫Ω

∫Ω

[uε(y, t)− u(y, t)

]u(x, t)

|x− y|d−2dxdydt

∣∣∣∣∣≤ C

∫ T

0

‖uε − u‖L

2dd+2‖uε‖

L2dd+2

dt+ C

∫ T

0

‖uε − u‖L

2dd+2‖u‖

L2dd+2

dt. (97)

For 2dd+2 ≤ p < 3d

d+1 , by using the interpolation inequality and Holder’s in-equality, we obtain∫ T

0

‖uε − u‖L

2dd+2 (Ω)

dt ≤∫ T

0

‖uε − u‖dp+2p−2d2d(p−1)

L1(Ω) ‖uε − u‖dp−2p

2d(p−1)

Lp(Ω) dt

≤ C(T )

(∫ T

0

‖uε − u‖pLp(Ω) dt

) d−22d(p−1)

→ 0, as ε→ 0. (98)

For 1 < p < 2dd+2 , since ‖uε(·, t)‖L∞(Rd) ≤ C and ‖u(·, t)‖L∞(Rd) ≤ C, we have∫ T

0

‖uε − u‖L

2dd+2 (Ω)

dt =

∫ T

0

(∫Ω

|uε − u|p|uε − u|2dd+2−pdx

) d+22d

dt

≤∫ T

0

(∫Ω

|uε − u|p(‖uε‖L∞(Ω) + ‖u‖L∞(Ω)

) 2dd+2−p

dx

) d+22d

dt

≤ C(T )

(∫ T

0

‖uε − u‖pLp(Ω) dt

) d+22d

→ 0, as ε→ 0. (99)

Combining (98) and (99) shows that, for all 1 < p < 3dd+1 ,

uε → u, in L1(

0, T ;L2dd+2 (Ω)

). (100)

Then we have I2 → 0, as ε→ 0. Until now, we obtain∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)

|x− y|2 + ε2uε(x, t)uε(y, t)(|x− y|2 + ε2

) d−22

dxdydt

→∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)

|x− y|2u(x, t)u(y, t)

|x− y|d−2dxdydt, (101)

as ε→ 0.


(iii) Finally, we will prove∫ T

0

∫Rd|∇uε(x, t)|p−2∇uε(x, t) · ∇ψ(x, t) dxdt

→∫ T

0

∫Rd|∇u(x, t)|p−2∇u(x, t) · ∇ψ(x, t) dxdt, (102)

as ε→ 0. Since ‖∇uε‖Lp(R+;Lp(Rd)

) ≤ C in (88), we have

∫R+

(∫Rd

∣∣∣∣|∇uε|p−2∇uε∣∣∣∣pp−1

dx

)dt =

∫R+

∫Rd|∇uε|p dxdt ≤ C.

There exists a χ such that

|∇uε|p−2∇uε χ, in Lpp−1

(R+;L

pp−1 (Rd)

). (103)

Letting ε→ 0 in (92), we have∫ T

0

∫Rduψt dxdt−

∫ T

0

∫Rdχ · ∇ψ dxdt+

∫ T

0

∫Rdu∇v · ∇ψ dxdt

+

∫Rdu0ψ(0)dx = 0. (104)

Then we will prove∫ T

0

∫Rd|∇u|p−2∇u · ∇ψ dxdt =

∫ T

0

∫Rdχ · ∇ψ dxdt, (105)

to finish the proof of the existence of a weak solution for (1). Choosingφ(x, t) ∈ C∞c

([0, T )× Rd

)with 0 ≤ φ ≤ 1, multiplying the first equation in

(68) by uεφ and integrating, we obtain

1

2

∫ T

0

∫Rdu2εφt dxdt−

∫ T

0

∫Rd

(|∇uε|p−2

+ ε)|∇uε|2φ dxdt

−∫ T

0

∫Rduε

(|∇uε|p−2

+ ε)∇uε · ∇φ dxdt+

1

2

∫ T

0

∫Rdu2ε∇vε · ∇φ dxdt

+1

2

∫ T

0

∫Rdu2εJε ∗ uε φ dxdt+

1

2

∫Rdu2

0ε(x)φ(x, 0)dx = 0. (106)

Since(|∇uε|p−2

+ ε)|∇uε|2 ≥ |∇uε|p, then (106) turns to

1

2

∫ T

0


∫ T

0

∫Rd|∇uε|pφ dxdt

−∫ T

0

∫Rduε

(|∇uε|p−2


1

2

∫ T

0


+1

2

∫ T

0


1

2

∫Rdu2

0ε(x)φ(x, 0)dx ≥ 0. (107)

For any ω ∈ Lp(0, T ;W 1,p(Rd)

)to be determined later, we can obtain the

following inequality by using Lemma 5.1∫ T

0

∫Rd

(|∇uε|p−2∇uε − |∇ω|p−2∇ω

)· ∇(uε − ω)φ(x, t) dxdt ≥ 0, (108)


i.e.∫ T

0

∫Rd|∇uε|pφ dxdt−

∫ T

0

∫Rd|∇ω|p−2∇ω · ∇uεφ dxdt

−∫ T

0

∫Rd|∇uε|p−2∇uε · ∇ωφ dxdt+

∫ T

0

∫Rd|∇ω|pφ dxdt ≥ 0. (109)

Combining (107) and (109) together, we have

1

2

∫ T

0


∫ T

0

∫Rd|∇ω|p−2∇ω · ∇(uε − ω)φ dxdt

−∫ T

0

∫Rd|∇uε|p−2∇uε · ∇ωφ dxdt

−∫ T

0

∫Rduε

(|∇uε|p−2


1

2

∫ T

0


+1

2

∫ T

0


1

2

∫Rdu2

0ε(x)φ(x, 0)dx ≥ 0. (110)

Next we estimate terms in (110) one by one. Since u0 ∈ L1 ∩ L∞(Rd),‖u0ε‖L1(Rd) = ‖u0‖L1(Rd), ‖u0ε‖L∞(Rd) ≤ C and uε → u in Lp (0, T ;Lp(Ω)),we have∣∣∣∣ ∫ T

0


∫ T

0

∫Rdu2φt dxdt

∣∣∣∣ ≤ C ∫ T

0

∫Ω

|uε − u|(uε + u) dxdt

≤ C(Ω, T )‖uε − u‖Lp(

0,T ;Lp(Ω)) → 0, as ε→ 0, (111)

and∣∣∣∣ ∫ T

0

∫Rdu2εJε ∗ uεφ dxdt−

∫ T

0

∫Rdu3φ dxdt

∣∣∣∣≤∫ T

0

∫Ω

(uε + u)|uε − u|Jε ∗ uε dxdt+

∫ T

0

∫Ω

u2(Jε ∗ uε − u) dxdt

≤ C∫ T

0

‖uε − u‖Lp‖Jε ∗ uε‖Lpp−1

dt+ C(Ω, T )

∫ T

0

‖Jε ∗ uε − u‖pLp dt

≤ C(Ω, T )‖uε − u‖Lp(

0,T ;Lp(Ω)) → 0, as ε→ 0. (112)

Since

∇uε ∇u, in Lp(R+;Lp(Rd)

),

|∇uε|p−2∇uε χ, in Lpp−1

(R+;L

pp−1 (Rd)

),

it is easy to see that∣∣∣∣ ∫ T

0

∫Rd|∇ω|p−2∇ω ·

[∇(uε − ω)−∇(u− ω)

]φ dxdt

∣∣∣∣=

∣∣∣∣ ∫ T

0

∫Rd|∇ω|p−2∇ω · ∇(uε − u)φ dxdt

∣∣∣∣→ 0, as ε→ 0, (113)

and∣∣∣∣ ∫ T

0

∫Rd|∇uε|p−2∇uε · ∇ωφ dxdt−

∫ T

0

∫Rdχ · ∇ωφ dxdt

∣∣∣∣→ 0, (114)


as ε→ 0, and∣∣∣∣ ∫ T

0

∫Rduε

(|∇uε|p−2

+ ε)∇uε · ∇φ dxdt−

∫ T

0

∫Rduχ · ∇φ dxdt

∣∣∣∣≤ Cε

∫ T

0

∫Ω

uε|∇uε| dxdt+

∣∣∣∣ ∫ T

0

∫Rduε

(|∇uε|p−2∇uε − χ

)· ∇φ dxdt

∣∣∣∣+

∣∣∣∣ ∫ T

0

∫Rd

(uε − u)χ · ∇φ dxdt∣∣∣∣→ 0, as ε→ 0, (115)

where (115) holds from ‖uε‖L∞(Rd) ≤ C. From (101) and ‖uε‖L∞(Rd) ≤ C,we can easily obtain∣∣∣∣ ∫ T

0

∫Rdu2ε∇vε · ∇φ dxdt−

∫ T

0

∫Rdu2∇v · ∇φ dxdt

∣∣∣∣→ 0, as ε→ 0. (116)

Then from (111)-(116), letting ε→ 0, (110) turns to

1

2

∫ T

0

∫Rdu2φt dxdt−

∫ T

0

∫Rd|∇ω|p−2∇ω · ∇(u− ω)φ dxdt

−∫ T

0

∫Rdχ · ∇ωφ dxdt−

∫ T

0

∫Rduχ · ∇φ dxdt+

1

2

∫ T

0

∫Rdu2∇v · ∇φ dxdt

+1

2

∫ T

0

∫Rdu3φ dxdt+

1

2

∫Rdu2

0φ(0)dx ≥ 0. (117)

Taking ψ = uφ in (104), we have

1

2

∫ T

0

∫Rdu2φt dxdt−

∫ T

0

∫Rdχ · ∇uφ dxdt−

∫ T

0

∫Rduχ · ∇φ dxdt

+1

2

∫ T

0

∫Rdu2∇v · ∇φ dxdt+

1

2

∫ T

0

∫Rdu3φ dxdt+

1

2

∫Rdu2

0φ(0)dx = 0.

(118)

Combining (117) and (118) together, we obtain∫ T

0

∫Rdχ · ∇(u− ω)φ dxdt−

∫ T

0

∫Rd|∇ω|p−2∇ω · ∇(u− ω)φ dxdt ≥ 0, (119)

i.e. ∫ T

0

∫Rd

(χ− |∇ω|p−2∇ω

)· ∇(u− ω)φ dxdt ≥ 0. (120)

Taking ω = u− λψ with λ ≥ 0 yields that∫ T

0

∫Rd

(χ− |∇(u− λψ)|p−2∇(u− λψ)

)· ∇ψ φ dxdt ≥ 0. (121)

Choosing φ such that supp ψ ⊂ supp φ and φ = 1 on supp ψ, letting λ→ 0,we obtain ∫ T

0

∫Rd

(χ− |∇u|p−2∇u

)· ∇ψ dxdt ≥ 0. (122)

Using the same method with λ ≤ 0, we have∫ T

0

∫Rd

(χ− |∇u|p−2∇u

)· ∇ψ dxdt ≤ 0. (123)

Then for any ψ ∈ C∞c([0, T )× Rd

), we have∫ T

0

∫Rd|∇u|p−2∇u · ∇ψ dxdt =

∫ T

0

∫Rdχ · ∇ψ dxdt, (124)

i.e. (102) holds.

Then combining (i)-(iii) and letting ε→ 0, for any 0 < T <∞, we have∫ T

0

∫Rdu(x, t)ψt(x, t) dxdt =

∫ T

0

∫Rd|∇u(x, t)|p−2∇u(x, t) · ∇ψ(x, t) dxdt

− 1

2dα(d)

∫ T

0

∫Rd

∫Rd

[∇ψ(x, t)−∇ψ(y, t)

]· (x− y)

|x− y|2u(x, t)u(y, t)

|x− y|d−2dxdydt

−∫Rdu0(x)ψ(x, 0)dx, (125)

which means that (u, v) is a global weak solution of (1).

For the subcritical case, we have the following theorem of the existence of a globalweak solution. Since the proof is almost identical as that for the supercritical case,we omit details.

Theorem 5.3. Let d ≥ 3 and p > 3dd+1 . If u0 ∈ L1

+(Rd)∩L∞(Rd), then there exists

a non-negative global weak solution (u, v) of (1).

6. Local existence of a weak solution and a blow-up criterion. In thissection, we prove that for u0 ∈ L1

+ ∩ L∞(Rd), a weak solution of (1) exists locallywithout any restriction for the size of initial data. Furthermore, we also prove thatif a weak solution blow up in finite time, then all Lh-norms of the weak solutionblow up at the same time for h > q.

Theorem 6.1. (Local existence of a weak solution) Let d ≥ 3, 1 0, such that (1) has a

weak solution in 0 < t < T .

Proof. Take any fixed r > q. Using the same way of obtaining (30) and takingh = r > q, we have

d

dt‖u(·, t)‖rLr(Rd) = − r(r − 1)pp

(r − 2 + p)p

∥∥∥∇u r−2+pp

∥∥∥pLp(Rd)

+ (r − 1)‖u‖r+1Lr+1(Rd)

≤ − r(r − 1)pp

2(r − 2 + p)p

∥∥∥∇u r−2+pp

∥∥∥pLp(Rd)

+ C(r)(‖u‖rLr(Rd)

)1+ 1r−q

,

i.e.d

dt‖u(·, t)‖rLr(Rd) ≤ C(r)

(‖u‖rLr(Rd)

)1+ 1r−q

. (126)

Solving the inequality (126) shows that

‖u(·, t)‖rLr(Rd) ≤

r−qC(r)

r−qC(r)

(‖u0‖rLr(Rd)

) 1q−r − t

r−q

. (127)

Denoting Tr := r−qC(r)

(‖u0‖rLr(Rd)

) 1q−r

, then for any fixed r, we choose 0 < T < Tr.

Next by the same way of proving Theorem 5.2, (1) has a local in time weak solutionin 0 < t < T .

Proposition 2. (Blow-up criterion) Under the same assumptions as Theorem 6.1and r = q + ε where ε is small enough, let T rmax be the largest Lr-norm existencetime of a weak solution, i.e.

‖u(·, t)‖Lr(Rd) <∞, for all 0 < t < T rmax,

lim supt→T rmax

‖u(·, t)‖Lr(Rd) =∞,

and Thmax be the largest Lh-norm existence time of a weak solution for h ≥ r > q.Then if Thmax <∞ for any h,

Thmax = T rmax, for all h ≥ r.

Proof. Since ‖u(·, t)‖L1(Rd) ≤ ‖u0‖L1(Rd), by interpolation inequality, we know that

for h ≥ r, Thmax ≤ T rmax. If Thmax < T rmax for any h ≥ r, then we will havecontradiction arguments. Thmax < T rmax implies

lim supt→Thmax

‖u(·, t)‖Lr(Rd) =: A <∞.

Then for h ≥ r > q, using (30), we have

d

dt‖u(·, t)‖hLh(Rd) ≤ C(h, r)

(‖u‖rLr(Rd)

)1+h−r+1r−q ≤ C(h, r,A), (128)

i.e.

‖u(·, t)‖Lh(Rd) ≤ C(h, r,A, ‖u0‖Lh(Rd), T

hmax

), for t ∈

(0, Thmax

),

which contradicts with

lim supt→Thmax

‖u(·, t)‖Lh(Rd) =∞.

Thus we have the conclusion that Thmax = T rmax for all h ≥ r > q, i.e. Lh-normsblow up at the same time.

Acknowledgments. The work of J.-G. Liu is partially supported by KI-Net NSFRNMS grant No. 1107291 and NSF grant DMS 1514826. Wenting Cong is partiallysupported by National Natural Science Foundation of China grant No. 11271154and wishes to express her gratitude to the China Scholarship Council for the schol-arship and Professor Wenjie Gao for his support and encouragement.

REFERENCES

[1] T. Aubin, Problemes isoperimetriques et espaces de Sobolev, J. Differential Geometry, 11(1976), 573–598.

[2] S. Bian and J.-G. Liu, Dynamic and steady states for multi-dimensional Keller-Segel modelwith diffusion exponent m > 0, Comm. Math. Phys., 323 (2013), 1017–1070.

[3] S. Bian, J.-G. Liu and C. Zou, Ultra-contractivity for Keller-Segel model with diffusion ex-ponent m > 1− 2/d, Kinet. Relat. Models, 7 (2014), 9–28.

[4] A. Blanchet, J. A. Carrillo and P. Laurencot, Critical mass for a Patlak-Keller-Segel modelwith degenerate diffusion in higher dimensions, Calc. Var. Partial Differential Equations, 35(2009), 133–168.

[5] F. E. Browder, Nonlinear elliptic boundary value problems, Bull. Amer. Math. Soc., 69

(1963), 862–874.[6] F. E. Browder, Non-linear equations of evolution, Ann. of Math., 80 (1964), 485–523.[7] L. Chen, J.-G. Liu and J. Wang, Multidimensional degenerate Keller-Segel system with critical

diffusion exponent 2n/(n + 2), SIAM J. Math. Anal., 44 (2012), 1077–1102.[8] L. Chen and J. Wang, Exact criterion for global existence and blow up to a degenerate

Keller-Segel system, Doc. Math., 19 (2014), 103–120.

http://www.ams.org/mathscinet-getitem?mr=MR0448404&return=pdf


http://dx.doi.org/10.1007/s00220-013-1777-z

http://dx.doi.org/10.1007/s00220-013-1777-z


http://dx.doi.org/10.3934/krm.2014.7.9

http://dx.doi.org/10.3934/krm.2014.7.9


http://dx.doi.org/10.1007/s00526-008-0200-7

http://dx.doi.org/10.1007/s00526-008-0200-7



http://dx.doi.org/10.2307/1970660


http://dx.doi.org/10.1137/110839102

http://dx.doi.org/10.1137/110839102

[9] L. Damascelli, Comparison theorems for some quasilinear degenerate elliptic operators and ap-plications to symmetry and monotonicity results, Ann. Inst. H. Poincare Anal. Non Lineaire,

15 (1998), 493–516.

[10] E. DiBenedetto, Degenerate Parabolic Equations, Springer-Verlag, New York, 1993.[11] E. DiBenedetto and M. A. Herrero, Non-negative solutions of the evolution p-Laplacian equa-

tion. Initial traces and Cauchy problem when 1 < p < 2, Arch. Rational Mech. Anal., 111(1990), 225–290.

[12] P. Hartman and G. Stampacchia, On some non-linear elliptic differential-functional equations,

Acta Math., 115 (1966), 271–310.[13] D. Horstmann, From 1970 until present: the Keller-Segel model in chemotaxis and its conse-

quences. I, Jahresber. Deutsch. Math.-Verein., 105 (2003), 103–165.

[14] E. F. Keller and L. A. Segel, Initiation of slime mold aggregation viewed as an instability, J.Theoret. Biol., 26 (1970), 399–415.

[15] I. Kim and Y. Yao, The Patlak-Keller-Segel model and its variations: Properties of solutions

via maximum principle, SIAM J. Math. Anal., 44 (2012), 568–602.[16] O. A. Ladyzenskaja, V. A. Solonnikov, N. N. Ural’ceva and S. Smith, Linear and Quasilinear

Equations of Parabolic Type, American Mathematical Society, Providence, R.I., 1968.

[17] J. Leray and J.-L. Lions, Quelques resultats de Visik sur les problemes elliptiques nonlineairespar les methodes de Minty-Browder, Bull. Soc. Math. France, 93 (1965), 97–107.

[18] E. H. Lieb and M. Loss, Analysis, Second edition, Graduate Studies in Mathematics, 14,American Mathematical Society, Providence, RI, 2001.

[19] J.-G. Liu and J. Wang, A note on L∞-bound and uniqueness to a degenerate Keller-Segel

model, Acta Appl. Math., 142 (2016), 173–188.[20] G. J. Minty, On a monotonicity method for the solution of nonlinear equations in Banach

spaces, Proc. Nat. Acad. Sci. U.S.A., 50 (1963), 1038–1041.

[21] G. J. Minty, Monotone (nonlinear) operators in Hilbert space, Duke Math. J., 29 (1962),341–346.

[22] B. Perthame, PDE models for chemotactic movements: Parabolic, hyperbolic and kinetic,

Appl. Math., 49 (2004), 539–564.[23] B. Perthame, Transport Equations in Biology, Birkhauser Verlag, Basel, 2007.

[24] Y. Sugiyama, Time global existence and asymptotic behavior of solutions to degenerate quasi-

linear parabolic systems of chemotaxis, Differential Integral Equations, 20 (2007), 133–180.[25] Y. Sugiyama and Y. Yahagi, Extinction, decay and blow-up for Keller–Segel systems of fast

diffusion type, J. Differential Equations, 250 (2011), 3047–3087.[26] G. Talenti, Best constant in Sobolev inequality, Ann. Mat. Pura Appl., 110 (1976), 353–372.

[27] J. L. Vazquez, The Porous Medium Equation: Mathematical Theory, The Clarendon Press,

Oxford University Press, Oxford, 2007.[28] Z. Wu, J. Zhao, J. Yin and H. Li, Nonlinear Diffusion Equations, World Scientific Publishing

Co., Inc., River Edge, NJ, 2001.

Received February 2016; revised April 2016.

E-mail address: [email protected]

E-mail address: [email protected]


http://dx.doi.org/10.1016/S0294-1449(98)80032-2

http://dx.doi.org/10.1016/S0294-1449(98)80032-2


http://dx.doi.org/10.1007/978-1-4612-0895-2


http://dx.doi.org/10.1007/BF00400111

http://dx.doi.org/10.1007/BF00400111


http://dx.doi.org/10.1007/BF02392210


http://dx.doi.org/10.1016/0022-5193(70)90092-5


http://dx.doi.org/10.1137/110823584

http://dx.doi.org/10.1137/110823584




http://dx.doi.org/10.1090/gsm/014


http://dx.doi.org/10.1007/s10440-015-0022-5

http://dx.doi.org/10.1007/s10440-015-0022-5


http://dx.doi.org/10.1073/pnas.50.6.1038

http://dx.doi.org/10.1073/pnas.50.6.1038


http://dx.doi.org/10.1215/S0012-7094-62-02933-2


http://dx.doi.org/10.1007/s10492-004-6431-9




http://dx.doi.org/10.1016/j.jde.2011.01.016

http://dx.doi.org/10.1016/j.jde.2011.01.016


http://dx.doi.org/10.1007/BF02418013



http://dx.doi.org/10.1142/9789812799791

mailto:[email protected]

mailto:[email protected]

wenting cong jian-guo liu

Documents