A DEGENERATE P-LAPLACIAN KELLER-SEGEL MODEL Wenting Cong Jian-Guo

Kinetic and Related Models doi:10.3934/krm.2016012 c American Institute of Mathematical Sciences Volume 9, Number 4, December 2016 pp. 687–714

A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL

Wenting Cong∗ School of Mathematics Jilin University Changchun 130012, China and Department of Physics and Department of Mathematics Duke 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- 3d d percritical diﬀusion exponent 1 < p < d+1 in the multi-dimensional space R d(3−p) under the condition that the L 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-Segel model in d ≥ 3:

 p−2 d  ∂tu = ∇ · |∇u| ∇u − ∇ · (u∇v) , x ∈ R , t > 0, d (1) −∆v = u, x ∈ R , t > 0,  d  u(x, 0) = u0(x), x ∈ R , where p > 1. 1 2 is called the slow p-Laplacian diﬀusion. Especially, the p-Laplacian Keller-Segel model turns to the original model when p = 2. The Keller-Segel model was ﬁrstly presented in 1970 to describe the chemotaxis of cellular slime molds [13, 14]. The original model was considered in 2D,  2  ∂tu = ∆u − ∇ · (u∇v), x ∈ R , t > 0, −∆v = u, x ∈ R2, t > 0, (2) 2  u(x, 0) = u0(x), x ∈ R .

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.

687 688 WENTING CONG AND JIAN-GUO LIU u(x, t) represents the cell density, and v(x, t) represents the concentration of the chemical substance which is given by the fundamental solution v(x, t) = Φ(x) ∗ u(x, t), where 1 − 2π log |x|, d = 2, Φ(x) = 1 1 , d ≥ 3, d(d−2)α(d) |x|d−2 α(d) is the volume of the d-dimensional unit ball. In this model, cells are attracted by 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,  m d  ∂tu = ∆u − ∇ · (u∇v), x ∈ R , t > 0, −∆v = u, x ∈ Rd, t > 0, (3) d  u(x, 0) = u0(x), x ∈ R , which has been widely studied [2,4,7,8, 15, 22, 23, 24, 25]. Another natural extension is the degenerate p-Laplacian Keller-Segel model in the multi-dimension since the porous medium equation and the p-Laplacian equation are all called nonlinear diffusion 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 porous medium equation and the p-Laplacian equation are different territories with some important traits in common. The evolution p-Laplacian equation is also called the non-Newtonian filtration equation which describes the diffusion with the diffusiv- ity depending on the gradient of the unknown. The comprehensive and systematic study for these two equations can be found in Vázquez[27], DiBenedetto [10] and Wu, Zhao, Yin and Li [28]. In the p-Laplacian Keller-Segel model, the exponent p plays an important role. 3d When p = d+1 , if (u, v) is a solution of (1), constructing the following mass invariant scaling for u and a corresponding scaling for v

1 uλ(x, t) = λu λ d x, λt , (4) 1− 2 1 vλ(x, t) = λ d v λ d x, λt ,

3d then (uλ, vλ) is also a solution for (1) and hence p = d+1 is referred to the critical exponent. For the general exponent p,(uλ, vλ) satisﬁes the following equation

( 1+ 1 p−3 p−2 u = λ( d ) ∇ · |∇u| ∇u − ∇ · (u∇v) , t (5) −∆v = u.

1 If 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, and the diffusion dominates the aggregation for low density(small λ) which leads to the 1 infinite-time spreading. If 1 + d p − 3 > 0 which is called the subcritical case, the aggregation dominates the diffusion for low density(small λ) which prevents spreading, 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 the existence of a global weak solution for (1) in the subcritical case. q d(3−p) In the supercritical case, there is a L space, where q = 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 ku0(·)k q d . Also L (R ) considering model (1), if (u, v) is a solution, then

3−p uλ(x, t) = λu λ p x, λt ,

3− 6 3−p vλ(x, t) = λ p v λ p x, λt , is also a solution of (1). Furthermore, the scaling of u(x, t) preserves the Lq norm 3d ku k q = kuk q . For 1 < p < , if ku k q d < C , where C is a universal λ L L d+1 0 L (R ) d,p d,p constant depending on d and p, then we will show that there exists a global weak 1 ∞ d solution. Since the initial condition u0 ∈ L+ ∩ L (R ), we can prove that weak solutions 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 local existence of a weak solution. This result also provides a natural blow-up criterion for 3d 1 d+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 theory of monotone operators was proposed by Minty [20, 21]. Then the theory was used to obtain the existence results for quasi-linear elliptic and parabolic partial diﬀerential equations 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, we give the a priori estimates of our weak solution. In Section 4, we prove the theorem about uniformly in time L∞ bound of weak solutions using a bootstrap iterative method. In Section 5, we construct a regularized problem to prove the existence of a global weak solution. Finally, in Section 6, we discuss the local existence of weak solutions and a blow-up criterion.

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

1 ∞ d Deﬁnition 2.1. (Weak solution) Let u0 ∈ L+ ∩ L (R ) be initial data and T ∈ (0, ∞). v(x, t) is given by the fundamental solution 1 Z u(y, t) v(x, t) = d−2 dy. d(d − 2)α(d) d R |x − y| Then (u, v) is a weak solution to (1) if u satisﬁes (i) Regularity:

2d ∞ 1 d p 1,p d 2 d+2 d u ∈ L 0,T ; L+(R ) ∩ L 0,T ; W (R ) ∩ L 0,T ; L (R ) ,

p −2, p d ∂tu ∈ L p−1 0,T ; W p−1 (R ) . ∞ d (ii) ∀ ψ(x, t) ∈ Cc [0,T ) × R , Z T Z Z T Z p−2 u(x, t)ψt(x, t) dxdt = |∇u(x, t)| ∇u(x, t) · ∇ψ(x, t) dxdt d d 0 R 0 R 690 WENTING CONG AND JIAN-GUO LIU

1 Z T Z Z ∇ψ(x, t) − ∇ψ(y, t) · (x − y) u(x, t)u(y, t) − 2 d−2 dxdydt 2dα(d) d d 0 R R |x − y| |x − y| Z − u0(x)ψ(x, 0)dx. (6) d R The following lemma is a Sobolev inequality with the best constant which was identiﬁed by Talenti [26] and Aubin [1].

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

kuk p∗ d ≤ K(d, p)k∇uk p d , (7) L (R ) L (R ) dp where p∗ = d−p and

1 1 " # d 1− p d − 1 − 1 p − 1 Γ(1 + 2 )Γ(d) K(d, p) = π 2 d p . (8) d − p d d Γ( p )Γ(1 + d − p ) 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 y0(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:

1 1 γ a 1 a−1 y(t) ≤ + , t > 0, β β(a − 1)t

(ii) for a = 1, y(t) decays as γ y(t) ≤ + y(0)e−βt, β

(iii) for a < 1, γ = 0, y(t) has the ﬁnite time extinction, which means that there y1−a(0) exists a Text satisfying 0 < Text ≤ 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 is a C1 function for t > 0 and satisﬁes y0(t) ≤ f(t) − βy(t)a for some constants a > 1 and β > 0, then for any t0 > 0 one has

1 − 1 f(t ) a a−1 y(t) ≤ 0 + β(a − 1)(t − t ) , for t > t . β 0 0

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

Lemma 2.5. Assume y(t) ≥ 0 is a C1 function for t > 0 satisfying y0(t) ≤ γ − βy(t)a for γ > 0, β > 0 and a > 0. If y(0) is bounded, then

1 ! γ a y(t) ≤ max y(0), , t > 0. β A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 691

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

3d d(3−p) Theorem 3.1. Let d ≥ 3, 1 0, where Cp,d = Kp(d,p)(q−2+p)p is a universal constant, let (u,v) be a non-negative weak solution of (1). Then ∞ q d q+1 q+1 d q−2+p p p d u ∈ L R+; L (R ) , u ∈ L R+; L (R ) and ∇u p ∈ L R+; L (R ) . Furthermore, following a priori estimates hold true: 2d (i) For 1 < p < , ku(·, t)k q d has the ﬁnite time extinction. The extinct d+1 L (R ) time Text satisﬁes

0 < Text ≤ T0,

where T depends on d, p, A(d, p), ku k 1 d and ku k q d . 0 0 L (R ) 0 L (R ) 2d (ii) For p = , ku(·, t)k q d decays exponentially in time d+1 L (R )

−Ct ku(·, t)k q d ≤ ku k q e , L (R ) 0 L

where C is a constant depending on d, p, A(d, p), ku k 1 d and ku k q d . 0 L (R ) 0 L (R ) 2d 3d (iii) For < p < , ku(·, t)k q d decays in time d+1 d+1 L (R )

ku0kLq ku(·, t)k q d ≤ , L (R ) q−1 p q p−2+ (1 + Ct) ( d )

where C depends on d, p, A(d, p), ku k 1 d and ku k q d . 0 L (R ) 0 L (R ) And for any 1 ≤ h ≤ q, ku(·, t)k h d decays in time L (R )

q(h−1) q−h h(q−1) h(q−1) ku0kLq ku0kL1 ku(·, t)k h d ≤ . L (R ) h−1 p h p−2+ (1 + Ct) ( d )

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

(q+−1)(h−q+1)(h−1) h−1 ! − p p p−2+ h+p−3+ − p h ( d )( d ) p−2+ ku(·, t)k h d ≤ C t + t d , L (R )

where C is a constant depending on h, d, p, A(d, p) and ku0kL1 , > 0 satisﬁes (q+)pp 3−p A(d,p) Kp(d,p)(q+−2+p)p − ku0kLq ≥ 2 .

q 3d Proof. Step 1. (The L estimate for 1 < p < d+1 ). Multiplying the ﬁrst equation in problem (1) by quq−1 and integrating it over Rd, we obtain Z Z d q p−2 q−1 q−1 ku(·, t)kLq ( d) = ∇ · |∇u| ∇u qu dx − ∇ · (u∇v)qu dx dt R d d R Z Z R = −q(q − 1) uq−2|∇u|p dx + (q − 1) ∇uq · ∇v dx d d R R p p q(q − 1)p q−2+p q+1 p = − ∇u (t) + (q − 1)kuk q+1 d . (9) p p d L (R ) (q − 2 + p) L (R ) 692 WENTING CONG AND JIAN-GUO LIU

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

d(q−2+p) q(pd+pq+p−3d) q+1 pd+pq−2d pd+pq−2d kuk q+1 d ≤ kuk kuk q L (R ) (q−2+p)d L L d−p dp q(pd+pq+p−3d) q−2+p pd+pq−2d p pd+pq−2d = u dp kukLq L d−p p q−2+p 3−p p = u dp kukLq , (10) L d−p d q(pd+pq+p−3d) where the last equality holds since pd+pq−2d = 1 and pd+pq−2d = 3 − p from d(3−p) q = p . Then using the Sobolev inequality (7), (10) turns to p q+1 q−2+p 3−p p p kukLq+1( d) ≤ K (d, p) ∇u kukLq , (11) R Lp where K(d, p) is given by (8). Substituting (11) into (9), we have

p p d q qp 3−p q−2+p p p kukLq + (q − 1) − K (d, p)kukLq ∇u ≤ 0. (12) dt (q − 2 + p)p Lp

1 h qpp i 3−p Since ku0(·)k q d < p p =: Cp,d, following two estimates hold L (R ) K (d,p)(q−2+p) true

ku(·, t)k q d < ku0(·)k q d < Cp,d, (13) L (R ) L (R ) Z ∞ p 3−p q−2+p p 3−p p (q − 1)K (d, p) C − ku0k q ∇u ds ≤ Cp,d. p,d L p 0 L Combining (11) with two estimates above, we obtain ∞ q d u(x, t) ∈ L R+; L (R ) , (14) q+1 q+1 d u(x, t) ∈ L R+; L (R ) , (15) q−2+p p p d ∇u p (x, t) ∈ L R+; L (R ) . (16) Step 2. (The Lq decay estimate). By using the interpolation inequality and (11), we have

(q+1)(q−1) 1 q q q ku(·, t)k q d ≤ kuk kuk 1 L (R ) Lq+1 L q−1 p q 1 q−2+p 3−p p p q ≤ K (d, p) ∇u kukLq kukL1 , (17) Lp i.e. p 2 p−2+ q −3+p 1+ d p q−1 q q−1 q−2+p kuk q kukLq p L ∇u ≥ 1 = 1 , (18) p L p q−1 p q−1 K (d, p)ku0kL1 K (d, p)ku0kL1 since ku(·, t)kL1 ≤ ku0kL1 . Substituting (18) into (12) yields that p p−2+ d d q (q − 1)A(d, p) q 1+ q−1 ku(·, t)kLq + 1 kukLq ≤ 0, (19) dt q−1 ku0kL1 3−p 3−p where we denote A(d, p) := Cp,d − ku0kLq . Next we discuss the inequality (19) in three diﬀerent situations. A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 693

p p−2+ d 2d 3d (a) If 1+ > 1, i.e. < p < , we can prove that ku(·, t)k q d decays q−1 d+1 d+1 L (R ) in time

ku0kLq ku(·, t)k q d ≤ , (20) L (R ) q−1 p q p−2+ (1 + Ct) ( d ) p p−2+ d p q q−1 A(d,p)(p−2+ d )(ku0kLq ) where C = 1 . q−1 ku0k L1 p p−2+ d 2d (b) If 1 + = 1, i.e. p = , ku(·, t)k q d decays exponentially in time q−1 d+1 L (R ) −Ct ku(·, t)k q d ≤ ku k q d e , L (R ) 0 L (R ) (q−1)A(d,p) where C = 1/(q−1) . qku0k 1 d L (R ) p p−2+ d 2d (c) If 0 < 1 + < 1, i.e. 1 < p < , ku(·, t)k q d has the ﬁnite q−1 d+1 L (R ) time extinction. The extinct time Text satisﬁes 0 < Text ≤ T0, where T0 = p q p−2+ ( d ) − q−1 1/(q−1) ku0k q d ku0k 1 d L (R ) L (R ) p . −A(d,p)(p−2+ d )

h 2d 3d Step 3. (The L decay estimate for any 1 ≤ h ≤ q when d+1 < p < d+1 ). Using the interpolation inequality and (20), we have

q(h−1) q−h q(h−1) q−h h(q−1) h(q−1) h(q−1) h(q−1) ku0kLq ku0kL1 ku(·, t)k h d ≤ ku(·, t)k ku(·, t)k ≤ . (21) L (R ) Lq ( d) L1( d) h−1 R R p h p−2+ (1 + Ct) ( d )

2d 3d Step 4. (The hyper-contractive property for any q < h < ∞ when d+1 0 such that (q + )pp A(d, p) − ku k3−p ≥ . (22) Kp(d, p)(q + − 2 + p)p 0 Lq 2 In the same way of obtaining (9)-(11), we obtain p p d r(r − 1)p r−2+p r+1 r p ku(·, t)kLr ( d) = − ∇u (t) + (r − 1)kukLr+1 , (23) dt R (r − 2 + p)p Lp and d(r−2+p)(r+1−q) q(pd+pr+p−3d) r+1 rd+pd+pq−qd−2d rd+pd+pq−qd−2d kuk r+1 d ≤ kuk kuk q L (R ) (r−2+p)d L L d−p dp(r+1−q) q(pd+pr+p−3d) r−2+p rd+pd+pq−qd−2d p rd+pd+pq−qd−2d = u dp kukLq L d−p p r−2+p 3−p p = u dp kukLq L d−p p r−2+p 3−p p p ≤ K (d, p) ∇u kukLq , (24) Lp d(r+1−q) q(pd+pr+p−3d) where the third equality holds since rd+pd+pq−qd−2d = 1 and rd+pd+pq−qd−2d = 3 − p, and the last inequality holds from the Sobolev inequality. Then combining 694 WENTING CONG AND JIAN-GUO LIU

(22), (23) and (24) together, we have p d (r − 1)K (d, p)A(d, p) r−2+p p r p ku(·, t)kLr + ∇u ≤ 0. (25) dt 2 Lp By using the interpolation inequality and (24), we have

(r+1)(r−1) 1 r r r ku(·, t)k r d ≤ kuk kuk L (R ) Lr+1 L1 r−1 p r r−2+p 3−p 1 p p r ≤ K (d, p) ∇u kukLq kukL1 Lp r−1 r(3−p)(q−1) (3−p)(r−q) r−2+p p r 1 q(r−1) q(r−1) p p r ≤ K (d, p) ∇u kukLr ku0kL1 ku0kL1 , Lp i.e. p p−2+ 1+ d p r r−1 r−2+p (kuk r ) ∇u p ≥ L , (26) Lp 1 1+ p(r−q) p r−1 d K (d, p)ku0kL1 since kuk 1 d ≤ ku0k 1 d . Substituting (26) into (25) yields that L (R ) L (R ) p p−2+ d d r r 1+ r−1 (r − 1)A(d, p) ku(·, t)k r + β1 kuk r ≤ 0, β1 := . (27) L L 1 p(r−q) dt r−1 1+ d 2ku0kL1 Solving this inequality by using Lemma 2.3, we have

r−1 − p r p−2+ ku(·, t)kLr ≤ C(r)t d . (28) Hyper-contractive estimates of Lh norm for h ≥ r. For h ≥ r > q, using the interpolation inequality, Sobolev inequality and Young’s inequality together, we obtain d(h−2+p)(h+1−r) r(pd+ph+p−3d) h+1 hd+pd+pr−rd−2d hd+pd+pr−rd−2d kuk h+1 d ≤ kuk kuk r L (R ) (h−2+p)d L L d−p dp(h+1−r) r(pd+ph+p−3d) h−2+p hd+pd+pr−rd−2d p hd+pd+pr−rd−2d = u dp kukLr L d−p dp(h+1−r) r(pd+ph+p−3d) dp(h+1−r) h−2+p hd+pd+pr−rd−2d hd+pd+pr−rd−2d p hd+pd+pr−rd−2d ≤ K (d, p) ∇u kukLr Lp p hp h−2+p p 1+ h−r+1 p r r−q ≤ ∇u + C(h, r) kukLr , (29) 2(h − 2 + p)p Lp where dp(h + 1 − r) pd(h + 1 − r) = < p. hd + pd + pr − rd − 2d d(h + 1 − r) + p(r − q) Considering (9) with h = q, we have p p d h(h − 1)p h−2+p h+1 h p ku(·, t)k h d = − ∇u (t) + (h − 1)kuk h+1 d L (R ) p p d L (R ) dt (h − 2 + p) L (R ) p h(h − 1)p h−2+p p 1+ h−r+1 p r r−q ≤ − ∇u (t) + C(h, r) kuk r . (30) p p d L 2(h − 2 + p) L (R ) Substituting (28) into (30) yields that

p (r−1)(h−q+1) h−2+p p − p d h(h − 1)p p−2+ (r−q) h p ( d ) ku(·, t)k h d ≤ − ∇u +C(h, r)t . (31) L (R ) p p d dt 2(h − 2 + p) L (R ) A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 695

By the same way of obtaining (26), we obtain p p−2+ 1+ d h h−1 h−2+p p kukLh ∇u p ≥ . (32) p 1 1+ p(h−q) L p h−1 ( d ) K (d, p)ku0kL1 Then (31) turns to p p−2+ d (r−1)(h−q+1) 1+ − p d h−1 p−2+ (r−q) h h ( d ) ku(·, t)k h d ≤ −β2 kuk h + C(h, r)t , (33) dt L (R ) L h(h−1)pp where β2 = p(h−q) . 1 1+ p p h−1 ( d ) 2(h−2+p) K (d,p)ku0k L1 p h p−2+ d Using Lemma 2.4 with y(t) = ku(·, t)k h d , a = 1 + > 1, β = β2 > 0 L (R ) h−1 (r−1)(h−q+1) − p (p−2+ )(r−q) and f(t) = C(h, r)t d , for any t > t0 > 0, we have

(q+−1)(h−q+1)(h−1) − p p h−1 p−2+ h+p−3+ − p h ( d )( d ) p−2+ ku(·, t)k h d ≤ C(h, r)t + C(h)(t − t0) d . (34) L (R ) 0 t By choosing t0 = 2 , we obtain that for any t > 0 (q+−1)(h−q+1)(h−1) h−1 ! − p p p−2+ h+p−3+ − p h ( d )( d ) p−2+ ku(·, t)k h d ≤ C t + t d , (35) L (R ) where C is a constant depending on h, d, p, A(d, p) and ku0kL1 , 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 by using a bootstrap iterative method. At the beginning of this section, we prove the following proposition concerning Lh norm estimates of weak solutions for 1 < h < ∞.

3d d(3−p) 1 d Proposition 1. Let d ≥ 3, 1 0, where Cp,d = 1 h qpp i 3−p Kp(d,p)(q−2+p)p is a universal constant, let (u, v) be a non-negative weak solution of (1). Then u(x, t) satisﬁes for any t > 0

q(h−1) h q−1 ku(·, t)k h d ≤ Cku0k q d , 1 < h ≤ q, (36) L (R ) L (R ) where C depends on h, q, and ku0kL1 , and h h ku(·, t)k h d ≤ C , q < h < ∞, (37) L (R ) u h where Cu is a constant depending on d, p, h, ku0kL1 and ku0kLh , > 0 satisﬁes (q+)pp 3−p A(d,p) Kp(d,p)(q+−2+p)p − ku0kLq ≥ 2 . Actually, the proof of Proposition1 is almost the same as the proof of Theorem 1 d h d 3.1, except for the diﬀerent initial condition u0 ∈ L+(R ) ∩ L (R ) for 1 < h < ∞. Proof. Using the same method in Step 1 of Theorem 3.1, we have for all t > 0

ku(·, t)k q d < ku0(·)k q d < Cp,d. L (R ) L (R ) Then we discuss in two diﬀerent situations with respect to h. 696 WENTING CONG AND JIAN-GUO LIU

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

q−h q(h−1) h q−1 q−1 ku(·, t)k h d ≤ ku0(·)k 1 d ku0(·)k q d . (38) L (R ) L (R ) L (R ) For q < h < ∞, letting r := q + ≤ h < ∞, there exists > 0 small enough such that (q + )pp A(d, p) − ku k3−p ≥ . Kp(d, p)(q + − 2 + p)p 0 Lq 2 Then (25) also holds true, i.e. p d (r − 1)K (d, p)A(d, p) r−2+p p r p ku(·, t)kLr + ∇u ≤ 0. dt 2 Lp r d Since q < r ≤ h, we have u0 ∈ L (R ) and

ku(·, t)k r d ≤ ku0(·)k r d , (39) L (R ) L (R ) for all t > 0. Combining (30), (32) and (39) together, we obtain

p p−2+ d h−r+1 1+ h−1 d h h r 1+ r−q ku(·, t)k h d ≤ −β3 kuk h + C(h, r) ku0k r , (40) dt L (R ) L L h(h−1)pp where β3 := p(h−q) > 0. Using Lemma 2.5 with y(t) = 1 1+ p p h−1 ( d ) 2(h−2+p) K (d,p)ku0k L1 p 1+ h−r+1 h p−2+ d r r−q ku(·, t)kLh , a = 1 + h−1 > 0, β = β3 > 0 and γ = C(h, r) ku0kLr > 0, for any t > 0, we have (h−q+1)(h−1) p h h r (h+p−3+ ) ku(·, t)kLh ≤ max ku0kLh ,C(h, r) ku0kLr d

( (h−q+1)(q+−1) ) p (h+p−3+ ) h h d h ≤ max ku0kLh ,C(h) ku0kLh =: Cu , (41)

(q+)pp 3−p A(d,p) where satisﬁes Kp(d,p)(q+−2+p)p − ku0kLq ≥ 2 . Next, we prove the uniformly in time L∞ boundness of u(x, t) by using a bootstrap iterative technique [3, 19] with Proposition1 and an additional initial condi- ∞ d tion u0 ∈ L (R ). 3d d(3−p) 1 d ∞ d Theorem 4.1. Let d ≥ 3, 1 0, where Cp,d = Kp(d,p)(q−2+p)p is a universal constant, let (u, v) be a non-negative weak solution of (1). Then for any t > 0,

ku(·, t)k ∞ d ≤ C(d, p, K ), L (R ) 0 where K = max 1, ku k 1 d , ku k ∞ d . 0 0 L (R ) 0 L (R )

Proof. We denote d(3 − p) h = 3k + + 1, for k ≥ 1. k p h −1 Multiplying the ﬁrst equation in (1) by hku k and integrating, we have p p d h hk(hk − 1)p hk−2+p h +1 ku(·, t)k k = − ∇u p (t) + (h − 1)kuk k . (42) Lhk ( d) p k Lhk+1 dt R (hk − 2 + p) Lp A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 697

p hk hk(hk−1)p Step 1. (The L estimate for 1 < p ≤ 2) Taking 0 < C1 ≤ p is a ﬁxed 2(hk−2+p) constant, then (42) turns to p d h hk−2+p h +1 k p k ku(·, t)k h d ≤ −2C1 ∇u (t) + hkkuk hk+1 . (43) dt L k (R ) Lp L Using the interpolation inequality and Sobolev inequality together, we obtain

hk+1 (hk+1)θ (hk+1)(1−θ) kuk h +1 d ≤ kuk kuk h L k (R ) (hk−2+p)d L k−1 L d−p p(hk+1)θ h −2+p k hk−2+p (h +1)(1−θ) p k = u dp kuk h L d−p L k−1 p(h +1)θ p(h +1)θ k k hk−2+p h −2+p k (hk+1)(1−θ) hk−2+p p ≤ K (d, p) ∇u kuk h , (44) Lp L k−1 where d(hk − 2 + p)(hk − hk−1 + 1) θ = , (hk + 1) (hk − 2 + p)d − hk−1(d − p)

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

k−1 d(3−p) d(3−p) p(hk+1)θ Since hk−1 = 3 + + 1 > , it is easy to see that < p. Then p p hk−2+p using Young’s inequality and (44), we have

p(h +1)θb hk−2+p p k hk+1 1 a 1 −b b (hk+1)(1−θ)b p hk−2+p hkkuk hk+1 d ≤ δ1 ∇u + δ1 K (d, p)hkkuk h L (R ) a Lp b L k−1 p hk−2+p (h +1)(1−θ)b p b k ≤ C1 ∇u + C2(hk)hkkuk h , (45) Lp L k−1 where h − 2 + p d(h − h + 1) + h p + pd − 3d a = k = k k−1 k−1 > 1, (hk + 1)θ d(hk − hk−1 + 1) h − 2 + p d(h − h + 1) + h p + pd − 3d b = k = k k−1 k−1 > 1, hk − 2 + p − (hk + 1)θ hk−1p + pd − 3d p(h +1)θb 1 1 − b k δ = (C a) a ,C (h ) = (C a) a K hk−2+p (d, p). 1 1 2 k b 1 2d+p 2d+p We can see that C2(hk) is uniformly bounded since a → 2d and b → p as k → ∞. Substituting (45) into (43) yields to

h −2+p p γ1 d h k hk−1 k p b ku(·, t)k h d ≤ −C1 ∇u (t) + C2(hk)hk kuk h , (46) dt L k (R ) Lp L k−1 where (hk + 1)(1 − θ)b hkp + pd − 3d + p γ1 = = < 3. hk−1 hk−1p + pd − 3d hk−2+p p Next, we estimate ∇u p (t) . By using the interpolation inequality and Lp Sobolev inequality, we have

hk hkβ hk(1−β) kuk h d ≤ kuk kuk h L k (R ) (hk−2+p)d L k−1 L d−p phkβ h −2+p k hk−2+p h (1−β) p k = u dp kuk h L d−p L k−1 ph β ph β k k hk−2+p h −2+p k hk(1−β) hk−2+p p ≤ K (d, p) ∇u kuk h , (47) Lp L k−1 698 WENTING CONG AND JIAN-GUO LIU where d(hk − 2 + p)(hk − hk−1) β = , hk (hk − 2 + p)d − hk−1(d − p)

hk−1(hkp + pd − 2d) 1 − β = . hk (hk − 2 + p)d − hk−1(d − p) Since it is easy to see that phkβ < p, then using Young’s inequality, we have hk−2+p

ph βb0 0 hk−2+p p 0 k 0 hk 1 a 1 −b hk(1−β)b p hk−2+p kuk hk d ≤ δ2 ∇u + δ2 K (d, p)kuk h L (R ) a0 Lp b0 L k−1 p γ hk−2+p h 2 p k−1 ≤ C1 ∇u + C3(hk) kuk h , (48) Lp L k−1 where h − 2 + p d(h − h ) + h p + pd − 2d a0 = k = k k−1 k−1 > 1, hkβ d(hk − hk−1) h − 2 + p d(h − h ) + h p + pd − 2d b0 = k = k k−1 k−1 > 1, hk − 2 + p − hkβ hk−1p + pd − 2d 0 ph βb0 0 1 1 0 − b k δ = (C a ) a0 ,C (h ) = (C a ) a0 K hk−2+p (d, p), 2 1 3 k b0 1 0 hk(1 − β)b hkp + pd − 2d γ2 = = < 3. hk−1 hk−1p + pd − 2d We can also check that C3(hk) is uniformly bounded as k → ∞. Combining (46) and (48) together, we have γ γ d h 1 h 2 kukhk ≤ −kukhk + C (h )hb kuk k−1 + C (h ) kuk k−1 . (49) dt Lhk Lhk 2 k k Lhk−1 3 k Lhk−1

Since C2(hk) and C3(hk) are both uniformly bounded as k → ∞, we can choose a constant C4 > 1 which is an upper bound of C2(hk) and C3(hk). Then by hk > 1 and b > 1, we have for any t > 0, γ γ d h 1 h 2 kukhk ≤ −kukhk + C hb kuk k−1 + kuk k−1 . (50) dt Lhk Lhk 4 k Lhk−1 Lhk−1

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

p h −2+p p d h p−2 h i hk(hk − 1)p k (h − 2 + p) kuk k = − ∇u p (t) k Lhk 2 dt (hk − 2 + p) Lp

p−2 hk+1 + (hk − 1)(hk − 2 + p) kukLhk+1 p hk−2+p h +1 0 p 2 k ≤ −2C1 ∇u (t) + C5hkkuk hk+1 , (51) Lp L p 0 hk(hk−1)p where 0 < C ≤ 2 is a ﬁxed constant and C5 is also a ﬁxed constant 1 2(hk−2+p) p−2 2 satisfying (hk − 1)(hk − 2 + p) ≤ C5hk since hk > 1 and p < 3. Using Young’s inequality and (44), we have

p(h +1)θb hk−2+p p k 2 hk+1 1 a 1 −b b 2b (hk+1)(1−θ)b p hk−2+p C5hkkuk hk+1 ≤ δ3 ∇u + δ3 (C5) K hk kuk h L a Lp b L k−1 p hk−2+p (h +1)(1−θ)b 0 p 0 2b k ≤ C1 ∇u (t) + C2(hk)hk kuk h , (52) Lp L k−1 where h − 2 + p d(h − h + 1) + h p + pd − 3d a = k = k k−1 k−1 > 1, (hk + 1)θ d(hk − hk−1 + 1) A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 699

h − 2 + p d(h − h + 1) + h p + pd − 3d b = k = k k−1 k−1 > 1, hk − 2 + p − (hk + 1)θ hk−1p + pd − 3d

p(h +1)θb 0 1 0 1 0 − b b k δ = (C a) a ,C (h ) = (C a) a (C ) K hk−2+p (d, p). 3 1 2 k b 1 5 0 We can see that C2(hk) is uniformly bounded as k → ∞. Since (hk−1 − 2 + (p−2)(hk+1)(1−θ)b p) hk−1 ≥ 1, (52) turns to

p γ h +1 hk−2+p h h i 1 2 k 0 p 0 2b p−2 k−1 C5hkkuk hk+1 ≤ C1 ∇u + C2(hk)hk (hk−1 − 2 + p) kuk h , L Lp L k−1 (53) where (hk + 1)(1 − θ)b hkp + pd − 3d + p γ1 = = < 3. hk−1 hk−1p + pd − 3d Substituting (53) into (51), we obtain

p d h h i hk−2+p p−2 k 0 p (hk − 2 + p) kuk h ≤ −C1 ∇u (t) dt L k Lp h iγ1 + C0 (h )h2b (h − 2 + p)p−2kukhk−1 . (54) 2 k k k−1 Lhk−1 Next, by using Young’s inequality and (47), we have

p h 1 0 hk−2+p p−2 k a p (hk − 2 + p) kuk h ≤ δ4 ∇u L k a0 Lp 0 0 phkβb 0 0 1 −b (p−2)b hk(1−β)b + δ K hk−2+p (d, p)(h − 2 + p) kuk b0 4 k Lhk−1 p hk−2+p 0 h (1−β)b0 0 p 0 b k ≤ C1 ∇u (t) + C3(hk)hk kuk h , (55) Lp L k−1 where h − 2 + p d(h − h ) + h p + pd − 2d a0 = k = k k−1 k−1 > 1, hkβ d(hk − hk−1)

h − 2 + p d(h − h ) + h p + pd − 2d b0 = k = k k−1 k−1 > 1, hk − 2 + p − hkβ hk−1p + pd − 2d

0 ph βb0 0 0 1 0 C6 0 0 − b k δ = (C a ) a0 ,C (h ) = (C a ) a0 K hk−2+p (d, p), 4 1 3 k b0 1 (p−2)b0 b0 and C6 is a ﬁxed constant such that (hk − 2 + p) ≤ C6hk . We can see that 0 (p−2)hk(1−β)b 0 hk−1 C3(hk) is uniformly bounded as k → ∞. Also since (hk−1 − 2 + p) ≥ 1, (55) turns to

p h hk−2+p p−2 k 0 p (hk − 2 + p) kuk hk ≤C1 ∇u (t) L Lp 0 h iγ2 + C0 (h )hb (h − 2 + p)p−2kukhk−1 , (56) 3 k k k−1 Lhk−1 where 0 hk(1 − β)b hkp + pd − 2d γ2 = = < 3. hk−1 hk−1p + pd − 2d 700 WENTING CONG AND JIAN-GUO LIU

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

d h i (h − 2 + p)p−2kukhk dt k Lhk γ h h i 1 ≤ −(h − 2 + p)p−2kukhk + C0 (h )h2b (h − 2 + p)p−2kuk k−1 k Lhk 2 k k k−1 Lhk−1 0 h iγ2 + C0 (h )hb (h − 2 + p)p−2kukhk−1 . (57) 3 k k k−1 Lhk−1

0 0 Since C2(hk) and C3(hk) are both uniformly bounded as k → ∞, we can choose a 0 0 constant C7 > 1 which is an upper bound of C2(hk) and C3(hk). Then by hk > 1 and 2b > b0 > 1, we have for any t > 0,

d h i (h − 2 + p)p−2kukhk ≤ −(h − 2 + p)p−2kukhk dt k Lhk k Lhk nh iγ1 h iγ2 o + C h2b (h − 2 + p)p−2kukhk−1 + (h − 2 + p)p−2kukhk−1 . 7 k k−1 Lhk−1 k−1 Lhk−1 (58)

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

d y (t) ≤ −y (t) + C h2b yγ1 + yγ2 . (59) dt k k 8 k k−1 k−1 Multiplying et to both sides of (59), we have d t 2b γ1 γ2 t 2b 3 t e yk(t) ≤ C8hk yk−1 + yk−1 e ≤ 2C8hk max 1, sup yk−1(t) e . (60) dt t≥0

Solving this ODE, we obtain for t ≥ 0, −t 2b 3 −t yk(t) ≤ e yk(0) + 2C8hk max 1, sup yk−1(t) 1 − e t≥0 2b 3 ≤ 2C8hk max 1, yk(0), sup yk−1(t) . (61) t≥0

It is easy to see that

d(3 − p) 2b d(3 − p) 2b h 2b = 3k + + 1 ≤ C 32bk + 1 , (62) k p 0 p where C0 is an appropriate positive constant. Combining (61) and (62) together, we can see 2b 2bk d(3 − p) 3 yk(t) ≤ C93 + 1 max 1, yk(0), sup yk−1(t) , p t≥0 A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 701

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

3k−1 2b! 2 k+1 d(2 − m) 2b( 3 − k − 3 ) y (t) ≤ C + 1 3 4 2 4 k 9 2

( k−1 ) X 3i 3k · max 1, sup yk−i(0), sup y0 (t) . (63) i=0 t≥0 t≥0 Denote K = max 1, ku k 1 d , ku k ∞ d , then 0 0 L (R ) 0 L (R ) ( n o ku khk ≤ max ku khk , ku khk , 1 < p ≤ 2, 0 Lhk 0 L1 0 L∞ yk(0) = (h − 2 + p)p−2ku khk , 2 < p < 3d , k 0 Lhk d+1 i.e. hk K0 , 1 < p ≤ 2, yk(0) ≤ p−2 hk 3d (64) (hk − 2 + p) K0 , 2 < p < d+1 , 3d and for any 1 < p < d+1 , 1 hk lim yk (0) ≤ K0, (65) k→∞ p−2 h 3d since lim (hk − 2 + p) k = 1 for 2 < p < . Furthermore, we also have k→∞ d+1

( k−1 ) i X 3 qk max 1, sup yk−i(0) ≤ kK0 . i=0 t≥0 Taking the power 1 to both sides of (63) and letting k → ∞, we obtain hk ku(·, t)k ∞ d ≤ C max sup y (t),K , (66) L (R ) 0 0 t≥0

3(2d+p) 1 d+1 2p 2 d(2−m) 2d+p where C = 3 C9 2 + 1 since b → p as k → ∞. Recalling (37) in Proposition1, it shows that q+2 q+2 y0(t) = ku(·, t)k q+2 d ≤ C . (67) L (R ) u Then (66) turns to

ku(·, t)k ∞ d ≤ C(d, p, K ). L (R ) 0

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 the supercritical case.

Lemma 5.1. For any η, η0 ∈ Rd, there exists 0 p−2 0 2 p−2 0 p−2 0 0 C1 (|η| + |η |) |η − η | , p > 1, |η| η − |η | η · (η − η ) ≥ 0 p C2|η − η | , p ≥ 2, where C1 and C2 are two positive constants only depending on p. 702 WENTING CONG AND JIAN-GUO LIU

3d d(3−p) 1 d ∞ d Theorem 5.2. Let d ≥ 3, 1 0, where Cp,d = Kp(d,p)(q−2+p)p is a universal constant. Then there exists a non-negative global weak solution (u, v) of (1), such that all a priori estimates in Theorem 3.1 and the uniform L∞ estimate in Theorem 4.1 hold true. Proof. We separate the proof of Theorem 5.2 into four steps. In Step 1, we construct the regularized problem of (1) and show that all a priori estimates in Theorem 3.1 and the uniform L∞ estimate in Theorem 4.1 hold true. Furthermore, we obtain the uniform estimate of ∇u. In Step 2 and 3, by applying the Aubin-Lions Lemma, we prove that a non-negative weak solution of the regularized problem (68) converges strongly to a non-negative weak solution of (1) in a bounded domain. Finally, in Step 4, using the weak convergence and strong convergence estimates obtained in Step 1-3, we prove the existence of a global weak solution of (1) with monotone operators. Step 1. (The regularized problem and a priori estimates) We consider the regularized problem of (1) for > 0,

 p−2 d  ∂tu = ∇ · |∇u| ∇u + ∆u − ∇ · (u∇v) , x ∈ R , t > 0, d (68) −∆v = J ∗ u, x ∈ R , t > 0,  d  u(x, 0) = u0(x), x ∈ R , − d+2 3d 1 x 1 2 2 where d ≥ 3, 1 < p < d+1 and J(x) = d J , J(x) = α(d) 1 + |x| R satisfying d J(x) dx = 1. A simple computation shows that v can be expressed R by Z 1 u(y, t) v(x, t) = d−2 dy, (69) d(d − 2)α(d) d 2 R |x − y|2 + 2 where α(d) is the volume of the d-dimensional unit ball. The initial condition ∞ d u0(x) ∈ C (R ) is a sequence of approximation for u0(x), which satisﬁes that there exists δ > 0 such that for all 0 < < δ, d u0(x) > 0, x ∈ R , r d u0(x) ∈ L (R ), for all r ≥ 1,

ku (·)k 1 d = ku (·)k 1 d , 0 L (R ) 0 L (R )

ku0(·)k ∞ d ≤ C. L (R ) According to the classical theory for parabolic equations [16], the regularized problem has a global smooth non-negative solution (u, v) with the regularity for all r ≥ 1, ∞ r d r+1 r+1 d u ∈ L R+; L (R ) ∩ L R+; L (R ) . Then we want to show that all a priori estimates in Theorem 3.1 hold true for our regularized problem. we take a cut-oﬀ function 0 ≤ ψ1(x) ≤ 1, satisfying 1, if |x| ≤ 1, ψ (x) = 1 0, if |x| ≥ 2, ∞ d x where ψ1(x) ∈ Cc (R ). Deﬁne ψR(x) := ψ1( ), then we know that lim ψR(x) = R R→∞ C1 C2 d 1, |∇ψR(x)| ≤ R and |∆ψR(x)| ≤ R2 for x ∈ R , C1,C2 are constants. We can A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 703

1 1 1 p p x p C3 also deﬁne ψ (x) := ψ ( ) and choose a constant C3, such that ∇ψ (x) ≤ . R 1 R R R q−1 d Multiplying the ﬁrst equation of (68) by qu ψR(x) and integrating over R , we obtain Z p Z q−2+p p d q q(q − 1)p p u ψR(x) dx + p ∇u ψR(x) dx dt d (q − 2 + p) d R R Z q 2 4(q − 1) 2 + ∇u ψR(x) dx q d Z R Z q p−2 q−1 = −(q − 1) u ∆vψR(x) dx − q |∇u| u ∇u · ∇ψR(x) dx d d Z R Z R q q + u ∇v · ∇ψR(x) dx + u ∆ψR(x) dx. (70) d d R R Integrating (70) from 0 to t yields that

Z p Z t Z q−2+p p q q(q − 1)p p u (x, t)ψR(x) dx + p ∇u ψR(x) dxds d (q − 2 + p) d R 0 R t Z Z q 2 4(q − 1) 2 + ∇u ψR(x) dxds q d 0 R Z Z t Z q q = u0(x)ψR(x) dx − (q − 1) u ∆vψR(x) dxds d d R 0 R Z t Z p−2 q−1 − q |∇u| u ∇u · ∇ψR(x) dxds d 0 R Z t Z Z t Z q q + u ∇v · ∇ψR(x) dxds + u ∆ψR(x) dxds. (71) d d 0 R 0 R For the second term on the right hand side of (71), by using H¨older’sinequality, we have Z t Z Z t Z q q −(q − 1) u ∆vψR(x) dxds ≤ (q − 1) u J ∗ u dxds d d 0 R 0 R Z t q ≤ (q − 1) kukLq+1 kJ ∗ ukLq+1 ds 0 Z t q+1 ≤ (q − 1) kukLq+1 ds ≤ C(), (72) 0 r+1 r+1 d since u ∈ L R+; L (R ) for any r ≥ 1. Then we can use the dominated convergence theorem as R → ∞ for any small > 0 later. Next, we want to prove that last three terms on the RHS of (71) go to 0 as R → ∞. First, by using H¨older’sinequality and Young’s inequality of convolution [18, pp.107], we obtain Z t Z Z t Z q C1 q u ∇v · ∇ψR(x) dxds ≤ u |∇v| dxds d R d 0 R 0 R Z t C1 q ≤ kuk d(q+1) k∇vk d(q+1) ds R 0 L d+1 L d−q C Z t x q ≤ kuk d(q+1) d d kuk d(q+1) ds R d+1 |x| d−1 L d+1 0 L Lw 704 WENTING CONG AND JIAN-GUO LIU

Z t C q+1 ≤ kuk d(q+1) ds. (73) R 0 L d+1 Then using the interpolation inequality, (73) yields to

Z t Z Z t q+1 (q+1)(dq−1) q C dq dq u ∇v · ∇ψR(x) dxds ≤ kukL1 kukLq+1 ds d R 0 R 0 dq−1 Z t dq C ku0kL1 q+1 C() ≤ kukLq+1 ds ≤ , (74) R 0 R q+1 q+1 d since u ∈ L R+; L (R ) for q > 1. ∞ r d Second, from u ∈ L R+; L (R ) for all r ≥ 1, we have Z t Z q C(t, ) u ∆ψR(x) dxds ≤ 2 . (75) d R 0 R Third, by using H¨older’sinequality, we have Z t Z p−2 q−1 − q |∇u| u ∇u · ∇ψR(x) dxds d 0 R Z t Z p−1 q−1 ≤ q |∇u| u |∇ψR(x)| dxds d 0 R Z t Z (q−2)(p−1) p−1 q−2+p p−1 p−1 p p p − p = q |∇u| u ψR (x)u ψR (x) |∇ψR(x)| dxds d 0 R p−1 Z t Z q−2+p p−1 q−2+p 1 p p p p = C ∇u ψR (x)u ∇ψR (x) dxds d 0 R p−1 1 Z t Z q−2+p p p Z t p C p q−2+p ≤ ∇u ψR(x) dxds kukLq−2+p ds . (76) R d 0 R 0

q−2+p p R t R p Then we should prove d ∇u ψR(x) dxds is bounded in order to show 0 R (76) goes to 0 as R → ∞. Using Young’s inequality for (76), we obtain Z t Z p−2 q−1 − q |∇u| u ∇u · ∇ψR(x) dxds d 0 R Z t Z q−2+p p Z t C p 1 q−2+p ≤ p ∇u ψR(x) dxds + kukLq−2+p ds p−1 d p R 0 R 0 Z t Z q−2+p p C p ≤ p ∇u ψR(x) dxds + C(t, ), (77) p−1 d R 0 R since q − 2 + p ≥ 1. Combining (71), (72), (74), (75) and (77) together, we have

Z p Z t Z q−2+p p q q(q − 1)p C p u (x, t)ψR(x) dx + p − p ∇u ψR(x) dxds d (q − 2 + p) p−1 d R R 0 R t Z Z q 2 4(q − 1) 2 + ∇u ψR(x) dxds q d 0 R Z q C(t, ) C(t, ) ≤ u0(x)ψR(x) dx + C(t, ) + + 2 . (78) d R R R A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 705

q(q−1)pp C Taking R large enough, we can see that (q−2+p)p − p > 0. Then (78) shows R p−1 that Z t Z q−2+p p p ∇u ψR(x) dxds ≤ C(t, ), (79) d 0 R q d since u0 ∈ L (R ) when R is large enough. Substituting (79) into (76), we have Z t Z p−2 q−1 C(t, ) − q |∇u| u ∇u · ∇ψR(x) dxds ≤ . (80) d R 0 R Until now, we have proved that last three terms on the RHS of (71) go to 0 as R → ∞. Using the dominated convergence theorem, when R → ∞,(71) turns to

Z p Z t Z q−2+p p q q(q − 1)p p u (x, t) dx + p ∇u dxds d (q − 2 + p) d R 0 R t Z Z q 2 4(q − 1) 2 + ∇u dxds q d 0 R Z Z t Z q q = u0(x)dx − (q − 1) u ∆v dxds, (81) d d R 0 R i.e., for any t > 0, p d Z q(q − 1)pp Z q−2+p 4(q − 1) Z q 2 q p 2 u (x, t) dx + p ∇u dx + ∇u dx dt d (q − 2 + p) d q d R R R Z q q+1 = (q − 1) u J ∗ u dx ≤ (q − 1)kukLq+1 d R q−2+p p p p 3−p ≤ (q − 1)K (d, p) ∇u kukLq , (82) Lp where last two inequalities can be obtained by the same method of (72) and (11). Then we have

p q−2+p p d q qp p 3−p p kukLq + (q − 1) p − K (d, p)kukLq ∇u ≤ 0, (83) dt (q − 2 + p) Lp which is same to (12), and all a priori estimates in Theorem 3.1 hold true for our solution of the regularized problem. We also have following uniformly bounded estimates,

kuk ∞ 1 q d ≤ C, (84) L (R+;L+∩L (R ))

kuk q+1 q+1 d ≤ C, (85) L (R+;L (R ))

r−2+p p ∇u ≤ C, 1 < r ≤ q. (86) p p d L (R+;L (R )) ∞ d Additionally, since u0 ∈ L ( ), we let u0(x) also satisfy ku0k ∞ d ≤ C, where R L (R ) C is a positive constant independent of . Then from the Theorem 4.1, we have the uniformly bounded estimate

ku(·, t)k ∞ d ≤ C. (87) L (R ) 3d For q ≥ 2, i.e. 1 < p ≤ d+2 , by taking r = 2 in (86), we have

k∇uk p p d ≤ C. L (R+;L (R )) 706 WENTING CONG AND JIAN-GUO LIU

3d 3d For 1 < r ≤ q < 2, i.e. d+2 < p ≤ d+1 , by using (87), we obtain

Z Z r−2+p p Z Z p r−2 p C ≥ ∇u dxdt = C u |∇u| dxdt d d R+ R R+ R Z Z p ≥ C kukL∞ |∇u| dxdt, d R+ R where C is a positive constant. From two estimates above, we have

k∇uk p p d ≤ C, (88) L R+;L (R ) 3d for all 1 < p < d+1 .

Step 2. (The time regularity of u) In this step, we want to estimate ∂tu in any bounded domain in order to use the Aubin-Lions Lemma. For any test function 2,p ϕ(x) which satisﬁes ϕ ∈ W (Ω) and kϕkW 2,p(Ω) ≤ 1, we have p−2 |h∂tu, ϕi| ≤ |h|∇u| ∇u, ∇ϕi| + |hu, ∆ϕi| + |hu∇v, ∇ϕi| p−1 p p ≤ k∇uk p + kuk + ku∇vk . (89) L (Ω) L p−1 (Ω) L p−1 (Ω)

Then for any T > 0, since ku(·, t)k ∞ d ≤ C and k∇uk p p d ≤ C, using L (R ) L (R+;L (R )) Sobolev inequality, we obtain T T T Z p Z p Z p p−1 p p−1 p p−1 p k∂tuk −2, dt ≤ C k∇ukLp(Ω) dt + kuk dt 0 W p−1 (Ω) 0 0 L p−1 (Ω) Z T p p−1 + ku∇vk p dt 0 L p−1 (Ω) T p Z p p−1 ≤ C(T )(1 + p−1 ) + C k∇vk p dt 0 L p−1 (Ω) Z T p p−1 ≤ 2C(T ) + C k∆vk dp dt 0 L dp−d+p (Ω) ≤ C(T ). (90)

p p Then we have k∂tuk −2, ≤ C. L p−1 0,T ;W p−1 (Ω) Step 3. (The application of the Aubin-Lions Lemma) It is easy to see that

kukLp(0,T ;Lp(Ω)) ≤ C(Ω,T ), from ku(·, t)k ∞ d ≤ C, where Ω is any bounded domain. Then we obtain L (R ) that kuk ≤ C. By the Sobolev Embedding Theorem, we have Lp0,T ;W 1,p(Ω) 1,p p dp W (Ω) ,→,→ L (Ω) since p < d and p < d−p . Until now, we already have

kuk ≤ C, Lp0,T ;W 1,p(Ω)

p p k∂tuk −2, ≤ C, L p−1 0,T ;W p−1 (Ω)

−2, p and W 1,p(Ω) ,→,→ Lp(Ω) ,→ W p−1 (Ω). By the Aubin-Lions Lemma, there existes a subsequence of {u} without relabeling such that p p u → u, in L 0,T ; L (Ω) . (91) A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 707

Step 4. (The existence of a global weak solution) Next, we will show that (u, v) is a weak solution of the problem (1). The crucial idea in this step follows the proof of Theorem 2.2.1 in [28, p171]. The weak formulation for u is that for any test ∞ d function ψ(x) ∈ Cc [0,T ) × R and any 0 < T < ∞, Z T Z Z T Z u(x, t)ψt(x, t) dxdt + u(x, t)∆ψ(x, t) dxdt d d 0 R 0 R Z T Z Z p−2 = |∇u(x, t)| ∇u(x, t) · ∇ψ(x, t) dxdt − u0(x)ψ(x, 0)dx d d 0 R R 1 Z T Z Z [∇ψ(x, t) − ∇ψ(y, t)] · (x − y) u (x, t)u (y, t) − dxdydt. 2 2 d−2 2dα(d) d d |x − y| + 2 0 R R |x − y|2 + 2 (92) Next, we separate the proof of this step into three parts. p p (i) Since u → u in L (0,T ; L (Ω)), using H¨older’sinequality, we have Z T Z Z T |u − u| dxdt ≤ C(Ω) ku − ukLp(Ω) dt 0 Ω 0 ≤ C(Ω,T )ku − uk Lp0,T ;Lp(Ω) → 0, as → 0, i.e. 1 1 u → u, in L 0,T ; L (Ω) . (93) (ii) Next we have

Z T Z Z ∇ψ(x, t) − ∇ψ(y, t) · (x − y) u (x, t)u (y, t) dxdydt 2 2 d−2 d d |x − y| + 2 0 R R |x − y|2 + 2

Z T Z Z ∇ψ(x, t) − ∇ψ(y, t) · (x − y) u(x, t)u(y, t)

− 2 d−2 dxdydt d d |x − y| |x − y| 0 R R Z T Z Z ≤ ∇ψ(x, t) − ∇ψ(y, t) · (x − y)u(x, t)u(y, t) d d 0 R R ! 1 1

· d − d dxdydt |x − y| 2 |x − y|2 + 2

Z T Z Z [∇ψ(x, t) − ∇ψ(y, t)] · (x − y)

+ d d d |x − y| 0 R R

· u(x, t)u(y, t) − u(x, t)u(y, t) dxdydt

=: I1 + I2. (94)

In order to estimate I1, we have 1 1 d2 d − d = d+2 |x − y| 2 2 |x − y|2 + 2 |x − y|2 + 2 708 WENTING CONG AND JIAN-GUO LIU

d d ≤ ≤ , (95) |x − y|d+1 |x − y| |x − y|d+1

since is small enough. Then using Hardy-Littlewood-Sobolev inequality, I1 satisﬁes

Z T Z Z ∇ψ(x, t) − ∇ψ(y, t) · (x − y)u (x, t)u (y, t) I1 ≤ d d+1 dxdydt d d |x − y| 0 R R Z T Z Z Z T u(x, t)u(y, t) 2 ≤ C dxdydt ≤ C kuk 2d dt d−1 d+1 0 Ω Ω |x − y| 0 L (Ω) ≤ CT → 0, as → 0. (96)

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

Z T Z Z u (x, t) − u(x, t)u (y, t) I ≤ C dxdydt 2 d−2 0 Ω Ω |x − y|

Z T Z Z u (y, t) − u(y, t)u(x, t) + C dxdydt d−2 0 Ω Ω |x − y| Z T Z T ≤ C ku − uk 2d kuk 2d dt + C ku − uk 2d kuk 2d dt. (97) d+2 d+2 d+2 d+2 0 L L 0 L L 2d 3d For d+2 ≤ p < d+1 , by using the interpolation inequality and H¨older’sinequality, we obtain

Z T Z T dp+2p−2d dp−2p 2d(p−1) 2d(p−1) ku − uk 2d dt ≤ ku − uk 1 ku − uk p dt d+2 L (Ω) L (Ω) 0 L (Ω) 0 d−2 Z T 2d(p−1) p ≤ C(T ) ku − ukLp(Ω) dt → 0, as → 0. (98) 0 2d For 1 < p < , since ku(·, t)k ∞ d ≤ C and ku(·, t)k ∞ d ≤ C, we have d+2 L (R ) L (R ) T T d+2 Z Z Z 2d 2d p d+2 −p ku − uk 2d dt = |u − u| |u − u| dx dt d+2 0 L (Ω) 0 Ω d+2 Z T Z 2d −p 2d p d+2 ≤ |u − u| kukL∞(Ω) + kukL∞(Ω) dx dt 0 Ω d+2 Z T 2d p ≤ C(T ) ku − ukLp(Ω) dt → 0, as → 0. (99) 0 3d Combining (98) and (99) shows that, for all 1 < p < d+1 , 1 2d u → u, in L 0,T ; L d+2 (Ω) . (100)

Then we have I2 → 0, as → 0. Until now, we obtain Z T Z Z ∇ψ(x, t) − ∇ψ(y, t) · (x − y) u (x, t)u (y, t) dxdydt 2 2 d−2 d d |x − y| + 2 0 R R |x − y|2 + 2 Z T Z Z ∇ψ(x, t) − ∇ψ(y, t) · (x − y) u(x, t)u(y, t) → 2 d−2 dxdydt, (101) d d |x − y| |x − y| 0 R R as → 0. A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 709

(iii) Finally, we will prove Z T Z p−2 |∇u(x, t)| ∇u(x, t) · ∇ψ(x, t) dxdt d 0 R Z T Z → |∇u(x, t)|p−2∇u(x, t) · ∇ψ(x, t) dxdt, (102) d 0 R

as → 0. Since k∇uk p p d ≤ C in (88), we have L R+;L (R ) p ! Z Z p−1 Z Z p−2 p |∇u| ∇u dx dt = |∇u| dxdt ≤ C. d d R+ R R+ R There exists a χ such that

p−2 p p d |∇u| ∇u * χ, in L p−1 R+; L p−1 (R ) . (103) Letting → 0 in (92), we have Z T Z Z T Z Z T Z uψt dxdt − χ · ∇ψ dxdt + u∇v · ∇ψ dxdt 0 d 0 d 0 d R Z R R + u0ψ(0)dx = 0. (104) d R Then we will prove Z T Z Z T Z |∇u|p−2∇u · ∇ψ dxdt = χ · ∇ψ dxdt, (105) d d 0 R 0 R to ﬁnish the proof of the existence of a weak solution for (1). Choosing ∞ d φ(x, t) ∈ Cc [0,T ) × R with 0 ≤ φ ≤ 1, multiplying the ﬁrst equation in (68) by uφ and integrating, we obtain Z T Z Z T Z 1 2 p−2 2 u φt dxdt − |∇u| + |∇u| φ dxdt 2 d d 0 R 0 R Z T Z Z T Z p−2 1 2 − u |∇u| + ∇u · ∇φ dxdt + u ∇v · ∇φ dxdt d 2 d 0 R 0 R Z T Z Z 1 2 1 2 + u J ∗ u φ dxdt + u0(x)φ(x, 0)dx = 0. (106) 2 d 2 d 0 R R p−2 2 p Since |∇u| + |∇u| ≥ |∇u| , then (106) turns to Z T Z Z T Z 1 2 p u φt dxdt − |∇u| φ dxdt 2 d d 0 R 0 R Z T Z Z T Z p−2 1 2 − u |∇u| + ∇u · ∇φ dxdt + u ∇v · ∇φ dxdt d 2 d 0 R 0 R Z T Z Z 1 2 1 2 + u J ∗ u φ dxdt + u0(x)φ(x, 0)dx ≥ 0. (107) 2 d 2 d 0 R R p 1,p d For any ω ∈ L 0,T ; W (R ) to be determined later, we can obtain the following inequality by using Lemma 5.1 Z T Z p−2 p−2 |∇u| ∇u − |∇ω| ∇ω · ∇(u − ω)φ(x, t) dxdt ≥ 0, (108) d 0 R 710 WENTING CONG AND JIAN-GUO LIU

i.e. Z T Z Z T Z p p−2 |∇u| φ dxdt − |∇ω| ∇ω · ∇uφ dxdt d d 0 R 0 R Z T Z Z T Z p−2 p − |∇u| ∇u · ∇ωφ dxdt + |∇ω| φ dxdt ≥ 0. (109) d d 0 R 0 R Combining (107) and (109) together, we have Z T Z Z T Z 1 2 p−2 u φt dxdt − |∇ω| ∇ω · ∇(u − ω)φ dxdt 2 d d 0 R 0 R Z T Z p−2 − |∇u| ∇u · ∇ωφ dxdt d 0 R Z T Z Z T Z p−2 1 2 − u |∇u| + ∇u · ∇φ dxdt + u ∇v · ∇φ dxdt d 2 d 0 R 0 R Z T Z Z 1 2 1 2 + u J ∗ u φ dxdt + u0(x)φ(x, 0)dx ≥ 0. (110) 2 d 2 d 0 R R 1 ∞ d Next we estimate terms in (110) one by one. Since u0 ∈ L ∩ L (R ), p p ku k 1 d = ku k 1 d , ku k ∞ d ≤ C and u → u in L (0,T ; L (Ω)), 0 L (R ) 0 L (R ) 0 L (R ) we have Z T Z Z T Z Z T Z 2 2 u φt dxdt − u φt dxdt ≤ C |u − u|(u + u) dxdt d d 0 R 0 R 0 Ω ≤ C(Ω,T )ku − uk → 0, as → 0, (111) Lp0,T ;Lp(Ω) and Z T Z Z T Z 2 3 u J ∗ uφ dxdt − u φ dxdt d d 0 R 0 R Z T Z Z T Z 2 ≤ (u + u)|u − u|J ∗ u dxdt + u (J ∗ u − u) dxdt 0 Ω 0 Ω Z T Z T p ≤ C ku − uk p kJ ∗ u k p dt + C(Ω,T ) kJ ∗ u − uk p dt L p−1 L 0 L 0 ≤ C(Ω,T )ku − uk → 0, as → 0. (112) Lp0,T ;Lp(Ω) Since p p d ∇u * ∇u, in L R+; L (R ) ,

p−2 p p d |∇u| ∇u * χ, in L p−1 R+; L p−1 (R ) , it is easy to see that Z T Z p−2 |∇ω| ∇ω · ∇(u − ω) − ∇(u − ω) φ dxdt d 0 R Z T Z p−2 = |∇ω| ∇ω · ∇(u − u)φ dxdt → 0, as → 0, (113) d 0 R and Z T Z Z T Z p−2 |∇u| ∇u · ∇ωφ dxdt − χ · ∇ωφ dxdt → 0, (114) d d 0 R 0 R A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 711 as → 0, and Z T Z Z T Z p−2 u |∇u| + ∇u · ∇φ dxdt − uχ · ∇φ dxdt d d 0 R 0 R Z T Z Z T Z p−2 ≤ C u|∇u| dxdt + u |∇u| ∇u − χ · ∇φ dxdt d 0 Ω 0 R Z T Z

+ (u − u)χ · ∇φ dxdt → 0, as → 0, (115) d 0 R where (115) holds from ku k ∞ d ≤ C. From (101) and ku k ∞ d ≤ C, L (R ) L (R ) we can easily obtain Z T Z Z T Z 2 2 u ∇v · ∇φ dxdt − u ∇v · ∇φ dxdt → 0, as → 0. (116) d d 0 R 0 R Then from (111)-(116), letting → 0, (110) turns to Z T Z Z T Z 1 2 p−2 u φt dxdt − |∇ω| ∇ω · ∇(u − ω)φ dxdt 2 d d 0 R 0 R Z T Z Z T Z 1 Z T Z − χ · ∇ωφ dxdt − uχ · ∇φ dxdt + u2∇v · ∇φ dxdt d d 2 d 0 R 0 R 0 R Z T Z Z 1 3 1 2 + u φ dxdt + u0φ(0)dx ≥ 0. (117) 2 d 2 d 0 R R Taking ψ = uφ in (104), we have Z T Z Z T Z Z T Z 1 2 u φt dxdt − χ · ∇uφ dxdt − uχ · ∇φ dxdt 2 d d d 0 R 0 R 0 R Z T Z Z T Z Z 1 2 1 3 1 2 + u ∇v · ∇φ dxdt + u φ dxdt + u0φ(0)dx = 0. 2 d 2 d 2 d 0 R 0 R R (118) Combining (117) and (118) together, we obtain Z T Z Z T Z χ · ∇(u − ω)φ dxdt − |∇ω|p−2∇ω · ∇(u − ω)φ dxdt ≥ 0, (119) d d 0 R 0 R i.e. Z T Z χ − |∇ω|p−2∇ω · ∇(u − ω)φ dxdt ≥ 0. (120) d 0 R Taking ω = u − λψ with λ ≥ 0 yields that Z T Z χ − |∇(u − λψ)|p−2∇(u − λψ) · ∇ψ φ dxdt ≥ 0. (121) d 0 R Choosing φ such that supp ψ ⊂ supp φ and φ = 1 on supp ψ, letting λ → 0, we obtain Z T Z χ − |∇u|p−2∇u · ∇ψ dxdt ≥ 0. (122) d 0 R Using the same method with λ ≤ 0, we have Z T Z χ − |∇u|p−2∇u · ∇ψ dxdt ≤ 0. (123) d 0 R 712 WENTING CONG AND JIAN-GUO LIU

∞ d Then for any ψ ∈ Cc [0,T ) × R , we have Z T Z Z T Z |∇u|p−2∇u · ∇ψ dxdt = χ · ∇ψ dxdt, (124) d d 0 R 0 R i.e. (102) holds. Then combining (i)-(iii) and letting → 0, for any 0 < T < ∞, we have Z T Z Z T Z p−2 u(x, t)ψt(x, t) dxdt = |∇u(x, t)| ∇u(x, t) · ∇ψ(x, t) dxdt d d 0 R 0 R 1 Z T Z Z ∇ψ(x, t) − ∇ψ(y, t) · (x − y) u(x, t)u(y, t) − 2 d−2 dxdydt 2dα(d) d d 0 R R |x − y| |x − y| Z − u0(x)ψ(x, 0)dx, (125) d R 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 global weak solution. Since the proof is almost identical as that for the supercritical case, we omit details. 3d 1 d ∞ d Theorem 5.3. Let d ≥ 3 and p > d+1 . If u0 ∈ L+(R )∩L (R ), 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 this 1 ∞ d section, we prove that for u0 ∈ L+ ∩ L (R ), a weak solution of (1) exists locally without any restriction for the size of initial data. Furthermore, we also prove that if a weak solution blow up in finite time, then all Lh-norms of the weak solution blow up at the same time for h > q. 3d 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 taking h = r > q, we have p p d r r(r − 1)p r−2+p r+1 p ku(·, t)k r d = − ∇u + (r − 1)kuk r+1 d L (R ) p p d L (R ) dt (r − 2 + p) L (R ) p 1 r(r − 1)p r−2+p p 1+ r−q p r ≤ − ∇u + C(r) kuk r d , p p d L (R ) 2(r − 2 + p) L (R ) i.e. 1+ 1 d r r r−q ku(·, t)k r d ≤ C(r) kuk r d . (126) dt L (R ) L (R ) Solving the inequality (126) shows that  r−q r−q r C(r) ku(·, t)k r d ≤   . (127) L (R )  1  r−q r q−r ku0k r d − t C(r) L (R ) 1 r−q r q−r Denoting Tr := ku0k r d , then for any fixed r, we choose 0 < T < Tr. C(r) L (R ) Next by the same way of proving Theorem 5.2,(1) has a local in time weak solution in 0 < t < T . A DEGENERATE p-LAPLACIAN KELLER-SEGEL MODEL 713

Proposition 2. (Blow-up criterion) Under the same assumptions as Theorem 6.1 r r and r = q + where is small enough, let Tmax be the largest L -norm existence time of a weak solution, i.e. r ku(·, t)k r d < ∞, for all 0 < t < T , L (R ) max

lim sup ku(·, t)kLr ( d) = ∞, r R t→Tmax h h and Tmax be the largest L -norm existence time of a weak solution for h ≥ r > q. h Then if Tmax < ∞ for any h, h r Tmax = Tmax, for all h ≥ r.

Proof. Since ku(·, t)k 1 d ≤ ku k 1 d , by interpolation inequality, we know that L (R ) 0 L (R ) h r h r for h ≥ r, Tmax ≤ Tmax. If Tmax < Tmax for any h ≥ r, then we will have h r contradiction arguments. Tmax < Tmax implies

lim sup ku(·, t)kLr ( d) =: A < ∞. h R t→Tmax Then for h ≥ r > q, using (30), we have

1+ h−r+1 d h r r−q ku(·, t)k h d ≤ C(h, r) kuk r d ≤ C(h, r, A), (128) dt L (R ) L (R ) i.e. h h ku(·, t)k h d ≤ C h, r, A, ku0k h d ,T , for t ∈ 0,T , L (R ) L (R ) max max which contradicts with

lim sup ku(·, t)kLh( d) = ∞. h R t→Tmax h r h Thus we have the conclusion that Tmax = Tmax for all h ≥ r > q, i.e. L -norms blow up at the same time.

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

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