Blow-up and global existence for a Chemotaxis-Growthsystem

Kyungkeun Kang(Joint work with A. Stevens )

2016 CMAC WorkshopDepartment of Mathematics

Yonsei UniversitySeoul, KoreaApril 1, 2016

Department of MathematicsYonsei University

Seoul, Korea

Plan

Overview of the talk

• Blow-up and Global existence in Chemotaxis-Growth system

(I) Motivation and review

(II) Main result

(III) Sketch of proofs

1

Introduction

Keller-Segel equations (KSE)

The classical model suggested by Patlak (1953) and Keller-Segel (1970) consists

in the system of the dynamics of the cell density n = n(t, x) and the

concentration of chemical attractant substance c = c(t, x), that isut = ∆u−∇ · (χu∇v),

vt = α∆v − βv + γu,

u(x, 0) = u0(x), v(x, 0) = v0(x).

If Ω is bounded, no-flux boundary conditions are assigned.

∂u

∂ν=

∂v

∂ν= 0 on ∂Ω.

Blow-up results have been known in 2D if initial mass is beyond critical mass (See

e.g. W. Jager & S. Luckhaus (1992) and A. Herrero & J. Velazquez (1997)).

2

Life cycle of Dictyostelium

Figure 1: Life cycle of Dictyostelium Discoideum (from Wikipedia).

3

Introduction

Figure 2: Aggregation (from Wikipedia, the free encyclopedia).

4

Introduction

Chemotaxis-Growth system

Growth and death of a biological cell species are taken into account

∂tu = ε∆u−∇ · (u∇χ(v)) + f(u, v),

τ∂tv = η∆v + g(u, v).

• (Hyperbolic-elliptic case I) We study one of special case of the above equation.

More precisely, we consider ϵ = τ = 0, η = 1, f(u, v) = ru− µu2 and

g(u, v) = −v + u, i.e.

∂tu = −∇ · (u∇v) + ru− µu2,

−∆v = −v + u

Winkler very recently proved global existence of weak solution in 1D case (2014,

JNS). We generalize the results in higher dimensions n ≥ 2.

5

Introduction

• (Hyperbolic-elliptic case II) We also consider more general types of decay

terms, namely f(u, v) = ru− µu1+β , where β > 0 is a constant, that is

∂tu = −∇ · (u∇v) + ru− µu1+β , −∆v = −v + u

As may be expected, we show that if β > 1, solutions globally exist. If β < 1,

solution can develop a singularity in finite time.

• (Parabolic-elliptic case) We also study another special case.

∂tu−∆u = −∇ · (u∇v) + ru− µu2,

−∆v = −v + u.

It was proved by Tello and Winkler (CPDE 2010) that unique bounded solutions

exists globally in time for µ > n−2n with n ≥ 3, in case Ω is bounded. For the

case of n = 2 Osaki, Tsujikawa, Yagi and Mimura (NA;TMA 2002) prove

existence of bounded solutions globally in time for any µ > 0.

6

Main results

Main Result I Let r ≥ 0, µ > 0 and Ω ⊂ Rn for n ≥ 2 either be the whole

space Rn or a bounded convex domain with smooth boundary. Suppose that

non-negative initial datum u0 ∈ W 2,p(Ω) ∩ L1(Ω) for all p with 1 < p < ∞.

Then, there is a maximal time T0 ∈ (0,∞] such that unique non-negative

solutions u ∈ Lp([0, t),W 1,p(Ω)) ∩ L∞([0, t)× Ω) and

v ∈ Lp([0, t),W 2,p(Ω)) ∩ L∞([0, t)× Ω) of (5) exist for t < T0 and the

following holds:

(i) Let µ ≥ 1. Then T0 = ∞. Furthermore, if either r = 0 and Ω = Rn, or if

µ > 1 and Ω is a bounded convex domain, then u is uniformly bounded, i.e.

∥u(t)∥L∞ < C = C(∥u0∥L∞).

(ii) Let 0 < µ < 1. If ∥u0∥Lp is sufficiently large for some 11−µ < p < ∞,

then T0 < ∞ and limt→T0

∥u(t)∥L∞(Ω) = ∞.

7

Main results

• We also consider more general types of decay terms, namely

f(u, v) = ru− µu1+β , where β > 0 is a constant, that is

∂tu = −∇ · (u∇v) + ru− µu1+β , −∆v = −v + u. (1)

Main Result II Let r ≥ 0, µ > 0, β > 0 with β = 1 and Ω ⊂ Rn for n ≥ 2

either be the whole space Rn or a bounded convex domain with smooth

boundary. Suppose that non-negative initial datum u0 ∈ W 2,p(Ω) ∩ L1(Ω) for

all p with 1 < p < ∞. Then, there is a maximal time T1 ∈ (0,∞] such that

unique non-negative solutions u ∈ Lp([0, t),W 1,p(Ω)) ∩ L∞([0, t)× Ω)

and v ∈ Lp([0, t),W 2,p(Ω)) ∩ L∞([0, t)× Ω) of (1) exist for any time

t < T1 and the following is satisfied:

(i) If β > 1, then T1 = ∞.

(ii) If 0 < β < 1 and if ∥u0∥Lp is sufficiently large for any 1 < p < ∞, then

T1 < ∞ and limt→T1

∥u(t)∥L∞(Ω) = ∞.

8

Main results

• Thirdly, we consider the parabolic-elliptic case, i.e. ϵ = 1

∂tu = ∆u−∇ · (u∇v) + ru− µu2,

−∆v = −v + u.

with initial condition u(x, 0) = u0(x). As mentioned earlier, it was proved by

Tello and Winkler (CPDE 2010) that unique bounded solutions exists globally in

time for µ > n−2n with n ≥ 3, in case Ω is bounded.

Main Result III Let r ≥ 0, µ > 0 and Ω ⊂ Rn for n ≥ 3 be either the whole

space Rn or a bounded domain with smooth boundary. Suppose that

non-negative initial datum u0 ∈ W 2,p(Ω) ∩L1(Ω) for all p with 1 < p < ∞. If

µ ≥ n−2n , then non-negative classical solutions exist globally.

9

Sketch of proof

Hyperbolic-Elliptic Case

We consider the chemotaxis-growth system

∂tu− ϵ∆u = −∇ · (u∇v) + ru− µu2, (2)

0 = ∆v − α1v + α2u, (3)

where ϵ > 0, r ≥ 0, µ > 0, α1 > 0 and α2 > 0 are constants. In the sequel,

without loss of generality, we assume that α1 = α2 = 1. As ϵ → 0, the limiting

equation formally becomes

∂tu = −∇ · (u∇v) + ru− µu2, (4)

0 = ∆v − v + u. (5)

• We say that u and v are regular solutions if u, v ∈ L∞x,t ∩ L∞

t W 2,px for any

1 < p < ∞, up to the time that these solutions exist.

10

Sketch of proof

• (A priori estimates) Let u and v be non-negative regular solutions If r ≥ 0 and

µ ≥ 1, then u is bounded for any time and satisfies

∥u(t)∥Lp ≤ ∥u0∥Lp exp(rt), 1 ≤ p ≤ ∞.

Moreover, if r = 0, µ ≥ 1, and Ω = Rn or if r ≥ 0, µ > 1, and Ω is a bounded

domain, then u is uniformly bounded. For r ≥ 0 and 0 < µ < 1, there exists

p0 > 1 such that ∥u(t)∥Lp blows up at a finite time for any p with p0 < p ≤ ∞.

▷ Indeed, Testing up−1, we note that

1

p

d

dt

∫Ω

up = −p− 1

p

∫Ω

∆vup + r

∫Ω

up − µ

∫Ω

up+1

= (p− 1

p− µ)

∫Ω

|u|p+1 − p− 1

p

∫Ω

vup + r

∫Ω

up.

11

Sketch of proof

Therefore, in case that µ ≥ 1, we obtain

1

p

d

dt

∫Ω

up ≤ r

∫Ω

up.

Gronwall’s inequality implies that

∥u(t)∥Lp ≤ ∥u0∥Lp exp(rt), 1 ≤ p < ∞,

which is also valid for p = ∞. Hence, we obtain (11). We also note that if r = 0,

then the Lp norm of u is uniformly bounded.

In the case of bounded domains, since µ > 1, we note that p−1p − µ is strictly

negative for any p. Indeed, say µ = 1 + δ with δ > 0, we see thatp−1p − µ = −( 1p + δ) ≤ −δ < 0. Since a domain is bounded, we see due to

Holder’s inequality that for any ϵ = ϵ(µ, r) < δ2r∫

Ω

up ≤ ϵ

∫Ω

up+1 + Cϵ |Ω| .

12

Sketch of proof

Therefore, if Ω is bounded, µ > 1 and r ≥ 0, we have

1

p

d

dt

∫Ω

up ≤(p− 1

p− µ

)∫Ω

|u|p+1+r

∫Ω

up ≤ −δ

2

∫Ω

up+1+rCϵ |Ω| .

Hence, there exists C = C(µ, r,Ω) such that (6) becomes

1

p

d

dt

∫Ω

up ≤ C =⇒ ∥u(t)∥Lp ≤ (Cpt+ ∥u0∥pLp)1p .

Passing p → ∞, we have ∥u(t)∥L∞(Ω) ≤ 1 + ∥u0∥L∞(Ω).

(Blow-up case) If µ < 1, then there exists q > 1 such that a := q−1q − µ > 0

and thus we have

1

q

d

dt

∫Ω

uq = a

∫Ω

uq+1 + r

∫Ω

uq − q − 1

q

∫Ω

vuq. (6)

13

Sketch of proof

We note first that via Sobolev embedding and elliptic estimates of v we get

∥v∥Lq+1(Ω) ≤ C ∥v∥W 1,q(Ω) ≤ C ∥u∥Lq(Ω) ,1

q=

1

q + 1+

1

n.

Using interpolation,

∥v∥Lq+1(Ω) ≲ ∥u∥Lq(Ω) ≲ ∥u∥1−θL1(Ω) ∥u∥

θLq+1(Ω) , θ =

q(n− 1)− 1

qn.

Via above estimate, we estimate (6) as follows:

1

q

d

dt∥u∥qLq(Ω) ≥ a ∥u∥q+1

Lq+1(Ω) + r ∥u∥qLq(Ω) − C ∥u∥1−θL1(Ω) ∥u∥

q+θLq+1(Ω)

≥ a

2∥u∥q+1

Lq+1(Ω) + r ∥u∥qLq(Ω) − C ∥u∥q+1L1(Ω) , (7)

where C = C(a, q) and Young’s inequality is used.

14

Sketch of proof

If Ω is bounded, we note that total mass of u is uniformly bounded, i.e.

∥u(t)∥L1(Rn) ≤ C , because

d

dt

∫Ω

u = r

∫Ω

u− µ

∫Ω

u2 ≤ r

∫Ω

u− µ

|Ω|

(∫Ω

u

)2

,

where we used (∫Ωu)2 ≤ |Ω|

∫Ωu2. Since

∫Ωuq+1 ≥ CΩ(

∫Ωuqdx)

q+1q ,

we obtain

1

q

d

dt∥u∥qLq(Ω) ≥

aCΩ

2∥u∥q+1

Lq(Ω) + r ∥u∥qLq(Ω) − C.

Since a > 0, and if ∥u0∥Lq(Ω) is sufficiently large, standard ODE theory implies

that the Lq norm of u blows up in finite time. This automatically implies that u

becomes unbounded in finite time.

15

Sketch of proof

In case Ω = Rn, we observe that ∥u(t)∥L1(Rn) ≤ ∥u0∥L1(Rn) ert, because

d

dt

∫Ω

u = r

∫Ω

u− µ

∫Ω

u2 ≤ r

∫Ω

u .

Again, we suppose that r > 0. If r = 0, then ∥u(t)∥L1(Rn) ≤ ∥u0∥L1(Rn),

which is uniformly bounded. In this case, blow-up can be more easily seen, than

in the case r > 0. From (7) we have

1

q

d

dt∥u∥qLq(Rn) ≥

a

2∥u∥q+1

Lq+1(Rn)+r ∥u∥qLq(Rn)−C ∥u0∥q+1L1(Rn) e

r(q+1)t.

Via interpolation, we note that

∥u∥Lq(Rn) ≤ ∥u∥θL1(Rn) ∥u∥1−θLq+1(Rn) , θ =

1

q2. (8)

16

Sketch of proof

We obtain

1

q

d

dt∥u∥qLq(Ω) ≥

a

2∥u∥−

1q−1

L1(Rn) ∥u∥q2

q−1

Lq(Rn)+r ∥u∥qLq(Rn)−C ∥u0∥q+1L1(Rn) e

r(q+1)t

≥ Ce−r

q−1 t

(∫Rn

uqdx

) qq−1

+ r ∥u∥qLq(Rn) − Cer(q+1)t

≥ Ce−r

q−1 t

((∫Rn

uqdx

) qq−1

− Cerq2

q−1 t

)+ r ∥u∥qLq(Rn) . (9)

If ∥u0∥Lq is sufficiently large, i.e. ∥u0∥Lq > C(∥u0∥L1), then one can show

that ∥u(t)∥Lq grows faster than ∥u0∥Lq ert. Define y(t) := ∥u(t)∥qLq(Rn).

We then observe that((∫Rn

uqdx

) qq−1

− Cerq2

q−1 t

)= (y(t))

qq−1 − (Cerqt)

qq−1

≥ (y(t))− Cerqt)(y(t))1

q−1 .

17

Sketch of proof

Then from inequality (9) we can derive:

d

dty(t) ≥ Ce−

rq−1 t

(y(t)− Cerqt

)(y(t))

1q−1 + rqy(t)

≥ Ce−r

q−1 t(y(t))q

q−1 + rqy(t). (10)

Therefore, we have

d

dt∥u∥Lq ≥ Ce−

rq−1 t ∥u∥

2q−1q−1

Lq + r ∥u∥Lq ≥ C ∥u∥q

q−1

Lq + r ∥u∥Lq .

Since q/(q − 1) > 1, this estimate gives blow up in finite time, which completes

the proof.

18

Sketch of proof

• (A priori gradient estimates) Let r ≥ 0, µ > 0, n ≥ 2 and u and v be regular

solutions of (4)-(5) in Ω× [0, T ]. Suppose that sup0≤t≤T

∥u(t)∥L∞(Ω) < ∞.

Then ∇u belongs to L∞([0, T ];Lp(Ω)) for all p < ∞ and for all times

T < ∞, and it satisfies

∥∇u(t)∥Lp ≤ exp (log (1 + ∥∇u0∥Lp) exp(Ct)) , t ≤ T, (11)

where C = C(∥u(t)∥L∞).

▷ First we consider the case Ω = Rn. It was proven by Kozono and Taniuchi

(CMP 2000) that for 1 < p < ∞ and s < p/n we have

∥f∥L∞(Rn) ≤ C(1 + ∥f∥BMO (1 + log+ ∥f∥W s,p(Rn))

), (12)

where log+ k = log k for k > 1, and log+ k = 0 otherwise.

19

Sketch of proof

Since ∇ · (u∇v) = ∇u · ∇v + u∆v = ∇u · ∇v+ u(v− u), we can rewrite

(4) as follows:

ut = −∇u · ∇v + (r − v)u+ (1− µ)u2.

For convenience let w = uxk, k = 1, 2, · · · , n. Then w solves

wt = −∇w · ∇v −∇u · ∇vxk− vxk

u+ (r − v)w + 2(1− µ)uw. (13)

Testing (13) with |w|q−2w, we obtain for q > n that

1

q

d

dt∥w∥qLq ≤ q − 1

q

∫Rn

|∆v| |w|q +∫Rn

|∇u|∣∣∇2v

∣∣ |w|q−1

+

∫Rn

|u| |∇v| |w|q−1+ C

∫Rn

(1 + |u|+ |v|) |w|q .

Adding the above inequality for all k = 1, 2, · · · , n, we get

1

q

d

dt∥∇u∥qLq ≤ C

∫Rn

∣∣∇2v∣∣ |∇u|q+C(1+∥u∥L∞(Rn)+∥v∥L∞(Rn))

∫Rn

|∇u|q .

20

Sketch of proof

Since v is bounded we obtain via standard elliptic estimates that for all t ≤ T

1

q

d

dt∥∇u∥qLq ≤ C

∫Rn

∣∣∇2v∣∣ |∇u|q + C

∫Rn

|∇u|q

≤ C(∥∥∇2v

∥∥L∞ + 1

)∥∇u∥qLq . (14)

Using (12), we have that∥∥∇2v∥∥L∞ ≤ C

(1 +

∥∥∇2v∥∥BMO

(1 + log∥∥∇2v

∥∥W 1,q )

), q > n.

(15)

Via BMO and Lp estimates of singular integrals, we have∥∥∇2v∥∥BMO

≤ C ∥u∥BMO ≤ C ∥u∥L∞ ,∥∥∇2v

∥∥W 1,q ≤ C ∥∇u∥Lq .

(16)

Combining all estimates, we obtain

1

q

d

dt∥∇u∥qLq ≤ C

(1 + ∥u∥L∞ (1 + log+ ∥∇u∥Lq )

)∥∇u∥qLq . (17)

21

Sketch of proof

Gronwall’s inequality gives ∇u ∈ L∞([0, T ];Lp(Rn)) and estimate (11). In

case of a bounded domain, Ogawa and Taniuchi (JDE2003) that (12) is also valid.

Thus, following the computations given above, we obtain the same result.

Next, we construct solutions satisfying the a priori estimates and show that the

constructed solution is unique. We recall the chemotaxis-growth system with

diffusion

∂tuϵ − ϵ∆uϵ = −∇ · (uϵ∇vϵ) + ruϵ − µu2ϵ , (18)

0 = ∆vϵ − vϵ + uϵ. (19)

• (Local existence of regular solutions of (18)-(19)) Let ϵ > 0, r ≥ 0 and µ > 0.

Suppose that u0 ∈ W 2,p(Ω) ∩ L1(Ω) for all p with 1 < p < ∞. Then there is

a maximal time Tϵ ∈ (0,∞], such that unique classical solutions

u, v ∈ Lp([0, t),W 2,p(Ω)) ∩ L∞([0, t)× Ω) of (18)-(19) exist for any time

t < Tϵ. If Tϵ < ∞, then limt→Tϵ

∥u(t)∥L∞(Ω) = ∞.

22

Sketch of proof

• Sketch of proof of main result I It suffices to show local existence and

uniqueness, since the other parts were already proved above. With the aid of

local existence of regular solutions of (18)-(19), we have a sequence of solutions

uϵ, vϵ of (18)-(19). For given ϵ > 0, a bound for ∥uϵ(t)∥L∞ can be

established locally in time, independently of ϵ > 0. In other words, there exists

T1, independent of ϵ, such that ∥uϵ(t)∥L∞ < ∞ for all t ≤ T1. This can be

shown by ∥∇uϵ(t)∥Lp < ∞ for all t ≤ T1 and p < ∞. This is also

independent of ϵ, since the estimate is depending on ∥uϵ(t)∥L∞ and the initial

data. By means of Lp−estimates of elliptic equations, we note due to (19) that

∥vϵ(t)∥W 3,p < ∞ for all t ≤ T1 and p < ∞. We then observe from the

equation (18) that ∂tuϵ is, independent of ϵ, in L1([0, T1];W−1,p(Ω)) for any

p < ∞. Therefore, we have

uϵ −→ u weakly∗ in L∞([0, T1]× Ω), (20)

uϵ −→ u weakly in Lp([0, T1];W1,p(Ω)), p < ∞. (21)

23

Sketch of proof

By Aubin-Lions compactness result, we can conclude that

uϵ −→ u strongly in Lp([0, T1];Lploc(Ω)), p < ∞. (22)

Via elliptic regularity, we also note that

vϵ −→ v strongly in Lp([0, T1];W2,ploc (Ω)), p < ∞. (23)

Let ξ be a smooth function and compactly supported in QT1 := (0, T1)× Ω.

Since uϵ and vϵ solve (18)-(19) weakly as well as classically at least up to time

T1, we have∫QT1

uϵξt−ϵ

∫QT1

∇uϵ·∇ξ+

∫QT1

uϵ∇vϵ·∇ξ+

∫QT1

ruϵξ−∫QT1

µu2ϵξ = 0,

−∫QT1

∆vϵξ +

∫QT1

vϵξ −∫QT1

uϵξ = 0.

24

Sketch of proof

For ϵ → 0, we obtain∫QT1

uξt +

∫QT1

u∇v · ∇ξ +

∫QT1

ruξ −∫QT1

µu2ξ = 0,

−∫QT1

∆vξ +

∫QT1

vξ −∫QT1

uξ = 0,

where we used the weak and strong convergence results (21)-(22). Collecting all

T1 satisfying the above equations and taking their supremum, we can establish

the maximal time of existence, say T0.

We can also show the uniqueness of solutions as well as non-negativity of

solutions. Uniqueness is proved as follows: Suppose there are two solutions

(u1, v1) and (u2, v2). For convenience define δu := u1 − u2 and

δv := v1 − v2. Then (δu, δv) solve the following equation in a weak sense:

25

Sketch of proof

δut = −∇ · (δu∇v1) +∇ · (u2∇δv) + rδu− µ(u1 + u2)δu, (24)

0 = ∆δv − δv + δu. (25)

In case n ≥ 3,

1

2

d

dt∥δu∥2L2 ≤

(1

2∥u1∥L∞ + C ∥∇u2∥Ln + ∥u2∥L∞ + r

)∥δu∥2L2 .

Similarly, for n = 2 we have

1

2

d

dt∥δu∥2L2 ≤

(1

2∥u1∥L∞ + C ∥∇u2∥L4 + ∥u2∥L∞ + r

)∥δu∥2L2 .

Gronwall’s inequality implies that for any t < T0

∥δu(t)∥2L2 ≤ ∥δu(0)∥2L2 exp

(C

∫ t

0

(∥u1∥L∞ + ∥∇u2∥L4 + ∥u2∥L∞ + r)dτ

).

26

Sketch of proof

• Sketch of proof of main result II We first consider the case β > 1. In this

case, we can show that solutions exist globally. Indeed, testing up−1, we have

1

p

d

dt

∫Ω

up =p− 1

p

∫Ω

up+1 − µ

∫Ω

up+β − p− 1

p

∫Ω

vup + r

∫Ω

up

≤ p− 1

p

∫Ω

(Cϵu

p + ϵup+β)− µ

∫Ω

up+β + r

∫Ω

up ≤ C

∫Ω

up.

Thus, for any time T < ∞ we obtain that u ∈ L∞([0, T ];Lp(Ω)) for all

1 ≤ p ≤ ∞. Following similar computations as before, one can also prove that

∇u ∈ L∞([0, T ];Lp(Ω)) for all 1 < p < ∞. We omit details here.

On the other hand, for 0 < β < 1, we can show that blow-up in finite time can

happen, depending on the size of the initial data. Indeed, first consider the case

that Ω is a bounded domain. We note that total mass is uniformly bounded, i.e.

∥u(t)∥L1 ≤ C for any t > 0, (26)

27

Sketch of proof

because

d

dt

∫Ω

u = r

∫Ω

u− µ

∫Ω

u1+β ≤ r

∫Ω

u− µ

|Ω|β

(∫Ω

u

)1+β

,

where we used that (∫Ωu)1+β ≤ |Ω|β

∫Ωu1+β . We then have

1

p

d

dt

∫Ω

up =p− 1

p

∫Ω

up+1 − p− 1

p

∫Ω

vup − µ

∫Ω

up+β + r

∫Ω

up.

Via interpolation, we observe that ∥u∥Lp+β ≤ ∥u∥θL1 ∥u∥1−θLp+1 , where

θ = 1−βp(p+β) . Thus for any η > 0, with the aid of (14) equation (28) becomes

d

dt∥u∥pLp ≥ (p−1)(1−2η) ∥u∥p+1

Lp+1−Cη ∥u∥p+1L1 −Cη ∥u∥L1+rp ∥u∥pLp

≥ (p− 1)(1− 2η) ∥u∥p+1Lp+1 − C + rp ∥u∥pLp ,

28

Sketch of proof

where we used ∥u∥(p+β)θL1 ∥u∥(1−θ)(p+β)

Lp+1 ≤ Cη ∥u∥L1 +ηpµ ∥u∥p+1

Lp+1 via

Young’s inequality. The above estimate leads to blow-up in finite time, provided

that ∥u0∥Lp is sufficiently large. Since the verification of this result is a repetition

of the arguments as in previous case, details are omitted here.

The case Ω = Rn can be treated in a similar way. Hence, we just indicate the

differences, compared to the case for bounded domains. First, we have

∥u(t)∥L1 ≤ ∥u0∥L1 ert for any t > 0 . (27)

Since ∥u∥Lp is not smaller or equal than CΩ ∥u∥Lp+1 in Rn, we use the

interpolation argument (8) by replacing q by p. This leads to

1

p

d

dt∥u∥pLp(Ω) ≥ (p− 1)(1− 2η) ∥u∥−

1p−1

L1(Rn) ∥u∥p2

p−1

Lp(Rn)

+r ∥u∥pLp(Rn) − C(1 + ∥u0∥L1(Rn))p+1er(p+1)t

29

Sketch of proof

≥ Ce−r

q−1 t

(∫Rn

uqdx

) qq−1

+ r ∥u∥qLq(Rn) − Cer(q+1)t

≥ Ce−r

q−1 t

((∫Rn

uqdx

) qq−1

− Cerq2

q−1 t

)+ r ∥u∥qLq(Rn) .

This estimate is exactly the same as before, and thus blow up in finite time occurs.

This completes the proof.

30

Sketch of proof

Parabolic-Elliptic Case

We study the following parabolic-elliptic model

∂tu− ϵ∆u = −χ∇ · (u∇v) + f(u), (28)

−∆v = −v + u. (29)

Here we assume the following structure for f :

For bounded domains, we suppose that f(s) ≤ C1 − µs2, s ≥ 0.

If Ω = Rn, we suppose that f(s) ≤ rs− µs2, s ≥ 0.

As mentioned earlier, it was proved by Tell and Winkler (CPDE 2010) that unique

bounded solutions exists globally in time for µ > n−2n with n ≥ 3, in case Ω is

bounded. For the case of n = 2 Osaki, Tsujikawa, Yagi and Mimura (NA;TMA

2002) prove existence of bounded solutions globally in time for any µ > 0.

We treat the limiting case µ = n−2n , n ≥ 3 for bounded domains as well as for

Rn and we prove that regular solutions exist globally in time.

31

Sketch of proof

• Sketch of proof of main result III

For simplicity, we assume f(s) = rs− µu2. We only consider the case of

bounded domains. For Rn, arguments work similarly with slight modifications.

Suppose that solutions blow up in finite time, that is Tϵ < ∞. First we observe

that for all t < Tϵ

∥u(t)∥Lp < ∥u0∥Lp + C for any 1 ≤ p ≤ p0 :=1

1− µ=

n

2, (30)

where C = C(Tϵ). Indeed, testing (28) with up−1, we get

1

p

d

dt∥u(·, t)∥pLp+

4(p− 1)

p2

∥∥∥∇up2

∥∥∥2L2

=

∫Ω

(p− 1

p(u− v)u+ ru− µu2

)up−1

= r

∫Ω

up − p− 1

p

∫Ω

vup +

∫Ω

(p− 1

p− µ

)up+1. (31)

Note that for 1 ≤ p ≤ p0 := 11−µ , the last term becomes non-positive.

32

Sketch of proof

Next, we show that there exists δ = δ(Tϵ, ∥u0∥Lp0 ) > 0 such that∫Ω

up(x, t)dx < C = C(Tϵ, ∥u0∥Lp), for any p0 < p ≤ p0+δ. (32)

Let p > p0, which will be specified later. Set w := up2 and rewrite (31) as

1

p

d

dt∥w(·, t)∥2L2 +

4(p− 1)

p2∥∇w∥2L2

= r

∫Ω

w2 − p− 1

p

∫Ω

vw2 + Cp,µ

∫Ω

w2(p+1)

p ,

where Cp,µ = p−1p − µ > 0. We then estimate the last term in (33) by∫

Ω

w2(p+1)

p = ∥w∥2(p+1)

p

L2(p+1)

p (Ω)

≤ C

(∥w∥

1p+1

L2p0p

∥w∥p

p+1

L2n

n−2

) 2(p+1)p

≤ C ∥u∥Lp0 ∥w∥2

L2n

n−2≤ C ∥u∥Lp0

(∥∇w∥2L2 + ∥w∥2L2

). (33)

33

Sketch of proof

Therefore, combining (30) and (33), we observe that

∥w∥2(p+1)

p

L2(p+1)

p (Ω)

≤ C1

(∥∇w∥2L2 + ∥w∥2L2

), (34)

where C1 = C(Tϵ, ∥u0∥Lp). Using (34), we estimate (33) as follows:

1

p

d

dt∥w(·, t)∥2L2 +

4(p− 1)

p2∥∇w∥2L2 +

p− 1

p

∫Ω

vw2

≤ C

∫Ω

w2 + Cp,µC1

(∥∇w∥2L2 + ∥w∥2L2

).

Taking δ > 0 sufficiently small with p = p0 + δ and thus Cp,µ as small as

necessary, such that Cp,µC1 < 2(p−1)p2 we obtain, by shifting ∥∇w∥2L2 to the

left hand side, that

1

p

d

dt∥w(·, t)∥2L2 +

2(p− 1)

p2∥∇w∥2L2 ≤ (C + Cp,µC1) ∥w∥2L2 .

Gronwall’s inequality implies assertion (32), returning back to original variable u.

34

Sketch of proof

Next we show that for any p with p0 + δ ≤ p < ∞ we have∫Ω

up(·, t)dx < C = C(Tϵ, ∥u0∥Lp). (35)

Let p1 := p0 + δ = n2 + δ. Performing similar computations as in (33), we

obtain for any p > p1 that∫Ω

w2(p+1)

p ≤ C ∥u∥(1−θ)(p+1)Lp1 ∥w∥θ

2(p+1)p

L2n

n−2, θ =

pn(p+ 1− p1)

(p+ 1)(pn− p1n+ 2p1).

Recalling the Sobolev embedding ∥w∥L

2nn−2

≤ C ∥w∥W 1,2 , we get∫Ω

w2(p+1)

p ≤ C ∥w∥2n(p+1−p1)

pn−p1(n−2)

L2n

n−2≤ C (∥w∥L2 + ∥∇w∥L2)

2n(p+1−p1)

pn−p1(n−2) .

35

Sketch of proof

We note that 2n(p+1−p1)pn−p1(n−2) < 2, since p1 > n

2 . Hence, we obtain∫Ω

w2(p+1)

p ≤ C ∥w∥2n(p+1−p1)

pn−p1(n−2)

L2n

n−2≤ C + C ∥w∥2L2 + σ ∥∇w∥2L2 , (36)

where σ is small and specified later. Combining estimate (33) and (36), we have

1

p

d

dt∥w(·, t)∥2L2+

4(p− 1)

p2∥∇w∥2L2 ≤ C+(r+C) ∥w∥2L2+σCp,µ ∥∇w∥2L2 .

Taking σ sufficiently small so that σCp,µ < 2(p−1)p2 , we get

1

p

d

dt∥w(·, t)∥2L2 +

2(p− 1)

p2∥∇w∥2L2 ≤ C + (r + C) ∥w∥2L2 . (37)

This estimate automatically implies assertion (35). In addition, since the constant

C in (37) is independent of p, we also have a uniform bound for the L∞-norm of

u for all t < Tϵ. This is in contrast to our hypothesis, and therefore, Tϵ = ∞.

This completes the proof.

36

THANK YOU !

37