Blow-up

for all t≥0, and this is a contradiction. Thus, our initial assumption is false, and we have established the Lemma.

Given the above-stated Lemmas, we can therefore prove the following Lemma.

Lemma 5.6. For almost allβ >4π, σβ is a critical value forEβ. Moreover, problem (5.4) admits a nontrivial solution for almost all β >4π.

Proof. This Lemma is established directly by Lemma 5.3.

Then, using (5.22), we have







∆v_n^∗ −αv_n^∗+µne^κvn∗ = 0 inΩ, R

Ωv_n^∗ dx= βn

α , ∂v_n^∗

∂~n = 0 on ∂Ω = Γ, (5.23)

whereµ_n →0. As in the previous section, for simplicity, we will drop the ∗ on v_n. By the Maximum Principle for elliptic operator, vn>0 inΩ. We want to show, following [37, 83], that asn→ ∞,

µ_n Z

Ω

e^κvn dx→4mπ, (5.24)

for some integer m. This means that β_n → 4mπ and β₀₀= 4mπ, which is a contradiction since we assumed earlier that β₀₀ 6= 4mπ. It remains to verify (5.24). We will be following an approach in [37, 83, 84]. We start by stating the following Lemma which was proven by Chanillo and Yangan Li in [14].

Lemma 5.7. Let L=

2

P

i,j=1

aij ∂²

∂xi∂xj be a uniformly elliptic operator, namely ν0I ≤(aij)1≤i,j≤2 ≤ν1I.

Then there exists a constant φ=φ(ν0, ν1) such that for any solution v of the problem

Lv=f(x) in Ω, v= 0 on ∂Ω, we have

Z

Ω

exp φ|v(x)|

kfk_L1(Ω)

!

dx≤C.

Next, we recall that by elliptic estimates, we have Z

Ω

|∇v_n|^q dx+ Z

Ω

|v_n|^qdx≤C, (5.25)

for any 1< q <2.

To analyse v_n, we rst introduce the following set. Let

S :=





 x∈Ω¯

there existsµn→0,solutionsvn of (5.23), xn∈Ω¯ such thatvn(xn)→ ∞, x_n→x







. (5.26)

Let Λ_n=R

Ωµ_ne^κvn dx.² Now, sincev_n satises Z

Ω

µ_ne^κvn

Λ_n dx= 1,

we can extract a subsequence ofv_n, still denoted byv_n(for simplicity) such that there exists a positive nite measure µin the set of all real bounded Borel measures onΩ¯,M( ¯Ω), such that asn→ ∞,

Z

Ω

µ_ne^κvn Λn

ϕ dx→ Z

Ω

ϕ dµ, (5.27)

for allϕ∈C₀^∞(R²). Let

v_n^∗ = v_n Λn

, g_n= µ_ne^κvn Λn

.

For each x0 ∈ ∂Ω, we can nd a smooth function Φx0 and a small positive constant r_x0 >0 such that

Φx0(x) :Brx0(x0)∩Ω¯ →Brx0(0)∩R²+, (5.28) in which R²+ = {(x₁, x₂)|x₂ > 0}. Then the Laplace operator ∆ becomes L_x0 +

2

P

l=1

b_l∂x^∂

l, where L_x0 =

2

P

i,j=1

a_ij∂x^∂2

i∂xj is a uniformly elliptic operator and |b_l| ≤ C = const. By the compactness of the boundary, we can choose a uniform φ = φ₀ in Lemma 5.7 for all Lx0, x0 ∈∂Ω.

For the sake of clarity, we now state the denition of δ−regular points of Ω¯ [37].

2Note thatΛn=βn.

Denition 5.2. For any δ > 0, we say that x₀ is a δ−regular point if there is a function ϕ∈C₀^∞(R²), 0≤ϕ≤1, withϕ= 1 in a neighbourhood ofx0, such that

Z

Ω

ϕ dµ < φ0

1 + 3δ. (5.29)

Let's denote the set of all points inΩ¯ which are notδ−regular by

Λ(δ) ={x₀|x₀ is not a δ−regular point}. (5.30) If no confusion is foreseen in the following, we will use the reference `regular', `irregular' and

`Λ' without mentioningδ.

Also, in the following, we notice that, by assumption, the set S dened in (5.26), is not empty. We therefore have the following Lemma which we will be using later in the proof for Lemma 5.11. See [37] among others.

Lemma 5.8. Let1< q <2. Then there exists a constant Cq, not dependent of n, such that k∇v_nk_Lq(Ω) ≤C_q.

Proof. Letq⁰ = _q−1^q >2. Then we know that k∇v_nk_Lq(Ω)≤sup

| Z

Ω

∇v_n· ∇ϕ dx| : ϕ∈L^q₁⁰(Ω), Z

Ω

ϕ dx= 0, kϕk

L^q₁⁰(Ω)= 1

. By using the Sobolev embedding theorem, we get that

kϕk_L∞(Ω) ≤C₁. Using the fact thatvn>0, we see that

| Z

Ω

∇v_n· ∇ϕ dx| = | Z

Ω

∆vnϕ dx|

= | Z

Ω

(αv_n−µ_ne^κvn)ϕ dx|

≤ C₁ Z

Ω

(v_n+µ_ne^κvn)dx

≤ C₂.

We thus have the following Lemmas from [83].

Lemma 5.9. If x0 is a δ−regular point, then {v_n} is bounded in L^∞(BR0(x0)) for some R₀ >0.

Proof. Let x0 be a regular point. We will give the proof for the case when x0 ∈ ∂Ω. The case for when x₀ is an interior point is simpler and can be proven in a similar way.

It follows from the denition of a regular point that there exists R₁ >0 such that Z

B_R1(x0)∩Ω¯

gndx < φ₀ 1 + 3δ.

Let's pick r < R₁, a small number. At x₀, since ^∂v_∂~nⁿ = 0, we can strengthen the boundary near B_R1(x0)∩Ω¯ by Φx0 dened in (5.28), and then extend vn by even extension (still denoted byvn) to

Lx0v_n^∗+

2

X

l=1

bl

∂v_n^∗

∂x_l −αv^∗_n+µne^κvn

Λ_n = 0 inBρ(0), where Lx0 =

2

P

i,j=1

aij ∂²

∂xi∂xj is a uniformly elliptic operator, |b_l| ≤ C and ρ = ρ(r) → 0 as r→0. We can pickr so small that

Z

Bρ(0)

gndx < φ0

1 + 3δ and, by (5.25)

Z

Bρ(0)

|

2

X

l=1

b_l∂v_n^∗

∂xl

|+αv_n^∗ ≤ Cαkv^∗_nk_W1,q(Ω)ρ

q−1 q

< Cρ

q−1 q

< φ0δ (1 + 2δ)(1 +δ). Thus

Z

Bρ(0)

|

2

X

l=1

b_l∂v_n^∗

∂x_l|+αv_n^∗ +gn

!

dx < φ0

1 +δ. (5.31)

Next, we split v^∗_n into two parts such that

v^∗_n=v_1n^∗ +v^∗_2n, in which v_1n^∗ is the solution for







Lx0v^∗_1n=−

2

P

l=1

bl∂v^∗_1n

∂xl +αv_1n^∗ −gn inBρ(0)

v_1n^∗ = 0 on ∂Bρ(0),

(5.32)

whilev_2n^∗ is the solution for







Lx0v_2n^∗ = 0 inBρ(0) v_2n^∗ =v^∗_n on ∂Bρ(0),

(5.33) Note that by the Maximum principle,v^∗_2n>0, and by (5.30) and (5.31),kv_1n^∗ k_L1(Bρ(0))≤ C. Thus, we get that R

Bρ(0)v_2n^∗ dx <R

Bρ(0)(v_n^∗ +|v^∗_1n|) dx≤C. Using Harnack inequality [19], we obtain that

kv_2n^∗ k_L^∞_(B

ρ/2(0))≤Ckv^∗_2nk_L1(Bρ(0)) ≤Ckv_n^∗k_L1(Ω)≤C.

Therefore, we only need to considerv_1n^∗ .

Now, using (5.31) and Lemma 5.7, we get that Z

Bρ(0)

exp

1 +δ 2

|v^∗_1n|

dx≤C. (5.34)

Sincev_2n^∗ is uniformly bounded, we thus have Z

B_ρ/2(0)

|

2

X

l=1

bl

∂v_n^∗

∂xl

|+αv_n^∗+gn

!1+δ

dx≤C (5.35)

on B_ρ/2(0). From the elliptic estimates, we get kv^∗_1nk_L∞(Bρ/4(0)) ≤ C, and the proof is complete.

From above, the following result is immediate.

Lemma 5.10. S = Λ(δ) ∀δ >0.

Proof. Let δ > 0 and suppose that x0 ∈/ Λ(δ). Then x0 is a regular point, so that, by Lemma 5.9, {v^∗_n} is bounded inL^∞(BR(x0))∩Ω¯ for some R >0. That is,x0∈/ S and thus S ⊂Λ(δ).

Conversely, suppose thatx₀∈Λ(δ). Then for everyR >0, we have that

x→∞lim kv_nk_L^∞_(BR(x0)) =∞. (5.36) Otherwise, there would be someR0>0and a subsequence, still denoted by vn, such that

kv_nk_L^∞_(BR(x0))< C,

for some constantC, not dependent onn. This would imply that µ_ne^vn ≤Cµ_n

uniformly asn→ ∞ on BR0(x0))∩Ω¯, so that Z

BR(x0)∩Ω¯

µne^vn dx≤Cµn≤ε0 < φ0

1 + 3δ.

This implies thatx₀ is a regular point, and sox₀ ∈/Λ(δ), which is a contradiction. Equation (5.36), by denition ofS, (5.26), implies then thatx₀ ∈S.

Hence, S = Λ(δ), and the proof is complete.

The statements in Lemmas 5.9 and 5.10 give that 1 ≤n(S) <∞, where n(S) denotes the cardinality of setS. Let's now decomposeS intoS1 =S∩∂ΩandS2 =S∩Ω. Let S= {p₁, . . . , p_N},rbe a small constant andθⁿ_j(r) =R

Br(pj)µ_ne^vn dx. Then lim

n→+∞

R

Ωµ_ne^vn dx=

N

P

j=1

n→+∞lim θ_jⁿ(r), for all smallr, which implies that

n→+∞lim Z

Ω

µ_ne^vn dx=

N

X

j=1

r→0lim lim

n→+∞θⁿ_j(r).

By Lemma 5.9,θⁿ_j(r)≥ φ0

1 + 3δ. In fact, it can be proven that

Lemma 5.11. Ifp_j ∈S₁, thenlim

r→0 lim

n→+∞θ_jⁿ(r) = 4π, and ifp_j ∈S₂, thenlim

r→0 lim

n→+∞θ_jⁿ(r) = 8π. In particular, β0 = 4mπ for some integer m >0.

Proof. We rst consider the case for when p ∈ S1. We will make use of the following Pohozaev's identity. Recall that for v satisfying

∆v−αv+f(v) = 0, inU ⊂R², we have the Pohozaev's identity [74]

Z

U

(−αv²+ 2F(v))dx

= Z

∂U

(x· ∇v)∂v

∂~n−(x·~n)|∇v|²

2 + (x·~n)(−αv²

2 +F(v))

dS,

(5.37)

whereF(v) =Rv

0 f(s)ds.

Let f(v) = µe^κv, where µ = ^{R β}

Ωe^κvdx. Without loss of generality, we can assume that p= 0. LetUr=B^r(0)∩Ω¯ and consider the functionwnwhich is a solution for the problem







∆w−αw= 0 inU_r

∂w

∂~n = ^∂v_∂~nⁿ on∂Ur,

(5.38)

Then it is trivial to see thatw_n =O(1) inC²(U_r), since|^∂v_∂~nⁿ| ≤C on ∂U_r. Now, x ω_n=

(vn−w_n)

θⁿ_j(r) . Then by regularity theory [20], we have thatω_n→G(·,0)inC_loc² (B_r(0)∩Ω¯\ {0}), whereG(·,0)satises







−∆G+αG=δ0 inUr

∂G

∂~n = 0 on∂U_r.

By potential theory [65], we see that for|x|small G(·,0) = 1

πlog|x|+O(1).

Thus, we have

vn= θⁿ_j(r)

π log|x|+O(1)

inC¹(∂U_r). Note that here O(1)may depend onr, but is uniform inn. Now, by Pohozaev's identity, we have

Z

Ur

(−αv_n²+ 2µ_n(e^κvn−1))dx

= Z

∂Ur

(x· ∇v_n)∂v_n

∂~n −(x·~n)|∇v_n|²

2 + (x·~n)(−αv²_n

2 +µn(e^κvn−1))

dS.

(5.39)

We now use Lemma 5.8 to estimate each term on both sides of (5.39) as follows; For the rst term on the left-hand side, we have

Z

Ur

v_n² dx=O(r^1/2kv_nk_L4(Ur)) =O(r^1/2kv_nk_W1,3/2(Ω)) =O(r^1/2).

For the second term on the left-hand side, Z

Ur

2µ_n(e^κvn−1)dx = 2µ_n Z

Ur

e^κvn dx+O(µ_n)

= 2θ_jⁿ(r) +O(µn).

Looking at the rst term on the right-hand side, we have Z

∂Ur

(x· ∇v_n)∂vn

∂~n dS =

θⁿ_j(r) π

²Z

∂Ur

(x·~n)

|x|² +O(1)

=

θⁿ_j(r) π

2

(π+O(r)).

For the second term on the right-hand side, Z

∂Ur

(x·~n)|∇v_n|² 2 dS=

θⁿ_j(r) π

² π

2 +O(r) . From the third term on the right-hand side,

Z

∂Ur

v_n² dS=O(r).

Lastly, for the last term on the right-hand side, we have Z

∂Ur

(x·~n)µn(e^κvn−1)dS=O

µn max

x∈∂Ur

e^κvn

=O(µn).

Now, if we rst let n→+∞, and then we letr →0, we get that 2 lim

r→0 lim

n→+∞θ_jⁿ(r) = 1 π²

π 2

r→0lim lim

n→+∞θⁿ_j(r) 2

, which implies that

r→0lim lim

n→+∞θⁿ_j(r) = 4π.

The interior blow up case (p∈S2) is proven in a similar way, with the following modi- cations. Instead of (5.38), we will considerw_n satisfying the problem







∆w−αw= 0 inUr

w=v_n on∂U_r.

(5.40)

We x ω_n = ^(v_θⁿn^−wn)

j(r) and assume that p = 0 ∈ Ω. Then, similarly, ω_n → G(·,0) in C_loc² (Br(0))/{0}, whereGis now a Green function with Dirichlet boundary data







−∆G+αG=δ0 inBr

G= 0 on∂Ur.

In this case, the Green function has the following expansion near0;

G(·,0) =− 1

2π log|x|+O(1).

We then obtain the same estimates as in the case when p ∈S2, except with the following two. Looking at the rst term on the right-hand side, we have

Z

∂Ur

(x· ∇v_n)∂vn

∂~n dS =

θⁿ_j(r) 2π

2Z

∂Ur

(x·~n)

|x|² +O(1)

=

θⁿ_j(r) 2π

2

(2π+O(r)), and for the second term on the right-hand side,

Z

∂Ur

(x·~n)|∇v_n|² 2 dS=

θⁿ_j(r) 2π

²

(π+O(r)).

Now, applying Pohozaev's identity again, we obtain in this case that 2 lim

r→0 lim

n→+∞θ_jⁿ(r) = 1 4π²π

r→0lim lim

n→+∞θ_jⁿ(r) 2

, which implies that

r→0lim lim

n→+∞θⁿ_j(r) = 8π.

The proof of the Lemma is complete.

We now note that, by Lemma 5.11, the statement in (5.24) holds true. We have therefore proven the following Theorem.

Theorem 5.12. Suppose that _|Ω|^β −β₀ < α, β > 4π, and β ∈ R\ {4mπ|m = 1, 2, . . .}. Then (5.4) has a non-constant solution.

In the following discussions, we will use a Lyapunov functional. Let (u, v) be a solution for (5.3), withu≥0. We introduce the following Lyapunov functional [26, 37].

F(u, v) = Z

Ω

1

2γ d2|∇v|²+λv²

+u(logu−1) + 1−(u−1)v

dx. (5.41)

By Lemma 4.7 in [26], we know that if we let f(v) = 1

2γ Z

Ω

d2|∇v|²+λv²

dx− |Ω|log R

Ωe^κvdx

|Ω| , (5.42)

then fort≥0,

f(v(t))≤F(u(t), v(t)). (5.43) From Lemma 5.11 and Theorem 5.12, we obtain the following Lemma.

Lemma 5.13. Assume that β >4π and β∈R\ {4mπ|m= 1, 2, . . .}. Then there exists a constant Kˆ ≤0 such that

f(v)≥K >ˆ −∞

holds for all solutions v of (5.23).

We therefore have the following Theorem, which summarizes some known facts about blow-up solutions.

Theorem 5.14. LetΩ⊂R² be a smooth domain, andKˆ be the constant from Lemma 5.13.

Suppose further thatβ >4π andβ∈R\ {4mπ|m= 1, 2, . . .}. Then there exist initial data (u₀, v₀) such that

F(u₀, v₀)<K,ˆ

and the corresponding solution of (5.3) blows up in nite or innite time. For these blow-up solutions, the following statements hold;

1. lim

t→Tmax

ku(x, t)k_L2(Ω)=∞ 2. lim

t→Tmax

R

Ωu(x, t)v(x, t)dx=∞ 3. lim

t→Tmax

k∇v(x, t)k_L2(Ω)=∞ 4. lim

t→T_max

R

Ωe^v(x,t)dx=∞ 5. lim

t→T_maxku(x, t)k_L∞(Ω)= lim

t→T_maxkv(x, t)k_L∞(Ω)=∞ 6. If 4π < β <8π andΩ is a simply connected domain, then

t→Tlimmax

Z

∂Ω

e^v(x,t)2 dS=∞

Proof. We rst note that for v_ε as dened in (5.7), it is clear that asε→0,

f(vε)→ −∞ (5.44)

and

kv_εk_L2(Ω)→ ∞. (5.45)

Thus, by Lemma 5.13, (5.44) and (5.45), the existence of a blow-up solution is established.

Next, suppose thatKˆ is a constant as in Lemma 5.13, and chooseε₀ arbitrary but xed³, and a xed x0 ∈∂Ωsuch that if we let

vε0(x) = log 32ε²₀

(ε²₀+|x−x₀|²)² − 1

|Ω|

Z

Ω

log 32ε²₀

(ε²₀+|x−x₀|²)² dx, then

f(vε0(x))<K.ˆ We note that

v_ε0(x)∈W^1,∞(Ω).

Now, set

u_ε(x) = |Ω|e^vε0(x) R

Ωe^vε0(x)dx. Thenuε(x) belongs toL^∞₊(Ω), and

F(uε(x), vε0(x)) =f(vε0(x))<K.ˆ

If we then choose our initial data such thatu₀(x) =u_ε0(x) and v₀(x) =v_ε0(x), then we see that the corresponding solution for the Keller-Segel model has to blow up in nite or innite time.

For the remaining statements of the Theorem, we recall the Lyapunov function (5.41).

Let S_u and S_v+ denote the blow-up sets for u(x, t) and the positive part of v respectively.

Then it is known, [33], that if there are initial data (u0, v0) such that the solution of (5.3) blows up, then

S_u∩ S_v+ 6=∅, and

t→Tlimmax

Z

Ω

|∇v|²dx=∞ and lim

t→Tmax

Z

Ω

e^vdx=∞.

This establishes 3. and 4. above for a blow-up solution of (5.3).

3The existence of thisε0 is guaranteed by (5.44)

Moreover, by the properties of (5.41),

F(u(x, t), v(x, t))≤F(u₀(x), v₀(x))≤K,ˆ so that

Z

Ω

1

2γ d₂|∇v|²+λv² dx≤

Z

Ω

(u−1)v dx+ ˆK (5.46) is true. From this inequality, statement 2. is established, while statement 1. follows by employing Cauchy's inequality. Combining statements 1. and 4. establishes statement 5.

It remains to prove statement 6. To this end, we remark that, Horstmann [33] showed in Lemma 3 that if β <8π, v∈H¹(Ω), andp∈(1,^8π_β ) is arbitrary but xed, then

log 1

|Ω|

Z

Ω

e^vdx

≤ p 16π

Z

Ω

|∇v|² dx+2 plog

Z

∂Ω

e^p

0v 2 dS

+K(p, p⁰, β),

(5.47)

wherep⁰ is the conjugate exponent ofp, andK(p, p⁰, β)is a constant dependent ofp, p⁰ and β. If we then use this inequality, we can estimate (5.41) from below, forp∈(1,^8π_β)arbitrary but xed, by

F(u, v)≥ Z

Ω

1

2γ d₂|∇v|²+λv²

dx− |Ω|log 1

|Ω|

Z

Ω

e^vdx

≥ Z

Ω

d2

2γ −p|Ω|

16π

|∇v|²+ λ 2γv²

dx

−2|Ω|

p⁰ log Z

∂Ω

e^p

0v 2 dS

+K₀(p, p⁰, β),

where K₀(p, p⁰, β) is a constant dependent of p, p⁰ and β. With statement 3. in mind, we see that

t→Tlimmax

Z

∂Ω

e^p

0v

2 dS=∞,

for every p⁰ ∈(_8πd^8πd2

2−γ|Ω|, ∞).

Furthermore, it has been proven in [69] that if Ω ⊂ R² is a simply connected, smooth

domain, andv∈H¹(Ω), then log

1

|Ω|

Z

Ω

e^vdx

≤ 1 16π

Z

Ω

|∇v|² dx+ 1 2|∂Ω|

Z

∂Ω

v dS + log

1

|∂Ω|

Z

∂Ω

e^v2 dS

+K₁,

(5.48)

whereK1 is an absolute constant. If we use this inequality instead of (5.47), we get that

F(u, v)≥ Z

Ω

1

2γ d₂|∇v|²+λv²

dx− |Ω|log 1

|Ω|

Z

Ω

e^vdx

≥ Z

Ω

d₂ 2γ − |Ω|

16π

|∇v|²+ λ 2γv²

dx

− |Ω|

2|∂Ω|

Z

∂Ω

v dS− |Ω|log Z

∂Ω

e^v2 dS

−K1|Ω|, and this gives us statement 6., and the Theorem is established.

The following Lemma gives us another result for a blow-up solution.

Lemma 5.15. If the solution (u(t), v(t)) of system (5.3) blows up, then we have that

t→Tlimmax

ku(t) logu(t)k_L1(Ω)=∞. (5.49) Proof. Let(u(t), v(t))be the blow-up solution for (5.3). Since

Z

Ω

u(t) logu(t)dx≥ −|Ω|

e we see that

F(u(t), v(t))≥ −|Ω|

e − Z

Ω

(u(t)−1)v(t)dx+ d₂

2γk∇v(t)k²_L2(Ω). Moreover, it is known, [17], that

kv(t)k²_LΦ(Ω)≤K˜k∇v(t)k²_L2(Ω),

whereΦ(s) =e^s−s−1, keeping in mind thatR

Ωv(t)dx= 0. Here and in what follows,L^Φ(Ω) denotes the Orlicz space which corresponds to the Young functionΦ(s), andk · k_LΦ(Ω)as its norm. Let's denote the Young function complementary to Φ by Ψ, so thatL^Ψ(Ω) denotes the Orlicz space, withk·k_LΨ(Ω)is its norm. It is also known thatΨ(s) = (s+1) log(s+1)−s.

Using Hölder's inequality for Orlicz spaces [2, 78], we see that F(u(t), v(t))≥ −|Ω|

e − Z

Ω

(u(t)−1)v(t)dx+ d2

2γk∇v(t)k²_L2(Ω)

≥ −|Ω|

e − kv(t)k_LΦ(Ω)ku(t)−1k_LΨ(Ω)+ d2

2γk∇v(t)k²_L2(Ω)

≥ −|Ω|

e − K˜

4εku(t)−1k_LΨ(Ω)+ d₂

2γ −ε

k∇v(t)k²_L2(Ω), whereε < _2γ^d2. Combining this with Lemma 6.3 of [26] and the fact that

Z

Ω

u(t)dx=|Ω| ∀t≥0, we get the result.

Remark 5.1. We remark that, except statement 6., all the statements of Theorem 5.14 and Lemma 5.15 are also true for blow-up solutions of (5.3) when Ω ⊂R² has a piecewise C² boundary.