Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift

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98 T. Cie´slak and Ph. Lauren¸cot

2. Preliminaries

In this section we summarize some results and methods introduced in [4]. Let a ∈

((0, ∞)) be a positive function and consider an initial condition u

∈ C([0, 1])

such that u

≥ m

> 0andu

 = M for some M>0andm

∈ (0,M). By

[4, Propositions 2 and 3] there is a unique maximal classical solution (u, v)to

(1.1)–(1.4) deﬁned on [0,T

max

) which satisﬁes

min

x∈[0,1]

u(t, x) > 0 , u(t) :=



u(t, x) dx = M, and (2.1)

v(t) :=



v(t, x) dx =0

for t ∈ (0,T

max

). In addition, T

max

= ∞ or T

max

< ∞ with u(t)

∞

(0,M)

→∞

as t → T

max

We next recall the approach introduced in [4] which will be used herein as

well. Owing to the positivity (2.1) and the regularity of u, the indeﬁnite integral

U(t, x):=



u(t, z)dz , x ∈ [0, 1] ,

is a smooth increasing function from [0, 1] onto [0,M]foreacht ∈ [0,T

max

)and

has a smooth inverse F deﬁned by

U(t, F (t, y)) = y, (t, y) ∈ [0,T

max

) × [0,M] . (2.2)

Introducing f(t, y):=∂

F (t, y), we have

f(t, y) u(t, F (t, y)) = 1 , (t, y) ∈ [0,T

max

) × [0,M] , (2.3)

and it follows from (1.1)–(1.4) that f solves

∂

f = ∂

Ψ(f) − 1+Mf , (t, y) ∈ (0,T

max

) × (0,M) , (2.4)

∂

f(t, 0) = ∂

f(t, M)=0,t∈ (0,T

max

) , (2.5)

f(0,y)=f

(y):=

(F (0,y))

,y∈ (0,M) , (2.6)

where



(r):=





for any r>0 , Ψ(1) := 0 . (2.7)

Moreover the conservation of mass (2.1) yields



f(t, y)dy = F (t, M) − F (t, 0) = 1 ,t∈ [0,T

max

) . (2.8)

At this point, the crucial observation is that, thanks to (2.3), ﬁnite time blowup

of u is equivalent to the vanishing (or touch-down) of f in ﬁnite time. In other

words, u exists globally if the minimum of f(t) is positive for each t>0. We refer

to [4, Proposition 1] for a more detailed description.

A salient property of (1.1)–(1.4) is the existence of a Liapunov function [4,

Lemma 8] which we recall now:

Global Existence vs. Blowup 99

Lemma 2.1. The function

(t):=



|∂

Ψ(f(t, y))|

dy +



(Ψ(f(t, y)) − M Ψ

(f(t, y))) dy

is a non-increasing function of time on [0,T

max

), the function Ψ

being deﬁned by

(1) := 0 and Ψ



(r):=rΨ



(r)=





,r∈ (0, ∞) . (2.9)

3. Finite time blowup

In this section we prove the blowup assertion of Theorem 1.1. To this end we

ﬁrst prove that the condition (1.5) allows us to construct a concave function B

satisfying (1.7) and (1.8) so that [4, Theorem 10] can be applied.

Lemma 3.1. Let a ∈ C

((0, ∞)) be a positive function such that a ∈ L

(1, ∞) and

(1.5) holds. Then there exists a concave function B ∈ C([0, ∞)) such that for all

r ≥ 0

B(r) ≥ r



∞

a(s)ds (3.1)

and

lim

r→∞

B(r)

=0. (3.2)

Proof of Lemma 3.1. We construct B :[0, ∞) → [0, ∞) in the following way: we

put



∞

a(s)ds , i ≥ 0 ,

and notice that {b

}

i≥0

is a decreasing sequence converging to zero as i →∞.We

next deﬁne

B(r)=

⎧

⎪

⎨

⎪

⎩

r + γ if r ∈ [0, 2],

r +

i−1

j=0

−b

j+1

+ γ if r ∈ (2

, 2

i+1

]andi ≥ 1.

(3.3)

Clearly, B ∈ C([0, ∞)) and



(r)=



if r ∈ (0, 2),

if r ∈ (2

, 2

i+1

)andi ≥ 1.

(3.4)

Hence B is concave as a consequence of the fact that the sequence {b

}

i≥0

decreasing. Furthermore, for r ∈ [0, 1], we have

B(r) ≥ γ ≥ r



∞

a(s)ds,

100 T. Cie´slak and Ph. Lauren¸cot

and for r ∈ [2

, 2

i+1

], i ≥ 0,

B(r) ≥ b

r = r



∞

a(s)ds ≥ r



∞

a(s)ds.

Therefore, B satisﬁes (3.1).

Finally, let k ≥ 1. If i ≥ k +1andr ∈ (2

, 2

i+1

], then

B(r)

= b

i−1

j=0

−b

j+1

)

j+1

≤ b

i−1

j=k

−b

j+1

k−1

j=0

−b

j+1

)

j+1

≤ b



γ +2

k−1

j=0

−b

j+1

)



+(b

−b

)

≤ b



γ +2



Consequently,

lim sup

r→∞

B(r)

≤ b

for all k ≥ 1 .

Letting k →∞, we obtain (3.2) since b

→ 0ask →∞and Lemma 3.1 is

proved. 

ProofofTheorem1.1 (i),Part1. When a belongs to L

(1, ∞) and satisﬁes (1.5),

it follows from Lemma 3.1 that the conditions (1.7) and (1.8) are satisﬁed so that

Theorem 1.1 (i) follows from [4, Theorem 10]. 

To handle the other case, we proceed in a diﬀerent way by showing an upper

bound for the function f deﬁned in Section 2. We ﬁrst observe that the function

Ψdeﬁnedin(2.7)satisﬁes

Ψ(r)=







ds =



1/r

a(s) ds , r ∈ (0, ∞) ,

so that, if a ∈ L

(1, ∞), Ψ(r) has a ﬁnite limit Ψ(0) := −a

(1,∞)

as r → 0. We

then deﬁne

Ψ(r):=Ψ(r) − Ψ(0) =







ds =



∞

1/r

a(s) ds , r ∈ (0, ∞) . (3.5)

Lemma 3.2. Let a ∈ C

((0, ∞)) be a positive function such that a ∈ L

(1, ∞).

There exists a positive constant μ

> 0 depending only on M and a such that, for

any non-negative function g ∈ H

(0,M) satisfying g

(0,M)

=1, we have



Ψ(g)

∞

(0,M)

≤ 32M L

(g)+μ

, (3.6)

Global Existence vs. Blowup 101

with

(g):=

∂

Ψ(g)

(0,M)



(Ψ(g) − M Ψ

(g)) (y) dy , (3.7)

the functions Ψ and Ψ

being deﬁned in (2.7) and (2.9), respectively.

Proof of Lemma 3.2. We set G := g

∞

(0,M)

which is ﬁnite owing to the con-

tinuous embedding of H

(0,M)inL

∞

(0,M). Assume ﬁrst that G>1. Then, for

y ∈ (0,M)andz ∈ (0,M), we have

Ψ(g(y)) =

Ψ(g(z)) +



∂

Ψ(g(x)) dx ≤

Ψ(g(z)) + M

1/2

∂

Ψ(g)

(0,M)

Integrating the above inequality over (0,M) with respect to z gives

Ψ(g(y)) ≤



Ψ(g(z)) dz + M

3/2

∂

Ψ(g)

(0,M)

≤



[0,2/M ]

(g(z))

Ψ(g(z))dz



(2/M,∞)

(g(z))

Ψ(g(z))dz + M

3/2

∂

Ψ(g)

(0,M)

≤ M





Ψ(G)



(2/M,∞)

(g(z))g(z) dz + M

3/2

∂

Ψ(g)

(0,M)

≤ M





Ψ(G)

+ M

3/2

∂

Ψ(g)

(0,M)

wherewehaveusedthepropertyg

(0,M)

= 1 to obtain the last inequality. Tak-

ing the supremum over y ∈ (0,M) and using the monotonicity and non-negativity

Ψ, we deduce that

Ψ(G) ≤ 2





+2M

1/2

∂

Ψ(g)

(0,M)

. (3.8)

We next observe that the integrability of a at inﬁnity also ensures that Ψ

(0) >

−∞,sothat

:= Ψ

− Ψ

(0) is well deﬁned and satisﬁes

(r)=



sΨ



(s) ds ≤ r

Ψ(r) ,r∈ (0, ∞) . (3.9)

Since g

(0,M)

= 1, it follows from (3.8) and (3.9) that



(g) dy ≤



Ψ(g) dy ≤

Ψ(G)



gdy

≤ 2





+2M

1/2

∂

Ψ(g)

(0,M)

(3.10)

102 T. Cie´slak and Ph. Lauren¸cot

We next infer from (3.10) and the non-negativity of

Ψthat

(g) ≥

∂

Ψ(g)

(0,M)



Ψ(g) dy + M Ψ(0) −M



(g) dy

≥

∂

Ψ(g)

(0,M)

+ MΨ(0) −2M





− 2M

3/2

∂

Ψ(g)

(0,M)

≥

∂

Ψ(g)

(0,M)



∂

Ψ(g)

(0,M)

− 2M

3/2



− 4M

+ MΨ(0) −2M





≥

∂

Ψ(g)

(0,M)

− 4M

+ MΨ(0) −2M





whence

∂

Ψ(g)

(0,M)

≤ 4L

(g)+16M

− 4MΨ(0) + 8M





It then follows from (3.8) and the above inequality that

Ψ(G)

≤ 8





+8M∂

Ψ(g)

(0,M)

≤ 8





+32ML

(g) + 128M

−32M

Ψ(0) + 64M





≤ 32M L

(g)+μ

with

:= 1 + 128M

−32M

Ψ(0) + 64M









+Ψ(0)

−32M Ψ(0).

We have thus shown Lemma 3.2 when G = g

∞

(0,M)

> 1. To complete the

proof, we ﬁnally consider the case G ∈ [0, 1] and notice that, in that case,

0 ≤

Ψ(G) ≤−Ψ(0) and L

(g) ≥



Ψ(g) dy + MΨ(0) ≥ M Ψ(0),

since Ψ

≤ 0in(0, 1) and

Ψ ≥ 0. Consequently,

Ψ(G)

≤ Ψ(0)

=32M Ψ(0) + Ψ(0)

−32MΨ(0) ≤ 32M L

(g)+μ

and the proof of Lemma 3.2 is complete. 

As an obvious consequence of Lemmas 2.1 and 3.2 we have the following

result:

Corollary 3.3. Let a ∈ C

((0, ∞)) be a positive function such that a ∈ L

(1, ∞).

For t ∈ [0,T

max

) and y ∈ [0,M], we have

0 ≤

Ψ(f(t, y)) ≤ (32M max {L

), 0} + μ

)

1/2

Global Existence vs. Blowup 103

Proof of Corollary 3.3. Clearly

(f(t)) = L

(t) ≤ L

(0) = L

) ≤ max {L

), 0}

for t ∈ [0,T

max

) by Lemma 2.1 and Corollary 3.3 readily follows from Lemma 3.2.



Remark 3.4. Corollary 3.3 provides an L

∞

-bound on f only if Ψ(r) →∞as

r →∞,thatis,ifa ∈ L

(0, 1). In that case, it gives a positive lower bound for u

by (2.3).

We next turn to the proof of the second part of Theorem 1.1 for which we

develop further the arguments from [4, Theorem 10].

ProofofTheorem1.1 (i),Part2. Assume now that a ∈ L

(1, ∞) and satisﬁes

(1.6). We ﬁx M>0, q>2, and ε

∈ (0, 1) such that

q>max



3+ϑ,

5+3ϑ

α(ϑ +1)− ϑ



and

q(q +1)



∞

1/ε

a(s) ds ≤

, (3.11)

the existence of ε

being guaranteed by the integrability of a at inﬁnity.

For

δ ∈



0, min



1, 2M,(2M)

−1/q



, (3.12)

we put

(y):=

2(1 − Mδ

)

(δ − y)

+ δ

≥ δ

> 0 ,y∈ [0,M] . (3.13)

Then



(y) dy =1, f



∞

(0,M)

2(1 −Mδ

)

+ δ

≤

. (3.14)

Denoting the corresponding solution to (2.4)–(2.6) by f we next introduce

(t):=



f(t, y) dy , t ∈ [0,T

max

) ,

and we have

(0) =



2(1 −Mδ

)

(q +1)(q +2)

q+1

q +1



≤ C

(3.15)

with



2+(q +2)M

q+1

(q +1)(q +2)



It follows from (2.4), (2.5), and the non-negativity of

Ψthat

= −q



q−1

∂

Ψ(f) dy + Mm

−

q+1

q +1

≤ q(q − 1)



q−2

Ψ(f) dy + Mm

−

q+1

q +1

. (3.16)

104 T. Cie´slak and Ph. Lauren¸cot

We shall now estimate the integral on the right-hand side of (3.16): to this end,

we split the domain of integration into three parts which we handle diﬀerently. As

a preliminary step, we notice that, by (1.6),



(r) ≤ γ

and Ψ(r) ≤

ϑ +1

ϑ+1

≤ γ

ϑ+1

,r≥ 1 . (3.17)

We next deﬁne

:= (32M max {L

), 0} + μ

)

1/2(2+ϑ)

> 1 ,

and consider (t, y) ∈ [0,T

max

) × [0,M].

• If f(t, y) ∈ (0,ε

], it follows from (3.11) and the monotonicity of

Ψthat

Ψ(f(t, y)) ≤

Ψ(ε



∞

1/ε

a(s) ds ≤

2q(q +1)

. (3.18)

• If f(t, y) ∈ (ε

), then (3.17) and the monotonicity of

Ψ yield

Ψ(f(t, y)) =

Ψ(f(t, y))

f(t, y)

f(t, y) ≤

Ψ(K

) − Ψ(0)

f(t, y)

≤

ϑ+1

− Ψ(0)

f(t, y) .

(3.19)

• If f(t, y) ≥ K

, Corollary 3.3 ensures that

Ψ(f(t, y)) =

Ψ(f(t, y))

f(t, y)

f(t, y) ≤

ϑ+2

f(t, y) ≤ K

ϑ+1

f(t, y) . (3.20)

Consequently, recalling that K

> 1andΨ(0)< 0, we deduce from (3.16) and

(3.18)–(3.20) that

≤ q(q − 1)



q−2

Ψ(f) 1

(0,ε

]

(f) dy

+ q(q − 1)



q−2

Ψ(f) 1

(ε

)

(f) dy

+ q(q − 1)



q−2

Ψ(f) 1

,∞)

(f) dy + Mm

−

q+1

q +1

≤

(q − 1)M

2(q +1)



q−2

dy + q(q − 1)

ϑ+1

−Ψ(0)



q−2

fdy

+ q(q − 1)K

ϑ+1



q−2

fdy+ Mm

−

q+1

q +1

≤ C

ϑ+1



q−2

fdy+ Mm

−

q+1

2(q +1)

Global Existence vs. Blowup 105

with C

:= q(q − 1)(γ

− Ψ(0) + ε

)/ε

.WenextuseH¨older’s inequality and

(2.8) to conclude that

≤ C

ϑ+1

(q−2)/q

+ Mm

−

q+1

2(q +1)

. (3.21)

It remains to estimate K

andinfactL

). Since Ψ is negative on (0, 1) and Ψ

is bounded from below by Ψ

(0), it follows from (3.12) and (3.13) that

) ≤

(1 − Mδ

)



|Ψ



dy +



Ψ(f

) dy − M

(0)

≤



|Ψ



dy +



Ψ(f

) dy − M

(0).

On the one hand, we infer from (3.14), (3.17), and the monotonicity of Ψ that



Ψ(f

) dy ≤ δ Ψ





≤ γ

ϑ+1

−ϑ

On the other hand, we have

(y) ≥ 1fory ∈ [0,y

]withy

:= δ −

1 − δ

2(1 − Mδ

)

> 0 ,

(y) ∈ [δ

, 1] for y ∈ [y

,δ] ,

so that, if y ∈ [0,y



(y))

≤ γ

(y)

2ϑ

≤ γ

−2ϑ

by (3.14) and (3.17), while, if y ∈ (y

,δ],



(y))

≤

(y)



(y)



≤ C

∞

(y)

2(α−2)

≤ C

∞

−2q(2−α)

by (1.6) since α ≤ 2. Therefore,

) ≤



−2ϑ

dy +



∞

−2q(2−α)

+ γ

ϑ+1

−ϑ

− M

(0)

≤ γ

ϑ+1

−3−2ϑ

+ C

∞

1 − δ

2(1 −Mδ

)

−2−2q(2−α)

+ γ

ϑ+1

−ϑ

− M

(0)

≤ γ

ϑ+1

−2(2+ϑ)

+ C

∞

−2−2q(2−α)

+ γ

ϑ+1

−ϑ

− M

(0)

≤ C



−2(2+ϑ)

+ δ

−2−2q(2−α)



with C

:= γ

ϑ+2

+ C

∞

− M

(0). Therefore,

ϑ+1

≤ C



−(ϑ+1)

+ δ

−(ϑ+1)(1+q(2−α))/(ϑ+2)



(3.22)

for some constant C

> 0 depending only on M and a.

106 T. Cie´slak and Ph. Lauren¸cot

For C

:= C

we deﬁne

):=C



−(ϑ+1)

+ δ

−(ϑ+1)(1+q(2−α))/(ϑ+2)



(q−2)/q

+ Mm

−

q+1

2(q +1)

Combining (3.21) and (3.22) yields

≤ Λ

) (3.23)

for t ∈ [0,T

max

). At this point, we note that the monotonicity of Λ

and (3.23)

imply that Λ

(t)) ≤ Λ

(0)) for t ∈ [0,T

max

)ifΛ

(0)) < 0, the latter

condition being satisﬁed for δ small enough since

(0))≤C

(q−2)/q



q−3−ϑ

+δ

(q(α(ϑ+1)−ϑ)−3ϑ−5)/(ϑ+2)



+MC

−

q+1

2(q +1)

by (3.11) and (3.15).

Summarizing, we have shown that, if δ satisﬁes (3.12) and

(q−2)/q



q−3−ϑ

+ δ

(q(α(ϑ+1)−ϑ)−3ϑ−5)/(ϑ+2)



+ MC

q+1

2(q +1)

, (3.24)

we have

(t) ≤ Λ

(t)) ≤ Λ

(0)) < 0 ,t∈ [0,T

max

) ,

an inequality which can only be true on a ﬁnite time interval owing to the non-

negativity of m

. Therefore, T

max

< ∞ in that case and, for any M>0, we have

found an initial condition u

given by (2.2), (2.3), and (3.13) (for δ small enough

according to the above analysis) such that u

 = M and the ﬁrst component u of

the corresponding solution to (1.1)–(1.4) blows up in ﬁnite time. 

4. Global existence

The proof of Theorem 1.1 (ii) also relies on the study of the function L

deﬁned in

Lemma 2.1. For that purpose, we ﬁrst recall another property from [4]. We deﬁne

the function E

(h):=

∂

h

L2(0,M)



(−∞,0)

(h(y)) h(y) dy , h ∈ H

(0,M) , (4.1)

for which we have the following lower bound.

Lemma 4.1. [4, Lemma 9] For M>0, we have

(h) ≥

∂

h

L2(0,M)

−M

− M





, (4.2)

and

h

L1(0,M)

≤ M

3/2

∂

h

L2(0,M)

+ M





(4.3)

Global Existence vs. Blowup 107

for every h ∈ H

(0,M) satisfying



−1

(h)(y) dy =1. (4.4)

We now show that the non-integrability of a at inﬁnity allows us to show

that T

max

= ∞. To this end, we use the alternative formulation (2.4)–(2.6) as in

[4] and prove that f cannot vanish in ﬁnite time.

Proof of Theorem 1.1 (ii). Owing to (2.6) and the assumptions made on u

,we

have

0 <f

(y) ≤

,y∈ [0,M].

Introducing Σ(t):=M

−1

+ e



−1

−M

−1



for t ≥ 0wehave

∂

Σ −∂

Ψ(Σ) − M Σ+1=M



Σ −



− MΣ+1=0,

Σ(0) =

≥ f

(y),y∈ (0,M),

and the comparison principle warrants that

f(t, y) ≤ Σ(t), (t, y) ∈ [0,T

max

) ×[0,M]. (4.5)

We now follow the strategy of the proof of [4, Theorem 5] and ﬁrst use the

properties of Ψ, Ψ

, and (4.5) to estimate the function L

(deﬁned in Lemma 2.1)

from below. Indeed, since Ψ ≥ 0on(1, ∞)andΨ

≤ 0on(0, 1) we arrive at

(0) ≥ L

(t)=

∂

Ψ(f(t))

(0,M)



(0,1)

(f(t, y))(Ψ − M Ψ

)(f(t, y)) dy



(1,∞)

(f(t, y))(Ψ − M Ψ

)(f(t, y)) dy

≥

∂

Ψ(f(t))

(0,M)



(−∞,0)

(Ψ(f(t, y)))Ψ(f(t, y)) dy

− M



(1,∞)

(f(t, y))Ψ

(f(t, y)) dy

≥ E

(Ψ(f(t))) − M

(Σ(t)),

where E

is deﬁned in (4.1) and we have used (4.5) to obtain the last inequality.

Next, by Lemma 4.1 and (2.8), we have

(0) ≥

∂

Ψ(f(t))

(0,M)

− M





− M

(Σ(t)),