GLOBAL EXISTENCE AND BLOW-UP

27,2 (2000), pp. 203–218

J. R E N C L A W O W I C Z (Warszawa)

GLOBAL EXISTENCE AND BLOW-UP

FOR A COMPLETELY COUPLED FUJITA TYPE SYSTEM

Abstract. The Fujita type global existence and blow-up theorems are proved for a reaction-diffusion system of m equations (m > 1) in the form

u

_it

= ∆u

_i

+ u

^p_i+1ⁱ

, i = 1, . . . , m − 1, u

_mt

= ∆u

_m

+ u

^p₁^m

, x ∈ R

^N

, t > 0,

with nonnegative, bounded, continuous initial values and positive num- bers p

_i

. The dependence on p

_i

of the length of existence time (its finiteness or infinitude) is established.

1. Introduction. The main concern of this paper is to examine the following system of reaction-diffusion equations:

(1.1)  u

it

− ∆u

_i

= u

^p_i+1ⁱ

, i = 1, . . . , m − 1, u

mt

− ∆u

m

= u

^p₁^m

,

for x ∈ R

^N

, t > 0, p

_i

> 0; and

u

i

(0, x) = u

0i

(x), i = 1, . . . , m, x ∈ R

^N

, where u

0i

are nonnegative, continuous, bounded functions.

We consider the behaviour of classical nonnegative solutions for (1.1) as regards their maximal existence time. We describe the cases of global existence and finite blow-up time for every solution of (1.1) (in the second situation, a nontrivial solution) in terms of p

i

and N .

This work generalizes results of Fujita [F1], [F2] (for the scalar Cauchy problem), Escobedo and Herrero [EH] (for the system (1.1) with m = 2) and the author [R] (for the system (1.1) with m = 3). We want to mention that the method used here is strictly connected with the special form of the

2000 Mathematics Subject Classification: 35B30, 35B50, 35K55, 35K57.

Key words and phrases : blow-up, global existence, reaction-diffusion system.

Research supported by KBN grant no. 2PO3A 038 16.

[203]

system (1.1). In the case of weakly coupled systems proofs are much more complicated (we refer to [EL] for the case of two equations).

We introduce some notations to formulate the main theorems. Let A

m

be a matrix of the form

(1.2) A

m

=



















0 p

1

0 p

2

0 . ..

. ..

0 p

m−1

p

_m

0

















 .

We denote by α = (α

1

, . . . , α

m

) the unique solution of (1.3) (A

m

− I)α

^t

= (1, . . . , 1)

^t

. Let

(1.4) δ = det(A

m

− I) = (−1)

^m+1

 Y

^m

j=1

p

j

− 1  . Whenever δ 6= 0, we have

α

1

= 1 + p

1

+ p

1

p

2

+ . . . + Q

m−1 j=1

p

j

Q

m

j=1

p

_j

− 1 , (1.5)

α

i

= 1 + p

i

+ p

i

p

i+1

+ . . . + Q

j6=i−1

p

j

Q

m

j=1

p

j

− 1 , i = 2, . . . , m.

Theorem 1.1. Suppose det(A

m

− I) = 0. Then every solution of (1.1) is global.

Theorem 1.2. Suppose det(A

m

− I) 6= 0 and max

i=1,...,m

α

i

< 0. Then all the solutions of (1.1) are global.

Theorem 1.3. Suppose det(A

m

− I) 6= 0 and max

i=1,...,m

α

i

≥ N/2.

Then every nontrivial solution of (1.1) blows up in a finite time.

The plan of this paper is the following: we prove some auxiliary lemmas in Section 2, global existence theorems is the content of the last section, whereas the global nonexistence theorem is proved in Section 3.

2. Preliminaries. Let S(t) be the semigroup operator for the heat equation, i.e.

S(t)ξ

0

(x) =

\

R^N

(4πt)

^−N/2

exp



− |x − y|

²

4t



ξ

0

(y) dy.

A classical solution of (1.1) satisfies the following variation of constants formulas:

(2.1)

u

i

(t) = S(t)u

0i

+

t

\

0

S(t − s)u

^p_i+1ⁱ

(s) ds, i = 1, . . . , m − 1,

u

m

(t) = S(t)u

0m

+

t

\

0

S(t − s)u

^p₁^m

(s) ds.

It follows that

u

i

(τ ) = S(τ − t

i

)u

i

(t

i

) +

τ −t\i

0

S(τ − t

i

− s)u

^p_i+1ⁱ

(s) ds

≥ S(τ − t

i

)u

i

(t

i

), i = 1, . . . , m − 1, u

m

(τ ) = S(τ − t

m

)u

m

(t

m

) +

τ −tm

\

0

S(τ − t

m

− s)u

^p₁^m

(s) ds

≥ S(τ − t

m

)u

m

(t

m

).

We remark that if (x

i

, t

i

) are such that u

i

(x

i

, t

i

) > 0, i = 1, . . . , m, then by positivity of S(t), u

i

(τ ) > 0 for τ > t

i

. Taking t

0

= max

i=1,...,m

t

i

we have u

_i

(x, τ ) > 0 for x ∈ R

^N

, τ > t

₀

, i = 1, . . . , m.

Lemma 2.1. Let u = (u

1

, . . . , u

m

) with u

0

= (u

01

, . . . , u

0m

) 6≡ 0 be a solution of (1.1). Then we can choose τ = τ (u

01

, . . . , u

0m

) and some constants c > 0, a > 0 such that min

i

(u

i

(τ )) ≥ ce

^−a|x|2

.

P r o o f. We can assume that for instance u

01

6= 0. Then there exists R > 0 satisfying

ν

1

= inf{u

01

(x) : |x| < R} > 0.

By formula (2.1) we have

u

1

(t) ≥ S(t)u

01

≥ ν

1

(4πt)

^−N/2

exp  −|x|

²

4t



\

|y|≤R

exp  −|y|

²

4t

 dy.

We put

u

1

(t) = u

1

(t + τ

0

) for some τ

0

> 0, a

1

= 1

4τ

0

, c

1

= ν

1

(4πτ

0

)

^−N/2

\

|y|≤R

exp  −|y|

²

4τ

0

 dy.

Hence

u

₁

(0) = u

₁

(τ

₀

) > c

₁

exp(−a

₁

|x|

²

).

To obtain the assertion we use an inductive argument. By (2.1), u

i

(τ

0

) ≥

τ\0

0

S(τ

0

− s)(S(s)u

0(i+1)

)

^pi

ds.

If p

i

≥ 1 then u

_i

(τ ) ≥

τ\

0

(S(τ − s)S(s)u

_0(i+1)

)

^pi

ds = τ (S(τ )u

_0(i+1)

)

^pi

. Otherwise,

u

i

(τ ) ≥

τ

\

0

S(τ − s)S(s)(u

_0(i+1)

)

^pi

ds = τ S(τ )u

^p_0(i+1)ⁱ

.

It is clear that if we assume that u

_0(i+1)

satisfies u

_0(i+1)

≥ c exp(−a|x|

²

) for some constants c and a, then an analogous estimate holds true for u

i

(τ ).

Replacing u

_0(i+1)

by u

_i+1

(τ

₀

) for some τ

₀

if necessary, we get the conclu- sion.

To fix ideas, we assume henceforth that max

i

α

i

= α

1

if

m

Y

j=1

p

j

> 1.

Lemma 2.2. Let (u

1

(t), . . . , u

m

(t)) be a bounded solution of (1.1) in some strip [0, T ) × R

^N

, 0 < T ≤ ∞. Assume that Q

m

j=1

p

_j

> 1. Let n be an integer such that

(2.2) p

i

≥ 1 for 1 ≤ i ≤ n, p

i

< 1 for n < i ≤ m.

Then there exists a positive constant C, depending on p

i

only, such that (2.3) t

^Qmj=n+1pjα1

kS(t)u

Qm j=n+1pj

01

k

_∞

≤ C if n < m,

t

^α1

kS(t)u

₀₁

k

_∞

≤ C if n = m, t ∈ [0, T ).

P r o o f. Using (2.1) for i = 1 in (2.1) for i = m we get u

_m

(t) ≥

t

\

0

S(t − s)(S(s)u

₀₁

)

^pm

ds.

Consider first n = m. Then by the Jensen inequality for p

m

≥ 1 we have

(2.4) u

m

(t) ≥

t

\

0

(S(t)u

01

)

^pm

ds = t(S(t)u

01

)

^pm

.

Using (2.4) in (2.1) for i = m − 1 and the Jensen inequality for p

m

p

m−1

≥ 1 we get

u

m−1

(t) ≥

t

\

0

S(t − s)(s(S(s)u

01

)

^pm

)

^pm−1

ds

≥

t

\

0

s

^pm−1

(S(t)u

01

)

^pmpm−1

ds

≥ 1

p

m−1

+ 1 (S(t)u

01

)

^pmpm−1

t

^pm−1+1

.

We now reason by induction. Assume that

(2.5) u

m−j

(t) ≥ c

j

(S(t)u

01

)

^πj

t

^rj

, where

(2.6)

π

j

= p

m−j

π

j−1

, π

0

= p

m

, r

j

= p

m−j

r

j−1

+ 1, r

0

= 1, c

j

= 1

r

j

c

^p_j−1^m−j

, c

0

= 1.

Substituting (2.5) in (2.1) for i = m − j − 1, by the Jensen inequality we obtain

u

m−(j+1)

(t) ≥

t

\

0

S(t − s)[c

j

(S(s)u

01

)

^πj

s

^rj

]

^pm−(j+1)

ds

≥ c

^p_j^m−(j+1)

t

\

0

s

^rjpm−j−1

(S(t)u

01

)

^{pm−j−1πj}

ds

≥ c

^p_j^m−j−1

r

j

p

m−j−1

+ 1 (S(t)u

01

)

^{pm−j−1πj}

t

^{rjpm−j−1+1}

, whence, by (2.6),

u

m−j−1

(t) ≥ c

j+1

(S(t)u

01

)

^πj+1

t

^rj+1

and for j = m − 1, (2.5) gives

(2.7) u

1

(t) ≥ c

m−1

(S(t)u

01

)

^πm−1

t

^rm−1

, where

(2.8)

π

_m−1

=

m

Y

j=1

p

_j

,

r

_m−1

= (. . . (((p

_m−1

+ 1)p

_m−2

+ 1)p

_m−3

+ 1) . . .)p

₁

+ 1

= α

1

 Y

^m

j=1

p

j

− 1  ,

c

⁻¹_m−1

= (p

_m−1

+ 1)

^p1...pm−2

(p

_m−2

(p

_m−1

+ 1) + 1)

^p1...pm−3

. . . . . . [(. . . ((p

_m−1

+ 1)p

_m−2

+ 1) . . .)p

₁

+ 1].

Iterating this scheme and setting p = π

m−1

, r = r

m−1

, we obtain an estimate for u

₁

(t) from below:

(2.9) u

1

(t) ≥

m

Y

i=1

A

^k_i

(S(t)u

01

)

^pk

t

^{r(1+p+p2+...+pk−1)}

, where

1/A

^k₁

= r

^pk−1

[r(p + 1)]

^pk−2

[r(p

²

+ p + 1)]

^pk−3

. . . (2.10)

. . . [r(p

^k−1

+ . . . + p

²

+ p + 1)],

1/A

^k_m

= (p

_m

r + 1)

^pk−1/pm

[p

_m

r(p + 1) + 1]

^pk−2/pm

× [p

m

r(p

²

+ p + 1) + 1]

^pk−3/pm

. . .

. . . [p

m

r(p

^k−2

+ . . . + p

²

+ p + 1) + 1]

^p/pm

,

1/A

^k_m−1

= (p

m−1

+ 1)

^{pk/(pm−1pm)}

(p

m−1

p

m

r + p

m−1

+ 1)

^{pk−1/(pm−1pm)}

× [p

m−1

p

m

r(p + 1) + p

m−1

+ 1]

^{pk−2/(pm−1pm)}

. . . (2.10)

[cont.]

. . . [p

m−1

p

m

r(p

^k−2

+. . .+p

²

+p+1)+p

m−1

+1]

^p/(pm−1pm)

, 1/A

^k_m−i

= [p

m−i

(. . . (p

m−1

+ 1) + . . . + 1)]

^{pk/(pm−i...pm)}

× [p

_m−i

(p

_m−i+1

(. . . (p

_m

r + 1)

+ . . . + 1) + 1) + 1]

^{pk−1/(pm−i...pm)}

× [p

_m−i

(. . . (p

_m

r(p+1)+1)+. . .+1)+1]

^{pk−2/(pm−i...pm)}

. . . . . . [p

m−i

(. . . (p

m

r(p

^k−2

+ . . . + p + 1) + 1)

+ . . . + 1) + 1]

^{p/(pm−i...pm)}

. We can rewrite 1/A

^k₁

in (2.10) as

(2.11) 1/A

^k₁

= r

^{(pk−1)/(p−1)}

k−1

Y

j=1

 p

^j+1

− 1 p − 1



p^k−j−1

.

By assumption and p > 1, we have max α

i

= α

1

, so the following inequalities remain true for j ≥ 0:

r(1 + p + . . . + p

^j

) = α

₁

(p − 1)(1 + p + . . . + p

^j

) > 1, r(1 + p + . . . + p

^j

) > 1 + p

m−1

,

r(1 + p + . . . + p

^j

) > 1 + p

_m−2

p

_m−1

+ p

_m−2

, . . . ,

r(1 + p + . . . + p

^j

) > 1 + p

m−i

. . . p

m−1

+ p

m−i

. . . p

m−2

+ . . . + p

m−i

, for i = 1, . . . , m − 2.

Using this in (2.10) for A

^k₂

to A

^k_m

we obtain

1/A

^k_m

≤ [r(p

_m

+ 1)]

^pk−1/pm

[r(p + 1)(p

_m

+ 1)]

^pk−2/pm

. . . (2.12)

. . . [r(p

^k−2

+ . . . + p

²

+ p + 1)(p

m

+ 1)]

^p/pm

= [r(p

m

+ 1)]

^{pm1 pk −pp−1}



k−2

Y

j=1

 p

^j+1

− 1 p − 1



p^k−j−1



1/pm

, 1/A

^k_m−1

≤ [r(p

m−1

p

m

+ 1)]

^{pk−1/(pm−1pm)}

× [r(p + 1)(p

_m−1

p

_m

+ 1)]

^{pk−2/(pm−1pm)}

. . . . . . [r(p

^k−2

+ . . . + p

²

+ p + 1)

× (p

m−1

p

m

+ 1)]

^p/(pm−1pm)

r

^{pk/(pm−1pm)}

= r

^{pk/(pm−1pm)}

[r(p

m−1

p

m

+ 1)]

^pm−1pm1

pk −p p−1

×



k−2

Y

j=1

 p

^j+1

− 1 p − 1



p^k−j−1



1/(pm−1pm)

, . . . , (2.13) (A

^k_m−i

)

^{−Qij=0pm−j}

≤ h Y

ⁱ

j=0

p

m−j

+ 1 

r i

p^k−1

h

r(p + 1)



i

Y

j=0

p

m−j

+ 1 i

p^k−2

. . .

. . . h

r(p

^k−2

+ . . . + p + 1)  Y

ⁱ

j=0

p

m−j

+ 1 i

p

r

^pk

= r

^pk

h Y

ⁱ

j=0

p

_m−j

+ 1 i

^{pk −p}_p−1 ^k−2

Y

j=1

 p

^j+1

− 1 p − 1



p^k−j−1

, for i = 1, . . . , m − 2.

Substituting (2.11), (2.12) and (2.13) for i = 1, . . . , m − 2 into (2.9) we infer

(2.14) (S(t)u

01

)

^pk

t

^{rpk −1p−1}

≤ r

pk −1

p−1[1+_pm^p + ^p

pm−1pm+...+Qm−2^p j=0 pm−j]

×

m−2

Y

i=0

h

1 +

i

Y

j=0

p

m−j



^{pk −p}_p−1(Qi

j=0pm−j)⁻¹

i u(t)

×



k−1

Y

j=1

 p

^j+1

− 1 p − 1



p^k−j−1



1+_pm¹ + ¹

pm−1pm+...+(Qm−2

j=0 pm−j)⁻¹

. Using (1.5) and (2.2) we observe that

1 + p p

m

+ . . . + p p

m

. . . p

2

= 1 + p

1

+ p

1

p

2

+ . . . + p

1

. . . p

m−1

= r, 1 + 1

p

_m

+ . . . + 1

p

_m

. . . p

₂

= 1 + r − 1 p ,

i

Y

j=0

p

m−j

≤

m−1

Y

j=0

p

m−j

= p for all i = 0, 1, . . . , m − 2, whence by (2.14) we obtain

(2.15) (S(t)u

01

)

^pk

t

^{rpk −1p−1}

≤ r

^{pk −1p−1r}

(1 + p)

^{pk −pp−1 r−1p}



k−1

Y

j=1

 p

^j+1

− 1 p − 1



k−j−1



1+^r−1_p

u(t).

Now, we have to estimate B

k

=

k−1

Y

j=1

 p

^j+1

− 1 p − 1



k−j−1

=

k

Y

j=1

 p

^j

− 1 p − 1



k−j

. Thus

ln B

k

=

k

X

j=1

p

^k−j

[ln(p

^j

− 1) − ln(p − 1)]

(2.16)

≤ p ln p p

^k

− 1

(p − 1)

²

− ln(p − 1) p

^k

− 1

p − 1 = p

^k

− 1

p − 1 ln  p

^p−1p

p − 1



= ln  p

^p−1p

p − 1



^{pk −1}_p−1



= ln[b

^{pk −1p−1}

], where b = p

^p/(p−1)

/(p − 1) > 0.

Employing this in (2.15) we finally have S(t)u

01

t

^r

pk −1 pk(p−1)

≤ r

(pk−1)r

pk(p−1)

(1 + p)

pk −p pk(p−1)

r−1 p

× b

pk −1

pk(p−1)(1+^r−1_p )

ku(t)k

^1/p_∞ ^k

. Using ku(t)k

∞

< ∞ and letting k → ∞, we get

(2.17) t

^r/(p−1)

kS(t)u

01

k

∞

≤ c < ∞, where c = c(p, r, b) = c(p

i

), i = 1, . . . , m.

Noting r = α

1

(p − 1) we get the assertion.

Now we assume that n < m, i.e. p

i

< 1 for n < i ≤ m. Then instead of (2.4) we get, using the Jensen inequality for p

m

< 1,

u

m

(t) ≥

t

\

0

S(t)u

^p₀₁^m

ds = tS(t)u

^p₀₁^m

and, analogously, for p

_m−j

< 1 (i.e. for m − j > n) we can prove inductively, exactly as (2.5),

(2.18) u

_m−j

(t) ≥ c

_j

S(t)u

^π₀₁^j

t

^rj

, where c

j

, π

j

, r

j

are given by (2.6).

Using (2.18) for j = m−n−1 in (2.1) for i = n, and the Jensen inequality for p

n

≥ 1, we get

u

n

(t) ≥ c

m−n

(S(t)u

^π₀₁^m−n−1

)

^pn

t

^rm−n

where π

m−n−1

=

m

Y

j=n+1

p

j

. Again, arguing by induction as in the proof of formula (2.5), we conclude

u

m−j

(t) ≥ c

j

(S(t)u

^π₀₁^m−n−1

)

^pn...pm−j

t

^rj

for j ≥ m − n, and

u

₁

(t) ≥ c

_m−1

(S(t)u

^π₀₁^m−n−1

)

^pn...p1

t

^rm−1

(2.19)

= c

m−1

(S(t)u

^π₀₁^m−n−1

)

^{πm−1/πm−n−1}

t

^rm−1

, where π

m−n−1

= Q

m

i=n+1

p

i

by (2.6) and π

m−1

, r

m−1

, c

m−1

are given by (2.8).

Let us remark that π

m−1

/π

m−n−1

= Q

n

i=1

p

i

≥ 1 and since π

m−1

> 1, also Q

l

i=1

p

i

≥ π

m−1

> 1 for any n ≤ l ≤ m. It follows that we apply hence- forth the Jensen inequality with an exponent greater than 1. Therefore, repeating the above considerations, we get instead of (2.9),

(2.20) u

₁

(t) ≥

m

Y

i=1

A

^k_i

(S(t)u

^π₀₁^m−n−1

)

^{pk/πm−n−1}

t

^{rpk −1p−1}

, where the A

^k_i

are defined by (2.10). To estimate Q

m

i=1

A

^k_i

, we proceed in the same way as before (because we have not used n there), whence

kS(t)u

^π₀₁^m−n−1

k

∞

t

^{πm−n−1α1}

≤ c < ∞ after letting k → ∞. Thus the proof is complete.

Lemma 2.3. Let the assumptions of Lemma 2.2 be satisfied. Then we can find a constant c > 0, c = c(p

i

), i = 1, . . . , m, such that for t > 0, (2.21) kS(t)u

^π₁^m−n−1

k

∞

t

^{πm−n−1α1}

≤ c < ∞.

P r o o f. For τ, t ≥ 0 we have, by (2.1) for i = 1, u

1

(t + τ ) = S(t + τ )u

01

+

t+τ

\

0

S(t + τ − s)u

^p₂¹

(s) ds

= S(t)u

1

(τ ) +

t

\

0

S(t − s)(u

2

(τ + s))

^p1

ds.

It now follows that we can replace u

01

by u

1

(τ ) in (2.3), whence we have the conclusion by Lemma 2.1 and setting t = τ .

3. Proof of Theorem 1.3. We prove Theorem 1.3 by contradiction.

Assuming that (2.2) holds and max α

i

≥ N/2 we see by (1.5) that α

1

≥ N/2 and p = Q

m

j=1

p

j

> 1. We shall derive a lower bound for a solution which contradicts the bounds obtained in Lemmas 2.2 and 2.3.

By Lemma 2.1 we have, for nontrivial initial data of (1.1), u

01

(x) ≥ c

∗

e

^−a|x|2

for some constants c

∗

> 0, a > 0. Employing this in (2.1) for i = 1, since S(t)e

^−a|x|2

= (1 + 4at)

^−N/2

exp  −a|x|

²

1 + 4at



,

we get

(3.1) u

1

(t) ≥ S(t)u

01

≥ c

∗

(1 + 4at)

^−N/2

exp  −a|x|

²

1 + 4at

 . Using this bound in (2.1) for i = m, we have

u

m

(t) ≥

t

\

0

S(t − s)(u

1

(s))

^pm

ds (3.2)

≥ c

^p_∗^m

t

\

0

S(t − s)(1 + 4as)

^{−N pm/2}

exp  −ap

m

|x|

²

1 + 4as

 ds

= c

^p_∗^m

t

\

0

(1 + 4as)

^{−N (pm−1)/2}

(1 + 4as + 4ap

_m

(t − s))

^−N/2

× exp

 −ap

_m

|x|

²

1 + 4as + 4ap

_m

(t − s)

 ds.

Recall the definition of the number n in Lemma 2.2. By induction, we prove that the following estimate holds for m − j > n and some constant C:

(3.3) u

m−j

(t) ≥ c(1 + 4at)

^−N/2

(4at − ̺

j

)

^βj

exp  −aπ

j

|x|

²

1 + 4aπ

j

t

 , where ̺

j

= ̺

^2j−1

for j > 0, ̺

0

= 0, ̺ > 1,

(3.4)

π

_j

= p

_m−j

π

_j−1

=

m

Y

k=m−j

p

_j

, π

₀

= p

_m

,

β

j

= p

m−j

β

j−1

+ N

2 (1 − p

m−j

) + 1, β

−1

= 0, and t > ̺

j

/(4a).

Let n < m. Then p

m

< 1, p

n+1

< 1 and p

n

≥ 1 by definition, and for the function f

_m

(s) = 1 + 4as + 4ap

_m

(t − s), we have f

^′

(s) = 4a(1 − p

_m

) > 0.

Thus f

m

(0) ≤ f

m

(s) ≤ f

m

(t) for s ∈ [0, t]. Using this, we estimate (3.2) as follows:

u

m

(t) ≥ c

^p_∗^m

(1 + 4at)

^−N/2

exp  −ap

m

|x|

²

1 + 4ap

m

t



t\

0

(1 + 4as)

^{−N (pm−1)/2}

ds

≥ c(1 + 4at)

^−N/2

(4at)

^{1−N (pm−1)/2}

exp  −ap

m

|x|

²

1 + 4ap

m

t



,

so we have (3.3) for j = 0.

Substituting (3.3) in (2.1) for i = m − j − 1, we have u

_m−(j+1)

(t)

≥

t

\

̺j/(4a)

S(t − s)(u

m−j

(s))

^pm−j−1

ds

≥ c

t

\

̺j/(4a)

S(t − s)(1 + 4as)

^{−N pm−j−1/2}

(4as − ̺

j

)

^{pm−j−1βj}

× exp  −ap

m−j−1

π

j

|x|

²

1 + 4aπ

j

s

 ds

= c

t

\

̺j/(4a)

(1 + 4as)

^{−N pm−j−1/2}

(1 + 4aπ

j

s)

^N/2

× (1 + 4aπ

j

s + 4aπ

j

p

m−j−1

(t − s))

^−N/2

× exp

 −ap

_m−j−1

π

_j

|x|

²

1 + 4aπ

j

s + 4aπ

j

p

m−j−1

(t − s)



(4as − ̺

j

)

^{pm−j−1βj}

ds.

Consider f

m−j−1

(s) = 1+4aπ

j

s+4aπ

j+1

(t−s). From f

_m−j−1^′

(s) = 4aπ

j

(1−

p

_m−j−1

) > 0, we can conclude u

m−(j+1)

(t)

≥ c

^p_j^m−j−1

(1 + 4aπ

_j

t)

^−N/2

exp  −aπ

j+1

|x|

²

1 + 4aπ

j+1

t



×

t

\

̺j/(4a)

(1 + 4as)

^{−N pm−j−1/2}

(4as − ̺

j

)

^{pm−j−1βj}

(1 + 4aπ

j

s)

^N/2

ds.

To estimate the integral, notice that

(1 + 4aπ

j

t)

^N/2

≥ π

_j^N/2

(1 + 4at)

^N/2

since π

j

< 1, (3.5)

(1 + 4aπ

j

t)

^−N/2

≥ (1 + 4aπ

m−1

t)

^−N/2

≥ π

^−N/2_m−1

(1 + 4at)

^−N/2

, (3.6)

and

(3.7) 4at − ̺

_j

≥ c

₀

(1 + 4at) for t > ̺

²_j

/(4a) with c

₀

= ̺

_j

̺

j

− 1

̺

²_j

+ 1 . Thus, integrating we get

u

_m−j−1

(t) ≥ c(1 + 4at)

^−N/2

exp  −aπ

j+1

|x|

²

1 + 4aπ

j+1

t



× (4at − ̺

²_j

)

^{1+N (1−pm−j−1)/2+pm−j−1βj}

for t > ̺

²_j

/(4a). This, in view of (3.4), gives (3.3) with j replaced by j + 1.

Analogously, for m − j ≤ n we have the following estimate for any t > ̺

_j

/(4a):

(3.8) u

m−j

(t) ≥ c(1 + 4at)

^−N/2

(4at − ̺

j

)

^βj

exp

 −aπ

j

|x|

²

1 + 4aπ

m−n−1

t

 , with ̺

_j

, β

_j

, π

_j

defined in (3.4).

We proceed in a similar way. Assuming (3.8) for some j ≥ m − n and using (2.1) for i = m − (j + 1) we obtain

u

m−j−1

(t) ≥ c

t

\

̺j/(4a)

(1 + 4as)

^{−N pm−j−1/2}

(4as − ̺

j

)

^{pm−j−1βj}

× (1 + 4aπ

m−n−1

s)

^N/2

× (1 + 4aπ

m−n−1

s + 4aπ

j

p

m−j−1

(t − s))

^−N/2

× exp

 −ap

m−j−1

π

j

|x|

²

1 + 4aπ

_m−n−1

s + 4aπ

_j

p

_m−j−1

(t − s)

 ds.

Therefore, we have g

m−j−1

(s) = 1 + 4a(π

m−n−1

s + π

j

p

m−j−1

(t − s)) so g

^′_m−j−1

(s) = 4aπ

_m−n−1



1 − π

_j+1

π

m−n−1



= 4aπ

_m−n−1

 1 −

n

Y

k=m−j−1

p

_k



and since p

k

≥ 1 for k ≤ n we get g

_m−j−1^′

(s) ≤ 0. Thus u

m−j−1

(t) ≥ c(1 + 4aπ

j+1

t)

^−N/2

exp

 −aπ

j+1

|x|

²

1 + 4aπ

_m−n−1

t



×

t

\

̺j/(4a)

(1 + 4as)

^{−N pm−j−1/2}

(4as − ̺

j

)

^{pm−j−1βj}

(1 + 4aπ

m−n−1

s)

^N/2

ds.

Using (3.5) with π

j

replaced by π

m−n−1

< 1, (3.6) with π

j+1

and integrating we obtain (3.8) with j replaced by j + 1 and t > ̺

_j+1

/(4a).

We need a lower bound for u

1

(t). By assumption Q

m

j=1

p

j

> 1, whence by (2.2), p

1

> 1, but we can have p

2

< 1 or p

2

≥ 1. In the former case we have m − j = 2 > n = 1, therefore (3.3) gives

(3.9) u

2

(t) ≥ c(1 + 4at)

^−N/2

(4at − ̺

m−2

)

^βm−2

exp  −aπ

m−2

|x|

²

1 + 4aπ

m−2

t

 . Otherwise, from m − j = 2 ≤ n and (3.8) we get

(3.10) u

2

(t) ≥ c(1 + 4at)

^−N/2

(4at − ̺

m−2

)

^βm−2

exp

 −aπ

m−2

|x|

²

1 + 4aπ

m−n−1

t



.

Defining M = π

m−n−1

(with π

m−2

= M for n = 1), we conclude (3.11) u

2

(t) ≥ c(1 + 4at)

^−N/2

(4at − ̺

m−2

)

^βm−2

exp  −aπ

m−2

|x|

²

1 + 4aM t

 . Using (2.1) for i = 1 we obtain, as p

₁

> 1,

(3.12) u

1

(t) ≥

t

\

̺m−2/(4a)

S(t − s)(u

2

(s))

^p1

ds

≥ c

t

\

̺m−2/(4a)

(1 + 4as)

^{−N p1/2}

(1 + 4aM s)

^N/2

× (1 + 4aM s + 4aπ

m−1

(t − s))

^−N/2

× exp

 −aπ

m−1

|x|

²

1 + 4aM s + 4aπ

m−1

(t − s)



(4as − ̺

m−2

)

^p1βm−2

ds

≥ c(1 + 4aπ

_m−1

t)

^−N/2

exp  −aπ

m−1

|x|

²

1 + 4aM t



×

t

\

̺m−2/(4a)

(4as − ̺

m−2

)

^p1βm−2

(1 + 4aM s)

^N/2

(1 + 4as)

^{−N p1/2}

ds

≥ c(1 + 4at)

^−N/2

exp  −aπ

m−1

|x|

²

1 + 4aM t



\t

̺m−1/(4a)

(1 + 4as)

^βm−1−1

ds.

By recursive definition (3.4) and a routine inductive calculation we get β

_j

= 1 + p

_m−j

+ p

_m−j

p

_m−j−1

+ . . .

. . . + p

m−j

. . . p

m−1

+ N

2 (1 − p

m−j

. . . p

m

), therefore, using the definition of p, r, α

₁

(see (1.5), (2.8)), we obtain

β

m−1

= 1 + p

1

+ p

1

p

2

+ . . . + p

1

. . . p

m−1

+ N

2 (1 − p

1

. . . p

m

)

= r + N

2 (1 − p).

Since α

1

≥ N/2, p > 1 and r = α

1

(p − 1) we see that β

m−1

≥ 0, so (3.12) gives

u

1

(t) ≥ (1 + 4at)

^−N/2

exp

 −ap|x|

²

1 + 4aM t

 log

 1 + 4at 1 + ̺

m−1



for t > ̺

m−1

/(4a). It follows that S(t)(u

1

(t))

^M

≥ (1 + 4at)

^{−N M /2}

 log

 1 + 4at 1 + ̺

m−1



M

S(t) exp  −apM |x|

²

1 + 4aM t



= c(1 + 4at)

^{−N M /2}

 log

 1 + 4at 1 + ̺

m−1



M

× (1 + 4aM (p + 1)t)

^−N/2

(1 + 4aM t)

^N/2

exp

 −apM |x|

²

1 + 4aM (p + 1)t



≥ c(1 + 4at)

^{−N M /2}

 1 + 4aM t (1 + 4aM t)(p + 1)



N/2

exp

 −apM |x|

²

1 + 4aM (p + 1)t



×

 log

 1 + 4at 1 + ̺

m−1



M

.

Putting x = 0 in the last estimate, we have (1 + 4at)

^{N M /2}

S(t)(u

1

(t, 0))

^M

≥ c

 log

 1 + 4at 1 + ̺

m−1



M

.

Therefore, for t > max{1, ̺

m−1

/(4a)} we obtain, using α

1

≥ N/2 and M = π

_m−n−1

,

(3.13) t

^{πm−n−1α1}

S(t)(u

1

(t, 0))

^πm−n−1

≥ c

 log

 1 + 4at 1 + ̺

_m−1

 

πm−n−1

. It is clear that for t large enough (3.13) is incompatible with the bound (2.21). This implies that u

1

(t) blows up in a finite time, which concludes the proof of the theorem in this case.

If n = m, i.e. p

i

≥ 1, i = 1, . . . , m, then instead of (3.8) we obtain (3.14) u

m−j

(t) ≥ c(1 + 4at)

^−N/2

(4at − ̺

j

)

^βj

exp  −aπ

j

|x|

²

1 + 4at



for t > ̺

j

/(4a) and with ̺

j

, β

j

, π

j

defined as above. Then we can repeat our considerations concerning the lower bound for u

1

(t) in the case n < m, simply using (3.14) instead of (3.3) and (3.8) and, starting from (3.11), replacing everywhere M = π

_m−n−1

by 1. Finally, we conclude that for t > max{1, ̺

m−1

/(4a)}

(3.15) t

^α1

S(t)u

1

(t, 0) ≥ c log

 1 + 4at 1 + ̺

m−1

 .

This contradicts (2.21) in the case n = m and the proof is complete.

4. Proofs of Theorems 1.1 and 1.2 Proof of Theorem 1.1. By assumption, Q

m

j=1

p

_j

= 1. We are looking for a global supersolution to (1.1) of the form

(4.1)



 u

1

.. . u

_m



 =





A

1

e

^θ1t

.. . A

_m

e

^θmt



 .

We choose A

i

, i = 1, . . . , m, so large that ku

0i

k

L^∞

≤ A

i

. Then (4.1) is a supersolution to (1.1) if for all t ≥ 0,

(4.2) u

it

− ∆u

i

≥ u

^p_i+1ⁱ

, i = 1, . . . , m − 1, u

_mt

− ∆u

_m

≥ u

^p₁^m

.

This is equivalent to

(4.3) θ

i

> A

⁻¹_i

A

^p_i+1ⁱ

exp((p

i

θ

i+1

− θ

i

)t), i = 1, . . . , m − 1, θ

m

> A

⁻¹_m

A

^p₁^m

exp((p

m

θ

1

− θ

m

)t).

If we take θ

i+1

= θ

i

/p

i

, i = 1, . . . , m − 1, then from Q

m

j=1

p

j

= 1 we get θ

m

= θ

1

/ Q

m−1

j=1

p

j

= θ

1

p

m

, so (4.3) holds for θ

1

large enough. Thus, the solution of (1.1) is global.

Proof of Theorem 1.2. In this case, from p

j

≥ 0, (1.4), (1.5), we have Q

m

j=1

p

j

< 1. We assert that (u

i

)

^m_i=1

of the form (4.4)



 u

1

.. . u

m



 =







A

1

(t + t

0

)

^θ1

.. . A

m

(t + t

0

)

^θm







is a global supersolution to (1.1) for some positive constants A

i

, θ

i

. We have to choose t

0

such that u

i

(x, 0) ≥ u

0i

for x ∈ R

^N

. The inequalities (4.2) imply that

(4.5) θ

i

− p

i

θ

i+1

≥ 1, i = 1, . . . , m − 1, θ

m

− p

m

θ

1

≥ 1.

Noting that (4.5) has the form (A − I)(−θ) ≥ 1 where θ = (θ

1

, . . . , θ

m

), 1 = (1, . . . , 1), A is given by (1.2), we assume θ

i

= −α

i

, so θ

i

> 0 as max α

_i

< 0. Then (4.4) satisfies (4.2) provided that

(4.6) A

i

θ

i

≥ A

^p_i+1ⁱ

, i = 1, . . . , m − 1, A

_m

θ

_m

≥ A

^p₁^m

,

and t

0

is large enough. Thus, the proof is complete.

References

[EH] M. E s c o b e d o and M. A. H e r r e r o, Boundedness and blow up for a semilinear reaction-diffusion system, J. Differential Equations 89 (1991), 176–202.

[EL] M. E s c o b e d o and H. A. L e v i n e, Critical blow up and global existence numbers for a weakly coupled system of reaction-diffusion equations, Arch. Rational Mech.

Anal. 129 (1995), 47–100.

[F1] H . F u j i t a, On the blowing up of solutions of the Cauchy problem for u

t

= ∆u + u

^1+α

, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 13 (1966), 109–124.

[F2] —, On some nonexistence and nonuniqueness theorems for nonlinear parabolic equations, in: Proc. Sympos. Pure Math. 18, Amer. Math. Soc., 1970, 105–113.

[R] J. R e n c l a w o w i c z, Global existence and blow up of solutions for a completely coupled Fujita type system of reaction-diffusion equations, Appl. Math. (Warsaw) 25 (1998), 313–326.

Joanna Renc lawowicz Institute of Mathematics Polish Academy of Sciences

´ Sniadeckich 8

00-950 Warszawa, Poland E-mail: jr@impan.gov.pl

Received on 4.8.1999