First, we note that if A is reducible, then the full system (1.1) can be reduced to several sub-systems, independent of each other. Therefore, in the following, we assume that A is irreducible. In addition, we suppose that k1 - m1 ≤ k2 - m2 ≤ · · · k
n
- m
n
.
Let
be the first eigenfunction of
with the first eigenvalue
, normalized by
, then
,
in Ω and
and
on ∂Ω (see [14–16]).
Thus, there exist some positive constants
,
,
, and
such that
We also have
provided
with
and some positive constant
. For the fixed
, there exists a positive constant
such that
if
.
Proof of the sufficiency. We divide this proof into three different cases.
Case 1. (k
i
< m
i
(i = 1,..., n)). Let
where Q
i
satisfies
, and constants P
i
, α
i
(i = 1,..., n) remain to be determined. Since
, by performing direct calculations, we have
in Ω × ℝ+. By setting
if m
i
≥ 1,
if m
i
< 1, we have one the boundary that
we have
if
and
Note that k
i
< m
i
(i = 1,..., n). From Lemmas 2.2 and 2.3, we know that inequalities (3.4) and (3.5) hold for suitable choices of P
i
, α
i
(i = 1,..., n). Moreover, if we choose P
i
, α
i
to be large enough such that
then
,
. Therefore, we have proved that
is a global upper solution of the system (1.1). The global existence of solutions to the problem (1.1) follows from the comparison principle.
Case 2. (k
i
≥ m
i
(i = 1,..., n)). Let
where
if m
i
≥ 1,
if m
i
< 1,
,
,
,
are defined in (3.1) and (3.2), α
i
(i = 1,..., n) are positive constants that remain to be determined, and
Since -ye-y≥ -e-1 for any y > 0, we know that
. Thus, for (x, t) ∈ Ω × ℝ+, a simple computation shows that
In addition, we have
Noting
on ∂Ω, we have on the boundary that
Then, we have
if
From Lemma 2.2, we know that inequalities (3.7) hold for suitable choices of α
i
(i = 1,..., n). Moreover, if we choose ∞
i
to be large enough such that
then
. Therefore, we have shown that
is an upper solution of (1.1) and exists globally. Therefore,
, and hence the solution (u1,..., u
n
) of (1.1) exists globally.
Case 3. (k
i
< m
i
(i = 1,..., s); k
i
≥ m
i
(i = s + 1,..., n)). Let
be as in (3.3) and
where
, and A
i
are as in case 2. By Lemma 2.3, we choose P
i
≥ (log Q
i
)-1||ui 0||∞ (i = 1,..., s) and M
i
≥ max{1, ||ui 0||∞} (i = s + 1,..., n) such that
Set
By similar arguments, in cases 1 and 2, we have on the boundary that
Therefore employing (3.8), we see that
if we knew
We deduce from Lemma 2.2 that (3.9) holds for suitable choices of α
i
(i = 1,..., n). Moreover, we can choose α
i
large enough to assure that
Then, as in the calculations of cases 1 and 2, we have
. We prove that
is an upper solution of (1.1), so (u1,..., u
n
) exists globally.
Proof of the necessity.
Without loss of generality, we first assume that all the lower-order principal minor determinants of A are non-negative, and |A| < 0, for, if not, there exists some l th-order (1 ≤ l < n) principal minor determinant detAl × lof A = (a
ij
)n×nwhich is negative. Without loss of generality, we may consider that
and all of the sth-order (1 ≤ s ≤ l - 1) principal minor determinants detAs × sof Al × lare non-negative. Then, we consider the following problem:
Note that
. If we can prove that the solution (w1,..., w
l
) of (3.10) blows up in finite time, then (w1,... w
l
, δ,..., δ) is a lower solution of (1.1) that blows up in finite time. Therefore, the solution of (1.1) blows up in finite time.
We will complete the proof of the necessity of our theorem in three different cases.
Case 1. (k
i
< m
i
(i = 1,..., n)). Let
where
,
,
,
, the α
i
are as given in Lemma 2.4 and satisfy
,
By direct computation for
, we have
For
, we have
Thus, by (3.12) and Lemma 2.4, we have
We confirm that (u1,..., u
n
) is a lower solution of (1.1), which blows up in finite time. We know by the comparison principle that the solution (u1,..., u
n
) blows up in finite time.
Case 2. (k
i
≥ m
i
(i = 1,..., n)). Let
if m
i
< 1,
if m
i
≥ 1. for k
i
≥ m
i
(i = 1,..., n), set
where α
i
(i = 1,..., n) are to determined later and
By a direct computation, for x ∈ Ω, 0 < t < c/b, we obtain that
If
, we have
, and thus
On the other hand, since -ye-y≥ -e-1 for any y > 0, we have
We have by (3.16), (3.18), and (3.19) that
.
If
, then
, and then
It follows from (3.16), (3.17), and (3.20) that
.
We have on the boundary that
Moreover, by (3.14) and Lemma 2.4, we have that
(3.15), (3.21), and (3.22) imply that
. Therefore, (u1,..., u1) is a lower solution of (1.1).
For k
i
= m
i
(i = 1,..., n), let
For k
i
= m
i
(i = 1,..., s) and k
i
> m
i
(i = s + 1,..., n), let
as in (3.13) and (3.23). Using similar arguments as above, we can prove that (u1,..., u
n
) is a lower solution of (1.1). Therefore, (u1,..., u
n
) ≤ (u1,..., u
n
). Consequently, (u1,..., u
n
) blows up in finite time.
Case 3. (k
i
< m
i
(i = 1,..., s); k
i
≥ m
i
(i = s + 1,..., n)). Let
be as in (3.11) and
where α
i
's are to determined later and
Based on arguments in cases 1 and 2, we have
for
. Furthermore, for
, we have
Thus,
holds if
From Lemma 2.4, we know that inequalities (3.24) hold for suitable choices of α
i
(i = 1,..., n). We show that (u1,..., u
n
) is a lower solution of (1.1). Since (u1,..., u
n
) blows up in finite time, it follows that the solution of (1.1) blows up in finite time.