dx1
Form the first equation, we have
(t1 )  c1 x2 (t1 )  0
dt
since x2 (0)  0, this implies that ther e is a time t 2  t1
such that
dx2
x2 (t 2 )  0 and
(t 2 )  0
dt
dx2
Form the forth equation, we have
(t 2 )  c1 x1 (t 2 )  0
dt
which implies that t 2  t1 and x1 (t 2 )  0,
This is a contraditi on.
Theorem 2 :
Solutions of the coupled three - level food chains are bounded
Proof :
We consider first the boundednes s of x1 and x2.
dx1 dx2
x
x

 rx1 (1  1 )  rx2 (1  2 )
dt
dt
k
k
r
 r ( x1  x2 )  ( x1  x2 ) 2
2k
d
r 2
Let   x1  x2 , then we have
 r  
dt
2k
A standard comparison argument shows that x1 and x2 are bounded
From the coupled system, we have
dx1 dx2 dy1 dy2 dz1 dz2





dy
dt
dt
dt
dt
dt
x1
x2
 rx1 (1  )  rx2 (1  )  d1 ( y1  y2 )  d 2 ( z1  z2 )
k
k
Let l  min d1 , d 2 , then there is M  0 such that
x1
x2
M  rx1 (1  )  rx2 (1  )  l ( x1  x2 )
k
k
Thus
dx1 dx2 dy1 dy2 dz1 dz2





 M  l ( x1  x2  y1  y2  z1  z2 )
dt dt dt
dt dt dt
Let  x1  x2  y1  y2  z1  z2 , it follows that
d
 M  l .
dt
By the comparison theorem, it follows that y1 , y2 , z1 and z2
are bounded
‧Main result
Define an unbounded region W of the coupling strength
c1 , c2 and c3 by the following
W ( A, B, C )  (c1, c2 , c3 ) c1  A, c2  Bc1, c3  Cc2
