past year exam q ch3
TRANSCRIPT
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Case 2
Chew T S MA1506-14 Chapter 3 1
2
2
2
4( )( ) 14
2( ) 2 2
B B s E B BB sE
s s s s
2( )F N sN BN E
2
4
BE
s
where1 2
B
s
2
2
1
4( )( ) 14
2( ) 2 2
B B s E BB sE
s s s
3.4 Harvesting
Warning:1 2
2 1NOT
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2
1 2( ) ( )( )( )
dNF N s N N
dt
1
2
2( )F N sN BN E
0
dN
dt 0
dN
dt 0
dN
dt
N
3.4 Harvesting
22 10 12 ( 2 6)( 2) ( 2)( 3)( 2)N N N N N N
2 22 10 12 ( 2)( 5 6) ( 2)( 3)( 2)N N N N N N
WHY?
See the last
two examples
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Chew T S MA1506-14 Chapter 3 3
N
t
1
2
equi soln
stable
equi soln
unstable
2
4
BE
s
3.4 Harvesting
B/s -
Good region
able to bounce back
Region 1
Region 2
Region 3
We have to ensure that the # of fish is not in the region 1,
since in this case , eventually, the population of fish will be zero.
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/B s
1
2
2
1.512
dN NN
dt
Initial population
under harvesting modelis B/s
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2( ) ( )dN
B D N B sN N BN sNdt
2dN BN sN Edt
1
2
B
s
2
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Chew T S MA1506-14 Chapter 3
3.4 Harvesting
6
1 2 2
2
0.2
0.8
B/S
2
1
2
4
BE
s
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2
0ax bx c
Recall
Suppose that one of the root is known, and a, b
are also given. How to find the constant c
It is known that
2
1 2
b
a
Hence we can find1
Furthermore, 1 2ca
Hence we can find c
Certainly, we can use formula
2
1 2
4,
2
b b ac
a
to find the value c.
But the computation
of the 1st
method is easier
2
1 2[ ( / ) ( / )] ( )( )a x b a x c a a x x
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2100000 ( 100000) 4(1)(100000 )68002
E
0.1B 0.1B
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NOT a harvesting model
1 2 0.0116 40.0029
ba
1 2/ / ( 0.0029) ( 2)(6)c a E
2 6
1 2
6
2
( 0.0029)(N 2)(N 6)dN
dt
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2012
10
20
3
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05
10
1sinh ( )
2
x xx e e
sinh 0 0x iff x
sinh 0 iff 0x x sinh 0 iff 0x x
(20 2 )(5 )
2 (10 )(5 )
similar to
dxx x x
dt
x x x
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