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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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496
Chapter 9. Nonlinear Differential Equations and Stability
7 i
2 *-/ / / / / / / / / / / / >/ s'
/ s s'
s' ? ? ? s s s'
t ^ s' '' ^ ? s s s'
t N s' ? ? ? ? ? s s ? ? s'
1.51 N --- s s ? sS s' s'
t \ 4 v" s s' ? s' s' s' S'
t \ \ s s' ? s' sS s" S' S'
t N \ s' s'
t t \ v --- ? s' s'
1 - \ --- ? ? ? ? ? ? s' s'
t t t \ V ? ? ? ? ? s' s'
t t t \ --- ? ? s' s'
t 1 / t t N s' s' s' *- s- s- s- s- s-
t t / / / / s- s- S- *- *-
0.5 --- / / / / --- --- *- --- ---
t t / / 1 s' s' ---- -- --- ---- *- ---- ---
t / / ------ \ ....... s' sS s' ~- *- --- --- *- *- ---
t / ^ ---* X S' s'
1 1 1 1
0 0.25 0.50 0 .75 1 1 25 x
FIGURE 9.4.3 Critical points and direction field for the system (21).
x = 0, y = 0. Neglecting the nonlinear terms in Eqs. (21), we obtain the linear system
dt,(y)=(j )- <*>
which is valid near the origin. The eigenvalues and eigenvectors of the system (22) are
1)
,0,
ri = i- ^(1) = ft); r2 = J-5- ^(2) = (!) - (23)
so the general solution is
y) = J + Mil eJ (24)
Therefore the origin is an unstable node of the linear system (22) and also of the nonlinear system (21). All trajectories leave the origin tangent to the y-axis except for one trajectory that lies along the x-axis.
x = 1, y = 0. The corresponding linear system is
d (u\ /1 A /u
dt w V J -J.25) \vy (25)
9.4 Competing Species
497
Its eigenvalues and eigenvectors are
r1 = -1, g(1) =
and its general solution is
r2 = -0.25,
g(2) =
4
3
c
e + c2
43 e-0.25t.
(26)
(27)
The point (1, 0) is an asymptotically stable node of the linear system (25) and of the nonlinear system (21). If the initial values of x and j are sufficiently close to (1,0), then the interaction process will lead ultimately to that state, that is, to the survival of species x and the extinction of species j. There is one pair of trajectories that approaches the critical point along the x-axis. All other trajectories approach (1, 0) tangent to the line with slope -3/4 that is determined by the eigenvector g(2).
x = 0, y = 2. The analysis in this case is similar to that for the point (1, 0). The appropriate linear system is
( u
dt \v
1
0
1.5 0.5
The eigenvalues and eigenvectors of this system are
r1 = -
and its general solution is
g(1) =
r2 = -0.5,
g(2) =
J = c1(3) e-t + c2U) e-05t.
(28)
(29)
(30)
Thus the critical point (0, 2) is an asymptotically stable node of both the linear system (28) and the nonlinear system (21). All trajectories approach the critical point along the j-axis except for one trajectory that approaches along the line with slope 3.
x = 0.5, y = 0.5. The corresponding linear system is
d (u
dt \v
The eigenvalues and eigenvectors are
-0.5 -0.5
0.375 0.125
-5 + V57
r1 =------- = 0.1594,
1 16
-5 - V57
r2 =---------- = -0.7844,
2 16
so the general solution is
g(1) =
g(2) =
1
(-3 -V57)/8/ V-1.3187
(-3 + V57)/8/ I 0.5687
-1.3187 j 0594' + 0.5687 1 e
0.7844t
(31)
(32)
(33)
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