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

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496

Chapter 9. Nonlinear Differential Equations and Stability

7 i

2 *-/ / / / / / / / / / / / >/ s'

/ s s'

s' ? ? ? s s s'

t ^ s' æ'' ^ ? 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 / / è 1 s' s' ---- -- ---¦ ---- *- ---- ---

t / / ------ \ »....... s' sS s' ~- *- --- --- *- *- ---

t / Ó ^ ---* 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 ÓÑ)- <*>

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) = ÑË 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 º0Ë594' + 0.5687 1 e

—0.7844t

(31)

(32)

(33)

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