微分方程（Blanchard Differential Equations 4th）中文版Exercise 5.1
- For the competing species population model
d x d t = 2 x ( 1 − x / 2 ) − x y , d y d t = 3 y ( 1 − y / 3 ) − 2 x y . \begin{align} \frac{dx}{dt} &= 2x(1-x/2) - xy, \\ \frac{dy}{dt} &= 3y(1 - y/3) -2xy. \end{align} dtdxdtdy=2x(1−x/2)−xy,=3y(1−y/3)−2xy.
studied in this section, we showed that the equilibrium point ( 1 , 1 ) (1, 1) (1,1) is a saddle.
(a) Find the linearized system near each of the other equilibrium points.
(b) Classify each equilibrium point (as either a source, a sink, a saddle, . . . ).
( c ) (c) (c) Sketch the phase portrait of each linearized system.
(d) Give a brief description of the phase portrait near each equilibrium point of the
nonlinear system.
ANS:
(a) Rewrite
0
=
2
x
(
1
−
x
/
2
)
−
x
y
,
0
=
3
y
(
1
−
y
/
3
)
−
2
x
y
.
\begin{align} 0 &= 2x(1-x/2) - xy, \\ 0 &= 3y(1 - y/3) -2xy. \end{align}
00=2x(1−x/2)−xy,=3y(1−y/3)−2xy.
as
0
=
2
x
(
2
−
x
−
y
)
,
0
=
y
(
3
−
y
−
2
x
)
.
\begin{align} 0 & = 2x(2-x-y), \\ 0 & = y(3 - y-2x). \end{align}
00=2x(2−x−y),=y(3−y−2x).
solve this nonlinear eqns, one can get four pairs of equilibrium vector solutions(ES)
(
0
0
2
0
0
3
1
1
)
\left(\begin{array}{cc} 0 & 0\\ 2 & 0\\ 0 & 3\\ 1 & 1 \end{array}\right)
02010031
(b) Directly computing, the Jacobi Matrix(JM) of the system is
(
2
−
y
−
2
x
−
x
−
2
y
3
−
2
y
−
2
x
)
\left(\begin{array}{cc} 2-y-2\,x & -x\\ -2\,y & 3-2\,y-2\,x \end{array}\right)
(2−y−2x−2y−x3−2y−2x)
Substituting E.S. into J.M., we have
(
2
0
−
2
−
2
−
1
0
−
1
−
1
0
3
0
−
1
−
6
−
3
−
2
−
1
)
\left(\begin{array}{cc} 2 & 0\\ -2 & -2\\ -1 & 0\\ -1 & -1\\ 0 & 3\\ 0 & -1\\ -6 & -3\\ -2 & -1 \end{array}\right)
2−2−1−100−6−20−20−13−1−3−1
The corresponding eigenvalues and eigenvectors are
(
2
0
0
3
)
\left(\begin{array}{cc} 2 & 0\\ 0 & 3 \end{array}\right)
(2003)
(
−
2
0
0
−
1
)
\left(\begin{array}{cc} -2 & 0\\ 0 & -1 \end{array}\right)
(−200−1)
(
−
3
0
0
−
1
)
\left(\begin{array}{cc} -3 & 0\\ 0 & -1 \end{array}\right)
(−300−1)
(
−
2
−
1
0
0
2
−
1
)
\left(\begin{array}{cc} -\sqrt{2}-1 & 0\\ 0 & \sqrt{2}-1 \end{array}\right)
(−2−1002−1)
(
1
0
0
1
)
\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)
(1001)
(
1
−
2
0
1
)
\left(\begin{array}{cc} 1 & -2\\ 0 & 1 \end{array}\right)
(10−21)
(
0
−
1
3
1
1
)
\left(\begin{array}{cc} 0 & -\frac{1}{3}\\ 1 & 1 \end{array}\right)
(01−311)
(
2
2
−
2
2
1
1
)
\left(\begin{array}{cc} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\\ 1 & 1 \end{array}\right)
(221−221)
Therefore, the four pairs of E.S. are Source, Sink, Sink and Saddle respectively.
©,(d) the global description of the phase portrait of the nonlinear system is as following.
原文地址：https://blog.csdn.net/weixin_42794212/article/details/143886321
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