Math 21b, Spring 2017. Worksheet 26, Study notes of Differential Equations

Two problems related to first order nonlinear systems and phase space approach to analyze second order differential equations. The first problem requires finding a first order nonlinear system in R2 such that the x- and y-nullclines are y = n and x = n, respectively, for all integers n. The second problem requires using the phase space approach to analyze the second order differential equation. solutions to both problems.

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Math 21b, Spring 2017. Worksheet 26.
Problem 1.
1. Find a first order nonlinear system in R2such that the x- and y-nullclines are y=nand x=n,
respectively, for all integers n.
2. Given a 3×3matrix Aand a vector v0R3, find a first order system of differential equations
where v0is a stationary point and Ais J(v0).
Solution. (1) One example is dx
dt = sin(2πy),dy
dt = sin(2πx).
(2) The system x0(t) = A(x(t)v0) is one possibility.
Problem 2. Use the phase space approach to analyze the second order differential equation
d2x
dt2= (x1)dx
dt .
Solution. We expand the equation to phase space as:
x0=y, y0= (x1)y.
The nullclines are y= 0 and (x1)y= 0. Every point on the line y= 0 is stationary and this corresponds
to the fact that xafor aRsolves our differential equation. We compute
J=0 1
y x 1
So
J(a, 0) = 0 1
0a1
which has eigenvales 0,(a1). So the stationary points to the left of x= 1 are stable while those to the
right are unstable.

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Math 21b, Spring 2017. Worksheet 26.

Problem 1.

  1. Find a first order nonlinear system in R^2 such that the x- and y-nullclines are y = n and x = n, respectively, for all integers n.
  2. Given a 3 × 3 matrix A and a vector v 0 ∈ R^3 , find a first order system of differential equations where v 0 is a stationary point and A is J(v 0 ).

Solution. (1) One example is dxdt = sin(2πy), dydt = sin(2πx).

(2) The system x′(t) = A(x(t) − v 0 ) is one possibility.

Problem 2. Use the phase space approach to analyze the second order differential equation

d^2 x dt^2

= (x − 1) dx dt

Solution. We expand the equation to phase space as:

x′^ = y, y′^ = (x − 1)y.

The nullclines are y = 0 and (x−1)y = 0. Every point on the line y = 0 is stationary and this corresponds to the fact that x ≡ a for a ∈ R solves our differential equation. We compute

J =

[

y x − 1

]

So

J(a, 0) =

[

0 a − 1

]

which has eigenvales 0, (a − 1). So the stationary points to the left of x = 1 are stable while those to the right are unstable.