Question

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Construct a one-dimension dynamical systems F(x) so that it has 2 three equilibrium points. Moreover, two of them are semi-stable and the third is a sink which sits in-between the two semi-stable ones. Sketch its phase line.

Answer 1

Solution: 2. Definition: An equilibrium solution is any constant (horizontal) function that is a solution to the differential equation.

Stable: The equilibrium solution is stable if all solutions with initial conditions $x_0$ `near' x = c approach c as $t \to \infty$.
Unstable: The equilibrium solution is unstable if all solutions with initial conditions $x_0$ `near' x= c do n aotpproach c as $t \to \infty$.
Semi-stable: The equilibrium solution is semi-stable if initial conditions $x_0$ on one side of c lead to solutions x(t) that approach c as $t \to \infty$ , while initial conditions $x_0$ on the other side of c do not approach c.

Construction of one-dimension dynamical system that has three equilibrium points:

(a) Let

d.x lt

(i) Here equilibrium points are given by

$\frac{dx}{dt}=x'=-(x-5)(x-8)^2=F(x)=0$

$\Rightarrow(x-5)(x-8)^2=0$

$\Rightarrow x=5, x=8,8$

Therefore,
r(t) 5
and
(t)-8, 8
are three equilibrium solutions.

(ii) For,
d.x lt
is negative, so x(t) is decreasing.

For,
dt
is negative, so x(t) is decreasing.

For,
dt
is positive, so x(t) is increasing.

Thus x(t) = 5 is stable and x(t)= 8 is semi-stable.

(b) Let

d.x lt

Here x=4 is stable and x=6 and x=9 are semistable