Solutions to Differential Equations

SOLUTIONS TO DIFFERENTIAL EQUATIONS
1. a) ẋ=-x1-x1x22
ẋ2=-2x2-2x12x2
x=0 is the equilibrium point for this system. If we introduce a
quadratic Lyapunov function
V=x12+ax22
Where a is a positive constant to be determined. v is positive
definite on the entire state space ℝ2. V is also radially unbounded
that is| v(x)|→∞ as‖ x‖→∞. The derivative of v along the
trajectories is given by
V' = 2x1 2ax2 -x1-x1x22
-2x2-2x12x2
=2x1(-x1-x12)+2ax2(-2x2-2x12x2)
=-2x12+(-2-4a)x1x22-4ax22
If we choose a =-1/2 then we eliminate x1x2 and v' becomes
v'=-2(x12-x22) .Therefore x=0 is globally
asymptotically stable equilibrium point if x1>x2 or -x1<-x2
since v(x) is a continuous differentiable function i.e.
v'(x)≤-k‖x‖a for all x∊ℝ2 and where k, a>0
hence the origin is exponentially stable .
b) x1'=x2
x2'=-x1-x23
X=0 is the equilibrium point for this system. If we introduce a
quadratic Lyapunov function
V=x12+ax22

SOLUTIONS TO DIFFERENTIAL EQUATIONS
Where a is a positive constant to be determined. v is positive
definite on the entire state space ℝ2. V is also radially unbounded
that is| v(x)|→∞ as‖ x‖→∞. The derivative of v along the
trajectories is given by
V'= 2x1 2ax2 x2
-x1-x23
=2x1x2+2ax2(-x1-x23)
=(2-2a)x1x2-2ax24)
If we choose a=1 we then eliminate the cross terms and we then
have v'=-2x24 therefore x=0 is a globally
asymptotically stable equilibrium point
since v(x) is a continuous differentiable function i.e.
v'(x)≤-k‖x‖a for all x∊ℝ2 and where k,a>0
hence the origin is globally exponentially stable.
c) x1'=-x2-x1(1-x12-x22)
x2'=x1-x2(1-x12-x22)
X=0 is the equilibrium point for this system. If we introduce a
quadratic Lyapunov function
V=x12+ax22
Where a is a positive constant to be determined. v is positive
definite on the entire state space ℝ2. V is also radially unbounded
that is| v(x)|→∞ as‖ x‖→∞. The derivative of v along the
trajectories is given by

SOLUTIONS TO DIFFERENTIAL EQUATIONS
V'= 2x1 2ax2 -x2-x1(1-x12-x22)
X1-x2(1-x12-x22)
Multiplying this matrix we obtain
v' = -2x12+2x14-2ax22+2ax24+(-2+2a)x1x2+(2+2a)x12x22
If we choose a=1 so that we eliminate the cross term we have
v' = -2x12+2x14-2x22+2x24+4x12x22
Hence x=0 is not a globally asymptotically stable equilibrium
point. since v(x) is a continuous differentiable function i.e.
v'(x)≥-k‖x‖a for all x∊ℝ2 and where k, a>0
hence the origin is neither exponentially stable nor globally
exponentially stable.
2) x'=σ(y-x)
y'=rx-y-xz
z'=xy-bz
where σ, r, b are positive constants and again we have that 0<r≤1
using the Lyapunov function
v(x,y,z)=x2/σ+y2+z2 where σ is to be determined
It is clear that v is positive definite on ℝ3 and is radially
unbounded. The derivative of v along the trajectories of the system
is given by
v'= 2x/σ 2y 2z σ(y-z)
rx-y-xz
xy-bz
=-2x2-2y2-2bz2+2xy 2rxy