MATH 2421 Introduction to Differential Equation For Positive Definite

=(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??² and where k,a>0

hence the origin is globally exponentially stable. 

1. 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 ?². 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(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??² 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 ?³ 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

=-2[(x²+y²+bz²-(1+r)xy)]

=-2[(x-1/2(1+r)y)²+(1-(1+r/2)²)y²+1/2bz²)

Since 0<r≤1 it follows that 0<(1+r)/2<1 and therefore v' is negative definite on the entire state space ?³ hence the origin is globally asymptotically stable.

3.a) x₁^'=x₂

X₂^'=-h₁(x₁)-x₂-h₂(x₃)

X₃^'=x₂-x₃

Where h₁ and h₂ are locally Lipschitz function that satisfy h_i(0)=0 and yh_i(y)>0 for all y?0.

X₂=0

  X₃=0

h₁(x₁)-h₂(0)=0

Since h₂(0)=0 implies that h₁(x₁)=0 and therefore x₁=0

Hence the system has a unique equilibrium point at the origin

b) v(x)=  ₁(y)dy +x₂²/2₂(y)dy

If we define v(x) the way it has been defined above, then it is clear that v(x) is locally positive definite.

1. c) v'(x) =-h₁(x₁)²+x₂-h₂(x₃)² 

Which is locally negative definite implying that x=0 or the origin is asymptotically stable equilibrium point.

1. d) The function v(x) should be positive definite on the entire state space and has the property |v(x)|→∞ as ?x?→∞ and again

v^'(x) should be negative definite on the entire state space and this is only achieved by h₁ and h₂ satisfying h_i(0)=0 and yh_i(y)>0 for all y?0

4) x₁^'=x₂

X₂^'=-sinx₁-g(t)x₂ 

Where g(t) is continuously differentiable and satisfies 0<a≤g(t)≤β<∞,

g^'(t)≤γ<2,for all t≥0

Lyapunov function

V(t,x)=1/2(asinx₁ +x₂)² +[1 +ag(t) -a²](1-cosx₁)

 To show that v(t,x) is positive definite and decrescent on G_r or origin. We assume that a>0

V(t,x)=1/2(asinx₁+x₂)² +[1+ag(t)-a²]2(sin²x₁/2)

Now ?2asin²x₁/2?≤?x?²max(a/2,1/2) where a/2 condition in the max function arises from

|sinx₁/2|≤|x₁/2| for all x₁

And 1/2 condition in the max function is for the case where x₂²/2 dominate if a is too small. Therefore w(?x?)=max(a/2,1/2)?x?² is a class w function that bound v from above showing that v is decrescent on G_r.

To show that v is positive definite on G_r, we use

b) v^'(t,x)=(asinx₁+x₂)(1+acosx₁)x₂+ag^'(t)sinx₁(-sinx₁-g(t)x₂)

=  a²x₂/2sin2x₁   +x₂asinx₁+ax₂cosx₁+x₂²-ag'(t)sin² x₁-a(g^'(t))²x₂

≤-(a-a)x₂²-a(2-γ)(1-cosx₁)+0(|x|³) where 0(|x|³)is a term bounded by k|x|³ (k>0) in some neighbourhood of the origin.

c) From the above (b) we have shown that v^'≤0 hence this suffices to show that the origin is uniformly asymptotically stable.

5) x₁^'=h(t)x₂-g(t)x₁³

X₂^'=-h(t)x₁-g(t)x₂³

X=0 is an equilibrium point of this system. If we introduce Lyapunov function  v(x)=x₁²+ax₂² then we have

v^'=2h(t)x₁²-2g(t)x₁⁴-2ah(t)x₁x₂-2ag(t)x₂⁴

if we choose a=0 then we have

v^'=2(h(t)-g(t)x₁²)x₁² and therefore v^'≤0 if and only if g(t)>h(t) hence x=0 is uniformly asymptotically stable

since the Lyapunov function above is positive definite on the entire space such that |v(x)|→∞ as ?x?→∞ in addition v^' is negative definite on the entire state space then x=0 which is the equilibrium point is globally uniformly asymptotically stable.

since v(x)  is a continuous differentiable function ie

v'(x)≤-k?x?^a for all x??² and where k, a>0

hence the origin is exponentially stable. If this happens in the entire state space then x=0 is said to be globally uniformly exponentially stable.

6)x^'=x³-x⁵

0=x³-x⁵

X=1    hence x=0 is unstable equilibrium point.

To show that all solutions are bounded and defined on [0,∞) we integrate the equation so as to obtain

X=1/4x⁴-1/6x⁶ or       

 x-1/4x⁴+1/6x⁶=0  if we solve for the values of x we realize that they are bounded on .

8)x₁^'=-6x₁/u²+2x₂₎

X₂^'=-2(x₁+x₂)/u²

Where u =1+x₁²  Let v(x)= x₁²/1+x₁² +x₂²

=x₁²+x₂²+x₁²x₂²/u Therefore it is clear to see that v(x) >o for all x??²/{0} 

v^'=   2x₁(1+x₁²)-x₁²(2x₁)/(1+x₁²)²    2x₂  -6x₁/1+x₁²+2x₂ 

=-12x₁²+4x₁x₂+4x₁³/(1+x₁²)³-4x₁x₂-4x₂²/1+x₁²

<-(4x₁x₂+4x₂² /1+x₁² which clearly shows that v^' (x)<0 for all x??²/{0}

b) x₂=2/x₁-√2 The vector field on the boundary of this hyperbola, also the trajectories to the right of the branch in the first quadrant cannot cross that branch because the trajectories in the direction of the vector fields are approaching the boundary of ?²asymptotically and therefore they cannot cross them. 

c)This is simply because the hyperbola does not pass through the origin and for it to be globally asymptotically stable the function should be positive definite on the entire state space and in addition its derivative  should also be negative definite on the entire space that’s this conditions also apply to Lyapunov stability theorem and there it cannot contradict. 

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