Write the Introduction to Differential Equation For Positive Definite.

## System 1: Global Asymptotically Stable Equilibrium Point

x=0_{ }is the equilibrium point for this system. If we introduce a quadratic Lyapunov function

V=x_{1}^{2}+ax_{2}^{2}

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^{' }=^{ } 2x_{1} 2ax_{2} -x_{1}-x_{1}x_{2}^{2}

-2x_{2}-2x_{1}^{2}x_{2}

_{ }

=2x_{1}(-x_{1}-x_{1}^{2})+2ax_{2}(-2x_{2}-2x_{1}^{2}x_{2})

=-2x_{1}^{2}+(-2-4a)x_{1}x_{2}^{2}-4ax_{2}^{2}

If we choose a =-1/2 then we eliminate x_{1}x_{2 }and v' becomes

v'=-2(x_{1}^{2}-x_{2}^{2}) .Therefore x=0 is globally asymptotically stable equilibrium point if x_{1}>x_{2} or -x_{1}<-x_{2 }

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 exponentially stable

- b) x
_{1}^{'}=x_{2}

x_{2}^{'}=-x_{1}-x_{2}^{3 }

X=0 is the equilibrium point for this system. If we introduce a quadratic Lyapunov function

V=x_{1}^{2}+ax_{2}^{2}

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^{'}= 2x_{1} 2ax_{2} x_{2}

-x_{1}-x_{2}^{3}

=2x_{1}x_{2}+2ax_{2}(-x_{1}-x_{2}^{3})

=(2-2a)x_{1}x_{2}-2ax_{2}^{4})

If we choose a=1 we then eliminate the cross terms and we then have v'=-2x_{2}^{4} 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.

- c) x
_{1}^{'}=-x_{2}-x_{1}(1-x_{1}^{2}-x_{2}^{2})

x_{2}'=x_{1}-x_{2}(1-x_{1}^{2}-x_{2}^{2})

X=0 is the equilibrium point for this system. If we introduce a quadratic Lyapunov function

V=x_{1}^{2}+ax_{2}^{2}

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^{'}= 2x_{1} 2ax_{2} -x_{2}-x_{1}(1-x_{1}^{2}-x_{2}^{2})

X_{1}-x_{2}(1-x_{1}^{2}-x_{2}^{2})

Multiplying this matrix we obtain

v^{'} = -2x_{1}^{2}+2x_{1}^{4}-2ax_{2}^{2}+2ax_{2}^{4}+(-2+2a)x_{1}x_{2}+(2+2a)x_{1}^{2}x_{2}^{2 }

If we choose a=1 so that we eliminate the cross term we have

v' = -2x_{1}^{2}+2x_{1}^{4}-2x_{2}^{2}+2x_{2}^{4}+4x_{1}^{2}x_{2}^{2 }

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)=x^{2}/σ+y^{2}+z^{2} 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

=-2x^{2}-2y^{2}-2bz^{2}+2xy 2rxy

=-2[(x^{2}+y^{2}+bz^{2}-(1+r)xy)]

=-2[(x-1/2(1+r)y)^{2}+(1-(1+r/2)^{2})y^{2}+1/2bz^{2})

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) x_{1}^{'}=x_{2}

X_{2}^{'}=-h_{1}(x_{1})-x_{2}-h_{2}(x_{3})

X_{3}^{'}=x_{2}-x_{3}

Where h_{1} and h_{2} are locally Lipschitz function that satisfy h_{i}(0)=0 and yh_{i}(y)>0 for all y?0.

X_{2}=0

X_{3}=0

h_{1}(x_{1})-h_{2}(0)=0

Since h_{2}(0)=0 implies that h_{1}(x_{1})=0 and therefore x_{1}=0

Hence the system has a unique equilibrium point at the origin

b) v(x)= _{1}(y)dy +x_{2}^{2}/2_{2}(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.

- c) v'(x) =-h
_{1}(x_{1})^{2}+x_{2}-h_{2}(x_{3})^{2}

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

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

## System 2: Globally Asymptotically Stable Equilibrium Point

v^{'}(x) should be negative definite on the entire state space and this is only achieved by h_{1} and h_{2} satisfying h_{i}(0)=0 and yh_{i}(y)>0 for all y?0

4) x_{1}^{'}=x_{2}

X_{2}^{'}=-sinx_{1}-g(t)x_{2}

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(asinx_{1} +x_{2})^{2} +[1 +ag(t) -a^{2}](1-cosx_{1})

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(asinx_{1}+x_{2})^{2} +[1+ag(t)-a^{2}]2(sin^{2}x_{1}/2)

Now ?2asin^{2}x_{1}/2?≤?x?^{2}max(a/2,1/2) where a/2 condition in the max function arises from

|sinx_{1}/2|≤|x_{1}/2| for all x_{1}

And 1/2 condition in the max function is for the case where x_{2}^{2}/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)=(asinx_{1}+x_{2})(1+acosx_{1})x_{2}+ag^{'}(t)sinx_{1}(-sinx_{1}-g(t)x_{2})

= a^{2}x_{2}/2sin2x_{1} +x_{2}asinx_{1}+ax_{2}cosx_{1}+x_{2}^{2}-ag'(t)sin^{2} x_{1}-a(g^{'}(t))^{2}x_{2}

≤-(a-a)x_{2}^{2}-a(2-γ)(1-cosx_{1})+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) x_{1}^{'}=h(t)x_{2}-g(t)x_{1}^{3}

X_{2}^{'}=-h(t)x_{1}-g(t)x_{2}^{3}

X=0 is an equilibrium point of this system. If we introduce Lyapunov function v(x)=x_{1}^{2}+ax_{2}^{2} then we have

v^{'}=2h(t)x_{1}^{2}-2g(t)x_{1}^{4}-2ah(t)x_{1}x_{2}-2ag(t)x_{2}^{4}

if we choose a=0 then we have

v^{'}=2(h(t)-g(t)x_{1}^{2})x_{1}^{2} 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^{'}=x^{3}-x^{5}

0=x^{3}-x^{5}

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=1/4x^{4}-1/6x^{6} or

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

8)x_{1}^{'}=-6x_{1}/u^{2}+2x_{2)}

X_{2}^{'}=-2(x_{1}+x_{2})/u^{2}

Where u =1+x_{1}^{2} Let v(x)= x_{1}^{2}/1+x_{1}^{2} +x_{2}^{2}

=x_{1}^{2}+x_{2}^{2}+x_{1}^{2}x_{2}^{2}/u Therefore it is clear to see that v(x) >o for all x??^{2}/{0}

v^{'}= 2x_{1}(1+x_{1}^{2})-x_{1}^{2}(2x_{1})/(1+x_{1}^{2})^{2} 2x_{2 } -6x_{1}/1+x_{1}^{2}+2x_{2 }

=-12x_{1}^{2}+4x_{1}x_{2}+4x_{1}^{3}/(1+x_{1}^{2})^{3}-4x_{1}x_{2}-4x_{2}^{2}/1+x_{1}^{2}

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

b) x_{2}=2/x_{1}-√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 ?^{2}asymptotically 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.

References

V. M. ALEKSEEV, An estimate for the perturbations of the solutions of ordinary differen tial equations, Vestnik Moskou. Univ. Ser. I. Math. Mekh, 2 (1961), 28-36. [Russian]

Z. S. ATHANASSOV, Perturbation theorems for nonlinear systems of ordinary differentialequations, J. Math. Anal. Appl. 86 (1982), 194-207.

I. BIHARI, A generalization of a lemma of Bellman and its application to uniqueness problems of differential equations, Acta Math. Hungar. 7 (1956), 71-94.

G. BIRKHOFF AND G.-C. ROTA, “Ordinary Differential Equations,” 3rd ed., Wiley, New York, 1978.

F. BRAUER, Perturbations of nonlinear systems of differential equations, J. Math. Anal. Appl. 14 (1967), 198-206.

F. BRAUER, Perturbations of nonlinear systems of differential equations, II, J. Mad Anal. Appl. 17 (1967), 418434.

LIPSCHITZ STABILITY OF NONLINEAR SYSTEMS 577 F. BRAUER AND STRAUSS,Perturbations of nonlinear systems of differential equations, III, J. Math. Anal. Appl. 31 (1970), 3748.

F. BFCALJER, Perturbations of nonlinear systems of differential equations, IV, J. Math. Anal. Appl. 31 (1972), 214-222

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