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 -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??2 and where k, a>0
hence the origin is exponentially stable
x2‘=-x1-x23
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-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??2 and where k,a>0
hence the origin is globally exponentially stable.
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 hi(0)=0 and yhi(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.
Which is locally negative definite implying that x=0 or the origin is asymptotically stable equilibrium point.
v‘(x) should be negative definite on the entire state space and this is only achieved by h1 and h2 satisfying hi(0)=0 and yhi(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 Gr 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 Gr.
To show that v is positive definite on Gr, 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.
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
Essay Writing Service Features
Our Experience
No matter how complex your assignment is, we can find the right professional for your specific task. Contact Essay is an essay writing company that hires only the smartest minds to help you with your projects. Our expertise allows us to provide students with high-quality academic writing, editing & proofreading services.
Free Features
Free revision policy
$10Free bibliography & reference
$8Free title page
$8Free formatting
$8How Our Essay Writing Service Works
First, you will need to complete an order form. It's not difficult but, in case there is anything you find not to be clear, you may always call us so that we can guide you through it. On the order form, you will need to include some basic information concerning your order: subject, topic, number of pages, etc. We also encourage our clients to upload any relevant information or sources that will help.
Complete the order form
Once we have all the information and instructions that we need, we select the most suitable writer for your assignment. While everything seems to be clear, the writer, who has complete knowledge of the subject, may need clarification from you. It is at that point that you would receive a call or email from us.
Writer’s assignment
As soon as the writer has finished, it will be delivered both to the website and to your email address so that you will not miss it. If your deadline is close at hand, we will place a call to you to make sure that you receive the paper on time.
Completing the order and download