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Proving the existence of a center or a focus in a system of ODEs

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1

$begingroup$

I'm asked to prove that for any odd value of $ninmathbb{N}$ there exist homogeneous polynomials $P_n,Q_ninmathbb{R}[x,y]$ such that the point $C=(0,0)inmathbb{R}^2$ is either a center or a focus of the system of ODEs
$$dot{x}=P_n(x,y),quad dot{y}=Q_n(x,y)$$
Here's my approach: I've shown that the origin is the only possible isolated critical point of the system. In this kind of problems, we usually consider the linear part of the system, which in this case would be the system
$$dot{x}=langlenabla P_n,(x,y)rangle,quaddot{y}=langlenabla Q_n,(x,y)rangle $$
in the origin. We would like to show that $C$ is a center or a focus of this new system. However, this only seems possible when $n=1$, since all the partial derivatives of $P_n$ and $Q_n$ are $0$ when evaluated at the origin if $n>1$, so I don't know how to proceed. Any help? Thank you in advance.

ordinary-differential-equations dynamical-systems homogeneous-equation

share|cite|improve this question

edited yesterday

J. W. Tanner

2,8421217

asked yesterday

Ray BernRay Bern

17110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try the following. Go to polar coordinates, you will get a system $dot{r} = R(r, varphi), ; dot{varphi} = Phi(r, varphi)$. For a focus or center you need at least $Phi(r, varphi)$ to be non-zero close to the origin: it should provide a constant increase or decrease of polar angle along the trajectories, excluding asymptotic directions. I'm not 100% sure, but to me this is already a strong guarantee that an equilibrium is either a center or a non-linear focus. If you can engineer $Phi(r, varphi)$ with such properties using $P_n$ and $Q_n$, you are done.
  $endgroup$
  – Evgeny
  22 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm asked to prove that for any odd value of $ninmathbb{N}$ there exist homogeneous polynomials $P_n,Q_ninmathbb{R}[x,y]$ such that the point $C=(0,0)inmathbb{R}^2$ is either a center or a focus of the system of ODEs
$$dot{x}=P_n(x,y),quad dot{y}=Q_n(x,y)$$
Here's my approach: I've shown that the origin is the only possible isolated critical point of the system. In this kind of problems, we usually consider the linear part of the system, which in this case would be the system
$$dot{x}=langlenabla P_n,(x,y)rangle,quaddot{y}=langlenabla Q_n,(x,y)rangle $$
in the origin. We would like to show that $C$ is a center or a focus of this new system. However, this only seems possible when $n=1$, since all the partial derivatives of $P_n$ and $Q_n$ are $0$ when evaluated at the origin if $n>1$, so I don't know how to proceed. Any help? Thank you in advance.

ordinary-differential-equations dynamical-systems homogeneous-equation

share|cite|improve this question

edited yesterday

J. W. Tanner

2,8421217

asked yesterday

Ray BernRay Bern

17110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try the following. Go to polar coordinates, you will get a system $dot{r} = R(r, varphi), ; dot{varphi} = Phi(r, varphi)$. For a focus or center you need at least $Phi(r, varphi)$ to be non-zero close to the origin: it should provide a constant increase or decrease of polar angle along the trajectories, excluding asymptotic directions. I'm not 100% sure, but to me this is already a strong guarantee that an equilibrium is either a center or a non-linear focus. If you can engineer $Phi(r, varphi)$ with such properties using $P_n$ and $Q_n$, you are done.
  $endgroup$
  – Evgeny
  22 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I'm asked to prove that for any odd value of $ninmathbb{N}$ there exist homogeneous polynomials $P_n,Q_ninmathbb{R}[x,y]$ such that the point $C=(0,0)inmathbb{R}^2$ is either a center or a focus of the system of ODEs
$$dot{x}=P_n(x,y),quad dot{y}=Q_n(x,y)$$
Here's my approach: I've shown that the origin is the only possible isolated critical point of the system. In this kind of problems, we usually consider the linear part of the system, which in this case would be the system
$$dot{x}=langlenabla P_n,(x,y)rangle,quaddot{y}=langlenabla Q_n,(x,y)rangle $$
in the origin. We would like to show that $C$ is a center or a focus of this new system. However, this only seems possible when $n=1$, since all the partial derivatives of $P_n$ and $Q_n$ are $0$ when evaluated at the origin if $n>1$, so I don't know how to proceed. Any help? Thank you in advance.

ordinary-differential-equations dynamical-systems homogeneous-equation

share|cite|improve this question

edited yesterday

J. W. Tanner

2,8421217

asked yesterday

Ray BernRay Bern

17110

$endgroup$

share|cite|improve this question

edited yesterday

J. W. Tanner

2,8421217

asked yesterday

Ray BernRay Bern

17110

share|cite|improve this question

edited yesterday

J. W. Tanner

2,8421217

asked yesterday

Ray BernRay Bern

17110

edited yesterday

J. W. Tanner

2,8421217

edited yesterday

J. W. Tanner

2,8421217

asked yesterday

Ray BernRay Bern

17110

asked yesterday

Ray BernRay Bern

17110

1 Answer
1

active

oldest

votes

2

$begingroup$

The question asks us to show the existence of such polynomials. Hence, we simplify the problem to the following system

$$dot{x}=P(y)$$
$$dot{y}=-Q(x).$$

We assume that $P$ and $Q$ only consist of odd powers. Now, we define the Lyapunov candidate function

$$V(x,y)=int_{0}^{x}Q(bar{x})dbar{x} + int_{0}^{y}P(bar{y})dbar{y}.$$

$P$ and $Q$ are odd the integrals are even, hence $V(x,y)>0$ for all $x,y in mathbb{R}-{(0,0)}$. For $dot{V}$ we have by the fundamental theorem of calculus.
$$dot{V}=Q(x)dot{x}+P(y)dot{y}=Q(x)P(y)-P(y)Q(x)equiv 0.$$
Hence, the origin is a center. Which can be seen as a constructive proof for the claim with the center.

Now, to the second part. We simplify the initial system from the question to

$$dot{x}=alpha ,x^{2n-1}$$
$$dot{y}=alpha ,y^{2n-1}.$$

Then we construct a Lyapunov candidate function
$$V(x,y)=1/2x^2+1/2y^2$$
$$implies dot{V}=xdot{x}+ydot{y}=alpha(x^{2n}+y^{2n}).$$

Depending on the sign of $alpha$ we can show that $dot{V}<0$ or $dot{V}>0$. This is equivalent to showing that the origin is an asymptotically stable or unstable focus.

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

MachineLearnerMachineLearner

5077

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why is there a minus sign in $dot{V}$? Thank you for your answer!
  $endgroup$
  – Ray Bern
  16 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for my imprecision. I mean the minus sign in the first case, when looking for a center, when applying the chain rule differentiating $V$, right?
  $endgroup$
  – Ray Bern
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RayBern: Sorry, I had mixed up some equations that I wrote down.
  $endgroup$
  – MachineLearner
  15 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

The question asks us to show the existence of such polynomials. Hence, we simplify the problem to the following system

$$dot{x}=P(y)$$
$$dot{y}=-Q(x).$$

We assume that $P$ and $Q$ only consist of odd powers. Now, we define the Lyapunov candidate function

$$V(x,y)=int_{0}^{x}Q(bar{x})dbar{x} + int_{0}^{y}P(bar{y})dbar{y}.$$

$P$ and $Q$ are odd the integrals are even, hence $V(x,y)>0$ for all $x,y in mathbb{R}-{(0,0)}$. For $dot{V}$ we have by the fundamental theorem of calculus.
$$dot{V}=Q(x)dot{x}+P(y)dot{y}=Q(x)P(y)-P(y)Q(x)equiv 0.$$
Hence, the origin is a center. Which can be seen as a constructive proof for the claim with the center.

Now, to the second part. We simplify the initial system from the question to

$$dot{x}=alpha ,x^{2n-1}$$
$$dot{y}=alpha ,y^{2n-1}.$$

Then we construct a Lyapunov candidate function
$$V(x,y)=1/2x^2+1/2y^2$$
$$implies dot{V}=xdot{x}+ydot{y}=alpha(x^{2n}+y^{2n}).$$

Depending on the sign of $alpha$ we can show that $dot{V}<0$ or $dot{V}>0$. This is equivalent to showing that the origin is an asymptotically stable or unstable focus.

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

MachineLearnerMachineLearner

5077

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why is there a minus sign in $dot{V}$? Thank you for your answer!
  $endgroup$
  – Ray Bern
  16 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for my imprecision. I mean the minus sign in the first case, when looking for a center, when applying the chain rule differentiating $V$, right?
  $endgroup$
  – Ray Bern
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RayBern: Sorry, I had mixed up some equations that I wrote down.
  $endgroup$
  – MachineLearner
  15 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

The question asks us to show the existence of such polynomials. Hence, we simplify the problem to the following system

$$dot{x}=P(y)$$
$$dot{y}=-Q(x).$$

We assume that $P$ and $Q$ only consist of odd powers. Now, we define the Lyapunov candidate function

$$V(x,y)=int_{0}^{x}Q(bar{x})dbar{x} + int_{0}^{y}P(bar{y})dbar{y}.$$

$P$ and $Q$ are odd the integrals are even, hence $V(x,y)>0$ for all $x,y in mathbb{R}-{(0,0)}$. For $dot{V}$ we have by the fundamental theorem of calculus.
$$dot{V}=Q(x)dot{x}+P(y)dot{y}=Q(x)P(y)-P(y)Q(x)equiv 0.$$
Hence, the origin is a center. Which can be seen as a constructive proof for the claim with the center.

Now, to the second part. We simplify the initial system from the question to

$$dot{x}=alpha ,x^{2n-1}$$
$$dot{y}=alpha ,y^{2n-1}.$$

Then we construct a Lyapunov candidate function
$$V(x,y)=1/2x^2+1/2y^2$$
$$implies dot{V}=xdot{x}+ydot{y}=alpha(x^{2n}+y^{2n}).$$

Depending on the sign of $alpha$ we can show that $dot{V}<0$ or $dot{V}>0$. This is equivalent to showing that the origin is an asymptotically stable or unstable focus.

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

MachineLearnerMachineLearner

5077

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why is there a minus sign in $dot{V}$? Thank you for your answer!
  $endgroup$
  – Ray Bern
  16 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for my imprecision. I mean the minus sign in the first case, when looking for a center, when applying the chain rule differentiating $V$, right?
  $endgroup$
  – Ray Bern
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RayBern: Sorry, I had mixed up some equations that I wrote down.
  $endgroup$
  – MachineLearner
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The question asks us to show the existence of such polynomials. Hence, we simplify the problem to the following system

$$dot{x}=P(y)$$
$$dot{y}=-Q(x).$$

We assume that $P$ and $Q$ only consist of odd powers. Now, we define the Lyapunov candidate function

$$V(x,y)=int_{0}^{x}Q(bar{x})dbar{x} + int_{0}^{y}P(bar{y})dbar{y}.$$

$P$ and $Q$ are odd the integrals are even, hence $V(x,y)>0$ for all $x,y in mathbb{R}-{(0,0)}$. For $dot{V}$ we have by the fundamental theorem of calculus.
$$dot{V}=Q(x)dot{x}+P(y)dot{y}=Q(x)P(y)-P(y)Q(x)equiv 0.$$
Hence, the origin is a center. Which can be seen as a constructive proof for the claim with the center.

Now, to the second part. We simplify the initial system from the question to

$$dot{x}=alpha ,x^{2n-1}$$
$$dot{y}=alpha ,y^{2n-1}.$$

Then we construct a Lyapunov candidate function
$$V(x,y)=1/2x^2+1/2y^2$$
$$implies dot{V}=xdot{x}+ydot{y}=alpha(x^{2n}+y^{2n}).$$

Depending on the sign of $alpha$ we can show that $dot{V}<0$ or $dot{V}>0$. This is equivalent to showing that the origin is an asymptotically stable or unstable focus.

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

MachineLearnerMachineLearner

5077

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 16 hours ago

MachineLearnerMachineLearner

5077

answered 16 hours ago

MachineLearnerMachineLearner

5077

answered 16 hours ago

MachineLearnerMachineLearner

5077