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Using a Lyapunov function to classify stability and sketching a phase portrait

Lyapunov stability question from Arnold's triviumNon linear phase portraitNonlinear phase portrait and linearizationSystem of differential equations, phase portraitDynamical Systems- Plotting Phase PortraitPhase portrait of ODE in polar coordinatesQuestions about stability in the sense of LyapunovLinearization method or Lyapunov function - examplestability using linearization instead of Lyapunov failsLyapunov function instead of linearization

3

$begingroup$

Consider the system
$$x' = -x^3-xy^{2k}$$
$$y' = -y^3-x^{2k}y$$
Where $k$ is a given positive integer.

a.) Find and classify according to stability the equilibrium solutions.

$it{Hint:}$ Let $V(x,y) = x^2 + y^2$

b.) Sketch a phase portrait when $k = 1$

$it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?

a.)
Using $V$, we get $frac{d}{dt}V=2xx'+2yy'$

Plugging in our system , we get:

$$frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
$$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
$$y^{2k}=-x^2$$
Which only works for $x=y=0$

Therefore our system is asymptotically stable at the origin.

I am having trouble with b.), mostly because the hint is confusing me.

Let $y=ax$, then our system becomes
$$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
$$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.

ordinary-differential-equations stability-in-odes lyapunov-functions

share|cite|improve this question

edited 1 hour ago

hkj447

asked 2 hours ago

hkj447hkj447

978

$endgroup$

add a comment | 

3

$begingroup$

Consider the system
$$x' = -x^3-xy^{2k}$$
$$y' = -y^3-x^{2k}y$$
Where $k$ is a given positive integer.

a.) Find and classify according to stability the equilibrium solutions.

$it{Hint:}$ Let $V(x,y) = x^2 + y^2$

b.) Sketch a phase portrait when $k = 1$

$it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?

a.)
Using $V$, we get $frac{d}{dt}V=2xx'+2yy'$

Plugging in our system , we get:

$$frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
$$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
$$y^{2k}=-x^2$$
Which only works for $x=y=0$

Therefore our system is asymptotically stable at the origin.

I am having trouble with b.), mostly because the hint is confusing me.

Let $y=ax$, then our system becomes
$$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
$$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.

ordinary-differential-equations stability-in-odes lyapunov-functions

share|cite|improve this question

edited 1 hour ago

hkj447

asked 2 hours ago

hkj447hkj447

978

$endgroup$

add a comment | 

$begingroup$

Consider the system
$$x' = -x^3-xy^{2k}$$
$$y' = -y^3-x^{2k}y$$
Where $k$ is a given positive integer.

a.) Find and classify according to stability the equilibrium solutions.

$it{Hint:}$ Let $V(x,y) = x^2 + y^2$

b.) Sketch a phase portrait when $k = 1$

$it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?

a.)
Using $V$, we get $frac{d}{dt}V=2xx'+2yy'$

Plugging in our system , we get:

$$frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
$$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
$$y^{2k}=-x^2$$
Which only works for $x=y=0$

Therefore our system is asymptotically stable at the origin.

I am having trouble with b.), mostly because the hint is confusing me.

Let $y=ax$, then our system becomes
$$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
$$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.

ordinary-differential-equations stability-in-odes lyapunov-functions

share|cite|improve this question

edited 1 hour ago

hkj447

asked 2 hours ago

hkj447hkj447

978

$endgroup$

share|cite|improve this question

edited 1 hour ago

hkj447

asked 2 hours ago

hkj447hkj447

978

share|cite|improve this question

edited 1 hour ago

hkj447

asked 2 hours ago

hkj447hkj447

978

asked 2 hours ago

hkj447hkj447

978

asked 2 hours ago

hkj447hkj447

978

2 Answers
2

active

oldest

votes

2

$begingroup$

Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.

So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.

share|cite|improve this answer

answered 2 hours ago

aghostinthefiguresaghostinthefigures

1,4391318

$endgroup$

add a comment | 

2

$begingroup$

Phase portraits - a partial offering

Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $dot{y}=0$ and $dot{y}=0$.

$k = 1$

The linear system is

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$

The lone critical point is the origin.

When $y = a x$, $ainmathbb{R}$, we have
$$begin{align}
begin{split}
dot{x} &= -x^{3}left( 1 + a^{2} right) \
dot{y} &= -a y^{3}left( 1 + a^{2} right)
end{split}
end{align}$$

$k = 2$

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$

The bounding curves for $dot{r}$ are when $cos 4theta = 1$

$$dot{r} = -r^{3}$$

and when $cos 4theta = -1$

$$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$

The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ ever positive.

share|cite|improve this answer

edited 50 mins ago

answered 1 hour ago

dantopadantopa

6,76442345

$endgroup$

add a comment | 

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2

$begingroup$

Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.

So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.

share|cite|improve this answer

answered 2 hours ago

aghostinthefiguresaghostinthefigures

1,4391318

$endgroup$

add a comment | 

2

$begingroup$

Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.

So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.

share|cite|improve this answer

answered 2 hours ago

aghostinthefiguresaghostinthefigures

1,4391318

$endgroup$

add a comment | 

$begingroup$

Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.

So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.

share|cite|improve this answer

answered 2 hours ago

aghostinthefiguresaghostinthefigures

1,4391318

$endgroup$

share|cite|improve this answer

answered 2 hours ago

aghostinthefiguresaghostinthefigures

1,4391318

answered 2 hours ago

aghostinthefiguresaghostinthefigures

1,4391318

answered 2 hours ago

aghostinthefiguresaghostinthefigures

1,4391318

2

$begingroup$

Phase portraits - a partial offering

Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $dot{y}=0$ and $dot{y}=0$.

$k = 1$

The linear system is

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$

The lone critical point is the origin.

When $y = a x$, $ainmathbb{R}$, we have
$$begin{align}
begin{split}
dot{x} &= -x^{3}left( 1 + a^{2} right) \
dot{y} &= -a y^{3}left( 1 + a^{2} right)
end{split}
end{align}$$

$k = 2$

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$

The bounding curves for $dot{r}$ are when $cos 4theta = 1$

$$dot{r} = -r^{3}$$

and when $cos 4theta = -1$

$$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$

The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ ever positive.

share|cite|improve this answer

edited 50 mins ago

answered 1 hour ago

dantopadantopa

6,76442345

$endgroup$

add a comment | 

2

$begingroup$

Phase portraits - a partial offering

Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $dot{y}=0$ and $dot{y}=0$.

$k = 1$

The linear system is

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$

The lone critical point is the origin.

When $y = a x$, $ainmathbb{R}$, we have
$$begin{align}
begin{split}
dot{x} &= -x^{3}left( 1 + a^{2} right) \
dot{y} &= -a y^{3}left( 1 + a^{2} right)
end{split}
end{align}$$

$k = 2$

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$

The bounding curves for $dot{r}$ are when $cos 4theta = 1$

$$dot{r} = -r^{3}$$

and when $cos 4theta = -1$

$$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$

The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ ever positive.

share|cite|improve this answer

edited 50 mins ago

answered 1 hour ago

dantopadantopa

6,76442345

$endgroup$

add a comment | 

$begingroup$

Phase portraits - a partial offering

Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $dot{y}=0$ and $dot{y}=0$.

$k = 1$

The linear system is

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{2} = -x left( x^{2} + y^{2} right) \
dot{y} &= -y^{3} - x^{2}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = frac{x dot{x} + y dot{y}}{r} = -r^{3} $$

The lone critical point is the origin.

When $y = a x$, $ainmathbb{R}$, we have
$$begin{align}
begin{split}
dot{x} &= -x^{3}left( 1 + a^{2} right) \
dot{y} &= -a y^{3}left( 1 + a^{2} right)
end{split}
end{align}$$

$k = 2$

$$begin{align}
begin{split}
dot{x} &= -x^{3} - xy^{4} = -x left( x^{2} + y^{4} right) \
dot{y} &= -y^{3} - x^{4}y = -y left( x^{2} + y^{2} right)
end{split}
end{align}$$

$$ dot{r} = tfrac{1}{8} r^3 left(left(r^2-2right) cos (4 theta )-r^2-6right) $$

The bounding curves for $dot{r}$ are when $cos 4theta = 1$

$$dot{r} = -r^{3}$$

and when $cos 4theta = -1$

$$dot{r} = -tfrac{1}{4} r^3 left(r^2+2right)$$

The bounding curves cross at $r=sqrt{2}$. At no point is $dot{r}$ ever positive.

share|cite|improve this answer

edited 50 mins ago

answered 1 hour ago

dantopadantopa

6,76442345

$endgroup$

share|cite|improve this answer

edited 50 mins ago

answered 1 hour ago

dantopadantopa

6,76442345

answered 1 hour ago

dantopadantopa

6,76442345

answered 1 hour ago

dantopadantopa

6,76442345