Mixing
Computing the stable manifold for a fixed point
$begingroup$
Compute the stable manifold of $(0,0)$ for the system
$$dot{x}=x-y^3$$
$$dot{y}=-y$$
Here in my proposed solution
My first step isn't necessary for the solution. Although, it is helpful in seeing whether or not the x-axis and y-axis are stable.
The Jacobian is $J(x,y)=begin{pmatrix} frac{partial{f}}{partial{x}} & frac{partial{f}}{partial{y}} \ frac{partial{g}}{partial{x}} & frac{partial{g}}{partial{y}} end{pmatrix}=begin{pmatrix} 1 & -3y^2 \ 0 & -1 end{pmatrix}$.
Therefore, $begin{pmatrix} frac{dx}{dt} \ frac{dy}{dt} end{pmatrix}=J(x_0,y_0)begin{pmatrix} x-x_0 \ y-y_0 end{pmatrix}=begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}begin{pmatrix} x \ y end{pmatrix}=begin{pmatrix} x \ -y end{pmatrix}$.
So, the x-axis is unstable while the y-axis is stable.
To compute the stable manifold, we need to apply the stable manifold theorem. By the definition of $dot{x}$ and $dot{y}$,
$$frac{dy}{dx}=frac{-y}{x-y^3}$$
Therefore,
$$ydx + (x-y^3)dy=0$$
We can now apply the theory of integrating factors which states that $M(x,y)dx
+N(x,y)dy= 0$. We are given that $M(x,y)=y$ and $N(x,y)=x-y^3$. If we follow the first example in this pdf, we see that the equations are exact since
$$M_y(x,y)=1=N_x(x,y)$$
So, we will therefore find a general solution such that $f(x,y)=C$. Hence,
$$f = int Mdx + g(y)= int ydx + g(y)= xy + g(y)$$
But, we know that
$$f_y=N=x-y^3=x+g'(y)$$
which implies that
$$g'(y)=-y^3$$
and so
$$g(y) = int -y^3dy =frac{-y^4}{4}+C$$
Hence,
$$f(x,y)=xy+frac{-y^4}{4}=C$$
is the general solution. To avoid singularities in the orbits of $y(x)$ which pass through $(0,0)$, we should set $C=0$. Then,
$$xy=frac{y^4}{4}$$
$$4x=y^3$$
$$sqrt[3]{4x}=y$$
Therefore, $y(x)=sqrt[3]{4x}$ is the stable manifold. I'm not sure if there is a shorter approach to compute the stable manifold. The solution inside this question doesn't appear to work for this problem.
Is this approach correct? Please let me know if there are any better alternatives.
ordinary-differential-equations proof-verification manifolds
asked Jan 2 at 18:27
Axion004Axion004
314212
$endgroup$
add a comment |
$begingroup$
Compute the stable manifold of $(0,0)$ for the system
$$dot{x}=x-y^3$$
$$dot{y}=-y$$
Here in my proposed solution
My first step isn't necessary for the solution. Although, it is helpful in seeing whether or not the x-axis and y-axis are stable.
The Jacobian is $J(x,y)=begin{pmatrix} frac{partial{f}}{partial{x}} & frac{partial{f}}{partial{y}} \ frac{partial{g}}{partial{x}} & frac{partial{g}}{partial{y}} end{pmatrix}=begin{pmatrix} 1 & -3y^2 \ 0 & -1 end{pmatrix}$.
Therefore, $begin{pmatrix} frac{dx}{dt} \ frac{dy}{dt} end{pmatrix}=J(x_0,y_0)begin{pmatrix} x-x_0 \ y-y_0 end{pmatrix}=begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}begin{pmatrix} x \ y end{pmatrix}=begin{pmatrix} x \ -y end{pmatrix}$.
So, the x-axis is unstable while the y-axis is stable.
To compute the stable manifold, we need to apply the stable manifold theorem. By the definition of $dot{x}$ and $dot{y}$,
$$frac{dy}{dx}=frac{-y}{x-y^3}$$
Therefore,
$$ydx + (x-y^3)dy=0$$
We can now apply the theory of integrating factors which states that $M(x,y)dx
+N(x,y)dy= 0$. We are given that $M(x,y)=y$ and $N(x,y)=x-y^3$. If we follow the first example in this pdf, we see that the equations are exact since
$$M_y(x,y)=1=N_x(x,y)$$
So, we will therefore find a general solution such that $f(x,y)=C$. Hence,
$$f = int Mdx + g(y)= int ydx + g(y)= xy + g(y)$$
But, we know that
$$f_y=N=x-y^3=x+g'(y)$$
which implies that
$$g'(y)=-y^3$$
and so
$$g(y) = int -y^3dy =frac{-y^4}{4}+C$$
Hence,
$$f(x,y)=xy+frac{-y^4}{4}=C$$
is the general solution. To avoid singularities in the orbits of $y(x)$ which pass through $(0,0)$, we should set $C=0$. Then,
$$xy=frac{y^4}{4}$$
$$4x=y^3$$
$$sqrt[3]{4x}=y$$
Therefore, $y(x)=sqrt[3]{4x}$ is the stable manifold. I'm not sure if there is a shorter approach to compute the stable manifold. The solution inside this question doesn't appear to work for this problem.
Is this approach correct? Please let me know if there are any better alternatives.
ordinary-differential-equations proof-verification manifolds
asked Jan 2 at 18:27
Axion004Axion004
314212
$endgroup$
add a comment |
$begingroup$
Compute the stable manifold of $(0,0)$ for the system
$$dot{x}=x-y^3$$
$$dot{y}=-y$$
Here in my proposed solution
My first step isn't necessary for the solution. Although, it is helpful in seeing whether or not the x-axis and y-axis are stable.
The Jacobian is $J(x,y)=begin{pmatrix} frac{partial{f}}{partial{x}} & frac{partial{f}}{partial{y}} \ frac{partial{g}}{partial{x}} & frac{partial{g}}{partial{y}} end{pmatrix}=begin{pmatrix} 1 & -3y^2 \ 0 & -1 end{pmatrix}$.
Therefore, $begin{pmatrix} frac{dx}{dt} \ frac{dy}{dt} end{pmatrix}=J(x_0,y_0)begin{pmatrix} x-x_0 \ y-y_0 end{pmatrix}=begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}begin{pmatrix} x \ y end{pmatrix}=begin{pmatrix} x \ -y end{pmatrix}$.
So, the x-axis is unstable while the y-axis is stable.
To compute the stable manifold, we need to apply the stable manifold theorem. By the definition of $dot{x}$ and $dot{y}$,
$$frac{dy}{dx}=frac{-y}{x-y^3}$$
Therefore,
$$ydx + (x-y^3)dy=0$$
We can now apply the theory of integrating factors which states that $M(x,y)dx
+N(x,y)dy= 0$. We are given that $M(x,y)=y$ and $N(x,y)=x-y^3$. If we follow the first example in this pdf, we see that the equations are exact since
$$M_y(x,y)=1=N_x(x,y)$$
So, we will therefore find a general solution such that $f(x,y)=C$. Hence,
$$f = int Mdx + g(y)= int ydx + g(y)= xy + g(y)$$
But, we know that
$$f_y=N=x-y^3=x+g'(y)$$
which implies that
$$g'(y)=-y^3$$
and so
$$g(y) = int -y^3dy =frac{-y^4}{4}+C$$
Hence,
$$f(x,y)=xy+frac{-y^4}{4}=C$$
is the general solution. To avoid singularities in the orbits of $y(x)$ which pass through $(0,0)$, we should set $C=0$. Then,
$$xy=frac{y^4}{4}$$
$$4x=y^3$$
$$sqrt[3]{4x}=y$$
Therefore, $y(x)=sqrt[3]{4x}$ is the stable manifold. I'm not sure if there is a shorter approach to compute the stable manifold. The solution inside this question doesn't appear to work for this problem.
Is this approach correct? Please let me know if there are any better alternatives.
ordinary-differential-equations proof-verification manifolds
asked Jan 2 at 18:27
Axion004Axion004
314212
$endgroup$
Compute the stable manifold of $(0,0)$ for the system
$$dot{x}=x-y^3$$
$$dot{y}=-y$$
Here in my proposed solution
My first step isn't necessary for the solution. Although, it is helpful in seeing whether or not the x-axis and y-axis are stable.
The Jacobian is $J(x,y)=begin{pmatrix} frac{partial{f}}{partial{x}} & frac{partial{f}}{partial{y}} \ frac{partial{g}}{partial{x}} & frac{partial{g}}{partial{y}} end{pmatrix}=begin{pmatrix} 1 & -3y^2 \ 0 & -1 end{pmatrix}$.
Therefore, $begin{pmatrix} frac{dx}{dt} \ frac{dy}{dt} end{pmatrix}=J(x_0,y_0)begin{pmatrix} x-x_0 \ y-y_0 end{pmatrix}=begin{pmatrix} 1 & 0 \ 0 & -1 end{pmatrix}begin{pmatrix} x \ y end{pmatrix}=begin{pmatrix} x \ -y end{pmatrix}$.
So, the x-axis is unstable while the y-axis is stable.
To compute the stable manifold, we need to apply the stable manifold theorem. By the definition of $dot{x}$ and $dot{y}$,
$$frac{dy}{dx}=frac{-y}{x-y^3}$$
Therefore,
$$ydx + (x-y^3)dy=0$$
We can now apply the theory of integrating factors which states that $M(x,y)dx
+N(x,y)dy= 0$. We are given that $M(x,y)=y$ and $N(x,y)=x-y^3$. If we follow the first example in this pdf, we see that the equations are exact since
$$M_y(x,y)=1=N_x(x,y)$$
So, we will therefore find a general solution such that $f(x,y)=C$. Hence,
$$f = int Mdx + g(y)= int ydx + g(y)= xy + g(y)$$
But, we know that
$$f_y=N=x-y^3=x+g'(y)$$
which implies that
$$g'(y)=-y^3$$
and so
$$g(y) = int -y^3dy =frac{-y^4}{4}+C$$
Hence,
$$f(x,y)=xy+frac{-y^4}{4}=C$$
is the general solution. To avoid singularities in the orbits of $y(x)$ which pass through $(0,0)$, we should set $C=0$. Then,
$$xy=frac{y^4}{4}$$
$$4x=y^3$$
$$sqrt[3]{4x}=y$$
Therefore, $y(x)=sqrt[3]{4x}$ is the stable manifold. I'm not sure if there is a shorter approach to compute the stable manifold. The solution inside this question doesn't appear to work for this problem.
Is this approach correct? Please let me know if there are any better alternatives.
ordinary-differential-equations proof-verification manifolds
ordinary-differential-equations proof-verification manifolds
asked Jan 2 at 18:27
Axion004Axion004
314212
asked Jan 2 at 18:27
Axion004Axion004
314212
edited Jan 3 at 2:29
Axion004
asked Jan 2 at 18:27
Axion004Axion004
314212
asked Jan 2 at 18:27
Axion004Axion004
314212
asked Jan 2 at 18:27
Axion004Axion004
314212
314212
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
$require{cancel}$
A simpler way is to realize that on the stable manifold the asymptotic values of $x$ and $y$ are 0
$$
lim_{tto +infty} x(t) = lim_{tto +infty} y(t) = 0
$$
With this in mind, note that the solution for $y$ is fairly trivial
$$
frac{{rm d}y}{{rm d}t} = -y ~~Rightarrow~~~ y(t) = e^{-t} tag{1}
$$
which makes the solution for $x$ even simpler
$$
frac{{rm d}x}{{rm d}t} = x - e^{-3t} ~~~Rightarrow~~~ x(t) = frac{1}{4}e^{-3t} + cancelto{0}{strut c} e^{t} = frac{1}{4}e^{-3t} tag{2}
$$
And from here it you can conclude that the equation of the manifold is
$$
x = frac{1}{4}(e^{-t})^3 = frac{1}{4}y^3 tag{3}
$$
Here's a simple sketch that shows this solution, just to make sure everything checks out. The curve (3) is the solid black line
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
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add a comment |
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$begingroup$
$require{cancel}$
A simpler way is to realize that on the stable manifold the asymptotic values of $x$ and $y$ are 0
$$
lim_{tto +infty} x(t) = lim_{tto +infty} y(t) = 0
$$
With this in mind, note that the solution for $y$ is fairly trivial
$$
frac{{rm d}y}{{rm d}t} = -y ~~Rightarrow~~~ y(t) = e^{-t} tag{1}
$$
which makes the solution for $x$ even simpler
$$
frac{{rm d}x}{{rm d}t} = x - e^{-3t} ~~~Rightarrow~~~ x(t) = frac{1}{4}e^{-3t} + cancelto{0}{strut c} e^{t} = frac{1}{4}e^{-3t} tag{2}
$$
And from here it you can conclude that the equation of the manifold is
$$
x = frac{1}{4}(e^{-t})^3 = frac{1}{4}y^3 tag{3}
$$
Here's a simple sketch that shows this solution, just to make sure everything checks out. The curve (3) is the solid black line
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
$endgroup$
add a comment |
$begingroup$
$require{cancel}$
A simpler way is to realize that on the stable manifold the asymptotic values of $x$ and $y$ are 0
$$
lim_{tto +infty} x(t) = lim_{tto +infty} y(t) = 0
$$
With this in mind, note that the solution for $y$ is fairly trivial
$$
frac{{rm d}y}{{rm d}t} = -y ~~Rightarrow~~~ y(t) = e^{-t} tag{1}
$$
which makes the solution for $x$ even simpler
$$
frac{{rm d}x}{{rm d}t} = x - e^{-3t} ~~~Rightarrow~~~ x(t) = frac{1}{4}e^{-3t} + cancelto{0}{strut c} e^{t} = frac{1}{4}e^{-3t} tag{2}
$$
And from here it you can conclude that the equation of the manifold is
$$
x = frac{1}{4}(e^{-t})^3 = frac{1}{4}y^3 tag{3}
$$
Here's a simple sketch that shows this solution, just to make sure everything checks out. The curve (3) is the solid black line
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
$endgroup$
add a comment |
$begingroup$
$require{cancel}$
A simpler way is to realize that on the stable manifold the asymptotic values of $x$ and $y$ are 0
$$
lim_{tto +infty} x(t) = lim_{tto +infty} y(t) = 0
$$
With this in mind, note that the solution for $y$ is fairly trivial
$$
frac{{rm d}y}{{rm d}t} = -y ~~Rightarrow~~~ y(t) = e^{-t} tag{1}
$$
which makes the solution for $x$ even simpler
$$
frac{{rm d}x}{{rm d}t} = x - e^{-3t} ~~~Rightarrow~~~ x(t) = frac{1}{4}e^{-3t} + cancelto{0}{strut c} e^{t} = frac{1}{4}e^{-3t} tag{2}
$$
And from here it you can conclude that the equation of the manifold is
$$
x = frac{1}{4}(e^{-t})^3 = frac{1}{4}y^3 tag{3}
$$
Here's a simple sketch that shows this solution, just to make sure everything checks out. The curve (3) is the solid black line
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
$endgroup$
$require{cancel}$
A simpler way is to realize that on the stable manifold the asymptotic values of $x$ and $y$ are 0
$$
lim_{tto +infty} x(t) = lim_{tto +infty} y(t) = 0
$$
With this in mind, note that the solution for $y$ is fairly trivial
$$
frac{{rm d}y}{{rm d}t} = -y ~~Rightarrow~~~ y(t) = e^{-t} tag{1}
$$
which makes the solution for $x$ even simpler
$$
frac{{rm d}x}{{rm d}t} = x - e^{-3t} ~~~Rightarrow~~~ x(t) = frac{1}{4}e^{-3t} + cancelto{0}{strut c} e^{t} = frac{1}{4}e^{-3t} tag{2}
$$
And from here it you can conclude that the equation of the manifold is
$$
x = frac{1}{4}(e^{-t})^3 = frac{1}{4}y^3 tag{3}
$$
Here's a simple sketch that shows this solution, just to make sure everything checks out. The curve (3) is the solid black line
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
edited Jan 2 at 20:12
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
answered Jan 2 at 20:04
caveraccaverac
14.2k21130
14.2k21130
add a comment |
add a comment |
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