Does this system of equations have attractors and periodic solutions? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solving linear system of differential equations of 2nd order$x'=Ax$ has one periodic solution. Prove that all solutions are periodic.Proving Nonexistence of Periodic Solutions in a Planar Systemcharacterising attractors for master equationsVerifying if system has periodic solutionsSolve system inhomogeneous differential equations with variable coefficientsPeriods of periodic solutions of the (Hamiltonian) system $dotx=y$, $doty=-x-x^2$Proving solutions of differential equations are periodicAre multiple kinds of attractors (chaotic and otherwise) possible within a single system of differential equations?How can I show that the system of non-linear differential equations does not have periodic orbits?
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Does this system of equations have attractors and periodic solutions?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solving linear system of differential equations of 2nd order$x'=Ax$ has one periodic solution. Prove that all solutions are periodic.Proving Nonexistence of Periodic Solutions in a Planar Systemcharacterising attractors for master equationsVerifying if system has periodic solutionsSolve system inhomogeneous differential equations with variable coefficientsPeriods of periodic solutions of the (Hamiltonian) system $dotx=y$, $doty=-x-x^2$Proving solutions of differential equations are periodicAre multiple kinds of attractors (chaotic and otherwise) possible within a single system of differential equations?How can I show that the system of non-linear differential equations does not have periodic orbits?
$begingroup$
I want to solve the following exercise:
Determine the critical points of the system
beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign
Are there attractors in this system? Determine a first integral. Do periodic solutions exist?
What I've tried so far:
I think that I've found the critical points of this system. If I inspect
beginalign
x^2 - y^3 = 0\
2x(x^2 - y) = 0
endalign
I find that $x = 0$ or $x = pm sqrty$.
Case $x = 0$:
$-y^3 = 0$ so that $y = 0$. Hence the first critical point that I find is $(0,0)$.
Case $x = pmsqrty$:
$y - y^3 = 0$ so $y=1$ or $y = -1$. If $y = -1$ then $x = pm i$. If $y = 1$ then $x = pm1$. Hence I find the critical points $(1, 1), (i, -1), (-1,1),$ and $(-i,-1)$. I think that I've found a first integral by looking at the equation
$$
dfracdfracdydtdfracdxdt = dfracdydx = dfrac2x^3 - 2xyx^2 - y^3
$$
Integrating this final equation gives
$$
int dfrac2x^3 - 2xyx^2 - y^3dx = y(y^2 - 1)log(y^3 - x^2) + x^2 + C
$$
Hence a first integral of this system is
$$
F(x,y) = y(y^2 - 1)log(y^3 - x^2) + x^2
$$
I tried plotting this function to find out if the system of equations has periodic solutions but I don't get any wiser looking at the plots.
I have the following definitions of an attractor:
A critical point $x = a$ of the equation $dotx = f(x)$ in $mathbbR^n$ is called a positive attractor if there exists a neighborhood $Omega_asubset mathbbR^n$ of $x = a$ such that $x(t_0)inOmega_a$ implies $lim_ttoinfty x(t) = a$. If a critical point $x = a$ has this property for $t to - infty$ then $x = a$ is called a negative attractor.
I think that in order to determine whether either of the critical points is an attractor I need to know what the solution $x(t)$ looks like. I'm not sure how to solve this system though and if I enter the equations in wolfram then I don't get a solution either (I might be missing something). So therefore I tried to linearize the equations. I have:
$$
f(x, y) = beginpmatrixx^2 - y^3\2x(x^2 - y)endpmatrix, dfracpartial f(x,y)partial(x,y) = beginpmatrix2x & -3y^2\6x^2& -2xendpmatrix, dfracpartial^2f(x,y)partial (x,y)^2 = beginpmatrix2 & -6y\12x & 0endpmatrix, \dfracpartial^3 f(x,y)partial (x,y)^3 = beginpmatrix0 & -6\12 & 0endpmatrix
$$
Hence, using the Taylor expansion around $(x,y) = (0,0)$ I get
beginalign
dotx &= x^2 - y^3 + ldots\
doty &= 2x^3 + ldots
endalign
but I'm not really sure how I can use this to learn anything new..
Question: How can I learn more about this system so that I will be able to say more about the existence of attractors and periodic solutions? Is my reasoning thus far correct?
ordinary-differential-equations taylor-expansion approximation
asked Apr 8 at 13:58
TitusTitus
1045
$endgroup$
add a comment |
$begingroup$
I want to solve the following exercise:
Determine the critical points of the system
beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign
Are there attractors in this system? Determine a first integral. Do periodic solutions exist?
What I've tried so far:
I think that I've found the critical points of this system. If I inspect
beginalign
x^2 - y^3 = 0\
2x(x^2 - y) = 0
endalign
I find that $x = 0$ or $x = pm sqrty$.
Case $x = 0$:
$-y^3 = 0$ so that $y = 0$. Hence the first critical point that I find is $(0,0)$.
Case $x = pmsqrty$:
$y - y^3 = 0$ so $y=1$ or $y = -1$. If $y = -1$ then $x = pm i$. If $y = 1$ then $x = pm1$. Hence I find the critical points $(1, 1), (i, -1), (-1,1),$ and $(-i,-1)$. I think that I've found a first integral by looking at the equation
$$
dfracdfracdydtdfracdxdt = dfracdydx = dfrac2x^3 - 2xyx^2 - y^3
$$
Integrating this final equation gives
$$
int dfrac2x^3 - 2xyx^2 - y^3dx = y(y^2 - 1)log(y^3 - x^2) + x^2 + C
$$
Hence a first integral of this system is
$$
F(x,y) = y(y^2 - 1)log(y^3 - x^2) + x^2
$$
I tried plotting this function to find out if the system of equations has periodic solutions but I don't get any wiser looking at the plots.
I have the following definitions of an attractor:
A critical point $x = a$ of the equation $dotx = f(x)$ in $mathbbR^n$ is called a positive attractor if there exists a neighborhood $Omega_asubset mathbbR^n$ of $x = a$ such that $x(t_0)inOmega_a$ implies $lim_ttoinfty x(t) = a$. If a critical point $x = a$ has this property for $t to - infty$ then $x = a$ is called a negative attractor.
I think that in order to determine whether either of the critical points is an attractor I need to know what the solution $x(t)$ looks like. I'm not sure how to solve this system though and if I enter the equations in wolfram then I don't get a solution either (I might be missing something). So therefore I tried to linearize the equations. I have:
$$
f(x, y) = beginpmatrixx^2 - y^3\2x(x^2 - y)endpmatrix, dfracpartial f(x,y)partial(x,y) = beginpmatrix2x & -3y^2\6x^2& -2xendpmatrix, dfracpartial^2f(x,y)partial (x,y)^2 = beginpmatrix2 & -6y\12x & 0endpmatrix, \dfracpartial^3 f(x,y)partial (x,y)^3 = beginpmatrix0 & -6\12 & 0endpmatrix
$$
Hence, using the Taylor expansion around $(x,y) = (0,0)$ I get
beginalign
dotx &= x^2 - y^3 + ldots\
doty &= 2x^3 + ldots
endalign
but I'm not really sure how I can use this to learn anything new..
Question: How can I learn more about this system so that I will be able to say more about the existence of attractors and periodic solutions? Is my reasoning thus far correct?
ordinary-differential-equations taylor-expansion approximation
asked Apr 8 at 13:58
TitusTitus
1045
$endgroup$
$begingroup$
Your integration for the first integral is not correct, you would need to consider $y$ as a function of $x$, not as a constant.
$endgroup$
– LutzL
Apr 8 at 16:11
$begingroup$
The Jacobian is wrong in the lower left entry, you need the $x$ derivative of both terms in $2x^3-2xy$. The higher order derivatives are tensors, as pure tensor the second derivative has 8 entries, as symmetric tensor this reduces to 6 independent entries.
$endgroup$
– LutzL
Apr 8 at 16:13
$begingroup$
The Taylor expansion of any polynomial around the origin is that same polynomial again.
$endgroup$
– LutzL
Apr 8 at 16:14
add a comment |
$begingroup$
I want to solve the following exercise:
Determine the critical points of the system
beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign
Are there attractors in this system? Determine a first integral. Do periodic solutions exist?
What I've tried so far:
I think that I've found the critical points of this system. If I inspect
beginalign
x^2 - y^3 = 0\
2x(x^2 - y) = 0
endalign
I find that $x = 0$ or $x = pm sqrty$.
Case $x = 0$:
$-y^3 = 0$ so that $y = 0$. Hence the first critical point that I find is $(0,0)$.
Case $x = pmsqrty$:
$y - y^3 = 0$ so $y=1$ or $y = -1$. If $y = -1$ then $x = pm i$. If $y = 1$ then $x = pm1$. Hence I find the critical points $(1, 1), (i, -1), (-1,1),$ and $(-i,-1)$. I think that I've found a first integral by looking at the equation
$$
dfracdfracdydtdfracdxdt = dfracdydx = dfrac2x^3 - 2xyx^2 - y^3
$$
Integrating this final equation gives
$$
int dfrac2x^3 - 2xyx^2 - y^3dx = y(y^2 - 1)log(y^3 - x^2) + x^2 + C
$$
Hence a first integral of this system is
$$
F(x,y) = y(y^2 - 1)log(y^3 - x^2) + x^2
$$
I tried plotting this function to find out if the system of equations has periodic solutions but I don't get any wiser looking at the plots.
I have the following definitions of an attractor:
A critical point $x = a$ of the equation $dotx = f(x)$ in $mathbbR^n$ is called a positive attractor if there exists a neighborhood $Omega_asubset mathbbR^n$ of $x = a$ such that $x(t_0)inOmega_a$ implies $lim_ttoinfty x(t) = a$. If a critical point $x = a$ has this property for $t to - infty$ then $x = a$ is called a negative attractor.
I think that in order to determine whether either of the critical points is an attractor I need to know what the solution $x(t)$ looks like. I'm not sure how to solve this system though and if I enter the equations in wolfram then I don't get a solution either (I might be missing something). So therefore I tried to linearize the equations. I have:
$$
f(x, y) = beginpmatrixx^2 - y^3\2x(x^2 - y)endpmatrix, dfracpartial f(x,y)partial(x,y) = beginpmatrix2x & -3y^2\6x^2& -2xendpmatrix, dfracpartial^2f(x,y)partial (x,y)^2 = beginpmatrix2 & -6y\12x & 0endpmatrix, \dfracpartial^3 f(x,y)partial (x,y)^3 = beginpmatrix0 & -6\12 & 0endpmatrix
$$
Hence, using the Taylor expansion around $(x,y) = (0,0)$ I get
beginalign
dotx &= x^2 - y^3 + ldots\
doty &= 2x^3 + ldots
endalign
but I'm not really sure how I can use this to learn anything new..
Question: How can I learn more about this system so that I will be able to say more about the existence of attractors and periodic solutions? Is my reasoning thus far correct?
ordinary-differential-equations taylor-expansion approximation
asked Apr 8 at 13:58
TitusTitus
1045
$endgroup$
I want to solve the following exercise:
Determine the critical points of the system
beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign
Are there attractors in this system? Determine a first integral. Do periodic solutions exist?
What I've tried so far:
I think that I've found the critical points of this system. If I inspect
beginalign
x^2 - y^3 = 0\
2x(x^2 - y) = 0
endalign
I find that $x = 0$ or $x = pm sqrty$.
Case $x = 0$:
$-y^3 = 0$ so that $y = 0$. Hence the first critical point that I find is $(0,0)$.
Case $x = pmsqrty$:
$y - y^3 = 0$ so $y=1$ or $y = -1$. If $y = -1$ then $x = pm i$. If $y = 1$ then $x = pm1$. Hence I find the critical points $(1, 1), (i, -1), (-1,1),$ and $(-i,-1)$. I think that I've found a first integral by looking at the equation
$$
dfracdfracdydtdfracdxdt = dfracdydx = dfrac2x^3 - 2xyx^2 - y^3
$$
Integrating this final equation gives
$$
int dfrac2x^3 - 2xyx^2 - y^3dx = y(y^2 - 1)log(y^3 - x^2) + x^2 + C
$$
Hence a first integral of this system is
$$
F(x,y) = y(y^2 - 1)log(y^3 - x^2) + x^2
$$
I tried plotting this function to find out if the system of equations has periodic solutions but I don't get any wiser looking at the plots.
I have the following definitions of an attractor:
A critical point $x = a$ of the equation $dotx = f(x)$ in $mathbbR^n$ is called a positive attractor if there exists a neighborhood $Omega_asubset mathbbR^n$ of $x = a$ such that $x(t_0)inOmega_a$ implies $lim_ttoinfty x(t) = a$. If a critical point $x = a$ has this property for $t to - infty$ then $x = a$ is called a negative attractor.
I think that in order to determine whether either of the critical points is an attractor I need to know what the solution $x(t)$ looks like. I'm not sure how to solve this system though and if I enter the equations in wolfram then I don't get a solution either (I might be missing something). So therefore I tried to linearize the equations. I have:
$$
f(x, y) = beginpmatrixx^2 - y^3\2x(x^2 - y)endpmatrix, dfracpartial f(x,y)partial(x,y) = beginpmatrix2x & -3y^2\6x^2& -2xendpmatrix, dfracpartial^2f(x,y)partial (x,y)^2 = beginpmatrix2 & -6y\12x & 0endpmatrix, \dfracpartial^3 f(x,y)partial (x,y)^3 = beginpmatrix0 & -6\12 & 0endpmatrix
$$
Hence, using the Taylor expansion around $(x,y) = (0,0)$ I get
beginalign
dotx &= x^2 - y^3 + ldots\
doty &= 2x^3 + ldots
endalign
but I'm not really sure how I can use this to learn anything new..
Question: How can I learn more about this system so that I will be able to say more about the existence of attractors and periodic solutions? Is my reasoning thus far correct?
ordinary-differential-equations taylor-expansion approximation
ordinary-differential-equations taylor-expansion approximation
asked Apr 8 at 13:58
TitusTitus
1045
asked Apr 8 at 13:58
TitusTitus
1045
asked Apr 8 at 13:58
TitusTitus
1045
asked Apr 8 at 13:58
TitusTitus
1045
asked Apr 8 at 13:58
TitusTitus
1045
1045
$begingroup$
Your integration for the first integral is not correct, you would need to consider $y$ as a function of $x$, not as a constant.
$endgroup$
– LutzL
Apr 8 at 16:11
$begingroup$
The Jacobian is wrong in the lower left entry, you need the $x$ derivative of both terms in $2x^3-2xy$. The higher order derivatives are tensors, as pure tensor the second derivative has 8 entries, as symmetric tensor this reduces to 6 independent entries.
$endgroup$
– LutzL
Apr 8 at 16:13
$begingroup$
The Taylor expansion of any polynomial around the origin is that same polynomial again.
$endgroup$
– LutzL
Apr 8 at 16:14
add a comment |
$begingroup$
Your integration for the first integral is not correct, you would need to consider $y$ as a function of $x$, not as a constant.
$endgroup$
– LutzL
Apr 8 at 16:11
$begingroup$
The Jacobian is wrong in the lower left entry, you need the $x$ derivative of both terms in $2x^3-2xy$. The higher order derivatives are tensors, as pure tensor the second derivative has 8 entries, as symmetric tensor this reduces to 6 independent entries.
$endgroup$
– LutzL
Apr 8 at 16:13
$begingroup$
The Taylor expansion of any polynomial around the origin is that same polynomial again.
$endgroup$
– LutzL
Apr 8 at 16:14
$begingroup$
Your integration for the first integral is not correct, you would need to consider $y$ as a function of $x$, not as a constant.
$endgroup$
– LutzL
Apr 8 at 16:11
$begingroup$
Your integration for the first integral is not correct, you would need to consider $y$ as a function of $x$, not as a constant.
$endgroup$
– LutzL
Apr 8 at 16:11
$begingroup$
The Jacobian is wrong in the lower left entry, you need the $x$ derivative of both terms in $2x^3-2xy$. The higher order derivatives are tensors, as pure tensor the second derivative has 8 entries, as symmetric tensor this reduces to 6 independent entries.
$endgroup$
– LutzL
Apr 8 at 16:13
$begingroup$
The Jacobian is wrong in the lower left entry, you need the $x$ derivative of both terms in $2x^3-2xy$. The higher order derivatives are tensors, as pure tensor the second derivative has 8 entries, as symmetric tensor this reduces to 6 independent entries.
$endgroup$
– LutzL
Apr 8 at 16:13
$begingroup$
The Taylor expansion of any polynomial around the origin is that same polynomial again.
$endgroup$
– LutzL
Apr 8 at 16:14
$begingroup$
The Taylor expansion of any polynomial around the origin is that same polynomial again.
$endgroup$
– LutzL
Apr 8 at 16:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign$$
$$fracdydx=frac2x(x^2 - y)x^2- y^3$$
Let $X=x^2$
$$fracdydX=fracX - yX- y^3$$
$$fracdXdy=fracX- y^3X-y$$
Let $X=y+u$
$$fracdXdy=1+fracdudy=fracy+u- y^3u$$
$$fracdudy=fracy- y^3u$$
$$2udu=2(y-y^3)dy$$
$$u^2=y^2-frac12 y^4+c$$
$$(X-y)^2=(x^2-y)^2=y^2-frac12 y^4+c$$
$$x^4-2x^2y=-frac12 y^4+c$$
The equation of the trajectory is :
$$boxedy^4-4x^2y+2x^4=2c$$
The analytical study of this equation shows that $cgeq -frac12$ and that the trajectory is a closed curve.
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
$endgroup$
1
$begingroup$
+1, I adore when someone sees all these changes of variables that lead to a final solution, but one observation just for the sake of simplification. The original system can be rewritten as an equivalent first order equation $lbrack 2x^3 - 2xy rbrack , dx + lbrack y^3 - x^2 rbrack , dy = 0$. This equation happens to be exact and thus system has a first integral $Phi(x, y) = fracx^42 - x^2y + fracy^44$, which is exactly the same that you've found.
$endgroup$
– Evgeny
Apr 10 at 8:29
add a comment |
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$begingroup$
$$beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign$$
$$fracdydx=frac2x(x^2 - y)x^2- y^3$$
Let $X=x^2$
$$fracdydX=fracX - yX- y^3$$
$$fracdXdy=fracX- y^3X-y$$
Let $X=y+u$
$$fracdXdy=1+fracdudy=fracy+u- y^3u$$
$$fracdudy=fracy- y^3u$$
$$2udu=2(y-y^3)dy$$
$$u^2=y^2-frac12 y^4+c$$
$$(X-y)^2=(x^2-y)^2=y^2-frac12 y^4+c$$
$$x^4-2x^2y=-frac12 y^4+c$$
The equation of the trajectory is :
$$boxedy^4-4x^2y+2x^4=2c$$
The analytical study of this equation shows that $cgeq -frac12$ and that the trajectory is a closed curve.
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
$endgroup$
1
$begingroup$
+1, I adore when someone sees all these changes of variables that lead to a final solution, but one observation just for the sake of simplification. The original system can be rewritten as an equivalent first order equation $lbrack 2x^3 - 2xy rbrack , dx + lbrack y^3 - x^2 rbrack , dy = 0$. This equation happens to be exact and thus system has a first integral $Phi(x, y) = fracx^42 - x^2y + fracy^44$, which is exactly the same that you've found.
$endgroup$
– Evgeny
Apr 10 at 8:29
add a comment |
$begingroup$
$$beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign$$
$$fracdydx=frac2x(x^2 - y)x^2- y^3$$
Let $X=x^2$
$$fracdydX=fracX - yX- y^3$$
$$fracdXdy=fracX- y^3X-y$$
Let $X=y+u$
$$fracdXdy=1+fracdudy=fracy+u- y^3u$$
$$fracdudy=fracy- y^3u$$
$$2udu=2(y-y^3)dy$$
$$u^2=y^2-frac12 y^4+c$$
$$(X-y)^2=(x^2-y)^2=y^2-frac12 y^4+c$$
$$x^4-2x^2y=-frac12 y^4+c$$
The equation of the trajectory is :
$$boxedy^4-4x^2y+2x^4=2c$$
The analytical study of this equation shows that $cgeq -frac12$ and that the trajectory is a closed curve.
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
$endgroup$
1
$begingroup$
+1, I adore when someone sees all these changes of variables that lead to a final solution, but one observation just for the sake of simplification. The original system can be rewritten as an equivalent first order equation $lbrack 2x^3 - 2xy rbrack , dx + lbrack y^3 - x^2 rbrack , dy = 0$. This equation happens to be exact and thus system has a first integral $Phi(x, y) = fracx^42 - x^2y + fracy^44$, which is exactly the same that you've found.
$endgroup$
– Evgeny
Apr 10 at 8:29
add a comment |
$begingroup$
$$beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign$$
$$fracdydx=frac2x(x^2 - y)x^2- y^3$$
Let $X=x^2$
$$fracdydX=fracX - yX- y^3$$
$$fracdXdy=fracX- y^3X-y$$
Let $X=y+u$
$$fracdXdy=1+fracdudy=fracy+u- y^3u$$
$$fracdudy=fracy- y^3u$$
$$2udu=2(y-y^3)dy$$
$$u^2=y^2-frac12 y^4+c$$
$$(X-y)^2=(x^2-y)^2=y^2-frac12 y^4+c$$
$$x^4-2x^2y=-frac12 y^4+c$$
The equation of the trajectory is :
$$boxedy^4-4x^2y+2x^4=2c$$
The analytical study of this equation shows that $cgeq -frac12$ and that the trajectory is a closed curve.
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
$endgroup$
$$beginalign
dotx &= x^2- y^3\
doty &= 2x(x^2 - y)
endalign$$
$$fracdydx=frac2x(x^2 - y)x^2- y^3$$
Let $X=x^2$
$$fracdydX=fracX - yX- y^3$$
$$fracdXdy=fracX- y^3X-y$$
Let $X=y+u$
$$fracdXdy=1+fracdudy=fracy+u- y^3u$$
$$fracdudy=fracy- y^3u$$
$$2udu=2(y-y^3)dy$$
$$u^2=y^2-frac12 y^4+c$$
$$(X-y)^2=(x^2-y)^2=y^2-frac12 y^4+c$$
$$x^4-2x^2y=-frac12 y^4+c$$
The equation of the trajectory is :
$$boxedy^4-4x^2y+2x^4=2c$$
The analytical study of this equation shows that $cgeq -frac12$ and that the trajectory is a closed curve.
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
edited Apr 9 at 11:33
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
answered Apr 8 at 21:48
JJacquelinJJacquelin
45.6k21857
45.6k21857
1
$begingroup$
+1, I adore when someone sees all these changes of variables that lead to a final solution, but one observation just for the sake of simplification. The original system can be rewritten as an equivalent first order equation $lbrack 2x^3 - 2xy rbrack , dx + lbrack y^3 - x^2 rbrack , dy = 0$. This equation happens to be exact and thus system has a first integral $Phi(x, y) = fracx^42 - x^2y + fracy^44$, which is exactly the same that you've found.
$endgroup$
– Evgeny
Apr 10 at 8:29
add a comment |
1
$begingroup$
+1, I adore when someone sees all these changes of variables that lead to a final solution, but one observation just for the sake of simplification. The original system can be rewritten as an equivalent first order equation $lbrack 2x^3 - 2xy rbrack , dx + lbrack y^3 - x^2 rbrack , dy = 0$. This equation happens to be exact and thus system has a first integral $Phi(x, y) = fracx^42 - x^2y + fracy^44$, which is exactly the same that you've found.
$endgroup$
– Evgeny
Apr 10 at 8:29
1
1
$begingroup$
+1, I adore when someone sees all these changes of variables that lead to a final solution, but one observation just for the sake of simplification. The original system can be rewritten as an equivalent first order equation $lbrack 2x^3 - 2xy rbrack , dx + lbrack y^3 - x^2 rbrack , dy = 0$. This equation happens to be exact and thus system has a first integral $Phi(x, y) = fracx^42 - x^2y + fracy^44$, which is exactly the same that you've found.
$endgroup$
– Evgeny
Apr 10 at 8:29
$begingroup$
+1, I adore when someone sees all these changes of variables that lead to a final solution, but one observation just for the sake of simplification. The original system can be rewritten as an equivalent first order equation $lbrack 2x^3 - 2xy rbrack , dx + lbrack y^3 - x^2 rbrack , dy = 0$. This equation happens to be exact and thus system has a first integral $Phi(x, y) = fracx^42 - x^2y + fracy^44$, which is exactly the same that you've found.
$endgroup$
– Evgeny
Apr 10 at 8:29
add a comment |
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$begingroup$
Your integration for the first integral is not correct, you would need to consider $y$ as a function of $x$, not as a constant.
$endgroup$
– LutzL
Apr 8 at 16:11
$begingroup$
The Jacobian is wrong in the lower left entry, you need the $x$ derivative of both terms in $2x^3-2xy$. The higher order derivatives are tensors, as pure tensor the second derivative has 8 entries, as symmetric tensor this reduces to 6 independent entries.
$endgroup$
– LutzL
Apr 8 at 16:13
$begingroup$
The Taylor expansion of any polynomial around the origin is that same polynomial again.
$endgroup$
– LutzL
Apr 8 at 16:14