Sketch phase portray of $ x'' + 3x^2 + - 3 = 0 $
$begingroup$
I am sketching the phase portray of $ x'' + 3x^2 + - 3 = 0 $.
First of all I found the corresponding ODE system:
begin{align*}
x' = y \
y' = -3x^2 + 3
end{align*}
The critical points are (-1,0) and (1,0). However I am going to center it in the origin so I will ignore the +3, the solution is the same anyway.
I tried to use the method of linearization but I an eigenvalue of real part equals to 0 and this does not give information.
Then I tried using Liaupnov method, so I need to find a function $V(x,y)$ such that $V(0,0)=0$, $V(x,y) geq 0$ and $V'(x,y) leq 0$, this would mean that the origin is stable. However I am struggling to find such a function here. Any hints?
ordinary-differential-equations
asked Jan 12 at 17:01
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
I am sketching the phase portray of $ x'' + 3x^2 + - 3 = 0 $.
First of all I found the corresponding ODE system:
begin{align*}
x' = y \
y' = -3x^2 + 3
end{align*}
The critical points are (-1,0) and (1,0). However I am going to center it in the origin so I will ignore the +3, the solution is the same anyway.
I tried to use the method of linearization but I an eigenvalue of real part equals to 0 and this does not give information.
Then I tried using Liaupnov method, so I need to find a function $V(x,y)$ such that $V(0,0)=0$, $V(x,y) geq 0$ and $V'(x,y) leq 0$, this would mean that the origin is stable. However I am struggling to find such a function here. Any hints?
ordinary-differential-equations
asked Jan 12 at 17:01
qcc101qcc101
627213
$endgroup$
$begingroup$
None of the objects/tools you mention is needed to draw a phase portrait. One only needs to know the signs of $x'$ and $y'$, depending on the point $(x,y)$. In your case, this is direct, no?
$endgroup$
– Did
Jan 12 at 17:30
$begingroup$
@Did Yes you are right I was totally overthinking it.
$endgroup$
– qcc101
Jan 12 at 17:41
add a comment |
$begingroup$
I am sketching the phase portray of $ x'' + 3x^2 + - 3 = 0 $.
First of all I found the corresponding ODE system:
begin{align*}
x' = y \
y' = -3x^2 + 3
end{align*}
The critical points are (-1,0) and (1,0). However I am going to center it in the origin so I will ignore the +3, the solution is the same anyway.
I tried to use the method of linearization but I an eigenvalue of real part equals to 0 and this does not give information.
Then I tried using Liaupnov method, so I need to find a function $V(x,y)$ such that $V(0,0)=0$, $V(x,y) geq 0$ and $V'(x,y) leq 0$, this would mean that the origin is stable. However I am struggling to find such a function here. Any hints?
ordinary-differential-equations
asked Jan 12 at 17:01
qcc101qcc101
627213
$endgroup$
I am sketching the phase portray of $ x'' + 3x^2 + - 3 = 0 $.
First of all I found the corresponding ODE system:
begin{align*}
x' = y \
y' = -3x^2 + 3
end{align*}
The critical points are (-1,0) and (1,0). However I am going to center it in the origin so I will ignore the +3, the solution is the same anyway.
I tried to use the method of linearization but I an eigenvalue of real part equals to 0 and this does not give information.
Then I tried using Liaupnov method, so I need to find a function $V(x,y)$ such that $V(0,0)=0$, $V(x,y) geq 0$ and $V'(x,y) leq 0$, this would mean that the origin is stable. However I am struggling to find such a function here. Any hints?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 12 at 17:01
qcc101qcc101
627213
asked Jan 12 at 17:01
qcc101qcc101
627213
asked Jan 12 at 17:01
qcc101qcc101
627213
asked Jan 12 at 17:01
qcc101qcc101
627213
asked Jan 12 at 17:01
qcc101qcc101
627213
627213
$begingroup$
None of the objects/tools you mention is needed to draw a phase portrait. One only needs to know the signs of $x'$ and $y'$, depending on the point $(x,y)$. In your case, this is direct, no?
$endgroup$
– Did
Jan 12 at 17:30
$begingroup$
@Did Yes you are right I was totally overthinking it.
$endgroup$
– qcc101
Jan 12 at 17:41
add a comment |
$begingroup$
None of the objects/tools you mention is needed to draw a phase portrait. One only needs to know the signs of $x'$ and $y'$, depending on the point $(x,y)$. In your case, this is direct, no?
$endgroup$
– Did
Jan 12 at 17:30
$begingroup$
@Did Yes you are right I was totally overthinking it.
$endgroup$
– qcc101
Jan 12 at 17:41
$begingroup$
None of the objects/tools you mention is needed to draw a phase portrait. One only needs to know the signs of $x'$ and $y'$, depending on the point $(x,y)$. In your case, this is direct, no?
$endgroup$
– Did
Jan 12 at 17:30
$begingroup$
None of the objects/tools you mention is needed to draw a phase portrait. One only needs to know the signs of $x'$ and $y'$, depending on the point $(x,y)$. In your case, this is direct, no?
$endgroup$
– Did
Jan 12 at 17:30
$begingroup$
@Did Yes you are right I was totally overthinking it.
$endgroup$
– qcc101
Jan 12 at 17:41
$begingroup$
@Did Yes you are right I was totally overthinking it.
$endgroup$
– qcc101
Jan 12 at 17:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write the Jacobian at the critical points ${bf x}^*$
$$
J({bf x}^*) = pmatrix{0 & 1 \ -6x^* & 0}
$$
with eigenvalues $lambda^2 = -6x^*$. That means that for
${bf x}_1^* = (+1,0)$ the eigenvalues are $lambda_1^{pm} = pm i sqrt{6}$
Solution is a cycle
${bf x}_2^* = (1,0)$ the eigenvalues are $lambda_2^{pm} = pm sqrt{6}$
Solution is a saddle point. You can find the directions of the stable and unstable manifolds by calculating the eigenvectors. I will leave that part to you
Here's a sketch
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
"Solution is a cycle" Not necessarily. Even if the eigenvalues of the Jacobian are purely complex, the solutions of the (original, non linearized) system need not be cycles.
$endgroup$
– Did
Jan 12 at 17:32
$begingroup$
@Did Thanks for the comment. I have, repeatedly, seen this type of clarifications form you. Are there any conditions for this to work at all? Because If I see the solutions, I can see them going around $(1, 0)$. Or should I abandon this for good? Thanks
$endgroup$
– caverac
Jan 12 at 17:51
$begingroup$
The Jacobian of a system $x'=f(x,y)$, $y'=g(x,y)$, near its fixed point $(0,0)$ describes $f(x,y)$ and $g(x,y)$ when $(x,y)to(0,0)$, but only at order $1$. A canonical example might be the system $$x'=-yqquad y'=x$$ with a fixed point at $(0,0)$ and a Jacobian matrix with eigenvalues $pm i$, indicating a center, and whose solutions indeed are the circles $$x^2+y^2=x_0^2+y_0^2$$ But consider its perturbation $$x'=-y+cx^3qquad y'=x+cy^3$$ for some fixed $cne0$. The Jacobian matrix at $(0,0)$ stays the same, again with eigenvalues $pm i$, but, ...
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
... by an elementary computation, $$(x^2+y^2)'=2c(x^4+y^4)$$ Thus, for every $cne0$, the solutions are not cycles anymore. Actually, for every $c<0$, $(0,0)$ is now a stable focus and the solutions are inwards spirals while, for every $c>0$, $(0,0)$ becomes an unstable focus and the solutions are outwards spirals.
$endgroup$
– Did
Jan 12 at 22:11
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Write the Jacobian at the critical points ${bf x}^*$
$$
J({bf x}^*) = pmatrix{0 & 1 \ -6x^* & 0}
$$
with eigenvalues $lambda^2 = -6x^*$. That means that for
${bf x}_1^* = (+1,0)$ the eigenvalues are $lambda_1^{pm} = pm i sqrt{6}$
Solution is a cycle
${bf x}_2^* = (1,0)$ the eigenvalues are $lambda_2^{pm} = pm sqrt{6}$
Solution is a saddle point. You can find the directions of the stable and unstable manifolds by calculating the eigenvectors. I will leave that part to you
Here's a sketch
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
"Solution is a cycle" Not necessarily. Even if the eigenvalues of the Jacobian are purely complex, the solutions of the (original, non linearized) system need not be cycles.
$endgroup$
– Did
Jan 12 at 17:32
$begingroup$
@Did Thanks for the comment. I have, repeatedly, seen this type of clarifications form you. Are there any conditions for this to work at all? Because If I see the solutions, I can see them going around $(1, 0)$. Or should I abandon this for good? Thanks
$endgroup$
– caverac
Jan 12 at 17:51
$begingroup$
The Jacobian of a system $x'=f(x,y)$, $y'=g(x,y)$, near its fixed point $(0,0)$ describes $f(x,y)$ and $g(x,y)$ when $(x,y)to(0,0)$, but only at order $1$. A canonical example might be the system $$x'=-yqquad y'=x$$ with a fixed point at $(0,0)$ and a Jacobian matrix with eigenvalues $pm i$, indicating a center, and whose solutions indeed are the circles $$x^2+y^2=x_0^2+y_0^2$$ But consider its perturbation $$x'=-y+cx^3qquad y'=x+cy^3$$ for some fixed $cne0$. The Jacobian matrix at $(0,0)$ stays the same, again with eigenvalues $pm i$, but, ...
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
... by an elementary computation, $$(x^2+y^2)'=2c(x^4+y^4)$$ Thus, for every $cne0$, the solutions are not cycles anymore. Actually, for every $c<0$, $(0,0)$ is now a stable focus and the solutions are inwards spirals while, for every $c>0$, $(0,0)$ becomes an unstable focus and the solutions are outwards spirals.
$endgroup$
– Did
Jan 12 at 22:11
add a comment |
$begingroup$
Write the Jacobian at the critical points ${bf x}^*$
$$
J({bf x}^*) = pmatrix{0 & 1 \ -6x^* & 0}
$$
with eigenvalues $lambda^2 = -6x^*$. That means that for
${bf x}_1^* = (+1,0)$ the eigenvalues are $lambda_1^{pm} = pm i sqrt{6}$
Solution is a cycle
${bf x}_2^* = (1,0)$ the eigenvalues are $lambda_2^{pm} = pm sqrt{6}$
Solution is a saddle point. You can find the directions of the stable and unstable manifolds by calculating the eigenvectors. I will leave that part to you
Here's a sketch
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
"Solution is a cycle" Not necessarily. Even if the eigenvalues of the Jacobian are purely complex, the solutions of the (original, non linearized) system need not be cycles.
$endgroup$
– Did
Jan 12 at 17:32
$begingroup$
@Did Thanks for the comment. I have, repeatedly, seen this type of clarifications form you. Are there any conditions for this to work at all? Because If I see the solutions, I can see them going around $(1, 0)$. Or should I abandon this for good? Thanks
$endgroup$
– caverac
Jan 12 at 17:51
$begingroup$
The Jacobian of a system $x'=f(x,y)$, $y'=g(x,y)$, near its fixed point $(0,0)$ describes $f(x,y)$ and $g(x,y)$ when $(x,y)to(0,0)$, but only at order $1$. A canonical example might be the system $$x'=-yqquad y'=x$$ with a fixed point at $(0,0)$ and a Jacobian matrix with eigenvalues $pm i$, indicating a center, and whose solutions indeed are the circles $$x^2+y^2=x_0^2+y_0^2$$ But consider its perturbation $$x'=-y+cx^3qquad y'=x+cy^3$$ for some fixed $cne0$. The Jacobian matrix at $(0,0)$ stays the same, again with eigenvalues $pm i$, but, ...
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
... by an elementary computation, $$(x^2+y^2)'=2c(x^4+y^4)$$ Thus, for every $cne0$, the solutions are not cycles anymore. Actually, for every $c<0$, $(0,0)$ is now a stable focus and the solutions are inwards spirals while, for every $c>0$, $(0,0)$ becomes an unstable focus and the solutions are outwards spirals.
$endgroup$
– Did
Jan 12 at 22:11
add a comment |
$begingroup$
Write the Jacobian at the critical points ${bf x}^*$
$$
J({bf x}^*) = pmatrix{0 & 1 \ -6x^* & 0}
$$
with eigenvalues $lambda^2 = -6x^*$. That means that for
${bf x}_1^* = (+1,0)$ the eigenvalues are $lambda_1^{pm} = pm i sqrt{6}$
Solution is a cycle
${bf x}_2^* = (1,0)$ the eigenvalues are $lambda_2^{pm} = pm sqrt{6}$
Solution is a saddle point. You can find the directions of the stable and unstable manifolds by calculating the eigenvectors. I will leave that part to you
Here's a sketch
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
$endgroup$
Write the Jacobian at the critical points ${bf x}^*$
$$
J({bf x}^*) = pmatrix{0 & 1 \ -6x^* & 0}
$$
with eigenvalues $lambda^2 = -6x^*$. That means that for
${bf x}_1^* = (+1,0)$ the eigenvalues are $lambda_1^{pm} = pm i sqrt{6}$
Solution is a cycle
${bf x}_2^* = (1,0)$ the eigenvalues are $lambda_2^{pm} = pm sqrt{6}$
Solution is a saddle point. You can find the directions of the stable and unstable manifolds by calculating the eigenvectors. I will leave that part to you
Here's a sketch
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
answered Jan 12 at 17:19
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
"Solution is a cycle" Not necessarily. Even if the eigenvalues of the Jacobian are purely complex, the solutions of the (original, non linearized) system need not be cycles.
$endgroup$
– Did
Jan 12 at 17:32
$begingroup$
@Did Thanks for the comment. I have, repeatedly, seen this type of clarifications form you. Are there any conditions for this to work at all? Because If I see the solutions, I can see them going around $(1, 0)$. Or should I abandon this for good? Thanks
$endgroup$
– caverac
Jan 12 at 17:51
$begingroup$
The Jacobian of a system $x'=f(x,y)$, $y'=g(x,y)$, near its fixed point $(0,0)$ describes $f(x,y)$ and $g(x,y)$ when $(x,y)to(0,0)$, but only at order $1$. A canonical example might be the system $$x'=-yqquad y'=x$$ with a fixed point at $(0,0)$ and a Jacobian matrix with eigenvalues $pm i$, indicating a center, and whose solutions indeed are the circles $$x^2+y^2=x_0^2+y_0^2$$ But consider its perturbation $$x'=-y+cx^3qquad y'=x+cy^3$$ for some fixed $cne0$. The Jacobian matrix at $(0,0)$ stays the same, again with eigenvalues $pm i$, but, ...
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
... by an elementary computation, $$(x^2+y^2)'=2c(x^4+y^4)$$ Thus, for every $cne0$, the solutions are not cycles anymore. Actually, for every $c<0$, $(0,0)$ is now a stable focus and the solutions are inwards spirals while, for every $c>0$, $(0,0)$ becomes an unstable focus and the solutions are outwards spirals.
$endgroup$
– Did
Jan 12 at 22:11
add a comment |
$begingroup$
"Solution is a cycle" Not necessarily. Even if the eigenvalues of the Jacobian are purely complex, the solutions of the (original, non linearized) system need not be cycles.
$endgroup$
– Did
Jan 12 at 17:32
$begingroup$
@Did Thanks for the comment. I have, repeatedly, seen this type of clarifications form you. Are there any conditions for this to work at all? Because If I see the solutions, I can see them going around $(1, 0)$. Or should I abandon this for good? Thanks
$endgroup$
– caverac
Jan 12 at 17:51
$begingroup$
The Jacobian of a system $x'=f(x,y)$, $y'=g(x,y)$, near its fixed point $(0,0)$ describes $f(x,y)$ and $g(x,y)$ when $(x,y)to(0,0)$, but only at order $1$. A canonical example might be the system $$x'=-yqquad y'=x$$ with a fixed point at $(0,0)$ and a Jacobian matrix with eigenvalues $pm i$, indicating a center, and whose solutions indeed are the circles $$x^2+y^2=x_0^2+y_0^2$$ But consider its perturbation $$x'=-y+cx^3qquad y'=x+cy^3$$ for some fixed $cne0$. The Jacobian matrix at $(0,0)$ stays the same, again with eigenvalues $pm i$, but, ...
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
... by an elementary computation, $$(x^2+y^2)'=2c(x^4+y^4)$$ Thus, for every $cne0$, the solutions are not cycles anymore. Actually, for every $c<0$, $(0,0)$ is now a stable focus and the solutions are inwards spirals while, for every $c>0$, $(0,0)$ becomes an unstable focus and the solutions are outwards spirals.
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
"Solution is a cycle" Not necessarily. Even if the eigenvalues of the Jacobian are purely complex, the solutions of the (original, non linearized) system need not be cycles.
$endgroup$
– Did
Jan 12 at 17:32
$begingroup$
"Solution is a cycle" Not necessarily. Even if the eigenvalues of the Jacobian are purely complex, the solutions of the (original, non linearized) system need not be cycles.
$endgroup$
– Did
Jan 12 at 17:32
$begingroup$
@Did Thanks for the comment. I have, repeatedly, seen this type of clarifications form you. Are there any conditions for this to work at all? Because If I see the solutions, I can see them going around $(1, 0)$. Or should I abandon this for good? Thanks
$endgroup$
– caverac
Jan 12 at 17:51
$begingroup$
@Did Thanks for the comment. I have, repeatedly, seen this type of clarifications form you. Are there any conditions for this to work at all? Because If I see the solutions, I can see them going around $(1, 0)$. Or should I abandon this for good? Thanks
$endgroup$
– caverac
Jan 12 at 17:51
$begingroup$
The Jacobian of a system $x'=f(x,y)$, $y'=g(x,y)$, near its fixed point $(0,0)$ describes $f(x,y)$ and $g(x,y)$ when $(x,y)to(0,0)$, but only at order $1$. A canonical example might be the system $$x'=-yqquad y'=x$$ with a fixed point at $(0,0)$ and a Jacobian matrix with eigenvalues $pm i$, indicating a center, and whose solutions indeed are the circles $$x^2+y^2=x_0^2+y_0^2$$ But consider its perturbation $$x'=-y+cx^3qquad y'=x+cy^3$$ for some fixed $cne0$. The Jacobian matrix at $(0,0)$ stays the same, again with eigenvalues $pm i$, but, ...
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
The Jacobian of a system $x'=f(x,y)$, $y'=g(x,y)$, near its fixed point $(0,0)$ describes $f(x,y)$ and $g(x,y)$ when $(x,y)to(0,0)$, but only at order $1$. A canonical example might be the system $$x'=-yqquad y'=x$$ with a fixed point at $(0,0)$ and a Jacobian matrix with eigenvalues $pm i$, indicating a center, and whose solutions indeed are the circles $$x^2+y^2=x_0^2+y_0^2$$ But consider its perturbation $$x'=-y+cx^3qquad y'=x+cy^3$$ for some fixed $cne0$. The Jacobian matrix at $(0,0)$ stays the same, again with eigenvalues $pm i$, but, ...
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
... by an elementary computation, $$(x^2+y^2)'=2c(x^4+y^4)$$ Thus, for every $cne0$, the solutions are not cycles anymore. Actually, for every $c<0$, $(0,0)$ is now a stable focus and the solutions are inwards spirals while, for every $c>0$, $(0,0)$ becomes an unstable focus and the solutions are outwards spirals.
$endgroup$
– Did
Jan 12 at 22:11
$begingroup$
... by an elementary computation, $$(x^2+y^2)'=2c(x^4+y^4)$$ Thus, for every $cne0$, the solutions are not cycles anymore. Actually, for every $c<0$, $(0,0)$ is now a stable focus and the solutions are inwards spirals while, for every $c>0$, $(0,0)$ becomes an unstable focus and the solutions are outwards spirals.
$endgroup$
– Did
Jan 12 at 22:11
add a comment |
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$begingroup$
None of the objects/tools you mention is needed to draw a phase portrait. One only needs to know the signs of $x'$ and $y'$, depending on the point $(x,y)$. In your case, this is direct, no?
$endgroup$
– Did
Jan 12 at 17:30
$begingroup$
@Did Yes you are right I was totally overthinking it.
$endgroup$
– qcc101
Jan 12 at 17:41