Solving linear system, finding equilibrium and bifurcation points
up vote
3
down vote
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For homework, I have to solve the following problem: consider the system of ODE's
begin{equation}
x'=-x^2+a \ y'=-y
end{equation}
where the parameter $a$ is a real number. I have to characterize the equilibrium points and draw a bifurcation diagram. I am asked to linearize the system first. I have missed a couple of classes and I am struggling to understand how I am supposed to work here.
If $a$ is positive, then there are two equilibrium points $(sqrt{a},0)$ and $(-sqrt{a},0)$, if $a$ is equal to zero, then there is one equilibrium point $ left ( 0,0 right )$ and finally if $a$ is negative, then no equilibrium points exist. The Jacobi matrix of the system is
begin{pmatrix}
- 2x & 0\
0 & -1
end{pmatrix}
which is diagonal, hence its eigenvalues are $-2x$ and $-1$. Then, I calculate this matrix at the equilibrium points:
If $a>0$, then the two Jacobi matrices are
begin{pmatrix}
-2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has two real and negative eigenvalues, so $(sqrt{a},0)$ is a stable point and
begin{pmatrix}
2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has one negative and one positive eigenvalue, so $(-sqrt{a},0)$ is a saddle point.
If $a=0$ the Jacobi matrix is
begin{pmatrix}
0 & 0\
0 & -1
end{pmatrix}
which has one zero eigenvalue so the linearization method does not provide any information about the stability of $left ( 0,0 right )$. As my notes suggest, for this point we find the eigenvectors of this matrix and the orbits of the solutions. But I do not understand why and what information we obtain from this. What can I say about the stability of $left ( 0,0 right )$ then? What about the saddle point, do I use a similar method? I understand that there is a bifurcation for $a=0$ but I don't know how am I supposed to draw the diagram without a computer. Another thing I find in the notes in another example is that we divide the equations and find a relationship between the solutions which does not involve time, for some reason.
Thanks in advance, any insight on this will be appreciated, right now I am lost.
differential-equations dynamical-systems stability-in-odes
asked Nov 13 at 14:08
kleinmeinpouts
1017
add a comment |
up vote
3
down vote
favorite
For homework, I have to solve the following problem: consider the system of ODE's
begin{equation}
x'=-x^2+a \ y'=-y
end{equation}
where the parameter $a$ is a real number. I have to characterize the equilibrium points and draw a bifurcation diagram. I am asked to linearize the system first. I have missed a couple of classes and I am struggling to understand how I am supposed to work here.
If $a$ is positive, then there are two equilibrium points $(sqrt{a},0)$ and $(-sqrt{a},0)$, if $a$ is equal to zero, then there is one equilibrium point $ left ( 0,0 right )$ and finally if $a$ is negative, then no equilibrium points exist. The Jacobi matrix of the system is
begin{pmatrix}
- 2x & 0\
0 & -1
end{pmatrix}
which is diagonal, hence its eigenvalues are $-2x$ and $-1$. Then, I calculate this matrix at the equilibrium points:
If $a>0$, then the two Jacobi matrices are
begin{pmatrix}
-2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has two real and negative eigenvalues, so $(sqrt{a},0)$ is a stable point and
begin{pmatrix}
2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has one negative and one positive eigenvalue, so $(-sqrt{a},0)$ is a saddle point.
If $a=0$ the Jacobi matrix is
begin{pmatrix}
0 & 0\
0 & -1
end{pmatrix}
which has one zero eigenvalue so the linearization method does not provide any information about the stability of $left ( 0,0 right )$. As my notes suggest, for this point we find the eigenvectors of this matrix and the orbits of the solutions. But I do not understand why and what information we obtain from this. What can I say about the stability of $left ( 0,0 right )$ then? What about the saddle point, do I use a similar method? I understand that there is a bifurcation for $a=0$ but I don't know how am I supposed to draw the diagram without a computer. Another thing I find in the notes in another example is that we divide the equations and find a relationship between the solutions which does not involve time, for some reason.
Thanks in advance, any insight on this will be appreciated, right now I am lost.
differential-equations dynamical-systems stability-in-odes
asked Nov 13 at 14:08
kleinmeinpouts
1017
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For homework, I have to solve the following problem: consider the system of ODE's
begin{equation}
x'=-x^2+a \ y'=-y
end{equation}
where the parameter $a$ is a real number. I have to characterize the equilibrium points and draw a bifurcation diagram. I am asked to linearize the system first. I have missed a couple of classes and I am struggling to understand how I am supposed to work here.
If $a$ is positive, then there are two equilibrium points $(sqrt{a},0)$ and $(-sqrt{a},0)$, if $a$ is equal to zero, then there is one equilibrium point $ left ( 0,0 right )$ and finally if $a$ is negative, then no equilibrium points exist. The Jacobi matrix of the system is
begin{pmatrix}
- 2x & 0\
0 & -1
end{pmatrix}
which is diagonal, hence its eigenvalues are $-2x$ and $-1$. Then, I calculate this matrix at the equilibrium points:
If $a>0$, then the two Jacobi matrices are
begin{pmatrix}
-2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has two real and negative eigenvalues, so $(sqrt{a},0)$ is a stable point and
begin{pmatrix}
2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has one negative and one positive eigenvalue, so $(-sqrt{a},0)$ is a saddle point.
If $a=0$ the Jacobi matrix is
begin{pmatrix}
0 & 0\
0 & -1
end{pmatrix}
which has one zero eigenvalue so the linearization method does not provide any information about the stability of $left ( 0,0 right )$. As my notes suggest, for this point we find the eigenvectors of this matrix and the orbits of the solutions. But I do not understand why and what information we obtain from this. What can I say about the stability of $left ( 0,0 right )$ then? What about the saddle point, do I use a similar method? I understand that there is a bifurcation for $a=0$ but I don't know how am I supposed to draw the diagram without a computer. Another thing I find in the notes in another example is that we divide the equations and find a relationship between the solutions which does not involve time, for some reason.
Thanks in advance, any insight on this will be appreciated, right now I am lost.
differential-equations dynamical-systems stability-in-odes
asked Nov 13 at 14:08
kleinmeinpouts
1017
For homework, I have to solve the following problem: consider the system of ODE's
begin{equation}
x'=-x^2+a \ y'=-y
end{equation}
where the parameter $a$ is a real number. I have to characterize the equilibrium points and draw a bifurcation diagram. I am asked to linearize the system first. I have missed a couple of classes and I am struggling to understand how I am supposed to work here.
If $a$ is positive, then there are two equilibrium points $(sqrt{a},0)$ and $(-sqrt{a},0)$, if $a$ is equal to zero, then there is one equilibrium point $ left ( 0,0 right )$ and finally if $a$ is negative, then no equilibrium points exist. The Jacobi matrix of the system is
begin{pmatrix}
- 2x & 0\
0 & -1
end{pmatrix}
which is diagonal, hence its eigenvalues are $-2x$ and $-1$. Then, I calculate this matrix at the equilibrium points:
If $a>0$, then the two Jacobi matrices are
begin{pmatrix}
-2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has two real and negative eigenvalues, so $(sqrt{a},0)$ is a stable point and
begin{pmatrix}
2sqrt{a} & 0\
0 & -1
end{pmatrix}
which has one negative and one positive eigenvalue, so $(-sqrt{a},0)$ is a saddle point.
If $a=0$ the Jacobi matrix is
begin{pmatrix}
0 & 0\
0 & -1
end{pmatrix}
which has one zero eigenvalue so the linearization method does not provide any information about the stability of $left ( 0,0 right )$. As my notes suggest, for this point we find the eigenvectors of this matrix and the orbits of the solutions. But I do not understand why and what information we obtain from this. What can I say about the stability of $left ( 0,0 right )$ then? What about the saddle point, do I use a similar method? I understand that there is a bifurcation for $a=0$ but I don't know how am I supposed to draw the diagram without a computer. Another thing I find in the notes in another example is that we divide the equations and find a relationship between the solutions which does not involve time, for some reason.
Thanks in advance, any insight on this will be appreciated, right now I am lost.
differential-equations dynamical-systems stability-in-odes
differential-equations dynamical-systems stability-in-odes
asked Nov 13 at 14:08
kleinmeinpouts
1017
asked Nov 13 at 14:08
kleinmeinpouts
1017
edited Nov 13 at 14:23
asked Nov 13 at 14:08
kleinmeinpouts
1017
asked Nov 13 at 14:08
kleinmeinpouts
1017
asked Nov 13 at 14:08
kleinmeinpouts
1017
1017
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Hint:
By integration, when $a=0$, either
$$x=0$$ or
$$frac1x=t+C.$$
answered Nov 13 at 14:16
Yves Daoust
121k668216
I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right?
– kleinmeinpouts
Nov 13 at 14:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint:
By integration, when $a=0$, either
$$x=0$$ or
$$frac1x=t+C.$$
answered Nov 13 at 14:16
Yves Daoust
121k668216
I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right?
– kleinmeinpouts
Nov 13 at 14:19
add a comment |
up vote
1
down vote
Hint:
By integration, when $a=0$, either
$$x=0$$ or
$$frac1x=t+C.$$
answered Nov 13 at 14:16
Yves Daoust
121k668216
I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right?
– kleinmeinpouts
Nov 13 at 14:19
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint:
By integration, when $a=0$, either
$$x=0$$ or
$$frac1x=t+C.$$
answered Nov 13 at 14:16
Yves Daoust
121k668216
Hint:
By integration, when $a=0$, either
$$x=0$$ or
$$frac1x=t+C.$$
answered Nov 13 at 14:16
Yves Daoust
121k668216
answered Nov 13 at 14:16
Yves Daoust
121k668216
answered Nov 13 at 14:16
Yves Daoust
121k668216
answered Nov 13 at 14:16
Yves Daoust
121k668216
121k668216
I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right?
– kleinmeinpouts
Nov 13 at 14:19
add a comment |
I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right?
– kleinmeinpouts
Nov 13 at 14:19
I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right?
– kleinmeinpouts
Nov 13 at 14:19
I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right?
– kleinmeinpouts
Nov 13 at 14:19
add a comment |
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