Mixing
Solution trajectories of a plane autonomous system
$begingroup$
I have the plane autonomous system
$dfrac{dx}{dt}=x(1-2x-y)$
$dfrac{dy}{dt}=y(1-x-2y)$
I need to show that the axes of the phase plane and the line $x=y$ are solution trajectories, but I don't know how to do this.
The other part of the question is as follows.
Use the Bendixson-Dulac theorem with $phi=dfrac{1}{xy}$ to show that there are no closed trajectories in $R={(x,y):x>0, y>0}$ . (I've done this bit) Comment on whether you can prove that there are no periodic functions in the entire phase plane including the origin.
For the last part, I have plotted the phase plane and there are no closed trajectories but how can i prove the last bit properly?
Thanks for any help.
ordinary-differential-equations systems-of-equations
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
asked Feb 6 '14 at 14:28
TomTom
282
$endgroup$
add a comment |
$begingroup$
I have the plane autonomous system
$dfrac{dx}{dt}=x(1-2x-y)$
$dfrac{dy}{dt}=y(1-x-2y)$
I need to show that the axes of the phase plane and the line $x=y$ are solution trajectories, but I don't know how to do this.
The other part of the question is as follows.
Use the Bendixson-Dulac theorem with $phi=dfrac{1}{xy}$ to show that there are no closed trajectories in $R={(x,y):x>0, y>0}$ . (I've done this bit) Comment on whether you can prove that there are no periodic functions in the entire phase plane including the origin.
For the last part, I have plotted the phase plane and there are no closed trajectories but how can i prove the last bit properly?
Thanks for any help.
ordinary-differential-equations systems-of-equations
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
asked Feb 6 '14 at 14:28
TomTom
282
$endgroup$
$begingroup$
Still interested in an answer?
$endgroup$
– Did
Dec 26 '14 at 10:17
add a comment |
$begingroup$
I have the plane autonomous system
$dfrac{dx}{dt}=x(1-2x-y)$
$dfrac{dy}{dt}=y(1-x-2y)$
I need to show that the axes of the phase plane and the line $x=y$ are solution trajectories, but I don't know how to do this.
The other part of the question is as follows.
Use the Bendixson-Dulac theorem with $phi=dfrac{1}{xy}$ to show that there are no closed trajectories in $R={(x,y):x>0, y>0}$ . (I've done this bit) Comment on whether you can prove that there are no periodic functions in the entire phase plane including the origin.
For the last part, I have plotted the phase plane and there are no closed trajectories but how can i prove the last bit properly?
Thanks for any help.
ordinary-differential-equations systems-of-equations
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
asked Feb 6 '14 at 14:28
TomTom
282
$endgroup$
I have the plane autonomous system
$dfrac{dx}{dt}=x(1-2x-y)$
$dfrac{dy}{dt}=y(1-x-2y)$
I need to show that the axes of the phase plane and the line $x=y$ are solution trajectories, but I don't know how to do this.
The other part of the question is as follows.
Use the Bendixson-Dulac theorem with $phi=dfrac{1}{xy}$ to show that there are no closed trajectories in $R={(x,y):x>0, y>0}$ . (I've done this bit) Comment on whether you can prove that there are no periodic functions in the entire phase plane including the origin.
For the last part, I have plotted the phase plane and there are no closed trajectories but how can i prove the last bit properly?
Thanks for any help.
ordinary-differential-equations systems-of-equations
ordinary-differential-equations systems-of-equations
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
asked Feb 6 '14 at 14:28
TomTom
282
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
asked Feb 6 '14 at 14:28
TomTom
282
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
edited Feb 6 '14 at 19:44
Harry Peter
5,48911439
5,48911439
asked Feb 6 '14 at 14:28
TomTom
282
asked Feb 6 '14 at 14:28
TomTom
282
asked Feb 6 '14 at 14:28
TomTom
282
282
$begingroup$
Still interested in an answer?
$endgroup$
– Did
Dec 26 '14 at 10:17
add a comment |
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Still interested in an answer?
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– Did
Dec 26 '14 at 10:17
$begingroup$
Still interested in an answer?
$endgroup$
– Did
Dec 26 '14 at 10:17
$begingroup$
Still interested in an answer?
$endgroup$
– Did
Dec 26 '14 at 10:17
add a comment |
1 Answer
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$begingroup$
The first part of this question is trivial. The axes of the phase plane correspond to $x=0$ and $y=0$. On substituting $x=0$ in the first equation, we get $frac{dx}{dt} = 0$. i.e the particle stays on the $y$-axis. Similarly for the $x$-axis.
For $x=y$ line, we find that $frac{dx}{dt} - frac{dy}{dt} = 0$. i.e. $v_x = v_y$ and hence, the particle continues on the line.
I am trying the last part of your problem.
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The first part of this question is trivial. The axes of the phase plane correspond to $x=0$ and $y=0$. On substituting $x=0$ in the first equation, we get $frac{dx}{dt} = 0$. i.e the particle stays on the $y$-axis. Similarly for the $x$-axis.
For $x=y$ line, we find that $frac{dx}{dt} - frac{dy}{dt} = 0$. i.e. $v_x = v_y$ and hence, the particle continues on the line.
I am trying the last part of your problem.
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
$endgroup$
add a comment |
$begingroup$
The first part of this question is trivial. The axes of the phase plane correspond to $x=0$ and $y=0$. On substituting $x=0$ in the first equation, we get $frac{dx}{dt} = 0$. i.e the particle stays on the $y$-axis. Similarly for the $x$-axis.
For $x=y$ line, we find that $frac{dx}{dt} - frac{dy}{dt} = 0$. i.e. $v_x = v_y$ and hence, the particle continues on the line.
I am trying the last part of your problem.
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
$endgroup$
add a comment |
$begingroup$
The first part of this question is trivial. The axes of the phase plane correspond to $x=0$ and $y=0$. On substituting $x=0$ in the first equation, we get $frac{dx}{dt} = 0$. i.e the particle stays on the $y$-axis. Similarly for the $x$-axis.
For $x=y$ line, we find that $frac{dx}{dt} - frac{dy}{dt} = 0$. i.e. $v_x = v_y$ and hence, the particle continues on the line.
I am trying the last part of your problem.
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
$endgroup$
The first part of this question is trivial. The axes of the phase plane correspond to $x=0$ and $y=0$. On substituting $x=0$ in the first equation, we get $frac{dx}{dt} = 0$. i.e the particle stays on the $y$-axis. Similarly for the $x$-axis.
For $x=y$ line, we find that $frac{dx}{dt} - frac{dy}{dt} = 0$. i.e. $v_x = v_y$ and hence, the particle continues on the line.
I am trying the last part of your problem.
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
answered Feb 7 '14 at 8:45
Sandeep ThilakanSandeep Thilakan
1,721715
1,721715
add a comment |
add a comment |
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– Did
Dec 26 '14 at 10:17