Fixed points of a system of differential equations
up vote
1
down vote
favorite
I'm wondering about how to find the fixed points for the following system:
$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$
$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$
I think the approach would be;
For $dot{x}$ I can state that either $x=0$ or the term in the parenthesis is
zero. For the term the parenthesis, consider $x=0$ and $y=0$ separately. This
gives the points $(0, k_1/i_1)$ when $x=0$ and $(k_1/c_1 , 0)$ when $y=0$.
The same approach is taken for $dot{y}$ which gives $(0, k_2/c_2)$ when $x=0$
and $(k_2/i_2, 0)$ when $y=0$.
This gives the fixed points
- $(0 , 0) $
$(0 , frac{k_1}{i_1}) $, (from $dot{x}$, where $x=0$)
$(frac{k_1}{c_1} , 0) $, (from $dot{x}$, where $y=0$)
$(0 , frac{k_2}{c_2}) $, (from $dot{y}$, where $x=0$)
$(frac{k_2}{c_2} , 0) $, (from $dot{y}$, where $y=0$)
differential-equations systems-of-equations
asked Nov 24 at 15:54
baxx
401310
add a comment |
up vote
1
down vote
favorite
I'm wondering about how to find the fixed points for the following system:
$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$
$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$
I think the approach would be;
For $dot{x}$ I can state that either $x=0$ or the term in the parenthesis is
zero. For the term the parenthesis, consider $x=0$ and $y=0$ separately. This
gives the points $(0, k_1/i_1)$ when $x=0$ and $(k_1/c_1 , 0)$ when $y=0$.
The same approach is taken for $dot{y}$ which gives $(0, k_2/c_2)$ when $x=0$
and $(k_2/i_2, 0)$ when $y=0$.
This gives the fixed points
- $(0 , 0) $
$(0 , frac{k_1}{i_1}) $, (from $dot{x}$, where $x=0$)
$(frac{k_1}{c_1} , 0) $, (from $dot{x}$, where $y=0$)
$(0 , frac{k_2}{c_2}) $, (from $dot{y}$, where $x=0$)
$(frac{k_2}{c_2} , 0) $, (from $dot{y}$, where $y=0$)
differential-equations systems-of-equations
asked Nov 24 at 15:54
baxx
401310
Hint: a fixed point is such that $dot x=dot y=0$ and this leaves a system of two equations in two unknowns.
– Yves Daoust
Nov 24 at 16:53
@YvesDaoust thanks - I'm a bit confused about the case where we have $x$ as part of the fixed point, rather than it being purely coefficients. Here : $$ y = frac{(k_1 - k_2) + x(i_2 - c_1) }{c_2 - i_1} $$ or $$ x = frac{(k_1 - k_2) + y(c_2 - i_1) }{i_2 - c_1} $$
– baxx
Nov 24 at 17:15
I mean, if I have a fixed point of the form $$ left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right) $$ Then when I evaulate it with the Jacobian it's not the same, usually the values are all constants that can be plugged in.
– baxx
Nov 24 at 17:20
No, write down the system of equations and solve it. If I am right, there are five distinct solutions.
– Yves Daoust
Nov 24 at 17:22
Er, I mean four solutions. You may not consider one equation at a time.
– Yves Daoust
Nov 25 at 17:50
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm wondering about how to find the fixed points for the following system:
$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$
$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$
I think the approach would be;
For $dot{x}$ I can state that either $x=0$ or the term in the parenthesis is
zero. For the term the parenthesis, consider $x=0$ and $y=0$ separately. This
gives the points $(0, k_1/i_1)$ when $x=0$ and $(k_1/c_1 , 0)$ when $y=0$.
The same approach is taken for $dot{y}$ which gives $(0, k_2/c_2)$ when $x=0$
and $(k_2/i_2, 0)$ when $y=0$.
This gives the fixed points
- $(0 , 0) $
$(0 , frac{k_1}{i_1}) $, (from $dot{x}$, where $x=0$)
$(frac{k_1}{c_1} , 0) $, (from $dot{x}$, where $y=0$)
$(0 , frac{k_2}{c_2}) $, (from $dot{y}$, where $x=0$)
$(frac{k_2}{c_2} , 0) $, (from $dot{y}$, where $y=0$)
differential-equations systems-of-equations
asked Nov 24 at 15:54
baxx
401310
I'm wondering about how to find the fixed points for the following system:
$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$
$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$
I think the approach would be;
For $dot{x}$ I can state that either $x=0$ or the term in the parenthesis is
zero. For the term the parenthesis, consider $x=0$ and $y=0$ separately. This
gives the points $(0, k_1/i_1)$ when $x=0$ and $(k_1/c_1 , 0)$ when $y=0$.
The same approach is taken for $dot{y}$ which gives $(0, k_2/c_2)$ when $x=0$
and $(k_2/i_2, 0)$ when $y=0$.
This gives the fixed points
- $(0 , 0) $
$(0 , frac{k_1}{i_1}) $, (from $dot{x}$, where $x=0$)
$(frac{k_1}{c_1} , 0) $, (from $dot{x}$, where $y=0$)
$(0 , frac{k_2}{c_2}) $, (from $dot{y}$, where $x=0$)
$(frac{k_2}{c_2} , 0) $, (from $dot{y}$, where $y=0$)
differential-equations systems-of-equations
differential-equations systems-of-equations
asked Nov 24 at 15:54
baxx
401310
asked Nov 24 at 15:54
baxx
401310
asked Nov 24 at 15:54
baxx
401310
asked Nov 24 at 15:54
baxx
401310
asked Nov 24 at 15:54
baxx
401310
401310
Hint: a fixed point is such that $dot x=dot y=0$ and this leaves a system of two equations in two unknowns.
– Yves Daoust
Nov 24 at 16:53
@YvesDaoust thanks - I'm a bit confused about the case where we have $x$ as part of the fixed point, rather than it being purely coefficients. Here : $$ y = frac{(k_1 - k_2) + x(i_2 - c_1) }{c_2 - i_1} $$ or $$ x = frac{(k_1 - k_2) + y(c_2 - i_1) }{i_2 - c_1} $$
– baxx
Nov 24 at 17:15
I mean, if I have a fixed point of the form $$ left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right) $$ Then when I evaulate it with the Jacobian it's not the same, usually the values are all constants that can be plugged in.
– baxx
Nov 24 at 17:20
No, write down the system of equations and solve it. If I am right, there are five distinct solutions.
– Yves Daoust
Nov 24 at 17:22
Er, I mean four solutions. You may not consider one equation at a time.
– Yves Daoust
Nov 25 at 17:50
add a comment |
Hint: a fixed point is such that $dot x=dot y=0$ and this leaves a system of two equations in two unknowns.
– Yves Daoust
Nov 24 at 16:53
@YvesDaoust thanks - I'm a bit confused about the case where we have $x$ as part of the fixed point, rather than it being purely coefficients. Here : $$ y = frac{(k_1 - k_2) + x(i_2 - c_1) }{c_2 - i_1} $$ or $$ x = frac{(k_1 - k_2) + y(c_2 - i_1) }{i_2 - c_1} $$
– baxx
Nov 24 at 17:15
I mean, if I have a fixed point of the form $$ left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right) $$ Then when I evaulate it with the Jacobian it's not the same, usually the values are all constants that can be plugged in.
– baxx
Nov 24 at 17:20
No, write down the system of equations and solve it. If I am right, there are five distinct solutions.
– Yves Daoust
Nov 24 at 17:22
Er, I mean four solutions. You may not consider one equation at a time.
– Yves Daoust
Nov 25 at 17:50
Hint: a fixed point is such that $dot x=dot y=0$ and this leaves a system of two equations in two unknowns.
– Yves Daoust
Nov 24 at 16:53
Hint: a fixed point is such that $dot x=dot y=0$ and this leaves a system of two equations in two unknowns.
– Yves Daoust
Nov 24 at 16:53
@YvesDaoust thanks - I'm a bit confused about the case where we have $x$ as part of the fixed point, rather than it being purely coefficients. Here : $$ y = frac{(k_1 - k_2) + x(i_2 - c_1) }{c_2 - i_1} $$ or $$ x = frac{(k_1 - k_2) + y(c_2 - i_1) }{i_2 - c_1} $$
– baxx
Nov 24 at 17:15
@YvesDaoust thanks - I'm a bit confused about the case where we have $x$ as part of the fixed point, rather than it being purely coefficients. Here : $$ y = frac{(k_1 - k_2) + x(i_2 - c_1) }{c_2 - i_1} $$ or $$ x = frac{(k_1 - k_2) + y(c_2 - i_1) }{i_2 - c_1} $$
– baxx
Nov 24 at 17:15
I mean, if I have a fixed point of the form $$ left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right) $$ Then when I evaulate it with the Jacobian it's not the same, usually the values are all constants that can be plugged in.
– baxx
Nov 24 at 17:20
I mean, if I have a fixed point of the form $$ left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right) $$ Then when I evaulate it with the Jacobian it's not the same, usually the values are all constants that can be plugged in.
– baxx
Nov 24 at 17:20
No, write down the system of equations and solve it. If I am right, there are five distinct solutions.
– Yves Daoust
Nov 24 at 17:22
No, write down the system of equations and solve it. If I am right, there are five distinct solutions.
– Yves Daoust
Nov 24 at 17:22
Er, I mean four solutions. You may not consider one equation at a time.
– Yves Daoust
Nov 25 at 17:50
Er, I mean four solutions. You may not consider one equation at a time.
– Yves Daoust
Nov 25 at 17:50
add a comment |
1 Answer
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up vote
2
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No, that's not correct.
For example, if $x=0$ then $dot x=0$ already, so you shouldn't look at the other factor in $dot x$ in that case.
The correct cases are:
$x = 0$ and $y = 0$
or
$x = 0$ and $k_2 - c_2 y - i_2 x = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $y = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $k_2 - c_2 y - i_2 x = 0$
Each of these cases gives you one fixed point $(x,y)$, so there are four fixed points in total (the origin, one on each axis, plus a nontrivial one).
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
Thanks, are the indices correct in what you've written?
– baxx
Nov 24 at 16:34
Oops, sorry, forgot to change $i_1$ to $i_2$ after copying. Now it should be OK.
– Hans Lundmark
Nov 24 at 16:36
1
It's a $2 times 2$ linear system (two equations, two unknowns). Just solve it as you would in linear algebra.
– Hans Lundmark
Nov 24 at 16:44
1
Surely linear algebra is a prerequisite for an ODE course where a problem like this would appear?!? Anyway, this is extremely basic stuff, just google “Gaussian elimination” and you will find a million explanations.
– Hans Lundmark
Nov 24 at 16:47
1
Just solve $$pmatrix{c_1&i_1\i_2&c_2}pmatrix{x\y}=pmatrix{k_1\k_2},$$ for instance using Cramer's rule or by multiplying with the adjoint matrix $$pmatrix{c_2&-i_1\-i_2&c_1}$$ from the left.
– LutzL
Nov 24 at 17:27
|
show 3 more comments
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1 Answer
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No, that's not correct.
For example, if $x=0$ then $dot x=0$ already, so you shouldn't look at the other factor in $dot x$ in that case.
The correct cases are:
$x = 0$ and $y = 0$
or
$x = 0$ and $k_2 - c_2 y - i_2 x = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $y = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $k_2 - c_2 y - i_2 x = 0$
Each of these cases gives you one fixed point $(x,y)$, so there are four fixed points in total (the origin, one on each axis, plus a nontrivial one).
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
Thanks, are the indices correct in what you've written?
– baxx
Nov 24 at 16:34
Oops, sorry, forgot to change $i_1$ to $i_2$ after copying. Now it should be OK.
– Hans Lundmark
Nov 24 at 16:36
1
It's a $2 times 2$ linear system (two equations, two unknowns). Just solve it as you would in linear algebra.
– Hans Lundmark
Nov 24 at 16:44
1
Surely linear algebra is a prerequisite for an ODE course where a problem like this would appear?!? Anyway, this is extremely basic stuff, just google “Gaussian elimination” and you will find a million explanations.
– Hans Lundmark
Nov 24 at 16:47
1
Just solve $$pmatrix{c_1&i_1\i_2&c_2}pmatrix{x\y}=pmatrix{k_1\k_2},$$ for instance using Cramer's rule or by multiplying with the adjoint matrix $$pmatrix{c_2&-i_1\-i_2&c_1}$$ from the left.
– LutzL
Nov 24 at 17:27
|
show 3 more comments
up vote
2
down vote
No, that's not correct.
For example, if $x=0$ then $dot x=0$ already, so you shouldn't look at the other factor in $dot x$ in that case.
The correct cases are:
$x = 0$ and $y = 0$
or
$x = 0$ and $k_2 - c_2 y - i_2 x = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $y = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $k_2 - c_2 y - i_2 x = 0$
Each of these cases gives you one fixed point $(x,y)$, so there are four fixed points in total (the origin, one on each axis, plus a nontrivial one).
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
Thanks, are the indices correct in what you've written?
– baxx
Nov 24 at 16:34
Oops, sorry, forgot to change $i_1$ to $i_2$ after copying. Now it should be OK.
– Hans Lundmark
Nov 24 at 16:36
1
It's a $2 times 2$ linear system (two equations, two unknowns). Just solve it as you would in linear algebra.
– Hans Lundmark
Nov 24 at 16:44
1
Surely linear algebra is a prerequisite for an ODE course where a problem like this would appear?!? Anyway, this is extremely basic stuff, just google “Gaussian elimination” and you will find a million explanations.
– Hans Lundmark
Nov 24 at 16:47
1
Just solve $$pmatrix{c_1&i_1\i_2&c_2}pmatrix{x\y}=pmatrix{k_1\k_2},$$ for instance using Cramer's rule or by multiplying with the adjoint matrix $$pmatrix{c_2&-i_1\-i_2&c_1}$$ from the left.
– LutzL
Nov 24 at 17:27
|
show 3 more comments
up vote
2
down vote
up vote
2
down vote
No, that's not correct.
For example, if $x=0$ then $dot x=0$ already, so you shouldn't look at the other factor in $dot x$ in that case.
The correct cases are:
$x = 0$ and $y = 0$
or
$x = 0$ and $k_2 - c_2 y - i_2 x = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $y = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $k_2 - c_2 y - i_2 x = 0$
Each of these cases gives you one fixed point $(x,y)$, so there are four fixed points in total (the origin, one on each axis, plus a nontrivial one).
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
No, that's not correct.
For example, if $x=0$ then $dot x=0$ already, so you shouldn't look at the other factor in $dot x$ in that case.
The correct cases are:
$x = 0$ and $y = 0$
or
$x = 0$ and $k_2 - c_2 y - i_2 x = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $y = 0$
or
$k_1 - c_1 x - i_1 y = 0$ and $k_2 - c_2 y - i_2 x = 0$
Each of these cases gives you one fixed point $(x,y)$, so there are four fixed points in total (the origin, one on each axis, plus a nontrivial one).
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
edited Nov 24 at 16:35
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
answered Nov 24 at 16:31
Hans Lundmark
34.9k564112
34.9k564112
Thanks, are the indices correct in what you've written?
– baxx
Nov 24 at 16:34
Oops, sorry, forgot to change $i_1$ to $i_2$ after copying. Now it should be OK.
– Hans Lundmark
Nov 24 at 16:36
1
It's a $2 times 2$ linear system (two equations, two unknowns). Just solve it as you would in linear algebra.
– Hans Lundmark
Nov 24 at 16:44
1
Surely linear algebra is a prerequisite for an ODE course where a problem like this would appear?!? Anyway, this is extremely basic stuff, just google “Gaussian elimination” and you will find a million explanations.
– Hans Lundmark
Nov 24 at 16:47
1
Just solve $$pmatrix{c_1&i_1\i_2&c_2}pmatrix{x\y}=pmatrix{k_1\k_2},$$ for instance using Cramer's rule or by multiplying with the adjoint matrix $$pmatrix{c_2&-i_1\-i_2&c_1}$$ from the left.
– LutzL
Nov 24 at 17:27
|
show 3 more comments
Thanks, are the indices correct in what you've written?
– baxx
Nov 24 at 16:34
Oops, sorry, forgot to change $i_1$ to $i_2$ after copying. Now it should be OK.
– Hans Lundmark
Nov 24 at 16:36
1
It's a $2 times 2$ linear system (two equations, two unknowns). Just solve it as you would in linear algebra.
– Hans Lundmark
Nov 24 at 16:44
1
Surely linear algebra is a prerequisite for an ODE course where a problem like this would appear?!? Anyway, this is extremely basic stuff, just google “Gaussian elimination” and you will find a million explanations.
– Hans Lundmark
Nov 24 at 16:47
1
Just solve $$pmatrix{c_1&i_1\i_2&c_2}pmatrix{x\y}=pmatrix{k_1\k_2},$$ for instance using Cramer's rule or by multiplying with the adjoint matrix $$pmatrix{c_2&-i_1\-i_2&c_1}$$ from the left.
– LutzL
Nov 24 at 17:27
Thanks, are the indices correct in what you've written?
– baxx
Nov 24 at 16:34
Thanks, are the indices correct in what you've written?
– baxx
Nov 24 at 16:34
Oops, sorry, forgot to change $i_1$ to $i_2$ after copying. Now it should be OK.
– Hans Lundmark
Nov 24 at 16:36
Oops, sorry, forgot to change $i_1$ to $i_2$ after copying. Now it should be OK.
– Hans Lundmark
Nov 24 at 16:36
1
1
It's a $2 times 2$ linear system (two equations, two unknowns). Just solve it as you would in linear algebra.
– Hans Lundmark
Nov 24 at 16:44
It's a $2 times 2$ linear system (two equations, two unknowns). Just solve it as you would in linear algebra.
– Hans Lundmark
Nov 24 at 16:44
1
1
Surely linear algebra is a prerequisite for an ODE course where a problem like this would appear?!? Anyway, this is extremely basic stuff, just google “Gaussian elimination” and you will find a million explanations.
– Hans Lundmark
Nov 24 at 16:47
Surely linear algebra is a prerequisite for an ODE course where a problem like this would appear?!? Anyway, this is extremely basic stuff, just google “Gaussian elimination” and you will find a million explanations.
– Hans Lundmark
Nov 24 at 16:47
1
1
Just solve $$pmatrix{c_1&i_1\i_2&c_2}pmatrix{x\y}=pmatrix{k_1\k_2},$$ for instance using Cramer's rule or by multiplying with the adjoint matrix $$pmatrix{c_2&-i_1\-i_2&c_1}$$ from the left.
– LutzL
Nov 24 at 17:27
Just solve $$pmatrix{c_1&i_1\i_2&c_2}pmatrix{x\y}=pmatrix{k_1\k_2},$$ for instance using Cramer's rule or by multiplying with the adjoint matrix $$pmatrix{c_2&-i_1\-i_2&c_1}$$ from the left.
– LutzL
Nov 24 at 17:27
|
show 3 more comments
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Hint: a fixed point is such that $dot x=dot y=0$ and this leaves a system of two equations in two unknowns.
– Yves Daoust
Nov 24 at 16:53
@YvesDaoust thanks - I'm a bit confused about the case where we have $x$ as part of the fixed point, rather than it being purely coefficients. Here : $$ y = frac{(k_1 - k_2) + x(i_2 - c_1) }{c_2 - i_1} $$ or $$ x = frac{(k_1 - k_2) + y(c_2 - i_1) }{i_2 - c_1} $$
– baxx
Nov 24 at 17:15
I mean, if I have a fixed point of the form $$ left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right) $$ Then when I evaulate it with the Jacobian it's not the same, usually the values are all constants that can be plugged in.
– baxx
Nov 24 at 17:20
No, write down the system of equations and solve it. If I am right, there are five distinct solutions.
– Yves Daoust
Nov 24 at 17:22
Er, I mean four solutions. You may not consider one equation at a time.
– Yves Daoust
Nov 25 at 17:50