Fixed points of a system of differential equations












1














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 $y=0$)










share|cite|improve this question






















  • 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


















1














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 $y=0$)










share|cite|improve this question






















  • 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
















1












1








1







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 $y=0$)










share|cite|improve this question













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






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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




















  • 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












1 Answer
1






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oldest

votes


















2














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).






share|cite|improve this answer























  • 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











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1 Answer
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1 Answer
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2














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).






share|cite|improve this answer























  • 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
















2














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).






share|cite|improve this answer























  • 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














2












2








2






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).






share|cite|improve this answer














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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 16:35

























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


















  • 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


















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