How to evaluate the Jacobian for a system of differential equations when the terms aren't constants












1












$begingroup$


For this system :



$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$



$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$



One of the fixed points is ( from $dot{x} = dot{y}$)



$$
left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right)
$$



Given that the Jacobian for this system is



$$
J =
begin{bmatrix}
r_1 - left(frac{2r_1c_1x}{k_1}right) - left( frac{r_1i_1 y}{k_1}right) &
- left(frac{r_1 i_1 x}{k_1}right) \
-left(frac{r_2 i_2 y}{k_2}right) &
r_2 - left( frac{2 r_2 c_2 y}{k_2} right) - left( frac{r_2 i_2 x}{k_2} right) \
end{bmatrix}
$$



I'm not sure how I should go about evaluating this point.



edit - evaulation



# these should both be zero for the fixed points

def f1(x,y,k1,c1,i1):

    return(k1 - c1*x - i1*y)

def f2(x,y,k2,c2,i2):

    return(k2 - c2*x - i2*y)



# just to assign values of x,y more easily

def xx(k1,k2,c1,c2,i1,i2):

    return((i2*k2 - c2*k1)/(i1*i2 - c1*c2))

def yy(k1,k2,c1,c2,i1,i2):

    return(( i2*k1 - c1*k2 )/(i1*i2 - c1*c2))



# some constants to test

k1 = 1

c1 = 2

i1 = 3



k2 = 3

c2 = 5

i2 = 7





x = xx(k1, k2, c1, c2, i1, i2)

y = yy(k1, k2, c1, c2, i1, i2)



# these should then be zero

print(f1(x, y, k1, c1, i1))

print(f2(y, y, k2, c2, i2))



# Output :

# -2.1818181818181817

# 1.9090909090909092





ordinary-differential-equations systems-of-equations jacobian







share|cite|improve this question











$endgroup$












  • $begingroup$
    You could use the complete and correct solution of the linear system for the fixed point, $$ (x_4,y_4)=frac{(k_1c_2-i_1k_2,,c_1k_2-i_2k_1)}{c_1c_2-i_1i_2} $$
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:34












  • $begingroup$
    If you pass the constants as parameters, you only need one f function. -- You have an index error in the xx computation, it should be (i1*k2 - c2*k1)/(i1*i2 - c1*c2).
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:43










  • $begingroup$
    Call f2 also with x,y. Change f2 to use c2 as coefficient of the main variable y, and i2 as coefficient of x. Then the result should be 0 in both cases.
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:54
















1












$begingroup$


For this system :



$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$



$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$



One of the fixed points is ( from $dot{x} = dot{y}$)



$$
left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right)
$$



Given that the Jacobian for this system is



$$
J =
begin{bmatrix}
r_1 - left(frac{2r_1c_1x}{k_1}right) - left( frac{r_1i_1 y}{k_1}right) &
- left(frac{r_1 i_1 x}{k_1}right) \
-left(frac{r_2 i_2 y}{k_2}right) &
r_2 - left( frac{2 r_2 c_2 y}{k_2} right) - left( frac{r_2 i_2 x}{k_2} right) \
end{bmatrix}
$$



I'm not sure how I should go about evaluating this point.



edit - evaulation



# these should both be zero for the fixed points

def f1(x,y,k1,c1,i1):

    return(k1 - c1*x - i1*y)

def f2(x,y,k2,c2,i2):

    return(k2 - c2*x - i2*y)



# just to assign values of x,y more easily

def xx(k1,k2,c1,c2,i1,i2):

    return((i2*k2 - c2*k1)/(i1*i2 - c1*c2))

def yy(k1,k2,c1,c2,i1,i2):

    return(( i2*k1 - c1*k2 )/(i1*i2 - c1*c2))



# some constants to test

k1 = 1

c1 = 2

i1 = 3



k2 = 3

c2 = 5

i2 = 7





x = xx(k1, k2, c1, c2, i1, i2)

y = yy(k1, k2, c1, c2, i1, i2)



# these should then be zero

print(f1(x, y, k1, c1, i1))

print(f2(y, y, k2, c2, i2))



# Output :

# -2.1818181818181817

# 1.9090909090909092





ordinary-differential-equations systems-of-equations jacobian







share|cite|improve this question











$endgroup$












  • $begingroup$
    You could use the complete and correct solution of the linear system for the fixed point, $$ (x_4,y_4)=frac{(k_1c_2-i_1k_2,,c_1k_2-i_2k_1)}{c_1c_2-i_1i_2} $$
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:34












  • $begingroup$
    If you pass the constants as parameters, you only need one f function. -- You have an index error in the xx computation, it should be (i1*k2 - c2*k1)/(i1*i2 - c1*c2).
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:43










  • $begingroup$
    Call f2 also with x,y. Change f2 to use c2 as coefficient of the main variable y, and i2 as coefficient of x. Then the result should be 0 in both cases.
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:54














1












1








1





$begingroup$


For this system :



$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$



$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$



One of the fixed points is ( from $dot{x} = dot{y}$)



$$
left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right)
$$



Given that the Jacobian for this system is



$$
J =
begin{bmatrix}
r_1 - left(frac{2r_1c_1x}{k_1}right) - left( frac{r_1i_1 y}{k_1}right) &
- left(frac{r_1 i_1 x}{k_1}right) \
-left(frac{r_2 i_2 y}{k_2}right) &
r_2 - left( frac{2 r_2 c_2 y}{k_2} right) - left( frac{r_2 i_2 x}{k_2} right) \
end{bmatrix}
$$



I'm not sure how I should go about evaluating this point.



edit - evaulation



# these should both be zero for the fixed points

def f1(x,y,k1,c1,i1):

    return(k1 - c1*x - i1*y)

def f2(x,y,k2,c2,i2):

    return(k2 - c2*x - i2*y)



# just to assign values of x,y more easily

def xx(k1,k2,c1,c2,i1,i2):

    return((i2*k2 - c2*k1)/(i1*i2 - c1*c2))

def yy(k1,k2,c1,c2,i1,i2):

    return(( i2*k1 - c1*k2 )/(i1*i2 - c1*c2))



# some constants to test

k1 = 1

c1 = 2

i1 = 3



k2 = 3

c2 = 5

i2 = 7





x = xx(k1, k2, c1, c2, i1, i2)

y = yy(k1, k2, c1, c2, i1, i2)



# these should then be zero

print(f1(x, y, k1, c1, i1))

print(f2(y, y, k2, c2, i2))



# Output :

# -2.1818181818181817

# 1.9090909090909092





ordinary-differential-equations systems-of-equations jacobian







share|cite|improve this question











$endgroup$




For this system :



$$
dot{x} = frac{xr_1}{k_1}left(k_1 - c_1 x - i_1 y right)
$$



$$
dot{y} = frac{y r_2}{k_2}left(k_2 - c_2 y - i_2 x right)
$$



One of the fixed points is ( from $dot{x} = dot{y}$)



$$
left( frac{k_1 - k_2}{c_1 - i_2} , frac{x(c_1 - i_2) - k_1 + k_2}{c_2 - i_1} right)
$$



Given that the Jacobian for this system is



$$
J =
begin{bmatrix}
r_1 - left(frac{2r_1c_1x}{k_1}right) - left( frac{r_1i_1 y}{k_1}right) &
- left(frac{r_1 i_1 x}{k_1}right) \
-left(frac{r_2 i_2 y}{k_2}right) &
r_2 - left( frac{2 r_2 c_2 y}{k_2} right) - left( frac{r_2 i_2 x}{k_2} right) \
end{bmatrix}
$$



I'm not sure how I should go about evaluating this point.



edit - evaulation



# these should both be zero for the fixed points

def f1(x,y,k1,c1,i1):

    return(k1 - c1*x - i1*y)

def f2(x,y,k2,c2,i2):

    return(k2 - c2*x - i2*y)



# just to assign values of x,y more easily

def xx(k1,k2,c1,c2,i1,i2):

    return((i2*k2 - c2*k1)/(i1*i2 - c1*c2))

def yy(k1,k2,c1,c2,i1,i2):

    return(( i2*k1 - c1*k2 )/(i1*i2 - c1*c2))



# some constants to test

k1 = 1

c1 = 2

i1 = 3



k2 = 3

c2 = 5

i2 = 7





x = xx(k1, k2, c1, c2, i1, i2)

y = yy(k1, k2, c1, c2, i1, i2)



# these should then be zero

print(f1(x, y, k1, c1, i1))

print(f2(y, y, k2, c2, i2))



# Output :

# -2.1818181818181817

# 1.9090909090909092





ordinary-differential-equations systems-of-equations jacobian




ordinary-differential-equations systems-of-equations jacobian






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 17:51







baxx

















asked Nov 24 '18 at 17:23









baxxbaxx

401310




401310












  • $begingroup$
    You could use the complete and correct solution of the linear system for the fixed point, $$ (x_4,y_4)=frac{(k_1c_2-i_1k_2,,c_1k_2-i_2k_1)}{c_1c_2-i_1i_2} $$
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:34












  • $begingroup$
    If you pass the constants as parameters, you only need one f function. -- You have an index error in the xx computation, it should be (i1*k2 - c2*k1)/(i1*i2 - c1*c2).
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:43










  • $begingroup$
    Call f2 also with x,y. Change f2 to use c2 as coefficient of the main variable y, and i2 as coefficient of x. Then the result should be 0 in both cases.
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:54


















  • $begingroup$
    You could use the complete and correct solution of the linear system for the fixed point, $$ (x_4,y_4)=frac{(k_1c_2-i_1k_2,,c_1k_2-i_2k_1)}{c_1c_2-i_1i_2} $$
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:34












  • $begingroup$
    If you pass the constants as parameters, you only need one f function. -- You have an index error in the xx computation, it should be (i1*k2 - c2*k1)/(i1*i2 - c1*c2).
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:43










  • $begingroup$
    Call f2 also with x,y. Change f2 to use c2 as coefficient of the main variable y, and i2 as coefficient of x. Then the result should be 0 in both cases.
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:54
















$begingroup$
You could use the complete and correct solution of the linear system for the fixed point, $$ (x_4,y_4)=frac{(k_1c_2-i_1k_2,,c_1k_2-i_2k_1)}{c_1c_2-i_1i_2} $$
$endgroup$
– LutzL
Nov 24 '18 at 17:34






$begingroup$
You could use the complete and correct solution of the linear system for the fixed point, $$ (x_4,y_4)=frac{(k_1c_2-i_1k_2,,c_1k_2-i_2k_1)}{c_1c_2-i_1i_2} $$
$endgroup$
– LutzL
Nov 24 '18 at 17:34














$begingroup$
If you pass the constants as parameters, you only need one f function. -- You have an index error in the xx computation, it should be (i1*k2 - c2*k1)/(i1*i2 - c1*c2).
$endgroup$
– LutzL
Nov 24 '18 at 18:43




$begingroup$
If you pass the constants as parameters, you only need one f function. -- You have an index error in the xx computation, it should be (i1*k2 - c2*k1)/(i1*i2 - c1*c2).
$endgroup$
– LutzL
Nov 24 '18 at 18:43












$begingroup$
Call f2 also with x,y. Change f2 to use c2 as coefficient of the main variable y, and i2 as coefficient of x. Then the result should be 0 in both cases.
$endgroup$
– LutzL
Nov 24 '18 at 18:54




$begingroup$
Call f2 also with x,y. Change f2 to use c2 as coefficient of the main variable y, and i2 as coefficient of x. Then the result should be 0 in both cases.
$endgroup$
– LutzL
Nov 24 '18 at 18:54










1 Answer
1






active

oldest

votes


















0












$begingroup$

The fixed points are the locations for which both $dot{x} = 0$ and $dot{y} = 0$. In your case these are the points



begin{eqnarray}
(x^*, y^*)_1 &=& (0, 0) \
(x^*, y^*)_2 &=& left(0, frac{k_2}{c_2} right)\
(x^*, y^*)_3 &=& left(frac{k_1}{c_1}, 0 right)\
(x^*, y^*)_4 &=& left(frac{i_2 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}, frac{i_1 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}right) \
end{eqnarray}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your second and third point are wrong. $x=0$, $yne 0$ implies $y=frac{k_2}{c_2}$.
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:36










  • $begingroup$
    @LutzL Thanks!!
    $endgroup$
    – caverac
    Nov 24 '18 at 17:44










  • $begingroup$
    @caverac are you sure about this answer? I've added some code to test it , and it doesn't seem to be returning zero
    $endgroup$
    – baxx
    Nov 24 '18 at 17:51










  • $begingroup$
    @baxx In your original post you have $dot{y} = cdots -c_2y cdots$, in your edit you have def f2 ... -c2 * x ...
    $endgroup$
    – caverac
    Nov 24 '18 at 17:55










  • $begingroup$
    There was no need to remove the middle fixed points, just correct them to $(0,k_2/c_2)$ and $(k_1/c_1,0)$. See the previous question math.stackexchange.com/q/3011709/115115
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:44













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The fixed points are the locations for which both $dot{x} = 0$ and $dot{y} = 0$. In your case these are the points



begin{eqnarray}
(x^*, y^*)_1 &=& (0, 0) \
(x^*, y^*)_2 &=& left(0, frac{k_2}{c_2} right)\
(x^*, y^*)_3 &=& left(frac{k_1}{c_1}, 0 right)\
(x^*, y^*)_4 &=& left(frac{i_2 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}, frac{i_1 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}right) \
end{eqnarray}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your second and third point are wrong. $x=0$, $yne 0$ implies $y=frac{k_2}{c_2}$.
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:36










  • $begingroup$
    @LutzL Thanks!!
    $endgroup$
    – caverac
    Nov 24 '18 at 17:44










  • $begingroup$
    @caverac are you sure about this answer? I've added some code to test it , and it doesn't seem to be returning zero
    $endgroup$
    – baxx
    Nov 24 '18 at 17:51










  • $begingroup$
    @baxx In your original post you have $dot{y} = cdots -c_2y cdots$, in your edit you have def f2 ... -c2 * x ...
    $endgroup$
    – caverac
    Nov 24 '18 at 17:55










  • $begingroup$
    There was no need to remove the middle fixed points, just correct them to $(0,k_2/c_2)$ and $(k_1/c_1,0)$. See the previous question math.stackexchange.com/q/3011709/115115
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:44


















0












$begingroup$

The fixed points are the locations for which both $dot{x} = 0$ and $dot{y} = 0$. In your case these are the points



begin{eqnarray}
(x^*, y^*)_1 &=& (0, 0) \
(x^*, y^*)_2 &=& left(0, frac{k_2}{c_2} right)\
(x^*, y^*)_3 &=& left(frac{k_1}{c_1}, 0 right)\
(x^*, y^*)_4 &=& left(frac{i_2 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}, frac{i_1 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}right) \
end{eqnarray}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Your second and third point are wrong. $x=0$, $yne 0$ implies $y=frac{k_2}{c_2}$.
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:36










  • $begingroup$
    @LutzL Thanks!!
    $endgroup$
    – caverac
    Nov 24 '18 at 17:44










  • $begingroup$
    @caverac are you sure about this answer? I've added some code to test it , and it doesn't seem to be returning zero
    $endgroup$
    – baxx
    Nov 24 '18 at 17:51










  • $begingroup$
    @baxx In your original post you have $dot{y} = cdots -c_2y cdots$, in your edit you have def f2 ... -c2 * x ...
    $endgroup$
    – caverac
    Nov 24 '18 at 17:55










  • $begingroup$
    There was no need to remove the middle fixed points, just correct them to $(0,k_2/c_2)$ and $(k_1/c_1,0)$. See the previous question math.stackexchange.com/q/3011709/115115
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:44
















0












0








0





$begingroup$

The fixed points are the locations for which both $dot{x} = 0$ and $dot{y} = 0$. In your case these are the points



begin{eqnarray}
(x^*, y^*)_1 &=& (0, 0) \
(x^*, y^*)_2 &=& left(0, frac{k_2}{c_2} right)\
(x^*, y^*)_3 &=& left(frac{k_1}{c_1}, 0 right)\
(x^*, y^*)_4 &=& left(frac{i_2 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}, frac{i_1 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}right) \
end{eqnarray}






share|cite|improve this answer











$endgroup$



The fixed points are the locations for which both $dot{x} = 0$ and $dot{y} = 0$. In your case these are the points



begin{eqnarray}
(x^*, y^*)_1 &=& (0, 0) \
(x^*, y^*)_2 &=& left(0, frac{k_2}{c_2} right)\
(x^*, y^*)_3 &=& left(frac{k_1}{c_1}, 0 right)\
(x^*, y^*)_4 &=& left(frac{i_2 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}, frac{i_1 k_2 - c_2 k_1}{i_1i_2 - c_1 c_2}right) \
end{eqnarray}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 '18 at 18:51

























answered Nov 24 '18 at 17:34









caveraccaverac

14.2k21130




14.2k21130












  • $begingroup$
    Your second and third point are wrong. $x=0$, $yne 0$ implies $y=frac{k_2}{c_2}$.
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:36










  • $begingroup$
    @LutzL Thanks!!
    $endgroup$
    – caverac
    Nov 24 '18 at 17:44










  • $begingroup$
    @caverac are you sure about this answer? I've added some code to test it , and it doesn't seem to be returning zero
    $endgroup$
    – baxx
    Nov 24 '18 at 17:51










  • $begingroup$
    @baxx In your original post you have $dot{y} = cdots -c_2y cdots$, in your edit you have def f2 ... -c2 * x ...
    $endgroup$
    – caverac
    Nov 24 '18 at 17:55










  • $begingroup$
    There was no need to remove the middle fixed points, just correct them to $(0,k_2/c_2)$ and $(k_1/c_1,0)$. See the previous question math.stackexchange.com/q/3011709/115115
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:44




















  • $begingroup$
    Your second and third point are wrong. $x=0$, $yne 0$ implies $y=frac{k_2}{c_2}$.
    $endgroup$
    – LutzL
    Nov 24 '18 at 17:36










  • $begingroup$
    @LutzL Thanks!!
    $endgroup$
    – caverac
    Nov 24 '18 at 17:44










  • $begingroup$
    @caverac are you sure about this answer? I've added some code to test it , and it doesn't seem to be returning zero
    $endgroup$
    – baxx
    Nov 24 '18 at 17:51










  • $begingroup$
    @baxx In your original post you have $dot{y} = cdots -c_2y cdots$, in your edit you have def f2 ... -c2 * x ...
    $endgroup$
    – caverac
    Nov 24 '18 at 17:55










  • $begingroup$
    There was no need to remove the middle fixed points, just correct them to $(0,k_2/c_2)$ and $(k_1/c_1,0)$. See the previous question math.stackexchange.com/q/3011709/115115
    $endgroup$
    – LutzL
    Nov 24 '18 at 18:44


















$begingroup$
Your second and third point are wrong. $x=0$, $yne 0$ implies $y=frac{k_2}{c_2}$.
$endgroup$
– LutzL
Nov 24 '18 at 17:36




$begingroup$
Your second and third point are wrong. $x=0$, $yne 0$ implies $y=frac{k_2}{c_2}$.
$endgroup$
– LutzL
Nov 24 '18 at 17:36












$begingroup$
@LutzL Thanks!!
$endgroup$
– caverac
Nov 24 '18 at 17:44




$begingroup$
@LutzL Thanks!!
$endgroup$
– caverac
Nov 24 '18 at 17:44












$begingroup$
@caverac are you sure about this answer? I've added some code to test it , and it doesn't seem to be returning zero
$endgroup$
– baxx
Nov 24 '18 at 17:51




$begingroup$
@caverac are you sure about this answer? I've added some code to test it , and it doesn't seem to be returning zero
$endgroup$
– baxx
Nov 24 '18 at 17:51












$begingroup$
@baxx In your original post you have $dot{y} = cdots -c_2y cdots$, in your edit you have def f2 ... -c2 * x ...
$endgroup$
– caverac
Nov 24 '18 at 17:55




$begingroup$
@baxx In your original post you have $dot{y} = cdots -c_2y cdots$, in your edit you have def f2 ... -c2 * x ...
$endgroup$
– caverac
Nov 24 '18 at 17:55












$begingroup$
There was no need to remove the middle fixed points, just correct them to $(0,k_2/c_2)$ and $(k_1/c_1,0)$. See the previous question math.stackexchange.com/q/3011709/115115
$endgroup$
– LutzL
Nov 24 '18 at 18:44






$begingroup$
There was no need to remove the middle fixed points, just correct them to $(0,k_2/c_2)$ and $(k_1/c_1,0)$. See the previous question math.stackexchange.com/q/3011709/115115
$endgroup$
– LutzL
Nov 24 '18 at 18:44




















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