System of linearly coupled ODE with quadratic term
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I am looking forward to solve the system of equations
$$frac{d x_i}{d t}=a_ix_i^2+b_ix_i-sumlimits_{jneq i}^nb_jx_j-d_i,$$
with $x_igeq 0$, $a_i,b_i,c_i,d_i>0$ and $iin[1,n]$.
Without the linear coupling term, the system reduces to a set of independent Riccati equations with constant coefficients that can be solved.
With the coupling, the system looks like a matrix equation but I don't know how to represent the quadratic term in matrix form.
Any kind of help for finding the solution, or link to literature would be very helpful.
Thank you!
ordinary-differential-equations systems-of-equations nonlinear-system
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
$endgroup$
add a comment |
$begingroup$
I am looking forward to solve the system of equations
$$frac{d x_i}{d t}=a_ix_i^2+b_ix_i-sumlimits_{jneq i}^nb_jx_j-d_i,$$
with $x_igeq 0$, $a_i,b_i,c_i,d_i>0$ and $iin[1,n]$.
Without the linear coupling term, the system reduces to a set of independent Riccati equations with constant coefficients that can be solved.
With the coupling, the system looks like a matrix equation but I don't know how to represent the quadratic term in matrix form.
Any kind of help for finding the solution, or link to literature would be very helpful.
Thank you!
ordinary-differential-equations systems-of-equations nonlinear-system
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
$endgroup$
$begingroup$
You can always write $boldsymbol{dot{x}} = boldsymbol{g}(boldsymbol{x}) + boldsymbol{underline{B}} boldsymbol{x} - boldsymbol{d}$, where $g_i(boldsymbol{x}) = a_i x_i^2$, $i=1,2,dots,n$.
$endgroup$
– Christoph
Jan 20 at 13:58
$begingroup$
Dear Christoph, Thank you for your reply! Although I am not sure I understand how to use your proposition to solve my system. Supose that $B$ is diagonalizable so $B=Vlambda V^{-1}$. Then system becomes $dot{x}=g(x)+Vlambda V^{-1}x−d$ then left multiplying both sides by $V^{-1}$ I obtain $V^{-1}dot{x}=V^{-1}g(x)+lambda V^{-1}x−V^{-1}d$ which is solving the system of independent equations $dot{x_i'}=g_i'(x)+lambda_i' x_i'−d_i'$ with $dot{x'}=V^{-1}dot{x}$, $lambda'=V^{-1}lambda$, $g'(x)=V^{-1}g(x)$ and $d'=V^{-1}d$. Am I correct ? Best regards
$endgroup$
– Yakari Dubois
Jan 23 at 0:21
$begingroup$
Yes, I think it doesn't help much for the solution, it's just a way to write the system in a more compact form. If you set $x = V x'$ you'll also note that the nonlinearity is no longer diagonal, as now $g_i'(x')$ depends on all of the $x_j'$, $j=1,2,dots,n$. I would recommend to go for a numerical solution.
$endgroup$
– Christoph
Jan 23 at 5:13
add a comment |
$begingroup$
I am looking forward to solve the system of equations
$$frac{d x_i}{d t}=a_ix_i^2+b_ix_i-sumlimits_{jneq i}^nb_jx_j-d_i,$$
with $x_igeq 0$, $a_i,b_i,c_i,d_i>0$ and $iin[1,n]$.
Without the linear coupling term, the system reduces to a set of independent Riccati equations with constant coefficients that can be solved.
With the coupling, the system looks like a matrix equation but I don't know how to represent the quadratic term in matrix form.
Any kind of help for finding the solution, or link to literature would be very helpful.
Thank you!
ordinary-differential-equations systems-of-equations nonlinear-system
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
$endgroup$
I am looking forward to solve the system of equations
$$frac{d x_i}{d t}=a_ix_i^2+b_ix_i-sumlimits_{jneq i}^nb_jx_j-d_i,$$
with $x_igeq 0$, $a_i,b_i,c_i,d_i>0$ and $iin[1,n]$.
Without the linear coupling term, the system reduces to a set of independent Riccati equations with constant coefficients that can be solved.
With the coupling, the system looks like a matrix equation but I don't know how to represent the quadratic term in matrix form.
Any kind of help for finding the solution, or link to literature would be very helpful.
Thank you!
ordinary-differential-equations systems-of-equations nonlinear-system
ordinary-differential-equations systems-of-equations nonlinear-system
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
edited Jan 23 at 1:19
Yakari Dubois
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
asked Jan 19 at 13:53
Yakari DuboisYakari Dubois
62
62
$begingroup$
You can always write $boldsymbol{dot{x}} = boldsymbol{g}(boldsymbol{x}) + boldsymbol{underline{B}} boldsymbol{x} - boldsymbol{d}$, where $g_i(boldsymbol{x}) = a_i x_i^2$, $i=1,2,dots,n$.
$endgroup$
– Christoph
Jan 20 at 13:58
$begingroup$
Dear Christoph, Thank you for your reply! Although I am not sure I understand how to use your proposition to solve my system. Supose that $B$ is diagonalizable so $B=Vlambda V^{-1}$. Then system becomes $dot{x}=g(x)+Vlambda V^{-1}x−d$ then left multiplying both sides by $V^{-1}$ I obtain $V^{-1}dot{x}=V^{-1}g(x)+lambda V^{-1}x−V^{-1}d$ which is solving the system of independent equations $dot{x_i'}=g_i'(x)+lambda_i' x_i'−d_i'$ with $dot{x'}=V^{-1}dot{x}$, $lambda'=V^{-1}lambda$, $g'(x)=V^{-1}g(x)$ and $d'=V^{-1}d$. Am I correct ? Best regards
$endgroup$
– Yakari Dubois
Jan 23 at 0:21
$begingroup$
Yes, I think it doesn't help much for the solution, it's just a way to write the system in a more compact form. If you set $x = V x'$ you'll also note that the nonlinearity is no longer diagonal, as now $g_i'(x')$ depends on all of the $x_j'$, $j=1,2,dots,n$. I would recommend to go for a numerical solution.
$endgroup$
– Christoph
Jan 23 at 5:13
add a comment |
$begingroup$
You can always write $boldsymbol{dot{x}} = boldsymbol{g}(boldsymbol{x}) + boldsymbol{underline{B}} boldsymbol{x} - boldsymbol{d}$, where $g_i(boldsymbol{x}) = a_i x_i^2$, $i=1,2,dots,n$.
$endgroup$
– Christoph
Jan 20 at 13:58
$begingroup$
Dear Christoph, Thank you for your reply! Although I am not sure I understand how to use your proposition to solve my system. Supose that $B$ is diagonalizable so $B=Vlambda V^{-1}$. Then system becomes $dot{x}=g(x)+Vlambda V^{-1}x−d$ then left multiplying both sides by $V^{-1}$ I obtain $V^{-1}dot{x}=V^{-1}g(x)+lambda V^{-1}x−V^{-1}d$ which is solving the system of independent equations $dot{x_i'}=g_i'(x)+lambda_i' x_i'−d_i'$ with $dot{x'}=V^{-1}dot{x}$, $lambda'=V^{-1}lambda$, $g'(x)=V^{-1}g(x)$ and $d'=V^{-1}d$. Am I correct ? Best regards
$endgroup$
– Yakari Dubois
Jan 23 at 0:21
$begingroup$
Yes, I think it doesn't help much for the solution, it's just a way to write the system in a more compact form. If you set $x = V x'$ you'll also note that the nonlinearity is no longer diagonal, as now $g_i'(x')$ depends on all of the $x_j'$, $j=1,2,dots,n$. I would recommend to go for a numerical solution.
$endgroup$
– Christoph
Jan 23 at 5:13
$begingroup$
You can always write $boldsymbol{dot{x}} = boldsymbol{g}(boldsymbol{x}) + boldsymbol{underline{B}} boldsymbol{x} - boldsymbol{d}$, where $g_i(boldsymbol{x}) = a_i x_i^2$, $i=1,2,dots,n$.
$endgroup$
– Christoph
Jan 20 at 13:58
$begingroup$
You can always write $boldsymbol{dot{x}} = boldsymbol{g}(boldsymbol{x}) + boldsymbol{underline{B}} boldsymbol{x} - boldsymbol{d}$, where $g_i(boldsymbol{x}) = a_i x_i^2$, $i=1,2,dots,n$.
$endgroup$
– Christoph
Jan 20 at 13:58
$begingroup$
Dear Christoph, Thank you for your reply! Although I am not sure I understand how to use your proposition to solve my system. Supose that $B$ is diagonalizable so $B=Vlambda V^{-1}$. Then system becomes $dot{x}=g(x)+Vlambda V^{-1}x−d$ then left multiplying both sides by $V^{-1}$ I obtain $V^{-1}dot{x}=V^{-1}g(x)+lambda V^{-1}x−V^{-1}d$ which is solving the system of independent equations $dot{x_i'}=g_i'(x)+lambda_i' x_i'−d_i'$ with $dot{x'}=V^{-1}dot{x}$, $lambda'=V^{-1}lambda$, $g'(x)=V^{-1}g(x)$ and $d'=V^{-1}d$. Am I correct ? Best regards
$endgroup$
– Yakari Dubois
Jan 23 at 0:21
$begingroup$
Dear Christoph, Thank you for your reply! Although I am not sure I understand how to use your proposition to solve my system. Supose that $B$ is diagonalizable so $B=Vlambda V^{-1}$. Then system becomes $dot{x}=g(x)+Vlambda V^{-1}x−d$ then left multiplying both sides by $V^{-1}$ I obtain $V^{-1}dot{x}=V^{-1}g(x)+lambda V^{-1}x−V^{-1}d$ which is solving the system of independent equations $dot{x_i'}=g_i'(x)+lambda_i' x_i'−d_i'$ with $dot{x'}=V^{-1}dot{x}$, $lambda'=V^{-1}lambda$, $g'(x)=V^{-1}g(x)$ and $d'=V^{-1}d$. Am I correct ? Best regards
$endgroup$
– Yakari Dubois
Jan 23 at 0:21
$begingroup$
Yes, I think it doesn't help much for the solution, it's just a way to write the system in a more compact form. If you set $x = V x'$ you'll also note that the nonlinearity is no longer diagonal, as now $g_i'(x')$ depends on all of the $x_j'$, $j=1,2,dots,n$. I would recommend to go for a numerical solution.
$endgroup$
– Christoph
Jan 23 at 5:13
$begingroup$
Yes, I think it doesn't help much for the solution, it's just a way to write the system in a more compact form. If you set $x = V x'$ you'll also note that the nonlinearity is no longer diagonal, as now $g_i'(x')$ depends on all of the $x_j'$, $j=1,2,dots,n$. I would recommend to go for a numerical solution.
$endgroup$
– Christoph
Jan 23 at 5:13
add a comment |
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$begingroup$
You can always write $boldsymbol{dot{x}} = boldsymbol{g}(boldsymbol{x}) + boldsymbol{underline{B}} boldsymbol{x} - boldsymbol{d}$, where $g_i(boldsymbol{x}) = a_i x_i^2$, $i=1,2,dots,n$.
$endgroup$
– Christoph
Jan 20 at 13:58
$begingroup$
Dear Christoph, Thank you for your reply! Although I am not sure I understand how to use your proposition to solve my system. Supose that $B$ is diagonalizable so $B=Vlambda V^{-1}$. Then system becomes $dot{x}=g(x)+Vlambda V^{-1}x−d$ then left multiplying both sides by $V^{-1}$ I obtain $V^{-1}dot{x}=V^{-1}g(x)+lambda V^{-1}x−V^{-1}d$ which is solving the system of independent equations $dot{x_i'}=g_i'(x)+lambda_i' x_i'−d_i'$ with $dot{x'}=V^{-1}dot{x}$, $lambda'=V^{-1}lambda$, $g'(x)=V^{-1}g(x)$ and $d'=V^{-1}d$. Am I correct ? Best regards
$endgroup$
– Yakari Dubois
Jan 23 at 0:21
$begingroup$
Yes, I think it doesn't help much for the solution, it's just a way to write the system in a more compact form. If you set $x = V x'$ you'll also note that the nonlinearity is no longer diagonal, as now $g_i'(x')$ depends on all of the $x_j'$, $j=1,2,dots,n$. I would recommend to go for a numerical solution.
$endgroup$
– Christoph
Jan 23 at 5:13