Coupled system of differential equations

  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1715477571200.png

1715477607785.png

Would it not make sense to use for ##x' = -sx - gx - ry## as a better version of ##x' = - gx - ry## since the ##sx## term connects the two DEs to form a coupled system (from what the author explains the ##sx## term represent insulin glucose transformation).

Thanks!
 
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  • #2
They are coupled because the function of one equation occurs in the equation of the other. The values ##g,r,s,d## are fixed parameters. What you wrote was ##x'=-sx-gx-ry=-\underbrace{(s+g)}_{=g'}x-ry## with simply another parameter ##g',## i.e. the same problem as the original one.
 
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  • #3
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

from what the author explains the sx term represent insulin glucose transformation
Apart from what was already said, glucose is not turned into insulin. If the body has high glucose levels, it produces insulin, which stimulates cells to absorb glucose.
 
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  • #4
Orodruin said:
Apart from what was already said, glucose is not turned into insulin. If the body has high glucose levels, it produces insulin, which stimulates cells to absorb glucose.
Yes.

And though somewhat tangential, there's a probability for the body to synthesize nonessential amino acid molecules from glucose. And the amino acids then have a probability of being assembled into insulin molecules. And alternatively, amino acids that are broken down in kidneys and liver have a probability to be used for gluconeogenesis. However, the timeframe for this pathway is probably longer than the insulin-glucose interaction, and its effects are negligible on the main glucose insulin populations, hence it's completely ignored in this study.
 
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  • #5
fresh_42 said:
They are coupled because the function of one equation occurs in the equation of the other. The values ##g,r,s,d## are fixed parameters. What you wrote was ##x'=-sx-gx-ry=-\underbrace{(s+g)}_{=g'}x-ry## with simply another parameter ##g',## i.e. the same problem as the original one.
Orodruin said:
Apart from what was already said, glucose is not turned into insulin. If the body has high glucose levels, it produces insulin, which stimulates cells to absorb glucose.
docnet said:
Yes.

And though somewhat tangential, there's a probability for the body to synthesize nonessential amino acid molecules from glucose. And the amino acids then have a probability of being assembled into insulin molecules. And alternatively, amino acids that are broken down in kidneys and liver have a probability to be used for gluconeogenesis. However, the timeframe for this pathway is probably longer than the insulin-glucose interaction, and its effects are negligible on the main glucose insulin populations, hence it's completely ignored in this study.
Thank you for your replies @fresh_42, @Orodruin and @docnet!

I've done some more thinking about the problem. I think it could be a good idea to draw a compartmental model of the problem to see what is going on with the coupling of the equations since I'm still slightly confused. So drawing a compartmental model of the problem I think would look like
1715729582470.png

or
1715729663870.png

However, I'm not sure which one is correct and why. Does someone please know?

Thanks!
 
  • #6
I'm not sure what you mean. The situation is
$$
\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}-g&-r\\s&-d\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}
$$
which you deal with in other threads, too. Furthermore, you are given the nullclines and the eigenvalues. What is the standard method according to your book in such cases?
 
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  • #7
fresh_42 said:
I'm not sure what you mean. The situation is
$$
\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}-g&-r\\s&-d\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}
$$
which you deal with in other threads, too. Furthermore, you are given the nullclines and the eigenvalues. What is the standard method according to your book in such cases?
Thank you for your reply @fresh_42! Sorry what do you mean by standard method?

In post #5, I am trying to find the compartmental model of the problem to better understand the relationship between the terms.

Thanks!
 
  • #8
I'm not used to this point of view, so I'm probably of little help here. Maybe these considerations help:
$$
A=\begin{pmatrix}-g&-r\\s&-d\end{pmatrix}\Longrightarrow A^{-1}=\dfrac{1}{gd+rs}\begin{pmatrix} -d&r\\ -s&-g\end{pmatrix}
$$
and therefore
$$
\begin{pmatrix}x\\y\end{pmatrix}=\dfrac{1}{gd+rs}\begin{pmatrix}-dx'+ry'\\sx'-gy'\end{pmatrix}.
$$
Now fit in your solutions for ##x'## and ##y'## and continue from there, probably by expressing ##y## as a function of ##x.##
 
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  • #9
Here is why it is the characteristic equation.
\begin{align*}
y&=r^{-1}(-x'-gx)\\
y'&=r^{-1}(-x''-gx')\\
&=sx-dy\\
&=sx-dr^{-1}(-x'-gx)\\
-x''-gx'&=rsx+dx'+dgx\\
x''&=-(g+d)x'-(rs+dg)x\\
0&=x''-\operatorname{trace}(A)x'+\det (A)x\\
0&=\lambda^2-\operatorname{trace}(A)\lambda^1+\det (A)\lambda^0
\end{align*}
 
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  • #10
Thank you for your replies @fresh_42 ! Do you please know whether when the solution is a stable node if rs is large and or when rs is small then origin is asymmetrically stable?

Thanks!
 
  • #11
ChiralSuperfields said:
Thank you for your replies @fresh_42 ! Do you please know whether when the solution is a stable node if rs is large and or when rs is small then origin is asymmetrically stable?

Thanks!
We have a trace ##-(g+d)=-3.68## and a determinant ##gd-rs=3.165## and two real eigenvalues
$$
\lambda =\dfrac{1}{2}\left(-3.68 \pm \sqrt{(2.12)^2-4\cdot 4.3\cdot 0.21}\,\right) \in \{-2.74\ ,\ -4.62\}.
$$
\begin{align*}
x'(t)&=0 \Longrightarrow y=-\dfrac{g}{r}\,x=-0.6744 \,x\\[6pt]
y'(t)&=0 \Longrightarrow y=\dfrac{s}{d}\,x=0.269\,x
\end{align*}
\begin{align*}
0&=x'' + 3.68x' +3.165=3.68\cdot \left(\dfrac{1}{3.68} x'' + x' + \dfrac{3.165}{3.68}\right)\\[6pt]
x(t)&=-\dfrac{1}{3.68} \cdot c_1 \cdot e^{-3.68\, t} +c_2 + \dfrac{3.165}{3.68} \cdot t\\[6pt]
x(t)&=-0.27174 \cdot c_1 \cdot e^{-3.68\, t} +c_2 + 0.86\cdot t\\[6pt]
x'(0)&=c_1-\dfrac{\det (A)}{\operatorname{trace}(A)}=c_1+\dfrac{gd-rs}{g+d}=c_1+0.86\\[6pt]
x(0)&=c_2 + \dfrac{1}{\operatorname{trace}(A)}\cdot c_1 =c_2-\dfrac{1}{g+d}\cdot c_1=c_2-\dfrac{c_1}{3.68}\\
&= c_2+\dfrac{1}{\operatorname{trace}(A)}\cdot \left(x'(0)+\dfrac{\det (A)}{\operatorname{trace}(A)}\right)
\end{align*}
This is in terms of the parameters
$$
x(t)=-\dfrac{1}{g+d} \cdot c_1 \cdot e^{-(g+d)\,t} +c_2 + \dfrac{gd-rs}{g+d} \cdot t
$$
If you solve the ODE for ##y(t)## in the same way, you should have all the information you can get. Now you can play around with different initial conditions, the variation of the parameters, and whatever you want to find. I would draw an ##(x,y)## coordinate system for various initial values and its vector field. The variation of parameters is probably a mess. We have ##4## parameters ##g,d,r,s## which can be either increased or decreased which are already ##16## cases, plus the fact that the eigenvalues (that also depend on the parameters) can become only one eigenvalue if the discriminant of the quadratic equation is ##0,## or become complex if the discriminant of the quadratic equation is negative.
 
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