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I am developing a model and my model equations are

${dA\over dt}=r_1A(1-{A+B\over k})-a_1 AE-m_1{AN\over (xN+A)}-m_4A$

${dB\over dt}=r_2B(1-{A+B\over k})+a_1 AE-m_2{BN\over (xN+B)}$

${dC\over dt}=a_2AE-m_eC$

${dE\over dt}=\gamma P -a_2AE-\gamma E$

${dN\over dt}= \alpha (R-N)(A+B)-m_3N$

Where P, R are constant.

I want to see what parameters are related with the initial exponential increase seen in the model. For this I thought of using a method similar to the question in where $r_0$ is the initial exponential growth rate and I am trying to come up with a relation similar to what was obtained in part 2 in that question. For this I linearized the above system. But, my question is if I set $f= r_1A(1-{A+B\over k})-a_1 AE-m_1{AN\over (xN+A)}-m_4A $ when linearizing I get terms such as $\frac{\partial f}{\partial A}=…..-d_1{xN^2\over {(xN+A)}^2}$

Therefore, can I linearize the above system and can I use a method similar to that used in the the other question in order to obtain a relationship between the parameters and the exponential growth rate.

  1. In this system I want to analyse whether there are linked parameters, that is if I increase one parameter there may be another parameter that I have to decrease in order to maintain the same behaviour. This I want to do because for some parameters in the long run the model starts oscillating and for some it decreases to zero. I want to analyse which parameter values or ranges give these different behaviours.
    The different behaviours of the model are
    enter image description here
    enter image description here

  2. Also, I tried to find the equilibrium solutions to the system, but Mathematica couldn’t come up with a solution for a long time. Why was it? Shouldn’t it be able to at least come up with the equilibrium point (0,0,0, P,0)

Solve [ { r1 * a [ t ] * ( 1 - ( ( ( a [ t ] + b [ t ] ) ) / k ) ) - a1 * a [ t ] * E1 [ t ] - m1 * a [ t ] * n [ t ] / ( ( x * n [ t ] + a [ t ] ) ) - m4 * a [ t ] ⩵ 0 , r2 * b [ t ] * ( 1 - ( ( ( a [ t ] + b [ t ] ) ) / k ) ) + a1 * a [ t ] * E1 [ t ] - m2 * b [ t ] * n [ t ] / ( ( x * n [ t ] + b [ t ] ) ) ⩵ 0 , a2 * a [ t ] * E1 [ t ] - de * C [ t ] ⩵ 0 , gamma * P - a2 * a [ t ] * E1 [ t ] - gamma * E1 [ t ] ⩵ 0 , alpha * ( R - n [ t ] ) * ( a [ t ] + b [ t ] ) - m3 * n [ t ] ⩵ 0 } , { a [ t ] , b [ t ] , C [ t ] , E1 [ t ] , n [ t ] } ]
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closed as too broad by JonMark Perry, José Carlos Santos, Siong Thye Goh, Xander Henderson, hardmath Sep 7 '17 at 4:36

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Please ask only one question per question. Also note the existence of Mathematica. $\endgroup$ – Wrzlprmft Sep 5 '17 at 12:11
  • $\begingroup$ when linearizing I get terms such as $\frac{\partial f}{\partial A}=…..-d_1{xN^2\over {(xN+A)}^2}$ – That’s obviously not linear. I don’t know what you did, but it’s almost certainly wrong. $\endgroup$ – Wrzlprmft Sep 5 '17 at 12:12
  • $\begingroup$ @Wrzlprmft Isn't $\frac{\partial ({AN\over xN+A})}{\partial A}={xN^2\over {(xN+A)}^2}$ $\endgroup$ – clarkson Sep 5 '17 at 12:34
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    $\begingroup$ @clarkson Don't forget that Jacobi matrix must be computed at equilibrium point. Your expressions for derivatives seem to be right (and of course they don't have to be linear), but you have to plug equilibrium coordinates into them. And so you have to find equilibria first. $\endgroup$ – Evgeny Sep 5 '17 at 13:47
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    $\begingroup$ This seems to be a nice write-up for progress in your project, but it surely exceeds the "bandwidth" for a "question" here. In part because there is no one issue to resolve, a comprehensive response to your post would take too much time and space to fit the site's framework. Please see How to Ask. $\endgroup$ – hardmath Sep 7 '17 at 4:36