IWhitematter
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How can a system of non-linear differential equations be solved using Mathematica?
More specifically, I understand the process that needs to be executed, but I don't understand how to make it work syntactically.
The process as I understand it:
Four expressions are given in four unknowns. Let the unknowns be expressed as x[n]. Each expression has three or less constant terms, which I'll just express as a[n]. These expressions are determined from an ideal, physical system which is being considered, so the constants can all be manipulated through experiment.
dx[2]/dt == a[1]x[3] - a[2]x[2] - a[3]x[2] == 0
dx[3]/dt == a[4]x[4]/(a[5] - x[4]) - a[6]x[3] == 0
dx[1]/dt == a[7]x[4]/(a[8] - x[4]) - a[9]x[1] == 0
x[4]/a[10] == (x[2]/a[11])^n/(1+(x[2]/a[11])^n
To make the situation simpler, the top three differential expressions are set equal to zero to indicate a steady state process, so now I'm interested in what constitutes steady state conditions.
In particular, I would like to get an expression in the form f(x[1]) = g(x[1]), where f(x[1]) is simply x[1]. Then, I can plot both f and g on a graph and find their points of intersection.
My problem is that this is a burdensome calculation and I have no idea how to set a system like this up to be solved in the way described. I've looked up several function definitions on Wolfram help forums, including solve, eliminate, reduce and a few others, but none of the calculations simplified anything.
Is there an alternative method?
More specifically, I understand the process that needs to be executed, but I don't understand how to make it work syntactically.
The process as I understand it:
Four expressions are given in four unknowns. Let the unknowns be expressed as x[n]. Each expression has three or less constant terms, which I'll just express as a[n]. These expressions are determined from an ideal, physical system which is being considered, so the constants can all be manipulated through experiment.
dx[2]/dt == a[1]x[3] - a[2]x[2] - a[3]x[2] == 0
dx[3]/dt == a[4]x[4]/(a[5] - x[4]) - a[6]x[3] == 0
dx[1]/dt == a[7]x[4]/(a[8] - x[4]) - a[9]x[1] == 0
x[4]/a[10] == (x[2]/a[11])^n/(1+(x[2]/a[11])^n
To make the situation simpler, the top three differential expressions are set equal to zero to indicate a steady state process, so now I'm interested in what constitutes steady state conditions.
In particular, I would like to get an expression in the form f(x[1]) = g(x[1]), where f(x[1]) is simply x[1]. Then, I can plot both f and g on a graph and find their points of intersection.
My problem is that this is a burdensome calculation and I have no idea how to set a system like this up to be solved in the way described. I've looked up several function definitions on Wolfram help forums, including solve, eliminate, reduce and a few others, but none of the calculations simplified anything.
Is there an alternative method?