here it is my question. I have a system of two equations in two variables, so I can get a solution. The issue is that those equations depend on three parameters, so my solution will depend on their values. How can I make Mathematica evaluating the solution for different values of the parameters?

Example: I want so find the solution of the following:

NSolve[x == (3 k + 6 m - 3 y)/(6 t) && y == (k + 2 t - 3 x + m)/(3 t), {x, y}]

Trivially, the solution depends on the values of m, t and k. Let's say each of them may assume any value in the interval [0,1]. I have already used Manipulate, just to get an insight:

Manipulate[NSolve[x == (3 k + 6 m - 3 y)/(6 t) &&  y == (k + 2 t - 3 x + m)/(3 t), {x, y}], {m, 0, 1}, {t, 0.1, 1}, {k, 0, 1}]

Of course, this is not a final solution. How can I solve the problem? P.s. if necessary, I can fix the value of t equal to 1. If so, how can I create a plot or a table of all the results?


1 Answer 1


You can write a function that takes the parameters,

s[k_,m_,t_] = {x, y} /. Solve[
    x == (3 k + 6 m - 3 y)/(6 t) && 
    y == (k + 2 t - 3 x + m)/(3 t), {x, y}][[1]]

where the [[1]] just drops one layer of {curly braces} (which is safe since your equations are linear).

Then you can plug numbers in for k, m, t:

Outer[s,(*k:*)Subdivide[0, 1, 2],(*m:*)Subdivide[0, 1, 3],(*t:*) Subdivide[0, 1, 4]]

which gives a table of shape {3,4,5,2} (the last 2 is for the {x,y} pair. You can replace Subdivide with Range or whatever other function you like to generate numbers. You can also

Table[s[k, m, t], {k, 0, 1, 1/2}, {m, 0, 1, 1/3}, {t, 0, 1, 1/4}]

to get the same result.

To plot at a fixed t you can do something like

witht[t_] :=  Flatten[Table[{{k, m}, s[k, m, t]}, 
                            {k, 0, 1, 0.01},
                            {m, 0, 1, 0.01}], 1]

ListVectorPlot[witht[1], FrameLabel -> {"k", "m"}]

which gives

2D stream plot

You can plot in 3D too,

vectors = Flatten[Table[{
    {k, m, t}, 
    s[k, m, t]~Join~{0}
    {k, 0, 1, 0.1}, {m, 0, 1, 0.1}, {t, 0, 1, 0.1}], 2]

ListVectorPlot3D[vectors, AxesLabel -> {"k", "m", "t"}]

(I turned down the number of points. The ~Join~{0} is required because ListVectorPlot3D needs a 3D vector to plot.

3D vector plot

You can also animate the 2D plot, using Animate (which is like Manipulate),

Animate[ListVectorPlot[witht[t], FrameLabel -> {"k", "m"}], {t, 0, 1, 0.01}]

animation changing t

  • $\begingroup$ As far as plotting goes... you have a 3-dimensional input space and a 2-dimensional output, so how to plot is not obvious. $\endgroup$
    – evanb
    Apr 21, 2020 at 16:07
  • $\begingroup$ Wow, that is a very good answer! Thanks a lot. $\endgroup$ Apr 21, 2020 at 17:57
  • $\begingroup$ Sure! Feel free to give it an upvote. After a day or two you can "Accept" the most helpful answer---waiting some time lets other people respond too. $\endgroup$
    – evanb
    Apr 21, 2020 at 18:00
  • $\begingroup$ Of course I have already upvoted, just my reputation is too low so it is not shown. Just to be sure, basically this code firstly creates a function called s which runs the command Solve for given values of k, m and t, right? From that part, I just don't understand this -> {x, y} /. and [[1]] (also, my bad, my functions are not linear like the ones in the example I set). $\endgroup$ Apr 21, 2020 at 18:04
  • $\begingroup$ Also, the Outer runs the function s for different combinations of k, m and t. There, I don't understand the division by 2, 3 and 4, and I do not get the meaning of the structure of the list. If it is possible, I would like to get a table giving me the final solution together with the values of the parameters that generate that solution. $\endgroup$ Apr 21, 2020 at 18:04

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