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So I have very little experience in Mathematica but would like to learn how to plot. I have seen a few threads on this for $2\times 2$ systems but not for $3\times 3$.

So I'm trying to plot the phase portrait of the solution to the following equation:

$$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}'=\begin{pmatrix}-2&1&0\\1&-2&1\\0&1&-2\end{pmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$

To make matters easier, I found a general solution:

$$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=c_1e^{2t}\begin{bmatrix}1\\0\\-1\end{bmatrix}+c_2e^{(2-\sqrt2)t}\begin{bmatrix}1\\-\sqrt2\\1\end{bmatrix}+c_3e^{(2+\sqrt 2)t}\begin{bmatrix}1\\\sqrt2\\1\end{bmatrix}$$

I considered $c_1=c_2=c_3=1$ and tried to plot a solution using the following with ParametricPlot3D:

ParametricPlot3D[{ E^(-2 u) + E^(-2 u + u Sqrt[2]) +
   E^(-2 u - u Sqrt[2]), 
  Sqrt[2] E^(-2 u + u Sqrt[2]) - 
   Sqrt[2] E^(-2 u - u Sqrt[2]), -E^(-2 u) + E^(-2 u + u Sqrt[2]) - 
   E^(-2 u - u Sqrt[2])}, {u, -1, 10}, PlotTheme -> "Marketing"]

I get a nice solution curve based on a single initial condition, but is there a nice way to plot many general solutions together?

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Here one way to plot many solutions for different C1, C[2], C[3].

dsol = DSolve[
         Thread[{x1'[t], x2'[t], 
                 x3'[t]} == {{-2, 1, 0}, {1, -2, 1}, {0, 1, -2}}.{x1[t], x2[t], 
                 x3[t]}], {x1, x2, x3}, t] // FullSimplify;

Since I work with MMA 8.0, I had to construct my own PlotLegend.

{ParametricPlot3D[
   Evaluate[({x1[t], x2[t], x3[t]} /. First@dsol) /. 
   Evaluate[
 tab = Flatten[
   Table[Thread[{C[1], C[2], C[3]} -> {i, j, k}], {i, -1, 
     1}, {j, -1, 1}, {k, -1, 1}], 2]]], {t, -1, 10}, 
  PlotStyle -> 
  Table[Hue[.8 (i - 1)/Length[tab]], {i, 1, Length[tab]}], 
  ImageSize -> 500], 
  Table[Graphics[{Hue[.8 (i - 1)/Length[tab]], 
Text[{C[1], C[2], C[3]} /. tab[[i]]]}, ImageSize -> 45], {i, 1, 
Length[tab]}]}

enter image description here

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  • $\begingroup$ Thank you very much! $\endgroup$ – Lanier Freeman Nov 21 '17 at 17:53

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