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Given the following system of differential equations (my actual system is $10$ times more complicated), I will need to find values of $x_1, x_2, x_3, \ldots, z_3$ at a given time, $t$.

$\dfrac{dx_1}{dt} = -k_1x_1+k_2x_2-(K_R)x_1y_1$

$\dfrac{dx_2}{dt} = k_1x_1-k_2x_2-k_3x_2-(K_R)x_2y_2$

$\dfrac{dx_3}{dt} = k_3x_3$

$\dfrac{dy_1}{dt} = -k_1y_1+k_2y_2-(K_R)x_1y_1$

$\dfrac{dy_2}{dt} = k_1y_1-k_2y_2-k_3y_2-(K_R)x_2y_2$

$\dfrac{dy_3}{dt} = k_3y_3$

$\dfrac{dz_1}{dt} = -k_1z_1+k_2z_2+(K_R)x_1y_1$

$\dfrac{dz_2}{dt} = k_1z_1-k_2z_2-k_3z_2+(K_R)x_2y_2$

$\dfrac{dz_3}{dt} = k_3z_3$

The initial conditions at $t = 0$, is $x_2 = 1$ and $y_2 = 10$. Everything else is zero at $t = 0$. The value of $K_R$ is 1e-3.

To help visualize, here's the system:

$$ \require{extpfeil} \Newextarrow{\xrightleftharpoons}{5,5}{0x21CC} \begin{array}{} x_1 & \xrightleftharpoons[k_2]{k_1} \enspace & x_2 &\xrightarrow[]{k_3} & x_3 \\ + & \ & + \\ y_1 & \xrightleftharpoons[k_5]{k_4} \enspace & y_2 &\xrightarrow[]{k_6} & y_3 \\ \downarrow^{k_R} & \ & \downarrow^{k_R} \\ z_1 & \xrightleftharpoons[k_8]{k_7} \enspace & z_2 &\xrightarrow[]{k_9} & z_3 \\ \end{array} $$

The values of $k_1, k_2, k_3, k_4, k_5, k_6, k_7, k_8, k_9,$ are $1, 0.25, 0.3, 1, 0.25, 0.3, 1, 0.25, 0.3.$

I have used Python to solve a system of first order differential equations using matrix methods, which does not seem to work for a system of first- and second-order differential equations.

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  • $\begingroup$ in $x_2$ are you sure it's $-k_2x_2- k_3x_2$ and not $-k_2x_2- k_3x_3$? the same for $y_2$ and $z_2$... $\endgroup$
    – Mick
    Mar 31, 2020 at 15:12
  • $\begingroup$ Yes, please see if the box-and-arrow system makes sense. $\endgroup$
    – DPdl
    Mar 31, 2020 at 15:23
  • $\begingroup$ @Moo, yes please. Really struggling with this. $\endgroup$
    – DPdl
    Mar 31, 2020 at 15:41

1 Answer 1

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Here is a numerical solution using Mathematica, but you should be able to do this with Python, Matlab, Maple V, SAGE and many others.

 k1 = 1;
 k2 = 0.25;
 k3 = 0.3;
 kr = 1 10^(-3);

 {a,b,c,d,e,f,g,h,i}=NDSolveValue[{x1'[t]==-k1 x1[t]+k2 x2[t]-(kr)x1[t]y1[t],x2'[t]==k1 x1[t]-k2 x2[t]-k3 x2[t]-(kr) x2[t] y2[t],x3'[t]==k3 x3[t],y1'[t]==-k1 y1[t]+k2 y2[t]-(kr) x1[t] y1[t],y2'[t]==k1 y1[t]-k2 y2[t]-k3 y2[t]-(kr)x2[t] y2[t],y3'[t]==k3 y3[t],z1'[t]==-k1 z1[t]+k2 z2[t]+(kr) x1[t] y1[t],z2'[t]==k1 z1[t]-k2 z2[t]-k3 z2[t]+(kr)x2[t] y2[t],z3'[t]==k3 z3[t],x1[0]==0,x2[0] == 1, x3[0]==0, y1[0]==0,y2[0]==10, y3[0]== 0, z1[0] ==  0, z2[0] == 0, z3[0]==0},{x1, x2, x3, y1, y2, y3, z1, z2, z3},{t,-5,5}]

   Plot[Evaluate[{a[t],b[t],c[t],d[t],e[t],f[t],g[t],h[t],i[t]}],{t,-5,5},PlotLegends->"Placeholder",PlotStyle->Thickness[0.01],  ImageSize->Large]

Here are the resulting graphs ($1 = x1, 2 = x2...$).

enter image description here

Note that three of the graphs are zero. The reason for this is that you have three of equations decoupled from the rest and they each result in something like

$$x_3' = \dfrac{3}{10} x_3, x_3(0) = 0 \implies x3(t) = 0$$

This is also true for $y_3(t)$ and $z_3(t)$.

Here is data from $(-5, 5)$ in steps of $0.5$ for each of the nine functions.The columns are $$(t, x_1(t), x_2(t), x_3(t), y_1(t), y_2(t), y_3(t), z_1(t), z_2(t), z_3(t))$$

$$\left( \begin{array}{cccccccccc} -5. & -102.864 & 750.148 & 0. & -1629.81 & 2752.71 & 0. & -66.7969 & -527.64 & 0. \\ -4.5 & -62.6326 & 200.726 & 0. & -848.138 & 1241.48 & 0. & -24.6458 & -85.0869 & 0. \\ -4. & -37.6416 & 80.0207 & 0. & -440.819 & 623.686 & 0. & -7.15583 & -19.6135 & 0. \\ -3.5 & -21.1229 & 37.2184 & 0. & -227.248 & 323.673 & 0. & -1.77987 & -5.39013 & 0. \\ -3. & -11.2592 & 18.6747 & 0. & -115.91 & 171.774 & 0. & -0.368699 & -1.66372 & 0. \\ -2.5 & -5.77925 & 9.86226 & 0. & -58.253 & 93.5828 & 0. & -0.0511651 & -0.559979 & 0. \\ -2. & -2.86044 & 5.46903 & 0. & -28.5741 & 52.8843 & 0. & 0.00335988 & -0.200667 & 0. \\ -1.5 & -1.34486 & 3.21088 & 0. & -13.3964 & 31.4409 & 0. & 0.00579917 & -0.0742138 & 0. \\ -1. & -0.5726 & 2.0186 & 0. & -5.70484 & 19.9456 & 0. & 0.00235167 & -0.026712 & 0. \\ -0.5 & -0.186919 & 1.36903 & 0. & -1.86521 & 13.6196 & 0. & 0.000442068 & -0.00785549 & 0. \\ 0. & 0. & 1. & 0. & 0. & 10. & 0. & 0. & 0. & 0. \\ 0.5 & 0.0856937 & 0.778327 & 0. & 0.85889 & 7.81397 & 0. & 0.000217037 & 0.00341098 & 0. \\ 1. & 0.12043 & 0.635708 & 0. & 1.20974 & 6.40144 & 0. & 0.000603757 & 0.00492809 & 0. \\ 1.5 & 0.129983 & 0.536819 & 0. & 1.30842 & 5.41844 & 0. & 0.000954418 & 0.00558321 & 0. \\ 2. & 0.1275 & 0.4632 & 0. & 1.28588 & 4.68431 & 0. & 0.001209 & 0.00581169 & 0. \\ 2.5 & 0.11964 & 0.405066 & 0. & 1.2087 & 4.10293 & 0. & 0.00136682 & 0.00580832 & 0. \\ 3. & 0.109736 & 0.357111 & 0. & 1.11037 & 3.62212 & 0. & 0.00144582 & 0.00566817 & 0. \\ 3.5 & 0.099422 & 0.316361 & 0. & 1.00742 & 3.2126 & 0. & 0.00146631 & 0.00544324 & 0. \\ 4. & 0.0894677 & 0.281071 & 0. & 0.907688 & 2.8572 & 0. & 0.00144561 & 0.00516559 & 0. \\ 4.5 & 0.0802052 & 0.250147 & 0. & 0.814626 & 2.54518 & 0. & 0.00139714 & 0.00485685 & 0. \\ 5. & 0.0717494 & 0.222855 & 0. & 0.729471 & 2.26935 & 0. & 0.00133081 & 0.00453248 & 0. \\ \end{array} \right)$$

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