Help interpreting behaviour of a simple system of differential equations using nullclines and direction fields

Question

I'm studying a system of differential equations using XPPAUT as a solver.

Here is the system

$$\frac{dx_1}{dt} = v - k_1x_1+x_2k_2+k_3x_1+k_4x_3 $$

$$\frac{dx_2}{dt} = k_1x_1 - k_2x_2 $$

$$\frac{dx_3}{dt} = k_3x_1 - k_4x_3 - k_5x_3 $$

with

$$ v = 0.0103, k1=0.33, k2=0.742, k3=0.33, k4=0.033, k5=0.005 $$ $$ x1_0=0.237, x2_0=2.092, x3_0=2.06 $$

Firstly I plotted $x_1$ on the x axis and $x_2$ on the y axis, integrated over $20$ time units. Then I plotted the x and y nullclines (red and green lines on the graphs below respectively) to find the equalibrium point and a direction field to qualitatively describe the system.

Here is the graph showing this information (XPPAUT screenshot):

Again, x1 is on the x and x2 on the y. Each line represents an integration starting from different initial conditons and the lines ending in a 'bump' are the direction field with lengths proportional to the gradient. As we can see, these two variables converge to a steady state where the null lines cross.

I now repeat this procedure with x1 on the x and x3 on the y axes.

My question regards the interpretation of such a plot. Is it right to say that since the nullclines here do not cross that there is no steady state? If you look at the time course of this system however, all variables tend toward a steady state which would be in direct contradiction to the 'no steady state interpretation'?

Since I am new to this sort of analysis (with no teacher/class) i'd really appreciate it if somebody with experience would give me their interpretation of the behaviour of this sytem.

Thanks

p.s. Here is the xpp code I used:

p   v= 0.0103, k1=0.33, k2=0.742, k3=0.33, k4=0.033, k5=0.005
dx1/dt = v - k1*x1 + k2*x2 - k3*x1 + k4*x3
dx2/dt = k1*x1 - k2*x2
dx3/dt = k3*x1 - k4*x3 - k5*x3
init x1=0.237, x2=2.092, x3=2.06
@ BACK=White
@ XP=x1
@ YP=x3
@ TOTAL=120
@ BUT=Fit:wf 

Answer 1

Note that your equation may be written in the form $x' = Ax + b$ where $x = [x_1,x_2,x_3]^{tr}$, $b = [v,0,0]^{tr}$ and $A$ is the matrix of the $k_i$ which, when multiplied with $x$ in the form above, are equivalent to your equations.

The solution to the system is $$ x(t) = e^{At}x_0 + \int_0^t e^{A(t-s)}ds\ b, $$ where $e^{At}$ is defined by $\sum_{n=0}^{\infty}A^n/n!$. The steady state of the system is $\lim_{n\rightarrow \infty}x(t)$, which you can find analytically by factoring $A$ into its Jordan canonical form and then exponentiating it (the Jordan blocks have known exponentials which you can look up). If the system happens to have three distinct eigenvalues, this is no different from diagonalizing the matrix and exponentiating the diagonal (resulting in the exponentials of each element). Another method is to take the Laplace transform and use the final value theorem; e.g.

$$ \lim_{n\rightarrow \infty}x(t) = \lim_{s\rightarrow 0} sX(s). $$

Note that this requires the you to compute $(sI-A)^{-1}$.

The equilibria of this system occur whenever $Ax = -b$ or $x = -A^{-1}b$ if $A$ is invertible. If this system has no solutions then there can be no equilibria. The system clearly only has nulclines if the linear system $Ax = -b$ possesses infinity many solutions. Otherwise there will be a finite number of equilibria or perhaps none at all.