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I have this ordinary differential equation (ODE)

$$\dot{x} = \dfrac{c}{1+\exp{(-x/d)}}-1, $$

where $c$ and $d$ are positive constants. Dot differentiation is with respected to time. I am not sure about an explicit solution for this, but this is not important as of now.

I am interested in the qualitative properties of this equation.

  1. Is there a name to this kind of equation?
  2. How would this equation behave in the long run? As ($t \rightarrow +\infty$).

I am not sure how to start finding what happens to $t \rightarrow +\infty$ because there is no $t$ term in the right hand side.

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The first step to asymptotic analysis is studying local behaviours around fixed points, i.e. given a point $x^*$ such that $\dot x=0$, determine if trajectories that start close to $x^*$ converge towards $x^*$ (stability) or escape a neighbourhood of $x^*$ (instability).

So let us start by computing fixed points by doing $\dot x=0$ to obtain: $$e^{-\frac{x^*}{d}}=c-1$$ or $$x^*=-d \ln(c-1)$$ Depending on the physical meaning of the problem, this may be nonsense. Let us assume that this equilibrium exists, so $c>1$. Other physical considerations could lead us to further restrictions, e.g. if you are only interested in positive equilibria, $c<2$ is required.

The most direct way to study stability is linearization, by analyzing the sign of the eigenvalues of the Jacobian at the fixed point. So let us derive: $$J=-\frac{c\, e^{-\frac{x}{d}}}{d (1+e^{-\frac{x}{d}})^2}$$ and substitute: $$J(x^*)=-\frac{c-1}{c\,d}<0$$ because we have already assumed $c>1$.

In conclusion, the fixed point $x^*=-d \ln(c-1)$ is stable and trajectories $x(t)$ with initial value close to $x^*$ fulfill $\lim_{t\to\infty}x(t)=x^*$.

The next step would be to do a global analysis, by determining the basin of attraction of $x^*$, but this is rather more complicated...

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$$\frac{\text{d}x}{\text{d}t} = \frac{c}{1+\exp{(-x/d)}}-1 =\frac{c-1-\exp{(-x/d)}}{1+\exp{(-x/d)}}$$ This equation is a so called "separable" ODE. $$\text{d}t=\frac{1+\exp{(-x/d)}}{c-1-\exp{(-x/d)}} \text{d}x$$ $$t=\int\frac{1+\exp{(-x/d)}}{c-1-\exp{(-x/d)}} \text{d}x$$ $$\text{If }\quad c\neq 1\qquad t=-x+\frac{cd}{c-1}\ln\left|1-(c-1)e^{x/d} \right|+\text{constant} $$ $$\text{If }\quad c= 1\qquad t=-x-d\:e^{x/d}$$ With initial condition $\quad x(0)=x_0$ : $$\text{If }\quad c\neq 1\qquad t=x_0-x+\frac{cd}{c-1}\ln\left| \frac{1-(c-1)e^{x/d}}{1-(c-1)e^{x_0/d}} \right|$$ $$\text{If }\quad c= 1\qquad t=x_0-x+d\:\left(e^{x_0/d}-e^{x/d}\right)$$ There is no closed form for the inverse function $x(t)$. This is not an obstacle to study the behavior of $x$ as a function of $t$. Study $t$ as a function of $x$ and deduce the inverse behavior.

HINT : One can use "reduced" variables : $\quad\begin{cases}X=\frac{x}{d}\\T=\frac{t}{d}\end{cases}$ $$T=X_0-X+\frac{c}{c-1}\ln\left| \frac{1-(c-1)e^{X}}{1-(c-1)e^{X_0}} \right|$$ If you want to plot $X(T)$, first plot $T(X)$ and then $X(T)$ is directly obtained by geometrical symmetry.

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