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Let $$\ddot{x}+c \dot{x}+x(1-x)=0 $$ Find sufficient conditions for $c$ such that the solution follows this properties:

$$\lim_{t \to -\infty}x(t)=0 \ \ \ \lim_{t \to \infty}x(t)=1 \ \ \ \dot{x}>0 \ \ \forall t$$

My attempt: I change the differential equation into a system of differential equations. I have the following:

\begin{align} \dot{x}_1 &= x_2\\ \dot{x}_2 &= -cx_2-x_1+x_1^2. \end{align}

Then, I calculate the equilibrium and get $P_1 = (0,0)$ and $P_2 = (1,0)$. The problem arises when I analyze the stability. For $P_1$ the point must be inestable and for $P_2$ stable. But I have some problems with the characteristic polynomial. For $P_1$ I got $$\lambda^2+c\lambda+1,$$ and for $P_2$ I got $$\lambda^2+c\lambda-1.$$ However, I do not know how I could assure both conditions in my problem.

Any help?

Bayesian guy
  • 922

1 Answers1

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I'll derive a necessary condition on $c$ for the existence of a solution to the differential equation $$ \ddot{x}+c\dot{x}+x(1-x)=0 \tag{1} $$ satisfying the conditions $$ \lim_{t\to-\infty}x(t)=0,\qquad \lim_{t\to\infty}x(t)=1, \qquad \dot{x}(t)>0\quad\forall t\in\mathbb{R}. \tag{2} $$ Multiplying $(1)$ by $\dot{x}$, we obtain the relation $$ \dot{E}=-c\dot{x}^2, \tag{3} $$ where $$ E:=\frac{1}{2}\dot{x}^2+\frac{1}{2}x^2-\frac{1}{3}x^3. \tag{4} $$ If a solution satisfies $\lim_{t\to-\infty}x(t)=0$ and $\lim_{t\to\infty}x(t)=1$, then $(4)$ implies $\lim_{t\to-\infty}E(t)=0$ and $\lim_{t\to\infty}E(t)=\frac{1}{6}$. According to $(3)$, this is not possible if $c\geq 0$, hence we must have $c<0$. This, however, is not the end of the story. In order to find a solution satisfying $\lim_{t\to-\infty}x(t)=0$, we can linearize $(1)$ around $x=0$: $$ \ddot{x}+c\dot{x}+x\approx 0. \tag{5} $$ The general solution to $(5)$ is $$ x(t)=A_+e^{\lambda_+t}+A_-e^{\lambda_-t},\quad\text{where}\quad \lambda_{\pm}=\frac{-c\pm\sqrt{c^2-4}}{2}. \tag{6} $$ Since we are assuming that $c<0$, both terms of $x(t)$ approach zero as $t\to-\infty$, as $\Re(\lambda_{\pm})>0$. However, the condition $\dot{x}>0$ $\forall t\in\mathbb{R}$ also requires that $\Im(\lambda_{\pm})=0$, or $|c|\geq 2$, otherwise the solution will oscillate around $x=0$. We conclude, therefore, that $c\leq -2$ is a necessary condition for the existence of a solution to $(1)$ satisfying the conditions $(2)$.

The argument given above does not exclude the possibility that $\dot{x}$ change sign in a region where the linearized ODE $(5)$ is no longer valid. However, it is possible to find an approximate solution to $(1)$ that satisfies $(2)$ if $-c\gg 1$. Indeed, defining $x(t)=z(\tau)$, where $\tau=-\frac{t}{c}$, we can rewrite $(1)$ as $$ \frac{1}{c^2}z''-z'+z(1-z)=0. \tag{7} $$ If $|c|\gg 1$, we can neglect the first term on the LHS of $(7)$. The resulting first order ODE $-z'+z(1-z)=0$ has a family of solutions that satisfy $(2)$: $$ z(\tau)=\frac{1}{1+e^{-(\tau-\tau_0)}}. \tag{8} $$

Gonçalo
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