I have a problem regarding differential equations, $f$ is increasing. $x,y$ are dependent on $t$ $$ f(0)=0 $$ $$x'=f(y-x)$$ $$y'=f(x-y)$$ $$x(0)=1$$ $$y(0)=0$$ prove that when $x,y$ are the solutions of this equation then $$ \lim_{t \to \infty} x = \lim_{t \to\infty}y $$ So far I have managed to see that $x'(0)=f(-1)$ and $y'(0)=f(1)$ which implies that $x'(0)<0$ and $y'(0)>0$ and basically $x'(t)<0$, $y'(t)>0$ as $x>y$ so as $f$ is increasing and $x$ is declining with $t$ and $y$ is increasing with $t$, the derivatives of $x$ and $y$ will be declining and increasing respectivly and they will have the boundary. Can anyone help from this on?


First we note, that $\lim_{t\rightarrow \infty} x = x_\infty$ indeed exists. This is because $x$ is decreasing monotone. On the other hand $y_\infty$ exists aswell. We have $x_\infty \geq y_\infty$, because if there is a $t_0$ with $x(t_0)=y(t_0)$ we have $x'(t_0)=y'(t_0)$ and this will be $x_\infty$.

Now assume that $x_\infty >y_\infty$. The genral solution of a differential equation is: $$x(t) = x(0)+ \int_0^t f(x(t)-y(t)dt$$ But we then would have $$\lim_{t\rightarrow \infty} x(t) =x(0) +\int_0^t f(y-x)dt \leq\lim_{t\rightarrow \infty} x(0) +\int_0^t f(y_\infty-x_\infty)dt =-\infty $$

This however contradicts that $x_\infty>y_\infty$


Let $u=y-x$. Then $$ \frac{du}{dt}=\frac{dy}{dt}-\frac{dx}{dt}=f(-u)-f(u) $$

Which is automatic with an equilibrium solution $u=0$. Note that your initial conditions become $u(0)=-1$

For $u<0$ since $f$ is increasing $f(-u)-f(u)>0$ giving $u$ is increasing (for all $u<0$). Assuming $u$ is a continuous solution it would need to pass through $u=0$ but cannot actually exceed this value because of the equilibrium there. so $\lim_{t\to \infty} u=0$ Thus $\lim_{t\to \infty} y-x =0$. So if either of the limits $\lim_{t \to 0} x$ or $\lim_{t \to 0} y$ exist (and are real) they must be equal. If they're $\pm \infty$ they also must be the same one.

  • $\begingroup$ I dont get why it would have to go through 0, why boundary can not be -1/2? for instance u would be an arcus tangens scaled properly $\endgroup$ – ryszard eggink Jun 11 '18 at 13:35
  • $\begingroup$ That argument is not sufficient, you first have to show, that 0 is the only possible limit $\endgroup$ – Jürg Merlin Spaak Jun 11 '18 at 13:50
  • $\begingroup$ Any other limit $L<0$ would have to also be an equilibrium solution (that is $\frac{du}{dt}=0$ ) and $\frac{du}{dt}$ is strictly greater than zero for negative values of $u$ $\endgroup$ – N8tron Jun 11 '18 at 14:08
  • $\begingroup$ @ryszardeggink By "boundary" you apparently mean "limit"? $\endgroup$ – user539887 Jun 12 '18 at 6:51
  • $\begingroup$ yes, but the problem is already solved :) $\endgroup$ – ryszard eggink Jun 12 '18 at 7:18

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