Given $$x'(t)=A-B\left(x(t)\right)^2, \quad x(0)=0.$$ Is it possible to find $\lim\limits_{t\to\infty}x(t)$ without solving the differential equation?

Assuming $\lim\limits_{t\to\infty}x'(t)=0$ gives $\lim\limits_{t\to\infty}x(t)=\sqrt{A/B}$ which is correct, but I can not manage to prove that the limit of the derivative is 0.

If it is of any help, the solution to the differential equations is $$x(t)=\sqrt{\frac AB}\tanh\left(\sqrt{AB}t\right).$$


You can use direction field. When $x=\sqrt{\frac{A}{B}}$, $x'=0$. Then show that it is the only stable equilibrium point so that $x\rightarrow\sqrt{\frac{A}{B}}$ as $t\rightarrow\infty$ with any initial value.

If $x>\sqrt{\frac{A}{B}}$, $A-Bx^2 <0$. If $-\sqrt{\frac{A}{B}}<x<\sqrt{\frac{A}{B}}$, $A-Bx^2>0$. This shows that $x=\sqrt{\frac{A}{B}}$ is a stable equilibrium.

  • $\begingroup$ You can use "\rightarrow" instead of "->" :). $\endgroup$ – Chinny84 Jan 28 '15 at 9:43
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    $\begingroup$ Thank you for pointing that out. Used MS word too much. :) $\endgroup$ – KittyL Jan 28 '15 at 9:44
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    $\begingroup$ Or just \to. I believe it is the same character, just shorter to write. $\endgroup$ – slo Jan 28 '15 at 10:08
  • $\begingroup$ Hold on! Am I missing something or should the inequalities be the other way around like "If $x>\sqrt{\frac{A}{B}}$, $A-Bx^2 <0$. If $-\sqrt{\frac{A}{B}}<x<\sqrt{\frac{A}{B}}$, $A-Bx^2>0$."? $\endgroup$ – slo Jan 28 '15 at 20:13
  • $\begingroup$ Sorry, typo. You are right. $\endgroup$ – KittyL Jan 28 '15 at 22:30

I was slower than @KittyL, but since I had drawn the actual figure, I'd like to add it. You can see this answer as a supplement to what KittyL writes, and I suggest you to accept KittyL's answer rather than mine.

In the figure/example I have let $A=1$ and $B=1$. What is drawn is the vector field $(1,1-x^2)$. We see that if we start a solution $x(0)>-\sqrt{A/B}=-1$ (this is your case, since $x(0)=0$) then it will tend to the stable equilibrium $\sqrt{A/B}=+1$. If we start with $x(0)<-\sqrt{A/B}$ the solution $x(t)$ will tend to $-\infty$ in finite time.

This is typically part of the subject of a first course in Ordinary differential equations.

direction field

  • $\begingroup$ Good point. Forgot the negative case. I'll revise my answer based on this. $\endgroup$ – KittyL Jan 28 '15 at 9:55

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