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From my lecture notes I found that the qualitative behavior of a solution can be determined by considering for each value of the state $x$ the sign of the derivative of $x$. Considering the dynamical system

$$\dot x(t) = -x^3(t) + x(t)$$

in the notes they recommend to find the roots, which are the equilibrium points of the system. But then it says

When the function is positive, the solution is increasing and when it is negative, the solution is decreasing.

I cannot understand this last statement. What's the relation of this with the roots? If I plot the function, how should I interpret the curve so as to see "the solution increases when the function is positive"?

BRabbit27
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2 Answers2

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If you plot the direction field, you see the equilibrium points ($x= 0 , \pm ~ 1$) and can also see the behaviors.

You are looking at points where the function is positive, negative or zero.

Here is the direction field (look at what happens with the function for $x$ positive, negative, zero).

enter image description here

Notice the behaviors of the function $x'(t)$ when $x \le 1$, $x \ge 1$ and $x = \pm 1$ and $x = 0$ and even in between the last three points.

Amzoti
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  • @amWhy: Good morning! Hope the visits are going great! Time to get cleaned up and get to work! Have a wonderful day! :-) – Amzoti Jan 03 '14 at 14:15
  • Thanks. This graph is plotting the solution of the differential equation, right? I see the behaviors, but the kind of graph I have in the lectures notes is a cubic, and there is where I don't understand how to qualitatively see the behavior. – BRabbit27 Jan 03 '14 at 14:45
  • Actually I was misinterpreting the my graph (which was the cubic I mentioned). If I'm correct, it shows the value of the derivative therefor telling me it increases/decreases which tells how the solution $x$ approaches to the equilibrium points. – BRabbit27 Jan 03 '14 at 14:51
  • The last comment is correct. When $x \ge 1$, the cubic is negative, hence the derivative is negative and the solution curves approach the equilibrium point. – Amzoti Jan 03 '14 at 15:03
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    The figure you presented is not a phase portrait. – Artem Jan 03 '14 at 17:52
  • @amWhy: Now that would be fun and MSE users can be doing statistical analyses real time! – Amzoti Jan 05 '14 at 14:38
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If the derivative of a function is positive, the function is increasing, if it is negative it is decreasing, if it is $0$ it is stationary.

So if you have a function which satisfies: $$ x'(t) = f(x(t)) $$ the function is increasing where $f>0$, decreasing where $f<0$ and stationary when $f=0$.

Emanuele Paolini
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