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I have a differential equation that looks like:

$$y'=y(4-y)$$ I have identified the critical points to be $0$ and $4$. I need to identify the behavior of the equation as $t \rightarrow \infty$ and this behavior it seems depends on the initial value.

Now depending on the differential equation I can say that

1.$y > 4$, y is decreasing

2.$0<y<4$, $y$ is increasing

3.$y<0$, $y$ is decreasing

And I infer this based on the slope values calculated from the $y'$ equation.

So based on this I was able to conclude that as $t \rightarrow \infty$, $y$ converges to $4$ and and diverges from $0$. But how do I identify this behavior from intial values? I feel like I'm missing something here.

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    $\begingroup$ Solve the equation. You get $$y=\frac{4 e^{4 t}}{e^{4C}+e^{4 t}}$$ and conclude what you need to conclude $\endgroup$ – Raffaele Feb 28 at 12:03
  • $\begingroup$ @Oepheus. Also considering its inverse function you get the same solution and insight ( easier too). $\endgroup$ – Narasimham Feb 28 at 19:49
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To go along with the other nice answer (+1), we can do some visual analyses using direction field and phase line plots.

A direction field plot is shown that highlights three initial conditions (pink, green, red). One is outside an equilibrium, one is inside the region between the two equilibria and one is outside the other equilibrium.

enter image description here

Notice the two equilibrium points and the directions that solutions go towards or away from them or trapped between them.

We can also draw a phase line to see this behavior (the open circle is unstable and the solid is stable). If you pick a point below zero, you go to minus infinity, if you pick a point between zero four, you go to four and if you pick a point above four, you go to four.

enter image description here

We can also solve this DEQ and do analyses

$$y(t) = \dfrac{4 e^{4 t}}{e^{4 t}+c}$$

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  • $\begingroup$ This was a really good answer!...But what is implied by the initial value?.....And what tool did ya use to generate the direction field? $\endgroup$ – Orpheus Feb 28 at 12:59
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    $\begingroup$ If you look at the solution curves, you can tell the limit as t approaches $\pm \infty$. for any initial condition. I just used three to demonstrate it. I used Mathematica and some code I wrote for the second and a built in command for the first. $\endgroup$ – Moo Feb 28 at 13:00
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    $\begingroup$ Thanks a lot @Moo $\endgroup$ – Orpheus Feb 28 at 13:51
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There are two equilibrium points (constant solutions) $y = 0$ and $y = 4$. By the uniqueness theorem other solutions cannot intersect with those two. The stable point $y = 4$ is a hyperbolic attractor and has a basin of attraction $(0, +\infty)$. The unstable point $y = 0$ is a hyperbolic repeller with basin $(-\infty, 0)$. For example, as you have figured out, $y$ is increasing on phase curves in $(0, 4)$ with some bounded below speed in $(\delta, 4-\delta)$ for any positive small enough $\delta$. But any trajectory with initial values in $(0,4)$ cannot leave the region. So it will converge to $4$ asymptotically.

Actually, it is possible to just find a general solution, so you can use it for anything you what to prove.

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