0
$\begingroup$

Let $f(t,y),$ be such that $f(t,3)=-1 $ is continuous for all $t$.

1. What information can we get from this fact for the direction field of the differential equation $\frac{dy}{dt}=f(t,y)?$

2. If $y(0) < 3$, can $y(t) \rightarrow \infty$ as $t\rightarrow \infty$?

My question:

  1. As far as I know, the only fact that I can deduce from the hypothesis is that the horizontal line $y=3$ is filled with points with constant slope ($\frac{dy}{dt}(t,3)=-1$) for all $t$.

Can something else be deduced from the fact that $f(t,3)$ is continuous?

  1. I believe $y(t) \not \rightarrow \infty,$ when $t \rightarrow \infty$ if $y(0) <3,$ because if that where so, since $y(0) <3$, by continuity, there would be a $t_0>0,$ such that $\frac{dy}{dt}(t_0,3) > 0,$ and since $\frac{dy}{dt}(t_0,3)=f(t, 3) = -1$ for all $t$, that cannot happen. Hence $y(t) \not \rightarrow \infty.$

Is this argument correct?

Thanks in advance for your help.

  • 0
    Since $f(t,3)<0$ and $f$ is continuous, essentially all you know is that if $y(0) \leq 3$ then $y(t) \leq 3$ for all $t \geq 0$. That's because whenever $y$ gets to $3$ (or even a little bit below) it must start decreasing. Your second argument is pretty much correct, just needs a clarification as to what $t_0$ actually is. – 2017-02-15
  • 0
    @Ian Thanks. What clarification do you mean? In order for $y(t) \rightarrow \infty$ as $t \rightarrow \infty $, $y(t)$ must intersect the line $y=3$ at some point $t_0 > 0$ if $y(t)$ is continuous, since $y(0) <3$, right? – 2017-02-15
  • 0
    Yes, the point was that $t_0$ was defined by $y(t_0)=3$. – 2017-02-15
  • 0
    @Ian Ok, I see now I missed that. Thank you for your help! – 2017-02-15

0 Answers 0