Author Topic: Problem 1.1.28  (Read 752 times)

Joseph Mandozzi

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Problem 1.1.28
« on: September 14, 2014, 08:23:11 PM »
Hello all!

Problem: For y'= e-t + y . Draw the direction field. And determine the behavior of y, as t approaches + infinity.

Here is the directional field:

If we notice as t-> infinity, if y>0 then y-> infinity, if y<0 y-> negative infinity.

If we notice the x-axis, y=0
Thus, all values (t,0), have positive slopes. Thus as t-> infinity y-> positive infinity if at some point y=0

However the solution says there exists an interval curve in which as t-> infinity y->0 . However, as I look at the directional fields as t-> infinity. If y>0 it goes to positive infinity eventually, if y=0 it goes to positive infinity eventually, and if y<0 it goes to negative infinity.

So what integral curve would exhibit such property that as t-> infinity y-> 0 ?

Also for this problem is there a way to analytically find which values of y(0), will give infinity, 0, or negative infinity as t-> infinity. Because graphically, we can see it is when y(0)>- 0.5 (approximate).  But is there a way to do this algebraically?

Thanks.
« Last Edit: September 14, 2014, 09:56:45 PM by Joseph Mandozzi »

Victor Ivrii

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Re: Problem 1.1.23
« Reply #1 on: September 14, 2014, 09:05:19 PM »
Observe that this is 1st order linear equation which could be solved analytically (not algebraically):
\begin{equation}
y'=e^{-t}+y.
\end{equation}
Solving it you will easily find a solution $y(t)$ s.t. $y(t)\to 0$ as $t\to +\infty$ (this solution separate solutions tending to $+\infty$ and to $-\infty$ as $t\to +\infty$.


Joseph Mandozzi

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Re: Problem 1.1.23
« Reply #2 on: September 14, 2014, 09:47:07 PM »
I understand the solution is y(t)=cet -0.5e-t

And if c=0... then y(t)-> 0 as t->infinity

Is there a way to tell this from the directional field? Because when y=0, or is near 0 (approaching 0 from the left or right). The directional field points away from y=0. So I don't understand how it can approach 0, if the direction of the field is sending it in a different direction?


Thanks for the reply!
« Last Edit: September 14, 2014, 09:54:23 PM by Joseph Mandozzi »

Victor Ivrii

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Re: Problem 1.1.23
« Reply #3 on: September 15, 2014, 09:21:58 AM »
I understand the solution is y(t)=cet -0.5e-t

And if c=0... then y(t)-> 0 as t->infinity

Is there a way to tell this from the directional field? Because when y=0, or is near 0 (approaching 0 from the left or right). The directional field points away from y=0. So I don't understand how it can approach 0, if the direction of the field is sending it in a different direction?


Thanks for the reply!

In some sense you are right: this solution is unstable. What does it mean we will learn later in Chapter 9.

Xinyi Li

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Re: Problem 1.1.28
« Reply #4 on: September 16, 2014, 01:44:47 PM »
Hi, I think you have mixed up something here.
For the directional field, when y = 0, if you let t goes from -infinity to +infinity,
the slope y' basically follow the function e^(-t) which finally approach 0 at some point.

That means for function y, although it keep increasing but the rate of increasing is keep getting smaller. Just like the -e^(-t) function, it keep increasing for t increases, but finally it approaches 0.You can definitely see it approaches 0 though the directional filed graph without any calculation of the solution. I don't see any confusion here.

Thanks

Victor Ivrii

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Re: Problem 1.1.28
« Reply #5 on: September 16, 2014, 07:24:24 PM »
Xinyi Li,
your explanation is superficial. One can note that if $y(t_0)=0$ then $y(t)\to +\infty $ as $t\to +\infty$. Everything was explained before: there is solution $y=-\frac{1}{2}e^{-t}$ tending to $0$ as $t\to +\infty$ and all other solutions tend to $\pm \infty$.

Xinyi Li

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Re: Problem 1.1.28
« Reply #6 on: September 17, 2014, 11:26:02 AM »
Thanks for explaining.
Yeah, i think i am using too easy theories to look at the questions.
Thanks for pointing out, now I can see the whole picture.
Big helps~

Chenxi Lai

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1.1 28
« Reply #7 on: September 26, 2014, 07:06:23 PM »
Hi, i've seen somebody post this question. while i dont think the graph is right. I think the y=-e^(-t) is a equilibrium solution. so the graph may look in following form.

Xinyi Li

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Re: 1.1 28
« Reply #8 on: September 27, 2014, 08:03:08 PM »
Based on my understanding, the directional field is purely described by this equation only: y^'=e^(-t)+y
From the equation we can see that, when y=0 and t→+infinity y’ will goes to 0. That is the only place that the directional field will have horizontal lines meaning zero slope. The answer should be something like the attachment.

For the general solution, clearly the prof has helped us confirmed is:
 
So if we set c = 0, then we will easily find an equilibrium solution for the ODE. I think you have a little misunderstanding with the directional field. You can look at the original post, professor Victor has given a pretty clear answers. Hope this can help.

Victor Ivrii

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Re: Problem 1.1.28
« Reply #9 on: September 27, 2014, 08:24:34 PM »
Chenxi Lai, if you saw that somebody already started this topic, you should not start a new one, but continue in the existing one! The question has been settled already: there is a solution which tends to $0$ at $+\infty$, all solutions above it tend to  $+\infty$ and all solutions below it — to $-\infty$.

PS Try not to post crappy snapshots from you cell