### Author Topic: MT, P4  (Read 6198 times)

#### Victor Ivrii

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##### MT, P4
« on: October 09, 2013, 07:22:47 PM »
Analyze the direction field and constant (equilibrium) solutions of the ODE
\begin{equation*}
y'=\frac{\sin y}{1+\sin^2 t}
\end{equation*}
to explain why the solution $y(t)$ of the initial value problem
\begin{equation*}
\end{equation*}
is defined for all values of $t$, is an increasing function  and  satisfies the inequality $0<y(t)<\pi$ for all values of $t$.

(Do not try to solve the initial value problem.)

#### Xiaozeng Yu

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##### Re: MT, P4
« Reply #1 on: October 09, 2013, 10:39:57 PM »
4

#### Xuewen Yang

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##### Re: MT, P4
« Reply #2 on: October 09, 2013, 10:49:45 PM »
Just wondering, for this question, do we need to draw the direction field?

#### Victor Ivrii

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##### Re: MT, P4
« Reply #3 on: October 10, 2013, 06:29:22 AM »
Solution is incorrect (it is based on presumption that $y$ takes values in $(0,\pi)$ instead of proving it).

#### Xiaozeng Yu

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##### Re: MT, P4
« Reply #4 on: October 10, 2013, 09:43:00 AM »
ahh...omg, because the rectangle must contain the initial value point (0,1) in order to have an unique solution of the initial value problem. sin1>0 make the function increasing, so the retangle (a<0<b, 0<y<pi) which containing the initial point contains the unique solution function of (0,1) which is increasing?
« Last Edit: October 10, 2013, 10:13:05 AM by Xiaozeng Yu »

#### Victor Ivrii

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##### Re: MT, P4
« Reply #5 on: October 10, 2013, 10:28:46 AM »
ahh...omg, because the rectangle must contain the initial value point (0,1) in order to have an unique solution of the initial value problem. sin1>0 make the function increasing, so the retangle (a<0<b, 0<y<pi) which containing the initial point contains the unique solution function of (0,1) which is increasing?

You need to prove that the solution while increasing as $t$ increases never goes above $\pi$ and also  that the solution while decreasing as $t$ decreases never goes below $0$. Nuff said. The solution remains pending.

#### Victor Ivrii

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##### Re: MT, P4
« Reply #6 on: October 15, 2013, 09:39:40 PM »
Since nobody posted a correct solution.

Observe first that the solution is unique since $f(t,y)= \frac{\sin{y}}{1+\sin^2(t)}$ satisfied conditions of  (Theorem 2.4.2 in the textbook). Since $|f(t,y)\le 1$
solutions exists for $t\in (-\infty,+\infty)$.

Observe also that $f(t,y)=0$ iff $\sin(y)=0 \iff y=n\pi$ with $n=0,\pm 1, \pm 2,\ldots$ Therefore all other solutions cannot cross these values and remain confined  between them; in particular, solution with $y(0)=1$ remains confined between $y=0$ and $y=2\pi$.

Since $y'=f(t,y)>0$ iff $2n\pi< y< (2n+1)\pi$ such solutions are monotone increasing. Since $y'=f(t,y)<0$ iff $(2n-1)\pi< y< 2n\pi$ such solutions are monotone decreasing. In particular, solution with $y(0)=1$ is monotone increasing.

Additional remarks: since $|f(t,y)|\ge \frac{1}{2}|\sin (\epsilon)|$ as $y \in (n\pi+\epsilon), (n+1)\pi-\epsilon)$ solutions will cross any other line and
* Any solution confined between $2n\pi$ and $(2n+1)\pi$, tends to $2n\pi$ as $t\to -\infty$ and to $(2n+1)\pi$ as $t\to +\infty$;
* Any solution confined between $(2n-1)\pi$ and $2n\pi$, tends to $2n\pi$ as $t\to -\infty$ and to $(2n-1)\pi$ as $t\to +\infty$.

Using language we learn in Chapter 9, $y=2n\pi$ are asymptotically unstable stationary solutions;  $y=(2n+1)\pi$ are asymptotically stable stationary solutions. See attached picture

« Last Edit: October 15, 2013, 09:48:53 PM by Victor Ivrii »