Toronto Math Forum
MAT2442013F => MAT244 MathTests => MidTerm => Topic started by: Victor Ivrii 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*}
y'=\frac{\sin y}{1+\sin^2 t},\qquad y(0)=1
\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.)

4

Just wondering, for this question, do we need to draw the direction field?

Solution is incorrect (it is based on presumption that $y$ takes values in $(0,\pi)$ instead of proving it).

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?

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.

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 $(2n1)\pi< y< 2n\pi$ 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 $(2n1)\pi $ and $2n\pi$, tends to $2n\pi$ as $t\to \infty$ and to $(2n1)\pi$ 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