MAT244-2013F > MidTerm

MT, P4

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Victor Ivrii:

--- Quote from: Xiaozeng Yu 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?

--- End quote ---

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:
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


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