Exercise 17, § 1

To study the equation $x' = \sin t x$ (IV, p. 174, Example 4) one puts $u = xy$ and considers the solutions of the corresponding equation $u' = \frac{u}{t} + t \sin u = F(t, u)$ which are $> 0$ for $t > 0$ (one has $u(0) = 0$ for every solution of this kind). Denote by $\Gamma_k$ the curve defined by the relations $(2k-1)\pi < u < 2k\pi, t \sin u + \frac{u}{t} = 0$ for each integer $k \geq 1$; by $D_k$ the open set defined by $(2k-1)\pi < u < 2k\pi, t \sin u + \frac{u}{t} < 0$; and by $E$ the complement of the union of the $\overline{D}_k$ in the set of $(t, u)$ such that $t > 0$ and $u > 0$.

a) Show that every integral curve which cuts a line $u = 2k\pi$ also cuts the line $u = (2k+1)\pi$.

b) If an integral curve $C$ cuts a curve $\Gamma_k$ at a point $(t_0, u_0)$, then the function $u$ is increasing for $0 < t < t_0$, decreasing for $t > t_0$, and, when $t$ tends to $+\infty$, $u(t)$ tends to $(2k-1)\pi$, the curve $C$ remaining in $D_k$ for $t > t_0$.

c) Show that there is no integral curve $C$ contained in $E$ and such that $u(t)$ tends to $+\infty$ when $t$ tends to $+\infty$. (Form the differential equation $dx/du = G(u, x)$ between $x$ and $u$ along $C$. If $C$ lies entirely in $E$, then $x^2 > u$ for $u = (2k-\frac{1}{2})\pi$; on the other hand, estimate $x^2(u)$ by using the preceding differential equation, and obtain a contradiction.)

d) Show that for every integer $k$ there exists one and only one integral curve $C$ contained in $E$ and such that $u(t)$ tends to $2k\pi$ when $t$ tends to $+\infty$. (On putting $v = 2k\pi - u$ one obtains the equation $v' = \frac{v - 2k\pi}{t} + t \sin v$; compare this equation to $v' = \frac{v - 2k\pi}{t} + t v$ with $t$ tending to $+\infty$ and $v$ to 0.)