Lemma 14.27.3C96C68E6BB33

Proof. The function $h_ s=sp$ solves (14.27.4) and lies strictly between $-x$ and $y$ throughout $(-1,1)$. Write $F_ s(r,h)$ for the right-hand side of the equation, which is smooth on the open domain $-x(r)<h<y(r)$.

Near $-1$ the coefficients are exactly

For a constant $c\in (-1,1)$, the product of coefficients for the trial slope $cp$ is

Fix $c_ L\in (0,s)$. At $p=0$, the factor preceding $pq$ for $c=c_ L$ exceeds that for $c=s$ by $s^2-c_ L^2>0$. Thus, on a sufficiently small fixed neighborhood of the left endpoint, $k_{c_ L}>k_ s$ and

Thus $c_ Lp$ is a lower barrier when the equation is solved backward, that is, toward $-1$.

Near $1$ the exact coefficients are

and the trial slope $cp$ has product of coefficients

Fix $s<t<1$. At $p=0$, the factor for $c=s$ exceeds that for $c=-t$ by $t^2-s^2>0$. On a sufficiently small fixed neighborhood of the right endpoint, therefore, $k_ s>k_{-t}$, so

Thus $-tp$ is a lower barrier when the equation is solved forward, toward $1$.

Choose $r_ L<0<r_ R$ in these respective neighborhoods. The graph of $h_ s$ over $[r_ L,r_ R]$ is a compact subset of the domain of $F_ s$. Existence and continuous dependence on initial values for smooth ordinary differential equations therefore give $\eta _0>0$ such that, for every $0<\eta <\eta _0$, the solution with initial value $sp(0)-\eta $ exists on $[r_ L,r_ R]$, stays in the domain, and satisfies

By local uniqueness, two solutions of the equation cannot cross. Since $h(0)<h_ s(0)$, we have $h<h_ s$ wherever $h$ exists.

Follow the solution backwards from $r_ L$; it cannot cross $c_ Lp$. Indeed, at a first crossing point the difference $h-c_ Lp$ would vanish and be positive immediately to its right, so its derivative there would be nonnegative, whereas (14.27.5) shows that this derivative is strictly negative. Thus $c_ Lp<h<sp$ on the part of the interval of existence to the left of $r_ L$. Likewise, (14.27.6) prevents a first downward crossing of $-tp$ to the right of $r_ R$, giving $-tp<h<sp$ there.

These bounds also show that the maximal interval of existence is all of $(-1,1)$. Indeed, near a possible endpoint inside $(-1,1)$, the factors $x+h$ and $y-h$ stay bounded away from zero, so the graph of $h$ stays in a compact subset of the domain of $F_ s$, and the continuation theorem for ordinary differential equations extends the solution beyond that endpoint. Hence the solution exists on $(-1,1)$ and satisfies the claimed bounds. $\square$