Lemma 14.28.3624B2CF7FBA2

Proof. The right-hand side $F(r,h)$ is smooth on $-x<h<y$, and $h=0$ is a solution. Near $-1$, $(x,y,z)=(1+p,p,1)$; near $1$, $(x,y,z)=(p,1+p,1)$. Fix $0<c<1$. For the trial function $-cp$, near the left endpoint

whereas near the right endpoint

as $r\to \pm 1$, since $p,p'\to 0$ at both endpoints. The strict inequalities therefore hold on fixed neighborhoods of the endpoints, and $-cp$ is a lower barrier for the equation solved backwards near $-1$ and forwards near $1$.

Choose a compact interval $[r_ L,r_ R]\ni 0$ with endpoints in those neighborhoods. Continuous dependence on initial values, around the solution $h=0$, gives $\eta _0>0$ such that for every $0<\eta <\eta _0$ the solution exists on $[r_ L,r_ R]$, stays in the domain of $F$, and lies above $-cp$ at $r_ L$ and $r_ R$. By uniqueness of solutions, it cannot meet the solution zero, so $h<0$. The barrier argument used in the proof of lemma 14.27.3 then gives $-cp<h<0$ near both ends. On each compact subinterval of $(-1,1)$ the trapped graph stays at a positive distance from the boundary of the domain of $F$; hence, by the continuation theorem, the solution exists on all of $(-1,1)$.

The equation factors as

Since $h<0$, we have $(\log (-h))'=L(r,h)$. Near the left endpoint, the trapping bounds and the explicit coefficients imply

after shrinking that neighborhood, with $C_1>0$. Near the right endpoint they give

Indeed, on the left, $1+h\geq 1-cp\geq 1/2$ and the denominator is at most $(1+p)(1+c)p$; on the right, the numerator is at most $-1$, and the denominator is at most $p(1+(1+c)p)$. Since $p$ is comparable to the square of the distance $d=1-|r|$ to the endpoint, integrating these inequalities backwards on the left and forwards on the right gives, for some $A>0$,

near either endpoint. The constants may depend on the solution; uniformity in $\eta $ is not needed for the smooth extension.

It remains to control all derivatives. On the trapped graph, all partial derivatives in $(r,h)$ of the rational function $L$ grow at most polynomially in $d^{-1}$. Indeed, near an endpoint $x,y$ are polynomials in $r$, $z=1$, $h$ is bounded, and of the two factors in the denominator, one is bounded below by a positive constant and the other by $(1-c)p$. Repeated differentiation of the quotient therefore only introduces finite powers of $p^{-1}$.

Inductively, for every $j\geq 0$ there are constants $C_ j,N_ j$ with

The case $j=0$ is (14.28.6). Differentiate $h'=hL(r,h(r))$ $j$ times. Every resulting term is the product of a derivative of the factor $h$, of order at most $j$, and a derivative of the composite function $L(r,h(r))$. The latter involves only partial derivatives of $L$ and derivatives of $h$ of order at most $j$, which are polynomially bounded by the induction hypothesis, while the first factor has the exponential bound. This proves the case $j+1$. All derivatives of $h$ therefore tend to zero at $\pm 1$. Extending them by zero and applying the mean value theorem to the difference quotients at the endpoints, as in lemma 14.27.7, shows that the extension of $h$ by zero is smooth and vanishes to infinite order at $\pm 1$. $\square$