14.27 Nonuniqueness with a zero set of empty interior5278ECB27C50

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. $\square$

Proof. Let $h$ be one of the solutions in lemma 14.27.3. On either endpoint neighborhood put $\xi =h/p$; since $p>0$ on $(-1,1)$, this quotient is smooth there, and the trapping bounds place $\xi $ in a fixed compact subinterval of $(-1,1)$. Define the rational functions

The explicit formulas for the coefficients near the endpoints give

Both rational functions are smooth, with all derivatives bounded, on a neighborhood of the relevant set of pairs $(p,\xi )$ with $p\geq 0$ small and $\xi $ in the compact interval above. Writing $R_{L/R}$ for the function corresponding to the endpoint under consideration, we obtain from $h'=p'\xi +p\xi '$ and $q=p'$

For the chosen bump,

This rational function and each of its derivatives grow at most like a power of the inverse distance to the endpoint. Moreover, each derivative of $p$ is $p$ times a rational function with the same property.

An induction using (14.27.8) shows that every derivative $\xi ^{(j)}$ grows at most polynomially in the inverse distance to the endpoint. The case $j=0$ is the trapping bound on $\xi $. For the induction step, differentiate the right-hand side of (14.27.8), including its term $-\xi $, $j$ times. Every resulting term is a product of a derivative of $p'/p$, a bounded partial derivative of $R_{L/R}$, and derivatives of $p$ and of $\xi $ of order at most $j$. All of these factors have polynomial bounds, by the induction hypothesis for the derivatives of $\xi $ and since every derivative of $p$ is bounded. This gives the bound for $\xi ^{(j+1)}$.

Leibniz' rule for $h=p\xi $ now gives, for each $j$, constants $C_ j,N_ j$ such that

near the endpoints. The right-hand side tends to zero at the endpoints, because $p$ decays exponentially in the inverse distance to the endpoint. Thus all derivatives of $h$ tend to zero at $\pm 1$. Extend each derivative by zero outside $(-1,1)$. By the mean value theorem, the difference quotient of $h^{(j)}$ at an endpoint tends to the limit of $h^{(j+1)}$, which is zero. Hence these extensions are the successive derivatives of the extension of $h$, which is therefore smooth and vanishes to infinite order at $\pm 1$. $\square$

of theorem 14.27.1. Let $\eta _0$ be as in lemma 14.27.3, and for $0<\eta <\eta _0$ denote by $h_\eta $ the solution with initial value $sp(0)-\eta $. Extend $h_\eta $ by zero, which gives a smooth function by the preceding lemma, and take a primitive $G_\eta $. It is constant near each end of the real line, so $G_\eta (r)$ extends smoothly to all of $\mathbb P^3$, including the two omitted projective lines. Set

On the open band,

The coefficients of $P$ and $Q$ are strictly positive by lemma 14.27.3, while the equation gives

Outside the band the form is $\omega _0$, and on the two boundary hypersurfaces it is also the rank-one form $\omega _0$, since $h_\eta $ vanishes to infinite order there. It is therefore semipositive on $X$, and by (14.26.4) its cube equals $6k_ sP\wedge Q\wedge R=(\omega _0+sH)^3$ on the band and $\omega _0^3=(\omega _0+sH)^3$ elsewhere. Thus its cube is everywhere the measure $g\omega _ X^3$ of (14.27.2). Finally, normalize by subtracting $\sup _ X\widetilde u_\eta $.

Distinct values of $\eta $ give different slopes at $r=0$, so the corresponding potentials do not differ by a constant, and their normalized versions are distinct. This gives an uncountable family. The background is that of theorem 14.26.1, so it has bounded mass. $\square$