9ABE80D68F2B
Infinite-order vanishing of the density is not needed for smooth nonuniqueness. We change the coefficients of $\omega _0$ so that its coefficient of $R$ stays positive and one of its coefficients of $P$ and $Q$ vanishes quadratically at each of the two hypersurfaces $M_\pm $. The resulting singular differential equation forces the slopes of differences of solutions to decay like $\exp (-A/d)$, where $d$ is the distance to the hypersurface and $A>0$. The slopes therefore extend smoothly by zero, even though the density vanishes only to finite order.
Theorem 14.28.1.9C11A9E88B60 On $\mathbb P^3$ there exist a smooth nonclosed Hermitian form $\omega $ and a smooth nonnegative density $g$ such that:
The zero set of $g$ consists of two disjoint smooth real hypersurfaces. Near either hypersurface, $g$ is the square of a smooth defining function times a smooth strictly positive function.
The equation $(\omega +\mathrm{dd}^{\mathrm{c}}u)^3=g\omega _ X^3$ has uncountably many distinct smooth $\omega $-plurisubharmonic solutions normalized by $\sup _ Xu=0$.
The background $\omega $ has bounded mass.
We keep the rank-one forms $P,Q,R$ and the function $r$ of 14.26.2, and use new scalar coefficients.
Choose a smooth function $a$ valued in $[0,1]$, equal to one for $r\leq -2/3$ and zero for $r\geq -1/3$, and put $b(r)=a(-r)$. Choose a smooth even cutoff $\chi $ valued in $[0,1]$, zero for $|r|\leq 5/4$ and one for $|r|\geq 3/2$. Set $p(r)=(1-r^2)^2$ and
ACD0920D4D78
14.28.2
\begin{equation} \begin{split} x& =(1-\chi )(p+a)+\chi \sigma ,\\ y& =(1-\chi )(p+b)+\chi (1-\sigma ),\\ z& =1-\chi +\chi \sigma ',\qquad \omega _0=xP+yQ+zR. \end{split} \end{equation}
Since these coefficients coincide with those of (14.26.6) for $|r|\geq 3/2$, the form $\omega _0$ extends by $\omega _ X$ across the two omitted projective lines. The coefficient $z$ is strictly positive, and $x,y$ are strictly positive except that $y(-1)=0$ and $x(1)=0$. Thus $\omega _0$ is positive off $M_\pm =\{ r=\pm 1\} $ and has rank two there.
Choose a smooth cutoff $\kappa \in C_ c^\infty ((-2,2))$ equal to one on $[-3/2,3/2]$. For a parameter $\varepsilon >0$ to be chosen, define the correction
\[ \rho =\varepsilon (r^2-1)\kappa (r),\qquad \omega =\omega _0+\mathrm{dd}^{\mathrm{c}}\rho . \]
For sufficiently small $\varepsilon >0$, the form $\omega $ is strictly positive. Indeed, on $[-3/2,3/2]$ its coefficients are $x+2\varepsilon r$, $y-2\varepsilon r$ and $z+2\varepsilon $. The first is positive for $r\geq 0$, and on $[-3/2,0]$ the function $x$ has a positive minimum, so it remains positive for small $\varepsilon $. Reflection gives the positivity of the second coefficient, and that of the third is immediate. On $3/2\leq |r|\leq 2$, the form $\omega _0$ is strictly positive on a compact set and the correction is $O(\varepsilon )$. Near the projective ends there is no correction. The form $\omega $ is not closed: at $r=0$, $x'=y'=0$ and $z=1$, so by (14.26.11)
\[ \mathrm{d}\omega =\mathrm{d}r\wedge \bigl ((x'-z)P+(y'+z)Q\bigr )\ne 0. \]
Define the density $g$ by
\[ g\omega _ X^3=\omega _0^3. \]
It is smooth and positive exactly off $M_-\cup M_+$. By (14.26.6) and (14.26.4), $g=xyz/(\sigma (r)(1-\sigma (r))\sigma '(r))$. Near either hypersurface, $xyz=(1+p)p$, and hence
\[ g(r)=\frac{(1+p(r))(1-r^2)^2}{\sigma (r)(1-\sigma (r))\sigma '(r)}. \]
The denominator is smooth and strictly positive at $\pm 1$, and $1-r^2$ is a defining function for $M_\pm $ near these hypersurfaces. This proves the asserted quadratic vanishing. The potential $-\rho $ is one smooth solution.
Lemma 14.28.3.624B2CF7FBA2 For every sufficiently small $\eta >0$, the equation
3D1CCE4952BA
14.28.4
\begin{equation} h'=\frac{xyz}{(x+h)(y-h)}-z,\qquad h(0)=-\eta \end{equation}
has a smooth solution on $(-1,1)$ with $-x<h<y$ and $h<0$. There are a constant $0<c<1$ and fixed neighborhoods of both endpoints, independent of $\eta $, on which
\[ -cp<h<0. \]
Every such solution, extended by zero, is a smooth function on $\mathbb R$ that vanishes to infinite order at $\pm 1$.
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
\[ F(r,-cp)+cp' =\frac{1+p}{(1+(1-c)p)(1+c)}-1+cp' \longrightarrow -\frac{c}{1+c}<0, \]
whereas near the right endpoint
\[ F(r,-cp)+cp' =\frac{1+p}{(1-c)(1+(1+c)p)}-1+cp' \longrightarrow \frac{c}{1-c}>0. \]
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
86148D5BD1E9
14.28.5
\begin{equation} h'=h\, L(r,h),\qquad L(r,h)=\frac{z(x-y+h)}{(x+h)(y-h)}. \end{equation}
Since $h<0$, we have $(\log (-h))'=L(r,h)$. Near the left endpoint, the trapping bounds and the explicit coefficients imply
\[ L(r,h)=\frac{1+h}{(1+p+h)(p-h)}\geq \frac{C_1}{p} \]
after shrinking that neighborhood, with $C_1>0$. Near the right endpoint they give
\[ L(r,h)=\frac{-1+h}{(p+h)(1+p-h)}\leq -\frac{C_2}{p}, \qquad C_2>0. \]
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 $A>0$,
94F5CC68CD22
14.28.6
\begin{equation} |h(r)|\leq C \mathrm{e}^{-A/d} \end{equation}
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
\[ |h^{(j)}(r)|\leq C_ j \mathrm{e}^{-A/d}d^{-N_ j}. \]
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 $\pm 1$.
$\square$
of theorem 14.28.1.
Let $\eta _0>0$ be such that lemma 14.28.3 applies for $0<\eta <\eta _0$, and let $h_\eta $ denote the solution with $h_\eta (0)=-\eta $. Take a primitive $G_\eta $ of its extension by zero. Since $G_\eta $ is constant near each end of the real line, $G_\eta (r)$ defines a smooth function on $X$. The potential $u_\eta =G_\eta (r)-\rho $ satisfies
\[ \omega +\mathrm{dd}^{\mathrm{c}}u_\eta =(x+h_\eta )P+(y-h_\eta )Q+(z+h_\eta ')R. \]
Inside the band, the first two coefficients are positive by the lemma, and the equation gives
\[ z+h_\eta '=\frac{xyz}{(x+h_\eta )(y-h_\eta )}>0. \]
Outside the band the form is $\omega _0$, and on the two boundary hypersurfaces it is also the rank-two form $\omega _0$, since $h_\eta $ vanishes to infinite order there. Thus it is semipositive on $X$, and by (14.26.4) its cube is $6xyzP\wedge Q\wedge R=\omega _0^3=g\omega _ X^3$. Different initial values give potentials that do not differ by constants, so after normalization by the supremum we obtain an uncountable family. Bounded mass follows from proposition 13.14.15, as for theorem 14.26.1.
$\square$
At each of the hypersurfaces $M_\pm $, the difference of two of these solutions agrees to infinite order with a constant, because its derivative is a difference of slopes that vanish to infinite order there. The constants at the two hypersurfaces need not agree. Thus the singular differential equation produces differences that are flat at $M_\pm $, even though the density has only quadratic zeros.
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