Without a logarithmic pole, the residual background is $\beta $, which degenerates along $E$. The radial construction still gives a bounded solution, and closedness of $\beta $ gives an energy argument for its uniqueness. Put
Here $R$ is smooth and positive, $R(1)=1$, and $Q$ is smooth at both endpoints.
Theorem 9.11.43.4CE6B71B5B25 For the same cone and prescribed measure, there is a unique pair $(c,u)$, with $c>0$ and $u$ a bounded $\theta _0$-plurisubharmonic function normalized by $\sup _ Xu=0$, such that $\mathrm{MA}^{\mathrm{ext}}_{X,\theta _0}(u)=c\mu $. The scalar is $c_ b$. The smooth function $\ell +Q(s)$ on $Y$ is constant on $E$ and descends to a continuous function on $X$; its normalization with supremum zero is $u$.
Every bounded $\theta _0$-plurisubharmonic function has mass $d$. Formula (9.11.35), with $\theta _0$ in place of $\theta _ H$, holds on every resolution and common refinement. If $\Psi =0$ and $k$ is nonconstant, or if $\psi $ is nonconstant, the solution is nonconstant.
Proof.
Set $P(s)=sR(s)$ with the data in (9.11.42). Then
BB19755D6BE9
9.11.44
\begin{equation} P(s)^ n=c_ bn\int _0^ s k(r)r^{n-1}\, \mathrm{d} r,\qquad P(0)=0,\qquad P(1)=1. \end{equation}
Smoothness and positivity of $R$ follow from its integral formula. Since $R(1)=1$, $Q$ is smooth at $s=1$, and its derivative is also smooth at zero. Put $U=\ell +Q(s)$. It is smooth on $Y$ and equals $Q(0)$ on $E$. On $0<x<1$, a local potential for $\beta +\mathrm{dd}^{\mathrm{c}}U$ is $F(\rho _\psi )$, where $F(r)=h(r)+Q(h'(r))$. Differentiation gives $F'=P(s)$ and $F''=c_ bk(s)s^ n(1-s)/P(s)^{n-1}$.
For $A=\beta +\mathrm{dd}^{\mathrm{c}}U$, it follows that
9D62C6B08002
9.11.45
\begin{equation} A=P(s)\sigma _\psi +\frac{c_ b k(s)}{R(s)^{n-1}}(B_\psi -s\sigma _\psi ),\qquad A^ n=c_ b\nu . \end{equation}
This formula shows smoothness at zero directly. At infinity the horizontal coefficient is one and the transverse coefficient is positive. Thus $A$ is semipositive on $Y$ and strictly positive on $Y\setminus E$. By theorem 9.11.5, $U=u_0\circ \pi $ for some $u_0\in \mathrm{PSH}(X,\theta _0)$. The proper surjection $\pi $ is a quotient map, so continuity of $U$ makes $u_0$ continuous. Subtracting $K_ b=\sup _ Xu_0$ gives the required solution.
Let $w$ be a bounded $\theta _0$-plurisubharmonic function. Its pullback $g=w\circ \pi $ is bounded and $\beta $-plurisubharmonic. Its bounded Monge–Ampère measure gives analytic sets zero mass and pushes forward to $\mathrm{MA}^{\mathrm{ext}}_{X,\theta _0}(w)$. Integrating the bounded expansion and using that $\beta $ is closed gives $\int _ Y(\beta +\mathrm{dd}^{\mathrm{c}}g)^ n=\int _ Y\beta ^ n=d$; this forces the scalar of any solution to be $c_ b=d/M$. The resolution identity follows from theorem 9.11.26. Now suppose that $w$ solves the equation, and put $q=U-K_ b$, $D=\beta +\mathrm{dd}^{\mathrm{c}}g$ and $v=g-q$. The bounded integration-by-parts calculus of section 5.27 gives
\[ 0=\int _ Yv(D^ n-A^ n)=-\sum _{j=0}^{n-1}\int _ Y\mathrm{d}v\wedge \mathrm d^{\mathrm c}v\wedge D^ j\wedge A^{n-1-j}. \]
Each integral in the sum is nonnegative, so all of them vanish; in particular, so does the one with the smooth factor $A^{n-1}$. In the weak-gradient interpretation, strict positivity of $A$ on each relatively compact chart of $Y\setminus E$ forces $\mathrm{d} v=0$ there. Since $Y\setminus E$ is connected, $v$ is constant almost everywhere on it. The divisor $E$ has volume zero, so lemma 5.24.3 gives equality everywhere on $Y$. Surjectivity of $\pi $ and the normalization give $w=u$ on $X$.
If $\Psi =0$ and the solution is constant, then $U=Q(x)$ is constant, so $Q'=0$ and $R=1$. Differentiating (9.11.44) then gives $c_ bk=1$, so $k$ is constant. If $\psi $ is nonconstant, the restriction of $U$ to the infinity section is $\psi $, so the solution is nonconstant for every $k$.
$\square$
Example 9.11.46.659B6A5D3749 Without the assumption $\Psi =0$, a nonconstant $k$ does not force a nonconstant solution. Take $\Psi =\log 2$ and $k(s)=2/(2-s)^{n+1}$. Then $s=2x/(1+x)$, $\ell =\log (1+x)$ and
\[ \frac{B_\psi ^ n}{\beta ^ n}=\frac{s^ n(1-s)}{x^ n(1-x)}=\frac{(2-s)^{n+1}}2. \]
Thus $\nu =\beta ^ n$. Since $\frac{\mathrm{d}}{\mathrm{d} s}(s/(2-s))^ n=nk(s)s^{n-1}$, one has $I=1/n$, $M=d$ and $c_ b=1$. The formulas (9.11.42) give $R=1/(2-s)$ and $Q(s)=\log (2-s)$, hence $U=\log 2$. The normalized solution is zero, although $k$ is nonconstant.
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