Inhomogeneous Dirichlet Problem in $\mathbb{R}^d$ Let $\mathcal B=\{x\in\mathbb R^d\,\,:\,\|x\|_2 \leq 1\}$
be the unit disk in $\mathbb{R}^d$,  assume that $d>2$ and find the continuous function $U : \mathbb{R}^d \to \mathbb{R}$ solving the following Dirichlet problem:
$$
\begin{cases}
ΔU(x) = C_d \sqrt{\frac{1}{\| x \|_2^{d-2}} - 1} & \text{ if }x\in\mathcal{B}\\
U = 0 & \text{ if }x\notin\mathcal{B},
\end{cases}
$$
where $C_d > 0$ is some positive constant and $Δ$ is the Laplacian. I would need to show that $U$ exists and that it satisfies:
$$
0 < U(x) < \frac{1}{\| x \|_2^{d-2}} - 1 \quad \mbox{ if } x \in \mathcal{B}.
$$
 How can the solution be characterised? (Perhaps can it be useful moving to spherical coordinates reducing the problem to a one dimensional one?)
 A: $\require{extpfeil}\def\R{\mathbb{R}}\def\d{\mathrm{d}}\def\Β{\mathrm{Β}}$The uniqueness of the solution to the equation can be readily proved by virtue of that of a Poisson equation as long as $U \in C^2(\R^d \setminus \{0\})$, so it remains to find one solution.
Define $f_d(r) = C_d \sqrt{\dfrac{1}{r^{d - 2}} - 1}$ ($r \in (0, 1]$) and assume that there exists $u \in C^2(\R_+)$ such that $U(x) = u(\|x\|)$ is a solution. Since for $1 \leqslant k \leqslant d$ and $x \in \R^d \setminus \{0\}$,\begin{gather*}
\frac{∂U}{∂x_k}(x) = u'(\|x\|) \frac{∂(\|x\|)}{∂x_k} = \frac{x_k}{\|x\|} u'(\|x\|),\\
\frac{∂^2 U}{∂x_k^2}(x) = \frac{∂}{∂x_k}\left( \frac{x_k}{\|x\|} u'(\|x\|) \right) = \frac{x_k^2}{\|x\|^2} u''(\|x\|) + \frac{\|x\|^2 - x_k^2}{\|x\|^3} u'(\|x\|),
\end{gather*}
plugging the above into the equation yields\begin{gather*}
\begin{cases}
u''(r) + (d - 1) \dfrac{u'(r)}{r} = f_d(r), & 0 < r \leqslant 1\\
u(r) = 0, & r > 1
\end{cases}.\tag{1}
\end{gather*}
Note that $u \in C^2(\R_+)$ implies $u(1) = u'(1) = 0$, thus the solution to (1) can be solved to be\begin{gather*}
u(r) = \begin{cases}
\displaystyle\int_r^1 \frac{1}{y^{d - 1}} \,\d y \int_y^1 z^{d - 1} f_d(z) \,\d z; & 0 < r < 1\\
0; & r \geqslant 1
\end{cases}.\tag{2}
\end{gather*}
Therefore, the solution is $U(x) = u(\|x\|)$, where $u$ is defined by (2).
Now, note that\begin{gather*}
\int_0^1 z^{d - 1} f_d(z) \,\d z = C_d \int_0^1 z^{d - 1} \sqrt{\frac{1}{z^{d - 2}} - 1} \,\d z\\
\xlongequal{w = z^{d - 2}} \frac{C_d}{d - 2} \int_0^1 w^{\frac{2}{d - 2} - \frac{1}{2}} \sqrt{1 - w} \,\d w = \frac{C_d}{d - 2} \Β\left( \frac{2}{d - 2} + \frac{1}{2}, \frac{3}{2} \right) := \widetilde{C}_d,
\end{gather*}
thus for $x \in B^\circ$,$$
0 < U(x) = u(\|x\|) = \int_{\|x\|}^1 \frac{1}{y^{d - 1}} \,\d y \int_y^1 z^{d - 1} f_d(z) \,\d z\\
< \widetilde{C}_d \int_{\|x\|}^1 \frac{1}{y^{d - 1}} \,\d y = \frac{\widetilde{C}_d}{d - 2} \left( \frac{1}{\|x\|^{d - 2}} - 1\right),
$$
and the inequality holds if $\widetilde{C}_d \leqslant d - 2$, i.e.$$
C_d \leqslant \frac{(d - 2)^2}{\Β\left( \dfrac{2}{d - 2} + \dfrac{1}{2}, \dfrac{3}{2} \right)}.
$$
