How to derive that after stereo-graphical projection, $\Delta u$ in $\mathbb{R}^n$ is transformed to
$$
\Delta\_{\mathbb{S}^n}u - \frac{n(n-2)}{4}u\ \text{in}\ \mathbb{S}^n.
$$
To be more precise, in this [paper](https://pdf.sciencedirectassets.com/272601/1-s2.0-S0022123610X00052/1-s2.0-S0022123609005114/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEDYaCXVzLWVhc3QtMSJHMEUCIFbXMf27WHEDMDXJGUvxxDCMyxaXJ4xeMrZK%2BDRCzhsSAiEAjh42q5rClB4mJ8%2FRHiv9Ke%2BIKa%2BkbwxotqzQ8yjf6j8q1QQIvv%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FARAFGgwwNTkwMDM1NDY4NjUiDIOTGR6hKf3HDGSdXiqpBMw5bPgD%2BWVLLv9a8PEZJsH%2BfwSTMx1Mzem2AXWpeMcZXy1yvUiBvqLDUFXJQn5KWtyo1AAV4EMkqeazJFtQL46RlRR7LdYUYmjGqOrB6q02wyYTqZVwh%2BlPpGF2Y7Ddr2YFzkcRCehhdCs7qAXfIn%2F8HIcD7P1w0pAVjsTcyPSx0BxctjlzYY%2FJTAE%2BMSjS65TAPyV7HpomETNJgpiwUTeWPQtteGhkExlU4xJyV6ERaRapeiW8JeYfZQUPR63l9wnjL4EBS6MeMXcVXHXnkQqUGVdNdRsUszNfksDz%2Bfe6Fx3OcBgLBbOpLE0zVS8vd9yNxDmWXyiUREhcELil2LcM9enRkpPvCr0k1qmtNgFA%2F1Sh3VlHO22FAHjcgLTtqb0dJtu%2BG%2FP0qWRdKdte44bOyN60CUKSZKl2qiRhX5A%2F7qGmaSYAF3GRgQdXz883O6xxa7NBQTO1DaOUrRNkHZ6LuwVvo0Lm7R96P5cxJIzw%2F3tqfRyRM3g28Gw09z7JX0EQPZG5cnCZr%2Fum14Oi83Es5HJf4PcPDaxC%2B3Cxl%2Ff8poNUFYhClYFm2Gf5hDdQUPgKanv13xfWqzMjLa21ruU4wiQ0GisDRuUCcmEpdBwwHSjYkmLcXUSreUkXhOyxPMD0GSmGQQzuEACTUCccWHB8wEweC52B569tJpx7mHxgAShcjWdsSIWsMN5B4yXc%2Bk2I9%2FfFjSFAoLAT6Z1MZ6QxqXNtQHfoyygwi8bSmAY6qQEe6sYbufDn8fQ3McnKb4rg1orNYcaILKZxUGjqP2KTZMbEXfffM32VjNHXXfEJlJmUnC1rdC7O%2FxUSAV8XAKH7un5KamQpZjoq8itiEbaZ%2FfriFWhYoXORsOeGvVJI4KVhcevTHmzl5pbmvIp69PQPNt5DZpWUXNpzjPUv0KrklyQy5SagKheWv8SDZRYnhqtIakiiby1d9DuOHeA4eFkbn7AB6hqd8iEO&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20220904T135030Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYXMPUFFST%2F20220904%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=325680fa17a0b4af2bcae827e53cec8582bfc80afbadbc688fbfb3332beacaa4&hash=6b0e2422a57f2bbdbf93bdf9474dce3953b63973b546dec6bb3204d69a8edcac&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0022123609005114&tid=spdf-4aca3989-b294-4b32-9a25-89d860a8ba3c&sid=d209ff5679cc14414928ebe4e4a1dd0b985egxrqa&type=client&ua=5152545955500657510c&rr=745731d14c25afc3), it is asserted that the equation
$$-\Delta\_{\mathbb{S}^n}\,u+\frac{n(n-2)}{2}u-\tilde{K}u^{\frac{n+2}{n-2}}=0\quad\text{on}\ \mathbb{S}^n$$
can be reduced to
$$-\Delta u = K(y)u^{\frac{n+2}{n-2}}\quad\text{in}\ \mathbb{R}^n.$$


I need some tips or relevant references.



Laplacian on sphere after stereographic projection
the relation between the two Laplacians is a bit more complicated. I follow the survey [paper](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-17/issue-1/The-Yamabe-problem/bams/1183553962.full) of Lee and Parker on the Yamabe problem.


The conformal Laplacian of a Riemannian metric $g$ is defined as
$$
L\_g = 4 \frac{n-1}{n-2} \Delta\_g + S\_g
$$
where $S\_g$ is the scalar curvature of the metric and $\Delta\_g = d\_g^\* d$ is the Hodge (positive) Laplacian. The main point is that if we do a conformal change of the metric $\tilde{g} = \varphi^2 g$ (where $\varphi>0$ is a smooth positive function) then
\begin{equation}\label{eq:conflap}
 L\_g (u) = \varphi^{\frac{n+2}{2}} L\_{\tilde{g}} (\varphi^{\frac{2-n}{2}} u) .
\end{equation}
The metric of the unit sphere in stereographic projection coordinates is
\begin{equation\*}
 g\_{S^n} = \varphi^2 g\_{\mathbb{R}^n} \hspace{2mm} \text{ with } \varphi = \frac{2}{(1+|x|^2)}.
\end{equation\*}
Using the equation for the conformal laplacian and that the scalar curvature of $S^n$ is $n(n-1)$ we obtain
\begin{equation}
\Delta\_{\mathbb{R}^n} u = \varphi^{\frac{n+2}{2}} 
\left(
\Delta\_{S^n} + \frac{n(n-2)}{4}
\right)
(\varphi^{\frac{2-n}{2}} u) .
\end{equation}


On the other hand, the conformal Laplacian relates to prescribing scalar curvature in conformal classes as follows.
Let $K$ be a given function,
then the metric $\tilde{g} = u^{4/(n-2)} g$ has scalar curvature equal to $K$ if and only if $u$ solves the PDE
\begin{equation\*}
 L\_g (u) = u^{\frac{n+2}{n-2}} K .
\end{equation\*}
In the case of the sphere, the metric $\tilde{g} = u^{4/(n-2)} g\_{S^n}$ has scalar curvature equal to $K$ if and only if
\begin{equation}\label{eq1}
 \Delta\_{S^n} u + \frac{n(n-2)}{4} u = u^{\frac{n+2}{n-2}} \tilde{K}
\end{equation}
where $\tilde{K}= \frac{n-2}{4(n-1)} K$. At the same time, $\tilde{g}= \tilde{u}^{4/(n-2)} g\_{\mathbb{R}^n}$ where $\tilde{u} = \varphi^{\frac{n-2}{2}} u$ with $\varphi = 2 (1+|x|^2)^{-1}$. We conclude that $u$ solves the above equation if and only if $\tilde{u}$ solves


\begin{equation}
 \Delta\_{\mathbb{R}^n} \tilde{u} = \tilde{u}^{\frac{n+2}{n-2}} \tilde{K} .
\end{equation}


