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Consider the following equation $-\Delta_{\mathbb{S}^n} u = u$ where the $(\mathbb{S}^n,g)$ where $g$ is the usual metric induced by the inverse stereographic projection $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $$g = \frac{4}{(1+|x|^2)^2} g_{\mathbb{R}^n},$$ where $$S(x) = \left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right).$$ Then consider $\phi(x) = u(S(x))$ where $\phi:\mathbb{R}^n\to \mathbb{R.}$ I wonder what is the equation satisfied by the function $\phi?$

My attempt: If we set $s=S(x)$ then first $-\Delta_{g} u(s) = u(s).$ Then following the equation below (1) in this link which states that for $\tilde{g}=e^{2f}g$ we have $$\Delta_{\tilde{g}}=e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i},$$ where $\tilde{g} = g_{\mathbb{S}^n}$, $g=g_{\mathbb{R}^n}$, $f=\ln\left(\frac{2}{1+|x|^2}\right)$ we get that with $s=S(x)$, $$\Delta_{\tilde{g}} u(s)=\left(\frac{1+|x|^2}{2}\right)^2\Delta_g u(S(x))+(n-2)\left(\frac{1+|x|^2}{2}\right)\nabla x \cdot \nabla u(S(x)).$$ Thus setting $\phi(x)=u(S(x))$ and using $-\Delta u (s) = u(s)$ we have that $$-\Delta_{\mathbb{R}^n} \phi(x) = \frac{2(n-2)}{(1+|x|^2)}x\cdot \nabla \phi(x)+\frac{4}{(1+|x|^2)^2}\phi(x).$$ Is this the right expression and if so, then is there a way to get rid of the term $x\cdot \nabla \phi?$

Edit 2: If we set $$F(x) = \sqrt{J_S(x)} f(S(x))$$ where $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $S(x)=\left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right)$ and the jacobian of the map $J_{S}(x) = \left(\frac{2}{1+|x|^2}\right)^{n}.$ Then since, $$\int_{\mathbb{R}^n}|\nabla F|^2 dx = \int_{\mathbb{S}^n} |\nabla f|^2 + \frac{n(n-2)}{4}|f|^2 d\omega$$ I guess we have that, $$-\Delta F(x) = -\Delta f(S(x)) + \frac{n(n-2)}{4} f(S(x)).$$ Is this a correct way to proceed?

Consider the following equation $-\Delta_{\mathbb{S}^n} u = u$ where the $(\mathbb{S}^n,g)$ where $g$ is the usual metric induced by the inverse stereographic projection $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $$g = \frac{4}{(1+|x|^2)^2} g_{\mathbb{R}^n},$$ where $$S(x) = \left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right).$$ Then consider $\phi(x) = u(S(x))$ where $\phi:\mathbb{R}^n\to \mathbb{R.}$ I wonder what is the equation satisfied by the function $\phi?$

My attempt: If we set $s=S(x)$ then first $-\Delta_{g} u(s) = u(s).$ Then following the equation below (1) in this link which states that for $\tilde{g}=e^{2f}g$ we have $$\Delta_{\tilde{g}}=e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i},$$ where $\tilde{g} = g_{\mathbb{S}^n}$, $g=g_{\mathbb{R}^n}$, $f=\ln\left(\frac{2}{1+|x|^2}\right)$ we get that with $s=S(x)$, $$\Delta_{\tilde{g}} u(s)=\left(\frac{1+|x|^2}{2}\right)^2\Delta_g u(S(x))+(n-2)\left(\frac{1+|x|^2}{2}\right)\nabla x \cdot \nabla u(S(x)).$$ Thus setting $\phi(x)=u(S(x))$ and using $-\Delta u (s) = u(s)$ we have that $$-\Delta_{\mathbb{R}^n} \phi(x) = \frac{2(n-2)}{(1+|x|^2)}x\cdot \nabla \phi(x)+\frac{4}{(1+|x|^2)^2}\phi(x).$$ Is this the right expression and if so, then is there a way to get rid of the term $x\cdot \nabla \phi?$

Edit 2: This is something that I came across Lieb's book on Analysis where he defines $$F(x) = \sqrt{J_S(x)} f(S(x))$$ where $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $S(x)=\left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right)$ and the jacobian of the map $J_{S}(x) = \left(\frac{2}{1+|x|^2}\right)^{n}.$ Then since, $$\int_{\mathbb{R}^n}|\nabla F|^2 dx = \int_{\mathbb{S}^n} |\nabla f|^2 + \frac{n(n-2)}{4}|f|^2 d\omega$$ I guess we have that, $$-\Delta F(x) = -\Delta f(S(x)) + \frac{n(n-2)}{4} f(S(x)).$$ Is this a correct way to proceed?

Consider the following equation $-\Delta_{\mathbb{S}^n} u = u$ where the $(\mathbb{S}^n,g)$ where $g$ is the usual metric induced by the inverse stereographic projection $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $$g = \frac{4}{(1+|x|^2)^2} g_{\mathbb{R}^n},$$ where $$S(x) = \left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right).$$ Then consider $\phi(x) = u(S(x))$ where $\phi:\mathbb{R}^n\to \mathbb{R.}$ I wonder what is the equation satisfied by the function $\phi?$

My attempt: If we set $s=S(x)$ then first $-\Delta_{g} u(s) = u(s).$ Then following the equation below (1) in this link which states that for $\tilde{g}=e^{2f}g$ we have $$\Delta_{\tilde{g}}=e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i},$$ where $\tilde{g} = g_{\mathbb{S}^n}$, $g=g_{\mathbb{R}^n}$, $f=\ln\left(\frac{2}{1+|x|^2}\right)$ we get that with $s=S(x)$, $$\Delta_{\tilde{g}} u(s)=\left(\frac{1+|x|^2}{2}\right)^2\Delta_g u(S(x))+(n-2)\left(\frac{1+|x|^2}{2}\right)\nabla x \cdot \nabla u(S(x)).$$ Thus setting $\phi(x)=u(S(x))$ and using $-\Delta u (s) = u(s)$ we have that $$-\Delta_{\mathbb{R}^n} \phi(x) = \frac{2(n-2)}{(1+|x|^2)}x\cdot \nabla \phi(x)+\frac{4}{(1+|x|^2)^2}\phi(x).$$ Is this the right expression and if so, then is there a way to get rid of the term $x\cdot \nabla \phi?$

Consider the following equation $-\Delta_{\mathbb{S}^n} u = u$ where the $(\mathbb{S}^n,g)$ where $g$ is the usual metric induced by the inverse stereographic projection $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $$g = \frac{4}{(1+|x|^2)^2} g_{\mathbb{R}^n},$$ where $$S(x) = \left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right).$$ Then consider $\phi(x) = u(S(x))$ where $\phi:\mathbb{R}^n\to \mathbb{R.}$ I wonder what is the equation satisfied by the function $\phi?$

My attempt: If we set $s=S(x)$ then first $-\Delta_{g} u(s) = u(s).$ Then following the equation below (1) in this link which states that for $\tilde{g}=e^{2f}g$ we have $$\Delta_{\tilde{g}}=e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i},$$ where $\tilde{g} = g_{\mathbb{S}^n}$, $g=g_{\mathbb{R}^n}$, $f=\ln\left(\frac{2}{1+|x|^2}\right)$ we get that with $s=S(x)$, $$\Delta_{\tilde{g}} u(s)=\left(\frac{1+|x|^2}{2}\right)^2\Delta_g u(S(x))+(n-2)\left(\frac{1+|x|^2}{2}\right)\nabla x \cdot \nabla u(S(x)).$$ Thus setting $\phi(x)=u(S(x))$ and using $-\Delta u (s) = u(s)$ we have that $$-\Delta_{\mathbb{R}^n} \phi(x) = \frac{2(n-2)}{(1+|x|^2)}x\cdot \nabla \phi(x)+\frac{4}{(1+|x|^2)^2}\phi(x).$$ Is this the right expression and if so, then is there a way to get rid of the term $x\cdot \nabla \phi?$

Edit 2: If we set $$F(x) = \sqrt{J_S(x)} f(S(x))$$ where $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $S(x)=\left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right)$ and the jacobian of the map $J_{S}(x) = \left(\frac{2}{1+|x|^2}\right)^{n}.$ Then since, $$\int_{\mathbb{R}^n}|\nabla F|^2 dx = \int_{\mathbb{S}^n} |\nabla f|^2 + \frac{n(n-2)}{4}|f|^2 d\omega$$ I guess we have that, $$-\Delta F(x) = -\Delta f(S(x)) + \frac{n(n-2)}{4} f(S(x)).$$ Is this a correct way to proceed?

Consider the following equation $-\Delta_{\mathbb{S}^n} u = u$ where the $(\mathbb{S}^n,g)$ where $g$ is the usual metric induced by the inverse stereographic projection $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $$g = \frac{4}{(1+|x|^2)^2} g_{\mathbb{R}^n},$$ where $$S(x) = \left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right).$$ Then consider $\phi(x) = u(S(x))$ where $\phi:\mathbb{R}^n\to \mathbb{R.}$ I wonder what is the equation satisfied by the function $\phi?$

My attempt: If we set $s=S(x)$ then first $-\Delta_{g} u(s) = u(s).$ Then by the conformal relation of the spherical and flat metrics we observe that, $$\Delta_g u = \frac{(1+|x|^2)^n}{2^{n}}\partial_j\left(\frac{2^n}{(1+|x|^2)^n}\frac{(1+|x|^2)^2}{4}\partial_j u\right)\\ =\frac{(1+|x|^2)^n}{4}\partial_j\left(\frac{1}{(1+|x|^2)^{n-2}}\partial_j u\right)\\ =\frac{(1+|x|^2)^n}{4}\left(2(2-n)(1+|x|^2)^{1-n}x_j \partial_j u + (1+|x|^2)^{2-n}\partial_j^2 u\right)\\ =\frac{(2-n)(1+|x|^2)}{2}x_j \partial_j u+ \frac{(1+|x|^2)^{2}}{4}\partial_j^2 u.$$ Thus I am guessing that the answer should be $$\Delta_{\mathbb{R}^n} \phi(x) = \frac{2(n-2)}{(1+|x|^2)}x\cdot \nabla \phi(x)-\frac{4}{(1+|x|^2)^2}\phi(x).$$ Is this the right expression and if so, then is there a way to get rid of the term $x\cdot \nabla \phi?$

Consider the following equation $-\Delta_{\mathbb{S}^n} u = u$ where the $(\mathbb{S}^n,g)$ where $g$ is the usual metric induced by the inverse stereographic projection $S:\mathbb{R}^n\to \mathbb{S}^n$ such that $$g = \frac{4}{(1+|x|^2)^2} g_{\mathbb{R}^n},$$ where $$S(x) = \left(\frac{2x}{1+|x|^2},\frac{1-|x|^2}{1+|x|^2}\right).$$ Then consider $\phi(x) = u(S(x))$ where $\phi:\mathbb{R}^n\to \mathbb{R.}$ I wonder what is the equation satisfied by the function $\phi?$

My attempt: If we set $s=S(x)$ then first $-\Delta_{g} u(s) = u(s).$ Then following the equation below (1) in this link which states that for $\tilde{g}=e^{2f}g$ we have $$\Delta_{\tilde{g}}=e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i},$$ where $\tilde{g} = g_{\mathbb{S}^n}$, $g=g_{\mathbb{R}^n}$, $f=\ln\left(\frac{2}{1+|x|^2}\right)$ we get that with $s=S(x)$, $$\Delta_{\tilde{g}} u(s)=\left(\frac{1+|x|^2}{2}\right)^2\Delta_g u(S(x))+(n-2)\left(\frac{1+|x|^2}{2}\right)\nabla x \cdot \nabla u(S(x)).$$ Thus setting $\phi(x)=u(S(x))$ and using $-\Delta u (s) = u(s)$ we have that $$-\Delta_{\mathbb{R}^n} \phi(x) = \frac{2(n-2)}{(1+|x|^2)}x\cdot \nabla \phi(x)+\frac{4}{(1+|x|^2)^2}\phi(x).$$ Is this the right expression and if so, then is there a way to get rid of the term $x\cdot \nabla \phi?$