I am looking at the following article: on page 5 the authors define a conformal flow of metrics, one of the conditions of which is that
$\Delta_{g_t} v_t =0 \: \: \text{on} \:\: M_t.$
Later on page 11 as an explicit example, the authors give as an example
$v_t = \frac{-e^{-t} + \frac{M}{2r}e^t}{e^{-t} + \frac{M}{2r}e^t}.$
I am not sure what I am missing here, but I don't see how the Laplace-Beltrami operator of this function would vanish to zero. One can compute for example
$\frac{\partial^2 v_t}{\partial x^2} + \frac{\partial^2v_t}{\partial y^2} + \frac{\partial^2 v_t}{\partial z^2} = \frac{16 M e^{2t}}{(2 r + M e^{2t})^3},$
as only some of the terms cancel out. Are we assuming that certain terms vanish? The metric for the Laplace-Beltrami operator is $g_t = u_t^4 g$, where in this explicit example $g$ is the spatial Schwarzchild metric in Cartesian coordinates (equation 3.13) and $u_t$ is given in equation 3.14
$u_t = \frac{e^{-t} + \frac{M}{2r}e^t}{1+ \frac{M}{2r}}.$
However, I still don't see how there is supposed to be a cancellation of terms to zero if one considers the Laplace-Beltrami operator as the trace of the Hessian tensor.