e^{-k^ 2 z^ 2 } f(z)  } { z}  \partial_z V_i (q,z) \Big) &+&
 \frac{ e^{-k^ 2 z^ 2 } }{ z } \Big(   \frac{\omega^ 2 }{ f(z) }  - \vert {\vec q}\vert^2 \Big)  
   V_i (q,z) \,=\, 0  \,.
       \label{EOMfiniteT}\end{aligned}$$

It is convenient [@Grigoryan:2010pj] to choose the momentum $q^\mu = (\omega , \vec 0 ) $ where the transversality of the current $ q^\mu J_\mu = 0 $ translates into the vanishing of the temporal component $J_0$. Then we just need to solve the equation for the spatial component: $ V_i (\omega ,z) \,=\, v (\omega ,z) V^0_i ( \omega ) $. In this case $ v (\omega ,z) $ satisfies the equation:

$$\partial_z \Big( \frac{ e^{-k^ 2 z^ 2 } f(z)  } { z}  \partial_z v (\omega ,z)  \Big) +
 \frac{ e^{-k^ 2 z^ 2 } }{ z }   \frac{\omega^ 2 }{ f(z) }    
   v (\omega ,z) \,=\, 0 \,.
   \label{eqmotionbtb}$$

The bulk to boundary propagator has to satisfy two boundary conditions. One is $$v (\omega, z=z_0) = 1\,,
\label{conditionbtb}$$ that was present in the zero temperature case and implies that the field components work as the sources of the correlation functions at $z= z_0$. The other is the 