to be given by $$\delta \mathcal{L}_{CFT} = -\frac{1}{16\pi \,G_{d+1}\, } \, \left(\hat{J}_\phi\,\hat{\phi}- \frac{1}{2(2\nu)\, r_D^{2\nu}}\, \hat{\phi}^2 \right) \propto \hat{J}_\phi \, {\cal O}_\Phi - \frac{(16\pi \,G_{d+1})}{2(2\nu)\, r_D^{2\nu}}\, {\cal O}_\Phi^2$$ which happens to be an irrelevant double-trace deformation [@Witten:2001ua; @Berkooz:2002ug] of the boundary CFT. Hence, at least in this simple setup the dual of the Dirichlet problem is to make the source of a primary ${\cal O}_\Phi$ a particular joint function of the vev of the primary in the given state and another fixed (state-independent) auxiliary source.

### The $k\neq0$ case : Derivative expansion up to $k^2$

Having seen the result for the translationally invariant case $k=0$, we now proceed with $k \neq 0$. It is well known that general solution to the wave equation is given in terms of Bessel functions which we parameterize as[^16] $$\Phi_k(r) = \frac{\phi_k}{r^\Delta}\times\frac{\Gamma(\nu)}{2(k/2r)^{\nu}}I_{\nu}(k/r) +  r^{\Delta-d} (J_\phi)_k  \times \frac{2(k/2r)^{\nu}}{\Gamma(\nu)}K_{\nu}(k/r)$$ Note that our previous result for $k=0$ follows from just keeping the leading $x^0$ terms in the expansions $$\begin{split}
\frac{2x^{\nu}}{\Gamma(\nu)}\;K_{\nu}(2x) &= \sum_{j=0}^{\infty} \frac{\Gamma(\nu-j)}{\Gamma(\nu)} \frac{(-x^2)^j}{j!} +x^{2\nu}\sum_{j=0}^{\infty} \frac{\Gamma(-\nu-j)}{\Gamma(\nu)} \frac{(-x^2)^j}{j!}\\
\frac{\Gamma(\nu)}{2x^{\nu}}I_{\nu}(2x) &=\sum_{j=0}^{\infty} \frac{\Gamma(\nu)}{(2\nu+2j)\Gamma(\nu+j)} \frac{x^{2j}}{j!}
\end{split}
\label{eq:expn}$$ For a general $k$, we can repeat the analysis of the previous section. While this can be done generally at all orders in $k$ with some 