Green's function of the biharmonic operator with symmetry n one direction

Question

This question is somehow related to this one. Knowing that $F(r)$ has symmetry respect to x and y-direction in Cartesian coordinate, Could any one find $G(r)$ from below equation?

$$\nabla^2 \nabla^2 G(r) = F(r)$$

Answer 1

As you have the symmetry in the $x-y$ plane, you might want to use the cylindrical coordinates. In those coordinates, $F$ is independent of angles due to the symmetry property. Now let $\nabla^{2}G=S$. Then $$\nabla^{2}S=F(\rho, z)$$ Or, using the Green's function method you have $$S(\rho,\varphi, z)=\int\mathcal{G}(\rho, \varphi, z, \rho', \varphi', z')F(\rho', z')d\rho'd\varphi'dz'$$ Where $$\nabla^{2}\mathcal{G}=\delta(\rho-\rho')\delta(z-z')\delta(\varphi-\varphi')$$ Then you also have $$\nabla^{2}G=S$$ I.e. $$G(\rho,\varphi, z)=\int\mathcal{G}(\rho, \varphi, z, \rho'', \varphi'', z'')S(\rho'', z'')d\rho''d\varphi''dz''$$ This is $$G(\rho,\varphi, z)=\int\int\mathcal{G}(\rho, \varphi, z, \rho'', \varphi'', z'')\mathcal{G}(\rho'', \varphi'', z, \rho', \varphi', z')F(\rho', \varphi'){d\rho'd\varphi'dz'}{d\rho''d\varphi''dz''}$$ The Green's function is known to be $$\mathcal{G}(\rho, \varphi, z, \rho', \varphi', z')=\frac{1}{2\pi^{2}}\sum_{m\in\mathbb{Z}}\int_{\mathbb{R}^{+}}I_{m}(k\rho)K_{m}(k\rho')e^{im(\varphi-\varphi')}\cos(k(z-z'))dk$$