Let $\Delta:=\partial_z\,\partial_{\overline {z}} $ be the Laplacian operator. I look for a particular non-trivial solution $u$ of $$\Delta u=\frac{a}{1-|z|^2}u$$ where $u\in C^2(\mathbb{D})$ and $a\in\mathbb{C} $.
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4$\begingroup$ Letting $u=f(r)$ be radial, your PDE becomes the ODE $f''+\frac{1}{r}f'-\frac{a}{1-r^2}f=0$. $\endgroup$– DispersionCommented Jan 11, 2023 at 16:27
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4$\begingroup$ Wolfram gives the solution as a linear combination of the Meijer G-function $G^{2,0}_{2,2}$ and the hypergeometric function $_{2}F_1$ in the radial case. $\endgroup$– DispersionCommented Jan 11, 2023 at 16:34
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1$\begingroup$ Thank you Zachary for your answer, but I want to know, what about the tangential derivative $\partial_\theta u(re^{i\theta})$? Because the function $u$ depends on the variables $z$ and $\overline{z}$. knowing that: $$\partial_z=\frac{1}{2}e^{-i\theta}\bigg(\partial_r-\frac{i}{r}\partial_\theta\bigg)$$ and $$\partial_{\overline{z}}=\frac{1}{2}e^{i\theta}\bigg(\partial_r+\frac{i}{r}\partial_\theta\bigg).$$ So the Laplacian operator can be $$\Delta=\frac{\partial^2_r}{\partial r^2}+\frac{\partial_r}{r\partial r}+\frac{\partial^2_\theta}{\partial\theta^2}.$$ $\endgroup$– Mhamdi MedCommented Jan 11, 2023 at 22:26
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1 Answer
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The equation is happily linear, so depending on the domain you may find separable analytical solutions to the Dirichlet problem thanks to Sturm–Liouville.
For instance, let's take the domain to be a unit circle and set $u(r=1,\theta)=f(\theta)$ as boundary value. Construct the solution as
\begin{align} u(r,\theta)=\sum_{n=0}^\infty\left[A_n\cos n\theta+B_n\sin n\theta\right]\frac{R^{(a)}_n(r)}{R^{(a)}_n(1)}, \end{align}
where
\begin{align} A_n=\frac{1}{\pi}\int_0^{2\pi}d\theta\: f(\theta)\cos n\theta\:,\:\:\:B_n=\frac{1}{\pi}\int_0^{2\pi}d\theta\: f(\theta)\sin n\theta, \end{align}
and $R^{(a)}_n(r)$ satisfies
\begin{align} \frac{d^2R^{(a)}_n}{dr^2}+\frac{1}{r}\frac{dR^{(a)}_n}{dr}-\left[\frac{a}{1-r^2}+n^2\right]R^{(a)}_n=0. \end{align}
This last equation has in general two linearly independent solutions, but only one of them is well behaved as $r\rightarrow 0 $.
For $n=0$ (which is the full solution if $f(\theta)$ is constant) one has, as rightly mentioned by Zachary, one solution that can be written in terms of Meijer G-functions, but has no $r\rightarrow 0 $ limit so we exclude it. The other one is
\begin{align} R^{(a)}_0=\, _2F_1\left(-\frac{i \sqrt{a}}{2},\frac{i \sqrt{a}}{2};1;r^2\right). \end{align}
