Poisson equation using complex analytic method

Question

I have the following question in a complex analysis text:

Find a particular solution to the following Poisson equation:

$$\nabla^2u(r, \theta) = r^2 \cos \theta.$$

The solution method outlined in the text uses Wirtinger derivatives to simplify the equation, then integrate twice. So here we would have:

$$4 \frac{\partial^2u}{\partial z \partial \bar{z}} = z \bar{z} \cos (\arg z).$$

Integrating with respect to $\bar{z}$ would then give:

$$4 \frac{\partial u}{\partial z} = z \left( \frac{\bar{z}^2}{2} \right) \cos (\arg z) + f(z).$$

Now, here I'm stuck. I can't get this integration to work, in part because I'm not sure you can integrate $\arg(z)$. Any pointers on where I'm going wrong would be greatly appreciated. Thanks in advance.

Answer 1

If $u$ is a solution to this one and $v$ is a solution to $\nabla^2 v(r,\theta) = r^2 \sin \theta)$, then $w = u + i v$ satisfies $$ 4 \dfrac{\partial^2 w}{\partial z \partial \overline{z}} = \nabla^2 w = r^2 \exp(i\theta) = r z= z^{3/2} \overline{z}^{1/2}$$ Integrate with respect to $z$ and $\overline{z}$, and you find one solution is $$ w = \frac{z^{5/2} \overline{z}^{3/2}}{15} = \frac{r^{3} z}{15} = \frac{r^4}{15} \exp(i\theta)$$ So taking the real part, $$u = \frac{r^4}{15} \cos(\theta)$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\cos\pars{\theta} = {x \over r} = {z + \bar{z} \over 2\root{\vphantom{\large a^{a}}z\bar{z}}}\quad}$ and $\ds{\quad\mrm{u}\pars{r,\theta} = \mrm{U}\pars{r\expo{\ic\theta},r\expo{-\ic\theta}}}$.

\begin{align} 4\,{\partial^{2}\mrm{U}\pars{z,\bar{z}} \over \partial z\,\partial\bar{z}} & = z\bar{z}\,{z + \bar{z} \over 2\root{\vphantom{\large a^{a}}z\bar{z}}} = {1 \over 2}\pars{z^{3/2}\,\bar{z}^{1/2} + z^{1/2}\,\bar{z}^{3/2}} \\[5mm] \partiald{\mrm{U}\pars{z,\bar{z}}}{z} & = {1 \over 8}\pars{z^{3/2}\,{2 \over 3}\,\bar{z}^{3/2} + z^{1/2}\,{2 \over 5}\,\bar{z}^{5/2}} + \mrm{f}\pars{z} \\ \mrm{U}\pars{z,\bar{z}} & = {1 \over 8}\pars{{2 \over 5}\,z^{5/2}\,{2 \over 3}\,\bar{z}^{3/2} + {2 \over 3}\,z^{3/2}\,{2 \over 5}\,\bar{z}^{5/2}}\ +\ \overbrace{\int\mrm{f}\pars{z}\dd z}^{\ds{\equiv\ \mrm{F}\pars{z}}} \\[5mm] \mrm{U}\pars{z,\bar{z}} & = {1 \over 15}\,\Re\pars{z^{5/2}\,\bar{z}^{3/2}} + \mrm{F}\pars{z} = {1 \over 15}\,\pars{z\,\bar{z}}^{3/2}\,\Re\pars{z} + \mrm{F}\pars{z} \\[2mm] & = \bbx{{1 \over 15}\,r^{4}\cos\pars{\theta} + \mrm{F}\pars{r\expo{\ic\theta}}} \end{align}