# Let $h:\mathbb{C}\to\mathbb{C}$ in $C^k(\mathbb{C})$ with compact support. Find solutions to the equation $f_x + if_y = h$.

Let $$h:\mathbb{C}\to\mathbb{C}$$ be a function in $$C^k(\mathbb{C})$$ and has compact support. What are the solutions in $$C^k(\mathbb{C})$$ to the equation $$\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y} = h?$$

Can this be solved without using techniques from PDE?

• I think this question should be reopened. The inhomogeneous CR equation is fundamental to all of complex analysis, and gathering many different perspectives on how to solve it in the 1 dimensional case with compact support is a valuable question. – Steven Gubkin Sep 23 '20 at 15:31

The answer is yes: the inhomogeneous Cauchy-Riemann equation can be solved by using only the theory of functions of a complex variable. However, this is not "free of charge" and the comparison of two different methods of solution solution proposed below shows this fact. The first one analyzed, according to what you asked, is based on complex variable techniques and Green's formula for planar domains, without techniques from the theory of PDEs, while the second one is based on the standard theory of distributions and thus it is based on techniques from the theory of PDEs.

Notation
Differentials and partial derivatives (Wirtinger derivatives) $$\begin{matrix} z=x+iy & \bar{z}=x-iy \\ \mathrm{d}z=\mathrm{d}x+i\mathrm{d}y & \mathrm{d}\bar{z}=\mathrm{d}x-i\mathrm{d}y\\ \dfrac{\partial f}{\partial z} = \dfrac{1}{2} \bigg(\dfrac{\partial f}{\partial x} - i\dfrac{\partial f}{\partial y}\bigg) & \dfrac{\partial f}{\partial\bar{z}} = \dfrac{1}{2} \bigg(\dfrac{\partial f}{\partial x} + i\dfrac{\partial f}{\partial y}\bigg)\\ \partial f= \dfrac{\partial f}{\partial z}\mathrm{d}z & \bar\partial f=\dfrac{\partial f}{\partial\bar z} \mathrm{d}\bar{z} \end{matrix}$$ The (multiple of the) laplacian as a product of complex partial derivatives $$\begin{split} \dfrac{\partial^2 f}{\partial z\partial\bar z}=\dfrac{\partial^2 f}{\partial\bar z\partial z}&=\dfrac{1}{4}\bigg(\dfrac{\partial}{\partial x} + i\dfrac{\partial}{\partial y}\bigg)\bigg(\dfrac{\partial}{\partial x} - i\dfrac{\partial}{\partial y}\bigg)f\\ &=\dfrac{1}{4} \bigg(\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2}\bigg)=\frac{1}{4}\Delta f \end{split}$$ From the theory of complex differential forms we can express the plane volume form as $$\begin{split} \frac{i}{2}\mathrm{d}z\wedge\mathrm{d}\bar{z}&=\frac{i}{2}\big(\mathrm{d}x\wedge\mathrm{d}x-i\mathrm{d}x\wedge\mathrm{d}y+i\mathrm{d}y\wedge\mathrm{d}x + \mathrm{d}y\wedge\mathrm{d}y\big)\\ &=\frac{i}{2}\big(-i\mathrm{d}x\wedge\mathrm{d}y+i\mathrm{d}y\wedge\mathrm{d}x \big)\\ &=\frac{i}{2}\big(-i\mathrm{d}x\wedge\mathrm{d}y-i\mathrm{d}x\wedge\mathrm{d}y\big)\\ &=\mathrm{d}x\wedge\mathrm{d}y=\mathrm{d}x\mathrm{d}y \end{split}$$ Finally, the equation we want to solve is the inhomogeneous Cauchy-Riemann equation in $$\Bbb C$$ $$\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y} = h\label{CR}\tag{CR}$$ for $$h\in C_c^k(\Bbb C)$$, $$k\in\Bbb N$$, $$k\ge 1$$: for the sake of precision, let's say for example $$\mathrm{supp}\,h=\{z=(x,y)\in\Bbb C|h(x,y)\neq 0\}\Subset D$$ for a given bounded open set $$D$$.

Complex variable proof
First of all we need this result (see for example [1], §1.1, theorem 1.1.1 pp. 9-10):
Theorem (Gauss-Green formula for planar domains). Let $$D\Subset \Bbb C$$ be an open set with $$C^1$$-boundary $$\partial D$$, and consider a function $$h\in C^0(\bar D)$$ such that $$\bar\partial h$$ is again in $$C^0(\bar D)$$. Then $$h(z)=\frac{1}{2\pi i}\int\limits_{\partial D}\dfrac{h(\zeta)\mathrm{d}\zeta}{\zeta-z}+\frac{i}{2\pi} \int\limits_{D}\dfrac{\bar\partial h(\zeta)\wedge\mathrm{d}\zeta}{\zeta-z}\quad z\in D. \label{1}\tag{1}$$ The result we are seeking for, a direct consequence of this theorem, is the following one:
Theorem. Let $$D\Subset C$$ a bounded open set in $$\Bbb C$$ and let $$h\in C_c^k(D)$$ for some $$k\in\Bbb N$$, $$k\ge 1$$. Then the function $$f$$ defined as $$f(z)=\frac{i}{2\pi} \int\limits_{D}\dfrac{h(\zeta)\mathrm{d}\bar\zeta\wedge\mathrm{d}{\zeta}}{\zeta-z} \label{2}\tag{2}$$ is a solution of the \ref{CR} equation.
Proof. Since $$h$$ has compact support contained in $$D$$ we can extend it by $$0$$ outside $$D$$ so the following function is correctly defined for all $$z\in\Bbb C$$ $$f(z)=\frac{i}{2\pi} \int\limits_{\Bbb C} \dfrac{h(\zeta)\mathrm{d}\bar\zeta\wedge\mathrm{d}{\zeta}}{\zeta-z}=\frac{i}{2\pi} \int\limits_{\Bbb C} \dfrac{h(\zeta+z)\mathrm{d}\bar\zeta\wedge\mathrm{d}{\zeta}}{\zeta}$$ But since $$\frac{\partial}{\partial\bar z} h(\zeta+z)\mathrm{d}\bar{\zeta}=\bar\partial h(\zeta+z)$$ we have $$\frac{\partial f(z)}{\partial\bar z}=\frac{i}{2\pi} \int\limits_{\Bbb C} \dfrac{\bar\partial h(\zeta+z)\wedge\mathrm{d}{\zeta}}{\zeta}=\frac{i}{2\pi} \int\limits_{\Bbb C} \dfrac{\bar\partial h(\zeta)\wedge\mathrm{d}{\zeta}}{\zeta-z} \quad z\in D,\label{3}\tag{3}$$ Now since $$h=0$$ outside $$D$$ (and on its boundary $$\partial D$$) we can apply the Gauss-Green formula \eqref{1} to the right side of \eqref{3} and obtain formula \eqref{2}. $$\blacksquare$$

Distribution/PDEs theory proof
Solving the \ref{CR} equation means finding its fundamental solution, i.e. the distributional solution $$\mathscr{E}$$ to the following equation $$\frac{\partial\mathscr{E}}{\partial\bar{z}}=\frac{\partial \mathscr{E}}{\partial x} + i\frac{\partial \mathscr{E}}{\partial y}=\delta(x,y)$$ Applying the operator $$\partial/\partial z$$ to both sides of this equation we have $$\dfrac{\partial^2 \mathscr{E}}{\partial z\partial\bar z}=\frac{\partial\delta}{\partial z}\iff \Delta \mathscr{E}(x,y) = 2\bigg(\dfrac{\partial }{\partial x} - i\dfrac{\partial }{\partial y}\bigg)\delta(x,y)\label{4}\tag{4}$$ Now, since the fundamental solution of the 2D laplacian is given by $$\mathscr{E}_{\!\Delta}(x,y)=\frac{1}{2\pi}\ln\sqrt{x^2+y^2},$$ (see for example [2], §2.3, example 2.3.8, pp. 33-35 or §15.4, example 15.4.7 p. 204), a solution of the right side equation in \eqref{4} is $$\begin{split} \mathscr{E}(x,y)&=2\mathscr{E}_{\!\Delta}\ast\bigg(\dfrac{\partial }{\partial x} - i\dfrac{\partial }{\partial y}\bigg)\delta(x,y)\\ &=2\bigg(\dfrac{\partial }{\partial x} - i\dfrac{\partial }{\partial y}\bigg)\mathscr{E}_{\!\Delta}\ast\delta(x,y)\\ &=2\bigg(\dfrac{\partial }{\partial x} - i\dfrac{\partial }{\partial y}\bigg)\mathscr{E}_{\!\Delta}(x,y)\\ &=\frac{1}{\pi}\bigg(\dfrac{\partial }{\partial x} - i\dfrac{\partial }{\partial y}\bigg)\ln\sqrt{x^2+y^2}\\ &= \frac{i}{2\pi}\dfrac{x -iy}{x^2+y^2}=\frac{1}{\pi z} \end{split}$$ Therefore the sought for solution of \eqref{CR} is, again as above, $$f(x,y)=\frac{1}{\pi}\int\limits_{D}\dfrac{h(t,s)}{(x-t)+i(y-s)}\mathrm{d}t\mathrm{d}s\iff f(z)=\frac{i}{2\pi} \int\limits_{D}\dfrac{h(\zeta)\mathrm{d}\zeta\wedge\mathrm{d}\bar{\zeta}}{z-\zeta}\label{5}\tag{5}$$

Comparison of the two methods and further notes

• Despite the apparent simplicity of the \ref{CR} equation, the two techniques of solution show that, to deal with it, some preparatory work must be done: the proof based on methods from the theory of function of a complex variable needs a, though elementary, knowledge of the theory of complex differential forms (including the Green-Gauss formula for plane domains), while the second method needs elementary knowledge of the theory of distribution, notably the use of convolution, fundamental solutions and its application to the theory of constant coefficients PDEs.
• Both the two methods proposed can be used to construct solutions to the \ref{CR} equation when $$\mathrm{supp}\,h$$ is not compact. However, in such case \eqref{2} represents the solution of the \ref{CR} equation only locally, i.e. it must be assumed that $$z\in D$$: this is accomplished for example in [1], §1.1, theorem 1.1.3 pp. 10-11.
• Bot the two methods can deal with non-smooth data $$h$$: however complex variable method can only deal with functions which are at least of class $$C^0$$ (see for example the reference given at the preceding point), while the distribution/PDEs theory based method can easily handle more complex situations, notably it works for all distributions for which the convolution with $$\frac{1}{\pi z}$$ is defined.

[1] Gennadi Khenkin Jürgen Leiterer (1984), Theory of functions on complex manifolds, Monographs in Mathematics, Vol. 7\9, Basel-Boston-Stuttgart: Birkhäuser Verlag, pp. 226, ISBN: 3-7643-1477-8, MR0774049, Zbl 0726.32001.

[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.