# Cauchy-Riemann conditions satisfying the Laplacian

How does one convert the Cauchy-Riemann conditions into the form:

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0, \qquad \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2v}{\partial y^2} = 0$$

from

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$

to show that a differentiable complex function has real and imaginary parts that separately satisfy the Laplace equation? The book that I'm working with says that by "differentiating first with respect to x and then with respect to y, one easily obtains (equation 2)", but my calc3 is weak and I fail to see it.

• I think this could be considered a duplicate of math.stackexchange.com/questions/848464/… Feb 18, 2015 at 20:49
• All analytic functions are harmonic. Just differentiate those two C.R. Equations and add them. Jul 5, 2016 at 9:53

Since $u_x = v_y$ and $u_y = -v_x$, then $u_{xx} = v_{yx}$ and $u_{yy} = -v_{xy}$. Hence $u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0$. Since $v_x = -u_y$, $v_{xx} = -u_{yx}$; since $v_y = u_x$, $v_{yy} = u_{xy}$. Therefore $v_{xx} + v_{yy} = -u_{yx} + u_{xy} = 0$.
We will use the Schwarz theorem for the second derivatives, which states that $$\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x}$$ for a function $f:\mathbb{R}^2\to\mathbb{R}$. The Cauchy-Riemann equations are defined as $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, ~~~~~(1) \\ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}. ~~~~~(2)$$ Differentiating $(1)$ with respect to $x$ and $(2)$ with respect to $y$ gives us $$\frac{\partial^2u}{\partial^2x}=\frac{\partial v}{\partial y\partial x},~\frac{\partial^2u}{\partial y^2}=-\frac{\partial v}{\partial x\partial y}.$$ Applying the Schwarz theorem to interchange the derivatives $\partial x\partial y$ and $\partial y\partial x$, we get $$\frac{\partial^2u}{\partial x^2}=-\frac{\partial^2u}{\partial y^2}\Rightarrow \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0.$$ Differentiating $(1)$ with respect to $y$ and $(2)$ with respect to $x$: $$\frac{\partial^2u}{\partial x\partial y}=\frac{\partial^2v}{\partial y^2},~\frac{\partial^2u}{\partial y\partial x}=-\frac{\partial^2v}{\partial x^2}.$$ Applying the Schwarz theorem to interchange the derivatives $\partial x\partial y$ and $\partial y\partial x$, we conclude $$\frac{\partial^2v}{\partial y^2}=-\frac{\partial^2v}{\partial x^2}\Rightarrow \frac{\partial^2v}{\partial x^2}+\frac{\partial^2v}{\partial y^2}=0.~_{\square}$$
• No. It is a known theorem for a function $f\in C^2$. en.wikipedia.org/wiki/Symmetry_of_second_derivatives