## 275- Harmonic functions and harmonic conjugates

December 23, 2008

Recall that a function $u(x,y)$ of two variables defined on an open domain $D$ is harmonic iff $u$ is $C^2$ (i.e., all four second order derivatives $u_{xx},u_{xy},u_{yx},u_{yy}$ exist and are continuous in $D$), and $u$ satisfies Laplace equation $u_{xx}+u_{yy}=0.$

As mentioned in problem 6 of the Fall 2008 Calculus III final exam, a function $v$ is a harmonic conjugate of $u$ iff $v$ is defined on $D$, $v_x$ and $v_y$ exist, and the Cauchy-Riemann equations hold: $u_x=v_y$ and $u_y=-v_x$.

It follows immediately from the Cauchy-Riemann equations that if $v$ is a harmonic conjugate of a harmonic function $u$, then $v$ is also $C^2$, with $v_{xx}=-u_{yx}$, $v_{xy}=-u_{yy}$, $v_{yx}=u_{xx}$ and $v_{yy}=u_{xy}$. It is also immediate that $v$ satisfies Laplace equation because $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$, since continuity guarantees that the mixed partial derivatives commute. Thus $v$ is also harmonic.

In fact, modulo continuity of the second order derivatives, the harmonic functions are precisely the functions that (locally) admit harmonic conjugates.

To see this, assume first that $u$ is $C^2$ in $D$ and that it admits a harmonic conjugate $v$. Then $u_{xx}=v_{yx}$ and $u_{yy}=-v_{xy}$ so $u_{xx}+u_{yy}=0$ and $u$ was harmonic to begin with.

Conversely, assume that $u$ is harmonic in $D$. Suppose first that $D$ is (connected and) simply connected. I claim that then $u$ admits a harmonic conjugate $v$ in $D$. To see this, letting ${\mathbf F}=(-u_y,u_x)$, notice that the existence of $v$ is equivalent to the claim that ${\mathbf F}$ is a gradient vector field, since ${\mathbf F}=\nabla v$ iff $v$ is a harmonic conjugate of $u.$ But, since $D$ is simply connected, then ${\mathbf F}$ is a gradient iff it is conservative, i.e., $\displaystyle \oint_\gamma {\mathbf F}\cdot d{\mathbf r}=0$ for any simple piecewise smooth loop $\gamma$ in $D$. Fix such a $\gamma$, and let $R$ denote its interior. Then, by Green’s theorem, $\displaystyle \oint_\gamma {\mathbf F}\cdot d{\mathbf r}=\pm\iint_R u_{xx}+u_{yy}\,dA=0,$

where the $\pm$ sign is to be chosen depending on the orientation of $\gamma$. It follows that ${\mathbf F}$ is indeed conservative and therefore a gradient, so $u$ admits a harmonic conjugate.

Finally, if $D$ is not simply connected, we cannot guarantee that such a $v$ exists in all of $D$, but the argument above shows that it does in any open (connected) simply connected subset of $D$, for example, any open ball contained in $D$.  That we cannot extend this to all of $D$ follows from considering, for example, $u(x,y)=\log(x^2+y^2)$ in $D={\mathbb R}^2\setminus\{(0,0)\}$. This is a harmonic function but it does not admit a harmonic conjugate in $D$, since there is no continuous $\arctan(y/x)$ in $D$. This example can be easily adapted (via a translation) to any non-simply connected $D$.

I close by remarking that, as mentioned in my previous post on average values of harmonic functions, one can use Green’s theorem to prove that harmonic functions $u$ satisfy the average (or mean) value property, and this property characterizes harmonicity as well, implies that $u$ is actually $C^{\infty}$ (i.e., $u$ admits partial derivatives of all orders, and they are all continuous) and has the additional advantage that it only requires that $u$ is continuous, rather than $C^2$. Similarly, one can show that the Cauchy-Riemann equations on $D$ suffice to guarantee that $u$ and $v$ are harmonic (and in particular, $C^\infty$). However, one needs to require that the equations hold everywhere on $D$. A pointwise requirement would not suffice. But I won’t address this issue here (I mention it in the notes in complex analysis that I hope to post some day).