Happiness is when what you think, what you say, and what you do are in harmony, Gandhi

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Harmonic functions

Definition. Let $D \subseteq \mathbb{R}^2$ be a domain (open and connected set). A function $\phi (x, y): D \to \mathbb{R}$ is said to be $C^2$ on D if $\phi$ and its partial derivatives $\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial^2 \phi}{\partial x^2}, \frac{\partial^2 \phi}{\partial y^2}$ are all continuous on D. According to this definition, the following conditions are met.

D is open and connected.
The first partial derivatives $\frac{\partial \phi}{\partial x} \text{ and } \frac{\partial \phi}{\partial y}$ exist and are continuous on D.
The second partial derivatives $\frac{\partial^2 \phi}{\partial x^2} \text{ and } \frac{\partial^2 \phi}{\partial y^2}$ exist and are continuous on D.
Notably, this definition does not require the existence or continuity of the mixed second partial derivatives $\frac{\partial^2 \phi}{\partial x \partial y} \text{ or } \frac{\partial^2 \phi}{\partial y \partial x}$. This is a notable departure (a weaker definition) from the standard $C^2$ definition.
The standard (stronger) meaning of $C^2$ on D requires all partial derivatives up to order 2 — including the mixed partials — to exist and be continuous on D: $\phi, \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial^2 \phi}{\partial x^2}, \frac{\partial^2 \phi}{\partial y^2}, \frac{\partial^2 \phi}{\partial y \partial x}, \frac{\partial^2 \phi}{\partial x \partial y}$.

Definition. The Laplacian of a function $\phi (x, y)$ is the sum of its unmixed second partial derivatives $\nabla^2 \phi := \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial \phi^2}{\partial y^2} ↭ \nabla^2 \phi := \phi_{xx}+ \phi_{yy}$.

The Laplace’s equation is the condition $\nabla^2 \phi = 0$, i.e., $\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial x^2} = 0$.

The standard notations for the Laplacian of a function $\phi (x, y)$ are $\nabla^2 \phi \text{ or } \Delta \phi $. The standard notations for partial derivatives: $\frac{\partial \phi}{\partial x} \text{ or } \phi_x$ for first partials, and for second partials, $\frac{\partial^2 \phi}{\partial x²} \text{ or } \phi_{xx}$. The superscript 2 on the numerator should precede the variable (applies to the differential operator $\partial$) to denote the second derivative; the notation $\frac{\partial^2 \phi}{\partial x^2}$ means “apply $\frac{\partial}{\partial x}$ twice to $\phi$.”

Definition. A real-valued function $\phi (x, y)$ is harmonic on a domain $D \subseteq \mathbb{R}^2$ if it is $C^2$ on D and satisfies Laplace’s equation: $\nabla^2 \phi = \frac{\partial \phi^2}{\partial x^2} + \frac{\partial \phi^2}{\partial y^2} = 0$.

Theorem. Let $f(z) = u(x,y) + iv(x,y)$ be analytic on a domain D (with z = x + iy). Then:

u, v are $C^\infty$ (indeed, real-analytic).
They satisfy the Cauchy–Riemann equations: $u_x = v_y, u_y = -v_x$.
Both it real part, u, and its imaginary part, v, are harmonic functions on D.

Definition. If u(x, y) is a harmonic function on a domain D, a function v(x, y) is called a harmonic conjugate of u if:

v is also harmonic on D.
The pair (u, v) satisfies the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$.

Theorem. Harmonic conjugates yield analytic functions. If u(x, y) is a harmonic function on a domain D and v is a harmonic conjugate of u, then f(z) = u(x, y) + iv(x, y) is analytic on D, where z = x + iy.

Theorem. Existence of Harmonic Conjugates. If $u \in C^2(D)$ is harmonic and D is simply connected, then there exists v with $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$, and $f(z) = u(x, y) + iv(x, y)$ is analytic on D. The conjugate v is unique up to an additive constant.

Examples of Harmonic Functions

Cross Terms. Consider the function $u(x, y) = xy$. This is a simple example of a harmonic function in two variables.

To verify, we compute its second partial derivatives. $\frac{\partial^2 u}{\partial x^2} = 0, \frac{\partial^2 u}{\partial y^2} = 0$. The seconds derivatives cancel out, so $ \nabla^2 u = 0$, confirming that u is harmonic everywhere.

A harmonic conjugate of u is a function v(x, y) such that u and v satisfy the Cauchy-Riemann equations $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$, making u + iv analytic.
$\frac{\partial u}{\partial x} = y \leadsto \frac{\partial v}{\partial y} = y$. Integrating this expression with respect to y: $u(x, y) = \int ydy = \frac{1}{2}y^2 + g(x)$.
Taking the partial derivative with respect to x and using $\frac{\partial u}{\partial y} = x = -\frac{\partial v}{\partial x}$, we get: $ g'(x) = -x \leadsto g(x) = -\frac{1}{2}x^2 + C$.
Thus, setting C = 0 for simplicity, a conjugate is $v(x, y) = \frac{1}{2}(y^2 - x^2)$.

The complex function f(z) = u + iv is analytic and can be expressed in terms of z = x + iy, $f(z) = xy + i\frac{1}{2}(y^2 - x^2)$. $z² = x² - y² + 2ixy, iz² = i(x² - y²) -2xy, -\frac{i}{2}z² = i\frac{1}{2}(y² - x²) + xy, \boxed{f(z) = -\frac{i}{2} z^2}$.

This show that the harmonic function $u(x, y) = xy$ and its conjugate $v(x, y) = \frac{1}{2}(y^2 - x^2)$ combine to form the simple analytic function $f(z) = -\frac{i}{2} z^2$. This example illustrates how cross terms like xy naturally arise from complex squaring.

Laplacian in Polar Coordinates

This is not for the faint of heart.

Polar to Cartesian: $x = r\cos(\theta), y = r\sin(\theta)$. Cartesian to Polar: $r = \sqrt{x^2+y^2}, \theta = \arctan\left(\frac{y}{x}\right)$
Partial Derivatives of r and θ: $\frac{\partial r}{\partial x} = \frac{\partial}{\partial x}\sqrt{x^2+y^2} = \frac{1}{2\sqrt{x^2+y^2}} \cdot 2x = \frac{x}{r} = \cos(\theta)$, $\frac{\partial r}{\partial y} = \frac{\partial}{\partial y}\sqrt{x^2+y^2} = \frac{1}{2\sqrt{x^2+y^2}} \cdot 2y = \frac{y}{r} = \sin(\theta), \frac{\partial\theta}{\partial x} = \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{\partial}{\partial x}\left(\frac{y}{x}\right) = \frac{1}{1+(\frac{y}{x})^2} \cdot \left(-\frac{y}{x^2}\right) = \frac{x^2}{x^2+y^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2+y^2} = -\frac{y}{r^2} = -\frac{\sin(\theta)}{r}, \frac{\partial\theta}{\partial y} = \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{\partial}{\partial y}\left(\frac{y}{x}\right) = \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{1}{x} = \frac{x^2}{x^2+y^2} \cdot \frac{1}{x} = \frac{x}{x^2+y^2} = \frac{x}{r^2} = \frac{\cos(\theta)}{r}$. $\frac{\partial r}{\partial x} = \cos(\theta), \frac{\partial r}{\partial y} = \sin(\theta), \frac{\partial\theta}{\partial x} = -\frac{\sin(\theta)}{r}, \frac{\partial\theta}{\partial y} = \frac{\cos(\theta)}{r}$
First-Order Partial Derivatives. Using the chain rule: $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x} = \cos\theta \frac{\partial f}{\partial r} - \frac{\sin\theta}{r} \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial y} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y} = \sin\theta \frac{\partial f}{\partial r} + \frac{\cos\theta}{r} \frac{\partial f}{\partial \theta}$.
Second-Order Partial Derivatives. Let’s compute $\frac{\partial^2 f}{\partial x^2}$: $\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(\cos\theta \frac{\partial f}{\partial r}\right) - \frac{\partial}{\partial x}\left(\frac{\sin\theta}{r} \frac{\partial f}{\partial \theta}\right)$

First term: Using the product rule: $\frac{\partial}{\partial x}\left(\cos\theta \frac{\partial f}{\partial r}\right) = \frac{\partial(\cos \theta)}{\partial x}\frac{\partial f}{\partial r} + \cos \theta \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial r}\right) = -\sin \theta \frac{\partial \theta}{\partial x}\frac{\partial f}{\partial r} + \cos \theta \left[\frac{\partial^2 f}{\partial r^2}\frac{\partial r}{\partial x} + \frac{\partial^2 f}{\partial r\partial \theta}\frac{\partial \theta}{\partial x}\right] = -\sin \theta \cdot \left(-\frac{\sin \theta}{r}\right)\frac{\partial f}{\partial r} + \cos \theta \frac{\partial^2 f}{\partial r^2}\cos \theta + \cos \theta \frac{\partial^2 f}{\partial r \partial \theta}\left(-\frac{\sin \theta}{r}\right) = \frac{\sin^2 \theta}{r}\frac{\partial f}{\partial r} + \cos^2 \theta \frac{\partial^2 f}{\partial r^2} - \frac{\cos \theta \sin \theta}{r}\frac{\partial^2 f}{\partial r \partial \theta}$

Second term: $-\frac{\partial}{\partial x}\left(\frac{\sin\theta}{r} \frac{\partial f}{\partial \theta}\right) = -\frac{\partial}{\partial x}\left(\frac{\sin \theta}{r}\right)\frac{\partial f}{\partial \theta} - \frac{\sin \theta}{r}\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial \theta}\right)$

For $\frac{\partial}{\partial x}\left(\frac{\sin \theta}{r}\right)$: $\frac{\partial}{\partial x}\left(\frac{\sin \theta}{r}\right) = \frac{1}{r}\frac{\partial(\sin \theta)}{\partial x} - \frac{\sin \theta}{r^2}\frac{\partial r}{\partial x}$ $= \frac{\cos \theta}{r} \cdot \left(-\frac{\sin \theta}{r}\right) - \frac{\sin \theta}{r^2} \cdot \cos \theta = -\frac{2\cos \theta \sin \theta}{r^2}$

For $\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial \theta}\right)$: $\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial \theta}\right) = \frac{\partial^2 f}{\partial r \partial \theta}\cos(\theta) + \frac{\partial^2 f}{\partial \theta^2}\left(-\frac{\sin \theta}{r}\right)$

So the second term becomes: $-\left(-\frac{2\cos \theta \sin \theta}{r^2}\right)\frac{\partial f}{\partial \theta} - \frac{\sin \theta}{r}\left[\frac{\partial^2 f}{\partial r \partial \theta}\cos(\theta) - \frac{\partial^2 f}{\partial \theta^2}\frac{\sin \theta}{r}\right] = \frac{2\cos \theta \sin \theta}{r^2} \frac{\partial f}{\partial \theta} - \frac{\sin \theta \cos \theta}{r} \frac{\partial^2 f}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2 f}{\partial \theta^2}$

Combining both terms: $\frac{\partial^2 f}{\partial x^2} = \frac{\sin^2 \theta}{r}\frac{\partial f}{\partial r} + \cos^2 \theta \frac{\partial^2 f}{\partial r^2} - \frac{\cos \theta \sin \theta}{r}\frac{\partial^2 f}{\partial r \partial \theta} + \frac{2\cos \theta \sin \theta}{r^2} \frac{\partial f}{\partial \theta} - \frac{\sin \theta \cos \theta}{r} \frac{\partial^2 f}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2 f}{\partial \theta^2}$

Computing $\frac{\partial^2 f}{\partial y^2}$: $\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\left(\sin\theta \frac{\partial f}{\partial r}\right) + \frac{\partial}{\partial y}\left(\frac{\cos\theta}{r} \frac{\partial f}{\partial \theta}\right)$

First term: $\frac{\partial}{\partial y}\left(\sin\theta \frac{\partial f}{\partial r}\right) = \frac{\partial(\sin \theta)}{\partial y}\frac{\partial f}{\partial r} + \sin \theta \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial r}\right)= \cos \theta \frac{\partial \theta}{\partial y}\frac{\partial f}{\partial r} + \sin \theta \left[\frac{\partial^2 f}{\partial r^2}\frac{\partial r}{\partial y} + \frac{\partial^2 f}{\partial r\partial \theta}\frac{\partial \theta}{\partial y}\right] = \cos \theta \cdot \frac{\cos \theta}{r}\frac{\partial f}{\partial r} + \sin \theta \frac{\partial^2 f}{\partial r^2}\sin \theta + \sin \theta \frac{\partial^2 f}{\partial r \partial \theta}\frac{\cos \theta}{r}= \frac{\cos^2 \theta}{r}\frac{\partial f}{\partial r} + \sin^2 \theta \frac{\partial^2 f}{\partial r^2} + \frac{\sin \theta \cos \theta}{r}\frac{\partial^2 f}{\partial r \partial \theta}$

Second term: $\frac{\partial}{\partial y}\left(\frac{\cos\theta}{r} \frac{\partial f}{\partial \theta}\right) = \frac{\partial}{\partial y}\left(\frac{\cos \theta}{r}\right)\frac{\partial f}{\partial \theta} + \frac{\cos \theta}{r}\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial \theta}\right)$

For $\frac{\partial}{\partial y}\left(\frac{\cos \theta}{r}\right)$: $\frac{\partial}{\partial y}\left(\frac{\cos \theta}{r}\right) = \frac{1}{r}\frac{\partial(\cos \theta)}{\partial y} - \frac{\cos \theta}{r^2}\frac{\partial r}{\partial y}= \frac{(-\sin \theta)}{r} \cdot \frac{\cos \theta}{r} - \frac{\cos \theta}{r^2} \cdot \sin \theta = -\frac{2\sin \theta \cos \theta}{r^2}$

For $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial \theta}\right)$: $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial \theta}\right) = \frac{\partial^2 f}{\partial r \partial \theta}\sin(\theta) + \frac{\partial^2 f}{\partial \theta^2}\frac{\cos \theta}{r}$

So the second term becomes: $-\frac{2\sin \theta \cos \theta}{r^2} \frac{\partial f}{\partial \theta} + \frac{\cos \theta}{r}\left[\frac{\partial^2 f}{\partial r \partial \theta}\sin(\theta) + \frac{\partial^2 f}{\partial \theta^2}\frac{\cos \theta}{r}\right]= -\frac{2\sin \theta \cos \theta}{r^2} \frac{\partial f}{\partial \theta} + \frac{\sin \theta \cos \theta}{r} \frac{\partial^2 f}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2 f}{\partial \theta^2}$

Combining both terms: $\frac{\partial^2 f}{\partial y^2} = \frac{\cos^2 \theta}{r}\frac{\partial f}{\partial r} + \sin^2 \theta \frac{\partial^2 f}{\partial r^2} + \frac{\sin \theta \cos \theta}{r}\frac{\partial^2 f}{\partial r \partial \theta}$ $- \frac{2\sin \theta \cos \theta}{r^2} \frac{\partial f}{\partial \theta} + \frac{\sin \theta \cos \theta}{r} \frac{\partial^2 f}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2 f}{\partial \theta^2}$

Final Result: The Laplacian. Adding $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\sin^2 \theta}{r}\frac{\partial f}{\partial r} + \cos^2 \theta \frac{\partial^2 f}{\partial r^2} - \frac{\cos \theta \sin \theta}{r}\frac{\partial^2 f}{\partial r \partial \theta} + \frac{2\cos \theta \sin \theta}{r^2} \frac{\partial f}{\partial \theta} - \frac{\sin \theta \cos \theta}{r} \frac{\partial^2 f}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2 f}{\partial \theta^2} + \frac{\cos^2 \theta}{r}\frac{\partial f}{\partial r} + \sin^2 \theta \frac{\partial^2 f}{\partial r^2} + \frac{\sin \theta \cos \theta}{r}\frac{\partial^2 f}{\partial r \partial \theta}- \frac{2\sin \theta \cos \theta}{r^2} \frac{\partial f}{\partial \theta} + \frac{\sin \theta \cos \theta}{r} \frac{\partial^2 f}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2 f}{\partial \theta^2}$:

Coefficients of $\frac{\partial^2 f}{\partial r^2}$: $\cos^2 \theta + \sin^2 \theta = 1$
Coefficients of $\frac{\partial f}{\partial r}$: $\frac{\sin^2 \theta}{r} + \frac{\cos^2 \theta}{r} = \frac{1}{r}$
Coefficients of $\frac{\partial^2 f}{\partial r \partial \theta}$: $-\frac{\cos \theta \sin \theta}{r} - \frac{\sin \theta \cos \theta}{r} + \frac{\sin \theta \cos \theta}{r} + \frac{\sin \theta \cos \theta}{r} = 0$
Coefficients of $\frac{\partial f}{\partial \theta}$: $\frac{2\cos \theta \sin \theta}{r^2} - \frac{2\sin \theta \cos \theta}{r^2} = 0$
Coefficients of $\frac{\partial^2 f}{\partial \theta^2}$: $\frac{\sin^2 \theta}{r^2} + \frac{\cos^2 \theta}{r^2} = \frac{1}{r^2}$

Conclusion: $\boxed{\nabla^2 f = \frac{\partial^2 f}{\partial r^2} + \frac{1}{r}\frac{\partial f}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2}}$
The function u(x, y) = $log(r) = log(\sqrt{x²+y²}) = \frac{1}{2}log(x²+y²)$ is a classic example of a harmonic function with a singularity. It is harmonic on the punctured plane $\mathbb{R}^2 \setminus \{ (0, 0) \}$.

Define $v = x^2 + y^2$, so u $= \frac{1}{2}log(v)$
First partial derivatives: $\frac{\partial u}{\partial x} = \frac{1}{2}\frac{1}{v}2x = \frac{x}{v}, \frac{\partial u}{\partial y} = \frac{1}{2}\frac{1}{v}2y = \frac{y}{v}.$
Second partial derivatives: $\frac{\partial^2 u}{\partial x^2} = \frac{v·1-x·2x}{v^2} = \frac{v -2x^2}{v^2}, \frac{\partial^2 u}{\partial y^2} = \frac{v·1-y·2y}{v^2} = \frac{v -2y^2}{v^2}$
Laplacian: $\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{v -2x^2 + v -2y^2}{v^2} = \frac{2v -2(x²+y²)}{v^2} = \frac{2v -2v}{v^2} = 0$. Thus, $\nabla^2 u = 0, \forall (x, y) \ne (0, 0)$, confirming harmonicity.
However, it fails to be harmonic at the origin due to the singularity there. At the origin, u is undefined because as $r = \sqrt{x^2+y^2} \to 0, \log(r) \to -\infty$, making u undefined at the origin.

For polar coordinates $(r, \theta)$, the Laplacian of a scalar function $u(r, \theta)$ is $\boxed{\nabla ^2u = \frac{\partial ^2u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial ^2u}{\partial \theta ^2}}.$

In polar coordinates $(r, \theta), u = \log(r)$. First derivatives: $u_r=\frac{\partial }{\partial r}(\log r)=\frac{1}{r},\qquad u_{\theta }=0.$

Second derivatives: $u_{rr}=\frac{\partial }{\partial r}\left( \frac{1}{r}\right) =-\frac{1}{r^2},\qquad u_{\theta \theta }=0.$

Plug into the polar Laplacian: $\nabla ^2u=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta \theta }=\left( -\frac{1}{r^2}\right) +\frac{1}{r}\left( \frac{1}{r}\right) +\frac{1}{r^2}\cdot 0=-\frac{1}{r^2}+\frac{1}{r^2}=0,$ for all r>0.

So in polar form, $\nabla ^2(\log r)=0\quad \mathrm{for\ }r>0,$ again showing harmonicity on $\mathbb{R^{\mathnormal{2}}}\setminus \{ 0\}$.

Real and imaginary parts of zⁿ: For each $n \in \mathbb{N}$, the functions $u_n(r,\theta) = r^n\cos(n\theta), \qquad v_n(r,\theta) = r^n\sin(n\theta)$ are harmonic on $\mathbb{R}^2$. These arise as the real and imaginary parts of the holomorphic function $f(z) = z^n$, since $z^n = (re^{i\theta})^n = r^n e^{in\theta} = r^n\cos(n\theta) + i\,r^n\sin(n\theta).$

By the theorem relating analytic functions to harmonic functions, both components are automatically harmonic.
To verify, consider the Laplacian in polar coordinates, $\boxed{\nabla^2 f = \frac{\partial^2 f}{\partial r^2} + \frac{1}{r}\frac{\partial f}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2}}$
Verification for $u_n(r,\theta) = r^n\cos(n\theta)$.
First partial derivatives: $\frac{\partial u_n}{\partial r} = nr^{n-1}cos(n\theta), \frac{\partial u_n}{\partial \theta} = -nr^{n}sin(n\theta)$
Second partial derivatives: $\frac{\partial^2 u_n}{\partial r^2} = n(n-1)r^{n-2}cos(n\theta), \frac{\partial^2 u_n}{\partial \theta^2} = -n^2r^ncos(n\theta)$

Assembling the Laplacian:

$$ \begin{aligned} \nabla^2 u_n &=n(n-1)\,r^{n-2}\cos(n\theta) + \frac{1}{r}\cdot n\,r^{n-1}\cos(n\theta) + \frac{1}{r^2}\cdot\bigl(-n^2\,r^n\cos(n\theta)\bigr) \\[2pt] &=r^{n-2}\cos(n\theta)\bigl[n(n-1) + n - n^2\bigr]\\[2pt] &= r^{n-2}\cos(n\theta)\bigl[n^2 - n + n - n^2\bigr]\\[2pt] &=0. \end{aligned} $$

Mutatis mutandis, $\nabla^2 vₙ = 0.$ The Laplacian vanishes for all $r > 0$ and all $\theta$.

Behavior at the origin. Although the polar Laplacian formula requires r > 0, the functions themselves extend smoothly to r = 0:

$r^n\cos(n\theta)$ and $r^n\sin(n\theta)$ are homogeneous polynomials of degree n in Cartesian coordinates.
Polynomials are smooth everywhere.
The Laplacian of a polynomial is again a polynomial, so it is defined at the origin.

Examples in Cartesian coordinates

n = 1, $u_1=r\cos \theta =x,\qquad v_1=r\sin \theta =y.$ Both are linear polynomials. Their Laplacians: $\nabla^2 x = 0,\qquad \nabla^2 y =0.$
n = 2, $u_2=r^2\cos (2\theta ) = r^2(\cos^2\theta - \sin^2\theta) = x^2 - y^2,$ $v_2=r^2\sin (2\theta )= 2(r\cos\theta)(r\sin\theta) =2xy$.
Both are quadratic polynomials. Their Laplacians: $\nabla^2(x^2 - y^2) = \frac{\partial^2}{\partial x^2}(x^2) - \frac{\partial^2}{\partial y^2}(y^2) = 2 - 2 = 0,\qquad \nabla^2(2xy) = 0 + 0 = 0.$

General n. For each $n\geq 1$, the expressions become homogeneous polynomials of degree n. $r^{n}\cos(n\theta) \text{ and } v_n(r,\theta) = r^{n}\sin(n\theta)$ correspond to real/imaginary parts of $z^n = (x + iy)^n$, which expands to a homogeneous polynomial of degree n via the binomial theorem.

Because $z^n$ is holomorphic on all of $\mathbb{C}$, its real and imaginary parts satisfy the Cauchy–Riemann equations everywhere, including at the origin.

A fundamental theorem states that the real and imaginary parts of any holomorphic function are harmonic on the entire domain of holomorphy. Since $z^n$ is entire, $\boxed{u_n(r,\theta) = r^{n}\cos(n\theta) \text{ and } v_n(r,\theta) = r^{n}\sin(n\theta) \text{are harmonic on } \mathbb{R}^2 \quad \text{for each } n \in \mathbb{N}.}$

Contrast with Non-Harmonic Functions. Not all smooth functions are harmonic. Paraboloid: $f(x, y) = x^2 + y^2$ satisfied $\nabla^2 f = \frac{\partial^2}{\partial x^2}(x^2 + y^2) + \frac{\partial^2}{\partial y^2}(x^2 + y^2) = 2 + 2 = 4 \neq 0,$ so it’s not harmonic. It possesses a strict global minimum at the origin. As we will see later, harmonic functions cannot have strict local extrema in their domain, so this verifies f is not harmonic.
Connection to Complex Analysis. Let $f(z) = u(x,y) + i\,v(x,y)$ be analytic on a domain D (with z = x + iy). Then:

u, v are $C^\infty$.
They satisfy the Cauchy–Riemann equations: $u_x = v_y, u_y = -v_x$.
Both u and v are harmonic functions on D.

Key Properties of Harmonic Functions

Harmonic functions possess several important properties that make them useful in various applications:

Linearity and Homogeneity: The sum of two harmonic functions is also harmonic. The set of harmonic functions on a domain $D$ forms a real vector space: it is closed under addition and scalar multiplication.
Addition. If $\nabla^2 u = 0$ and $\nabla^2 v = 0$, then by linearity of the Laplacian, $\nabla^2(u + v) = \nabla^2 u + \nabla^2 v = 0 + 0 = 0.$
Homogeneity (scalar multiplication). For any $c \in \mathbb{R}$: $\nabla^2(c \cdot u) = c \cdot \nabla^2 u = c \cdot 0 = 0.$
More generally, any finite linear combination $\sum_{k=1}^{N} c_k u_k$ of harmonic functions is harmonic.
However, the product of two harmonic functions is generally not harmonic, e.g., $u = v = x$ gives $u \cdot v = x^2$, and $\nabla^2(x^2) = 2 \neq 0$.
Mean Value Property: The value of a harmonic function at any point $u(z_0)$ is the average of its values over any circle centered at that point. More formally, if u is harmonic on a domain D, then for any closed disk $\overline{B_r(z_0)} \subset D, u(z_0) = \frac{1}{2\pi} \int_{0}^{2\pi} u(z₀+re^{i\theta})d\theta \qquad \text{(circular mean)} = \frac{1}{\pi r^2}\int \int_{|z-z₀| < r} u(z)dA \qquad \text{(areal mean)}$

Intuition. A harmonic function has no sources or sinks ($\nabla^2 u = 0$ means no net “flux creation”). Consequently, the value at any point is perfectly balanced by its surroundings — there is no tendency to be pushed up or down relative to nearby values.

Example: Consider $f(x,y) = x^2 - y^2$. A function u(x, y) is harmonic if it satisfies Laplace’s equation, which states that the sum of its second partial derivatives is zero: $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} = 2 - 2 = 0$, so $f(x, y) = x^2 - y^2$ is indeed a harmonic function.

The Mean Value Property states that the value of the function at the center of a circle of radius r centered at (0, 0) is equal to the average of its values over that circle. u(0, 0) = $\frac{1}{2\pi} \int_{0}^{2\pi} f(rcos(\theta), rsin(\theta))d\theta = \frac{1}{2\pi} \int_{0}^{2\pi} (rcos(\theta))^2 - (rsin(\theta))^2d\theta = \frac{1}{2\pi} \int_{0}^{2\pi} r^2(cos(\theta)^2 - sin(\theta)^2)d\theta = \frac{1}{2\pi} \int_{0}^{2\pi} r^2cos(2\theta)d\theta = \frac{r^2}{2\pi} \frac{1}{2}sin(2\theta)\bigg|_{0}^{2\pi}$

Average Value = $ \frac{r^2}{2\pi}\left[\frac{\sin(2\theta)}{2}\right]_0^{2\pi} = \frac{r^2}{2\pi}\cdot\frac{1}{2}\bigl(\sin 4\pi - \sin 0\bigr) = 0$ The average value of the function over the circle is 0. Now, let’s find the value of the function at the center, (0, 0): $f(0,0) = 0^2 - 0^2 = 0$. The average value (0) matches the value of the function at the center (0), confirming the Mean Value Property. This property is a fundamental characteristic of harmonic functions and plays a crucial role in complex analysis.

Counterexample. Consider the non-harmonic function $g(x,y) = x^2 + y^2$ with $\nabla^2 g = 4$. $\frac{1}{2\pi}\int_0^{2\pi} g(r\cos\theta, r\sin\theta)\,d\theta = \frac{1}{2\pi}\int_0^{2\pi} r^2d\theta = r^2$.

However, $g(0,0) = 0 \neq r^2$ for $r > 0$. The mean value property fails — the average over the circle exceeds the center value.

Maximum Principle: Let $u$ be a harmonic function on a connected open set (domain) $D \subseteq \mathbb{C}$. If $u$ attains a global maximum or global minimum at an interior point of $D$, then $u$ is constant on $D$. Equivalently, for a non-constant harmonic function on a bounded domain $D$ with continuous extension to $\overline{D}$: $\min_{\partial D}\, u \;\leq\; u(z) \;\leq\; \max_{\partial D}\, u \qquad \text{for every } z \in D,$ with strict inequalities in the interior.

Proof:

Assume, for contradiction, that u is non-constant and attains a local maximum at an interior point $z_0 \in D$: that is, $u(z) \leq u(z_0)$ for all $z$ in some neighborhood of $z_0$.
By continuity there is a radius r > 0 such that $\overline{B_r(z₀)} \subset D$ and $u(z) \leq u(z_0), \forall z \in \overline{B_r(z₀)}$.
Apply the Mean Value Property on the circle of radius r: $u(z_0) = \frac{1}{2\pi}\int_0^{2\pi} u(z_0 + re^{i\theta})\,d\theta.$
Because each integrand satisfies $u(z_0+re^{i\theta}) ≤ u(z_0)$, equality in the integral forces $u(z_0+re^{i\theta}) = u(z_0)$.
Vary r arbitrarily small; the same argument shows u is constant on every concentric circle, hence u is constant on the entire disk $\overline{B_r(z₀)}$. In words, Mean Value Property forces local constancy.
Spreading the constancy via connectedness (a domain D is an open and connected subset of $\mathbb{C}$). Define the sets: $E = \{z \in D \mid u(z) = u(z_0)\}, \qquad F = \{z \in D \mid u(z) \neq u(z_0)\}$.
$E$ is open: For any $z_1 \in E$, the function $u$ attains its maximum at $z_1$ as well, so the previous reasoning applies at $z_1$ as well (mean value property forces local constancy), giving an open disk around $z_1$ contained in $E$.
$F$ is open: Since $\{u(z_0)\}$ is closed in $\mathbb{R}$ and $u$ is continuous, $F = u^{-1}(\mathbb{R} \setminus \{u(z_0)\})$ is the preimage of an open set under a continuous map, hence open.
Obviously $D = E \sqcup F$ (disjoint union of two open sets), and $D$ is connected. Since $E \neq \emptyset$ (it contains $z_0$), connectedness (A topological space is connected if it cannot be written as the union of two non-empty disjoint open sets) forces $F = \emptyset$, so $E = D$. Thus $u \equiv u(z_0)$ on $D$, contradicting the initial assumption that $u$ is non-constant. $\quad\square$
The same argument applied to $-u$ (which is also harmonic) yields the Minimum Principle: a non-constant harmonic function cannot attain an interior minimum either.
Corollary: If $D$ is bounded and $u$ is continuous on the closure $\bar{D}$, the maximum and minimum values of $u$ occur on the boundary $\partial D$. The corollary for bounded domains follows directly: extrema must occur on $\partial D$ since interior extrema imply constancy.
u is a harmonic function on $D \subseteq \mathbb{C}$. A singleton set (just one real number) {$x \in \mathbb{R} \mid x = u(z_0)$} is closed in $\mathbb{R}$. The complement of a closed set is open, so $\mathbb{R} \setminus \{ u(z_0) \}$ is open in $\mathbb{R}$. $F = u^{-1}(\mathbb{R} \setminus \{ u(z_0) \})$ is the pre-image of an open set under the continuous map u, therefore open in the domain D.

Example: $f(x,y) = x$.

A function u(x, y) is harmonic if it satisfies Laplace’s equation, which states that the sum of its second partial derivatives is zero: $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} = 0 + 0 = 0$, so f(x, y) = x is indeed harmonic on $\mathbb{R}^2$.

Consider this function on the closed unit square $\overline{D} = [0,1] \times [0,1]$, the set of all points (x, y) where 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.

Interior: In the open rectangle 0 < x < 1 and 0 < y < 1, the function’s value ranges from 0 to 1, but there are no “peaks” or “valleys.” The function steadily increases as x increases.
Boundary: The minimum value of the function occurs when x = 0, which is on the left edge of the rectangle, and the maximum value occurs when x = 1, which is on the right edge.

The extrema of the function are entirely determined by the boundary of the domain. This beautifully demonstrates the Maximum Principle. The function f(x, y) = x on this domain has a maximum of 1 and a minimum of 0, both of which are achieved on the boundary and not in the interior.

Theorem (Regularity): Every harmonic function on a domain $D \subseteq \mathbb{R}^n$ is:

$C^\infty$, infinitely differentiable, which means all their partial derivatives of all orders exist and are continuous.
Real-analytic. It can be represented by a convergent power series in a neighborhood of every point.

This is a striking regularity result: the hypothesis $\nabla^2 u = 0$ involves only second derivatives, yet it forces the existence and continuity of derivatives of all orders. Harmonic functions are as smooth and “well-behaved” as functions can possibly be — no sharp corners, cusps, or discontinuities in their derivatives.

In two dimensions, this analyticity is intimately connected to complex analysis: a harmonic function $u$ is locally the real part of a holomorphic function $f = u + iv$, and holomorphic functions are analytic.

Uniqueness for the Dirichlet problem. If $u$ and $v$ are both harmonic on a bounded domain $D$ and continuous on $\overline{D}$, and if $u = v$ on $\partial D$, then $u \equiv v$ on $D$. The Dirichlet problem is a type of boundary value problem that seeks to find a harmonic function in a given domain D that takes on specified values on the boundary ∂D. This uniqueness theorem proves that if you find one such harmonic function, it’s the only one. In other words, a harmonic function is completely determined by its boundary values.

Proof.

Let’s prove this by contradiction. Assume that u and v are two different harmonic functions on a domain D that both satisfy the same boundary conditions, meaning $u = v$ on $\partial D$.
Define a new function: Let’s create a new function, w(x,y) = u(x, y) − v(x, y).
Harmonicity of w: Since u and v are both harmonic, their difference, w, must also be harmonic. We can show this by checking if w satisfies Laplace’s equation: $\nabla^2 w = \nabla^2 u - \nabla^2 v = 0 - 0 = 0$, so $w$ is harmonic on $D$.
Boundary conditions for w: On the boundary $\partial D$, $w = u - v = 0$
Applying the Maximum Principle: Since w is a harmonic on $D$ and continuous on $\overline{D}$, the Maximum Principle states that its maximum and minimum values over $\overline{D}$ must occur on the boundary $\partial D$. $\max_{\overline{D}}\, w = \max_{\partial D}\, w = 0, \qquad \min_{\overline{D}}\, w = \min_{\partial D}\, w = 0.$
Because w = 0 on the entire boundary, the maximum value of w on the domain must be 0, and its minimum value must also be 0. $0 \leq w(z) \leq 0$ for all $z \in \overline{D}$, so $w \equiv 0$, i.e., $u \equiv v$ on $D$.

In words, if the maximum and minimum values of w are both 0, then the function w must be identically zero everywhere in the domain D. This means w(x, y) = u(x, y) − v(x, y) = 0, which implies u(x, y) = v(x, y) for all points in D. This contradicts our initial assumption that u and v were two different functions.

Key Examples and Properties of Harmonic Functions

Harmonic functions

Examples of Harmonic Functions

Laplacian in Polar Coordinates

Key Properties of Harmonic Functions