Harmonic Functions and Their Relationship to Analytic Functions

If you find yourself in a hole, stop digging, Will Rogers

The Complex Exponential Function

The complex exponential function, denoted as eᶻ or exp(z), is one of the most important functions in all of mathematics. It is the unique entire function that extends the real exponential function eˣ to the complex plane. This function is not only entire but also periodic.

To define the complex exponential function, we seek a function f: ℂ→ℂ that satisfies two key properties:

1. f(z₁ + z₂) = f(z₁)·f(z₂) for all z₁, z₂ ∈ ℂ
2. Consistency with the Real Exponential: f(x) = eˣ where x is a real number.

Let z = x + iy, where x, y ∈ ℝ. f(z) = f(x + iy) =[1] f(x)·f(iy) =[2] eˣ·f(iy)

Writing f(iy) = A(y) + iB(y), f(z) = eˣ·A(y) + ieˣ·B(y), where A(y) and B(y) are real-valued functions of y.

For f(z) to be a well-behaved function, we want it to be differentiable everywhere, which means it must satisfy the Cauchy-Riemann equations. The real and imaginary parts of f(z) are: u(x, y) = eˣ·A(y), v(x, y) = eˣ·B(y). The partial derivatives are uₓ = eˣ·A(y), u_y = eˣ·A’(y), vₓ = eˣ·B(y), and v_y = eˣ·B’(y).

Applying the Cauchy-Riemann equations, uₓ = v_y and u_y = - vₓ, we get: eˣ·A(y) = eˣ·B’(y), then, A(y) = B’(y). Similarly, eˣ·A’(y) = - eˣ·B(y), then A’(y) = -B(y).

From these two equations, we can derive a second-order differential equation for B(y): B’’(y) = -B(y) ↭ B’’(y) + B(y) = 0. This linear, homogeneous, constant-coefficient ODE has characteristic equation r² + 1 = 0 ⇒ r = ±i, the general solution is $y(t) = e^{at}(αcos(bt) + βsin(bt))$, where ‘α’ and ‘β’ are real numbers.

From A(0) = 1 and B(0) = 0 ⇒ β = 1, α = 0. Thus, we have uniquely determined the functions A(y) and B(y): B(y) = sin(y), A(y) = cos(y). Substituting these back into the expression for f(z), we arrive at the definition of the complex exponential function, f(z) = eˣ·cos(y) + ieˣ·sin(y) = eˣ(cos(y) + isin(y)).

The exponential function is well-defined for all z ∈ ℂ, entire, consistent with the real exponential, non-zero, periodic with period 2πi ($e^{z+2\pi i}=e^z $), $\frac{d}{dz}e^z = e^z$, and can be represented as a power series $e^z = \sum_{n=0}^∞ \frac{z^n}{n!}$ (it has infinite radius of convergence).

Harmonic functions

Definition. Let D ⊆ ℝ² be a domain (open and connected set). A function $\phi (x, y): D \to \real$ is C² on D if $\phi, \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial^2 \phi}{\partial x^2}, \frac{\partial^2 \phi}{\partial y^2}$ are all continuous on D. According to this definition, the following conditions are met.

1. D is open and connected.
2. The first partial derivatives $\frac{\partial \phi}{\partial x} \text{ and } \frac{\partial \phi}{\partial y}$ exist and are continuous on D.
3. 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.
4. The 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 C² definition, which typically includes all second-order partial derivatives (both mixed and unmixed).

   The standard (the previous definition is a weaker variant) meaning of C² on D requires all partial derivatives up to order 2 exist and are continuous on D: $\phi, \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial^2 \phi}{\partial x^2}, \frac{\partial^2 \phi}{\partial y^2}, \frac{\partial^2 \phi}{\partial y \partial x}, \frac{\partial^2 \phi}{\partial x \partial y}$ are continuous.

Definition. The Laplacian of a function $\phi (x, y)$ is $\nabla^2 \phi := \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial \phi^2}{\partial y^2} ↭ \nabla^2 \phi := \phi_{xx}+ \phi_{yy}$. The Laplace’s equation is $\nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial x^2} = 0$.

The standard notations for the Laplacian of a function $\phi (x, y)$ are $\nabla^2 \phi \text{ or } \Delta \phi $, and for its partial derivatives, $\frac{\partial \phi}{\partial x} \text{ or } \phi_x$, and for second partials, $\frac{\partial^2 \phi}{\partial x²} \text{ or } \phi_{xx}$. The superscript 2 should precede the variable to denote the second derivative, not the variable itself.

Definition. A real-valued function $\phi (x, y)$ is harmonic on a domain D ⊆ ℝ if it is C² on D and satisfies Laplace’s equation: $\nabla^2 \phi = \frac{\partial \phi^2}{\partial x^2} + \frac{\partial \phi^2}{\partial y^2} = 0$.

Understanding Harmonic Functions: A Physical and Mathematical Perspective

Let’s try to understand what a harmonic function is. Imagine you have a thin, homogeneous metal plate with no heat sources or sinks:

1. Initial State: You apply different temperatures to the boundary of the plate.
2. Transient State: Heat flows from warmer regions to cooler ones.
3. Steady State: Eventually, the heat will spread out, the temperature distribution stabilizes and no longer changes with time.

At this steady state, the temperature function $ u(x,y) $ is harmonic - it satisfies Laplace’s equation: $\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$

A harmonic function u(x,y) describes this steady-state temperature at every point (x,y) on the plate.

Consider a small square region around a point (x, y):

• If $\frac{\partial^2 u}{\partial x^2} > 0$, the temperature curves upward in the x-direction (like a valley)
• If $\frac{\partial^2 u}{\partial x^2} < 0$, it curves downward (like a hill).
• Same for the y-direction.

This equation says that at every point, the function curves in the x-direction and y-direction in such a way that they perfectly cancel each other out. It requires that these curvatures exactly balance:

• Where the function curves up in one direction, it must curve down in the other
• This perfect thermal balance ensures no net heat flow into or out of any point.
• In physical terms: What flows in must flow out.

The mean value property has a beautiful physical explanation. If you have a steady-state temperature distribution, the temperature at any point must equal the average temperature of its immediate surroundings.

Analytic functions have harmonic real/imaginary parts

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

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

Proof that u is harmonic.

Differentiate the CR equations appropriately and use equality of mixed partials (Clairaut’s theorem):

1. From $u_x = v_y$, differentiate in x: $u_{xx} = v_{xy}$.
2. From $u_y = -v_x$, differentiate in y: $u_{yy} = -v_{yx}$.
3. Add and use $v_{yx} = v_{xy}$ (Clairaut’s theorem): $u_{xx} + u_{yy} =[\text{1, 2}] v_{xy} - v_{yx} =[\text{Clairaut’s theorem}] v_{xy} - v_{xy} = 0$. Thus, u is harmonic. The same argument (swap roles of u and v) gives Δv = 0.

The proof provided is correct under the standard definition of C². However, the proof remains valid for analytic functions.

1. Analyticity implies C^∞ smoothness. If f(z) = u(x, y) + iv(x, y) is analytic on a domain D, then f is infinitely differentiable (C^∞) on D. Consequently, its real and imaginary parts u and v are also C^∞ on D.
2. C^∞ satisfies all C^k definitions. Since u and v are C^∞, they are automatically C² in the standard sense (all second-order partials, including mixed derivatives, exist and are continuous). This satisfies the hypotheses of Clairaut’s theorem, ensuring $v_{yx} = v_{xy}$.
3. Proof validity under the weaker C² definition. Even though our definition of C² does not explicitly require continuity of mixed partials, the analyticity of f guarantees the stronger condition C^∞. Thus, the equality $v_{yx} = v_{xy}$ holds, and the proof is unaffected.

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:

1. v is also harmonic on D.
2. 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. 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.

Proof:

1. Existence and Continuity of First-Order Partial Derivatives: Since u and v are harmonic on D, they have continuous second-order partial derivatives. This implies that their first-order partial derivatives exist and are continuous on D (because continuity of higher-order derivatives implies continuity of lower-order derivatives).
2. Satisfaction of the Cauchy-Riemann Equations. By the definition of a harmonic conjugate, u and v satisfy the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$. This is a direct consequence of v being a harmonic conjugate of u.
3. The sufficient condition for analyticity is that the first-order partial derivatives of u and v exist, are continuous, and satisfy the Cauchy-Riemann equations in D.Therefore, by the sufficient condition for complex differentiability (C¹+ CR ⇒ holomorphic), f(z) = u + iv is analytic on D.

Example. Let u(x, y) = x² - y², which is harmonic since: $\frac{\partial u^2}{\partial x^2} + \frac{\partial u^2}{\partial x^2} = 2 -2 = 0$.

To find a harmonic conjugate v, we use the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = 2x = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -2y = - \frac{\partial v}{\partial x}$

Integrating the first equation $\frac{\partial v}{\partial y} = 2x$ with respect to y: v(x, y) = $\int 2xdy = 2xy + h(x)$ where h(x) is an arbitrary function of x.

Now, differentiate with respect to x and use the second Cauchy-Riemann equation: $\frac{\partial v}{\partial x} = 2y + h'(x) = 2y \text{ (since } -\frac{\partial v}{\partial x} = -2y \leadsto \frac{\partial v}{\partial x} = 2y \text{)}$ ⇒ h’(x) = 0 ⇒ h(x) = C (a constant). Setting C = 0 for simplicity, v(x, y) = 2xy.

Thus, f(z) = u + iv = x² - y² + i(2xy) = (x + iy)² = z². The function z² is analytic everywhere in ℂ (as it is a polynomial), confirming the theorem.

The theorem assumes the existence of a harmonic conjugate v on D. For a harmonic conjugate to exist, D must be simply connected (e.g., no holes). The harmonic conjugate v is unique up to an additive constant (as seen in the example, where h(x) = C). The continuity of the first-order partial derivatives (from harmonicity) ensures the applicability of the sufficient condition for analyticity.

Theorem. If u ∈ C²(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.

A harmonic conjugate exists for any harmonic function on a simply connected domain (a domain with no holes, such as a disk or the entire plane).

$u(x, y) = \frac{log(x² + y²)}{2}$ is harmonic on the domain ℝ²∖ {(0,0)}, but it does not have a single-valued harmonic conjugate on this domain because it is not simply connected.

1. First, let’s verify that $u(x, y) = \frac{\log(x^2 + y^2)}{2}$ is harmonic on $\mathbb{R}^2 \setminus \{(0,0)\}$. A function is harmonic if it satisfies Laplace’s equation: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$.
   Computing the first partial derivatives: $\frac{\partial u}{\partial x} = \frac{1}{2} \cdot \frac{1}{x^2 + y^2} \cdot 2x = \frac{x}{x^2 + y^2}, \frac{\partial u}{\partial y} = \frac{1}{2} \cdot \frac{1}{x^2 + y^2} \cdot 2y = \frac{y}{x^2 + y^2}$
   Now computing the second partial derivatives: $\frac{\partial^2 u}{\partial x^2} = \frac{(x^2 + y^2) - x \cdot 2x}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}, \frac{\partial^2 u}{\partial y^2} = \frac{(x^2 + y^2) - y \cdot 2y}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2}$
   Adding these together: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} + \frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2 + x^2 - y^2}{(x^2 + y^2)^2} = 0$.
2. The Issue with Harmonic Conjugate. A harmonic conjugate of $u$ is a function $v$ such that $f(z) = u + iv$ is analytic. For $v$ to exist, it must satisfy the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
   Using our computed partial derivatives of u: $\frac{\partial v}{\partial y} = \frac{x}{x^2 + y^2}, \frac{\partial v}{\partial x} = -\frac{y}{x^2 + y^2}$.
   Integrating the first equation with respect to y: v(x, y) = $\int \frac{x}{x^2 + y^2} dy = \arctan\left(\frac{y}{x}\right) + g(x)$
   [Last step] Let $u = \frac{y}{x}$ so $dy=xdu$. Then $\int \frac{x}{x^2+y^2}dy = \int \frac{x}{x^2+x^2u^2}xdu = \int \frac{x}{x^2(1+u^2)}xdu=\int \frac{1}{1+u^2}du=\arctan(u)+C=\arctan\left(\frac{y}{x}\right)+g(x).$
   Differentiating v(x, y) with respect to x. Let $u = \frac{y}{x}, \frac{\partial}{\partial x} \arctan(u) = \frac{1}{1+u^2} \cdot \frac{\partial u}{\partial x}; u = y \cdot x^{-1} \Rightarrow \frac{\partial u}{\partial x} = -y \cdot x^{-2} = -\frac{y}{x^2}$.
   Since $1 + \left(\frac{y}{x}\right)^2 = \frac{x^2 + y^2}{x^2}$, we have $\frac{\partial}{\partial x} \arctan\left(\frac{y}{x}\right) = \frac{x^2}{x^2 + y^2}·(-\frac{y}{x^2}) = -\frac{y}{x^2 + y^2}$.
   Then, using the second Cauchy-Riemann equation: $\frac{\partial v}{\partial x} = \frac{-y}{x^2 + y^2} + g'(x) = -\frac{y}{x^2 + y^2}$. This implies $g'(x) = 0$, so $g(x)$ is a constant $C$.
   Therefore: $v(x, y) = \arctan\left(\frac{y}{x}\right) + C$
   In polar coordinates, $\arctan\left(\frac{y}{x}\right)$ is the angle $\theta$, which is the argument of the complex number $z = x + iy$. So we can write: $v(x, y) = \arg(z) + C$
3. Why the Harmonic Conjugate is Not Single-Valued. The problem is that $\arg(z)$ is not single-valued on $\mathbb{R}^2 \setminus \{(0,0)\}$. As we traverse a closed curve around the origin, $\arg(z)$ increases by $2\pi$ (or a multiple of $2\pi$), depending on how many times we wind around the origin.

This is directly related to the topology of the domain $\mathbb{R}^2 \setminus \{(0,0)\}$. This domain is not simply connected because there exist closed curves (specifically, those loops encircling around the origin) that cannot be continuously contracted to a point without leaving the domain.

In topology, a space is simply connected if it’s both. 1. Path‑connected: you can get from any point to any other by a continuous path that stays in the space. 2. Loop‑shrinkable: every closed loop in the space can be continuously deformed (contracted) to a single point without leaving the space. Examples: ℂ itself, disks, half-planes. Counterexamples: ℂ∖{0} (punctured plane), a doughnut shape (torus) where the hole through the middle stops some loops from contracting.

In simply connected domains, every harmonic function has a single-valued harmonic conjugate. However, in domains that are not simply connected, this property is not guaranteed, as we see in this case.

If we write z = x + i y, then in polar form $z = re^{i\theta}$ with: $r = |z| = \sqrt{x^2 + y^2}, \theta = \arg(z).$ For z ≠ 0: $\log z = \log r + i \theta$ where:

• $\log r = log(\sqrt{x^2 + y^2}) = log((x^2 + y^2)^{\frac{1}{2}}) = \frac{1}{2}\log(x^2 + y^2)$ is the real part u(x, y) of the complex logarithm $\log(z) = \log|z| + i\arg(z)$.
• $\theta = \arg(z)$ is the imaginary part v(x,y), which is multi‑valued because going around the origin changes θ by 2π, $k\in\mathbb{Z}$. The complex logarithm is multi-valued precisely because $\arg(z)$ is multi-valued, and this is why we need to define branches of the logarithm when working with it.

The Problem: Finding the Harmonic Conjugate

The main question is: Given a harmonic function u, can we always find its harmonic conjugate v? And if so, is it unique?

In a simply connected domain, every harmonic function has a single-valued harmonic conjugate, unique up to an additive constant.

The existence of v relies on solving $dv = \frac{\partial v}{\partial x}dx + \frac{\partial v}{\partial y}dy =[\text{ u and v must satisfy the Cauchy–Riemann equations }] -\frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy$.

This is saying that if we move a tiny amount dx in the x-direction and dy in the y-direction, then v should change by the amount on the right side.

The expression $-\frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy$ is called a 1-form. This 1-form is closed (because u is harmonic -the proof is left as an exercise to the reader $\blacksquare$ QED, Quod Erat Demonstrandum), and in simply connected domains: every closed form is exact ($\blacksquare$ QED), i.e., it’s the differential of some function. Then, there exists a function v such that $dv = -\frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy$. This is exactly what we wanted! This means v exists.

When D is simply connected and a 1-form is exact, something wonderful happens: line integrals become path-independent. Thus, v can be defined unambiguously via: $v(z) = \int_{z_0}^z -\frac{\partial u}{\partial y}dx + \frac{\partial u}{\partial x}dy + C$ for any path from z₀ to z in D. In other words, this integral gives the same result regardless of which path you take from point z₀ to point z.

Why v is Unique Up to a Constant. Suppose we have two different harmonic conjugates of u, call them v₁ and v₂. This means both satisfy:

$\frac{\partial u}{\partial x} = \frac{\partial v_1}{\partial y} = \frac{\partial v_2}{\partial y}$; $\frac{\partial u}{\partial y} = -\frac{\partial v_1}{\partial x} = -\frac{\partial v_2}{\partial x}$ From the first equation: $\frac{\partial v_1}{\partial y} = \frac{\partial v_2}{\partial y}$. This means: $\frac{\partial (v_1 - v_2)}{\partial y} = 0$

From the second equation: $-\frac{\partial v_1}{\partial x} = -\frac{\partial v_2}{\partial x}$. This means: $\frac{\partial (v_1 - v_2)}{\partial x} = 0$

So the function (v₁ - v₂) has zero partial derivatives in both x and y directions. In a connected domain (which simply connected domains always are), this means v₁ - v₂ must be constant everywhere. Therefore: v₁ = v₂ + constant. This proves that the harmonic conjugate is unique up to adding a constant.

Theorem. Let u(x, y) be harmonic on a domain D, and let v be a harmonic conjugate of u. If the function f(z) = u(x, y) + iv(x, y) where z = x + iy, satisfies |f(z)| = k for all z ∈ D where k is a constant, then u and v are both constants on D.

Proof.

Since v is a harmonic conjugate of u, by the previous Theorem, the function f(z) = u(x, y) + iv(x, y) is analytic on D.

By assumption, |f(z)| = k for all z ∈ D. This implies u(x, y)² + v(x, y)² = k² ∀(x, y) ∈ D.

As u and v are harmonic (and thus have continuous first-order partial derivatives), we differentiate this equation with respect to x and y:

Differentiating with respect to x: $2u\frac{\partial u}{\partial x} + 2v\frac{\partial v}{\partial x} = 0$.
Differentiating with respect to y: $2u\frac{\partial u}{\partial y} + 2v\frac{\partial v}{\partial y} = 0$
Applying the Cauchy-Riemann equations $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$, we get: $u\frac{\partial u}{\partial x} - v\frac{\partial u}{\partial y} = 0, u\frac{\partial u}{\partial y} + v\frac{\partial u}{\partial x} = 0$.

These equations from a linear system in $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$.

$\begin{cases} u\frac{\partial u}{\partial x} - v\frac{\partial u}{\partial y} = 0 \\\\ v\frac{\partial u}{\partial x} + u\frac{\partial u}{\partial y} = 0 \end{cases}$

The determinant of the coefficient matrix is: u² + v² = k². We consider two cases:

Case 1. If k = 0, then u = v = 0 everywhere in D. Thus, u and v are constant.
Case 2. If k ≠ 0, the determinant k² is non-zero (this is a homogeneous linear system of two equations and two variables, and the determinant of the coefficient matrix is not zero, the system has only one solution; that one solution is the trivial solution), so the only solution is: $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0$

Since D is a domain (open and connected), “locally constant” implies “constant”, u is constant on D. In other words, Since D is connected and u has vanishing partial derivatives, u is constant on D (check out the proof). By the Cauchy-Riemann equations, v is also constant.

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Theorem. If u: D → ℝⁿ is a function on a connected domain D ⊆ ℝⁿ such that: $\frac{\partial u}{\partial x_i}$ for all i = 1, 2, …, n, then u is constant on D.

The crucial point is that connectedness allows us to bridge from local (in every neighborhood, the function doesn’t change) to global (the function has the same value throughout the entire domain) behavior.

Step 1. If ∇u = 0 at every point, then by the Mean Value Theorem, u is locally constant around each point.

Recall. The classical Mean Value Theorem. If f:[a,b]→ℝ is continuous on [a,b] and differentiable on (a,b), then there exists some c ∈ (a,b) such that: $f'(c) = \frac{f(b)-f(a)}{b-a} ↭ f(b) - f(a) = f'(c)(b - a)$.

1. Given $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0$ at every point in domain D.
2. For any two points P₁ = (x₁, y₁) and P₂ = (x₂, y₂) in a small neighborhood, consider the line segment connecting them.
3. Parametrize the line segment: $\gamma(t) = (1-t)P_1 + tP_2 = (x_1 + t(x_2-x_1), y_1 + t(y_2-y_1))$ for t ∈ [0,1].
4. Define the function along this path: $g(t) = u(\gamma(t)) = u(x_1 + t(x_2-x_1), y_1 + t(y_2-y_1))$
5. Apply the chain rule: $g'(t) = \frac{\partial u}{\partial x} \cdot (x_2-x_1) + \frac{\partial u}{\partial y} \cdot (y_2-y_1)$
6. Since ∇u = 0: $g'(t) = 0 \cdot (x_2-x_1) + 0 \cdot (y_2-y_1) = 0$
7. Apply the classical (ℝ) Mean Value Theorem to g(t): Since g′(t)=0 for all t∈[0,1], there exists c ∈ (0, 1) such that: $g(1) - g(0) = g'(c)(1-0) = 0 \cdot 1 = 0$.
8. Therefore: $u(P_2) = g(1) = g(0) = u(P_1)$.

More precisely, we’re using the Multivariable Mean Value Theorem:

If $u: D → \mathbb{R}$ is differentiable on a convex set D, then for any two points $\mathbf{a}, \mathbf{b} \in D$, there exists a point c on the line segment connecting a and b such that: $u(\mathbf{b}) - u(\mathbf{a}) = \nabla u(\mathbf{c}) \cdot (\mathbf{b} - \mathbf{a})$

If ∇u(c) = 0 for all points c, then: u(b) − u(a) = 0⋅(b − a) = 0. Therefore: u(b) = u(a).

Step 2. Global Extension via Connectedness

1. Fix any point a ∈ D and define S = {x ∈ D: u(x) = u(a)}.
2. S is open: For any x ∈ S, since u is locally constant around each point x, there exists a neighborhood where u(x) = u(a).
3. S is closed in D: If $\\{x_n\\} ⊂ S$ converges to x ∈ D, then by continuity, u(x) = $\lim u(x_n) = u(a) $, so x ∈ S.
4. S ≠ 0 since a ∈ S.
5. Since D is connected and S is both open and closed in D, we must have S = D. Therefore, u(x) = u(a) for all x ∈ D.

Recall. A topological space D is disconnected if it can be written as D = U ∪ V, where U and V are disjoint, non-empty open subsets of D. If D cannot be expressed in this way, it is connected.

If S is a clopen (it is both open and closed) subset of D, then its complement D\S is also clopen. If S is nonempty and not equal to D, then both S and D\S are nonempty open sets. Since S and D\S are disjoint and their union is D, this would mean D is disconnected. Therefore, in a connected space, no such proper nonempty clopen subset can exist.

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Theorem. Let u(x, y) be harmonic on a domain D and v a harmonic conjugate of u. If f’(z) = 0 for all z ∈ D, where f(z) = u(x, y) + iv(x, y), then u and v are both constants on D.

Proof.

Since v is a harmonic conjugate of u, the function f(z) = u(x, y) + iv(x, y) is analytic on D by the previous theorem.

The complex derivative is $f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}$

Given f′(z) = 0 for all z ∈ D, we have: $\frac{\partial u}{\partial x} = 0$ and $\frac{\partial v}{\partial x} = 0$

By the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = 0, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} = 0$. Thus, $\nabla u = (\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}) = (0, 0), \nabla u = (\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y})$ all partial derivatives of u and v are zero. Since D is connected and all partial derivatives vanish identically, u and v are constant on D.

Exercise. Show that u(x, y) = xy³ -x³y is a harmonic function and find the harmonic conjugate of u(x, y)

First, let’s demonstrate that u(x, y) = xy³ -x³y is indeed a harmonic function.

$\frac{\partial u}{\partial x} = y^3 -3x^2y, \frac{\partial u}{\partial y} = 3xy^2 -x^3, \frac{\partial^2 u}{\partial x^2} = -6xy, \frac{\partial^2 u}{\partial y^2} = 6xy, \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ at any (x, y) ∈ ℂ. Thus, $u(x,y)=xy^3-x^3y$ is harmonic.

Harmonic conjugate via Cauchy–Riemann. Use the Cauchy–Riemann equations for f=u+iv: $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} \text{ (i) } \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \text{ (ii) }$

From (i): $\frac{\partial v}{\partial y} = y^3 -3x^2y$. Integrate $\frac{\partial v}{\partial y}$ with respect to y: $v = \int (y^3 -3x^2y)dy + g(x) = \frac{y^4}{4}-\frac{3x^2y^2}{2}+ g(x)$ where g(x) is a function of x.

From (ii), differentiate v with respect to x and match $\frac{\partial v}{\partial x}$: $-3xy^2 + g'(x) = -3xy^2 +x^3 \leadsto g'(x) = x^3 \leadsto g(x) = \frac{x^4}{4} + C$ where C is a constant of integration.

Therefore, a harmonic conjugate is, $v = \frac{y^4}{4}-\frac{3x^2y^2}{2}+ \frac{x^4}{4} + C$

Exercise. Find an analytic function u + iv given that its imaginary part is v(x, y) = $e^ysin(x)$

Compute the partial derivatives of v(x, y): $\frac{\partial v}{\partial x} = e^ycos(x), \frac{\partial v}{\partial y} = e^ysin(x)$

Check for harmonicity: $\frac{\partial^2 v}{\partial x^2} = -e^ysin(x), \frac{\partial^2 v}{\partial x^2} = e^ysin(x), \nabla^2 v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = -e^ysin(x) + e^ysin(x) = 0.$ Use the Cauchy-Riemann equations to find u(x,y): For a function f(z) = u(x, y) + iv(x, y) to be analytic, it must satisfy the Cauchy-Riemann equations: $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} \text{ (i) } \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \text{ (ii) }$

From (i): $\frac{\partial u}{\partial x} = e^ysin(x)$. Integrate $\frac{\partial u}{\partial x}$ with respect to x: $u(x, y) = \int (e^ysin(x))dx + g(y) = -e^ycos(x) + g(y)$. Here, g(y) is an arbitrary function of y, arising from the integration.

From (ii) differentiate our expression with respect to y and match $\frac{\partial v}{\partial x}$: $\frac{\partial u}{\partial y} = -e^ycos(x) + g'(y) = -e^ycos(x) \leadsto g'(y) = 0 \leadsto g(y) = C$, so g(y) is a real constant.

Now we have: u(x, y) = $-e^ycos(x) + C$. Thus, the analytic function is: $f(z) = u(x, y) + iv(x, y) = -e^ycos(x) + C + i(e^ysin(x))$.

Or, more compactly: $f(z) = C + e^y(-\cos(x) + i \sin(x)) =[-\cos(x)+i\sin(x)=-e^{-ix}] C + e^y·(-e^{-ix})$. Since $e^y e^{-ix} = e^{y - ix} \leadsto f(z) = C - e^{y - ix}$. Since y -ix = -iz, we conclude: $\boxed{f(z) = C - e^{-iz}}$