Harmonic Functions and What Analyticity Forces
Show that the real and imaginary parts of an analytic function satisfy Laplace's equation, and construct harmonic conjugates with attention to the domain.
What a learner can do afterwards
- Prove that analytic components are harmonic and state the maximum principle with its hypotheses
- Construct a harmonic conjugate and state conditions for global existence
- Explain the connection to steady heat and electrostatic potential, including conjugate level curves
1 · Read
An open connected set D is called a domain. A real-valued function u with continuous second partial derivatives is harmonic on D if Δu = u_xx + u_yy = 0 throughout D. Compute Δu by differentiating twice in x, twice in y, then adding. At a point where u is harmonic, u_yy = -u_xx.
As in the Cauchy-Riemann lesson, write z = x + iy and use i² = -1. Expanding z² + 4z gives u = x² - y² + 4x and v = 2xy + 4y. The terms without i form u; the coefficient of i forms v. Here u_xx = 2 and u_yy = -2, while v_xx = v_yy = 0. Both Laplacians vanish.
Here analytic means holomorphic on an open domain. Analytic functions have smooth components, so their mixed partials agree. From u_x = v_y and u_y = -v_x, obtain u_xx = v_yx and u_yy = -v_xy. Thus Δu = 0. Also v_xx = -u_yx and v_yy = u_xy, so Δv = 0.
The maximum principle says that an interior local maximum or minimum forces a harmonic function to be constant on its connected domain. Constant harmonic functions can attain extrema inside. On a bounded domain, a harmonic function continuous on the closure attains its maximum and minimum on the boundary. Without those conditions, an extremum need not exist: u = x has no maximum on the whole plane.
A harmonic conjugate v makes u + iv analytic. Solve v_y = u_x and v_x = -u_y. Integrate u_x with respect to y, keeping an unknown function h(x): v = ∫u_x dy + h(x). Use v_x = -u_y to find h'(x), then integrate it. Check both equations throughout the domain. On a connected domain, two conjugates of the same u differ by a real constant.
For u = x² - y² + 5x, integrating v_y = 2x + 5 gives v = (2x + 5)y + h(x). Since v_x must equal 2y, h'(x) = 0. Thus v = 2xy + 5y + C and f(z) = z² + 5z + iC. A specified value of v at one point fixes C.
For u = ax + by, conjugate construction gives v = ay - bx + C and f(z) = (a - ib)z + iC. You can also recover the coefficient directly: if c = A + iB, then cz = Ax - By + i(Ay + Bx). Comparing coefficients of x and y in the real part determines A and B.
Two harmonic functions need not form an analytic pair. For u = x and v = -y, both Laplacians vanish, but u_x = 1 differs from v_y = -1. For a proposed pair, calculate the residuals u_x - v_y and u_y + v_x. Both must vanish throughout the domain. With continuous first partials, these equations establish analyticity.
For a real constant a, u = e^(ax) cos(ay) has u_xx = a²u and u_yy = -a²u, so Δu = 0. Differentiating cos(ay) in y gives -a sin(ay), then -a² cos(ay). A conjugate is v = e^(ax) sin(ay) + C: v_y = u_x and v_x = -u_y. Substitute a given value of v to find C before evaluating elsewhere. Angles are in radians.
A harmonic u has a conjugate near each point. A global single-valued conjugate exists if every closed loop in the domain can shrink to a point inside it. A hole can obstruct this. For r = sqrt(x² + y²), u = log r is harmonic on the punctured plane. A local conjugate is the angle θ in radians. One circuit about 0 changes θ by 2π. Adding a constant cannot remove this change, so no global conjugate exists there.
For heat diffusion with constant diffusivity κ > 0 and no internal sources, T_t = κΔT. Steady state means T_t = 0, so ΔT = 0. With a source term s, the model is T_t = κΔT + s. At steady state, ΔT = -s/κ. Thus steady temperature is not harmonic where s is nonzero.
In a charge-free planar region with constant permittivity, electrostatic potential u satisfies Δu = 0. The gradient ∇u = (u_x,u_y) is normal to a regular level curve. If v is a conjugate, CR gives ∇u·∇v = 0. Where ∇u ≠ 0, the two level curves cross at right angles. The field E = -∇u is tangent to v's level curve and normal to u's. At a zero gradient, this regular-curve description need not apply.
Analytic functions have harmonic components. Construct a conjugate with both Cauchy-Riemann equations, and check the domain before claiming a global one.
Original lesson, questions and flashcards written for this dataset, with revisions for mathematical accuracy.
Three original readings and nine original replacement questions written for this dataset. Flashcards revised for mathematical accuracy. Reading and question source identities are recorded in the file.
More to read
Why both analytic components are harmonic
The planar Laplacian of u is Δu = u_xx + u_yy. A real-valued function with continuous second partial derivatives is harmonic when this sum is zero throughout its open domain.
Suppose f = u + iv is analytic. Its components are smooth and satisfy the Cauchy-Riemann equations u_x = v_y and u_y = -v_x. Differentiate the first in x and the second in y. Adding gives Δu = v_yx - v_xy = 0 because the mixed partials agree. Reversing the roles gives Δv = -u_yx + u_xy = 0. Both components are harmonic.
The converse needs care. Two harmonic functions need not satisfy the Cauchy-Riemann equations together. For u = x and v = -y, both Laplacians vanish, but u_x = 1 and v_y = -1. This pair does not form an analytic function.
Harmonic functions satisfy a mean value property: the value at the centre of a circle equals the average around the circle when its closed disc lies inside the domain. An interior local maximum or minimum forces a harmonic function to be constant on a connected domain. Constant functions therefore are exceptions to an unconditional ban on interior extrema. Boundary attainment additionally needs a bounded domain and continuity on its closure.
Which steady physical models are harmonic?
A temperature profile is harmonic under specified physical assumptions. For diffusion with constant diffusivity κ > 0 and no internal sources, the heat equation is T_t = κΔT. At steady state the temperature stops changing with time, so T_t = 0 and ΔT = 0. Spatial variation can remain: steady does not mean constant in space.
An internal source changes the conclusion. For T_t = κΔT + s, steady state gives ΔT = -s/κ. The temperature is not harmonic where s is nonzero. Steadiness alone is therefore insufficient.
Electrostatic potential satisfies Laplace's equation in a charge-free region with constant permittivity. A planar model uses the same two-variable equation as complex analysis. Charges within the region change the equation. For incompressible, irrotational planar flow, a velocity potential is also harmonic wherever the flow is regular; an arbitrary steady flow need not meet those conditions.
Boundary values give uniqueness when the maximum principle applies. Suppose two harmonic functions on a bounded domain are continuous on its closure and agree on its boundary. Their difference is harmonic and zero on the boundary. Its maximum and minimum are zero, so the two functions agree throughout the domain. This proves uniqueness of a solution if one exists; it does not establish existence for arbitrary data.
Constructing a conjugate and checking its domain
To find a harmonic conjugate v of u, integrate v_y = u_x with respect to y. Keep an unknown function h(x), since differentiation in y would erase it. Then use v_x = -u_y to determine h'(x). Integrate that equation and check both Cauchy-Riemann equations. On a connected domain, any two conjugates differ by a real constant.
Conjugates also describe level curves. The equations give ∇u·∇v = u_x(-u_y) + u_yu_x = 0. The gradient is normal to a regular level curve. Where ∇u is nonzero, the level curves of u and v meet at right angles. If u is an electrostatic potential, E = -∇u is normal to u's level curve and tangent to v's. At zero gradients this regular-curve description may fail.
A harmonic function has a conjugate near each point. A global single-valued conjugate exists on a domain where every closed loop can shrink to a point within the domain. A hole can obstruct this extension.
For u = log r on the punctured plane, a local conjugate is the angle θ in radians. Following a loop once around the missing origin changes θ by 2π. Adding a constant leaves that change intact. A single-valued function must return to its original value at the same point, so log r has no global single-valued conjugate on the punctured plane.
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