advanced30 min studyLesson 369 in Mathematics

Harmonic functions and Dirichlet problems

Connect holomorphic real parts to Laplace's equation and solve the continuous boundary-value problem on a disk.

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What you’ll understand

  • Relate harmonic functions to local holomorphic conjugates.
  • State disk Dirichlet existence and uniqueness with boundary regularity.
  • Extend finite Fourier boundary data into harmonic disk modes.

The real and imaginary parts of a holomorphic function solve Laplace's equation. This connects complex differentiation with steady potentials, temperature distributions, and boundary-value problems. The converse is local: a harmonic real function has a conjugate nearby, but topology can obstruct one global conjugate. On a disk the Poisson kernel constructs the harmonic function with prescribed continuous boundary values, and the maximum principle proves uniqueness. These statements concern a particular domain and boundary regularity, rather than every irregular region or arbitrary boundary assignment.

Harmonicity is a differential equation with a local conjugate

A real-valued twice continuously differentiable function u is harmonic on an open set when its Laplacian vanishes. If u is the real part of a holomorphic f=u+iv, the Cauchy–Riemann equations and commuting mixed derivatives give that equation. Holomorphic functions are smooth by Cauchy theory, so the derivative interchange is justified. The imaginary part is harmonic as well. Harmonicity therefore captures part of the structure of a holomorphic function without recording the full complex-valued map.

Conversely, the equations for a conjugate are vx=−uyv_x=-u_y and vy=uxv_y=u_x. Harmonicity makes this differential form locally integrable, so a conjugate exists on small disks. On a simply connected domain the path integrals can define a global conjugate, unique up to an additive real constant. On the punctured plane, log|z| is harmonic but a global conjugate would provide a single-valued argument, which is impossible. A local derivative condition does not remove a global period obstruction.

Δu=uxx+uyy=0,dv=−uy dx+ux dy,f=u+iv holomorphic locally\Delta u=u_{xx}+u_{yy}=0,\qquad dv=-u_y\,dx+u_x\,dy,\qquad f=u+iv\text{ holomorphic locally}
For real C² harmonic u the displayed form is closed and has local primitives. A simply connected domain gives a global conjugate; a domain with holes requires checking its periods.↔ Scroll wide equations horizontally.

The disk Poisson kernel solves a precise boundary problem

Give a continuous real function F on the unit circle. Its Poisson integral defines a harmonic function for r<1, extends continuously to the closed disk, and equals F on the boundary. The kernel is positive and has integral 2π, so the construction averages boundary values with location-dependent weights. At the center those weights are uniform, recovering the mean-value property. Toward a boundary point they concentrate near that point, which explains why continuity yields the stated boundary limit.

Uniqueness follows from the maximum principle: the difference of two harmonic continuous-closure solutions has zero boundary values and therefore has neither a positive maximum nor a negative minimum inside. Boundedness of the domain and continuity up to the boundary matter for this argument. On other domains, existence can depend on boundary regularity and the meaning of boundary data. Do not import the disk's clean conclusion into a region with unexamined boundary behavior.

u(reiθ)=12π∫02π1−r21−2rcos⁡(θ−t)+r2F(eit) dt,0≤r<1u(re^{i\theta})=\frac1{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-t)+r^2}F(e^{it})\,dt,\qquad 0\le r<1
For continuous real boundary data on the unit circle, this is the unique harmonic solution continuous on the closed disk with those boundary values.↔ Scroll wide equations horizontally.

Boundary frequencies decay radially, not by heat time

A boundary cosine of frequency n extends as rⁿcos(nθ), and a boundary sine extends as rⁿsin(nθ). These are real or imaginary parts of zⁿ and hence harmonic. Finite Fourier data can therefore be extended term by term without a convergence complication. Higher angular frequencies are attenuated more strongly at a fixed interior radius, expressing a spatial smoothing effect. The constant mode remains constant and determines the center value.

This radial attenuation should not be confused with heat evolution, whose Fourier modes decay as the exponential of minus diffusivity times n² times time. Both favor low frequencies, but they solve different equations with different independent variables and factors. Checking the governing equation prevents one appealing smoothing animation from being mistaken for another solver. For a proposed disk solution, verify harmonicity, boundary matching, and continuity separately; uniqueness then certifies that no second solution with the same conditions has been overlooked.

Extend three boundary Fourier terms

Solve the unit-disk Dirichlet problem with boundary values F(eiθ)=2+3cos⁡θ−cos⁡2θF(e^{i\theta})=2+3\cos\theta-\cos2\theta. Evaluate the solution at (x,y)=(0.2,0.4) and find a harmonic conjugate.

  1. Extend each mode radially: u=2+3rcos⁡θ−r2cos⁡2θu=2+3r\cos\theta-r^2\cos2\theta. In Cartesian coordinates this is u(x,y)=2+3x−x2+y2u(x,y)=2+3x-x^2+y^2. Its Laplacian is −2+2=0.

  2. On r=1 the formula matches the continuous boundary data. It is polynomial and continuous on the closed disk, so uniqueness makes it the required solution. At (0.2,0.4), its value is 2+0.6−0.04+0.16=2.722+0.6-0.04+0.16=2.72.

  3. The holomorphic polynomial f(z)=2+3z−z2f(z)=2+3z-z^2 has this real part. Its imaginary part is v(x,y)=3y−2xyv(x,y)=3y-2xy, a conjugate with v(0,0)=0. At the chosen point, v=1.2−0.16=1.04.

The result

The harmonic solution is 2+3x−x²+y² and its normalized conjugate is 3y−2xy. The value 2.72 is fixed by the whole boundary data, not by a single boundary sample.

Explore the model

Interactive lab

Interactive lab · heat

Distinguish two kinds of mode smoothing

Pause and Reset; set Spatial mode n=1 and Diffusivity κ=1, then Step once to time 0.5 and predict amplitude e^(−0.5). For a fresh run Reset, set n=2,κ=1, and Step once: predict e^(−2). By contrast, a disk harmonic mode at radius r=e^(−0.5) has attenuation r²=e^(−1) for n=2. The lab computes periodic heat flow, not the Poisson integral; the comparison exposes their different n versus n² factors.

Distinguish two kinds of mode smoothingThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forSimilar-looking smoothing can arise from different equations, so connect each attenuation factor to its actual independent variable.

Check your understanding

Which function extends boundary data cos(3θ) harmonically into the unit disk and continuously onto its boundary?

Try it yourself

Find the harmonic disk extension of cos⁡3θ+sin⁡2θ\cos3\theta+\sin2\theta and evaluate it at r=1/2, θ=π/4. Verify the Cartesian expression is harmonic.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Extend each mode by its own power of r, then use Re(z³) and Im(z²) to check the Laplacian.

Compare with the solution

The solution is u=r3cos⁡3θ+r2sin⁡2θ=x3−3xy2+2xyu=r^3\cos3\theta+r^2\sin2\theta=x^3-3xy^2+2xy. Its second derivatives sum to 6x−6x=06x-6x=0, and it matches the continuous boundary data. At the specified point the value is −2/16+1/4=(4−2)/16≈0.16161-\sqrt2/16+1/4=(4-\sqrt2)/16\approx0.16161. Continuity on the closed disk and the maximum principle give uniqueness.

Take it with you

  • Harmonic functions have local holomorphic conjugates; global conjugates can face period obstructions.
  • Continuous circle data have a unique continuous-closure harmonic disk extension.
  • Poisson radial attenuation uses rⁿ, distinct from heat evolution's exponential n² factor.

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