advanced33 min studyLesson 395 in Mathematics

Green's functions and Poisson kernels

Represent source and boundary effects with kernels whose normalization, signs, and domains are explicit.

On this page

What you’ll understand

  • Verify a Dirichlet Green kernel through its derivative jump.
  • Use the normalized disk Poisson kernel to extend continuous boundary data.
  • Distinguish a whole-space fundamental solution from a boundary-adapted Green function.

A source at one location influences an elliptic solution throughout the domain. A Green function records that influence while respecting a chosen boundary condition. Boundary forcing has its own representation through a Poisson kernel. These formulas are powerful precisely because their signs, normalization, and geometry are fixed rather than left implicit. A whole-space fundamental solution does not automatically satisfy a wall condition, and a disk kernel cannot be reused unchanged on an arbitrary domain. We will verify simple kernels directly before interpreting broader representation formulas.

Adapt the point source to the boundary

For minus the Laplacian, a fundamental solution satisfies the distributional equation with a unit delta source in the whole space. In dimension at least three it is proportional to the distance raised to two minus the dimension; in two dimensions it is minus the logarithm divided by two pi. A Dirichlet Green function adds a harmonic correction so that its boundary values vanish. The correction depends on the domain, even though the local singularity near the source is inherited from the fundamental solution.

On the unit interval the boundary-adapted kernel is especially transparent. For fixed source location y, use one affine function to the left and another to the right, join them continuously, enforce zero at both endpoints, and arrange a derivative jump of minus one. The second distributional derivative then has a negative delta jump, so applying minus that derivative gives the required positive unit source. This direct sign check is safer than memorizing a representation with an unspecified Laplacian convention.

G(x,y)=min⁡(x,y)(1−max⁡(x,y)),−∂x2G(x,y)=δy,u(x)=∫01G(x,y)f(y) dyG(x,y)=\min(x,y)\bigl(1-\max(x,y)\bigr),\quad -\partial_x^2G(x,y)=\delta_y,\quad u(x)=\int_0^1G(x,y)f(y)\,dy
This is the zero-Dirichlet Green kernel for −d²/dx² on (0,1). The representation holds for suitable integrable sources, with weak interpretation when classical derivatives are unavailable.↔ Scroll wide equations horizontally.

Boundary averages with the correct weights

For continuous data on the unit circle, the disk Poisson formula integrates the boundary data against a positive normalized weight. At the center the weight is constant, so the value is the ordinary circle average. Away from the center it emphasizes nearby boundary angles while retaining total mass one. The resulting function is harmonic in the disk and extends continuously to the prescribed data on the circle. Positivity immediately keeps its values between the boundary minimum and maximum.

The disk formula separates angular frequencies: the boundary cosine mode of integer order m extends as radius to the m times cosine of m theta. High frequencies decay quickly away from the boundary, while the constant mode persists unchanged. This is spatial harmonic extension, not time-dependent heat decay. Near the boundary the derivatives of high-frequency modes can still be large, so interior smoothing should not be mistaken for a uniform derivative estimate independent of rough boundary data.

u(r,θ)=12π∫02π1−r21−2rcos⁡(θ−φ)+r2 g(φ) dφ,0≤r<1u(r,\theta)=\frac1{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\varphi)+r^2}\,g(\varphi)\,d\varphi,\qquad 0\leq r<1
The unit-disk kernel is normalized with the factor 1/(2π). For continuous circle data g it gives the harmonic extension; the geometry and dimension are part of the formula.↔ Scroll wide equations horizontally.

Know which representation is being used

For smooth bounded domains and sufficiently regular data, Green's identity represents a Dirichlet Poisson solution as a volume source integral plus a boundary term involving the normal derivative of the Green function. With the positive minus-Laplacian convention, the Poisson weight is minus the outward source-normal derivative of G. Regularity is needed to interpret that derivative classically; rough domains often require harmonic measure or weak formulations instead. A boundary integral written without its measure and normal orientation is incomplete.

A Green function can fail to be unique or exist in the desired form if the operator and boundary condition have a kernel. Pure Neumann problems, for example, need a compatibility condition and normalization, so their Green functions require a corresponding constant correction. Symmetry of the zero-Dirichlet Green function follows from the self-adjoint setting, not from every variable-coefficient operator. The interval and disk formulas illustrate specific invertible Laplacian problems rather than a universal formula for all elliptic equations.

Integrate an interval source in two pieces

Use the interval Green kernel to solve −u″=6x on (0,1) with zero endpoint values. Evaluate the two pieces of the integral explicitly.

  1. Split the source integral at y=x. For y≤x, G=y(1−x), giving ∫0x6y2(1−x)dy=2(1−x)x3\int_0^x6y^2(1-x)dy=2(1-x)x^3. This part accounts for sources to the left of the observation point.

  2. For y≥x, G=x(1−y), giving ∫x16xy(1−y)dy=x(1−3x2+2x3)\int_x^1 6xy(1-y)dy=x(1-3x^2+2x^3). The two pieces use different affine branches of the same continuous kernel.

  3. Adding yields u=x−x³. Its endpoint values vanish, and its second derivative is −6x, so it satisfies the original source equation. At x=1/2 the solution value is 3/8. The derivative jump condition is what guarantees the correct source sign in this representation.

The result

The kernel reconstructs the unique zero-boundary solution while separating the source location from the observation location.

Explore the model

Interactive lab

Interactive lab · spectral

Inspect boundary frequencies before harmonic extension

Pause, Reset, and set mode count N=3 and coefficient decay p=1. Compare from Reset with both N=9 and p=1. Predict the additional spatial oscillations in the finite sine sum at time zero. Each such sine mode would acquire a factor rⁿ under disk harmonic extension, but this instrument does not apply that factor: playback shifts phases and is not a radial or heat-evolution solver.

Inspect boundary frequencies before harmonic extensionA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. 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 forThe boundary's frequency decomposition is inspectable, while the specific elliptic kernel determines how those frequencies extend into the domain.

Check your understanding

What is the value at the disk center of the harmonic extension of continuous circle data g?

Try it yourself

For disk boundary data g(θ)=2+cos(2θ), find the harmonic extension and evaluate it at the center and at r=1/2 with θ=0 and θ=π/2.

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

Give me a hint

The constant mode stays fixed, and the order-two cosine mode gains a factor r².

Compare with the solution

The extension is u(r,θ)=2+r2cos⁡(2θ)u(r,\theta)=2+r^2\cos(2\theta). Its polar Laplacian vanishes, and at r=1 it gives the stated data. At the center it is two. At r=1/2 and θ=0 it is 9/4; at θ=π/2 it is 7/4. All values lie between the boundary bounds one and three, consistent with the positive normalized kernel and the maximum principle.

Take it with you

  • A Green function combines the correct point-source singularity with a particular boundary condition.
  • Poisson kernels are normalized geometric weights for boundary data.
  • Spatial harmonic attenuation, phase playback, and heat evolution are different operations.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.