Recover interior values from surrounding averages

Try

A steady source-free temperature field satisfies Δu=0, where Δ is the sum of second derivatives in the spatial coordinates. Such a function is harmonic. In two dimensions, u=x²−y² has curved level sets but zero Laplacian; u=x²+y² has positive Laplacian. Curvature in one direction can cancel curvature in another.

Calculate both second derivatives before deciding whether a saddle-shaped graph can be harmonic. A harmonic function need not be constant on a bounded region with varying boundary values.

Predict

Average x²−y² around a circle centered at the origin. Predict the result by symmetry, then parametrize the circle. Next move the center away from the origin and predict whether the average remains zero or equals the value at the new center.

Experiment

On a circle centered at (a,b), substitute x=a+r cosθ and y=b+r sinθ. The linear sine and cosine terms average to zero, while the squared terms cancel.

Term	Circular average
a²−b²	a²−b²
2ar cosθ−2br sinθ	0
r²(cos²θ−sin²θ)	0

Thus the average equals u(a,b), independent of r as long as the circle and its interior lie in the harmonic domain. Repeat with x²+y² and identify the extra r² term.

Understand

For a C² harmonic function on a ball, differentiate its normalized spherical average with respect to radius. The derivative equals the outward derivative flux divided by sphere area. By the divergence theorem, that flux is the integral of Δu over the ball, hence zero. The mean is constant in radius, and its radius-zero limit is the value at the center. Averaging these sphere identities radially gives the ball mean-value property as well.

This identity proves a strong maximum principle. At an attained interior maximum, every nearby value is at most that maximum, yet the surrounding average equals it. Continuity excludes any strictly smaller region, so a whole small ball shares the maximum. The set where the maximum is attained is both open and closed relative to a connected domain, forcing constancy.

For positive harmonic functions, the ball Poisson kernel is bounded above and below on a smaller concentric ball by constants depending on the radius ratio and dimension. Comparing the kernel representations gives a Harnack inequality relating the maximum and minimum there. Positivity is essential; sign-changing harmonic functions do not satisfy this form.

Bounded entire harmonic functions are constant. One proof uses the interior estimate |∇u(x)|≤C sup|u|/R on a ball of radius R. It follows from differentiating the Poisson formula. Since an entire bounded function permits arbitrarily large R, its gradient vanishes. The construction below proves the derivative estimate from the explicit ball kernel; it is not inferred from boundedness alone.

Apply

Test a polynomial for harmonicity, calculate one circular average and explain whether it can attain a strict interior maximum. Then identify which hypothesis fails for the nonconstant entire harmonic function u=x in the bounded Liouville theorem. State positivity and compact containment explicitly when applying Harnack.

Construct a classical solution in a ball, not just a finite mode

Let B_R be the radius-R ball centered at zero in Rᵈ, with d≥2. Write σ_d for the surface area of the unit sphere, so σ₂=2π and σ₃=4π. We will solve the Dirichlet problem for every continuous boundary function g, then derive the Harnack and Liouville claims from the same formula. This is a canonical multidimensional existence result; it is not a claim about every rough domain.

Check the source identity and the boundary correction

For the operator −Δ, define Φ(z)=|z|²⁻ᵈ/((d−2)σ_d) when d≥3, and Φ(z)=−log|z|/(2π) when d=2. Both are locally integrable and harmonic away from zero. Their radial derivative is −1/(σ_d rᵈ⁻¹). To prove −ΔΦ=δ₀, test against a compactly supported smooth φ and integrate by parts on its support with a radius-ε ball removed. On the inner sphere the outward normal for the punctured region points toward the origin. The boundary term involving Φ times ∂ₙφ tends to zero: it is O(ε) for d≥3 and O(ε|log ε|) for d=2. The other term is the spherical average of φ, tending to φ(0). Thus ∫Φ(−Δφ)=φ(0). This checks the distributional source, including its sign, rather than merely evaluating Δ away from its pole.

For a nonzero pole x inside B_R let x*=R²x/|x|², which lies outside the closed ball. In the source variable y define

$$G(x,y)=\begin{cases}\displaystyle\Phi(y-x)-\left(\frac R{|x|}\right)^{d-2}\Phi(y-x^*),&d\ge3,\\[4pt]\displaystyle\frac1{2\pi}\log\frac{(|x|/R)|y-x^*|}{|y-x|},&d=2.\end{cases}$$

On |y|=R the identity |y−x*|=(R/|x|)|y−x| makes G zero. The image-pole term is smooth and harmonic in the ball, so −Δ_yG=δ_x. For x=0 define G(0,y)=Φ(y)−Φ(R), interpreting Φ(R) as its radial value; this is also the limit of the formula. This is the missing boundary correction: the free-space fundamental solution alone would not vanish on the sphere.

Differentiate in the outward y-normal, y/R, at |y|=R. Substituting the distance identity and the radial derivative of Φ gives the Poisson kernel

$$P_R(x,y)=-\partial_{n_y}G(x,y)=\frac{R^2-|x|^2}{\sigma_d R|x-y|^d}.$$

For example, in d≥3 the first normal derivative is −(R²−x·y)/(σ_d R|x−y|ᵈ). The image derivative, including its coefficient, is −(|x|²−x·y)/(σ_d R|x−y|ᵈ). Their difference is the negative of the displayed P_R; the logarithmic case has the same radial-derivative computation. P_R is positive inside the ball.

Integrating Δ_yG=0 over the ball with a small sphere about x removed shows its total outer normal flux equals −1: the inner normal flux tends to +1 by the source calculation, while the smooth correction contributes a vanishing inner flux. Therefore

$$\int_{\partial B_R}P_R(x,y)\,dS_y=1.$$

Prove existence, boundary convergence and uniqueness

Define U(x)=∫∂B_R P_R(x,y)g(y)dS_y. On every compact subball the denominator is bounded away from zero, and all x-derivatives of the kernel are uniformly bounded on the sphere. Thus differentiation under the finite surface integral is justified to every order. The kernel is harmonic in x. To verify this directly, put q=|x−y|: Δ_xq⁻ᵈ=2d q⁻ᵈ⁻², and the product rule gives

$$\Delta_x\bigl[(R^2-|x|^2)|x-y|^{-d}\bigr]=2d|x-y|^{-d-2}(R^2-|y|^2)=0.$$

Consequently U is smooth and harmonic in the interior. It remains to prove, rather than assume, that U takes the prescribed data. Fix y₀ on the sphere. Since the kernel has unit mass,

U(x)−g(y₀)=∫∂B_R P_R(x,y)(g(y)−g(y₀))dS_y.

Given ε>0, uniform continuity makes |g(y)−g(y₀)|≤ε when |y−y₀| is less than δ. The near contribution is at most ε by positivity and mass one. For |x−y₀| is less than δ/2, the far denominator obeys |x−y|≥δ/2, so the far contribution has absolute value at most

$$2\|g\|_\infty\,|\partial B_R|\,\frac{R^2-|x|^2}{\sigma_d R(\delta/2)^d}\longrightarrow0\quad\text{as }x\to y_0.$$

This works for any interior approach, not just a radial one. Defining U=g on the boundary gives a continuous function on the closed ball. The maximum principle makes it the unique member of C²(B_R)∩C(closure B_R) with ΔU=0 and those data. In particular, any harmonic function smooth on a neighborhood of a closed ball equals its Poisson integral there. We have proved classical interior existence with continuous boundary attainment, not continuous second derivatives on the closed ball for arbitrary continuous g.

Obtain quantitative Harnack bounds

Suppose u is nonnegative and harmonic on a neighborhood of the closed ball. Its boundary values are nonnegative. Since P_R(0,y)=1/(σ_d Rᵈ⁻¹), for |x|≤ρR with ρ between zero (inclusive) and one (exclusive),

$$\frac{1-\rho^2}{(1+\rho)^d}\le\frac{P_R(x,y)}{P_R(0,y)}=\frac{1-|x|^2/R^2}{(|x-y|/R)^d}\le\frac1{(1-\rho)^d}.$$

Integrating against u(y) preserves these inequalities. Therefore cρu(0)≤u(x)≤Cρu(0), with the explicit positive lower and upper constants shown. Comparing two points in the smaller ball gives sup u≤(Cρ/cρ)inf u. If u(0)=0, positivity and the formula force u=0 throughout the ball; no division by zero is needed. For a harmonic function known only on an open domain, first choose the closed ball strictly inside it. Sign-changing functions do not permit integration of pointwise kernel bounds against their data, which identifies exactly where positivity is used.

Derive the derivative bound and Liouville’s theorem

Differentiating the kernel at the center gives ∇ₓP_R(0,y)=d y/(σ_d Rᵈ⁺¹). Its vector norm integrates to d/R. Hence, if |u|≤M on the sphere,

$$|\nabla u(0)|=\left|\int_{\partial B_R}\nabla_xP_R(0,y)u(y)\,dS_y\right|\le\frac{dM}{R}.$$

Translate the ball to any center a. For an entire harmonic u bounded in absolute value by one global M, every radius R is available, so |∇u(a)|≤dM/R for all R. Letting R tend to infinity gives ∇u(a)=0. Integrating the zero gradient along any line segment proves u is constant. This proves the intermediate estimate previously left as a dependency and shows why u(x)=x₁ is not a counterexample: it has no global bound M. The argument applies to C² harmonic functions; the interior regularity proof in the weak-elliptic lesson also upgrades H¹_loc weak harmonic solutions to this class.

Changed-case practice: certify the construction

On the unit disk, take g(cos θ,sin θ)=cos(2θ). Predict the value at the center and the harmonic extension, then justify it by uniqueness rather than assuming a truncated Fourier sum solves arbitrary data. Next use d=2 and ρ=1/2 in the Harnack bounds above. Finally explain why the constant boundary function 1 cannot be solved by integrating the free-space Φ alone without its correction.

Feedback. The extension is x²−y²=r²cos(2θ), with center value zero; it is harmonic and has the prescribed boundary values. It changes sign, so the nonnegative Harnack bound is inapplicable. For nonnegative data the displayed conservative constants are cρ=1/3 and Cρ=4, giving a sup/inf bound of 12 in the half-radius disk. The boundary integral of P is exactly one, so it extends constant 1 to constant 1. A free-space potential has the desired source identity but does not impose the sphere boundary condition. A complete proof identifies the normal’s variable and orientation, mass one, the near/far boundary limit and the positivity requirement; a picture of a smooth field does not establish these facts.

These original derivations address the ball Poisson formula, Harnack and Liouville topics explicitly listed in MIT 18.152 Fall 2011, lecture 9 and the disk problem in Stanford Math 220 Autumn 2018, public syllabus, checked 30 October 2026. They neither use restricted Stanford notes nor claim arbitrary-domain Green-function existence.

Continue

Maximum and comparison principles convert these local facts into bounds and uniqueness for boundary-value problems, including forced Poisson equations where the sign of the source matters.