Build forced solutions from point responses

Try

To solve −u″=f on [0,1] with zero endpoints, imagine first placing one unit source at a point ξ. Away from that point the response is affine. It must be continuous, vanish at both endpoints and have a slope jump that produces exactly one delta source under the operator −d²/dx².

The resulting Green function is G(x,ξ)=x(1−ξ) when x≤ξ and ξ(1−x) when x≥ξ. Both pieces agree at x=ξ.

Predict

Should the slope jump be +1 or −1? Integrate −G″ across a tiny interval around ξ and use the desired unit source. The leading minus sign determines the answer.

Experiment

The left slope is 1−ξ and the right slope is −ξ, so right minus left is −1.

Property	Verification
Left endpoint	G(0,ξ)=0
Right endpoint	G(1,ξ)=0
Continuity	Both pieces equal ξ(1−ξ) at the source
Derivative jump	−ξ−(1−ξ)=−1

For constant source f=4, integrate G against 4 or solve directly to get u=2x(1−x). Its midpoint value is 1/2. Check that this equals the integral of the triangular kernel at x=1/2 multiplied by 4.

Understand

The derivative jump means −Gₓₓ=δξ in distributions. Therefore u(x)=∫₀¹G(x,ξ)f(ξ)dξ solves the equation weakly and satisfies the zero endpoints. With sufficient regularity on f, it is the classical solution; uniqueness identifies the constructed response. Linearity lets us add point responses, but convergence and regularity determine which integrals and derivatives are justified.

In three dimensions, a fundamental solution for −Δ is 1/(4πr). In two dimensions it is −log r/(2π). Each is harmonic away from r=0. Its negative outward gradient flux through a small surrounding sphere or circle is 1, fixing both the sign and normalization. The singularity is interpreted distributionally, not by evaluating the classical Laplacian at zero.

A fundamental solution lives in free space. A domain Green function additionally satisfies the domain’s boundary condition, usually by subtracting a harmonic correction. Confusing the two loses the boundary information.

For a smooth domain with Dirichlet Green function and boundary data g, Green’s identity gives u(x)=∫ΩG(x,y)f(y)dy−∫∂Ωg(y)∂ₙG(x,y)dS. The normal derivative is taken in the source variable y and points outward. The minus sign is consistent with the positive Poisson kernel for the operator −Δ. The formula assumes sufficient regularity to justify the identity and singular limiting process; general rough-domain theory is beyond this derivation.

For a full multidimensional construction, the ball Poisson proof verifies the distributional source, constructs the image-pole Green correction, derives its outward-normal kernel and proves continuous boundary attainment. It does not assume a Green function exists on an arbitrary rough domain.

Complete the forced ball problem

Here is a classical existence result for nonzero smooth forcing, not just a harmonic extension. Let f be smooth on a neighborhood of the closed radius-R ball in Rᵈ, d≥2, and let g be continuous on its sphere. Choose a smooth compactly supported cutoff χ equal to one on a neighborhood of that closed ball, with support inside the region where f is defined. Extending χf by zero produces a compactly supported smooth function F on all of Rᵈ. Define the Newtonian potential N(x)=∫Φ(x−y)F(y)dy, using the fundamental solution whose distribution identity was proved in the linked ball construction.

The integral is finite: its only local singularity is integrable (the radial integral is proportional to ∫₀¹r dr for d≥3 and ∫₀¹r|log r|dr for d=2), and F has compact support. To justify derivatives of every order without differentiating the singular kernel, change variables to N(x)=∫Φ(z)F(x−z)dz. For x in any fixed compact set, all differentiated factors F(x−z) have support in one bounded z-set, and their bounds are multiplied by a locally integrable |Φ(z)|. Dominated convergence therefore gives DᵅN=∫Φ(z)DᵅF(x−z)dz for every multi-index α. Thus N is smooth.

For a compactly supported smooth test φ, Fubini is justified by the same local integrability on bounded sets. The identity −ΔΦ=δ yields ∫N(−Δφ)=∫Fφ. Since N and F are smooth, −ΔN=F pointwise. Let H be the Poisson extension in the ball of the continuous boundary function g−N. The proved kernel construction gives H harmonic and smooth inside, continuous up to the boundary. Then u=N+H satisfies −Δu=f in the ball and u=g on the sphere. The maximum principle applied to the difference of two solutions proves uniqueness in C²(B_R)∩C(closure B_R). Different cutoffs may change N and H separately but cannot change their sum by uniqueness.

Changed data. On the unit ball take f=2d and g=0. The candidate u=1−|x|² has −Δu=2d and zero boundary data, so uniqueness identifies it with the integral construction. If f is only L², the dominated differentiation of F of every order is unavailable. The weak-existence and interior H² theorems in the variational lesson give the appropriate different conclusions; one must not continue calling the answer C² without additional data regularity.

This supplies the canonical multidimensional Poisson existence mechanism accompanying the fundamental-solution and Green-function topics in MIT 18.152 Fall 2011, lectures 6–8, checked 30 October 2026. The assumptions deliberately separate an interior classical equation, continuous boundary attainment, and stronger regularity of derivatives at the boundary.

Apply

Construct the interval Green function for a different length L by matching two affine pieces and imposing a derivative jump of −1. Check its units and endpoint values. Then explain why adding an arbitrary harmonic function to a free-space fundamental solution preserves the source identity but changes boundary behavior.

Continue

Neumann problems require a compatibility condition and retain an additive constant. Their Green functions therefore cannot be specified by copying the Dirichlet construction unchanged.