Green's Functions for Boundary-Value Problems

From free space into a room with walls

You arrive here with two ideas already in hand. From the first guide of this rung you know the picture of a response to a point source — ring a single point of a medium and watch how it answers everywhere else. From the second you know the fundamental solution: the precise response when the source is a Dirac delta and there is nothing else around, no walls, just empty space stretching to infinity. For Laplace's operator in the plane that response is the logarithmic potential, and in three dimensions it is the familiar 1/(4*pi*r) of a point charge. These are the cleanest possible answers — because they ignore the boundary entirely.

Real problems are rarely set in empty space. You want the steady temperature inside a metal plate, the electric potential inside a grounded box, the displacement of a drumhead clamped to a rim. There is a boundary, and along it you nail down data — say u = 0 on the edge, the Dirichlet problem you met back in the Laplace rung. A bare fundamental solution will not obey that boundary condition; the logarithmic potential placed at an interior point is certainly not zero out on the rim. So the question of this guide is simple to state and rich to answer: how do we build a response-to-a-point-source that also respects the walls?

The recipe: fundamental solution plus a corrector

Here is the whole construction in one move. Fix a source point y inside the region D, and write the Green's function G(x, y) as two pieces: the free-space fundamental solution Phi(x, y), which already gets the singularity at x = y exactly right, plus a boundary corrector h(x, y) that is perfectly smooth and harmonic inside D — no source of its own. The corrector's only job is to bend the total down to zero on the boundary, cancelling whatever value the fundamental solution happens to leave there. Add them and the sum has the right point source inside and the right value on the edge.

G(x, y) = Phi(x, y) + h(x, y)

  Phi(x, y) : free-space fundamental solution
              Laplacian Phi = delta at x = y   (carries the singularity)

  h(x, y)   : the boundary corrector
              Laplacian h = 0 inside D          (no source of its own)
              h(x, y) = -Phi(x, y) on boundary  (cancels Phi at the wall)

  ==>  Laplacian_x G = delta(x - y) in D,   G = 0 on boundary

Notice what this quietly demands. Finding h means solving a Dirichlet problem of exactly the kind from the Laplace rung: find a harmonic function in D whose boundary values are prescribed (here, minus the fundamental solution restricted to the boundary). So constructing the Green's function is not free — it is as hard as solving a boundary-value problem. That sounds circular, and in a sense it is, but the payoff is enormous: once you have G for a region, you can solve the boundary-value problem for every right-hand side and every boundary data at once, just by integrating against it. You pay the boundary-value price once and reuse the answer forever.

It helps to read the corrector physically. Picture a point charge sitting at y inside a grounded conducting box. The charge alone would push the potential to nonzero values on the walls, but a grounded conductor must sit at zero potential everywhere on its surface. So charge rearranges on the walls until the field outside is killed — and that rearrangement is exactly the corrector h. It is the induced response of the boundary to the point source. The next guide, the method of images, will reveal a startlingly simple way to build h for special geometries by imagining a phantom charge outside the region.

Symmetry: source and observer can trade places

Now comes the most beautiful fact about Green's functions, and it costs almost nothing to prove. The Green's function is symmetric: G(x, y) = G(y, x). In words, the response felt at x from a unit source planted at y is exactly the response felt at y from the same source planted at x. This is the reciprocity principle, and it is genuinely surprising — there is no obvious reason a region built around source point y should feel the same the other way around. Yet for the symmetric (self-adjoint) operators like the Laplacian, it always holds.

The proof is a one-page application of Green's identities, the same integration-by-parts machinery you have already seen. Take Green's second identity for the two functions u = G(x, a) and v = G(x, b), integrate over D, and the interior terms collapse because each G is harmonic away from its own source; the delta singularities pick out exactly the two values G(b, a) and G(a, b); and the boundary terms vanish because both functions are zero on the boundary by construction. What is left is G(b, a) = G(a, b). The symmetry is inherited from the operator being self-adjoint — change to a non-self-adjoint operator and you must instead pair G with the Green's function of the adjoint problem.

Why this is worth the trouble: the representation formula

All this machinery earns its keep in one master formula. Once you have G for a region, the solution of the boundary-value problem Laplacian u = -f inside D with u = g on the boundary is written down explicitly: u at any point y is an integral of the source f against G, plus an integral of the boundary data g against the normal derivative of G along the boundary. This is the representation formula. It converts the verb "solve the PDE" into the noun "do two integrals," with the entire difficulty of the geometry packed once and for all into G.

Look closely and the two integrals tell two physical stories. The volume integral over f is superposition in action — chop the distributed source f into a swarm of tiny point sources, respond to each one with G (this is allowed precisely because the operator is linear, the superposition principle you have leaned on all along), and add up the responses. The boundary integral is how the prescribed edge data g leaks its influence inward. With f = 0 you are back to the pure Dirichlet problem and the formula reduces to the boundary term alone — which on a disk is nothing other than the Poisson integral formula you already met, now revealed as a Green's-function calculation in disguise.

One honest limit to carry forward. Green's functions are a triumph for linear equations, because the whole edifice rests on superposition — you may chop the source into points only because responses add. The moment the equation is nonlinear, this collapses: there is no Green's function for Burgers' equation or the Navier-Stokes equations, because doubling the source does not double the response. And even within the linear world, an explicit closed-form G exists only for special, symmetric geometries — a half-space, a ball, a slab. For a general lumpy region G certainly exists but usually has no formula, and you compute it numerically. The next guide shows the happy exceptions where a phantom "image" source hands you G by inspection.

A tiny worked example: the half-line

Strip everything down to one dimension to see the recipe run end to end. Work on the half-line x > 0, with the boundary condition G = 0 at the single wall x = 0, and put a unit source at a point y > 0. In one dimension the free-space fundamental solution of -G_xx = delta is the simple tent Phi(x, y) = -|x - y|/2 — a kink that points down at the source. By itself it is not zero at the wall: at x = 0 it equals -y/2. So we need a corrector h that is source-free (here just linear in x, since linear functions satisfy h_xx = 0) and cancels that leftover value on the wall.

1. Write the ansatz G(x, y) = -|x - y|/2 + h(x, y), with h harmonic (here, linear in x) so the source stays exactly the original delta at x = y.
2. Impose the wall condition G = 0 at x = 0: this forces h(0, y) = +y/2 to cancel the fundamental solution's value -y/2 there.
3. The clean source-free choice that does it is h(x, y) = +|x + y|/2, which is just the fundamental solution of an equal source reflected to the mirror point -y outside the wall.
4. Combine: G(x, y) = (|x + y| - |x - y|)/2. Check it — at x = 0 the two terms are equal and G = 0, and you can verify directly that G(x, y) = G(y, x).

Two things are worth savouring here. First, the symmetry G(x, y) = G(y, x) is plainly visible in the final formula — swap x and y and the two absolute values just trade names — so reciprocity passes its own check, exactly as promised. Second, the corrector turned out to be a fundamental solution placed at the mirror image of the source across the wall at -y. That is no accident; it is the entire idea of the method of images, where a cleverly placed phantom source outside the region automatically enforces the boundary condition. The half-line was a warm-up. The next guide turns that trick loose on half-spaces and balls.