Strip away the walls first
The previous guides taught the central idea of this rung: solve a linear problem L u = f by first answering one cleaner question — how does the system respond to a single, infinitely concentrated poke? That response is the Green's function G, defined by L G(x, s) = delta(x - s) together with the boundary conditions, and then the superposition integral u(x) = integral of G(x, s) f(s) ds builds the answer to ANY load by adding up point-source responses. But there is a sneaky difficulty hiding in that definition: G must satisfy both the delta on the right AND the boundary conditions, and the boundary conditions are usually the hard part. So we split the job in two.
The first half ignores the walls entirely. Ask: in infinite, featureless space — no boundaries anywhere, just the condition that the field decay sensibly far off — what does the operator do to a single point source? That bare answer is the [[fundamental-solution|fundamental solution]], and viewed as a function of source point s and field point x it is the [[free-space-greens-function|free-space Green's function]], written G_free. It is the purest building block in the whole subject: what an infinite, empty medium does when you poke it once. Build this once for each operator and you can reuse it forever. The second half — putting the walls back — is what the method of images handles, and it does so by cleverly stacking up copies of this same free-space object.
The Laplacian: the potential of a point charge
Start with the most famous of all. The operator is L = -nabla^2, the negative Laplacian, and its source equation -nabla^2 G = delta(x - s) is just Poisson's equation for a unit point charge sitting at s. Away from the charge there is nothing, so G must solve Laplace's equation nabla^2 G = 0 everywhere except at the source. By symmetry the field can only depend on the distance r = |x - s| from the charge, and the radial Laplacian forces a unique decaying shape. In three dimensions the answer is G_free = 1/(4 pi r) — the familiar Coulomb potential, the gravitational potential, the 1/r law of nature you have met a dozen times. Now you know where the 4 pi comes from: it is exactly the factor that makes a unit source come out, the surface area of the unit sphere.
Dimension matters, and it matters in a way worth picturing. In two dimensions the same logic gives G_free = -(1/(2 pi)) ln r — a logarithm, not a power. Notice it grows without bound as r increases instead of dying off; influence in flatland never fully fades, a genuine and slightly unsettling feature of two-dimensional potential theory. In one dimension the fundamental solution of -d^2/dx^2 is the tent function -(1/2)|x - s|, a straight rise to the source and a straight fall away from it. Three shapes — a tent, a logarithm, a 1/r decay — and each is exactly what the delta forces when you account for how a sphere's surface area scales with dimension.
The heat and wave operators: a spreading blob and an expanding shell
Now add time. For the heat operator L = d/dt - k nabla^2, the fundamental solution answers a vivid question: deposit a unit pulse of heat at one point at the instant t = 0, then watch. The answer is the heat kernel, the Gaussian G_free = (4 pi k t)^{-n/2} exp(-r^2/(4 k t)) for time t > 0, a bell-shaped blob centred on the source. At the very first instant it is an infinitely tall, infinitely narrow spike (the delta itself); an instant later it is a smooth, finite bump; and forever after it spreads and flattens, its width growing like sqrt(t). This is the precise mathematics of how a dropped bead of dye blurs out in still water, and the Gaussian shape is no coincidence — it is the same bell curve that governs random walks, because heat diffusion IS the limit of countless random microscopic jostles.
An honest wrinkle worth noticing: the Gaussian is strictly positive everywhere for every t > 0, however small. That means the instant you light a match here, the temperature at the far side of the galaxy rises by an absurdly tiny but nonzero amount. The heat equation propagates influence at infinite speed — a known idealization, fine for engineering but physically only approximate. The wave operator L = d^2/dt^2 - c^2 nabla^2 could not be more different. In three dimensions its fundamental solution is not a spreading blob but an infinitely thin expanding spherical shell: a flash at the origin is felt only on the sphere r = c t and nowhere inside it. A sharp clap reaches you, passes, and is gone, leaving silence behind — light and sound obey a strict speed limit and clean afterward.
There is a strange and lovely twist hiding here, and it is honest to flag it. The clean expanding-shell picture — a flash that arrives and departs sharply, leaving silence — holds in THREE space dimensions, and also in every odd dimension. In TWO dimensions it fails: a flash in flatland is felt sharply when the wavefront arrives but then leaves a slowly fading tail that never quite reaches zero, like ripples on a pond that keep trembling long after the leading ring has passed. This is the famous distinction between 'sharp' propagation in odd dimensions and the lingering 'wake' in even dimensions, and it is one reason we live comfortably in three dimensions: a world where echoes never cleanly stopped would make speech and vision a smeared mess.
The method of images: replacing a wall with a ghost
Now put a wall back. Suppose you have a single point charge sitting at height z above a flat, grounded metal plane (the plane held at potential zero — a Dirichlet boundary condition). You want the Green's function for the half-space above the plane: it must have the point source in the right place AND vanish on the plane. Solving that boundary-value problem head-on looks miserable. The [[method-of-images|method of images]] is a gorgeous shortcut. Pretend the wall is not there at all. Then, on the far side of where the wall was, place a fictitious mirror-image charge — same strength, opposite sign — at the reflected position, the same distance below the plane as the real charge is above it.
Why does this work? On the plane itself, every point is exactly equidistant from the real charge and its mirror image, so the positive potential of the real charge and the negative potential of the image cancel perfectly — the total is zero all along the plane, automatically, with no effort. The Dirichlet condition is satisfied for free. And here is the crucial bit of bookkeeping: the image charge lives BELOW the plane, outside the region we actually care about, so inside the half-space it contributes nothing but a smooth solution of Laplace's equation — it does not add any spurious source where we are looking. The real point source upstairs is untouched, the wall is satisfied, and we are done. The Green's function is simply G = 1/(4 pi |x - s|) - 1/(4 pi |x - s_image|), the free-space kernel for the real source minus the same kernel for its reflection.
Half-space z > 0, Dirichlet wall at z = 0 (potential = 0 on the wall)
real source s = (a, b, c), c > 0
* +1 field point x
| o
__________|__________________________________ wall z = 0
|
* -1 (image at s' = (a, b, -c), opposite sign)
G(x, s) = 1/(4 pi |x - s|) - 1/(4 pi |x - s'|)
On the wall, |x - s| = |x - s'| => G = 0 automatically.
Variants by reflecting the SAME free-space kernel:
Neumann wall (insulated, dG/dn = 0): use a +1 image (same sign) -> potentials ADD
Slab 0 < z < L (two parallel walls): infinitely many images, reflected again and again
Right-angle corner / wedge of angle pi/n: a finite ring of 2n alternating images
Sphere of radius R: one image of strength -R/d at the inverse point R^2/d (Kelvin)How far images reach — and where they stop
The same trick adapts to a whole family of symmetric situations, and the rule for adapting it is intuitive once you see the pattern. Change the boundary CONDITION and you change the image's SIGN: a grounded wall (Dirichlet, value zero) wants an opposite-sign image so the two potentials cancel on the wall; an insulated wall (Neumann, zero normal derivative — no heat or flux crosses it) wants a same-sign image, so the field arrives at the wall flat, with zero slope across it, like a ball bouncing straight back. Heat conduction with a perfectly insulated boundary, a vibrating string fixed or free at an end, a charge near a conductor — all are the same reflection idea wearing different clothes.
Change the GEOMETRY and you change how many images you need. A single flat wall needs one image. Two parallel walls (a slab, or a string clamped at both ends) need an infinite train: each image must itself be reflected in the other wall, bouncing back and forth forever, producing an endless line of images — which, satisfyingly, is exactly the source of the discrete harmonics of a guitar string. A wedge with a corner angle of pi/n (a right-angle corner is n = 2) needs a finite ring of 2n alternating images arranged around the corner. A sphere needs one image too, but placed by the Kelvin inversion at the point R^2/d along the radius and weighted by -R/d — a more subtle reflection, yet still a single ghost.
- Build the free-space Green's function once for your operator — 1/(4 pi r) for -nabla^2 in 3D, the Gaussian heat kernel, the expanding shell for the wave operator.
- Reflect the real source across each flat boundary to find the image positions; for a sphere use the Kelvin inverse point R^2/d.
- Choose each image's sign and strength by the boundary condition: opposite sign for a Dirichlet (value-zero) wall, same sign for a Neumann (insulated) wall.
- Add up the free-space kernel of the real source and all images; the singularities of the images lie outside the region, so inside you get exactly the point source you wanted with the boundary satisfied automatically.
Be honest about the ceiling, though, because it is real. The method of images is EXACT where it applies — there is no approximation, no truncation error, the closed-form answer is genuinely correct. But it applies only to boundaries symmetric enough that a finite (or at least a clean, summable infinite) set of reflected sources can reproduce the boundary condition: planes, spheres, special wedges, the disk. For a generic curved or lumpy boundary no such collection of images exists, and the trick simply does not apply. Then you fall back on the methods the rest of this rung builds — the eigenfunction expansion of the Green's function, which assembles G from the natural modes of the region, and ultimately numerical solvers. Images are the elegant special case; eigenfunction expansions are the general engine.