The corrector you had to find, and the one you can draw
You arrive here with the hard part already understood. From the previous guide you know that the Green's function G(x, y) for a region is the response felt at x when you poke a unit point source at the interior location y — but a response forced, on top of that, to vanish on the boundary so it can do the Dirichlet problem's job. You split it as G = (free fundamental solution) minus a corrector h, where h is itself harmonic inside the region and is rigged to cancel the fundamental solution exactly along the edge. That corrector is the whole difficulty: it is the solution of a boundary-value problem in its own right, and on a lumpy domain you have no formula for it.
The method of images is the happy news that, for a few perfectly symmetric regions, you do not have to solve for the corrector — you can draw it. The corrector must be a harmonic function inside the region that matches the fundamental solution on the boundary. Where would you find a ready-made harmonic function? A fundamental solution centred at a point is harmonic everywhere except at that point. So plant a second point source at a cleverly chosen point outside the region — a phantom, an image — strong and placed so that on the boundary its field equals the real source's field. Inside the region the phantom never blows up (its singularity is on the wrong side), so it is the harmonic corrector you needed, served for free.
The half-space: one mirror behind the wall
Take the cleanest case: the upper half-plane, every point with height above zero, with the wall being the x-axis. Put a real unit source at a point y sitting at height a > 0 above the wall. The free-space fundamental solution it radiates is the logarithmic potential in 2D, growing more negative as you near the source. Left alone, this field is not zero on the wall — so it is not yet the Green's function. Now do the only thing the picture suggests: reflect y straight down across the wall to its mirror point y*, the same horizontal position but at height minus a, and place there a source of the opposite sign.
Why does this nail it? Any point on the wall is, by construction, the same distance from y as from its mirror y* — the wall is the set of points equidistant from a point and its reflection. The real source contributes a field that depends only on that distance; the image contributes the same magnitude with a minus sign. Equal magnitude, opposite sign, summed on the wall: exactly zero. The combined field of source-minus-image is therefore zero all along the x-axis, automatically satisfying the Dirichlet condition. And the image's singularity sits at height minus a, safely outside the upper half-plane, so inside our region the correction is perfectly harmonic. That source-minus-image field is the Green's function for the half-space.
Upper half-plane, wall = x-axis, real source at y = (b, a), a > 0
real source (+) at y = (b, a)
image source (-) at y* = (b, -a) <-- reflect across wall, flip sign
G(x,y) = Phi(x - y) - Phi(x - y*)
^free soln ^the corrector, drawn not solved
on the wall: |x - y| = |x - y*| ==> G = 0 automatically
Neumann variant: use the SAME sign image (+), giving zero normal slope.Notice the small but real distinction packed into the last line of the sketch. For the Dirichlet problem you flip the image's sign so the values cancel on the wall. For the Neumann problem you want the normal slope to vanish, and a moment's symmetry shows you must instead use an image of the same sign — then the field crosses the wall flat, with zero outward derivative, an insulated boundary. The two boundary conditions you met in the Laplace rung are the very thing that decides whether your phantom is a mirror twin or an anti-twin. This is the same odd-versus-even reflection logic you saw in wave reflection at a fixed versus a free end — flip for one, keep for the other.
The ball: a mirror that also rescales
A flat wall is the easy mirror because reflection there is rigid — distances are preserved, so an equal-and-opposite image cancels perfectly. A sphere is the surprising case, and it is where the method earns its keep. Put a real source at a point y inside a ball of radius R. The natural guess is to reflect y across the sphere, but reflection in a curved surface is not rigid: the correct map is inversion, sending y to the point y* on the same ray from the centre but at distance R^2 over |y|, so a source near the centre throws its image far away and vice versa. Crucially, the image is no longer of unit strength — you must also rescale its charge.
Here is the small magic that makes inversion the right mirror. For a point P on the sphere, the two triangles (centre, y, P) and (centre, P, y*) share the angle at the centre, and because |centre to y| times |centre to y*| equals R^2 by construction, those triangles are similar. Similar triangles mean the ratio of distances |P - y| over |P - y*| is the same constant for every point P on the sphere — namely |y| over R. So if you scale the image charge by exactly that factor R over |y|, the real field and the image field have equal magnitude all over the sphere, and with opposite signs they cancel. The Dirichlet Green's function for the ball is the real source minus the rescaled, inverted image. Inversion is the curved-space cousin of the flat reflection.
Images in time: the heat equation on a half-line
The trick is not confined to Laplace. The same reflection idea tames the heat equation u_t = k u_xx on a half-line, say x > 0 with the end held at zero temperature for all time. The free response to a point of heat released at position a is the heat kernel, a spreading Gaussian bump centred at a. Left alone it warms the boundary point x = 0, violating the condition there. So reflect: imagine a phantom cold spike — a negative heat kernel — released at the mirror position minus a, on the forbidden side x < 0.
As both bumps spread and overlap, what happens at the wall x = 0? By symmetry the warm bump from a and the cold bump from minus a are always exactly equal in size and opposite in sign at the midpoint x = 0, for every instant of time. Their sum there is zero forever, so the boundary stays clamped at zero temperature without you ever enforcing it — symmetry holds the boundary for you. The solution in the region x > 0 is simply the difference of the two heat kernels. An odd reflection of the initial data across the wall gives a Dirichlet (zero-value) boundary; an even reflection, with a same-sign warm image, gives a Neumann (insulated) boundary.
Be honest about the limits, because they matter. Images work cleanly only for geometries that reflect onto themselves: a half-line, a full line, an interval (where you reflect endlessly in both walls, generating an infinite train of images — that infinite sum is one way the Fourier sine series is born). The infinite-speed smoothing of the heat equation does not break the method; both real and image bumps are smooth and the difference stays a genuine solution. But you cannot, in any of these settings, run the construction backwards in time to recover a past from a present — diffusion is irreversible, and no clever placement of images repairs an ill-posed backward problem.
What images buy, and where they stop
Step back and see what you have gained. The method of images is a way to construct an exact Green's function with nothing but a ruler and a sign convention — no infinite series, no integral to evaluate, no solving of an auxiliary boundary-value problem. When the geometry cooperates, the corrector you laboured over in the previous guide becomes a single reflected, possibly rescaled, copy of the source. And it transfers across the great operators: the Laplacian gives potential-theory images on walls and spheres, the heat equation gives images in space that persist for all time, and even the wave equation reflects pulses off boundaries the same way, sending a flipped echo back.
- Identify the boundary and the source. Pin down the region, where its boundary is, and the interior source point y you are responding to.
- Find the reflection that maps the region to its complement — across a flat wall it is ordinary reflection; across a sphere it is inversion to R^2 over |y|.
- Place the image source at the reflected point, rescaling its strength if the mirror is curved (the ball needs the factor R over |y|).
- Choose the sign: opposite for Dirichlet (values cancel on the boundary), same for Neumann (normal slope cancels). Sum real plus image — that is G.
But hold the honest line about reach. The whole edifice rests on a reflection symmetry that swaps the region with its outside, and most regions simply have none — an ellipse, an L-shaped room, a kidney-shaped blob admit no exact images. For those you fall back to the general theory: solve numerically, or expand the Green's function in eigenfunctions, or use the integral-formula machinery built from the few cases images do solve. So treat the method as a precious set of exactly solvable landmarks — half-space, slab, ball, wedge with the right angle — not as a universal algorithm. Its true value is half computational and half conceptual: it shows you, with a picture, what a boundary really is to a Green's function — a mirror.