An equation with too many answers
You arrive here already knowing the lead actor. From the first two guides of this rung you know that Laplace's equation Laplacian u = 0 picks out the harmonic functions — the equilibrium shapes a soap film or a steady temperature settles into — and you know they obey the mean-value property and the maximum principle. But notice a quiet problem. On a disk, u = 1 is harmonic. So is u = x. So is u = x^2 - y^2, and u = e^x cos y, and endlessly many others. Laplace's equation, taken alone over a region, has an infinite ocean of solutions.
This is not a flaw of Laplace's equation; it is the nature of every PDE. Recall the honest lesson from the foundations rung: the general solution of a PDE involves not arbitrary constants but arbitrary functions. An ODE for a falling ball needs two numbers — position and velocity — to pin it down. A PDE on a region needs a whole function's worth of data, and the natural place to nail that data down is the boundary of the region. The interior equation says how u must behave locally; the boundary data says which one of the infinitely many globally consistent shapes you actually mean.
The Dirichlet problem: prescribe the value
The most natural choice is to nail down u itself on the edge. The Dirichlet problem asks: given a region D and a function g defined on its boundary, find u that is harmonic inside D and equals g on the boundary. In symbols, Laplacian u = 0 in D, with the Dirichlet boundary condition u = g on the boundary. Picture a metal plate whose rim you clamp to a fixed temperature profile g — warm here, cold there. Heat redistributes inside until it stops changing; the steady temperature it settles into, with the rim held fast, is exactly the solution u.
There is the same soap-film picture too: bend a wire loop into some shape over the plane, dip it, and the film that spans it is (for gentle slopes) a harmonic graph whose height matches the wire at every point of the rim. The wire is the boundary data g; the film is the solution. Both pictures share the punchline that already feels right from the maximum principle: the film cannot poke above the highest point of the wire or sag below its lowest. A harmonic function is squeezed entirely between the extremes of its own boundary values.
That squeezing is exactly why the Dirichlet problem has at most one solution. Suppose u and v both solve it with the same boundary data g. Their difference w = u - v is harmonic (Laplace's equation is linear, so differences of solutions are solutions — the superposition principle at work) and on the boundary w = g - g = 0. By the maximum principle a harmonic function attains its maximum and minimum on the boundary, where w is zero; so w is squeezed between 0 and 0, forcing w = 0 throughout. Hence u = v. Uniqueness costs us nothing beyond the maximum principle you already proved.
The Neumann problem: prescribe the flux
There is a second natural thing to fix on the boundary, and it is not the value of u but its outward slope. The Neumann problem asks for u harmonic in D with the Neumann boundary condition u_n = h on the boundary, where u_n is the derivative of u in the direction of the outward normal — how steeply u changes as you step straight out across the edge. Physically, this is the natural condition when you control the flow across the boundary rather than the level at it. From Fourier's law, flux is proportional to the gradient, so u_n at the rim is exactly the heat flowing out per unit length there. An insulated edge — no heat crosses — is simply h = 0.
But here a surprise waits, and it is a deep one. You cannot prescribe the outflow h however you like. Add up the total heat crossing the entire boundary: in a steady state, with nothing being created or destroyed inside, the net flow out of the region must be exactly zero. Everything that leaves through one part of the rim must have entered through another. So the boundary data must satisfy the Neumann compatibility condition: the integral of h around the whole boundary has to vanish. If you try to specify a net outflow that does not balance, no harmonic solution exists at all — you would be demanding that an equilibrium leak heat forever from nowhere.
DIRICHLET NEUMANN
Laplacian u = 0 Laplacian u = 0 (interior: harmonic)
u = g on boundary u_n = h on boundary (boundary data)
unique solution solution unique only up to + constant
exists on nice D EXISTS only if (integral of h over boundary) = 0
^ compatibility conditionThe compatibility condition is not a technicality you can wave away with a careful proof; it is forced by the physics of conservation, and it falls out of Green's identities in one line (integrate Laplacian u = 0 over D and the divergence theorem turns it into the boundary integral of u_n). The lesson generalizes: when you change which data you prescribe, you change what well-posedness even means. Dirichlet data can be anything continuous; Neumann data must balance. The interior equation is the same harmonic one either way — but the question you are allowed to ask is not.
Why Neumann uniqueness loses a constant
Run the uniqueness argument again, this time for Neumann, and a second twist appears. Suppose u and v both solve the Neumann problem with the same flux h. Their difference w = u - v is harmonic and now has w_n = 0 all around the boundary. The maximum-principle trick fails here — it controls boundary values, but we have only fixed boundary slopes. The clean way through is an energy identity from Green's first identity: the integral of |gradient of w|^2 over D equals a boundary integral of w times w_n, which is zero because w_n = 0. A nonnegative integrand integrating to zero must be zero everywhere, so gradient of w = 0 throughout — meaning w is a constant.
So the Neumann problem pins down the solution only up to an additive constant: if u solves it, so does u + 7, and there is no way to tell them apart from the flux alone. This is not a defect either — it is exactly right physically. The Neumann data fixes how heat flows, which sets the shape of the temperature, but it never fixes the overall thermostat reading. You can offset every temperature by the same amount and no flux changes, because flux feels only differences. To pick out one solution you must add one more scrap of information — say, the average value of u, or its value at a single chosen point.
A tiny worked example on the disk
Concrete numbers fix the ideas. Work on the unit disk in the plane and prescribe boundary data using the angle: take g = cos(angle) on the rim. We want u harmonic inside with u = cos(angle) on the boundary. A guess that respects the symmetry is u = x, since x = r cos(angle) equals cos(angle) when r = 1, and u = x is harmonic (its second derivatives both vanish). So u = x solves this Dirichlet problem, and by the uniqueness we just proved, it is the only solution. No infinite ocean anymore — the boundary data drained it down to one.
Now the Neumann version on the same disk. The outward normal on the unit circle points radially, so u_n = u_r at the rim. Try to prescribe h = cos(angle): does it pass the compatibility test? Integrate cos(angle) once around the circle and you get zero, so a solution should exist. And indeed u = x works again, because for u = x the radial derivative on the rim is u_r = cos(angle) = h. But watch: u = x + 100 has exactly the same outward flux, since adding a constant changes no derivative. Both solve the Neumann problem — the promised constant ambiguity, made flesh. Had we instead asked for h = 1 (a constant net outflow), the integral around the circle would be 2*pi, not zero, and the compatibility condition would forbid any solution at all.
One honest caveat to carry forward: guessing the answer worked only because the boundary data was a single clean trigonometric function tuned to the disk's symmetry. For a general g — a jagged temperature profile around the rim — there is no lucky guess, and you need a genuine machine that turns any boundary function into the harmonic function inside. That machine is the Poisson integral formula, the subject of the very next guide; it will prove existence on the disk outright and recover this u = x example as the easy special case. The Dirichlet and Neumann problems are the questions; the next two guides build the answers.