Press ▶ Solve. The plate starts at a flat guess and relaxes, point by point, toward the field the boundaries demand. Watch the residual fall — when it's tiny, you've reached steady state.
Heat Transfer · interactive
A flat plate, four edges, and the question every heat-transfer student meets first: once everything settles down, what is the temperature everywhere inside? Set the boundary conditions, press play, and watch the field relax to steady state — the same way the computer actually solves it.
Press ▶ Solve. The plate starts at a flat guess and relaxes, point by point, toward the field the boundaries demand. Watch the residual fall — when it's tiny, you've reached steady state.
Lay a grid over the plate. Replace each second derivative with a finite difference, and Laplace's equation collapses to a beautifully simple rule: every interior node's temperature is the average of its four neighbours (plus a generation term). That can't be solved in one shot — each node depends on its neighbours, which depend on it. So we sweep the grid over and over, updating each node from the latest values around it. This is Gauss–Seidel relaxation (here accelerated with over-relaxation), and the animation above is literally those sweeps. The residual is how much the field still changed on the last sweep; when it stops changing, the discrete equation is satisfied.
A fixed-temperature edge (Dirichlet) pins the boundary — a wall held by a reservoir. An insulated edge (Neumann, zero-flux) lets no heat cross; we enforce it by mirroring the interior so the gradient at the wall is zero. Flip any edge between the two and watch how the isotherms bend to meet it — they always cross an insulated wall at a right angle.
Steady state means energy in = energy out. If you insulate all four edges while heat is being generated inside, that balance is impossible — the heat has nowhere to go, so the temperature just climbs forever and no steady state exists. The solver here detects that case and tells you, instead of quietly returning nonsense. (Insulate all four edges with no generation and you get the opposite problem: any uniform temperature is a valid answer — the solution is only defined up to a constant.) Try the Break it scenario.
Here's the one that surprises people. With fixed-temperature walls and no generation, k cancels out of the equation entirely — slide it from copper to firebrick and the temperature field never moves. What k does change is the heat flux: the rate q = −k∇T at which energy flows through. Same temperatures, very different watts. Turn on heat generation and k matters again, because now it sets how hot the interior gets relative to the walls. Try Does k matter? and watch the field stay frozen while the flux readout moves.
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