SlotLayer 0
Discretize a PDE into one linear system
Replace every continuous variable at once — space together with time, or space together with velocity — so that the whole problem becomes a single matrix equation. Nothing is left to march: the grid values at every recorded point are unknowns of one system, solved in one go.
A linear PDE with its conditions, a grid over every continuous variable the problem carries, and — where the problem is posed as a boundary-value problem rather than an initial-value one — the boundary treatment that makes the resulting matrix well posed.
One matrix and one right-hand side over all the grid unknowns together, with the condition number that the cost of solving it will be measured against, and the discretization error that fixes how fine the grid had to be.
This one, drawn
From Partial differential equation to Linear system Ax = b
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Why this is a layer
This is a different act from discretizing space and then discretizing time, and the clearest case is the one where that composition is not merely a different reading but unavailable. Novikau, Dodin and Startsev have no time axis to discretize at all: their problem is posed at a fixed drive frequency as a boundary-value problem, so what they grid is space and velocity, the matrix that comes out is the answer rather than a propagator, and there is no time-discretization step for a composition to pass through. Linden, Montanaro and Shao point the same way from the other side, though less decisively: their forward-time centre-space scheme produces one block system over every timestep at once, and its condition number is a property of that whole system rather than of either stage. What competes here is which continuous variables get replaced together, and the choice is forced by how the problem was posed rather than by a preference — which is why a reader arriving with a boundary-value problem cannot use the method of lines at all.
Ways to do this
2 methods recorded
- Forward-time, centre-space (FTCS)
Take the forward difference in time and the three-point central difference in space, then stack the resulting one-step relations for every timestep into a single block lower-bidiagonal system whose unknowns are the grid values at all recorded times together.
- Phase-space grid for a boundary-value problem
Grid the position and velocity coordinates together and take central differences in both, with the derivatives at the edges obtained from Lagrange interpolating polynomials. Because the problem is posed at a fixed drive frequency rather than as an evolution, what results is directly the matrix equation to be solved — there is no time axis left to march along.
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In the Atlas
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