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FDM basic steps; explicit and implicit schemes

Short answer

Basic steps: state PDE/IC/BC, discretize grid, approximate derivatives with finite differences, form algebraic equations, impose initial/boundary values, solve sequentially (explicit) or as simultaneous equations (implicit). Explicit example: Q_i^{n+1}=Q_i^n-C(Q_i^n-Q_{i-1}^n), stable C≤1. Implicit: Q_i^{n+1}=(Q_i^n+C Q_{i-1}^{n+1})/(1+C), unconditionally stable.

Describe the basic steps in the finite difference method. Explain explicit and implicit schemes in the finite difference method using suitable examples and expressions.

Basic steps in the finite difference method

  1. State the problem: write the governing PDE with its domain, initial conditions (at t=0t=0) and boundary conditions.
  2. Discretise the domain: draw a grid with spacing Δx\Delta x and time step Δt\Delta t; points are (xi,tn)(x_i,t^n).
  3. Approximate the derivatives: replace each derivative by a forward, backward or central difference obtained from the Taylor series.
  4. Write the algebraic equation at every grid point (explicit or implicit form).
  5. Insert the initial and boundary values into the equations.
  6. Solve: march forward in time (explicit), or solve the simultaneous equations at each time step (implicit, e.g. by the Thomas algorithm).
  7. Check consistency, accuracy (truncation error), stability (Courant number) and convergence by refining Δx,Δt\Delta x,\Delta t.
 t
 n+1 : o   o   o   o      unknown
 n   : o   o   o   o      known (initial / previous step)
 n-1 : .   .   .   .
       i-1 i  i+1 i+2    --> x

Explicit and implicit schemes (example: ∂Q/∂t+c ∂Q/∂x=0\partial Q/\partial t+c\,\partial Q/\partial x=0, C=cΔt/ΔxC=c\Delta t/\Delta x)

Explicit: the spatial derivative is evaluated at the known time level nn.

Qin+1−QinΔt+c Qin−Qi−1nΔx=0 ⇒ Qin+1=Qin−C (Qin−Qi−1n)\frac{Q_i^{n+1}-Q_i^n}{\Delta t}+c\,\frac{Q_i^n-Q_{i-1}^n}{\Delta x}=0\ \Rightarrow\ Q_i^{n+1}=Q_i^n-C\,(Q_i^n-Q_{i-1}^n)

Each unknown is found directly from known values. It is conditionally stable (needs C≤1C\le1).

Implicit: the spatial derivative is evaluated at the new time level n+1n+1.

Qin+1−QinΔt+c Qin+1−Qi−1n+1Δx=0 ⇒ Qin+1=Qin+C Qi−1n+11+C\frac{Q_i^{n+1}-Q_i^n}{\Delta t}+c\,\frac{Q_i^{n+1}-Q_{i-1}^{n+1}}{\Delta x}=0\ \Rightarrow\ Q_i^{n+1}=\frac{Q_i^n+C\,Q_{i-1}^{n+1}}{1+C}

The unknown at point ii depends on other unknowns at n+1n+1. For problems such as groundwater flow the result is a set of simultaneous equations (tridiagonal matrix). It is unconditionally stable.

Remark: with the implicit form the new value at ii also depends on the unknown Qi−1n+1Q_{i-1}^{n+1}; for the one-way kinematic wave this is solved by sweeping from the upstream boundary, while for diffusion-type equations a tridiagonal system must be solved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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