In the explicit scheme the spatial terms are evaluated at:

The old time level n, so each new value is found directly

Which statement about stability is correct?

Implicit and Crank-Nicolson are unconditionally stable; explicit needs r <= 0.5

Explicit is unconditionally stable

All three are conditionally stable

Practice · 5 marks

Explicit, implicit and Crank-Nicolson schemes

Short answer

Explicit (FTCS): T_i^{n+1}=T_i^n+rδ²T_i^n, O(Δt,Δx²), conditionally stable (r≤0.5). Implicit (BTCS): -rT_{i-1}^{n+1}+(1+2r)T_i^{n+1}-rT_{i+1}^{n+1}=T_i^n, O(Δt,Δx²), unconditionally stable. Crank-Nicolson: average of explicit/implicit, O(Δt²,Δx²), unconditionally stable. Explicit cheap per step but tiny Δt; implicit/CN require tridiagonal solves but allow larger Δt.

Differentiate between explicit, implicit and Crank-Nicolson time-marching schemes for the one-dimensional transient diffusion equation. Write the discretised form of each and compare accuracy, stability and cost.

For $\partial T/\partial t=\alpha\,\partial^2T/\partial x^2$ , let $r=\alpha\Delta t/\Delta x^2$ and $\delta^2T_i=T_{i+1}-2T_i+T_{i-1}$ .

Discretised forms

Explicit (FTCS): $T_i^{n+1}=T_i^n+r\,\delta^2T_i^{\,n}$ . The right side uses only old values, so each new value is found directly.
Fully implicit (BTCS): $T_i^{n+1}=T_i^n+r\,\delta^2T_i^{\,n+1}$ , or $-rT_{i-1}^{n+1}+(1+2r)T_i^{n+1}-rT_{i+1}^{n+1}=T_i^n$ . This needs a tridiagonal system solved each step (Thomas algorithm).
Crank-Nicolson: average of the two, $T_i^{n+1}=T_i^n+\dfrac r2\left(\delta^2T_i^{\,n}+\delta^2T_i^{\,n+1}\right)$ . Also tridiagonal.

Comparison

Property	Explicit	Implicit	Crank-Nicolson
Time level of spatial terms	$n$	$n+1$	average of $n$ and $n+1$
Accuracy	$O(\Delta t,\Delta x^2)$	$O(\Delta t,\Delta x^2)$	$O(\Delta t^2,\Delta x^2)$
Stability	Conditional: $r\le0.5$	Unconditional	Unconditional
Work per step	Very small	Matrix solve	Matrix solve
Time step	Restricted by stability	Chosen by accuracy	Chosen by accuracy
Weakness	Tiny $\Delta t$ on fine mesh	Only first-order in time; damps	May oscillate for large $r$

Remarks

With the explicit scheme, halving $\Delta x$ forces $\Delta t$ to be one quarter, so cost rises about 8 times in 1D.
Implicit methods are used in most commercial CFD solvers because stable large time steps are possible, especially for steady-state solutions and stiff problems.
Unconditional stability does not mean accuracy: too large a $\Delta t$ still gives a poor transient.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Explicit, implicit and Crank-Nicolson schemes

Discretised forms

Comparison

Remarks

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