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Original solutions to all three questions of the numerical solutions of partial differential equations course of the University of Oxford Part B examination, held on 6 June 2023. Source: original released PDF, bundle pages 292–295; paper pages 1–4. The numbering follows the paper. In 3(e), the growth constant must be allowed to depend on the mesh; the literal mesh-independent assertion in the native PDF is false. The corrected result and the intended failure of practical stability of a finite difference scheme are proved below.

For a grid sequence define its semidiscrete Fourier transform in the dimensionless frequency variable $\theta \in [-\pi ,\pi ]$ byFor general square-summable data the transform is interpreted as an limit in the L2 norm, rather than an absolutely convergent pointwise sum. The Parseval identity in this normalization isThe Fourier symbols of $D_x^0$ and $D_x^{+}{D}_x^{-}$ are $i\sin \theta /\Delta x$ and $-4{\sin }^2(\theta /2)/(\Delta x{)}^2$. Transforming the implicit equation givesIts amplification factor is consequentlyThe denominator has real part at least one because $a\ge 0$. It never vanishes, so multiplication by $G_I$ defines a unique next value in the l2 sequence space, andThe Parseval identity now givesThis is unconditional practical stability of a finite difference scheme: in the course's terminology the numerical evolution does not increase this norm. Linearity gives the same estimate for perturbations of the initial data. This contraction convention is stronger than a finite-time mesh-uniform bound allowing physical exponential growth.^[ref]

When $a=0$, the explicit amplification factor becomesA single Fourier mode has constant modulus in space and is not in the l2 sequence space. Instead choose a whole interval of frequencies on which there is definite growth. LetThe inverse Fourier transform here gives the real sequenceBy the Parseval identity, its squared norm is $\Vert U^0{\Vert }_{{\ell }_2}^2=\Delta x|E|/(2\pi )=\Delta x/2$, finite and strictly positive. These data are also compatible with sampling a bounded continuous function: useThe value at zero is the limit of the displayed quotient, and $u_0(j\Delta x)=U_j^0$.

On $E$ we have ${\sin }^2\theta \ge 1/2$. Therefore the transformed solution and the Parseval identity giveFor each fixed pair $\Delta x,\Delta t>0$, takeTaking square roots proves $\Vert U^m{\Vert }_{{\ell }_2}\ge (1+c{)}^m\Vert U^0{\Vert }_{{\ell }_2}$ for every $m\ge 1$. In particular the norm already increases at the first step, so no positive mesh pair gives practical stability in the contraction sense of (b).

The native PDF places the existence of $c>0$ before the universal quantifier over mesh pairs. If this means a single positive $c$ independent of $\Delta x,\Delta t$, that assertion is false for nonzero data. Indeed, for every datum the same Fourier symbol givesFor any proposed fixed $c>0$, choose $0<\nu <\sqrt{(1+c{)}^2-1}$. The upper bound then contradicts the requested lower bound at $m=1$. The correct quantifiers are $\forall \Delta x,\Delta t>0\exists c=c(\Delta x,\Delta t)>0$, with nonzero square-summable data as constructed above.

Finally, failure of this stronger practical stability of a finite difference scheme should not be confused with every possible refinement notion of numerical stability. If $\Delta t/(\Delta x{)}^2\le K$, then $m\Delta t\le T$ givesThus a finite-time mesh-uniform bound can hold under this severe refinement even though contraction fails on every positive mesh. This is the distinction recorded in centered-advection refinement-dependent stability.