Past exam of the numerical solutions of partial differential equations course of the University of Oxford / 2023 / 3

Use the Backward Euler method in time and the prescribed central finite differences in space. With $j\in \mathbb{Z}$ and $m\ge 0$, the implicit finite difference method for the advection-diffusion equation isEquivalently, for $\nu =\Delta t/\Delta x$ and $\mu =\Delta t/(\Delta x{)}^2$, the new time level solvesThe spatial terms are evaluated at the new time level, making this an implicit scheme. On the whole line there are no finite-endpoint boundary conditions; in the next part its solution is chosen in the l2 sequence space.

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]

Use the forward Euler method in time and evaluate both spatial finite difference operators at the old time level:With $\nu =\Delta t/\Delta x$ and $\mu =\Delta t/(\Delta x{)}^2$, the explicit update isEvery value on the new time level is computed directly from three old values. The sum of the stencil weights is one, as required to preserve constant solutions of the advection-diffusion equation.

Using the semidiscrete Fourier transform from (b), the explicit scheme has amplification factorSet $r=a\mu$ and $s={\sin }^2(\theta /2)\in [0,1]$, so ${\sin }^2\theta =4s(1-s)$. Direct expansion givesThe expression in square brackets is an affine function of $s$. At the two endpoints its values areusing precisely ${\nu }^2\le 2r\le 1$. Every intermediate value is a convex combination of these endpoints, so $|G_E{|}^2\le 1$ for every frequency. The Parseval identity provesThus the scheme is practically stable under the given condition. For $a>0$ it is equivalent to $\Delta t\le 2a$ and $\Delta t\le (\Delta x{)}^2/(2a)$. This is the Forward Euler stability for centered advection-diffusion calculation with the diffusion coefficient retained. The proof does not require all three stencil weights to be nonnegative.

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.