Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 69 / 1

Put $h=\Delta x$, let $C_h$ denote the sum of the four axial shifts, and let $K_h$ denote the sum of the four diagonal shifts. Reading both sides of the PDF stencil gives $M_h\dot{U}=L_{h}U$, where $M_h=2I/3+C_h/12$ and $L_h=(-10I/3+2C_h/3+K_h/6)/h^2$. The left-hand stencil acts on the time derivative; treating it as a scalar multiplier would lose the compact correction.

Write $D={\partial }_x^2+{\partial }_y^2$. Taylor expansion of the symmetric shifts givesandFor the exact diffusion solution $u_t=Du$, subtracting givesThe leading coefficient is not identically zero for general smooth data. Thus the normalized local residual is fourth order, and in the question's $h^{p+1}$ convention $\boxed{p=3}$. This is the compact semidiscrete nine-point diffusion scheme; its usual spatial numerical consistency order is four, rather than three.

Use the Fourier amplification symbol for the semidiscrete generator. At a mode $e^{i(k\xi +l\eta )}$, the two stencil symbols areThe mass symbol satisfies $1/3\le m\le 1$. The numerator of $\ell$ is increasing in each cosine, since each partial derivative is at least two, and its maximum is zero at $(1,1)$. Hence $\ell \le 0$ and the generator ${\lambda }_h=\ell /m$ is real and nonpositive.

Every exact-time mode is multiplied by $e^{t{\lambda }_h}$, of modulus at most one. Parseval identity therefore provesThe method is stable, uniformly in the spatial mesh. This is a semidiscrete conclusion: it imposes no time-step condition until a particular time integrator is applied. For the stated merely square-integrable initial data, cell averages give a well-defined stable initialization, with $\Vert U(0){\Vert }_{{\ell }_h^2}\le \Vert v{\Vert }_{L^2}$ by Cauchy–Schwarz. Arbitrary point sampling of an L2 equivalence class is not required.