Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 63 / 1

Put $d=\Delta x$, $k=\Delta t$ and use the diffusion Courant number $r=k/d^2$. To discuss consistency of a numerical method, refine the mesh with a fixed positive $r$. With ${\delta }^2{U}_m=U_{m-1}-2U_m+U_{m+1}$, the printed update is $U^{n+1}-U^n=\alpha {\delta }^2{U}^{n+1}$. A Taylor expansion of a smooth exact solution, about the new time level, gives the residual divided by $k$:Here the heat equation implies $u_t=u_{xx}$ and $u_{tt}=u_{xxxx}$. Since $\alpha$ depends only on $r$, the leading term can vanish for every smooth solution only if $\alpha (r)=r$. Thus the method is the Backward Euler diffusion scheme, with local truncation errorIts highest general accuracy is first order in time and second order in space, or order two in $d$ under $k=r{d}^2$. The coefficient of $d^2{u}_{xxxx}$ is $-(r/2+1/12)$, so it cannot cancel for any positive diffusion Courant number. This is a conclusion about the stencil actually printed: the new-time spatial difference fixes the sign of its temporal error. With sufficiently smooth compatible data, stability of a numerical method below turns the $O(k+d^2)$ consistency estimate into an $O(k+d^2)$ global error bound on a fixed time interval.

The zero Dirichlet boundary conditions permit an exact finite-interval calculation. Let $T=\mathrm{tridiag}(1,-2,1)$ on the $M$ interior grid points. The Dirichlet discrete Laplacian has an orthogonal eigenvector basisConsequently the update matrix $G=(I-rT{)}^{-1}$ has modal amplification factorsFor every physical $r\ge 0$, all denominators are at least one. Orthogonal diagonalization therefore provesuniformly in $M$, $r$ and $n$. The same bound controls initial perturbations; a forcing increment is propagated by contraction, so successive increments accumulate at most by their sum. The highest-order consistent method is unconditionally stable for all $r\ge 0$. This proof incorporates the boundary conditions, whereas a periodic Fourier mode calculation alone would not do so.

For completeness, if “range” is interpreted algebraically to include negative $r$ on a fixed grid, the exact power-stable range isIndeed, for $r<0$ every denominator is less than one, so $|g_j|\le 1$ requires $1+4r{s}_j\le -1$ for every $j$. The strongest restriction comes from $s_1$. Equality gives a simple amplification factor $-1$ and is allowed because the update is orthogonally diagonalizable. Other negative values either amplify a mode or make the update singular. This extra branch describes backward time stepping on a fixed spatial grid; it is not a positive-time diffusion discretization, and no fixed negative $r$ remains in that branch as $M\to \infty$.