Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 69 / Section I

Let $I=(-1,1)$ and use the clamped second-order Sobolev spaceHere weak derivatives up to order two are square integrable, with Sobolev norm $\Vert v{\Vert }_{H^2}^2=\Vert v{\Vert }_2^2+\Vert v^{\prime }{\Vert }_2^2+\Vert v^{\prime \prime }{\Vert }_2^2$. The boundary values are the continuous traces given by the Sobolev trace theorem, not arbitrary pointwise representatives. This space is the $H^2$ closure of $C_c^{\infty }(I)$, so it is a closed Hilbert space. The derivative boundary conditions are essential, and $H_0^1(I)$ alone is insufficient.

Define the symmetric bilinear form and bounded linear functionalThen the variational problem isFor complex data replace $a$ by its sesquilinear version, take the real part of the forcing term and use absolute squares in $J$; the same argument applies.

Its first variation is $J^{\prime }(u)[v]=a(u,v)-\ell (v)$. Thus stationary points satisfy $a(u,v)=\ell (v)$ for every $v\in H$, the weak formulation of a clamped fourth-order equation. For smooth coefficients and functions, twice integrating the highest term and once the middle term gives the differential equation, with all boundary terms zero because $v=v^{\prime }=0$ at both ends. With only the stated continuous coefficients, this manipulation is instead understood distributionally: compactly supported test functions give $(p{u}^{\prime \prime }{)}^{\prime \prime }-(q{u}^{\prime }{)}^{\prime }+ru=f$ as a distribution. This avoids assuming nonexistent classical derivatives of $p$ or $q$.

By the extreme value theorem, $p_0={\min }_{[-1,1]}p$ is strictly positive, and $p,q,r$ are bounded. The Cauchy-Schwarz inequality therefore givesFor completeness, the clamped interval derivative norm bound follows from $v^{\prime }(-1)=v(-1)=0$ and the one-dimensional Sobolev representative: integrate $v^{\prime \prime }$ to recover $v^{\prime }$, and then $v^{\prime }$ to recover $v$. Applying Cauchy-Schwarz and integrating over an interval of length two givesConsequently $\Vert v{\Vert }_{H^2}^2\le 21\Vert v^{\prime \prime }{\Vert }_2^2$. Nonnegativity of $q,r$ yields the precise coercive bilinear form boundThe Lax-Milgram theorem states that a bounded bilinear form on a real Hilbert space satisfying $a(v,v)\ge c\Vert v{\Vert }^2$ with $c>0$ represents every bounded linear functional uniquely: there is exactly one $u$ with $a(u,v)=\ell (v)$ for all $v$. All of its hypotheses have now been verified on $H$.

This weak solution is the unique global minimum, rather than merely a stationary point. The Ritz-Galerkin equivalence for a symmetric coercive form here is the exact identityIt is strictly positive for every nonzero $w\in H$. Conversely, a minimum has zero first variation in every direction, so it satisfies the same weak equation. Density of compactly supported test functions and boundedness of $a,\ell$ extend the distributional equation back to all $H$ test vectors. There is exactly one minimum, and it is exactly the clamped weak solution. No additional coefficient smoothness or classical fourth derivative is needed.

The coefficients are those of the three-stage Lobatto IIIA method. Set $e=(1,1,1{)}^T$, $c=Ae=(0,1/2,1{)}^T$ and $C=\mathrm{diag}(c)$. Direct calculation gives all eight fourth-order conditions for a Runge-Kutta method:Powers of $c$ are componentwise. These Butcher order conditions prove order at least four for a smooth general nonlinear vector field; matching a scalar test equation alone would not suffice. A necessary fifth-order condition fails:Hence the method has order exactly four, with one-step local truncation error $O(h^5)$ and, under stability and smoothness hypotheses, global error $O(h^4)$.

For the Dahlquist test equation with $z=h\lambda$, the stage equations give the stability function $R(z)=1+z{b}^T(I-zA{)}^{-1}e$. Substitution simplifies it toThe denominator zeros are $3\pm i\sqrt{3}$, so no pole lies in the closed left half-plane. Moreover,Thus $|R(z)|\le 1$ throughout that half-plane: the method is A-stable. On the imaginary axis its amplification modulus is one. Since $R(z)\to 1$ as $z\to -\infty$, it is not L-stable; linear stability does not guarantee strong damping of very stiff modes.

Algebraic stability of a Runge-Kutta method requires $b_i\ge 0$ and a positive-semidefinite symmetric matrixAlthough all weights are positive, its first diagonal entry isThe quadratic form on the first coordinate vector is negative, so $M$ cannot be positive semidefinite. The method is not algebraically stable. This distinguishes nonlinear contractivity from the A-stability proved in part 2.

Let $h=\Delta x$ and sample a sufficiently smooth periodic exact solution. The central finite differences have expansionsandThe semidiscrete right side therefore equals $u_{xx}+\alpha u_x+h^2(u_{xxxx}/12+\alpha u_{xxx}/6)+O(h^4)$. The spatial consistency of a numerical method is second order, for each fixed $\alpha$. There is no time discretization in the stated ordinary differential system, so no separate temporal approximation order is being assigned. The Taylor statement concerns smooth solutions; L2-compatible initialization of grid data handles nonsmooth initial data without undefined point sampling.

Use the discrete norm $\Vert U{\Vert }_h^2=h{\sum }_{m=1}^N|U_m{|}^2$, with periodic indices. Two independent arguments prove uniform stability of a numerical method.

For Fourier stability analysis, a discrete Fourier mode $U_m(t)={\hat{U}}_k(t)e^{im{\theta }_k}$ with ${\theta }_k=2\pi k/N$ has semidiscrete eigenvalueIts real part is nonpositive, so $|{\hat{U}}_k(t)|\le |{\hat{U}}_k(0)|$. The unitary discrete Fourier transform and discrete Parseval identity then give $\Vert U(t){\Vert }_h\le \Vert U(0){\Vert }_h$ for every $N$, $t\ge 0$ and fixed real $\alpha$.

For the energy method, denote the centered first difference by $D_0$ and the forward difference by $D_{+}$. Periodic discrete summation by parts makes $D_0$ skew-adjoint and gives $⟨U,D_{xx}U{⟩}_h=-\Vert D_{+}U{\Vert }_h^2$. ThereforeThe constant mode is preserved rather than strictly damped. Both proofs apply equally to differences of two solutions. The semidiscretization is contractive in the discrete L2 norm. A later choice of a time integrator would require its own stability analysis; no time-step condition is needed for this continuous-time result.

The characteristic polynomials of a linear multistep method areExpanding the exponential-symbol order criterion for a multistep method givesThus the formal local consistency order is two for $\alpha \ne 2$, and three at $\alpha =2$, since the fourth-degree coefficient there is $-1/12$. This exceptional formal order is degenerate: ${\rho }^{\prime }(1)=\sigma (1)=0$, so it is not a convergent third-order time integrator.

For zero-stability, factor $\rho (\zeta )=(\zeta -1)(\zeta -(\alpha -1))$. The root condition for a multistep method requires every root to have modulus at most one and every unit-modulus root to be simple. For real $\alpha$ it holds exactly forAt zero the second root is $-1$, simple and admissible; at two the root one is double and inadmissible. Outside the closed interval the second root has modulus greater than one. The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable, for a locally Lipschitz initial-value problem with consistent starting values and solvable implicit steps. Hence the displayed interval is also exactly the convergence range, and every convergent member has order two.

The common-factor cancellation defect in a multistep recurrence explains why the $\alpha =2$ exception cannot be repaired just by canceling $\zeta -1$. For $y^{\prime }=0$, its recurrence allows linear growth from a vanishing but insufficiently small starting perturbation: $y_0=0$, $y_1=\sqrt{h}$ gives $y_n=n\sqrt{h}$, diverging at a fixed positive time. Canceling the factor silently imposes an additional starting relation and changes the method.

Every convergent member is also A-stable. For $0<\alpha <2$, put $\zeta =e^{i\theta }$ and calculate its boundary-locus test for multistep A-stability:The denominator never vanishes on the unit circle in this parameter interval. It follows thatTherefore no amplification root can cross the unit circle as $z=h\lambda$ moves in the open left half-plane. For small negative real $z$, the root near one moves to $1+z+O(z^2)$, inside the circle; the other begins at $\alpha -1$, also strictly inside. The leading coefficient $1-(1-\alpha /4)z$ never vanishes in the left half-plane. Continuity of polynomial roots and connectedness of that half-plane now keep both roots inside everywhere. On its imaginary boundary the same locus has zero real part only at $\zeta =1$, corresponding to $z=0$; that root is simple because $2-\alpha >0$.

At $\alpha =0$, the two interlaced subsequences follow the trapezoidal rule with step $2h$:The right side has modulus at most one for $\mathrm{Re}z\le 0$. Its unit-circle roots are simple; the double zero at $z=-1$ lies strictly inside and is harmless. Thus the endpoint is also A-stable. Combining this with part 1, the requested range is $0\le \alpha <2$, including the BDF2 method at $\alpha =4/3$.

Take the endpoint values $u_0=u_M=0$ and evolve the interior indices $1,\dots ,M-1$. This matches $\Delta x=1/M$ on $[0,1]$. If the printed upper index $m=M$ is retained, the odd ghost extension $u_{M+1}=-u_{M-1}$ makes its boundary equation identically compatible; it does not add an interior unknown.

Writing $h=\Delta x$ and $\tau =\Delta t$, this is the theta method with parameter $a$:For a smooth exact heat solution, its normalized residual at $t_n$ isThus the generic consistency orders are first in time and second in space for $a\ne 1/2$, and second in both time and space for $a=1/2$, the Crank-Nicolson method. These are independent-step orders. Along the special parabolic refinement $\tau =\mu h^2$, the leading combined residual cancels if $a=1/2-1/(12\mu )$, giving a higher combined smooth-data order; it does not change the generic temporal order at independently varied meshes.

The given L2 initial condition need not have pointwise values or four derivatives. Initialize by a bounded cell-average or projection operator, as in L2-compatible initialization of grid data. The Taylor estimate applies to smooth data or sufficiently positive times after heat smoothing, rather than an unqualified uniform high-order estimate at the initial instant for arbitrary rough data. Stability and convergence for rough data are distinct from this consistency order.

The Dirichlet difference Laplacian has an orthogonal sine basis with eigenvalues $-{\eta }_j/h^2$, where ${\eta }_j=4{\sin }^2(j\pi /(2M))$ for $j=1,\dots ,M-1$. On a mode the theta method amplification isIf $a\ge 1/2$, the denominator is positive andThus $|G(s)|\le 1$ for every mode and every $\mu >0$. Orthogonality gives a uniform discrete L2 norm contraction, not merely individual eigenvalue estimates.

If $a<0$, a positive $s=-1/a$ makes the implicit matrix singular. If $a=0$, $G=1-s$ becomes arbitrarily large. If $0<a<1/2$, $G(s)$ tends to $-(1-a)/a$, whose magnitude is greater than one. Hence unconditional stability holds exactly for $a\ge 1/2$.

For completeness, $a<1/2$ is contractive under $\mu {\eta }_{\max }\le 2/(1-2a)$; the mesh-independent sufficient restriction is $\mu \le 1/[2(1-2a)]$. This follows from the same squared-modulus difference and keeps the denominator positive even for negative $a$. At $a=1/2$, very high frequencies approach amplification $-1$, so A-stability does not imply damping of rough heat data. At $a=1$ the Backward Euler method damps them strongly; this is the rough-data damping distinction for the theta method.