Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A

Substitute the exact solution and expand every value about $t_n$. The coefficient of $h^q{y}^{(q)}(t_n)$ in the defect is zero for $q=0,\dots ,p$ exactly whenThese are the order conditions for a linear multistep method, specialized to the two nonzero derivative coefficients. The first failed identity determines the leading local truncation error.

For $s=1$, imposing order two givesThe method is therefore the trapezoidal ruleIts first characteristic polynomial is $\rho (\zeta )=\zeta -1$, so it satisfies the root condition for a multistep method.

For $s=2$, the four conditions through order three giveand henceHereso the root condition for a multistep method again holds. Both methods are consistent and zero-stable, and the Dahlquist equivalence theorem therefore proves that both are convergent.

A highest-order method in this family has order $s+1$. The Second Dahlquist barrier says that an irreducible A-stable multistep method has order at most two, so $s+1\le 2$ and hence $s=1$. Part b then leaves only the trapezoidal rule. Its amplification factor isand $|R(z)|\le 1$ whenever $\mathrm{Re}z\le 0$. Thus the trapezoidal rule is the unique highest-order A-stable method of the stated form, apart from representations containing removable common factors.

For the nodes $c_1=\alpha$ and $c_2=1$, the Lagrange basis isDirect integration givesand thereforeThe Collocation Runge-Kutta method also requiresThis exposes a sign error in the printed tableau: its lower-right entry is shown as $(-1+2\alpha )/[2(1-\alpha )]$. With $1-2\alpha$ in that position, the tableau is exactly the claimed collocation method. Taken literally, the printed weights satisfy $b^{T}1=\alpha /(1-\alpha )$, so the method is not even consistent unless $\alpha =1/2$ and cannot be a collocation method for general $\alpha$.

For the intended collocation weights, interpolation makes the associated quadrature rule exact for every polynomial of degree at most one, so the method has order at least two. It has order at least three precisely when the node polynomial is orthogonal to constants:ThusAt $\alpha =1/3$ this is the two-stage Radau IIA method. Since one node is fixed at the endpoint, no value of $\alpha$ makes the two-node quadrature exact through degree three, so order four cannot occur.

For completeness, the tableau exactly as printed has order zero when $\alpha \ne 1/2$, since $b^{T}1\ne 1$. At $\alpha =1/2$ the two signs coincide because the second weight vanishes, and the resulting method has order two.

For algebraic stability of a Runge-Kutta method, the weights must be nonnegative andmust be positive semidefinite. With the intended collocation weights,Positive semidefiniteness is therefore possible only at $\alpha =1/3$; substitution gives nonnegative weights and a positive-semidefinite $M$. Hence the intended family is algebraically stable exactly whenWith the sign printed in the paper, one instead obtainsEquality forces $\alpha =1/2$, where $M$ still has nonzero off-diagonal entries and is indefinite. The literal printed tableau is consequently algebraically stable for no value of $\alpha$.

Write $H={\partial }_x^2-V(x)$. The real potential and periodic boundary condition make $H$ a self-adjoint operator on the periodic domain. Since $u_t=-iHu$,because the inner product $⟨u,Hu⟩$ is real. Thus the Schrodinger equation preserves the $L^2$ norm.

Let $H_h$ be the periodic centered second-difference matrix minus the real diagonal matrix containing $V(x_m)$. It is a Hermitian matrix, so the semidiscrete system iswith a skew-Hermitian matrix generator. ConsequentlyIts exact propagator $e^{-it{H}_h}$ is a unitary matrix, and therefore the semidiscretization is stable in the discrete $2$-norm, uniformly for all times and mesh sizes.

Apply the implicit midpoint rule, equivalently the Crank--Nicolson method, to the semidiscrete equation:It has order two. Its amplification matrix is the Cayley transformBecause $H_h$ is Hermitian, $Q$ is unitary. Hence $\Vert {\mathbf{u}}^{n+1}{\Vert }_2=\Vert {\mathbf{u}}^n{\Vert }_2$ exactly.

A real linear operator $\mathcal{L}$ on an inner-product space is a positive-definite operator when it is self-adjoint andfor every nonzero $u$ in its domain. In the variational setting one normally requires the stronger uniform estimate $⟨\mathcal{L}u,u⟩\ge c\Vert u{\Vert }_V^2$ for some $c>0$, which is coercivity.

Choose the Sobolev space $V$ encoding the homogeneous essential boundary conditions, setafter the appropriate integration by parts, and define the energy functionalIts first variation is $J^{\prime }(u)v=a(u,v)-\ell (v)$, so its stationary points are exactly the solutions of the weak formulationIf $a$ is bounded, symmetric, and coercive and $\ell \in V^{\prime }$, the Lax-Milgram theorem supplies a unique weak solution $u$. Moreover, for every $w\in V$,unless $w=0$. Thus $J$ is strictly convex, and $u$ is its unique global minimizer. This proves existence and uniqueness of the minimizer and of the weak solution simultaneously.

Let $\Omega =(-1,1{)}^2$ and use the clamped energy space $V=H_0^2(\Omega )$, whose traces satisfy $u={\partial }_{n}u=0$ on $\partial \Omega$. Two applications of integration by parts giveIf equality holds, then $\Delta u=0$. The maximum principle for harmonic functions and the zero Dirichlet boundary condition imply $u=0$. The biharmonic operator is therefore positive definite. The same identity and conclusion hold for the simply supported conditions $u=\Delta u=0$; either standard interpretation of the paper's phrase “zero boundary conditions” gives the result.

Near $t=0$, take $\Omega (t)=\log F(t)$ to be the continuous local matrix logarithm with $\Omega (0)=0$. The symmetry identity gives $F(-t)=F(t{)}^{-1}$, and uniqueness of this logarithm yieldsThus $\Omega$ is an odd function. The local-logarithm qualification is necessary because the matrix exponential is not globally injective; the statement is naturally understood either near $t=0$ or as an identity of formal power series.

For the Strang splittingreversing $t$ reverses all three factors, soIt is therefore a time-symmetric numerical method. Multiplication of the three exponential series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the Baker--Campbell--Hausdorff formula gives an odd modified generatorfor a matrix $D$ made from nested commutators. Exponentiating givesfor a matrix $C$ depending on $A$ and $B$.

The three substeps have total length $2\alpha +(1-2\alpha )=1$. Their leading cubic defects add, so cancellation requiresTaking the real cube root and solving givesThis is the coefficient in the higher-order composition of a symmetric splitting; the middle substep $1-2\alpha$ is negative.

The composition is palindromic, and henceThus $G_{\alpha }$ is symmetric. Part c cancels the cubic term in its odd formal logarithm; symmetry forbids a fourth-degree term, so the next possible defect has degree five. Thereforewhich makes the composition a fourth-order method.