Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 11 / Solution

Local errors of the Euler schemes. Let $h=\Delta t>0$, and evaluate $f$ and its partial derivatives at an exact starting point $(y(t),t)$. DefineThe exact Taylor expansion is $y(t+h)=y+hf+\frac{1}{2}{h}^{2}A+O(h^3)$. The forward Euler method retains only $y+hf$, so its one-step defect, exact minus computed value, isFor the Backward Euler method, substitution of the exact solution into the scheme gives the residualbecause $f(y(t+h),t+h)=f+hA+O(h^2)$. Under local smoothness, solving the implicit step changes this residual by a factor $1+O(h)$, so the one-step defect has the same leading term:These are unnormalized local truncation errors, as in the question. A convention dividing by $h$ instead gives $O(h)$. Both schemes have first-order global error under the usual smoothness and stability hypotheses.

Euler stability. For the Dahlquist test equation $y^{\prime }=-\Lambda y$, put $z=h\Lambda$, with $\mathrm{Re}z>0$. The numerical amplification factors areStrict decay requires $|R|<1$; the boundary $|R|=1$ gives bounded but generally nondecaying values. For Forward Euler method,For positive real $\Lambda$, this reduces to $0<h\Lambda <2$. The absolute stability region in the $z$ plane is the disk $|z-1|\le 1$. For Backward Euler method, $|1+z|>1$ whenever $\mathrm{Re}z>0$, so it decays for every step size. It is A-stable and also L-stable, since $R_{BE}(z)\to 0$ for large $z$ in this half-plane. Here $z$ has the opposite sign from the alternative convention $z=h\lambda$ for $y^{\prime }=\lambda y$.

Predictor-corrector accuracy. The printed two-stage construction is Heun's method, equivalentlyAlthough the second stage resembles a Backward Euler method evaluation, it uses the predicted value, so the combined method is explicit. Expand that stage around the exact starting point, puttingThenand the numerical step is $y+hf+\frac{1}{2}{h}^{2}A+\frac{1}{4}{h}^{3}B+O(h^4)$. Along the exact solution, $y^{\prime \prime \prime }=B+f_{y}A$, so comparison with the exact Taylor expansion givesThus the local defect improves by one power of $h$, giving second-order global error under the usual hypotheses. For the linear test equation the defect reduces to $-h^3{\Lambda }^{3}y/6+O(h^4)$.

Predictor-corrector stability. Applied to the test equation, the predictor is $y^{(1)}=(1-z)y_n$ and the second value is $y^{(2)}=(1-z+z^2)y_n$. Their average givesFor real positive $z$, $R_H(z)=\frac{1}{2}[(z-1{)}^2+1]$, so strict decay holds exactly for $0<z<2$, the same real-axis step bound as Forward Euler method. For complex $z=x+iy$ the precise condition isThe region is bounded and does not contain the full right half-plane: for example $z=3$ gives $R_H=5/2$. It is therefore not A-stable. Its imaginary-axis values satisfy $|R_H(iy){|}^2=1+y^4/4>1$ for $y\ne 0$. Better accuracy has not inherited the unconditional stability of an actual implicit Backward Euler method step.