= Solution
\b[Local errors of the Euler schemes.] Let $h=\Delta t>0$, and evaluate $f$ and its partial <derivatives> at an exact starting point $(y(t),t)$. Define
$$
A=f_t+f_yf=y''(t).
$$
The exact <Taylor expansion> is $y(t+h)=y+hf+\tfrac12h^2A+O(h^3)$. The <forward Euler method> retains only $y+hf$, so its one-step defect, exact minus computed value, is
$$
\boxed{\delta_E=\frac{h^2}2A+O(h^3)=O(h^2).}
$$
For the <Backward Euler method>, substitution of the exact solution into the scheme gives the residual
$$
y(t+h)-y(t)-h f(y(t+h),t+h)=-\frac{h^2}2A+O(h^3),
$$
because $f(y(t+h),t+h)=f+hA+O(h^2)$. 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:
$$
\boxed{\delta_{BE}=-\frac{h^2}2A+O(h^3)=O(h^2).}
$$
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.
\b[Euler stability.] For the <Dahlquist test equation> $y'=-\Lambda y$, put $z=h\Lambda$, with $\operatorname{Re}z>0$. The numerical <amplification factors> are
$$
\boxed{R_E(z)=1-z,\qquad R_{BE}(z)=\frac1{1+z}.}
$$
Strict decay requires $|R|<1$; the boundary $|R|=1$ gives bounded but generally nondecaying values. For <Forward Euler method>,
$$
|1-h\Lambda|<1\quad\Longleftrightarrow\quad0<h<\frac{2\operatorname{Re}\Lambda}{|\Lambda|^2}.
$$
For positive real $\Lambda$, this reduces to $0<h\Lambda<2$. The <absolute stability> region in the $z$ plane is the disk $|z-1|\leq1$. For <Backward Euler method>, $|1+z|>1$ whenever $\operatorname{Re}z>0$, so it decays for every step size. It is <A-stable> and also <L-stable>, since $R_{BE}(z)\to0$ for large $z$ in this half-plane. Here $z$ has the opposite sign from the alternative convention $z=h\lambda$ for $y'=\lambda y$.
\b[Predictor-corrector accuracy.] The printed two-stage construction is <Heun's method>, equivalently
$$
y_{n+1}=y_n+\frac h2\left[f(y_n,t_n)+f(y_n+hf(y_n,t_n),t_n+h)\right].
$$
Although 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, putting
$$
B=f_{tt}+2f f_{ty}+f^2f_{yy}.
$$
Then
$$
f(y+hf,t+h)=f+hA+\frac{h^2}2B+O(h^3),
$$
and the numerical step is $y+hf+\tfrac12h^2A+\tfrac14h^3B+O(h^4)$. Along the exact solution, $y'''=B+f_yA$, so comparison with the exact <Taylor expansion> gives
$$
\boxed{\delta_H=h^3\left(\frac{f_yA}6-\frac B{12}\right)+O(h^4)=O(h^3).}
$$
Thus 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 $-h^3\Lambda^3y/6+O(h^4)$.
\b[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 gives
$$
\boxed{R_H(z)=1-z+\frac{z^2}2,\qquad\text{stability region: }\left|1-z+\frac{z^2}2\right|\leq1.}
$$
For real positive $z$, $R_H(z)=\tfrac12[(z-1)^2+1]$, 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 is
$$
\left[1-x+\frac{x^2-y^2}2\right]^2+y^2(x-1)^2\leq1.
$$
The 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 $y\ne0$. Better accuracy has not inherited the unconditional stability of an actual implicit <Backward Euler method> step.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-1-time-stepping-stability.png]
{title=Absolute stability regions of the three time-stepping methods in the exam}
{description=Shading shows amplification factor of modulus at most one, using z equal to step size times Lambda for y prime equal to minus Lambda y. Positive real z therefore represents physical decay; Heun improves accuracy but retains a bounded stability region.}
Back to article page
