Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 5

Let ${\Phi }_h$ be the one-step map of a numerical method. It is a time-symmetric numerical method when reversing the step exactly reverses the update:

Let ${\phi }_h$ be the exact flow map and suppose the method has order $p$ with a nonzero leading local truncation error:Inverting this expansion changes the sign of its leading perturbation, sowhere transport by the exact flow only changes $d$ by $O(h)$ and hence does not affect the leading parity. On the other hand, replacing $h$ by $-h$ in the first expansion givesTime symmetry equates these expressions, so $(-1{)}^{p+1}=-1$. Hence $p+1$ is odd and

The method has step map ${\Phi }_h(y)=y+hf(y)+h^{2}g(y)/2$. It is not time symmetric. It suffices to test the scalar linear equation $y^{\prime }=\lambda y$, for which the stability function isTime symmetry would require $R(-z)R(z)=1$, butfor generic $z$. Thus $\boxed{\text{method 1 is not time symmetric}}$.

The trapezoidal rule isInterchanging $y_n$ and $y_{n+1}$ while replacing $h$ by $-h$ leaves the equation unchanged. Its backward step is therefore exactly its inverse, and $\boxed{\text{method 2 is time symmetric}}$.

The Butcher tableau has one stage with $a_{11}=1/2$ and $b_1=1$, so it is the implicit midpoint ruleAgain, interchanging the endpoints and changing $h$ to $-h$ leaves the equation invariant. Hence $\boxed{\text{method 3 is time symmetric}}$.