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

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 $\boxed{\text{method 3 is time symmetric}}$.