= Solution
Write $h=\Delta x$, $k=\Delta t$ and apply the recurrence to an exact smooth solution. The centred second difference is $h^2u_{xx}+h^4u_{xxxx}/12+O(h^6)$. Expanding the previous-time second difference and using $u_t=u_{xx}$ gives the one-step residual
$$
\begin{aligned}
&u(t+k)-u(t)-\tfrac32\mu\delta_h^2u(t)+\tfrac12\mu\delta_h^2u(t-k)\\
&=\frac{k^2}{2}(u_{tt}-u_{xxt})-\frac{kh^2}{12}u_{xxxx}
+O(k^3+k^2h^2+kh^4)\\
&=-\frac{k^2}{12\mu}u_{xxxx}+O(k^3),
\end{aligned}
$$
where $h^2=k/\mu$ with fixed positive $\mu$. Thus \b[the local one-step error is $O((\Delta t)^2)$] as stated. If <local truncation error> is defined after division by the step, it is instead $O(\Delta t)$ in this coupled refinement; the spatial error is only second order in $h$. This distinguishes the printed local-error convention from the second-order time accuracy of the underlying <Adams-Bashforth method> at fixed spatial discretization.
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