 Solution

With the supplied angles, $c_{13}^4{\sin }^{2}2{\theta }_{12}=0.70545$ and ${\sin }^{2}2{\theta }_{13}=0.11698$. At $E=3\mathrm{MeV}$ and with the two atmospheric-scale splittings approximated by a common $3\times 10^{-3}{\mathrm{eV}}^2$, the requested sketch is represented byThe fast period is $2.480\mathrm{km}$ and its first disappearance maximum is at $1.240\mathrm{km}$. The slow period is $92.99\mathrm{km}$, giving slow minima near $46.49$ and $139.48\mathrm{km}$. At each baseline the fast oscillations lie between $1-0.70545{\sin }^2(0.0337849L)$ and that value minus $0.11698$. Averaging over unresolved fast oscillations subtracts half that amplitude, rather than erasing the slow oscillation.

The equal atmospheric splittings are an approximation: exact squared masses require $\Delta m_{13}^2=\Delta m_{12}^2+\Delta m_{23}^2$. For example, choosing magnitudes $\Delta m_{31}^2=3.00\times 10^{-3}$ and $\Delta m_{21}^2=0.08\times 10^{-3}{\mathrm{eV}}^2$ gives $\Delta m_{32}^2=2.92\times 10^{-3}{\mathrm{eV}}^2$. This changes the fine fast pattern while retaining the two physical scales. The figure deliberately displays the source's common-frequency approximation and labels the fast-averaged curve separately from the monoenergetic result.