Appendix: Useful formulas
$$
\begin{align}
u'(t_n) &\approx
\lbrack D_tu\rbrack^n = \frac{u^{n+\half} - u^{n-\half}}{\Delta t} \tag{393}\\
u'(t_n) &a\approx
\lbrack D_{2t}u\rbrack^n = \frac{u^{n+1} - u^{n-1}}{2\Delta t} +
\tag{394}\\
u'(t_n) &=
\lbrack D_t^-u\rbrack^n = \frac{u^{n} - u^{n-1}}{\Delta t}
\tag{395}\\
u'(t_n) &\approx
\lbrack D_t^+u\rbrack^n = \frac{u^{n+1} - u^{n}}{\Delta t}
\tag{396}\\
u'(t_{n+\theta}) &=
\lbrack \bar D_tu\rbrack^{n+\theta} = \frac{u^{n+1} - u^{n}}{\Delta t}
\tag{397}\\
u'(t_n) &\approx
\lbrack D_t^{2-}u\rbrack^n = \frac{3u^{n} - 4u^{n-1} + u^{n-2}}{2\Delta t}
\tag{398}\\
u''(t_n) &\approx
\lbrack D_tD_t u\rbrack^n = \frac{u^{n+1} - 2u^{n} + u^{n-1}}{\Delta t^2}
\tag{399}\\
u(t_{n+\half}) &\approx \lbrack \overline{u}^{t}\rbrack^{n+\half}
= \half(u^{n+1} + u^n)
\tag{400}\\
u(t_{n+\half})^2 &\approx \lbrack \overline{u^2}^{t,g}\rbrack^{n+\half}
= u^{n+1}u^n
\tag{401}\\
u(t_{n+\half}) &\approx \lbrack \overline{u}^{t,h}\rbrack^{n+\half}
= 2\frac{\frac{1}{u^{n+1}} + \frac{1}{u^n}}
\tag{402}\\
u(t_{n+\theta}) &\approx \lbrack \overline{u}^{t,\theta}\rbrack^{n+\theta}
= \theta u^{n+1} + (1-\theta)u^n ,\quad t_{n+\theta}=\theta t_{n+1} + (1-\theta)t_{n-1}
\tag{403}
\end{align}
$$
$$
\begin{align}
\uex'(t_n) &=
[D_t\uex]^n + R^n = \frac{\uex^{n+\half} - \uex^{n-\half}}{\Delta t} +R^n\nonumber,\\
R^n &= -\frac{1}{24}\uex'''(t_n)\Delta t^2 + {\cal O}(\Delta t^4)
\tag{404}\\
\uex'(t_n) &=
[D_{2t}\uex]^n +R^n = \frac{\uex^{n+1} - \uex^{n-1}}{2\Delta t} +
R^n\nonumber,\\
R^n &= -\frac{1}{6}\uex'''(t_n)\Delta t^2 + {\cal O}(\Delta t^4)
\tag{405}\\
\uex'(t_n) &=
[D_t^-\uex]^n +R^n = \frac{\uex^{n} - \uex^{n-1}}{\Delta t}
+R^n\nonumber,\\
R^n &= -\half\uex''(t_n)\Delta t + {\cal O}(\Delta t^2)
\tag{406}\\
\uex'(t_n) &=
[D_t^+\uex]^n +R^n = \frac{\uex^{n+1} - \uex^{n}}{\Delta t}
+R^n\nonumber,\\
R^n &= -\half\uex''(t_n)\Delta t + {\cal O}(\Delta t^2)
\tag{407}\\
\uex'(t_{n+\theta}) &=
[\bar D_t\uex]^{n+\theta} +R^{n+\theta} = \frac{\uex^{n+1} - \uex^{n}}{\Delta t}
+R^{n+\theta}\nonumber,\\
R^{n+\theta} &= -\half(1-2\theta)\uex''(t_{n+\theta})\Delta t +
\frac{1}{6}((1 - \theta)^3 - \theta^3)\uex'''(t_{n+\theta})\Delta t^2 +
\nonumber\\
&\quad {\cal O}(\Delta t^3)
\tag{408}\\
\uex'(t_n) &=
[D_t^{2-}\uex]^n +R^n = \frac{3\uex^{n} - 4\uex^{n-1} + \uex^{n-2}}{2\Delta t}
+R^n\nonumber,\\
R^n &= \frac{1}{3}\uex'''(t_n)\Delta t^2 + {\cal O}(\Delta t^3)
\tag{409}\\
\uex''(t_n) &=
[D_tD_t \uex]^n +R^n = \frac{\uex^{n+1} - 2\uex^{n} + \uex^{n-1}}{\Delta t^2}
+R^n\nonumber,\\
R^n &= -\frac{1}{12}\uex''''(t_n)\Delta t^2 + {\cal O}(\Delta t^4)
\tag{410}
\end{align}
$$
$$
\begin{align}
\uex(t_{n+\theta}) &= [\overline{\uex}^{t,\theta}]^{n+\theta} +R^{n+\theta}
= \theta \uex^{n+1} + (1-\theta)\uex^n +R^{n+\theta},\nonumber\\
R^{n+\theta} &= -\half\uex''(t_{n+\theta})\Delta t^2\theta (1-\theta) +
{\cal O}(\Delta t^3)
\tp
\tag{411}
\end{align}
$$
Complex exponentials
Let \( u^n = \exp{(i\omega n\Delta t)} = e^{i\omega t} \).
$$
\begin{align}
[D_tD_t u]^n &= u^n \frac{2}{\Delta t}(\cos \omega\Delta t - 1) =
-\frac{4}{\Delta t}\sin^2\left(\frac{\omega\Delta t}{2}\right),
\tag{412}\\
[D_t^+ u]^n &= u^n\frac{1}{\Delta t}(\exp{(i\omega\Delta t)} - 1),
\tag{413}\\
[D_t^- u]^n &= u^n\frac{1}{\Delta t}(1 - \exp{(-i\omega\Delta t)}),
\tag{414}\\
[D_t u]^n &= u^n\frac{2}{\Delta t}i\sin{\left(\frac{\omega\Delta t}{2}\right)},
\tag{415}\\
[D_{2t} u]^n &= u^n\frac{1}{\Delta t}i\sin{(\omega\Delta t)}
\tag{416}
\tp
\end{align}
$$
Real exponentials
Let \( u^n = \exp{(\omega n\Delta t)} = e^{\omega t} \).
$$
\begin{align}
[D_tD_t u]^n &= u^n \frac{2}{\Delta t}(\cos \omega\Delta t - 1) =
-\frac{4}{\Delta t}\sin^2\left(\frac{\omega\Delta t}{2}\right),
\tag{417}\\
[D_t^+ u]^n &= u^n\frac{1}{\Delta t}(\exp{(i\omega\Delta t)} - 1),
\tag{418}\\
[D_t^- u]^n &= u^n\frac{1}{\Delta t}(1 - \exp{(-i\omega\Delta t)}),
\tag{419}\\
[D_t u]^n &= u^n\frac{2}{\Delta t}i\sin{\left(\frac{\omega\Delta t}{2}\right)},
\tag{420}\\
[D_{2t} u]^n &= u^n\frac{1}{\Delta t}i\sin{(\omega\Delta t)}
\tag{421}
\tp
\end{align}
$$
The following results are useful when checking if a polynomial term in a
solution fulfills the discrete equation for the numerical method.
$$
\begin{align}
\lbrack D_t^+ t\rbrack^n = 1,
\tag{422}\\
\lbrack D_t^- t\rbrack^n = 1,
\tag{423}\\
\lbrack D_t t\rbrack^n = 1,
\tag{424}
\lbrack D_{2t} t\rbrack^n = 1,
\tag{425}
\lbrack D_{t}D_t t\rbrack^n = 0
\tag{426}
\tp
\end{align}
$$
The next formulas concern the action of difference operators on a \( t^2 \) term.
$$
\begin{align}
\lbrack D_t^+ t^2\rbrack^n = (2n+1)\Delta t,
\tag{427}\\
\lbrack D_t^- t^2\rbrack^n = (2n-1)\Delta t,
\tag{428}\\
\lbrack D_t t^2\rbrack^n = 2n\Delta t,
\tag{429}
\lbrack D_{2t} t^2\rbrack^n = 2n\Delta t,
\tag{430}
\lbrack D_{t}D_t t^2\rbrack^n = 2,
\tag{431}
\end{align}
$$
Finally, we present formulas for a \( t^3 \) term: These must be controlled
against lib.py. Use \( t_n \) instead of \( n\Delta t \)??
$$
\begin{align}
\lbrack D_t^+ t^3\rbrack^n &= 3(n\Delta t)^2 + 3n\Delta t^2 + \Delta t^2,
\tag{432}\\
\lbrack D_t^- t^3\rbrack^n &= 3(n\Delta t)^2 - 3n\Delta t^2 + \Delta t^2,
\tag{433}\\
\lbrack D_t t^3\rbrack^n &= 3(n\Delta t)^2 + \frac{1}{4}\Delta t^2,
\tag{434}\\
\lbrack D_{2t} t^3\rbrack^n &= 3(n\Delta t)^2 + \Delta t^2,
\tag{435}\\
\lbrack D_{t}D_t t^3\rbrack^n &= 6n\Delta t,
\tag{436}
\end{align}
$$