$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\halfi}{{1/2}}
\newcommand{\xpoint}{\boldsymbol{x}}
\newcommand{\normalvec}{\boldsymbol{n}}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\It}{\mathcal{I}_t}
\newcommand{\setb}[1]{#1^0} % set begin
\newcommand{\sete}[1]{#1^{-1}} % set end
\newcommand{\setl}[1]{#1^-}
\newcommand{\setr}[1]{#1^+}
\newcommand{\seti}[1]{#1^i}
\newcommand{\Real}{\mathbb{R}}
$$
Verification: quadratic solution (2)
- \( [D_t D_t 1]^n=0 \)
- \( [D_t D_t t]^n=0 \)
- \( [D_t D_t t^2]=2 \)
- \( D_tD_t \) is a linear operator: \( [D_tD_t (au+bv)]^n = a[D_tD_t u]^n +
b[D_tD_t v]^n \)
$$
\begin{align*}
[D_xD_x \uex]^n_{i,j} &= [y(L_y-y)(1+{\half}t) D_xD_x x(L_x-x)]^n_{i,j}\\
&= y_j(L_y-y_j)(1+{\half}t_n)2
\end{align*}
$$
- Similar calculations for \( [D_yD_y\uex]^n_{i,j} \) and
\( [D_tD_t \uex]^n_{i,j} \) terms
- Must also check the equation for \( u^1_{i,j} \)
