$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\uexd}[1]{{u_{\small\mbox{e}, #1}}}
\newcommand{\vex}{{v_{\small\mbox{e}}}}
\newcommand{\vexd}[1]{{v_{\small\mbox{e}, #1}}}
\newcommand{\Aex}{{A_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\halfi}{{1/2}}
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\newcommand{\Ddt}[1]{\frac{D #1}{dt}}
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\newcommand{\xpoint}{\boldsymbol{x}}
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\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\X}{\boldsymbol{X}}
\renewcommand{\u}{\boldsymbol{u}}
\renewcommand{\v}{\boldsymbol{v}}
\newcommand{\w}{\boldsymbol{w}}
\newcommand{\V}{\boldsymbol{V}}
\newcommand{\e}{\boldsymbol{e}}
\newcommand{\f}{\boldsymbol{f}}
\newcommand{\F}{\boldsymbol{F}}
\newcommand{\stress}{\boldsymbol{\sigma}}
\newcommand{\strain}{\boldsymbol{\varepsilon}}
\newcommand{\stressc}{{\sigma}}
\newcommand{\strainc}{{\varepsilon}}
\newcommand{\I}{\boldsymbol{I}}
\newcommand{\T}{\boldsymbol{T}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
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\newcommand{\iz}{\boldsymbol{i}_z}
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\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\Iz}{\mathcal{I}_z}
\newcommand{\It}{\mathcal{I}_t}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\Ifb}{{I_b}} % for FEM
\newcommand{\setb}[1]{#1^0} % set begin
\newcommand{\sete}[1]{#1^{-1}} % set end
\newcommand{\setl}[1]{#1^-}
\newcommand{\setr}[1]{#1^+}
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\newcommand{\sinL}[1]{\sin\left((#1+1)\pi\frac{x}{L}\right)}
\newcommand{\xno}[1]{x_{#1}}
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$$
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 \( u^1_{i,j} \)