数学代写|Perturbations 数值分析代考
数值分析代写
Consider the special case, when $K_{3}$ is a small perturbation (order of $\varepsilon$ ) from $K_{1}$
$$
K_{3}=K_{1}+\varepsilon K_{2}
$$
Suppose $u_{1}$ is a solution to the conductivity problem with $K_{1}$
$$
\left{\begin{aligned}
\nabla \cdot K_{1} \nabla u_{1} &=0 & & \text { in } \Omega \
u_{1} &=f & & \text { on } \partial \Omega
\end{aligned}\right.
$$
Then there is a perturbation $v$, such that $u_{\varepsilon}:=u_{1}+\varepsilon v$ is an approximation solution to the problem with conductivity $K_{3}$
$$
\left{\begin{aligned}
\nabla \cdot K_{3} \nabla u_{\varepsilon} &=\mathcal{O}\left(\varepsilon^{2}\right) \quad \text { in } \Omega \
u_{\varepsilon} &=f \quad \text { on } \partial \Omega
\end{aligned}\right.
$$
DN map $\Lambda_{3}$ is approximately $\Lambda_{\varepsilon}$ which is a perturbation of $\Lambda_{1}$. For any Dirichlet data $f$ :
$$
\Lambda_{3}(f) \approx \Lambda_{\varepsilon}(f)=\mathbf{n} \cdot \nabla\left(u_{1}+\varepsilon v\right)=\Lambda_{1}(f)+\varepsilon(\mathbf{n} \cdot \nabla v)
$$
This can be shown by expanding the PDE in $\varepsilon$
$$
\begin{aligned}
\nabla \cdot K_{3} \nabla u_{\varepsilon}=& \nabla \cdot\left(K_{1}+\varepsilon K_{2}\right) \nabla\left(u_{1}+\varepsilon v\right) \
=& \nabla \cdot K_{1} \nabla\left(u_{1}+\varepsilon v\right)+\varepsilon \nabla \cdot K_{2} \nabla\left(u_{1}+\varepsilon v\right) \
=& \nabla \cdot K_{1} \nabla u_{1} \
& \quad+\varepsilon \nabla \cdot K_{1} \nabla v+\varepsilon \nabla \cdot K_{2} \nabla u_{1} \
& \quad+\varepsilon^{2} \nabla \cdot K_{1} \nabla v
\end{aligned}
$$
Thus the equation is satisfied up to $\mathcal{O}\left(\varepsilon^{2}\right)$ if $v$ is the solution to
$$
\left{\begin{aligned}
\nabla \cdot K_{1} \nabla v &=-\nabla \cdot K_{2} \nabla u_{1} \quad \text { in } \Omega, \
v &=0 \quad \text { on } \partial \Omega .
\end{aligned}\right.
$$
数值分析代考
考虑特殊情况，当 $K_{3}$ 是来自 $K_{1}$ 的小扰动（$\varepsilon$ 的阶）
$$
K_{3}=K_{1}+\varepsilon K_{2}
$$
假设 $u_{1}$ 是具有 $K_{1}$ 的电导率问题的解
$$
\left{\begin{对齐}
\nabla \cdot K_{1} \nabla u_{1} &=0 & & \text { in } \Omega \
u_{1} &=f & & \text { on } \partial \Omega
\end{对齐}\对。
$$
然后有一个扰动$v$，使得$u_{\varepsilon}:=u_{1}+\varepsilon v$ 是电导率$K_ {3}$ 问题的近似解
$$
\left{\begin{对齐}
\nabla \cdot K_{3} \nabla u_{\varepsilon} &=\mathcal{O}\left(\varepsilon^{2}\right) \quad \text { in } \Omega \
u_{\varepsilon} &=f \quad \text { on } \partial \Omega
\end{对齐}\对。
$$
DN 映射 $\Lambda_{3}$ 约为 $\Lambda_{\varepsilon}$，它是 $\Lambda_{1}$ 的扰动。对于任何 Dirichlet 数据 $f$ ：
$$
\Lambda_{3}(f) \近似 \Lambda_{\varepsilon}(f)=\mathbf{n} \cdot \nabla\left(u_{1}+\varepsilon v\right)=\Lambda_{1}( f)+\varepsilon(\mathbf{n} \cdot \nabla v)
$$
这可以通过扩展 $\varepsilon$ 中的 PDE 来显示
$$
\开始{对齐}
\nabla \cdot K_{3} \nabla u_{\varepsilon}=& \nabla \cdot\left(K_{1}+\varepsilon K_{2}\right) \nabla\left(u_{1}+\varepsilon v\右）\
=& \nabla \cdot K_{1} \nabla\left(u_{1}+\varepsilon v\right)+\varepsilon \nabla \cdot K_{2} \nabla\left(u_{1}+\varepsilon v \正确的） \
=& \nabla \cdot K_{1} \nabla u_{1} \
& \quad+\varepsilon \nabla \cdot K_{1} \nabla v+\varepsilon \nabla \cdot K_{2} \nabla u_{1} \
& \quad+\varepsilon^{2} \nabla \cdot K_{1} \nabla v
\end{对齐}
$$
因此，如果 $v$ 是
$$
\left{\begin{对齐}
\nabla \cdot K_{1} \nabla v &=-\nabla \cdot K_{2} \nabla u_{1} \quad \text { in } \Omega, \
v &=0 \quad \text { on } \partial \Omega 。
\end{对齐}\对。
$$
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