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数学代写|Perturbations 数值分析代考

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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