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# 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

$$K_{3}=K_{1}+\varepsilon K_{2}$$

$$\left{\begin{对齐} \nabla \cdot K_{1} \nabla u_{1} &=0 & & \text { in } \Omega \ u_{1} &=f & & \text { on } \partial \Omega \end{对齐}\对。$$

$$\left{\begin{对齐} \nabla \cdot K_{3} \nabla u_{\varepsilon} &=\mathcal{O}\left(\varepsilon^{2}\right) \quad \text { in } \Omega \ u_{\varepsilon} &=f \quad \text { on } \partial \Omega \end{对齐}\对。$$

DN 映射 $\Lambda_{3}$ 约为 $\Lambda_{\varepsilon}$，它是 $\Lambda_{1}$ 的扰动。对于任何 Dirichlet 数据 $f$ ：

$$\Lambda_{3}(f) \近似 \Lambda_{\varepsilon}(f)=\mathbf{n} \cdot \nabla\left(u_{1}+\varepsilon v\right)=\Lambda_{1}( f)+\varepsilon(\mathbf{n} \cdot \nabla v)$$

$$\开始{对齐} \nabla \cdot K_{3} \nabla u_{\varepsilon}=& \nabla \cdot\left(K_{1}+\varepsilon K_{2}\right) \nabla\left(u_{1}+\varepsilon v\右）\ =& \nabla \cdot K_{1} \nabla\left(u_{1}+\varepsilon v\right)+\varepsilon \nabla \cdot K_{2} \nabla\left(u_{1}+\varepsilon v \正确的） \ =& \nabla \cdot K_{1} \nabla u_{1} \ & \quad+\varepsilon \nabla \cdot K_{1} \nabla v+\varepsilon \nabla \cdot K_{2} \nabla u_{1} \ & \quad+\varepsilon^{2} \nabla \cdot K_{1} \nabla v \end{对齐}$$

$$\left{\begin{对齐} \nabla \cdot K_{1} \nabla v &=-\nabla \cdot K_{2} \nabla u_{1} \quad \text { in } \Omega, \ v &=0 \quad \text { on } \partial \Omega 。 \end{对齐}\对。$$

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