Posted on Categories:Ordinary Differential Equations, 常微分方程, 数学代写

# 数学代写|常微分方程代考Ordinary Differential Equations代写|Tikhonov regularisation

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## 数学代写|常微分方程代考Ordinary Differential Equations代写|Tikhonov regularisation

We have seen that an inverse problem may have or not a unique solution depending on the location where the state of the system is measured. We have also recognised that the inverse of $K$ becomes continuous if we restrict the function space where the solution is sought. In particular, this fact is also stated in the following theorem due to Andrey Nikolayevich Tikhonov, who has made many pioneering contribution in the field of inverse problems. We have the following Theorem.

Theorem 11.1 Let $X, Y$ normed vector spaces. If $K: D(K) \rightarrow Y, D(K) \subseteq X$ is a continuous one-to-one operator and $C \subseteq D(K)$ is a compact set, then the inverse of the restriction of the operator $K$ to $C,\left(\left.K\right|_C\right)^{-1}$ is continuous. $(D(K)$ denotes the domain of definition of $K$.)

Thus, in the case of the source problem for (11.1), we may restrict the function space where the source $q$ is sought to the set $|q|_1 \leq c, c>0$, which is compact in $C(I)$; see the Arzelá-Ascoli theorem. Therefore with this setting the inverse source problem is well defined.

Now, for a more general discussion, let $K: H_1 \rightarrow H_2$ be a bounded linear operator between the real Hilbert spaces $H_1$ and $\mathrm{H}_2$, and consider the following problem of determining $q$ for a given $y$ as follows:
$$K q=y$$
A solution to this problem exists if $y \in R(K)$ (the range of $K$ ); however, $R(K)$ may be only a subset of $\mathrm{H}_2$.

## 数学代写|常微分方程代考Ordinary Differential Equations代写|Inverse problems with nonlinear models

The formulation (11.27) can be taken as the starting point for solving inverse problems where $K$ is a nonlinear operator, $K: D(K) \subseteq H_1 \rightarrow H_2$. The corresponding so-called “penalised least-squares” problem is stated as follows:
$$\min J_\alpha(q)=\left|K(q)-y^\delta\right|_{H_2}^2+\alpha|q|_{H_1}^2$$
To solve this problem in the case where $K$ is Fréchet differentiable, we can consider an initial approximation for $q$, say $q_0$, and correspondingly use the linearisation
$$K\left(q_0+\delta q\right)=K\left(q_0\right)+\partial K\left(q_0\right) \delta q+r\left(q_0, \delta q\right)$$

where $\partial K\left(q_0\right)$ represents the Fréchet derivative of $K$ at $q_0$, and $\left|r\left(q_0, \delta q\right)\right|=$ $o(|\delta q|)$. Now, let $y_0=K\left(q_0\right)$, and require that $K\left(q_0+\delta q\right)=y^\delta$, then we have the following linear equation:
$$\partial K\left(q_0\right) \delta q=y^\delta-y_0$$
However, this equation may be ill-posed, and hence Tikhonov regularisation could be used to solve it. Thus, we consider the following normal equations:
$$\left(\partial K\left(q_0\right)^* \partial K\left(q_0\right)+\alpha I\right) \delta q_\alpha=\partial K\left(q_0\right)^*\left(y^\delta-y_0\right)$$
The solution $\delta q_\alpha$ to this problem provides a new approximation for $q$, i.e., $q_1=q_0+\delta q_\alpha$. With this new approximation the procedure is repeated assembling (11.37) at $q_1$ with $y_1=K\left(q_1\right)$, and so on iteratively until a given convergence criterion is met. This is the Levenberg-Marquardt method, and the linearisation strategy is called “output least-squares.”

# 常微分方程代写

## 数学代写|常微分方程代考Ordinary Differential Equations代写|Tikhonov regularisation

$$K q=y$$

## 数学代写|常微分方程代考Ordinary Differential Equations代写|Inverse problems with nonlinear models

$$\min J_\alpha(q)=\left|K(q)-y^\delta\right|{H_2}^2+\alpha|q|{H_1}^2$$

$$K\left(q_0+\delta q\right)=K\left(q_0\right)+\partial K\left(q_0\right) \delta q+r\left(q_0, \delta q\right)$$

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## MATLAB代写

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