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# 数学代写|数值分析代写Numerical analysis代考|AMATH352 STABILITY OF NUMERICAL METHODS

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## 数学代写|数值分析代写Numerical analysis代考|STABILITY OF NUMERICAL METHODS

The word stability has a multitude of meanings when solving differential equations, both with respect to the differential equation under consideration and to the numerical method for its solution. For the stability of the initial value problem, review the definitions in the latter part of Section 9.1, in which stability referred to the effect on the solution $Y(x)$ of a perturbation in the initial value $Y_0$. The result (9.14)
Chapter 9 THE NUMERICAL SOLUTION OF DIFFERENTIAL EQUATIONS stated that if $Y_0$ is changed to $Y_0+\epsilon$, then the solution $Y(x)$ is changed to $Y_\epsilon(x)$ with
$$\max {x_0 \leq x \leq b}\left|Y(x)-Y\epsilon(x)\right| \rightarrow 0 \quad \text { as } \epsilon \rightarrow 0$$
One of the meanings of numerical stability is that we would like the same type of property to hold for the numerical method being used to solve the initial value problem.

To aid in understanding what it would mean to lack numerical stability, we introduce and consider a special two step method. To begin, if $Y(x)$ is three times continuously differentiable, then it can be shown that
\begin{aligned} Y\left(x_{n+1}\right)=3 Y\left(x_n\right)-2 Y\left(x_{n-1}\right) &+\frac{h}{2}\left[Y^{\prime}\left(x_n\right)-3 Y^{\prime}\left(x_{n-1}\right)\right] \ &+\frac{7}{12} h^3 Y^{\prime \prime \prime}\left(\xi_n\right) \end{aligned}
for some $x_{n-1} \leq \xi_n \leq x_{n+1}$; see problem 1. Assuming further that $Y(x)$ satisfies the differential equation
$$Y^{\prime}(x)=f(x, Y(x))$$
and then dropping the truncation error term in $(9.95)$, we obtain the numerical method
$$y_{n+1}=3 y_n-2 y_{n-1}+\frac{h}{2}\left[f\left(x_n, y_n\right)-3 f\left(x_{n-1}, y_{n-1}\right)\right], \quad n \geq 1 \quad \text { (9.96) }$$
This is a numerical method whose truncation error is $O\left(h^3\right)$, but in general its solution does not converge to $Y(x)$ as $h \rightarrow 0$.

## 数学代写|数值分析代写Numerical analysis代考|Stability Regions

The stability result in Theorem $9.3$ states that the method is well behaved, provided that the stepsize $h$ is sufficiently small. We would like to further examine this restriction, as it has important practical consequences for the relative desirability of various numerical methods.
Examining the stability question for the general problem
$$Y^{\prime}(x)=f(x, Y(x)), \quad Y\left(x_0\right)=Y_0$$
is too complicated. Instead, we examine the stability of numerical methods for the model equation
$$Y^{\prime}(x)=\lambda Y(x)+g(x), \quad Y(0)=Y_0$$
whose exact solution can be found from (9.5). Questions regarding stability and convergence are more easily answered for this equation; and the answers to these questions can be shown to usually be the answers to those same questions for the more general equation (9.105).

Let $Y(x)$ be the solution of $(9.106)$, and let $Y_c(x)$ be the solution with the perturbed initial data $Y_0+\epsilon$,
$$Y_\epsilon^{\prime}(x)=\lambda Y_\epsilon(x)+g(x), \quad Y_\epsilon(0)=Y_0+\epsilon$$
Let $Z_\epsilon(x)$ denote the change in the solution
$$Z_\epsilon(x)=Y_\epsilon(x)-Y(x)$$
Then, subtracting the equation for $Y_\epsilon(x)$ from (9.106),
$$Z_\epsilon^{\prime}(x)=\lambda Z_\epsilon(x), \quad Z_\epsilon(0)=\epsilon$$
The solution is
$$Z_\epsilon(x)=\epsilon e^{\lambda x}$$

## 数学代写|数值分析代写Numerical analysis代考|STABILITY OF NUMERICAL METHODS

$$\max x_0 \leq x \leq b|Y(x)-Y \epsilon(x)| \rightarrow 0 \quad \text { as } \epsilon \rightarrow 0$$

$$Y\left(x_{n+1}\right)=3 Y\left(x_n\right)-2 Y\left(x_{n-1}\right)+\frac{h}{2}\left[Y^{\prime}\left(x_n\right)-3 Y^{\prime}\left(x_{n-1}\right)\right] \quad+\frac{7}{12} h^3 Y^{\prime \prime \prime}\left(\xi_n\right)$$

$$Y^{\prime}(x)=f(x, Y(x))$$

$$y_{n+1}=3 y_n-2 y_{n-1}+\frac{h}{2}\left[f\left(x_n, y_n\right)-3 f\left(x_{n-1}, y_{n-1}\right)\right], \quad n \geq 1 \quad(9.96)$$

## 数学代写|数值分析代写Numerical analysis代考|Stability Regions

$$Y^{\prime}(x)=f(x, Y(x)), \quad Y\left(x_0\right)=Y_0$$

$$Y^{\prime}(x)=\lambda Y(x)+g(x), \quad Y(0)=Y_0$$

$$Y_\epsilon^{\prime}(x)=\lambda Y_\epsilon(x)+g(x), \quad Y_\epsilon(0)=Y_0+\epsilon$$

$$Z_\epsilon(x)=Y_\epsilon(x)-Y(x)$$

$$Z_\epsilon^{\prime}(x)=\lambda Z_\epsilon(x), \quad Z_\epsilon(0)=\epsilon$$

$$Z_\epsilon(x)=\epsilon e^{\lambda x}$$

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