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# 数学代写|数值分析代写Numerical analysis代考|MATH345 Linear multistep methods: zero stability

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## 数学代写数值分析代写Numerical analysis代考|Linear multistep methods: zero stability

Does consistency imply convergence for linear multistep methods?
This is always the case for one-step methods, as proved in section 5.3, but the example at the end of the last lecture suggests the issue is less straightforward for multistep methods. By understanding the subtleties, we will come to appreciate one of the most significant themes in numerical analysis: stability of discretizations.

We are interested in the behavior of linear multistep methods as $h \rightarrow 0$. In this limit, the right hand side of the formula for the generic multistep method,
$$\sum_{j=0}^m \alpha_j x_{k+j}=h \sum_{j=0}^m \beta_j f\left(t_{k+j}, x_{k+j}\right),$$
makes a negligible contribution. This motivates our consideration of the trivial model problem $x^{\prime}(t)=0$ with $x(0)=0$. Does the linear multistep method recover the exact solution, $x(t)=0$ ?

When $x^{\prime}(t)=0$, clearly we have $f_{k+j}=0$ for all $j$. The condition $\alpha_m \neq 0$ allows us to write
$$x_m=-\frac{\left(\alpha_0 x_0+\alpha_1 x_1+\cdots+\alpha_{m-1} x_{m-1}\right)}{\alpha_m}$$
Hence if the method is started with exact data
$$x_0=x_1=\cdots=x_{m-1}=0,$$
then
$$x_m=-\frac{\left(\alpha_0 \cdot 0+\alpha_1 \cdot 0+\cdots+\alpha_{m-1} \cdot 0\right)}{\alpha_m}=0,$$
and this pattern will continue: $x_{m+1}=0, x_{m+2}=0, \ldots$ Any linear multistep method with exact starting data produces the exact solution for this special problem, regardless of the time-step.

## 数学代写|数值分析代写Numerical analysis代考|The Root Condition

The real trouble with the previous method was that the formula for $x_k$ involves the term $(-2)^k$. Since $|-2|>1$, this component of $x_k$ grows exponentially in $k$. This term is simply an artifact of the finite difference equation, and has nothing to do with the underlying differential equation. As $k$ increases, this $(-2)^k$ term swamps the other term in the solution. It is called a parasitic solution.
Let us review how we determined the general form of the solution. We assumed a solution of the form $x_k=\gamma^k$, then plugged this solution into the recurrence $x_{k+2}=2 x_k-x_{k+1}$. The possible values for $\gamma$ were roots of the equation $\gamma^2=2-\gamma$.

The process we just applied to one specific linear multistep method can readily be extended to the general linear multistep method
$(5 \cdot 6)$
$$\sum_{j=0}^m \alpha_j x_{k+j}=h \sum_{j=0}^m \beta_j f\left(t_{k+j}, x_{k+j}\right) .$$
by applying the method again to the trivial equation $x^{\prime}(t)=0$. For this special equation, the method (5.6) reduces to
$$\sum_{j=0}^m \alpha_j x_{k+j}=0 .$$

## 数学代写数值分析代写Numerical analysis代考|Linear multistep methods: zero stability

$$\sum_{j=0}^m \alpha_j x_{k+j}=h \sum_{j=0}^m \beta_j f\left(t_{k+j}, x_{k+j}\right),$$

$$x_m=-\frac{\left(\alpha_0 x_0+\alpha_1 x_1+\cdots+\alpha_{m-1} x_{m-1}\right)}{\alpha_m}$$

$$x_0=x_1=\cdots=x_{m-1}=0,$$

$$x_m=-\frac{\left(\alpha_0 \cdot 0+\alpha_1 \cdot 0+\cdots+\alpha_{m-1} \cdot 0\right)}{\alpha_m}=0,$$

## 数学代写|数值分析代写Numerical analysis代考|The Root Condition

$$\sum_{j=0}^m \alpha_j x_{k+j}=h \sum_{j=0}^m \beta_j f\left(t_{k+j}, x_{k+j}\right) .$$

$$\sum_{j=0}^m \alpha_j x_{k+j}=0$$

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