数学代写|数值分析代写numerical analysis代考|MATHS7104

数学代写|数值分析代写numerical analysis代考|Local truncation error

In this section, we derive the local truncation error in the application of the Euler method. To obtain the local truncation error, we consider the cases when $t_0$ is rightdense and right-scattered separately.

1. Case 1. $t_0$ is right-scattered.
In a single step of the Euler method, the computed result is
$$x_1=x_0+\left(\sigma^{l_0}\left(t_0\right)-t_0\right) f\left(t_0, x\left(t_0\right)\right)$$
and it differs from the exact answer $x\left(t_1\right)=x\left(\sigma^{l_0}\left(t_0\right)\right)$ by
\begin{aligned} x\left(t_1\right)-x_1 &=x\left(\sigma^{l_0}\left(t_0\right)\right)-x\left(t_0\right)-\left(\sigma^{l_0}\left(t_0\right)-t_0\right) f\left(t_0, x\left(t_0\right)\right) \ &=x\left(\sigma^{l_0}\left(t_0\right)\right)-x\left(t_0\right)-\left(\sigma^{l_0}\left(t_0\right)-t_0\right) x^{\Lambda}\left(t_0\right) . \end{aligned}
If $l_0=1$, then $x\left(t_1\right)=x_1$. Assuming that $x$ has continuous first and second order delta derivatives, this can be written, using Taylor formula, in the form
$$\int_{t_0}^{\rho\left(\sigma^{h_n}\left(t_0\right)\right)} h_1\left(\sigma^{l_0}\left(t_0\right), \sigma(\tau)\right) x^{\Lambda^2}(\tau) \Delta \tau .$$
Another way of writing the error, assuming that the third derivative $x^{\Delta^3}$ also exists and is bounded, is
$$h_2\left(\sigma^{l_0}\left(t_0\right), t_0\right) x^{\Delta^2}\left(t_0\right)+O\left(h_3\left(\sigma^{l_0}\left(t_0\right), t_0\right)\right) .$$
2. Case 2. $t_0$ is right-dense.
In a single step of the Euler method, the computed result
$$x_1=x_0+q_0 f\left(t_0, x\left(t_0\right)\right)$$
differs from the exact solution $x\left(t_1\right)$ by
\begin{aligned} x\left(t_1\right)-x_1 &=x\left(t_0+q_0\right)-x\left(t_0\right)-q_0 f\left(t_0, x\left(t_0\right)\right) \ &=x\left(t_0+q_0\right)-x\left(t_0\right)-q_0 x^{\Delta}\left(t_0\right) . \end{aligned}

数学代写|数值分析代写numerical analysis代考|Global truncation error

We continue with the derivation of the global truncation error of the Euler method. Let $\tilde{x}(t)$ denote the computed solution on the interval $\left[t_0, \bar{t}\right]{\mathbb{T}}$. That is, at step values $t_0$, $t_1, \ldots, t_N=\bar{t}$ defined by (4.2), $\tilde{x}$ is computed, using equation (4.3). For “offstep” points, $\tilde{x}(t)$ is defined by linear interpolation, or, equivalently, $\tilde{x}(t)$ is evaluated using a partial step from the most recently computed step values. That is, if $t \in\left[t{k-1}, t_k\right], k=1, \ldots, N$, then
$$\tilde{x}(t)=x_{k-1}+h_1\left(t, t_{k-1}\right) f\left(t_{k-1}, x_{k-1}\right) .$$
Define the maximum step size as
$$m=\max {1 \leq i \leq N}\left{t_i-t{i-1}\right} .$$
Also, let
\begin{aligned} &\alpha(t)=x(t)-\tilde{x}(t), \ &\beta(t)=f(t, x(t))-f(t, \tilde{x}(t)) . \end{aligned}
Suppose that
$$|f(t, x)-f(t, z)| \leq L|x-z| \quad \text { for all } t \in \mathbb{T} \text { and } x, z \in \mathbb{R} \text {, }$$
where $L>0$. From (4.5) and (4.6), we have
$$|\beta(t)| \leq L|\alpha(t)|, \quad t \in \mathbb{T} .$$
Define $E(t), t \in \mathbb{T}$, so that the exact solution satisfies
$$x(t)=x\left(t_{k-1}\right)+h_1\left(t, t_{k-1}\right) f\left(t_{k-1}, x\left(t_{k-1}\right)\right)+h_2\left(t, t_{k-1}\right) E(t), \quad t \in\left[t_{k-1}, t_k\right], t \in \mathbb{T},$$
and assume that $|E(t)| \leq p, t \in \mathbb{T}$. Subtracting (4.4) from (4.7), we get
\begin{aligned} x(t)-\tilde{x}(t)=& x\left(t_{k-1}\right)-x_{k-1}+h_1\left(t, t_{k-1}\right)\left(f\left(t_{k-1}, x\left(t_{k-1}\right)\right)-f\left(t_{k-1}, x_{k-1}\right)\right) \ &+h_2\left(t, t_{k-1}\right) E(t), \quad t \in \mathbb{T} . \end{aligned}

数值分析代考

数学代写|数值分析代写numerical analysis代考|Local truncation error

$t_0$ 分别是右密和右散。

1. 情况1。 $t_0$ 是右散布的。
在欧拉方法的一个步聚中, 计算结果为
$$x_1=x_0+\left(\sigma^{l_0}\left(t_0\right)-t_0\right) f\left(t_0, x\left(t_0\right)\right)$$
它与确切的答案不同 $x\left(t_1\right)=x\left(\sigma^{l_0}\left(t_0\right)\right)$ 经过
$$x\left(t_1\right)-x_1=x\left(\sigma^{l_0}\left(t_0\right)\right)-x\left(t_0\right)-\left(\sigma^{l_0}\left(t_0\right)-t_0\right) f\left(t_0, x\left(t_0\right)\right) \quad=x\left(\sigma^{l_0}\left(t_0\right)\right)-x\left(t_0\right)-\left(\sigma^{l_0}\left(t_0\right)\right.$$
如果 $l_0=1$ ，然后 $x\left(t_1\right)=x_1$. 假如说 $x$ 具有连续的一阶和二阶 delta 导数，这可以用泰勒公式写
成，形式为
$$\int_{t_0}^{\rho\left(\sigma^{h_n}\left(t_0\right)\right)} h_1\left(\sigma^{l_0}\left(t_0\right), \sigma(\tau)\right) x^{\Lambda^2}(\tau) \Delta \tau$$
写错误的另一种方式，假设三阶导数 $x^{\Delta}$ 也存在且有界，是
$$h_2\left(\sigma^{l_0}\left(t_0\right), t_0\right) x^{\Delta^2}\left(t_0\right)+O\left(h_3\left(\sigma^{l_0}\left(t_0\right), t_0\right)\right) .$$
2. 穼例 $2 。 t_0$ 是右密。
在欧拉方法的一个步㵵中，计算结果
$$x_1=x_0+q_0 f\left(t_0, x\left(t_0\right)\right)$$
与精确解不同 $x\left(t_1\right)$ 经过
$$x\left(t_1\right)-x_1=x\left(t_0+q_0\right)-x\left(t_0\right)-q_0 f\left(t_0, x\left(t_0\right)\right) \quad=x\left(t_0+q_0\right)-x\left(t_0\right)-q_0 x^{\Delta}\left(t_0\right) .$$

数学代写|数值分析代写numerical analysis代考|Global truncation error

$$\tilde{x}(t)=x_{k-1}+h_1\left(t, t_{k-1}\right) f\left(t_{k-1}, x_{k-1}\right) .$$

$$\alpha(t)=x(t)-\tilde{x}(t), \quad \beta(t)=f(t, x(t))-f(t, \tilde{x}(t))$$

$$|f(t, x)-f(t, z)| \leq L|x-z| \quad \text { for all } t \in \mathbb{T} \text { and } x, z \in \mathbb{R}$$

$$|\beta(t)| \leq L|\alpha(t)|, \quad t \in \mathbb{T} .$$

$$x(t)=x\left(t_{k-1}\right)+h_1\left(t, t_{k-1}\right) f\left(t_{k-1}, x\left(t_{k-1}\right)\right)+h_2\left(t, t_{k-1}\right) E(t), \quad t \in\left[t_{k-1}, t_k\right], t \in \mathbb{T},$$

$$x(t)-\tilde{x}(t)=x\left(t_{k-1}\right)-x_{k-1}+h_1\left(t, t_{k-1}\right)\left(f\left(t_{k-1}, x\left(t_{k-1}\right)\right)-f\left(t_{k-1}, x_{k-1}\right)\right) \quad+h_2\left(t, t_{k-1}\right) E(t)$$

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