# 数学代写|数值分析代写numerical analysis代考|CIVL5458

## 数学代写|数值分析代写numerical analysis代考|Numerical examples

In this section, we will apply the trapezoid rule to specific examples. The first example is a linear dynamic and the second one is a nonlinear dynamic equation. We use MATLAB for the numerical computations and employ the Newton method to find the solution of the implicit relation arising in the second example.

Example 5.7. Let $\mathbb{T}=\mathbb{N}0$. Consider the IVP associated with the linear dynamic equation $$\left{\begin{array}{l} x^{\Lambda}(t)=\frac{1}{t+1} x(t)+\frac{1}{t^2+1}, \quad t>0, \ x(0)=x_0 \end{array}\right.$$ where $t_0=0, t_f=20$. The exact solution of this equation has the form $$\chi(t)=x_0 e{\frac{1}{t+1}}(t, 0)+\int_0^t e_{\frac{1}{t+1}}(t, \sigma(\tau)) \frac{1}{\tau^2+1} \Delta \tau, \quad t \geq 0$$
On $\mathbb{T}=\mathbb{N}0$ we have $\sigma(t)=t+1, \mu(t)=1$, and $$e{\frac{1}{t+1}}(t, s)=\prod_{p=s}^{t-1}\left(1+\frac{1}{p+1}\right)=\frac{s+2}{s+1} \frac{s+3}{s+2} \cdots \frac{t}{t-1} \frac{t+1}{t}=\frac{t+1}{s+1}, \quad t \geq s .$$
The integral $\int_0^t e_{\frac{1}{t+1}}(t, \sigma(\tau)) \frac{1}{\tau^2+1} \Delta \tau, t \geq 0$ is evaluated as
\begin{aligned} \int_0^t e_{\frac{1}{t+1}}(t, \sigma(\tau)) \frac{1}{\tau^2+1} \Delta \tau &=\int_0^t e_{\frac{1}{t+1}}(t, \tau+1) \frac{1}{\tau^2+1} \Delta \tau \ &=\int_0^t \frac{t+1}{\tau+2} \frac{1}{\tau^2+1} \Delta \tau \ &=(t+1) \sum_{p=0}^{t-1} \frac{1}{(p+2)\left(p^2+1\right)}, \quad t \geq 0 \end{aligned}

## 数学代写|数值分析代写numerical analysis代考|Analyzing the order p Taylor series method

Suppose that $p \in \mathbb{N}, p \geq 2, t_0, t_f \in \mathbb{T}, t_00$ be such that $t, t+r \in\left[t_0, t_f\right]$. Consider the initial value problem (IVP)
$$\left{\begin{array}{l} x^{\Delta}(t)=f(t, x(t)), \quad t \in\left[t_0, t_f\right], \ x\left(t_0\right)=x_0, \end{array}\right.$$
where $x_0 \in \mathbb{R}$ is a given constant and the function $f$ satisfies the following conditions:
(H1) $\left{\begin{array}{l}|f(t, x)| \leq A, \quad t \in \mathbb{T}, x \in \mathbb{R}, \ \left.\quad \text { there exist } g_k\left(t, x(t), \ldots, x^{\Delta^k}(t)\right)=(f(t, x(t)))\right)^{\Delta^k}, \quad k \in{1, \ldots, p-1}, \ \quad \text { such that }\left|\frac{\partial f}{\partial y}(t, z)\right| \leq A, \quad\left|\Delta_1 g_k\left(t, y_1, \ldots, y_{k+1}\right)\right| \leq A, \ \quad \text { and }\left|\frac{\partial}{\partial y_j} g_k\left(t, y_1, \ldots, y_{k+1}\right)\right| \leq A, \quad j \in{1, \ldots, k+1}, \ \quad \text { for any } t \in \mathbb{T} \text { and for } z, y_j \in \mathbb{R}, j \in{1, \ldots, p-1}, \ \quad \text { where } p e^{t_f-t_0} A<1 \text { and } A>0 .\end{array}\right.$
By the Taylor formula on time scales, we get
\begin{aligned} \chi(t+r)=& x(t)+h_1(t+r, t) \chi^{\Delta}(t)+h_2(t+r, t) x^{\Delta^2}(t)+\cdots+h_p(t+r, t) x^{\Delta^p}(t) \ &+\int_i^{\rho^p(t+r)} h_p(t+r, \sigma(u)) x^{\Delta^{p+1}}(u) \Delta u . \end{aligned}
Let
$$R_p(t)=\int_t^{\rho^p(t+r)} h_p(t+r, \sigma(u)) x^{\Delta^{p+1}}(u) \Delta u$$

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|Numerical examples

$$x^{\Lambda}(t)=\frac{1}{t+1} x(t)+\frac{1}{t^2+1}, \quad t>0, x(0)=x_0$$

where $\$ t_0=0, t_f=20 \$$. Theexactsolutionofthisequationhastheform \backslash chi (\mathrm{t})=x_{-} 0 e { frac {1}{t+1}}(\mathrm{t}, 0)+\backslash int_ 0^{\wedge} \mathrm{t} e__ { frac {1}{\mathrm{t}+1}}(\mathrm{t}, \backslash sigma \backslash tau \left.)\right) \backslash frac \left.{1} \backslash \operatorname{tau}^{\wedge} 2+1\right} \backslash Delta \backslash tau, \backslash quad \mathrm{t} \backslash geq 0 On \ \mathbb{T}=\mathbb{N} 0 \$$ wehave $\$ \sigma(t)=t+1, \mu(t)=1 \$$, and \backslash c dots \backslash frac {t}{t-1} \backslash frac {t+1}{t}=\backslash frac {t+1}{s+1}, \quad t \backslash geq s 。 Theintegral\ \int_0^t e_{\frac{1}{l+1}}(t, \sigma(\tau)) \frac{1}{\tau^2+1} \Delta \tau, t \geq 0 \isevaluatedas \int_0^t e_{\frac{1}{t+1}}(t, \sigma(\tau)) \frac{1}{\tau^2+1} \Delta \tau=\int_0^t e_{\frac{1}{t+1}}(t, \tau+1) \frac{1}{\tau^2+1} \Delta \tau \quad=\int_0^t \frac{t+1}{\tau+2} \frac{1}{\tau^2+1} \Delta \tau=(t+1) \sum_{p=0}^{t-1} \ \$$

## 数学代写|数值分析代写numerical analysis代考|Analyzing the order p Taylor series method

$\$ \$$\backslash left}$$
x^{\Delta}(t)=f(t, x(t)), \quad t \in\left[t_0, t_f\right], x\left(t_0\right)=x_0,
$$址确的。 \ \$$

$(\mathrm{H} 1) \$ \mathrm{left}{|f(t, x)| \leq A, \quad t \in \mathbb{T}, x \in \mathbb{R}, \quad$there exist$\left.g_k\left(t, x(t), \ldots, x^{\Delta^k}(t)\right)=(f(t, x(t)))\right)^{\Delta^k}, \quad k \in 1, \ldots\$,
【正确的。BytheTaylor formulaontimescales, weget
\begin{aligned} &\chi(t+r)=x(t)+h_1(t+r, t) \chi^{\Delta}(t)+h_2(t+r, t) x^{\Delta^2}(t)+\cdots+h_p(t+r, t) x^{\Delta^p}(t) \quad+\int_i^{p^p(t+r)} h_p \ &\text { Let }_p(t)=\int_t^{\rho^{\prime}(t+r)} h_p(t+r, \sigma(u)) x^{\Delta^{p+1}}(u) \Delta u \ \end{aligned}

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