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## 数学代写|常微分方程代写ordinary differential equation代考|Fixed point theorems

Let $X$ be a real vector space. A norm on $X$ is a map $|\cdot|: X \rightarrow[0, \infty)$ satisfying the following requirements:
(i) $|0|=0,|x|>0$ for $x \in X \backslash{0}$.
(ii) $|\lambda x|=|\lambda||x|$ for $\lambda \in \mathbb{R}$ and $x \in X$.
(iii) $|x+y| \leq|x|+|y|$ for $x, y \in X$ (triangle inequality).
The pair $(X,||$.$) is called a normed vector space. Given a normed$ vector space $X$, we have the concept of convergence and of a Cauchy sequence in this space. The normed vector space is called complete if every Cauchy sequence converges. A complete normed vector space is called a Banach space.

Clearly $\mathbb{R}^n\left(\right.$ or $\mathbb{C}^n$ ) is a Banach space with the usual Euclidean norm
$$|x|=\sqrt{\sum_{j=1}^n\left|x_j\right|^2}$$
We will be mainly interested in the following example: Let $I$ be a compact interval and consider the continuous functions $C(I)$ on this interval. They form a vector space if all operations are defined pointwise. Moreover, $C(I)$ becomes a normed space if we define
$$|x|=\sup {t \in I}|x(t)| .$$ I leave it as an exercise to check the three requirements from above. Now what about convergence in this space? A sequence of functions $x_n(t)$ converges to $x$ if and only if $$\lim {n \rightarrow \infty}\left|x_n-x\right|=\lim {n \rightarrow \infty} \sup {t \in I}\left|x_n(t)-x(t)\right|=0 .$$
That is, in the language of real analysis, $x_n$ converges uniformly to $x$. Now let us look at the case where $x_n$ is only a Cauchy sequence. Then $x_n(t)$ is clearly a Cauchy sequence of real numbers for any fixed $t \in I$. In particular, by completeness of $\mathbb{R}$, there is a limit $x(t)$ for each $t$. Thus we get a limiting function $x(t)$. Moreover, letting $m \rightarrow \infty$ in
$$\left|x_n(t)-x_m(t)\right| \leq \varepsilon \quad \forall n, m>N_{\varepsilon}, t \in I$$
we see
$$\left|x_n(t)-x(t)\right| \leq \varepsilon \quad \forall n>N_{\varepsilon}, t \in I,$$
that is, $x_n(t)$ converges uniformly to $x(t)$.

## 数学代写|常微分方程代写ordinary differential equation代考|The basic existence and uniqueness result

Now we want to use the preparations of the previous section to show existence and uniqueness of solutions for the following initial value problem (IVP)
$$\dot{x}=f(t, x), \quad x\left(t_0\right)=x_0 .$$
We suppose $f \in C\left(U, \mathbb{R}^n\right)$, where $U$ is an open subset of $\mathbb{R}^{n+1}$ and $\left(t_0, x_0\right) \in$ $U$

First of all note that integrating both sides with respect to $t$ shows that (2.11) is equivalent to the following integral equation
$$x(t)=x_0+\int_{t_0}^t f(s, x(s)) d s .$$

At first sight this does not seem to help much. However, note that $x_0(t)=x_0$ is an approximating solution at least for small $t$. Plugging $x_0(t)$ into our integral equation we get another approximating solution
$$x_1(t)=x_0+\int_{t_0}^t f\left(s, x_0(s)\right) d s .$$
Iterating this procedure we get a sequence of approximating solutions
$$x_n(t)=K^n\left(x_0\right)(t), \quad K(x)(t)=x_0+\int_{t_0}^t f(s, x(s)) d s .$$
Now this observation begs us to apply the contraction principle from the previous section to the fixed point equation $x=K(x)$, which is precisely our integral equation $(2.12)$.

We will set $t_0=0$ for notational simplicity and consider only the case $t \geq$ 0 to avoid excessive numbers of absolute values in the following estimates.
First of all we will need a Banach space. The obvious choice is $X=$ $C\left(I, \mathbb{R}^n\right)$, where $I=[0, T]$ is some suitable interval containing $t_0=0$. Furthermore, we need a closed subset $C \subseteq X$ such that $K: C \rightarrow C$. We will try a closed ball of radius $\delta$ around $x_0$, where $\delta>0$ has to be determined.
Choose $V=[0, T] \times \overline{B_\delta\left(x_0\right)} \subset U$, where $B_\delta\left(x_0\right)=\left{x \in \mathbb{R}^n|| x-x_0 \mid<\delta\right}$. Then
$$\left|K(x)(t)-x_0\right| \leq \int_0^t|f(s, x(s))| d s \leq t \max {(t, x) \in V}|f(t, x)|$$ (here the maximum exists by continuity of $f$ and compactness of $V$ ) whenever the graph of $x$ lies within $V$, that is, ${(t, x(t)) \mid t \in[0, T]} \subset V$. Hence, for $t \leq T_0$, where $$T_0=\min \left(T, \frac{\delta}{M}\right), \quad M=\max {(t, x) \in V}|f(t, x)|,$$
we have $T_0 M \leq \delta$ and the graph of $K(x)$ is again in $V$.

# 常微分方程代写

## 数学代写|常微分方程代写ordinary differential equation代考|Fixed point theorems

(i) $|0|=0,|x|>0$ 为了 $x \in X \backslash 0$.
(二) $|\lambda x|=|\lambda||x|$ 为了 $\lambda \in \mathbb{R}$ 和 $x \in X$. (三) $|x+y| \leq|x|+|y|$ 为了 $x, y \in X$ (三角不等式)。

$$|x|=\sqrt{\sum_{j=1}^n\left|x_j\right|^2}$$

$$|x|=\sup t \in I|x(t)| .$$

$$\lim n \rightarrow \infty\left|x_n-x\right|=\lim n \rightarrow \infty \sup t \in I\left|x_n(t)-x(t)\right|=0 .$$

$$\left|x_n(t)-x_m(t)\right| \leq \varepsilon \quad \forall n, m>N_{\varepsilon}, t \in I$$

$$\left|x_n(t)-x(t)\right| \leq \varepsilon \quad \forall n>N_{\varepsilon}, t \in I,$$

## 数学代写|常微分方程代写ordinary differential equation代考|The basic existence and uniqueness result

$$\dot{x}=f(t, x), \quad x\left(t_0\right)=x_0 .$$

$$x(t)=x_0+\int_{t_0}^t f(s, x(s)) d s .$$

$$x_1(t)=x_0+\int_{t_0}^t f\left(s, x_0(s)\right) d s .$$

$$x_n(t)=K^n\left(x_0\right)(t), \quad K(x)(t)=x_0+\int_{t_0}^t f(s, x(s)) d s .$$

$$\left|K(x)(t)-x_0\right| \leq \int_0^t|f(s, x(s))| d s \leq t \max (t, x) \in V|f(t, x)|$$
(这里最大值存在于连续性 $f$ 和紧牶性 $V$ ) 每当图 $x$ 位于内 $V$ ，那是， $(t, x(t)) \mid t \in[0, T] \subset V$. 因此，对于 $t \leq T_0$ ，在哪里
$$T_0=\min \left(T, \frac{\delta}{M}\right), \quad M=\max (t, x) \in V|f(t, x)|,$$

## 有限元方法代写

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## MATLAB代写

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