### 数学代写|偏微分方程代写partial difference equations代考|MATH1470

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|偏微分方程代写partial difference equations代考|Well-posedness of the parabolic initial-boundary value problem

The parabolic maximum principle yields not only uniqueness of solutions of (3.56), but also the a priori estimate (3.58). This allows us to prove well-posedness of the problem in the case where $\Omega$ is an interval.

We consider the following initial-boundary value problem: given $u_0 \in C_0(0, \pi)$, we seek a solution $u$ of
\begin{aligned} u_t & =u_{x x}, & & \text { in }(0, \infty) \times(0, \pi), \ u(t, 0) & =u(t, \pi)=0, & & t \geq 0, \ u(0, x) & =u_0(x), & & x \in[0, \pi] . \end{aligned}
Theorem 3.36 Let $u_0 \in C([0, \pi])$ be such that $u_0(0)=u_0(\pi)=0$. Then (3.59) has a unique solution $u \in C^{\infty}((0, \infty) \times[0, \pi]) \cap C([0, \infty) \times[0, \pi])$. Moreover, for this solution we have $|u|_{C([0, \infty) \times[0, \pi])} \leq\left|u_0\right|_{C([0, \pi])}$.

Proof We have already proved uniqueness and continuous dependence on the data; we still need to establish existence. Let $u_{0 n} \in C([0, \pi])$ be trigonometric polynomials of the form
$$u_{0 n}(x)=\sum_{k=1}^{\infty} b_k^n \sin (k x)$$
such that $u_{0 n} \rightarrow u_0$ in $C([0, \pi])$ (see Corollary 3.19). Here, for fixed $n \in \mathbb{N}, b_k^n=0$ for all but finitely many $k \in \mathbb{N}$, while $\lim {n \rightarrow \infty} b_k^n=b_k$ (cf. the proof of Theorem 3.28). Now for the initial value $u{0 n}$ the solution of $(3.59)$ is given by
$$u_n(t, x)=\sum_{k=1}^{\infty} b_k^n e^{-k^2 t} \sin (k x)$$

## 数学代写|偏微分方程代写partial difference equations代考|The heat equation in Rd

We now wish to study the heat equation in the whole space $\mathbb{R}^d$. This will also be useful when we come to the Black-Scholes equation in a later section. We first consider the one-dimensional case $d=1$. Analogously to what we did above, we denote by $C^{1,2}((0, \infty) \times \mathbb{R})$ the space of those functions $u=u(t, x)$ whose partial derivatives $u_t, u_x, u_{x x}$ exist and are continuous in $(0, \infty) \times \mathbb{R}$.
Our goal is to find solutions $u \in C^{1,2}((0, \infty) \times \mathbb{R})$ of the heat equation
$$u_t=u_{x x}, \quad t>0, x \in \mathbb{R} .$$
The following arguments will allow us to construct a solution. Assume that $u \in C^{1,2}((0, \infty) \times \mathbb{R})$ is a solution of $(3.62)$ and let $a>0$. Then
$$v(t, x):=u(a t, \sqrt{a} x)$$
also defines a solution of (3.62), as can be checked by direct computation. We will attempt to find a solution $u$ which is invariant under the change of variables (3.63), that is, we want
$$u(t, x)=u(a t, \sqrt{a} x), \quad t>0, x \in \mathbb{R}$$
to hold for all $a>0$. If we make the particular choice $a=\frac{1}{t}$, then we obtain
$$u(t, x)=u\left(1, \frac{x}{\sqrt{t}}\right),$$

whence $u(t, x)=g\left(\frac{x}{\sqrt{t}}\right)$ for $g(y):=u(1, y)$. If such a function $u$ is a solution of (3.62), then $g \in C^2(\mathbb{R})$ and
$$u_t=g^{\prime}\left(\frac{x}{\sqrt{t}}\right)\left(-\frac{1}{2}\right) \frac{x}{t^{3 / 2}}, \quad u_x=g^{\prime}\left(\frac{x}{\sqrt{t}}\right) \frac{1}{\sqrt{t}}, \quad u_{x x}=g^{\prime \prime}\left(\frac{x}{\sqrt{t}}\right) \frac{1}{t} .$$
$$0=u_t-u_{x x}=-\frac{1}{t}\left(\frac{1}{2} p g^{\prime}(p)+g^{\prime \prime}(p)\right) \text {, }$$

# 偏微分方程代写

## 数学代写|偏微分方程代写partial difference equations代考|Well-posedness of the parabolic initial-boundary value problem

$$u_t=u_{x x}, \quad \text { in }(0, \infty) \times(0, \pi), u(t, 0) \quad=u(t, \pi)=0, \quad t \geq 0, u(0, x)=u_0(x)$$

$$u_{0 n}(x)=\sum_{k=1}^{\infty} b_k^n \sin (k x)$$

$$u_n(t, x)=\sum_{k=1}^{\infty} b_k^n e^{-k^2 t} \sin (k x)$$

## 数学代写|偏微分方程代写partial difference equations代考|The heat equation in Rd

$$u_t=u_{x x}, \quad t>0, x \in \mathbb{R} .$$

$$v(t, x):=u(a t, \sqrt{a} x)$$

$$u(t, x)=u(a t, \sqrt{a} x), \quad t>0, x \in \mathbb{R}$$

$$u(t, x)=u\left(1, \frac{x}{\sqrt{t}}\right),$$

$$u_t=g^{\prime}\left(\frac{x}{\sqrt{t}}\right)\left(-\frac{1}{2}\right) \frac{x}{t^{3 / 2}}, \quad u_x=g^{\prime}\left(\frac{x}{\sqrt{t}}\right) \frac{1}{\sqrt{t}}, \quad u_{x x}=g^{\prime \prime}\left(\frac{x}{\sqrt{t}}\right) \frac{1}{t} .$$

$$0=u_t-u_{x x}=-\frac{1}{t}\left(\frac{1}{2} p g^{\prime}(p)+g^{\prime \prime}(p)\right)$$

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