# 数学代写|随机过程统计代写Stochastic process statistics代考|STAT3021

## 数学代写|随机过程统计代写Stochastic process statistics代考|The heat equation

In Chapter 7 we have seen that the Laplacian $\left(\frac{1}{2} \Delta, \mathfrak{D}(\Delta)\right)$ is the infinitesimal generator of a $\mathrm{BM}^d$. In fact, applying Lemma $7.10$ to a Brownian motion yields
$$\frac{d}{d t} P_t f(x)=\frac{1}{2} \Delta_x P_t f(x) \text { for all } f \in \mathfrak{D}(\Delta) .$$
Setting $u(t, x):=P_t f(x)=\mathbb{E}^x f\left(B_t\right)$ this almost proves the following lemma.
8.1 Lemma. Let $\left(B_t\right){t \geqslant 0}$ be $a \operatorname{BM}^d, f \in \mathfrak{D}(\Delta)$ and $u(t, x):=\mathbb{E}^x f\left(B_t\right)$. Then $u(t, x)$ is the unique bounded solution of the heat equation with initial value $f$, \begin{aligned} \frac{\partial}{\partial t} u(t, x)-\frac{1}{2} \Delta_x u(t, x) &=0, \ u(0, x) &=f(x) . \end{aligned} Proof. It remains to show uniqueness. Let $u(t, x)$ be a bounded solution of (8.3). Its Laplace transform is given by $$v\lambda(x)=\int_0^{\infty} u(t, x) e^{-\lambda t} d t .$$
Since we may differentiate under the integral sign, we get
\begin{aligned} \lambda v_\lambda(x)-\frac{1}{2} \Delta_x v_\lambda(x) &=\lambda v_\lambda(x)-\int_0^{\infty} \frac{1}{2} \Delta_x u(t, x) e^{-\lambda t} d t \ &=\lambda v_\lambda(x)-\int_0^{\infty} \frac{\partial}{\partial t} u(t, x) e^{-\lambda t} d t . \end{aligned}
Integration by parts and ( $8.3 \mathrm{~b})$ yield
$$\left(\lambda \mathrm{id}-\frac{1}{2} \Delta_x\right) v_\lambda(x)=\lambda v_\lambda(x)+\int_0^{\infty} u(t, x) \frac{\partial}{\partial t} e^{-\lambda t} d t-\left.u(t, x) e^{-\lambda t}\right|{t=0} ^{t=\infty}=f(x) .$$ Since the equation $\left(\lambda\right.$ id $\left.-\frac{1}{2} \Delta_x\right) v\lambda=f, f \in \mathfrak{D}(\Delta), \lambda>0$, has the unique solution $v_\lambda=U_\lambda f$ where $U_\lambda$ is the resolvent operator, and since the Laplace transform is unique, cf. [170, Proposition 1.2], we conclude that $u(t, x)$ is unique.

## 数学代写|随机过程统计代写Stochastic process statistics代考|The inhomogeneous initial value problem

If we replace the right-hand side of the heat equation (8.3a) with some function $g(t, x)$, we get the following inhomogeneous initial value problem.
\begin{aligned} \frac{\partial}{\partial t} v(t, x)-\frac{1}{2} \Delta_x v(t, x) &=g(t, x) \ v(0, x) &=f(x) \end{aligned}
Assume that $u$ solves the corresponding homogeneous problem (8.3). If $w$ solves the inhomogeneous problem (8.6) with zero initial condition $f=0$, then $v=u+w$ is a solution of (8.6). Therefore, it is customary to consider
\begin{aligned} \frac{\partial}{\partial t} v(t, x)-\frac{1}{2} \Delta_x v(t, x) &=g(t, x), \ v(0, x) &=0 \end{aligned}
instead of (8.6). Let us use the probabilistic approach of Section $8.1$ to develop an educated guess about the solution to (8.7).

Let $v(t, x)$ be a solution to (8.7) which is bounded and sufficiently smooth $\left(e^{1,2}\right.$ and (8.1) will do). Then. for $s \in[0, t)$.
$$M_s^v:=v\left(t-s, B_s\right)-v\left(t, B_0\right)+\int_0^s \underbrace{\left(\frac{\partial}{\partial t} v\left(t-r, B_r\right)-\frac{1}{2} \Delta_x v\left(t-r, B_r\right)\right)}{=g\left(t-r, B_r\right) \text { by }(8.7 \mathrm{a})} d r$$ is a martingale, and so is $N_s^v:=v\left(t-s, B_s\right)+\int_0^s g\left(t-r, B_r\right) d r, s \in[0, t)$; in particular, $\mathbb{E}^x N_0^v=\mathbb{E}^x N_s^v$ for all $s \in[0, t)$. If we can use the martingale convergence theorem, e. g. if the martingale is uniformly integrable, we can let $s \rightarrow t$ and get $$v(t, x)=\mathbb{E}^x N_0^v=\lim {s \rightarrow t} \mathbb{E}^x N_s^v=\mathbb{E}^x N_t^v \stackrel{(8.7 \mathrm{~b})}{=} \mathbb{E}^x\left(\int_0^t g\left(t-r, B_r\right) d r\right)$$

## 数学代写|随机过程统计代写Stochastic process statistics代考|The heat equation

$$\frac{d}{d t} P_t f(x)=\frac{1}{2} \Delta_x P_t f(x) \text { for all } f \in \mathfrak{D}(\Delta) .$$

$8.1$ 引理。让 $\left(B_t\right) t \geqslant 0$ 是 $a \mathrm{BM}^d, f \in \mathfrak{D}(\Delta)$ 和 $u(t, x):=\mathbb{E}^x f\left(B_t\right)$. 然后 $u(t, x)$ 是热方程的唯一有界 解，具有初始值 $f$ ，
$$\frac{\partial}{\partial t} u(t, x)-\frac{1}{2} \Delta_x u(t, x)=0, u(0, x) \quad=f(x) .$$

$$v \lambda(x)=\int_0^{\infty} u(t, x) e^{-\lambda t} d t .$$

$$\lambda v_\lambda(x)-\frac{1}{2} \Delta_x v_\lambda(x)=\lambda v_\lambda(x)-\int_0^{\infty} \frac{1}{2} \Delta_x u(t, x) e^{-\lambda t} d t \quad=\lambda v_\lambda(x)-\int_0^{\infty} \frac{\partial}{\partial t} u(t, x) e^{-\lambda t} d t .$$

$$\left(\lambda \operatorname{id}-\frac{1}{2} \Delta_x\right) v_\lambda(x)=\lambda v_\lambda(x)+\int_0^{\infty} u(t, x) \frac{\partial}{\partial t} e^{-\lambda t} d t-u(t, x) e^{-\lambda t} \mid t=0^{t=\infty}=f(x) .$$

## 数学代写|随机过程统计代写Stochastic process statistics代考|The inhomogeneous initial value problem

$$\frac{\partial}{\partial t} v(t, x)-\frac{1}{2} \Delta_x v(t, x)=g(t, x) v(0, x) \quad=f(x)$$

$$\frac{\partial}{\partial t} v(t, x)-\frac{1}{2} \Delta_x v(t, x)=g(t, x), v(0, x) \quad=0$$

$$M_s^v:=v\left(t-s, B_s\right)-v\left(t, B_0\right)+\int_0^s \underbrace{\left(\frac{\partial}{\partial t} v\left(t-r, B_r\right)-\frac{1}{2} \Delta_x v\left(t-r, B_r\right)\right)}=g\left(t-r, B_r\right) \text { by (8.7a) } d r$$

$$v(t, x)=\mathbb{E}^x N_0^v=\lim s \rightarrow t \mathbb{E}^x N_s^v=\mathbb{E}^x N_t^v \stackrel{(8.7 \mathrm{~b})}{=} \mathbb{E}^x\left(\int_0^t g\left(t-r, B_r\right) d r\right)$$

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