### 数学代写|微分方程代写differential equation代考|MATHS 2102

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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|微分方程代写differential equation代考|Discrete Boxes

Suppose there are a number of boxes connected side-by-side along a one-dimensional line, with concentration of some chemical species $u_{j}$ in box $j,-\infty<j<\infty$. Now suppose that the chemical leaves box $j$ at rate $2 \lambda$, so that the concentration in box $j$ is governed by
$$\frac{d u_{j}}{d t}=-2 \lambda u_{j},$$
provided there is no inflow. This is exactly the decay process described in Section 1.3.1. However, here we assume that the particles that flow out of box $j$ are evenly split to go into the neighboring boxes $j-1$ and $j+1$. Consequently, half of the particles that leave boxes $j-1$ and $j+1$ enter box $j$, so that
$$\frac{d u_{j}}{d t}=\lambda u_{j-1}-2 \lambda u_{j}+\lambda u_{j+1}$$

It is a straightforward matter to simulate this system of ordinary differential equations. The Matlab file to do so is titled diffusion_via_MOL.m, and you are encouraged to run this code to see if what happens matches with your intuition.

Now suppose that $u_{j}$ is a sample of a smooth function $u(x, t)$ at points $x=j \Delta x$, i.e., $u_{j}=u(j \Delta x, t)$. Using Taylor’s theorem,
\begin{aligned} u_{j \pm 1} \equiv & u\left(x_{j} \pm \Delta x, t\right) \ =& u\left(x_{j}, t\right) \pm \Delta x \frac{\partial}{\partial x} u\left(x_{j}, t\right)+\frac{1}{2} \Delta x^{2} \frac{\partial^{2}}{\partial x^{2}} u\left(x_{j}, t\right) \ & \pm \frac{1}{6} \Delta x^{3} \frac{\partial^{3}}{\partial x^{3}} u\left(x_{j}, t\right)+O\left(\Delta x^{4}\right) \end{aligned}
Substituting this Taylor series into (3.4), It follows that
$$\frac{\partial u}{\partial t}=\lambda \Delta x^{2} \frac{\partial^{2} u}{\partial x^{2}}+O\left(\Delta x^{4}\right)$$
which, keeping only the largest terms in $\Delta x$, is the diffusion equation with diffusion constant $D=\lambda \Delta x^{2}$.

## 数学代写|微分方程代写differential equation代考|A Random Walk

Consider the problem where we take a number of random steps at discrete times, and for each step we make a decision to take a step of length $m \Delta x$ where $m=-1,0$, or 1 , with probability $\alpha, 1-2 \alpha$, and $\alpha$, respectively. Let $x_{n}$ be the position after $n$ steps, $x_{n}=\Delta x \sum_{j=1}^{n} m_{j}$.

The first thing to do here is to simulate this process. This is easy to do, and the Matlab code for this is entitled discrete_random_walk.m. (Or, with a group of friends or classmates, perform this experiment for yourselves, taking steps on a sidewalk to the left when a coin flip gives heads and a step to the right when a coin flip gives tails.) Examples of sample paths for this process are shown in Figure 3.1(a) and the mean squared displacement $\left\langle x_{n}^{2}\right\rangle$, defined as $\left\langle x_{n}^{2}\right\rangle=\frac{1}{N} \sum_{N \text { trials }} x_{n}^{2}$, as a function of time step $n$, averaged over $N=1000$ particle trajectories, is shown in Figure 3.1(b).

Lel’s nuw salculate the probability that $x_{n}$ has the value $k \Delta x$, denuted $p_{k, n}=$ $P\left(x_{n}=k \Delta x\right)$
$$p_{k, n}=\alpha p_{k-1, n-1}+(1-2 \alpha) p_{k, n-1}+\alpha p_{k+1, n-1} .$$
In words, the probability that $x_{n}$ is $k \Delta x$ is the sum of three terms, $\alpha$ times the probability that $x_{n-1}$ is $(k-1) \Delta x, \alpha$ times the probability that $x_{n-1}$ is $(k+1) \Delta x$, and $1-2 \alpha$ times the probability that $x_{n-1}$ is $k \Delta x$. Now, suppose that $p_{k, n}=P\left(x_{n}=k \Delta x\right)$ is the sampling of a smooth function $p(x, t)$, where $p_{k, n}=P\left(x_{n}=k \Delta x\right)=p(k \Delta x, n \Delta t)$. Again, using Taylor series, it follows that, to leading order in $\Delta t$ and $\Delta x$ (i.e., keeping only the largest terms in $\Delta t$ and $\Delta x$,
$$\frac{\partial p}{\partial t}=\alpha \frac{\Delta x^{2}}{\Delta t} \frac{\partial^{2} p}{\partial x^{2}}$$
which is, once again, the diffusion equation, with diffusion coefficient $D=\alpha \frac{\Delta x^{2}}{\Delta t}$. This is the same diffusion coefficient as above if we make the identification $\lambda=\frac{\alpha}{\Delta t}$.

## 数学代写|微分方程代写differential equation代考|The Cable Equation

The third derivation of the diffusion equation comes from a completely different, and perhaps surprising, consideration.

The membrane of a cell is a phospholipid bilayer that acts as a barrier to the movement of ions between the intracellular (inside) and extracellular (outside) spaces. As a barrier, it can store charge much like a capacitor. Further, the movement of ions across a membrane is carefully regulated and they flow through a variety of ion channels. This is true for many electrically active cells, including neurons, cardiac cells, and smooth muscle cells. For example, the neurons studied by Hodgkin and Huxley (see Exercise 1.11) have three different ion species that flow through ion channels. These are depicted in Figure $3.2$ as $I_{N a}, I_{K}$, and $I_{l}$, for sodium, potassium, and leak, respectively. Consequently, the electrical nature of these cells can be described by a capacitor (the membrane) and resistors (the ion channels) in parallel, as shown in the circuit diagram in Figure 3.2. For this diagram there are two transmembrane currents, the ionic currents $I_{\text {ion }}$, and the capacitive current. The fundamental law of capacitance states that the total charge on the capacitor is capacitance times voltage, $Q=C_{m} V$, where $C_{m}$ is the membrane capacitance, and $V=V_{i}-V_{e}$ is the transmembrane voltage potential, $V_{i}$ and $V_{e}$ are the intracellular and extracellular voltage potentials, respectively. This implies that the capacitive current is $I_{c}=\frac{d Q}{d t}=C_{m} \frac{d V}{d t}$. Thus, the total transmembrane current, $I_{t}$, is the sum of capacitive and ionic currents, i.e.,
$$C_{m} \frac{d V}{d t}+I_{\text {ion }}=I_{t} .$$
This model applies only for a small homogeneous patch of membrane. However, nerve cells, or neurons, have axons, that are long slender cylindrical projections that extend away from the neuron’s cell body, or soma, and can be quite long (cf. Figure 3.3). For example, the human sciatic nerve originates in the lower back and extends down the back of the thigh and leg, ending in the foot.

To incorporate the effects of an elongated membrane, we view the axon as a long cylindrical piece of membrane surrounding an interior of cytoplasm (called a cable), and suppose that everywhere along its length, the potential depends only on the length variable and not on radial or angular variables. We divide the cable into a number of short pieces of isopotential membrane each of length $d x$, two sections of which are depicted in Figure 3.4.

## 数学代写|微分方程代写differential equation代考|Discrete Boxes

d在jd吨=−2λ在j,

d在jd吨=λ在j−1−2λ在j+λ在j+1

∂在∂吨=λΔX2∂2在∂X2+○(ΔX4)

## 数学代写|微分方程代写differential equation代考|A Random Walk

Lel’s nuw 计算出以下概率Xn有价值ķΔX, 表示pķ,n= 磷(Xn=ķΔX)

pķ,n=一个pķ−1,n−1+(1−2一个)pķ,n−1+一个pķ+1,n−1.

∂p∂吨=一个ΔX2Δ吨∂2p∂X2

C米d在d吨+我离子 =我吨.

## 有限元方法代写

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## MATLAB代写

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