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# 数学代写|偏微分方程作业代写Partial Differential Equations代考|A Touch of Numerics, I: Computing the Solution of the Transport Equation

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## 数学代写|偏微分方程作业代写Partial Differential Equations代考|Three Consistent Schemes

We begin by discretizing space and time. This means that we replace the continuums of space and time with a finite, or at most countable, collection of positions and times. Specifically, we do the following:

• Fix values of $\Delta x>0$ and $\Delta t>0$ called the spatial and temporal step size.
• Consider “grid points” $x_{j}=j \Delta x$ and $t_{n}=n \Delta t$ with $j=0, \pm 1, \pm 2, \ldots$ and $n=$ $0,1,2, \ldots$.

Numerically computing the solution means that we find (approximate) values of the true solution at these particular values of $x$ and $t$. That is, we then attempt to find the solution at the grid points
$$U_{j}^{n}:=u(j \Delta x, n \Delta t) .$$
In the finite difference method, we approximate the values of the partial derivatives at a grid point, based upon values at neighboring grid points. For example, a natural way to approximate $u_{t}$ is by a forward finite difference
$$u_{t}(j \Delta x, n \Delta t) \approx \frac{u(j \Delta x, n \Delta t+\Delta t)-u(j \Delta x, n \Delta t)}{\Delta t}=\frac{U_{j}^{n+1}-U_{j}^{n}}{\Delta t} .$$

## 数学代写|偏微分方程作业代写Partial Differential Equations代考|von Neumann Stability Analysis

We are interested in the propagation of small errors at different grid points. If these errors are not controlled at each stage of the time iteration, they can grow and create a solution completely different from the desired solution at a later time (achieved by many time iterations). So what do we mean when we say “controlled”. To this end, we follow von Neumann ${ }^{13}$ stability analysis. We start with an initial pulse which has bounded size but oscillates with a frequency $k$, for example $\sin k x$ or $\cos k x$. From the perspective of the algebra, it is more convenient to just take a complex exponential (cf. Section 6.1) $e^{i k x}=\cos k x+i \sin k x$, for some $k \in \mathbb{R}$. The size of such a complex-valued pulse is given by its modulus, and note that $\left|e^{i k x}\right|=1$. We then look at the growth in size when applying the scheme once to $e^{i k x}$, captured by an amplification constant called the growth factor. Let us illustrate these steps with the backward difference in space scheme (2.50). At time step $n$, we take
$$U_{j}^{n}=e^{i k(j \Delta x)}, \quad j \in \mathbb{Z},$$
noting that
$$U_{j-1}^{n}=e^{i k(j \Delta x)} e^{-i k \Delta x}, \quad j \in \mathbb{Z} .$$
Applying scheme (2.50) we find that
\begin{aligned} U_{j}^{n+1} &=U_{j}^{n}-r\left(U_{j}^{n}-U_{j-1}^{n}\right) \ &=e^{i k(j \Delta x)}-r\left(e^{i k(j \Delta x)}-e^{i k(j \Delta x)} e^{-i k \Delta x}\right) \ &=\left(1-r+r e^{-i k \Delta x}\right) e^{i k(j \Delta x)} \end{aligned}

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|THREE CONSISTENT SCHEMES

• 固定值ΔX>0和Δ吨>0称为空间和时间步长。
• 考虑“网格点”Xj=jΔX和吨n=nΔ吨和j=0,±1,±2,…和n= 0,1,2,….

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|VON NEUMANN STABILITY ANALYSIS

\begin{aligned} U_{j}^{n+1} &=U_{j}^{n}-r\left(U_{j}^{n}-U_{j-1}^{n}\right) \ &=e^{i k(j \Delta x)}-r\left(e^{i k(j \Delta x)}-e^{i k(j \Delta x)} e^{-i k \Delta x}\right) \ &=\left(1-r+r e^{-i k \Delta x}\right) e^{i k(j \Delta x)} \end{aligned}