# 数学代写|应用数学代写applied mathematics代考|MTH103

## 数学代写|应用数学代写applied mathematics代考|A point source solution

Consider a solution $h(x, t)$ of the porous medium equation (2.30) that approaches an initial point source:
$$h(x, t) \rightarrow I \delta(x), \quad t \rightarrow 0^{+},$$
where $\delta(x)$ denotes the Dirac delta ‘function.’ Explicitly, this means that we require
\begin{aligned} &h(x, t) \rightarrow 0 \quad \text { as } t \rightarrow 0^{+} \text {if } x \neq 0 \ &\lim {t \rightarrow 0^{+}} \int{-\infty}^{\infty} h(x, t) d x=I \end{aligned}

The delta-function is a distribution, rather than a function. We will not discuss distribution theory here (see [44] for an introduction and [27] for a detailed account). Instead, we will define the delta-function formally as a ‘function’ $\delta$ with the properties that
\begin{aligned} &\delta(x)=0 \quad \text { for } x \neq 0 \ &\int_{-\infty}^{\infty} f(x) \delta(x) d x=f(0) \end{aligned}
for any continuous function $f$.
The solution of the porous medium with the initial data (2.31) describes the development of of wetting front due to an instantaneous ‘flooding’ at the origin by a volume of water $I$. It provides the long time asymptotic behavior of solutions of the porous medium equation with a concentrated non-point source initial condition $h(x, t)=h_0(x)$ where $h_0$ is a compactly supported function with integral $I$.

The dimensional parameters at our disposal in solving this problem are $k$ and I. A fundamental system of units is $L, T, H$ where $H$ is a unit for the pressure head. Since we measure the pressure head $h$ in units of length, it is reasonable to ask why we should use different units for $h$ and $x$. The explanation is that the units of vertical length used to measure the head play a different role in the model than the units used to measure horizontal lengths, and we should be able to rescale $x$ and $z$ independently.
Equating the dimension of different terms in $(2.30)$, we find that
$$[k]=\frac{L^2}{H T}, \quad[I]=L H$$

## 数学代写|应用数学代写applied mathematics代考|A pedestrian derivation

Let us consider an alternative method for finding the point source solution that does not require dimensional analysis, but is less systematic.

First, we remove the constants in (2.30) and (2.31) by rescaling the variables. Defining
$$u(x, \bar{t})=\frac{1}{I} h(x, t), \quad \bar{t}=k I t,$$
and dropping the bars on $\bar{t}$, we find that $u(x, t)$ satisfies
$$u_t=\left(u u_x\right)_x$$

The initial condition (2.31) becomes
\begin{aligned} &u(x, t) \rightarrow 0 \quad \text { as } t \rightarrow 0^{+} \text {if } x \neq 0 \ &\lim {t \rightarrow 0^{+}} \int{-\infty}^{\infty} u(x, t) d x=1 \end{aligned}
We seek a similarity solution of (2.36)-(2.37) of the form
$$u(x, t)=\frac{1}{t^m} f\left(\frac{x}{t^n}\right)$$
for some exponents $m, n$. In order to obtain such a solution, the PDE for $u(x, t)$ must reduce to an ODE for $f(\xi)$. As we will see, this is the case provided that $m$, $n$ are chosen appropriately.

Remark 2.12. Equation (2.38) is a typical form of a self-similar solution that is invariant under scaling transformations, whether or not they are derived from a change in units. Dimensional analysis of this problem allowed us to deduce that the solution is self-similar. Here, we simply seek a self-similar solution of the form (2.38) and hope that it works.

# 应用数学代考

## 数学代写|应用数学代写applied mathematics代考|A point source solution

$$h(x, t) \rightarrow I \delta(x), \quad t \rightarrow 0^{+},$$

$$h(x, t) \rightarrow 0 \quad \text { as } t \rightarrow 0^{+} \text {if } x \neq 0 \quad \lim t \rightarrow 0^{+} \int-\infty^{\infty} h(x, t) d x=I$$
delta 函数是一个分布，而不是一个函数。我们不会在这里讨论分布理论（参见 [44] 的介绍和 [27] 的详 细说明) 。相反，我们将 delta 函数正式定义为“函数“ $\delta$ 具有以下属性
$$\delta(x)=0 \quad \text { for } x \neq 0 \quad \int_{-\infty}^{\infty} f(x) \delta(x) d x=f(0)$$

$$[k]=\frac{L^2}{H T}, \quad[I]=L H$$

## 数学代写|应用数学代写applied mathematics代考|A pedestrian derivation

$$u(x, \bar{t})=\frac{1}{I} h(x, t), \quad \bar{t}=k I t,$$

$$u_t=\left(u u_x\right)_x$$

$$u(x, t) \rightarrow 0 \quad \text { as } t \rightarrow 0^{+} \text {if } x \neq 0 \quad \lim t \rightarrow 0^{+} \int-\infty^{\infty} u(x, t) d x=1$$

$$u(x, t)=\frac{1}{t^m} f\left(\frac{x}{t^n}\right)$$

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