# 0.6 Solution of the partial differential equations  (Page 6/13)

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$\begin{array}{c}{U}_{3}=\frac{{a}^{2}}{4\pi \phantom{\rule{0.166667em}{0ex}}R}erfc\left(\frac{R}{\sqrt{4t/{a}^{2}}}\right),\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}t>0\hfill \\ R={\left[{\left(x-{x}_{o}\right)}^{2}+{\left(y-{y}_{o}\right)}^{2}+{\left(z-{z}_{o}\right)}^{2}\right]}^{1/2}\hfill \end{array}$
The ${a}^{2}$ .factor has the units of $\text{time}/{L}^{2}$ . If time is made dimensionless with respect to ${a}^{2}/{R}_{o}^{2}$ and $R$ with respect to ${R}_{o}$ , then the factor will disappear from the argument of the erfc.

## Assignment 7.4

Plot the profiles of the response to a continuous source in 1, 2, and 3 dimensions using the MATLAB code contins.m and continf.m in the diffuse subdirectory. From the integral of the profiles as a function of time, determine the magnitude, spatial and time dependence of the source. Note: The exponential integral function, expint will give error messages for extreme values of the argument. It still computes the correct values of the function.

## Convective-diffusion equation

The convective-diffusion equation in one dimension will be expressed in terms of velocity and dispersion,

$\begin{array}{c}\frac{\partial u}{\partial t}+v\frac{\partial u}{\partial x}=K\frac{{\partial }^{2}u}{\partial {x}^{2}}\hfill \\ u\left(x,\right)=0,\phantom{\rule{1.em}{0ex}}x>0\hfill \\ u\left(0,t\right)=1,\phantom{\rule{1.em}{0ex}}t>0\hfill \end{array}$

The independent variables can be transformed from $\left(x,t\right)$ to a spatial coordinate that translates with the velocity of the wave in the absence of dispersion, $\left(y,t\right)$ .

$y=x-v\phantom{\rule{0.166667em}{0ex}}t$

This transforms the equation to the diffusion equation in the transformed coordinates.

$\frac{\partial u}{\partial t}=K\frac{{\partial }^{2}u}{\partial {y}^{2}}$

To see this, we will transform the differentials from $x$ to $y$ .

$\begin{array}{c}\frac{\partial y}{\partial t}=-v\hfill \\ \frac{\partial y}{\partial x}=1\hfill \end{array}$

The total differentials expressed as a function of $\left(x,t\right)$ or $\left(y,t\right)$ are equal to each other.

$\begin{array}{c}du={\left(\frac{\partial u}{\partial t}\right)}_{x}dt+{\left(\frac{\partial u}{\partial x}\right)}_{t}dx\hfill \\ du={\left(\frac{\partial u}{\partial t}\right)}_{y}dt+{\left(\frac{\partial u}{\partial y}\right)}_{t}dy\hfill \end{array}$

The total differentials expressed either way are equal. The partial derivatives in $t$ and $x$ can be expressed in terms of partial derivatives in $t$ and $y$ by equating the total differentials with either $dt$ or $dx$ equal to zero and dividing by the non-zero differential.

$\begin{array}{ccc}\hfill {\left(\frac{\partial u}{\partial t}\right)}_{x}& =& {\left(\frac{\partial u}{\partial t}\right)}_{y}+{\left(\frac{\partial u}{\partial y}\right)}_{t}{\left(\frac{\partial y}{\partial t}\right)}_{x}\hfill \\ & =& {\left(\frac{\partial u}{\partial t}\right)}_{y}-v\phantom{\rule{0.166667em}{0ex}}{\left(\frac{\partial u}{\partial y}\right)}_{t}\hfill \\ & & \mathrm{and}\hfill \\ \hfill {\left(\frac{\partial u}{\partial x}\right)}_{t}& =& {\left(\frac{\partial u}{\partial y}\right)}_{t}{\left(\frac{\partial y}{\partial x}\right)}_{t}\hfill \\ & =& {\left(\frac{\partial u}{\partial y}\right)}_{t}\hfill \\ \hfill {\left(\frac{{\partial }^{2}u}{\partial {x}^{2}}\right)}_{t}& =& {\left(\frac{{\partial }^{2}u}{\partial {y}^{2}}\right)}_{t}\hfill \end{array}$

Substitution into the original equation results in the transformed equation. This result could have been derived in fewer steps by using the chain rule but would not have been as enlightening.

The boundary condition at $x=0$ is now at changing values of $y$ . We will seek an approximate solution that has the boundary condition $u\left(y\to -\infty \right)=1$ . A simple solution can be found for the following initial and boundary conditions.

$\begin{array}{c}u\left(y,0\right)=\left\{\begin{array}{c}1,\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}y<0\\ 1/2,\phantom{\rule{1.em}{0ex}}y=0\\ 0,\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}y>0\end{array}\right)\hfill \\ u\left(y\to -\infty ,t\right)=1\hfill \\ u\left(y\to \infty ,t\right)=0\hfill \end{array}$

This system is a step with no dispersion at $t=0$ . Dispersion occurs for $t>0$ as the wave propagates through the system. The solution can be found with a similarity transform, which we will discuss later. For now, the approximate solution is given as

$u=\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}\mathrm{erfc}\left(\frac{y}{\sqrt{4K\phantom{\rule{0.166667em}{0ex}}t}}\right)=\frac{1}{2}\phantom{\rule{0.166667em}{0ex}}\mathrm{erfc}\left(\frac{x-v\phantom{\rule{0.166667em}{0ex}}t}{\sqrt{4K\phantom{\rule{0.166667em}{0ex}}t}}\right)$

The boundary condition at $x=0$ will be approximately satisfied after a small time unless the Peclet number is very small.

## Similarity transformation

In some cases a partial differential equation and its boundary conditions (and initial condition) can be transformed to an ordinary differential equation with boundary conditions by combining two independent variables into a single independent variable. We will illustrate the approach here with the diffusion equation. It will be used later for hyperbolic PDEs and for the boundary layer problems.

The method will be illustrated for the solution of the one-dimensional diffusion equation with the following initial and boundary conditions. The approach will follow that of the Hellums-Churchill method.

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