# 0.6 Solution of the partial differential equations  (Page 5/13)

 Page 5 / 13
${\nabla }^{2}u-{a}^{2}\frac{\partial u}{\partial t}=-\phantom{\rule{0.166667em}{0ex}}\rho \left(\mathbf{x},t\right)$

where the parameter ${a}^{2}$ represents the ratio of the storage capacity and the conductivity of the system and $\rho$ is a known distribution of sources in space and time. The infinite domain Green's function ${g}_{n}\left(x,t{x}_{o},{t}_{o}\right)$ is a solution of the following equation

${\nabla }^{2}{g}_{n}\left(\mathbf{x},t\left|{\mathbf{x}}_{o},{t}_{o}\right)\right)-{a}^{2}\frac{\partial {g}_{n}\left(\mathbf{x},t\left|{\mathbf{x}}_{o},{t}_{o}\right)\right)}{\partial t}=-\phantom{\rule{0.166667em}{0ex}}\delta \left(\mathbf{x}-{\mathbf{x}}_{o}\right)\phantom{\rule{0.166667em}{0ex}}\delta \left(t-{t}_{o}\right)$

The source term is an impulse in the spatial and time variables. The form of the Green's function for the infinite domain, for $n$ dimensions, is (Morse and Feshbach, 1953)

$\begin{array}{c}{g}_{n}\left(\mathbf{x},t\left|{\mathbf{x}}_{o},{t}_{o}\right)\right)={g}_{n}\left(R,\tau \right)=\left\{\begin{array}{c}\frac{1}{{a}^{2}}{\left(\frac{a}{2\sqrt{\pi \phantom{\rule{0.166667em}{0ex}}\tau }}\right)}^{n}{e}^{-\left({a}^{2}{R}^{2}/4\tau \right)},\phantom{\rule{1.em}{0ex}}\tau >0\\ 0\phantom{\rule{1.em}{0ex}},\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\tau <0\end{array}\right)\hfill \\ \text{where}\hfill \\ \tau =t-{t}_{o}\hfill \\ R=\left|\mathbf{x}-{\mathbf{x}}_{o}\right|\hfill \end{array}$

This Green's function satisfies an important integral property that is valid for all values of $n$ :

$\int {g}_{n}\left(R,\tau \right)\phantom{\rule{0.166667em}{0ex}}d{V}_{n}=\frac{1}{{a}^{2}},\phantom{\rule{1.em}{0ex}}\tau >0$

This expression is an expression of the conservation of heat energy. At a time ${t}_{o}$ at ${\mathbf{x}}_{o}$ , a source of heat is introduced. The heat diffuses out through the medium, but in such a fashion that the total heat energy is unchanged.

The properties of this Green's function can be more easily seen be expressing it in a standard form

${a}^{2}{g}_{n}\left(R,\tau \right)=\left\{\begin{array}{c}{\left(\frac{1}{\sqrt{2\pi \phantom{\rule{0.166667em}{0ex}}\left(2\tau /{a}^{2}\right)}}\right)}^{n}{e}^{-\left[{R}^{2}/2\left(2\tau /{a}^{2}\right)\right]},\phantom{\rule{1.em}{0ex}}\tau >0\\ 0\phantom{\rule{1.em}{0ex}},\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\tau <0\end{array}\right)$

The normalized function ${a}^{2}{g}_{n}$ for $n=3$ represent the probability distribution of the location of a Brownian particle that was at ${x}_{o}$ at time ${t}_{o}$ . The cumulative probability is equal to unity.

The same normalized function for $n=1$ , corresponds to the normal or Gaussian distribution with the standard deviation given by

$\sigma =\frac{\sqrt{2\tau }}{a}$

Observe the Green's function in one, two, and three dimensions by executing greens.m and the function, greenf.m in the diffuse subdirectory of CENG501. You may wish to use the code as a template for future assignments.

Step Response Function The infinite domain Green's function is the impulse response function in space and time. The response for a distribution of sources in space or as an arbitrary function of time can be determined by convolution. In particular the response to a constant source for $\tau >0$ is the step response function . It has classical solutions in one and two dimensions. The unit step function or Heaviside function is the integral of the Dirac delta function.

${\int }_{-\infty }^{t}\delta \left({t}^{\text{'}}-{t}_{o}\right)\phantom{\rule{0.166667em}{0ex}}d{t}^{\text{'}}=S\left(t-{t}_{o}\right)=\left\{\begin{array}{c}1,\phantom{\rule{1.em}{0ex}}t-{t}_{o}>0\\ 0,\phantom{\rule{1.em}{0ex}}t-{t}_{o}<0\end{array}\right)$

The response function to a unit step in the source can be determined by integrating the Greens function or the impulse response function in time.

$\begin{array}{c}\underset{-\infty }{\overset{t}{\int }}\left[{\nabla }^{2}{g}_{n}-{a}^{2}\frac{\partial {g}_{n}}{\partial t}\right]\phantom{\rule{0.166667em}{0ex}}d{t}^{\text{'}}=-\phantom{\rule{0.166667em}{0ex}}\delta \left(\mathbf{x}-{\mathbf{x}}_{o}\right)\underset{-\infty }{\overset{t}{\int }}\delta \left({t}^{\text{'}}-{t}_{o}\right)\phantom{\rule{0.166667em}{0ex}}d{t}^{\text{'}}\phantom{\rule{0.166667em}{0ex}}\hfill \\ {\nabla }^{2}\left(\underset{-\infty }{\overset{t}{\int }}{g}_{n}\phantom{\rule{0.166667em}{0ex}}d{t}^{\text{'}}\right)-{a}^{2}\frac{\partial \left(\underset{-\infty }{\overset{t}{\int }}{g}_{n}\phantom{\rule{0.166667em}{0ex}}d{t}^{\text{'}}\right)}{\partial t}=-\phantom{\rule{0.166667em}{0ex}}\delta \left(\mathbf{x}-{\mathbf{x}}_{o}\right)\phantom{\rule{0.166667em}{0ex}}S\left(t-{t}_{o}\right)\hfill \\ {\nabla }^{2}{U}_{n}-{a}^{2}\phantom{\rule{0.166667em}{0ex}}\frac{\partial {U}_{n}}{\partial t}=-\phantom{\rule{0.166667em}{0ex}}\delta \left(\mathbf{x}-{\mathbf{x}}_{o}\right)\phantom{\rule{0.166667em}{0ex}}S\left(t-{t}_{o}\right)\hfill \\ \text{where}\hfill \\ {U}_{n}=\underset{-\infty }{\overset{t}{\int }}{g}_{n}\phantom{\rule{0.166667em}{0ex}}d{t}^{\text{'}}\hfill \end{array}$

In one dimension, the step response function that has a unit flux at $x=0$ is (R. V. Churchill, Operational Mathematics , 1958) (note: source is $2\delta \left(x-0\right)\phantom{\rule{0.166667em}{0ex}}S\left(t-0\right)$ )

${U}_{1}^{\text{flux}=-1}=2{a}^{2}\sqrt{\frac{t}{\pi \phantom{\rule{0.166667em}{0ex}}{a}^{2}}}\phantom{\rule{0.166667em}{0ex}}{e}^{\frac{-{x}^{2}}{4t/{a}^{2}}}-{a}^{2}\left|x\right|\phantom{\rule{0.166667em}{0ex}}erfc\left(\frac{\left|x\right|}{\sqrt{4t/{a}^{2}}}\right),\phantom{\rule{1.em}{0ex}}t>0$

For comparison, the function that has a value of unity at $x=0$ (Dirichlet boundary condition) is

${U}_{1}^{=1}=erfc\left(\frac{\left|x\right|}{2\sqrt{t/{a}^{2}}}\right),\phantom{\rule{1.em}{0ex}}t>0$

In two dimensions, the unit step response function for a continuous line source is (H. S. Carslaw and J. C. Jaeger, Conduction of Heat in Solids, 1959)

$\begin{array}{ccc}\hfill {U}_{2}& =& \frac{-{a}^{2}}{4\pi }\phantom{\rule{0.166667em}{0ex}}Ei\left(\frac{-{R}^{2}}{4t/{a}^{2}}\right),\phantom{\rule{1.em}{0ex}}t>0\hfill \\ \hfill {R}^{2}& =& {\left(x-{x}_{o}\right)}^{2}+{\left(y-{y}_{o}\right)}^{2}\hfill \\ \hfill -Ei\left(-x\right)& =& {\int }_{x}^{\infty }\frac{{e}^{-u}}{u}\phantom{\rule{0.166667em}{0ex}}du\hfill \\ & =& expint\left(x\right),\phantom{\rule{1.em}{0ex}}\text{MATLAB}\phantom{\rule{0.277778em}{0ex}}\text{function}\hfill \end{array}$

For large times this function can be expressed as

$\begin{array}{c}{U}_{2}^{\text{approx}.}=\frac{{a}^{2}}{4\pi }ln\left(\frac{4t/{a}^{2}}{{R}^{2}}\right)-\frac{\gamma \phantom{\rule{0.166667em}{0ex}}{a}^{2}}{4\pi },\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\frac{4t}{{a}^{2}\phantom{\rule{0.166667em}{0ex}}{R}^{2}}>100\hfill \\ \gamma =0.5772...\phantom{\rule{1.em}{0ex}}\hfill \end{array}$

In three dimensions, the unit step response function for a continuous point source is (H. S. Carslaw and J. C. Jaeger, Conduction of Heat in Solids, 1959)

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learn
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da
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Bhagvanji
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Professor
I think
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biomolecules are e building blocks of every organics and inorganic materials.
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