# 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)

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply 12, 17, 22.... 25th term Alexandra Reply 12, 17, 22.... 25th term Akash College algebra is really hard? Shirleen Reply Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table. Carole I'm 13 and I understand it great AJ I am 1 year old but I can do it! 1+1=2 proof very hard for me though. Atone hi Adu Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily. Vedant hi vedant can u help me with some assignments Solomon find the 15th term of the geometric sequince whose first is 18 and last term of 387 Jerwin Reply I know this work salma The given of f(x=x-2. then what is the value of this f(3) 5f(x+1) virgelyn Reply hmm well what is the answer Abhi If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10 Augustine how do they get the third part x = (32)5/4 kinnecy Reply make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be AJ how Sheref can someone help me with some logarithmic and exponential equations. Jeffrey Reply sure. what is your question? ninjadapaul 20/(×-6^2) Salomon okay, so you have 6 raised to the power of 2. what is that part of your answer ninjadapaul I don't understand what the A with approx sign and the boxed x mean ninjadapaul it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared Salomon I'm not sure why it wrote it the other way Salomon I got X =-6 Salomon ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6 ninjadapaul oops. ignore that. ninjadapaul so you not have an equal sign anywhere in the original equation? ninjadapaul hmm Abhi is it a question of log Abhi 🤔. Abhi I rally confuse this number And equations too I need exactly help salma But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends salma Commplementary angles Idrissa Reply hello Sherica im all ears I need to learn Sherica right! what he said ⤴⤴⤴ Tamia hii Uday hi salma hi Ayuba Hello opoku hi Ali greetings from Iran Ali salut. from Algeria Bach hi Nharnhar A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers! By Naveen Tomar By By      By 