# 0.6 Solution of the partial differential equations  (Page 7/13)

 Page 7 / 13
$\begin{array}{c}\frac{\partial u}{\partial t}=K\frac{{\partial }^{2}u}{\partial {x}^{2}},\phantom{\rule{1.em}{0ex}}t>0,\phantom{\rule{0.277778em}{0ex}}x>0\hfill \\ u\left(x,0\right)={u}_{IC}\hfill \\ u\left(0,t\right)={u}_{BC}\hfill \end{array}$

The PDE, IC and BC are made dimensionless with respect to reference quantities.

$\begin{array}{ccc}\hfill u*& =& \frac{u-{u}_{IC}}{{u}_{o}}\hfill \\ \hfill t*& =& \frac{t}{{t}_{o}}\hfill \\ \hfill x*& =& \frac{x}{{x}_{o}}\hfill \\ \hfill \frac{\partial u*}{\partial t*}& =& \left[\frac{K\phantom{\rule{0.166667em}{0ex}}{t}_{o}}{{x}_{o}^{2}}\right]\phantom{\rule{0.166667em}{0ex}}\frac{{\partial }^{2}u*}{\partial x{*}^{2}}\hfill \\ \hfill u*\left(x*,0\right)& =& 0\hfill \\ \hfill u*\left(0,t*\right)& =& \left[\frac{{u}_{BC}-{u}_{IC}}{{u}_{o}}\right]=1,\phantom{\rule{1.em}{0ex}}⇒{u}_{o}={u}_{BC}-{u}_{IC}\hfill \end{array}$

There are three unspecified reference quantities and two dimensionless groups. The BC can be specified to equal unity. However, the system does not have a characteristic time or length scales to specify the dimensionless group in the PDE. This suggests that the system is over specified and the independent variables can be combined to specify the dimensionless group in the PDE to equal 1/4.

$\begin{array}{c}\left[\frac{K\phantom{\rule{0.166667em}{0ex}}{t}_{o}}{{x}_{o}^{2}}\right]=\frac{1}{4}\phantom{\rule{1.em}{0ex}}⇒\eta =\frac{x}{\sqrt{4K\phantom{\rule{0.166667em}{0ex}}t}}\phantom{\rule{1.em}{0ex}}\mathrm{is}\phantom{\rule{0.277778em}{0ex}}\mathrm{dimensionless}\hfill \\ u\left(x,t\right)=u\left(\eta \right)\hfill \end{array}$

The partial derivatives will now be expressed as a function of the derivatives of the transformed similarity variable.

$\begin{array}{ccc}\hfill \frac{\partial \eta }{\partial x}& =& \frac{1}{\sqrt{4K\phantom{\rule{0.166667em}{0ex}}t}}\hfill \\ \hfill \frac{\partial \eta }{\partial t}& =& \frac{-\eta }{2t}\hfill \\ \hfill \frac{\partial u}{\partial t}& =& \frac{du}{d\eta }\phantom{\rule{0.166667em}{0ex}}\frac{\partial \eta }{\partial t}=\frac{-\eta }{2t}\phantom{\rule{0.166667em}{0ex}}\frac{du}{d\eta }\hfill \\ \hfill \frac{\partial u}{\partial x}& =& \frac{1}{\sqrt{4K\phantom{\rule{0.166667em}{0ex}}t}}\phantom{\rule{0.166667em}{0ex}}\frac{du}{d\eta }\hfill \\ \hfill \frac{{\partial }^{2}u}{\partial {x}^{2}}& =& \frac{1}{4K\phantom{\rule{0.166667em}{0ex}}t}\frac{{d}^{2}u}{d{\eta }^{2}}\hfill \end{array}$

The PDE is now transformed into an ODE with two boundary conditions.

$\begin{array}{c}\frac{{d}^{2}u*}{d{\eta }^{2}}+2\eta \phantom{\rule{0.166667em}{0ex}}\frac{du*}{d\eta }=0\hfill \\ u*\left(\eta =0\right)=1\hfill \\ u*\left(\eta \to \infty \right)=0\hfill \end{array}$
$\begin{array}{c}\mathrm{Let}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{1.em}{0ex}}v=\frac{du*}{d\eta }\hfill \\ \frac{dv}{d\eta }+2\eta \phantom{\rule{0.166667em}{0ex}}v=0\hfill \\ \frac{dv}{v}=d\phantom{\rule{0.166667em}{0ex}}ln\phantom{\rule{0.166667em}{0ex}}v=-2\eta \phantom{\rule{0.166667em}{0ex}}d\eta \hfill \\ v={C}_{1}\phantom{\rule{0.166667em}{0ex}}{e}^{-{\eta }^{2}}\hfill \\ u*={C}_{1}\phantom{\rule{0.166667em}{0ex}}{\int }_{0}^{\eta }{e}^{-{\eta }^{2}}d\eta +{C}_{2}\hfill \\ u*\left(\eta =0\right)=1\phantom{\rule{1.em}{0ex}}⇒\phantom{\rule{1.em}{0ex}}{C}_{2}=1\hfill \\ u*\left(\eta \to \infty \right)=0={C}_{1}\phantom{\rule{0.166667em}{0ex}}{\int }_{0}^{\infty }{e}^{-{\eta }^{2}}d\eta +1\hfill \\ {C}_{1}\phantom{\rule{0.166667em}{0ex}}=\frac{-1}{{\int }_{0}^{\infty }{e}^{-{\eta }^{2}}d\eta }\hfill \\ u*\left(\eta \right)=\frac{-{\int }_{0}^{\eta }{e}^{-{\eta }^{2}}d\eta }{{\int }_{0}^{\infty }{e}^{-{\eta }^{2}}d\eta }+1=\mathrm{erfc}\left(\eta \right)\hfill \\ u*\left(x,t\right)=\mathrm{erfc}\left(\frac{x}{\sqrt{4K\phantom{\rule{0.166667em}{0ex}}t}}\right)\hfill \end{array}$

Therefore, we have a solution in terms of the combined similarity variable that is a solution of the PDE, BC, and IC.

## Complex potential for irrotational flow

Incompressible, irrotational flows in two dimensions can be easily solved in two dimensions by the process of conformal mapping in the complex plane. First we will review the kinematic conditions that lead to the PDE and boundary conditions. Because the flow is irrotational, the velocity is the gradient of a velocity potential. Because the flow is solenoidal, the velocity is also the curl of a vector potential. Because the flow is two dimensional, the vector potential has only one non-zero component that is identified as the stream function. The kinematic condition at solid boundaries is that the normal component of velocity is zero. No condition is placed on the tangential component of velocity at solid surfaces because the fluid must be inviscid in order to be irrotational.

$\begin{array}{ccc}\hfill \mathbf{v}& =& \nabla \varphi =\left(\frac{\partial \varphi }{\partial x},\frac{\partial \varphi }{\partial y},0\right)=\left({v}_{x},{v}_{y},{v}_{z}\right)\hfill \\ \hfill \nabla •\mathbf{v}& =& 0={\nabla }^{2}\varphi \hfill \\ \hfill \mathbf{v}& =& \nabla ×\mathbf{A}=\left(\frac{\partial {A}_{3}}{\partial y},\phantom{\rule{0.277778em}{0ex}}\frac{-\partial {A}_{3}}{\partial x},\phantom{\rule{0.277778em}{0ex}}0\right)\phantom{\rule{1.em}{0ex}}\hfill \\ \hfill & =& \left(\frac{\partial \psi }{\partial y},\phantom{\rule{0.277778em}{0ex}}\frac{-\partial \psi }{\partial x},\phantom{\rule{0.277778em}{0ex}}0\right)=\left({v}_{x},{v}_{y},{v}_{z}\right)\hfill \\ \hfill \nabla ×\mathbf{v}& =& w=0={\nabla }^{2}\psi \hfill \\ \hfill \mathbf{v}& =& \nabla \varphi =\nabla ×\mathbf{A}\hfill \\ \hfill \left(\frac{\partial \varphi }{\partial x},,,\phantom{\rule{0.277778em}{0ex}},\frac{\partial \varphi }{\partial y},,,0\right)& =& \left(\frac{\partial \psi }{\partial y},\phantom{\rule{0.277778em}{0ex}}\frac{-\partial \psi }{\partial x},\phantom{\rule{0.277778em}{0ex}}0\right)\hfill \\ \hfill \mathrm{or}\phantom{\rule{1.em}{0ex}}\frac{\partial \varphi }{\partial x}& =& \frac{\partial \psi }{\partial y}\hfill \\ \hfill \mathrm{and}\phantom{\rule{1.em}{0ex}}\frac{\partial \varphi }{\partial y}& =& \frac{-\partial \psi }{\partial x}\hfill \end{array}$

Both functions are a solution of the Laplace equation, i.e., they are harmonic and the last pair of equations corresponds to the Cauchy-Riemann conditions if $\varphi$ and $\psi$ are the real and imaginary conjugate parts of a complex function, $w\left(z\right)$ .

$\begin{array}{ccc}\hfill w\left(z\right)& =& \varphi \left(z\right)+i\phantom{\rule{0.166667em}{0ex}}\psi \left(z\right)\hfill \\ \hfill z& =& x+i\phantom{\rule{0.166667em}{0ex}}y\hfill \\ & =& r\phantom{\rule{0.166667em}{0ex}}{e}^{i\theta }=r\phantom{\rule{0.166667em}{0ex}}\left(cos\theta +i\phantom{\rule{0.166667em}{0ex}}sin\theta \right),\phantom{\rule{1.em}{0ex}}r=\left|z\right|={\left({x}^{2}+{y}^{2}\right)}^{1/2},\phantom{\rule{1.em}{0ex}}\theta =arctan\left(y/x\right)\hfill \\ & \mathrm{or}& \\ \hfill \varphi \left(z\right)& =& \mathrm{real}\left[w\left(z\right)\right]\hfill \\ \hfill \psi \left(z\right)& =& \mathrm{imaginary}\left[w\left(z\right)\right]\hfill \end{array}$

The Cauchy-Riemann conditions are the necessary and sufficient condition for the derivative of a complex function to exist at a point ${z}_{o}$ , i.e., for it to be analytical . The necessary condition can be illustrated by equating the derivative of $w\left(z\right)$ taken along the real and imaginary axis.

$\begin{array}{ccc}\hfill \frac{dw\left(z\right)}{dz}& =& \mathrm{Re}\left({w}^{\text{'}}\left(z\right)\right)+i\mathrm{Im}\left({w}^{\text{'}}\left(z\right)\right)\hfill \\ & =& lim\frac{\delta \varphi +i\delta \psi }{\delta x+i0}=\frac{\partial \varphi }{\partial x}+i\frac{\partial \psi }{\partial x}\hfill \\ & =& lim\frac{\delta \varphi +i\delta \psi }{0+i\delta y}\frac{i}{i}=\frac{\partial \psi }{\partial y}-i\frac{\partial \varphi }{\partial y}\hfill \\ \hfill & & \therefore \phantom{\rule{0.277778em}{0ex}}\frac{\partial \varphi }{\partial x}=\frac{\partial \psi }{\partial y}\hfill \\ \hfill \mathrm{and}\phantom{\rule{0.277778em}{0ex}}\frac{\partial \psi }{\partial x}& =& -\frac{\partial \varphi }{\partial y},\phantom{\rule{1.em}{0ex}}\mathrm{Q}.\mathrm{E}.\mathrm{D}.\hfill \end{array}$

Also, if the functions have second derivatives, the Cauchy-Riemann conditions imply that each function satisfies the Laplace equation.

Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
hi
Loga
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
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!