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

Page 7 / 13

\begin{matrix} \frac{\partial u}{\partial t} = K \frac{\partial^{2} u}{\partial x^{2}}, t > 0, x > 0 \\ u (x, 0) = u_{I C} \\ u (0, t) = u_{B C} \end{matrix}

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

\begin{matrix} u * & = & \frac{u - u_{I C}}{u_{o}} \\ t * & = & \frac{t}{t_{o}} \\ x * & = & \frac{x}{x_{o}} \\ \frac{\partial u *}{\partial t *} & = & [\frac{K t_{o}}{x_{o}^{2}}] \frac{\partial^{2} u *}{\partial x *^{2}} \\ u * (x *, 0) & = & 0 \\ u * (0, t *) & = & [\frac{u_{B C} - u_{I C}}{u_{o}}] = 1, \Rightarrow u_{o} = u_{B C} - u_{I C} \end{matrix}

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{matrix} [\frac{K t_{o}}{x_{o}^{2}}] = \frac{1}{4} \Rightarrow η = \frac{x}{\sqrt{4 K t}} is dimensionless \\ u (x, t) = u (η) \end{matrix}

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

\begin{matrix} \frac{\partial η}{\partial x} & = & \frac{1}{\sqrt{4 K t}} \\ \frac{\partial η}{\partial t} & = & \frac{- η}{2 t} \\ \frac{\partial u}{\partial t} & = & \frac{d u}{d η} \frac{\partial η}{\partial t} = \frac{- η}{2 t} \frac{d u}{d η} \\ \frac{\partial u}{\partial x} & = & \frac{1}{\sqrt{4 K t}} \frac{d u}{d η} \\ \frac{\partial^{2} u}{\partial x^{2}} & = & \frac{1}{4 K t} \frac{d^{2} u}{d η^{2}} \end{matrix}

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

\begin{matrix} \frac{d^{2} u *}{d η^{2}} + 2 η \frac{d u *}{d η} = 0 \\ u * (η = 0) = 1 \\ u * (η \to \infty) = 0 \end{matrix}

\begin{matrix} Let v = \frac{d u *}{d η} \\ \frac{d v}{d η} + 2 η v = 0 \\ \frac{d v}{v} = d ln v = - 2 η d η \\ v = C_{1} e^{- η^{2}} \\ u * = C_{1} \int_{0}^{η} e^{- η^{2}} d η + C_{2} \\ u * (η = 0) = 1 \Rightarrow C_{2} = 1 \\ u * (η \to \infty) = 0 = C_{1} \int_{0}^{\infty} e^{- η^{2}} d η + 1 \\ C_{1} = \frac{- 1}{\int_{0}^{\infty} e^{- η^{2}} d η} \\ u * (η) = \frac{- \int_{0}^{η} e^{- η^{2}} d η}{\int_{0}^{\infty} e^{- η^{2}} d η} + 1 = erfc (η) \\ u * (x, t) = erfc (\frac{x}{\sqrt{4 K t}}) \end{matrix}

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{matrix} v & = & \nabla ϕ = (\frac{\partial ϕ}{\partial x}, \frac{\partial ϕ}{\partial y}, 0) = (v_{x}, v_{y}, v_{z}) \\ \nabla • v & = & 0 = \nabla^{2} ϕ \\ v & = & \nabla \times A = (\frac{\partial A_{3}}{\partial y}, \frac{- \partial A_{3}}{\partial x}, 0) \\ = & (\frac{\partial ψ}{\partial y}, \frac{- \partial ψ}{\partial x}, 0) = (v_{x}, v_{y}, v_{z}) \\ \nabla \times v & = & w = 0 = \nabla^{2} ψ \\ v & = & \nabla ϕ = \nabla \times A \\ (\frac{\partial ϕ}{\partial x}, \frac{\partial ϕ}{\partial y}, 0) & = & (\frac{\partial ψ}{\partial y}, \frac{- \partial ψ}{\partial x}, 0) \\ or \frac{\partial ϕ}{\partial x} & = & \frac{\partial ψ}{\partial y} \\ and \frac{\partial ϕ}{\partial y} & = & \frac{- \partial ψ}{\partial x} \end{matrix}

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 $ϕ$ and $ψ$ are the real and imaginary conjugate parts of a complex function, $w (z)$ .

\begin{matrix} w (z) & = & ϕ (z) + i ψ (z) \\ z & = & x + i y \\ = & r e^{i θ} = r (cos θ + i sin θ), r = |z| = {(x^{2} + y^{2})}^{1 / 2}, θ = arctan (y / x) \\ or \\ ϕ (z) & = & real [w (z)] \\ ψ (z) & = & imaginary [w (z)] \end{matrix}

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 $w (z)$ taken along the real and imaginary axis.

\begin{matrix} \frac{d w (z)}{d z} & = & Re (w^{'} (z)) + i Im (w^{'} (z)) \\ = & lim \frac{δ ϕ + i δ ψ}{δ x + i 0} = \frac{\partial ϕ}{\partial x} + i \frac{\partial ψ}{\partial x} \\ = & lim \frac{δ ϕ + i δ ψ}{0 + i δ y} \frac{i}{i} = \frac{\partial ψ}{\partial y} - i \frac{\partial ϕ}{\partial y} \\ ∴ \frac{\partial ϕ}{\partial x} = \frac{\partial ψ}{\partial y} \\ and \frac{\partial ψ}{\partial x} & = & - \frac{\partial ϕ}{\partial y}, Q . E . D . \end{matrix}

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

Questions & Answers

what is phylogeny

Odigie Reply

evolutionary history and relationship of an organism or group of organisms

AI-Robot

Deng

what is biology

Hajah Reply

the study of living organisms and their interactions with one another and their environments

AI-Robot

what is biology

Victoria Reply

HOW CAN MAN ORGAN FUNCTION

Alfred Reply

the diagram of the digestive system

Assiatu Reply

allimentary cannel

Ogenrwot

How does twins formed

William Reply

They formed in two ways first when one sperm and one egg are splited by mitosis or two sperm and two eggs join together

Oluwatobi

what is genetics

Josephine Reply

Genetics is the study of heredity

Misack

how does twins formed?

Misack

What is manual

Hassan Reply

discuss biological phenomenon and provide pieces of evidence to show that it was responsible for the formation of eukaryotic organelles

Joseph Reply

what is biology

Yousuf Reply

the study of living organisms and their interactions with one another and their environment.

Wine

discuss the biological phenomenon and provide pieces of evidence to show that it was responsible for the formation of eukaryotic organelles in an essay form

Joseph Reply

what is the blood cells

Shaker Reply

list any five characteristics of the blood cells

Shaker

lack electricity and its more savely than electronic microscope because its naturally by using of light

Abdullahi Reply

advantage of electronic microscope is easily and clearly while disadvantage is dangerous because its electronic. advantage of light microscope is savely and naturally by sun while disadvantage is not easily,means its not sharp and not clear

Abdullahi

cell theory state that every organisms composed of one or more cell,cell is the basic unit of life

Abdullahi

is like gone fail us

DENG

cells is the basic structure and functions of all living things

Ramadan

What is classification

ISCONT Reply

is organisms that are similar into groups called tara

Yamosa

in what situation (s) would be the use of a scanning electron microscope be ideal and why?

Kenna Reply

A scanning electron microscope (SEM) is ideal for situations requiring high-resolution imaging of surfaces. It is commonly used in materials science, biology, and geology to examine the topography and composition of samples at a nanoscale level. SEM is particularly useful for studying fine details,

Hilary

Got questions? Join the online conversation and get instant answers!

Jobilize.com Reply

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Transport phenomena. OpenStax CNX. May 24, 2010 Download for free at http://cnx.org/content/col11205/1.1

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Transport phenomena' conversation and receive update notifications?

Ask

©flickr:	Anatomy Physiology By Jemekia Weeden Start Quiz
	8 Biology 08 Photosynthesis MCQ By OpenStax Start Quiz
	2 Muscular System MCQ By Nick Swain Start Quiz
	2 Sociology 02 Sociological Research MCQ By OpenStax Start Quiz
	4 CDL Quiz - General Knowledge Part 2 By Jazzycazz Jackson Start Quiz
	Anthropology Language Culture By Richley Crapo Start Assignment
	7 BOD Urinary Tract quiz By Brooke Delaney Start Exam
	9 Physiotherapy Modalities By Rhodes Start Exam
©flickr:	CTE/STEM Technology Literacy Test By Sebastian Sieczko... Start Quiz
	Fluid Mechanics MCQ By Stephanie Redfern Start Quiz