Introduction and Preliminary Concepts

Separation of Variables

Sturm-Liouville Theory

Fourier and Laplace Transforms

The Diffusion Equation and other Similar Equations

Preliminary Concepts

Linearity, Superposition and Classifications: A partial differential operator (PDO) L is considered linear if the following is satisfied

functionsarevuandscalaraiscwherecLvLucvuL & )(

Linearity, Superposition and Classification (3)

For example,

cLvLuvvcuu

cvuy

cvuL

yy

)2()2(

)(2)(

therefore, PDO L is linear.

It follows that for a linear PDO the equationfLu

is homogeneous if 0f and heterogeneous if f is any other function.The order of a PDE is determined based on the highest order derivative that is in it. For example, the inviscid Burgers’ equation below

0 xt uuu

is first-order while the Korteweg-deVries (KdV) equation

0 xxxxt uuuu

is third-order due to the additional term.If the linear homogeneous equation

0Lvis satisfied by it also follows that will satisfy it.nuu ,...1 nnucucu ...11

Linearity, Superposition and Classification (3)

In addition, PDEs can be classified based on the sign of the discriminant B2-AC using the following criteria.

0 if elliptic,0 if hyperbolic,0 if parabolic

2

2

2

ACBACBACB

The heat/diffusion equation, which we will be discussing, is parabolic since,

0)0)((0constant a is where

222

2

ACBuu txx

Examples of solutions of the equation are shown below.

Image modified from http://home.comcast.net/~sharov/PopEcol/lec12/diffus.htmlMathematical techniques for engineers and scientists (2)

Separation of Variables

Diffusion in chemical reaction*

X=0

X=R

Catalyst wallr=kA

First order chemical reaction takes place on the catalyst wall of cylindrical containerFind out the concentration change of A inside the containerSet up PDE:D*Axx=At, or D*∂2A/∂x2 =∂A/∂tA-concentration of AD-diffusivity

*from Chemical Engineering Kinetics (6)

Diffusion in chemical reaction

X=0

X=R

Catalyst wallr=kA

First order chemical reaction takes plane on the catalyst wallFind out the concentration change of A inside the cylindrical contatinerBoundary and initial conditions:I.C t=0, A=A0

B.C1 x=0, ∂A/∂x │x=0 =0, by symmetryB.C2 x=R, D*∂A/∂x │x=R =k*A, diffusion =reaction at wall

Diffusion in chemical reaction

X=0

X=R

Catalyst wallr=kA

Run dimensionless:u=A/A0, s=x/R, τ=t/tc (=tD/R2)Then we get:tc*D/R2∂2u/∂s2 =∂u/∂τ, let tc*D/R2=1, then tc=R2/D

BCs: ∂u/∂s =0 at s=0∂u/∂s =kR/D*u=φu, φ=kR/Dat s=1

Diffusion in chemical reaction

X=0

X=R

Catalyst wallr=kA

Assume: u=F(s)*G(τ)

G(τ)*dF2/ds2=F(s)*dG(τ)/ds, 1/F(s)*dF2/ds2=1/G(τ)*dG(τ)/ds=-λn

2

Solve 1/G(τ)*dG(τ)/ds=-λn2

we get: dG(τ)=exp(-λn2τ)

For 1/F(s)*dF2/ds2=-λn2

assume F(s)=∑(Ansinλns+ Bncosλns)

Diffusion in chemical reaction

X=0

X=R

Catalyst wallr=kA

Use BC1: ∂u/∂s =0 at s=0∂u/∂s=G(τ)*dF/ds=0∑(Anλncosλn*0- Bnλnsinλn*0)=0An=0F(s)= ∑Bncosλns

Use BC2: ∂u/∂s =φu at s=1-Bnλnsinλn=φBncosλn-λntanλn=φ, λncan be solved

Diffusion in chemical reaction

X=0

X=R

Catalyst wallr=kA

u(s, τ) =∑exp(-λn2 τ) Bncosλns

How to get Bn?

τ =0, u(s,0)=1= ∑1*Bncosλns

Use orthogonality principle∫0

1 *cos(λns)ds=Bn∫0

1cos2(λns)ds

Bn = ∫01

cos(λns)ds/∫01cos2(λns)ds

Bn =4sin(λn )/(2λn+sin2λn)

When ψ is large When ψ is small

0

5

10

15

20

25

30

35

40

45

0 2 4 6 8 10 12ti me(t)

Conc

entr

atio

n A

0

5

10

15

20

25

30

35

40

45

0 2 4 6 8 10 12ti me(t)

Conc

entr

atio

n A

Strum-Liouville Problems

Second order, homogenous, linear differential equation with boundary conditions.

Solutions to such problems contain an infinite set of eigenvalues with corresponding eigenfunction solutions

What are Strum-Liouville Problems?

 

Equations from Advanced Engineering Mathematics (9)

 A simple example:

Problem from Introduction to Differential Equations (7)

 

Problem from Introduction to Differential Equations (7)

 

Problem from Introduction to Differential Equations (7)

 

Problem from Introduction to Differential Equations (7)

 

Problem from Introduction to Differential Equations (7)

 

Problem from Introduction to Differential Equations (7)

Many differential equations are in the form of the Sturm-Liouville problem.

Sturm-Liouville analysis can help determine integration constants for equations with an infinite number of terms (eigenfunctions).

How is this useful for Separation of Variables?

 

A more complex example:

Problem from Heat Conduction (8)

The boundary conditions are such that

 

Figure from Heat Conduction (8)

 

Problem from Heat Conduction (8)

 

Problem from Heat Conduction (8)

 

Problem from Heat Conduction (8)

 

Problem from Heat Conduction (8)

 

Problem from Heat Conduction (8)

 

Problem from Heat Conduction (8)

 

Problem from Heat Conduction (8)

 

Problem from Heat Conduction (8)

• Other types of problems• Wave equation, diffusion equation…

• Other types of boundaries:• Fixed, Periodic, Moving

• Handled same as in regular Sturm-Liouville Problem

Other considerations considerations:

Fourier and Laplace Transforms

)( uuxhs

xxkAu

x

xx xxxkAu

Image modified from http://www.ceb.cam.ac.uk/pages/finite-element-simulations.html

The energy balance over the infinitesimal volume of a rod shown in the figure is as follows

Diffusion Equation:

tmcucuvAcuvAuuxhskAukAu

xxxxxxxx

)()(

where, m=AΔxσc. Dividing the full equation by AΔxσc and taking the limit as x goes to 0 yields

AchsH

ckuuHvuuu xtxx & e, wher)( 22

Variables:

tyconductivi thermaltcoefficienfer heat trans

area sectional -crossdifference etemperatur

rod of speedncecircumfere

density massheat specific

khA

uuvs

c

Heat Conduction and the Heat Equation (5)

Initial/Boundary Conditions: For all subsequent calculations we assume that diffusion in the rod is one-dimensional and thus set H and and v to zero.

txx uu 2

The initial condition is simply a set function for u(x,t) over the length of the rod at t=0 and is written thus,

)0( )()0,( Lxxfxu

The boundary condition can fall into one of three categories.Dirichlet boundary condition:

)(),0( ttu

Fundamental Solutions (4)

Meaning that the left end of the bar is kept at a set temperature.Neumann boundary condition:

)(),0(),0( ttkutq x

This boundary condition specifies the heat flow out of the rod. Particularly, setting 0)( t means insulating the left end of the rod.

Heat Conduction and the Heat Equation (5)

Initial/Boundary Conditions (cont`d):

Robin boundary condition:Starting with Newton’s Law of Cooling

fx

fx

f

hutLhutLkutLuuhtLku

tLuuhtL

),(),()],([),(

)],([),(

)( fs uuhAq

where, us and uf are temperatures of the solid and fluidrespectively. It then follows that,

Heat Conduction of an Infinite Rod (Fourier Transforms)

xx)(xf

)()0,( xfxu t

For non-periodic functions [f(x)] the heat equation can be solved using Fourier exponential transform and a known Gaussian function g(x,t) the steps follow…..

Mathematical techniques for engineers and scientists (2)

),(),();,( tsUdxtxuesxtxuF isx

The Fourier exponential transform can be written as

Using the above definition we can write

The heat equation can then be rewritten as

We can solve for U using the initial condition to yield

),();,(

),();,( 2

tsUsxtxuFtsUssxtxuF

tt

xx

sxfF)where, G(ssGsUICtUsut

);()()0,(:0,022

sesGtsU ts ,)(),(22

yields ))()()()(( n theoremconvolutio theand

)2

1( ansformFourier tr inverse theUsing2222 )2/(

dgxfsgsf

ee xs

txet

txgdtxgftxu22 4/

21),( where, )()(),(

deft

txu tx

22 4/)()(

)(21),(

Thus,

Mathematical techniques for engineers and scientists (2)

……(1)

Special Case 1:

The simplest solution to the heat equation is for Fxf constant)(

asrewritten becan (1)equation 2

settingby txa

datet

Ftxu a

22

),(2

Assuming that all terms except a are constant the function can be integrated to yield

FFtxu ),(

Thus, for the limiting case of f(x) = constant the heat will also equal that constant.

Advanced engineering mathematics (9)

Special Case 2:

In this case, assume the initial condition is as follows

Using these initial conditions the heat equation can be solved to yield

))2

(1(*504100),(

0

4)( 2

txerfde

ttxu t

x

since the integral can then be rewritten as

dezerfz

0

22)(

Beny Neta Presentation (1)

-400 -300 -200 -100 0 100 200 300 4000

102030405060708090

100t=10t=500t=1000t=10000100000100000001000000000

x

u(x,t)

From the chart we can see the function “smoothing” out at large values of t. This is due to the behavior of the error function, erf(∞)=1 and erf(~0)=0. Thus, for positive values of x at small times u(x,t) approaches F and at large times u(x,t) approaches F/2. For negative values of x the negative sign can be taken outside of the error function so the value of the error function is subtracted from 1. Thus at small times u(x,t) approaches 0 and at large times u(x,t) approaches F/2.

Typical Behavior

Images taken from http://en.wikipedia.org/wiki/File:Heatequation_exampleB.gif

The following curves are of temperature collected at different times. From the plots a “smoothing” effect can again be seen as the temperature along entire length of the rod approaches an equilibrium value.

t1

t2

x

Kto

figure modified from: http://mobjectivist.blogspot.com/2010_06_01_archive.html

Similar to the error function the integral of equation (1) can be rewritten as the kernel

Special Case 3:

tetxK

tx

2);(

22 4/)(

and the function becomes….

dtxKftxu );()(),(

shown that becan it 2

settingby txa

1);(),(

dtxKtxu

)( ,0 as hus xKtT

From the graph you can see that similar “smoothing” process occurs for the Kernel as for the temperature profile. It is also important to note that the Kernel is a response function to an initial temperature function described by the delta function

Heat Conduction of a Semi-Infinite Rod (Laplace Transforms)

x)(xf

)()0,( xfxu t

)(),0(

tgtu

)0,0(,2 txuu txx 0,0),(),(),0(: txastxutftuBC

)0( ,0)0,( : xxuIC

For non-periodic functions [f(x)] the heat equation can be solved in the semi-infinite domain using Laplace exponential transform the steps follow…..

Mathematical techniques for engineers and scientists (2)

2/3

4 2

2

2t

eex

t

xpx

………(2)

Using the inverse Laplace transform U(x,p) becomes,

t

dtgftgf0

)()())(*(

)(2)(),(

2/3

)4/( 22

txetgtxu

tx

Then using the convolution theorem

u(x,t) becomes,

detgxtxut x

0

2/3

4/ 22

)(2

),(

Mathematical techniques for engineers and scientists (2)

),(),();,( pxUdttxuepttxu ptL

The Laplace exponential transform can be written as

Using the above definition we can write

The heat equation can then be rewritten as

We can solve for U using the boundary condition to yield

),();,(

),();,(pxpUpttxupxUpttxu

t

xxxx

LL

)()(0),()(),0(:

0,02

tfpGwherexaspxUpGpUBC

xUpU xx

L

pepGpxUpx

,)(),(

Mathematical techniques for engineers and scientists (2)

Special Case :The simplest solution to the heat equation is for

Ftconsxg tan)(

By setting 2

22

4xa we can solve

for

22

2

4 ax

and da

axd

32

2

42

Thus equation (2) becomes

daeF tx

a

2

0

22

or otherwise

)2

(*),(t

xerfcFtxu

Advanced engineering mathematics (9)

For F=50 and α=1 u(x,t) is plotted below

0 100 200 300 400 500 600 700 800 90005

101520253035404550

t=10 t=500 t=1000 t=10000100000 10000000 1000000000

x

u(x,t)From the chart we can see a similar “smoothing” process as in the case of the infinite domain at large values of t. Again, erf(∞)=1 and erf(~0)=0. Thus at small times u(x,t) approaches F and at large times u(x,t) approaches 0.

References

1.Neta, Beny. "PARTIAL DIFFERENTIAL EQUATIONS MA 3132 LECTURE NOTES." math.nps.navy. Department of Mathematics Naval Postgraduate School, n.d. Web. 20 Oct. 2011. <www.math.nps.navy.mil/~bneta/pde.pdf>.

2.Andrews, Larry C., and Ronald L. Phillips. Mathematical techniques for engineers and scientists. Bellingham, Wash.: SPIE Press, 2003. Print.

3."Linearity, Superposition and Classification." math.unl. N.p., n.d. Web. 21 Oct. 2011. <www.math.unl.edu/~scohn1/8423/class.pdf>.

4."CHAPTER 2: The Diffusion Equation." mth.pdx. N.p., n.d. Web. 21 Oct. 2011. <www.mth.pdx.edu/~marek/mth510pde/notes%202.pdf>.

5."Heat Conduction and the Heat Equation." ncsu. N.p., n.d. Web. 21 Oct. 2011. <www4.ncsu.edu/~rsmith/MA573_F09/Heat_Equation.pdf>.

6.Platnaik. Introduction to Differential Equations. PHI Learning Pvt. Ltd.7.Jiji, Latif M. Heat Conduction. 3rd Ed. Springer Publishing. 2009.8.Greenberg, Michael D.. Advanced engineering mathematics. 2nd ed.

Upper Saddle River, N.J.: Prentice Hall, 1998. Print.