MATH 461 Fourier Series and Boundary ValueProblems

Chapter II Separation of Variables

Greg Fasshauer

Department of Applied MathematicsIllinois Institute of Technology

Fall 2015

fasshaueriitedu MATH 461 ndash Chapter 2 1

Outline

1 Model Problem

2 Linearity

3 Heat Equation for a Finite Rod with Zero End Temperature

4 Other Boundary Value Problems

5 Laplacersquos Equation

fasshaueriitedu MATH 461 ndash Chapter 2 2

Model Problem

For much of the following discussion we will use the following 1D heatequation with constant values of c ρK0 as a model problem

part

parttu(x t) = k

part2

partx2 u(x t) +Q(x t)

cρ for 0 lt x lt L t gt 0

with initial condition

u(x 0) = f (x) for 0 lt x lt L

and boundary conditions

u(0 t) = T1(t) u(L t) = T2(t) for t gt 0

fasshaueriitedu MATH 461 ndash Chapter 2 4

Linearity

Linearity will play a very important role in our work

DefinitionThe operator L is linear if

L(c1u1 + c2u2) = c1L(u1) + c2L(u2)

for any constants c1 c2 and functions u1u2

Differentiation and integration are linear operations

ExampleConsider ordinary differentiation of a univariate function ieL = d

dx Then

ddx

(c1f1 + c2f2) (x) = c1d

dxf1(x) + c2

ddx

f2(x)

fasshaueriitedu MATH 461 ndash Chapter 2 6

Linearity

ExampleThe same is true for partial derivatives

part

partt(c1u1 + c2u2) (x t) = c1

part

parttu1(x t) + c2

part

parttu2(x t)

In particular the heat operator partpartt minus k part2

partx2 is linearTherefore the heat equation

part

parttu(x t)minus k

part2

partx2 u(x t) = 0

is a linear PDE If the given right-hand side function is identicallyzero then the PDE is called homogeneous

RemarkA linear homogeneous equation Lu = 0 always has at least the trivialsolution u equiv 0

fasshaueriitedu MATH 461 ndash Chapter 2 7

Linearity

ExampleAre the following equations linear or nonlinear homogeneous ornonhomogeneous

part2

partx2 u(x y) +part2

party2 u(x y) = f (x y)

is linear and generally nonhomogeneous (Poissonrsquos equation)

part2

partx2 u(x y) +part2

party2 u(x y) = 0

is linear and homogeneous (Laplacersquos equation)

part

parttu(x t)minus κ part

partx

[u(x t)

part

partxu(x t)

]= 0

is nonlinear and homogeneous (nonlinear heat equation thermalconductivity depends on temperature)

fasshaueriitedu MATH 461 ndash Chapter 2 8

Linearity

Theorem (Superposition Principle)If u1 and u2 are both solutions of a linear homogeneous equationLu = 0 and c1 c2 are arbitrary constants then c1u1 + c2u2 is also asolution of Lu = 0

ProofWe are given a linear operator L and functions u1u2 such that

Lu1 = 0 Lu2 = 0

We need to show that

L (c1u1 + c2u2) = 0

Straightforward computation gives

L (c1u1 + c2u2)L linear

= c1 Lu1︸︷︷︸=0

+c2 Lu2︸︷︷︸=0

= 0

fasshaueriitedu MATH 461 ndash Chapter 2 9

Heat Equation for a Finite Rod with Zero End Temperature

We want to solve the PDE

part

parttu(x t) = k

part2

partx2 u(x t) for 0 lt x lt L t gt 0 (1)

with initial condition

u(x 0) = f (x) for 0 lt x lt L (2)

and boundary conditions

u(0 t) = u(L t) = 0 for t gt 0 (3)

This is a linear and homogeneous PDE with linear and homogeneousBCs mdash a perfect candidate for the technique of separation of variables

fasshaueriitedu MATH 461 ndash Chapter 2 11

Heat Equation for a Finite Rod with Zero End Temperature

Separation of VariablesThis technique often just ldquoworksrdquo especially for linear homogeneousPDEs and BCs by magically() converting the PDE to a pair of ODEsmdash and those we should be able to solve1The starting point is to take the unknown function u = u(x t) andldquoseparate its variablesrdquo ie to make the Ansatz

u(x t) = ϕ(x)G(t) (4)

In other words we just guess that the solution u is of this special formand hope for the best

RemarkYou may remember another form of separation of variables from MATH 152 orMATH 252 (separable ODEs) In that case the right-hand side of the DE isgiven with separated variables ie dy

dx = f (x)g(y) Now we assume (or hope)that the solution is separable

1If you donrsquot remember you might want to review Chapters 2 and 5 (maybe also 4)of something like [Zill]

fasshaueriitedu MATH 461 ndash Chapter 2 12

Heat Equation for a Finite Rod with Zero End Temperature

If u is of the form u(x t) = ϕ(x)G(t) then

part

parttu(x t) = ϕ(x)

ddt

G(t)

part2

partx2 u(x t) =d2

dx2ϕ(x)G(t)

Therefore the PDE (1) turns into

ϕ(x)ddt

G(t) = kd2

dx2ϕ(x)G(t)

Now we separate variables

1kG(t)

ddt

G(t)︸ ︷︷ ︸depends only on t

=1

ϕ(x)

d2

dx2ϕ(x)︸ ︷︷ ︸depends only on x

The only way for this equation to be true for all x and t is if both sidesare constant (independent of x and t)

fasshaueriitedu MATH 461 ndash Chapter 2 13

Heat Equation for a Finite Rod with Zero End Temperature

Therefore1

kG(t)ddt

G(t) =1

ϕ(x)

d2

dx2ϕ(x) = minusλ (5)

The constant λ is known as the separation constantThe ldquominusrdquo sign appears mostly for cosmetic purposes

Equations (5) give two separate ODEs

ϕprimeprime(x) = minusλϕ(x) (6)Gprime(t) = minusλkG(t) (7)

fasshaueriitedu MATH 461 ndash Chapter 2 14

Heat Equation for a Finite Rod with Zero End Temperature

Before we attempt to solve the two ODEs we note that from the BCs(3) and the Ansatz (4) we get (assuming G(t) 6= 0)

u(0 t) = ϕ(0)G(t) = 0=rArr ϕ(0) = 0 (8)

and

u(L t) = ϕ(L)G(t) = 0=rArr ϕ(L) = 0 (9)

Together (6) (8) and (9) form a two-point ODE boundary valueproblem

fasshaueriitedu MATH 461 ndash Chapter 2 15

Heat Equation for a Finite Rod with Zero End Temperature

RemarkNote that the initial condition (2) u(x 0) = f (x) does not become aninitial condition for (7)

Gprime(t) = minusλkG(t)

(since the IC provides spatial x information while (7) is an ODE intime t)Instead (7) provides us only with

G(t) = ceminusλkt

and we will use the initial condition (2) elsewhere later

fasshaueriitedu MATH 461 ndash Chapter 2 16

Heat Equation for a Finite Rod with Zero End Temperature

Solution of the Two-Point BVP

We now solve

ϕprimeprime(x) = minusλϕ(x)

ϕ(0) = ϕ(L) = 0

This kind of problem is discussed in detail in MATH 252 (see egChapter 5 of [Zill])The characteristic equation of this ODE is

r2 = minusλ

which is obtained from another Ansatz namely ϕ(x) = erx What are the roots r (and therefore the general solution ϕ)

fasshaueriitedu MATH 461 ndash Chapter 2 17

Heat Equation for a Finite Rod with Zero End Temperature

For a real separation constant λ there are three casesCase I λ gt 0 In this case r2 = minusλ gives us

r = plusmniradicλ

along with the general solution

ϕ(x) = c1 cos(radic

λx)

+ c2 sin(radic

λx)

Now we use the BCs

ϕ(0) = 0 = c1 cos 0︸ ︷︷ ︸=1

+c2 sin 0︸︷︷︸=0

=rArr c1 = 0

ϕ(L) = 0c1=0= c2 sin

(radicλL)

=rArr c2 = 0 or sinradicλL = 0

The solution c1 = c2 = 0 is not desirable (since it leads to the trivialsolution ϕ equiv 0) Therefore at this point we conclude

c1 = 0 and sinradicλL = 0

fasshaueriitedu MATH 461 ndash Chapter 2 18

Heat Equation for a Finite Rod with Zero End Temperature

Our conclusionsc1 = 0 and sin

radicλL = 0

do not yet specify the solution ϕ so we still have work to doNote that the equation sin

radicλL = 0 is true whenever

radicλL = nπ for any

positive integer nIn other words we get

λn =(nπ

L

)2 n = 123

Each eigenvalue λn =(nπ

L

)2 gives us an eigenfunction

ϕn(x) = c2 sinnπL

x n = 123

mdash each one of which is a solution to the BVP

fasshaueriitedu MATH 461 ndash Chapter 2 19

Heat Equation for a Finite Rod with Zero End Temperature

Case II λ = 0 In this case r2 = minusλ = 0 implies r = 0 or

ϕ(x) = c1 + c2x

The BCs lead toϕ(0) = 0 = c1

ϕ(L) = 0c1=0= c2L =rArr c2 = 0

Now we have only the solution c1 = c2 = 0 ie the trivial solutionϕ equiv 0Since by definition ϕ equiv 0 cannot be an eigenfunction this implies thatλ = 0 is not an eigenvalue for our BVPIn other words this case does not contribute to the solution

fasshaueriitedu MATH 461 ndash Chapter 2 20

Heat Equation for a Finite Rod with Zero End Temperature

Case III λ lt 0 Now r2 = minusλ︸︷︷︸gt0

implies r = plusmnradicminusλ or

ϕ(x) = c1eradicminusλx + c2eminus

radicminusλx

The BCs lead to

ϕ(0) = 0 = c1 + c2 =rArr c2 = minusc1

ϕ(L) = 0c2=minusc1= c1e

radicminusλL minus c1eminus

radicminusλL

or c1eradicminusλL = c1eminus

radicminusλL

The last row can only be true if c1 = 0 or L = 0 The latter does notmake any physical sense so we again have only the trivial solutionc1 = c2 = 0 (or ϕ equiv 0)

RemarkInstead of ϕ(x) = c1e

radicminusλx + c2eminus

radicminusλx we could have used the

alternate formulation ϕ(x) = c1 cosh(radicminusλx

)+ c2 sinh

(radicminusλx

)mdash to

the same effectfasshaueriitedu MATH 461 ndash Chapter 2 21

Heat Equation for a Finite Rod with Zero End Temperature

Summary (so far)

The two-point BVP

ϕprimeprime(x) = minusλϕ(x)

ϕ(0) = ϕ(L) = 0

has eigenvalues

λn =(nπ

L

)2 n = 123

and eigenfunctions

ϕn(x) = sinnπx

L n = 123

and together with the solution for G found above we have that

fasshaueriitedu MATH 461 ndash Chapter 2 22

Heat Equation for a Finite Rod with Zero End Temperature

Summary (cont)

The PDE-BVP

part

parttu(x t) = k

part2

partx2 u(x t) for 0 lt x lt L t gt 0

u(0 t) = u(L t) = 0 for t gt 0

has solutions

un(x t) = ϕn(x)Gn(t)

= sinnπx

Leminusλnkt

= sinnπx

Leminusk( nπ

L )2t n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 23

Heat Equation for a Finite Rod with Zero End Temperature

RemarkNote that so far we have not yet used the initial conditionu(x 0) = f (x)Physically the temperature should decrease to zero everywhere inthe rod ie

limtrarrinfin

u(x t) = 0

We see that each

un(x t) = sinnπx

Leminusk( nπ

L )2t

satisfies this property

fasshaueriitedu MATH 461 ndash Chapter 2 24

Heat Equation for a Finite Rod with Zero End Temperature

By the principle of superposition any linear combination of unn = 123 will also be a solution ie

u(x t) =sum

n

Bn sinnπx

Leminusk( nπ

L )2t (10)

for arbitrary constants Bn is also a solution

To get a solution u which also satisfies the initial condition we will haveto choose the Bns accordingly

Notice that the above solution implies

u(x 0) =sum

n

Bn sinnπx

L

for the initial condition u(x 0) = f (x)

fasshaueriitedu MATH 461 ndash Chapter 2 25

Heat Equation for a Finite Rod with Zero End Temperature

Fourier in ActionIf an air column a string or some other object vibrates at a specificfrequency it will produce a sound We illustrate this in the MATLAB

script SoundwavesmIf only one single frequency is present then we have a sine wave

Most of the time we hear a more complex sound (with overtones orharmonics) This corresponds to a weighted sum of sine waves withdifferent frequencies

fasshaueriitedu MATH 461 ndash Chapter 2 26

Heat Equation for a Finite Rod with Zero End Temperature

On March 27 2008 researchers announced that they had found asound recording made by Eacutedouard-Leacuteon Scott de Martinville on April9 1860 mdash 17 years before Thomas Edison invented the phonograph

Figure The phonautograph a device that scratched sound waves onto asheet of paper blackened by the smoke of an oil lamp

Figure A typical phonautogram

And this is what it sounds likefasshaueriitedu MATH 461 ndash Chapter 2 27

Au Clair de la Lune--French folk song (1860 Phonautogram)

Phonautogram by Eacutedouard-Leacuteon Scott de Martinville

Recorded on April 9 1860 1861 Deposit Acadeacutemie des Sciences (No 324) Number 5

2008

Other

10840803

eng - wwwfirstsoundsorg

Heat Equation for a Finite Rod with Zero End Temperature

fasshaueriitedu MATH 461 ndash Chapter 2 28

Fourier analysis can be used to take apart and analyze complexsoundsThe MATLAB GUI touchtone lets us analyze which buttons werepressed on a touch-tone phoneWave-like phenomena also play a fundamental role in

heat flow and other diffusion problems (eg the spreading ofpollutants)vibration problemssound and image file compression (eg MP3 or JPEG files)filtering of noisy audio or video

Heat Equation for a Finite Rod with Zero End Temperature

ExampleIf the initial temperature distribution f is of the form

f (x) = sinmπx

L

where m is fixed then

u(x t) = um(x t) = sinmπx

Leminusk( mπ

L )2t

will satisfy the entire heat equation problem ie the series solution

(10) collapses to just one term so Bn =

0 if n 6= m1 if n = m

fasshaueriitedu MATH 461 ndash Chapter 2 29

Heat Equation for a Finite Rod with Zero End Temperature

ExampleIf the initial temperature distribution f is of the form

f (x) =Msum

n=1

Bn sinnπx

L

then

u(x t) =Msum

n=1

Bnun(x t) =Msum

n=1

Bn sinnπx

Leminusk( nπ

L )2t

will satisfy the entire heat equation problem In this case the seriessolution (10) is finite

fasshaueriitedu MATH 461 ndash Chapter 2 30

Heat Equation for a Finite Rod with Zero End Temperature

What about other initial temperature distributions f

The general idea will be to use an infinite series (ie a Fourier series)to represent an arbitrary f ie we will show that any (with some mildrestrictions) function f can be written as

f (x) =infinsum

n=1

Bn sinnπx

L

and so the solution of the heat equation (1) (2) and (3) with arbitrary fis given by

u(x t) =infinsum

n=1

Bn sinnπx

Leminusk( nπ

L )2t

Remaining question How do the coefficients Bn depend on f

fasshaueriitedu MATH 461 ndash Chapter 2 31

Heat Equation for a Finite Rod with Zero End Temperature

Orthogonality (of vectors)

Earlier we noted that the angle θ between two vectors a and b isrelated to the dot product by

cos θ =a middot bab

and therefore the vectors a and b are orthogonal a perp b (orperpendicular ie θ = π

2 ) if and only if a middot b = 0In terms of the vector components this becomes

a perp b lArrrArr a middot b =nsum

i=1

aibi = 0

If AB are two sets of vectors then A is orthogonal to B if a middot b = 0for every a isin A and every b isin BA is an orthogonal set (or simply orthogonal) if a middot b = 0 for everyab isin A with a 6= b

fasshaueriitedu MATH 461 ndash Chapter 2 32

Heat Equation for a Finite Rod with Zero End Temperature

Orthogonality (of functions)We can let our ldquovectorsrdquo be functions f and g defined on some interval[ab] Then f and g are orthogonal on [ab] with respect to the weightfunction ω if and only if 〈f g〉 = 0 where the inner product is defined by

〈f g〉 =

int b

af (x)g(x)ω(x) dx

RemarkOrthogonality of vectors is usually discussed in linear algebrawhile orthogonality of functions is a topic that belongs to functionalanalysisNote that orthogonality of functions always is specified relative toan interval and a weight functionThere are many different classes of orthogonal functions such aseg orthogonal polynomials trigonometric functions or waveletsOrthogonality is one of the most fundamental (and useful)concepts in mathematics

fasshaueriitedu MATH 461 ndash Chapter 2 33

Heat Equation for a Finite Rod with Zero End Temperature

Example1 Show that the polynomials p1(x) = 1 and p2(x) = x are

orthogonal on the interval [minus11] with respect to the weightfunction ω(x) equiv 1

2 Determine the constants α and β such that a third polynomial p3of the form

p3(x) = αx2 + βx minus 1

is orthogonal to both p1 and p2

The polynomials p1p2p3 are known as the first three Legendrepolynomials

SolutionAltogether we need to show thatint 1

minus1pj(x)pk (x)ω(x) dx = 0 whenever j 6= k = 123

fasshaueriitedu MATH 461 ndash Chapter 2 34

Heat Equation for a Finite Rod with Zero End Temperature

Solution (cont)1 p1(x) = 1 and p2(x) = x are orthogonal sinceint 1

minus1p1(x)︸ ︷︷ ︸=1

p2(x)︸ ︷︷ ︸=x

ω(x)︸︷︷︸=1

dx =

int 1

minus1x dx =

x2

2

∣∣∣∣1minus1

= 0

Of course we also know that the integral is zero since we integratean odd function over an interval symmetric about the origin

fasshaueriitedu MATH 461 ndash Chapter 2 35

Heat Equation for a Finite Rod with Zero End Temperature

Solution (cont)2 We need to find α and β such that bothint 1

minus1p1(x)p3(x)ω(x) dx =

int 1

minus1p2(x)p3(x)ω(x) dx = 0

This leads toint 1

minus1

(αx2 + βx minus 1

)dx =

[α

x3

3+ β

x2

2minus x

]1

minus1=

23αminus 2

= 0

andint 1

minus1x(αx2 + βx minus 1

)dx =

[α

x4

4+ β

x3

3minus x2

2

]1

minus1=

12β

= 0

so that we have α = 3 β = 0 and

p3(x) = 3x2 minus 1

fasshaueriitedu MATH 461 ndash Chapter 2 36

Heat Equation for a Finite Rod with Zero End Temperature

Orthogonality of Sines

We now show that the functionssin

πxL sin

2πxL sin

3πxL

are orthogonal on [0L] with respect to the weight ω equiv 1

To this end we need to evaluateint L

0sin

nπxL

sinmπx

Ldx

for different combinations of integers n and m

We will discuss the cases m 6= n and m = n separately

fasshaueriitedu MATH 461 ndash Chapter 2 37

Heat Equation for a Finite Rod with Zero End Temperature

Case I m 6= n Using the trigonometric identity

sin A sin B =12

(cos(Aminus B)minus cos(A + B))

we getint L

0sin

nπxL

sinmπx

Ldx =

12

int L

0

[cos

((n minus m)

πxL

)minus cos

((n + m)

πxL

)]dx

=12

[L

(n minus m)πsin((n minus m)

πxL

)minus L

(n + m)πsin((n + m)

πxL

)]L

0

=12

[L

(n minus m)π

(sin (n minus m)︸ ︷︷ ︸

integer

π

︸ ︷︷ ︸=0

minus sin 0︸︷︷︸=0

)minus L

(n + m)π

(sin (n + m)︸ ︷︷ ︸

integer

π

︸ ︷︷ ︸=0

minus sin 0︸︷︷︸=0

)]= 0

fasshaueriitedu MATH 461 ndash Chapter 2 38

Heat Equation for a Finite Rod with Zero End Temperature

Case II m = n Using the trigonometric identity

sin2 A =12

(1minus cos 2A)

we get int L

0sin2 nπx

Ldx =

12

int L

0

(1minus cos

2nπxL

)dx

=12

[x minus L

2nπsin

2nπxL

]L

0

=12

[(Lminus 0

)minus L

2nπ(

sin 2nπ︸ ︷︷ ︸=0

minus sin 0︸︷︷︸=0

)]=

L2

fasshaueriitedu MATH 461 ndash Chapter 2 39

Heat Equation for a Finite Rod with Zero End Temperature

Orthogonality of Sines

In summary we haveint L

0sin

nπxL

sinmπx

Ldx =

0 if m 6= nL2 if m = n

and we have established that the set of functionssin

πxL sin

2πxL sin

3πxL

is orthogonal on [0L] with respect to the weight function ω equiv 1

RemarkLater we will also use other orthogonal sets such as cosines or sinesand cosines and other intervals of orthogonality

fasshaueriitedu MATH 461 ndash Chapter 2 40

Heat Equation for a Finite Rod with Zero End Temperature

We are now finally ready to return to the determination of thecoefficients Bn in the solution

u(x t) =infinsum

n=1

Bn sinnπx

Leminusk( nπ

L )2t

of our model problemRecall that we assumed that the initial temperature was representableas

f (x) =infinsum

n=1

Bn sinnπx

L

RemarkNote that the set of sines above was infinite This together with theorthogonality of the sines will allow us to find the Bn

fasshaueriitedu MATH 461 ndash Chapter 2 41

Heat Equation for a Finite Rod with Zero End Temperature

Start with

f (x) =infinsum

n=1

Bn sinnπx

L

multiply both sides by sin mπxL and integrate wrt x from 0 to L

=rArrint L

0f (x) sin

mπxL

dx =

int L

0

[ infinsumn=1

Bn sinnπx

L

]sin

mπxL

dx

Assume we can interchange integration and infinite summation2 thenint L

0f (x) sin

mπxL

dx =infinsum

n=1

Bn

int L

0sin

nπxL

sinmπx

Ldx︸ ︷︷ ︸

=

0 if n 6= mL2 if m = n

Therefore int L

0f (x) sin

mπxL

dx = BmL2

2This is not trivial It requires uniform convergence of the seriesfasshaueriitedu MATH 461 ndash Chapter 2 42

Heat Equation for a Finite Rod with Zero End Temperature

But int L

0f (x) sin

mπxL

dx = BmL2

is equivalent to

Bm =2L

int L

0f (x) sin

mπxL

dx m = 123

The Bm are known as the Fourier (sine) coefficients of f

fasshaueriitedu MATH 461 ndash Chapter 2 43

Heat Equation for a Finite Rod with Zero End Temperature

ExampleAssume we have a rod of length L whose left end is placed in an icebath and then the rod is heated so that we obtain a linear initialtemperature distribution (from u = 0C at the left end to u = LC at theother end) Now insulate the lateral surface and immerse both ends inan ice bath fixed at 0CWhat is the temperature in the rod at any later time tThis corresponds to the model problem

PDEpart

parttu(x t) = k

part2

partx2 u(x t) 0 lt x lt L t gt 0

IC u(x 0) = x 0 lt x lt LBCs u(0 t) = u(L t) = 0 t gt 0

fasshaueriitedu MATH 461 ndash Chapter 2 44

Heat Equation for a Finite Rod with Zero End Temperature

SolutionFrom our earlier work we know that

u(x t) =infinsum

n=1

Bn sinnπx

Leminusk( nπ

L )2t

This implies that (just plug in t = 0) u(x 0) =infinsum

n=1

Bn sinnπx

L

But we also know that u(x 0) = x so that we have the Fourier sineseries representation

x =infinsum

n=1

Bn sinnπx

L

or

Bn =2L

int L

0x sin

nπxL

dx n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 45

Heat Equation for a Finite Rod with Zero End Temperature

Solution (cont)

We now compute the Fourier coefficients of f (x) = x ie

Bn =2L

int L

0x sin

nπxL

dx n = 123

Integration by parts (with u = x dv = sin nπxL dx) yields

Bn =2L

[minusx

Lnπ

cosnπx

L

]L

0+

2L

int L

0

Lnπ

cosnπx

Ldx

=2L

[minusL

Lnπ

cos nπ + 0]

+2

nπ

[L

nπsin

nπxL

]L

0

= minus2Lnπ

cos nπ︸ ︷︷ ︸=(minus1)n

+2L

n2π2 (sin nπ minus sin 0)

ThereforeBn =

2Lnπ

(minus1)n+1 =

2Lnπ if n is oddminus 2L

nπ if n is evenfasshaueriitedu MATH 461 ndash Chapter 2 46

Heat Equation for a Finite Rod with Zero End Temperature

The solution of the previous example is illustrated in the Mathematicanotebook Heatnb

fasshaueriitedu MATH 461 ndash Chapter 2 47

Other Boundary Value Problems

A 1D Rod with Insulated Ends

We now solve the same PDE as before ie the heat equation

part

parttu(x t) = k

part2

partx2 u(x t) for 0 lt x lt L t gt 0

with initial condition

u(x 0) = f (x) for 0 lt x lt L

and new boundary conditions

partupartx

(0 t) =partupartx

(L t) = 0 for t gt 0

Since the PDE and its boundary conditions are still linear andhomogeneous we can again try separation of variablesHowever since the BCs have changed we need to go through a newderivation of the solution

fasshaueriitedu MATH 461 ndash Chapter 2 49

Other Boundary Value Problems

We again start with the Ansatz u(x t) = ϕ(x)G(t) which turns the heatequation into

ϕ(x)ddt

G(t) = kd2

dx2ϕ(x)G(t)

Separating variables with separation constant λ gives

1kG(t)

ddt

G(t) =1

ϕ(x)

d2

dx2ϕ(x) = minusλ

along with the two separate ODEs

Gprime(t) = minusλkG(t) =rArr G(t) = ceminusλkt

ϕprimeprime(x) = minusλϕ(x) (11)

fasshaueriitedu MATH 461 ndash Chapter 2 50

Other Boundary Value Problems

The ODE (11) now will have a different set of BCs We have (assumingG(t) 6= 0)

partupartx

u(0 t) = ϕprime(0)G(t) = 0

=rArr ϕprime(0) = 0

and

partupartx

u(L t) = ϕprime(L)G(t) = 0

=rArr ϕprime(L) = 0

fasshaueriitedu MATH 461 ndash Chapter 2 51

Other Boundary Value Problems

Next we find the eigenvalues and eigenfunctions As before there arethree cases to discussCase I λ gt 0 So ϕprimeprime(x) = minusλϕ(x) has characteristic equation r2 = minusλwith roots

r = plusmniradicλ

We have the general solution (using our Ansatz ϕ(x) = erx )

ϕ(x) = c1 cosradicλx + c2 sin

radicλx

Since the BCs are now given in terms of the derivative of ϕ we need

ϕprime(x) = minusradicλc1 sin

radicλx +

radicλc2 cos

radicλx

and the BCs give us

ϕprime(0) = 0 = minusradicλc1 sin 0︸︷︷︸

=0

+radicλc2 cos 0︸ ︷︷ ︸

=1

λgt0=rArr c2 = 0

ϕprime(L) = 0c2=0= minus

radicλc1 sin

radicλL λgt0

=rArr c1 = 0 or sinradicλL = 0

fasshaueriitedu MATH 461 ndash Chapter 2 52

Other Boundary Value Problems

As always the solution c1 = c2 = 0 is not desirable ( trivial solutionϕ equiv 0) Therefore we have

c2 = 0 and sinradicλL = 0 lArrrArr

radicλL = nπ

At the end of Case I we therefore haveeigenvalues

λn =(nπ

L

)2 n = 123

and eigenfunctions

ϕn(x) = cosnπL

x n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 53

Other Boundary Value Problems

Case II λ = 0 Now ϕprimeprime(x) = 0

ϕ(x) = c1x + c2

so thatϕprime(x) = c1

Both BCs give us

ϕprime(0) = 0 = c1ϕprime(L) = 0 = c1

=rArr c1 = 0

Thereforeϕ(x) = c2 = const

is a solution mdash in fact itrsquos an eigenfunction to the eigenvalue λ = 0

fasshaueriitedu MATH 461 ndash Chapter 2 54

Other Boundary Value Problems

Case III λ lt 0 Now ϕprimeprime(x) = minusλϕ(x) again has characteristicequation r2 = minusλ with roots r = plusmn

radicminusλ

But the general solution is (using our Ansatz ϕ(x) = erx )

ϕ(x) = c1 coshradicminusλx + c2 sinh

radicminusλx

withϕprime(x) =

radicminusλc1 sinh

radicminusλx +

radicminusλc2 cosh

radicminusλx

The BCs give us

ϕprime(0) = 0 =radicminusλc1 sinh 0︸ ︷︷ ︸

=0

+radicminusλc2 cosh 0︸ ︷︷ ︸

=1

λlt0=rArr c2 = 0

ϕprime(L) = 0c2=0=radicminusλc1 sinh

radicminusλL λlt0

=rArr c1 = 0 or sinhradicminusλL = 0

fasshaueriitedu MATH 461 ndash Chapter 2 55

Other Boundary Value Problems

As always the solution c1 = c2 = 0 is not desirableOn the other hand sinh A = 0 only for A = 0 and this would imply theunphysical situation L = 0Therefore Case III does not provide any additional eigenvalues oreigenfunctions

Altogether mdash after considering all three cases mdash we haveeigenvalues

λ = 0 and λn =(nπ

L

)2 n = 123

and eigenfunctions

ϕ(x) = 1 and ϕn(x) = cosnπL

x n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 56

Other Boundary Value Problems

Summarizing by the principle of superposition

u(x t) = A0 +infinsum

n=1

An cosnπx

Leminusk( nπ

L )2t

will satisfy the heat equation and the insulated ends BCs for arbitraryconstants An

Remark

Since A0 = A0 cos 0e0 we can also write

u(x t) =infinsum

n=0

An cosnπx

Leminusk( nπ

L )2t

fasshaueriitedu MATH 461 ndash Chapter 2 57

Other Boundary Value Problems

Finding the Fourier Cosine Coefficients

We now consider the initial condition

u(x 0) = f (x)

From our work so far we know that

u(x 0) =infinsum

n=0

An cosnπx

L

and we need to see how the coefficients An depend on f

fasshaueriitedu MATH 461 ndash Chapter 2 58

Other Boundary Value Problems

In HW problem 236 you should have shown

int L

0cos

nπxL

cosmπx

Ldx =

0 if m 6= nL2 if m = n 6= 0L if m = n = 0

and therefore the set of functions1 cos

πxL cos

2πxL cos

3πxL

is orthogonal on [0L] with respect to the weight function ω equiv 1

fasshaueriitedu MATH 461 ndash Chapter 2 59

Other Boundary Value Problems

To find the Fourier (cosine) coefficients An we start with

f (x) =infinsum

n=0

An cosnπx

L

multiply both sides by cos mπxL and integrate wrt x from 0 to L

=rArrint L

0f (x) cos

mπxL

dx =

int L

0

[ infinsumn=0

An cosnπx

L

]cos

mπxL

dx

Again assuming interchangeability of integration and infinitesummation we getint L

0f (x) cos

mπxL

dx =infinsum

n=0

An

int L

0cos

nπxL

cosmπx

Ldx︸ ︷︷ ︸

=

0 if n 6= mL2 if m = n 6= 0L if m = n = 0

fasshaueriitedu MATH 461 ndash Chapter 2 60

Other Boundary Value Problems

By looking at what remains ofint L

0f (x) cos

mπxL

dx =infinsum

n=0

An

int L

0cos

nπxL

cosmπx

Ldx︸ ︷︷ ︸

=

0 if n 6= mL2 if m = n 6= 0L if m = n = 0in the various cases we get

A0L =

int L

0f (x) dx

AmL2

=

int L

0f (x) cos

mπxL

dx m gt 0

But this is equivalent to

A0 =1L

int L

0f (x) dx

An =2L

int L

0f (x) cos

nπxL

dx n gt 0the Fourier cosine coefficients of f

fasshaueriitedu MATH 461 ndash Chapter 2 61

Other Boundary Value Problems

RemarkSince the solution in this problem with insulated ends is of the form

u(x t) = A0 +infinsum

n=1

An cosnπx

Leminusk( nπ

L )2t︸ ︷︷ ︸

rarr 0 for t rarrinfinfor any n

we see that

limtrarrinfin

u(x t) = A0 =1L

int L

0f (x) dx

the average of f (cf our steady-state computations in Chapter 1)

fasshaueriitedu MATH 461 ndash Chapter 2 62

Other Boundary Value Problems

Periodic Boundary Conditions

fasshaueriitedu MATH 461 ndash Chapter 2 63

Letrsquos consider a circular ring withinsulated lateral sides

The corresponding model for heatconduction in the case of perfectthermal contact at the (common)ends x = minusL and x = L is

PDEpart

parttu(x t) = k

part2

partx2 u(x t) for minus L lt x lt L t gt 0

IC u(x 0) = f (x) for minus L lt x lt L

and new periodic boundary conditions

u(minusL t) = u(L t) for t gt 0partupartx

(minusL t) =partupartx

(L t) for t gt 0

Other Boundary Value Problems

Everything is again nice and linear and homogeneous so we useseparation of variables

As always we use the Ansatz u(x t) = ϕ(x)G(t) so that we get thetwo ODEs

Gprime(t) = minusλkG(t) =rArr G(t) = ceminusλkt

ϕprimeprime(x) = minusλϕ(x)

where the second ODE has periodic boundary conditions

ϕ(minusL) = ϕ(L)

ϕprime(minusL) = ϕprime(L)

Now we look for the eigenvalues and eigenfunctions of this problem

fasshaueriitedu MATH 461 ndash Chapter 2 64

Other Boundary Value Problems

Case I λ gt 0 ϕprimeprime(x) = minusλϕ(x) has the general solution

ϕ(x) = c1 cosradicλx + c2 sin

radicλx

The BCs are given in terms of both ϕ and its derivative so we alsoneed

ϕprime(x) = minusradicλc1 sin

radicλx +

radicλc2 cos

radicλx

The first BC ϕ(minusL) = ϕ(L) is equivalent to

c1 cosradicλ(minusL)︸ ︷︷ ︸

=cosradicλL

+c2 sinradicλ(minusL)︸ ︷︷ ︸

=minus sinradicλL

= c1 cos

radicλL + c2 sin

radicλL

lArrrArr 2c2 sinradicλL

= 0

While ϕprime(minusL) = ϕprime(L) is equivalent tominusc1 sin

radicλ(minusL)︸ ︷︷ ︸

=minus sinradicλL

+c2 cosradicλ(minusL)︸ ︷︷ ︸

=cosradicλL

= minusc1 sin

radicλL + c2 cos

radicλL

lArrrArr 2c1 sinradicλL

= 0fasshaueriitedu MATH 461 ndash Chapter 2 65

Other Boundary Value Problems

Together we have

2c1 sinradicλL = 0 and 2c2 sin

radicλL = 0

We canrsquot have both c1 = 0 and c2 = 0 Therefore

sinradicλL = 0 or λn =

(nπL

)2 n = 123

This leaves c1 and c2 unrestricted so that the eigenfunctions are givenby

ϕn(x) = c1 cosnπx

L+ c2 sin

nπxL n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 66

Other Boundary Value Problems

Case II λ = 0 ϕprimeprime(x) = 0 implies ϕ(x) = c1x + c2 with ϕprime(x) = c1The BC ϕ(minusL) = ϕ(L) gives us

c1(minusL) + c2

= c1L + c2

lArrrArr 2c1L = 0

L 6=0=rArr c1 = 0

The BC ϕprime(minusL) = ϕprime(L) states

c1

= c1

which is completely neutral

Therefore λ = 0 is another eigenvalue with associated eigenfunctionϕ(x) = 1

fasshaueriitedu MATH 461 ndash Chapter 2 67

Other Boundary Value Problems

Similar to before one can establish that Case III λ lt 0 does notprovide any additional eigenvalues or eigenfunctions

Altogether mdash after considering all three cases mdash we haveeigenvalues

λ = 0 and λn =(nπ

L

)2 n = 123

and eigenfunctions

ϕ(x) = 1 and ϕn(x) = c1 cosnπL

x+c2 sinnπL

x n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 68

Other Boundary Value Problems

By the principle of superposition

u(x t) = a0 +infinsum

n=1

an cosnπx

Leminusk( nπ

L )2t +

infinsumn=1

bn sinnπx

Leminusk( nπ

L )2t

is a solution of the PDE with periodic BCs and the IC u(x 0) = f (x) issatisfied if

a0 +infinsum

n=1

an cosnπx

L+infinsum

n=1

bn sinnπx

L= f (x)

In order to find the Fourier coefficients an and bn we need to establishthat

1 cosπxL sin

πxL cos

2πxL sin

2πxL

is orthogonal on [minusLL] wrt ω(x) = 1

fasshaueriitedu MATH 461 ndash Chapter 2 69

Other Boundary Value Problems

We now look at the various orthogonality relations

Sinceint LminusL even fct = 2

int L0 even fct we have

int L

minusLcos

nπxL

cosmπx

Ldx =

0 if m 6= nL if m = n 6= 02L if m = n = 0

Since the product of two odd functions is even we haveint L

minusLsin

nπxL

sinmπx

Ldx =

0 if m 6= nL if m = n 6= 0

Sinceint LminusL odd fct = 0 and the product of an even and an odd

function is odd we haveint L

minusLcos

nπxL

sinmπx

Ldx = 0

fasshaueriitedu MATH 461 ndash Chapter 2 70

Other Boundary Value Problems

Now we can determine the coefficients an by multiplying both sides of

f (x) =infinsum

n=0

an cosnπx

L+infinsum

n=1

bn sinnπx

L

by cos mπxL and integrating wrt x from minusL to L

This givesint L

minusLf (x) cos

mπxL

dx =

int L

minusL

[ infinsumn=0

an cosnπx

L

]cos

mπxL

dx

+

int L

minusL

[ infinsumn=1

bn sinnπx

L

]cos

mπxL

dx︸ ︷︷ ︸=0

=rArr a0 =12L

int L

minusLf (x) dx

and an =1L

int L

minusLf (x) cos

nπxL

dx n ge 1

fasshaueriitedu MATH 461 ndash Chapter 2 71

Other Boundary Value Problems

If we multiply both sides of

f (x) =infinsum

n=0

an cosnπx

L+infinsum

n=1

bn sinnπx

L

by sin mπxL and integrate wrt x from minusL to L

we get int L

minusLf (x) sin

mπxL

dx =

int L

minusL

[ infinsumn=0

an cosnπx

L

]sin

mπxL

dx︸ ︷︷ ︸=0

+

int L

minusL

[ infinsumn=1

bn sinnπx

L

]sin

mπxL

dx

=rArr bn =1L

int L

minusLf (x) sin

nπxL

dx n ge 1

Together an and bn are the Fourier coefficients of f

fasshaueriitedu MATH 461 ndash Chapter 2 72

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

Recall that Laplacersquos equation corresponds to a steady-state heatequation problem ie there are no initial conditions to considerWe solve the PDE (Dirichlet problem) on a rectangle ie

part2upartx2 +

part2uparty2 = 0 0 le x le L 0 le y le H

subject to the BCs (prescribed boundary temperature)

u(x 0) = f1(x) 0 le x le Lu(x H) = f2(x) 0 le x le Lu(0 y) = g1(y) 0 le y le Hu(L y) = g2(y) 0 le y le H

RemarkNote that we canrsquot use separation of variables here since the BCs arenot homogeneous

fasshaueriitedu MATH 461 ndash Chapter 2 74

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

We can still salvage this approach by breaking the Dirichlet problem upinto four sub-problems ndash each of which has

one nonhomogeneous BC (similar to how we dealt with the ICearlier)and three homogeneous BCs

We then use the principle of superposition to construct the overallsolution from the solutions u1 u4 of the sub-problems

u = u1 + u2 + u3 + u4

fasshaueriitedu MATH 461 ndash Chapter 2 75

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

We solve the first problem (the other three are similar)If we start with the Ansatz

u1(x y) = ϕ(x)h(y)

then separation of variables requires

part2u1

partx2 = ϕprimeprime(x)h(y)

part2u1

party2 = ϕ(x)hprimeprime(y)

Therefore the Laplace equation becomes

nabla2u1(x y) = ϕprimeprime(x)h(y) + ϕ(x)hprimeprime(y) = 0

We separate1ϕ

d2ϕ

dx2 = minus1h

d2hdy2 = minusλ

fasshaueriitedu MATH 461 ndash Chapter 2 76

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

The two resulting ODEs are

ϕprimeprime(x) + λϕ(x) = 0 (12)

with BCs

u1(0 y) = 0 =rArr ϕ(0) = 0u1(L y) = 0 =rArr ϕ(L) = 0

andhprimeprime(y)minus λh(y) = 0 (13)

with BCs

u1(x 0) = f1(x) canrsquot use yetu1(x H) = 0 =rArr h(H) = 0

fasshaueriitedu MATH 461 ndash Chapter 2 77

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

We solve the ODE (12) as before Its characteristic equation isr2 = minusλ and we study the usual three cases

Case I λ gt 0 Then r = plusmniradicλ and

ϕ(x) = c1 cosradicλx + c2 sin

radicλx

From the BCs we haveϕ(0) = 0 = c1

ϕ(L) = 0 = c2 sinradicλL =rArr

radicλL = nπ

Thus our eigenvalues and eigenfunctions (so far) are

λn =(nπ

L

)2 ϕn(x) = sin

nπxL n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 78

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

Case II λ = 0 Then ϕ(x) = c1x + c2 and the BCs imply

ϕ(0) = 0 = c2

ϕ(L) = 0 = c1L

so that wersquore left with the trivial solution onlyCase III λ lt 0 Then r = plusmn

radicminusλ and

ϕ(x)c1 coshradicminusλ+ c2 sinh

radicminusλx

for which the eigenvalues imply

ϕ(0) = 0 = c1

ϕ(L) = 0 = c2 sinhradicminusλL =rArr

radicminusλL = 0

so that wersquore again only left with the trivial solution

fasshaueriitedu MATH 461 ndash Chapter 2 79

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

Now we use the eigenvalues in the second ODE (13) ie we solve

hprimeprimen(y) =(nπ

L

)2hn(y)

hn(H) = 0

Since r2 =(nπ

L

)2 (or r = plusmnnπL ) the solution must be of the form

hn(y) = c1enπy

L + c2eminusnπy

L

We can use the BC

hn(H) = 0 = c1enπH

L + c2eminusnπH

L

to derivec2 = minusc1e

nπHL + nπH

L = minusc1e2nπH

L

orhn(y) = c1

(e

nπyL minus e

nπ(2Hminusy)L

)

fasshaueriitedu MATH 461 ndash Chapter 2 80

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

Since the second ODE comes with only one homogeneous BC we cannow pick the constant c1 in

hn(y) = c1

(e

nπyL minus e

nπ(2Hminusy)L

)to get a nice and compact representation for hnThe choice

c1 = minus12

eminusnπH

L

gives us

hn(y) = minus12

enπ(yminusH)

L +12

enπ(Hminusy)

L

= sinhnπ(H minus y)

L

fasshaueriitedu MATH 461 ndash Chapter 2 81

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

Summarizing our work so far we know (using superposition) that

u1(x y) =infinsum

n=1

Bn sinnπx

Lsinh

nπ(H minus y)

L

satisfies the first sub-problem except for the nonhomogeneous BCu1(x 0) = f1(x)We enforce this as

u1(x 0) =infinsum

n=1

Bn sinhnπ(H)

L︸ ︷︷ ︸=bn

sinnπx

L

= f1(x)

From our discussion of Fourier sine series we know

bn =2L

int L

0f1(x) sin

nπxL

dx n = 12

Therefore

Bn =bn

sinh nπ(H)L

=2

L sinh nπ(H)L

int L

0f1(x) sin

nπxL

dx n = 12

fasshaueriitedu MATH 461 ndash Chapter 2 82

Laplacersquos Equation Laplacersquos Equation Inside a Rectangle

RemarkAs discussed at the beginning of this example the solution for theentire Laplace equation is obtained by solving the three similarproblems for u2 u3 and u4 and assembling

u = u1 + u2 + u3 + u4

The details of the calculations for finding u3 are given in the textbook[Haberman pp 68ndash71] (where this function is called u4) and u4 isdetermined in [Haberman Exercise 251(h)]

fasshaueriitedu MATH 461 ndash Chapter 2 83

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

Laplacersquos Equation for a Circular Disk

fasshaueriitedu MATH 461 ndash Chapter 2 84

Now we consider the steady-stateheat equation on a circular diskwith prescribed boundarytemperatureThe model for this case seems tobe (using the Laplacian incylindrical coordinates derived inChapter 1)

PDE nabla2u =1rpart

partr

(rpartupartr

)+

1r2part2upartθ2 = 0 for 0 lt r lt a minusπ lt θ lt π

BC u(a θ) = f (θ) for minus π lt θ lt π

Since the PDE involves two derivatives in r and two derivatives in θ westill need three more conditions How should they be chosen

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

Perfect thermal contact (periodic BCs in θ)

u(r minusπ) = u(r π) for 0 lt r lt apartupartθ

(r minusπ) =partupartθ

(r π) for 0 lt r lt a

The three conditions listed are all linear and homogeneous so we cantry separation of variables

We leave the fourth (and nonhomogeneous) condition open for now

RemarkThis is a nice example where the mathematical model we derive fromthe physical setup seems to be ill-posed (at this point there is no waywe can ensure a unique solution)However the mathematics below will tell us how to think about thephysical situation and how to get a meaningful fourth condition

fasshaueriitedu MATH 461 ndash Chapter 2 85

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

We begin with the separation Ansatz

u(r θ) = R(r)Θ(θ)

We can separate our PDE (similar to HW problem 231)

nabla2u(r θ) =1rpart

partr

(rpartupartr

)+

1r2part2upartθ2 = 0

lArrrArr 1r

ddr

(r

ddr

R(r)

)Θ(θ) +

R(r)

r2d2

dθ2 Θ(θ) = 0

lArrrArr rR(r)

ddr

(r

ddr

R(r)

)= minus 1

Θ(θ)

d2

dθ2 Θ(θ) = λ

Note that λ works better here than minusλ

fasshaueriitedu MATH 461 ndash Chapter 2 86

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

The two resulting ODEs are

rR(r)

(ddr R(r) + r d2

dr2 R(r))

= λ

lArrrArr r2Rprimeprime(r)R(r) + r

R(r)Rprime(r)minus λ = 0

orr2Rprimeprime(r) + rRprime(r)minus λR(r) = 0 (14)

andΘprimeprime(θ) = minusλΘ(θ) (15)

for which we have the periodic boundary conditions

Θ(minusπ) = Θ(π) Θprime(minusπ) = Θprime(π)

fasshaueriitedu MATH 461 ndash Chapter 2 87

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

Note that the ODE (15) along with its BCs matches the circular ringexample studied earlier (with L = π)Therefore we already know the eigenvalues and eigenfunctions

λ0 = 0 λn = n2 n = 12 Θ0(θ) = 1 Θn(θ) = c1 cos nθ + c2 sin nθ n = 12

fasshaueriitedu MATH 461 ndash Chapter 2 88

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

Using these eigenvalues in (14) we have

r2Rprimeprimen (r) + rRprimen(r)minus n2Rn(r) = 0 n = 012

This type of equation is called a Cauchy-Euler equation (and youshould have studied its solution in your first DE course)

We quickly review how to obtain the solution

Rn(r) =

c3 + c4 ln r if n = 0c3rn + c4rminusn for n gt 0

The key is to use the Ansatz R(r) = rp and to find suitable values of p

fasshaueriitedu MATH 461 ndash Chapter 2 89

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

If R(r) = rp then

Rprime(r) = prpminus1 and Rprimeprime(r) = p(p minus 1)rpminus2

so that the CE equation

r2Rprimeprimen (r) + rRprimen(r)minus n2Rn(r) = 0

turns into [p(p minus 1) + p minus n2

]rp = 0

Assuming rp 6= 0 we get the characteristic equation

p(p minus 1) + p minus n2 = 0 lArrrArr p2 = n2

so thatp = plusmnn

If n = 0 we need to introduce the second (linearly independent)solution R(r) = ln r

fasshaueriitedu MATH 461 ndash Chapter 2 90

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

We now look at the two casesCase I n = 0 We know the general solution is of the form

R(r) = c3 + c4 ln r

We need to find and impose the missing BCNote that

ln r rarr minusinfin for r rarr 0

This would imply that R(0) ndash and therefore u(0 θ) the temperature atthe center of the disk ndash would blow up That is completely unphysicaland we need to prevent this from happening in our modelWe therefore require a bounded temperature at the origin ie

|u(0 θ)| ltinfin =rArr |R(0)| ltinfin

This ldquoboundary conditionrdquo now implies that c4 = 0 and

R(r) = c3 = const

fasshaueriitedu MATH 461 ndash Chapter 2 91

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

Case II n gt 0 Now the general solution is of the form

R(r) = c3rn + c4rminusn

and we again impose the bounded temperature condition ie

|R(0)| ltinfin

Note that ∣∣rminusn∣∣ =

∣∣∣∣ 1rn

∣∣∣∣rarrinfin as r rarr 0

Therefore this condition implies c4 = 0 and

R(r) = c3rn n = 12

Summarizing (and using superposition) we have up to now

u(r θ) = A0 +infinsum

n=1

Anrn cos nθ + Bnrn sin nθ

fasshaueriitedu MATH 461 ndash Chapter 2 92

Laplacersquos Equation Laplacersquos Equation for a Circular Disk

Finally we use the boundary temperature distribution u(a θ) = f (θ) todetermine the coefficients An Bn

u(a θ) = A0 +infinsum

n=1

Anan cos nθ + Bnan sin nθ = f (θ)

From our earlier work we know that the functions

1 cos θ sin θ cos 2θ sin 2θ

are orthogonal on the interval [minusπ π] (just substitute L = π in ourearlier analysis)It therefore follows as before that

A0 =1

2π

int π

minusπf (θ) dθ

Anan =1π

int π

minusπf (θ) cos nθ dθ n = 123

Bnan =1π

int π

minusπf (θ) sin nθ dθ n = 123

fasshaueriitedu MATH 461 ndash Chapter 2 93

Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

The solution

u(r θ) = A0 +infinsum

n=1

Anrn cos nθ + Bnrn sin nθ

of the circular disk problem tells us that the temperature at the centerof the disk is given by

u(0 θ) = A0 =1

2π

int π

minusπf (θ) dθ

ie the average of the boundary temperatureIn fact a more general statement is true

The temperature at the center of any circle inside of which thetemperature is harmonic (ie nabla2u = 0) is equal to theaverage of the boundary temperature

This fact is reminiscent of the mean value theorem from calculus andis therefore called the mean value principle for Laplacersquos equation

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Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

Maximum Principle for Laplacersquos Equation

fasshaueriitedu MATH 461 ndash Chapter 2 95

TheoremBoth the maximum and the minimum temperature of the steady-stateheat equation on an arbitrary region R occur on the boundary of R

ProofAssume that the maximumminimum occurs atan arbitrary point P inside of R and show thisleads to a contradictionTake a circle C around P that lies inside of RBy the mean value principle the temperature at P is the average of thetemperature on CTherefore there are points on C at which the temperature isgreaterless than or equal the temperature at PBut this contradicts our assumption that the maximumminimumtemperature occurs at P (inside the circle)

Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

Well-posednessDefinitionA problem is well-posed if all three of the following statements are true

A solution to the problem existsThe solution is uniqueThe solution depends continuously on the data (eg BCs) iesmall changes in the data lead to only small changes in thesolution

RemarkThis definition was provided by Jacques Hadamard around 1900Well-posed problems are ldquonicerdquo problems However in practice manyproblems are ill-posed For example the inverse heat problem ietrying to find the initial temperature distribution or heat source from thefinal temperature distribution (such as when investigating a fire) isill-posed (see examples below)

fasshaueriitedu MATH 461 ndash Chapter 2 96

Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

Theorem

The Dirichlet problem nabla2u = 0 inside a given region R and u = f onthe boundary is well-posed

Proof

(a) Existence Compute the solution using separation of variables(details depend on the specific domain R)

(b) Uniqueness Assume u1 and u2 are solutions of the Dirichletproblem and show that w = u1 minus u2 = 0 ie u1 = u2

(c) Continuity Assume u is a solution of the Dirichlet problem withBC u = f and v is a solution with BC v = g = f minus ε and show thatmin ε le u minus v le max ε

Details for (b) and (c) now follow

fasshaueriitedu MATH 461 ndash Chapter 2 97

Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

(b) Uniqueness Assume u1 and u2 are solutions of the Dirichletproblem ie

nabla2u1 = 0 nabla2u2 = 0

Let w = u1 minus u2 Then by linearity

nabla2w = nabla2 (u1 minus u2) = 0

On the boundary we have for both

u1 = f u2 = f

So again by linearity

w = u1 minus u2 = f minus f = 0 on the boundary

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Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

(b) (cont) What does w look like inside the domainAn obvious inequality is

min(w) le w le max(w)

for which the maximum principle implies

0 le w le 0 =rArr w = 0

since the maximum and minimum are attained on the boundary(where w = 0)

fasshaueriitedu MATH 461 ndash Chapter 2 99

Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

(c) Continuity We assume

nabla2u = 0 and u = f on partRnabla2v = 0 and v = g on partR

where g is a small perturbation (by the function ε) of f ie

g = f minus ε

Now by linearity w = u minus v satisfies

nabla2w = nabla2 (u minus v) = 0 inside Rw = u minus v = f minus g = ε on partR

By the maximum principle

min(ε) le w︸︷︷︸=uminusv

le max(ε)

fasshaueriitedu MATH 461 ndash Chapter 2 100

Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

Example (An ill-posed problem)The problem

nabla2u = 0 in Rnablau middot n = 0 on partR

does not have a unique solution since u = c is a solution for anyconstant c

RemarkIf we interpret the above problem as the steady-state of atime-dependent problem with initial temperature distribution f then theconstant would be uniquely defined as the average of f

fasshaueriitedu MATH 461 ndash Chapter 2 101

Laplacersquos Equation Qualitative Properties of Laplacersquos Equation

Example (Another ill-posed problem)The problem

nabla2u = 0 in Rnablau middot n = f on partR

may have no solution at allThe definition of the Laplacian and Greenrsquos theorem give us

0 =

intintR

nabla2u︸︷︷︸=0

dA def=

intintR

nabla middot nablau dA Green=

intpartR

nablau middot n︸ ︷︷ ︸=f

ds

so that only very special functions f permit a solution

RemarkPhysically this says that the net flux through the boundary must bezero A non-zero boundary flux integral would allow for a change intemperature (which is unphysical for a steady-state equation)

fasshaueriitedu MATH 461 ndash Chapter 2 102

Appendix References

References I

R HabermanApplied Partial Differential EquationsPearson (5th ed) Upper Saddle River NJ 2012

J StewartCalculusCengage Learning (8th ed) Boston MA 2015

D G ZillA First Course in Differential EquationsBrooksCole (10th ed) Boston MA 2013

fasshaueriitedu MATH 461 ndash Chapter 2 103