math 461: fourier series and boundary value problemsfass/notes461_ch2.pdf · 2015-09-24 · math...

MATH 461: Fourier Series and Boundary ValueProblems

Chapter II: Separation of Variables

Greg Fasshauer

Department of Applied MathematicsIllinois Institute of Technology

Fall 2015

[email protected] MATH 461 – Chapter 2 1

http://math.iit.eduhttp://math.iit.edu/~fass

Outline

1 Model Problem

2 Linearity

3 Heat Equation for a Finite Rod with Zero End Temperature

4 Other Boundary Value Problems

5 Laplace’s Equation



Model Problem

Outline

1 Model Problem

2 Linearity






Model Problem

For much of the following discussion we will use the following 1D heatequation with constant values of c, ρ,K0 as a model problem:

∂

∂tu(x , t) = k

∂2

∂x2u(x , t) +

Q(x , t)cρ

, for 0 < x < L, t > 0

with initial condition

u(x ,0) = f (x) for 0 < x < L

and boundary conditions

u(0, t) = T1(t), u(L, t) = T2(t) for t > 0



Linearity

Outline

1 Model Problem

2 Linearity






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 u1,u2.

Differentiation and integration are linear operations.

ExampleConsider ordinary differentiation of a univariate function, i.e.,L = ddx . Then

ddx

(c1f1 + c2f2) (x) = c1d

dxf1(x) + c2

ddx

f2(x).



Linearity

ExampleThe same is true for partial derivatives:

∂

∂t(c1u1 + c2u2) (x , t) = c1

∂

∂tu1(x , t) + c2

∂

∂tu2(x , t).

In particular, the heat operator ∂∂t − k∂2

∂x2 is linear.Therefore, the heat equation

RemarkA linear homogeneous equation, Lu = 0, always has at least the trivialsolution u ≡ 0.



Linearity


∂

∂t(c1u1 + c2u2) (x , t) = c1

∂

∂tu1(x , t) + c2

∂

∂tu2(x , t).



∂

∂tu(x , t)− k ∂

2

∂x2u(x , t) =

Q(x , t)cρ︸︷︷︸

given function

is a linear PDE. .




Linearity


∂

∂t(c1u1 + c2u2) (x , t) = c1

∂

∂tu1(x , t) + c2

∂

∂tu2(x , t).



∂

∂tu(x , t)− k ∂

2

∂x2u(x , t) = 0

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




Linearity

ExampleAre the following equations linear or nonlinear, homogeneous ornonhomogeneous?

∂2

∂x2u(x , y) +

∂2

∂y2u(x , y) = f (x , y).

is linear and generally nonhomogeneous (Poisson’s equation).

∂2

∂x2u(x , y) +

∂2

∂y2u(x , y) = 0.

is linear and homogeneous (Laplace’s equation).

∂

∂tu(x , t)− κ ∂

∂x

[u(x , t)

∂

∂xu(x , t)

]= 0.

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



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.

Proof.We are given a linear operator L and functions u1,u2 such that

Lu1 = 0, Lu2 = 0.

We need to show that

L (c1u1 + c2u2) = 0.

Straightforward computation gives



Linearity



Lu1 = 0, Lu2 = 0.


L (c1u1 + c2u2) = 0.


L (c1u1 + c2u2)L linear

= c1Lu1 + c2Lu2



Linearity



Lu1 = 0, Lu2 = 0.


L (c1u1 + c2u2) = 0.



= c1 Lu1︸︷︷︸=0

+c2 Lu2︸︷︷︸=0



Linearity



Lu1 = 0, Lu2 = 0.


L (c1u1 + c2u2) = 0.



= c1 Lu1︸︷︷︸=0

+c2 Lu2︸︷︷︸=0

= 0.



Heat Equation for a Finite Rod with Zero End Temperature

Outline

1 Model Problem

2 Linearity







We want to solve the PDE

∂

∂tu(x , t) = k

∂2

∂x2u(x , t), for 0 < x < L, t > 0 (1)

with initial condition

u(x ,0) = f (x) for 0 < x < L (2)

and boundary conditions

u(0, t) = u(L, t) = 0 for t > 0 (3)

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




Separation of VariablesThis technique often just “works”, especially for linear homogeneousPDEs and BCs, by magically(?) converting the PDE to a pair of ODEs— and those we should be able to solve1.

The starting point is to take the unknown function u = u(x , t) and“separate its variables”, i.e., 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 form,and 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, i.e., dydx = f (x)g(y). Now we assume (or hope)that the solution is separable.

1If you don’t remember, you might want to review Chapters 2 and 5 (maybe also 4)of something like [Zill].




Separation of VariablesThis technique often just “works”, especially for linear homogeneousPDEs and BCs, by magically(?) converting the PDE to a pair of ODEs— and those we should be able to solve1.The starting point is to take the unknown function u = u(x , t) and“separate its variables”, i.e., 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 form,and 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, i.e., dydx = f (x)g(y). Now we assume (or hope)that the solution is separable.

1If you don’t remember, you might want to review Chapters 2 and 5 (maybe also 4)of something like [Zill].




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

∂

∂tu(x , t) = ϕ(x)

ddt

G(t)

∂2

∂x2u(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).




Therefore1

kG(t)ddt

G(t) =1

ϕ(x)d2

dx2ϕ(x) = −λ (5)

The constant λ is known as the separation constant.The “−” sign appears mostly for cosmetic purposes.

Equations (5) give two separate ODEs:

ϕ′′(x) = −λϕ(x) (6)G′(t) = −λkG(t) (7)




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

=⇒ ϕ(0) = 0 (8)

and

u(L, t) = ϕ(L)G(t) = 0=⇒ ϕ(L) = 0 (9)

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





u(0, t) = ϕ(0)G(t) = 0=⇒ ϕ(0) = 0 (8)

and

u(L, t) = ϕ(L)G(t) = 0=⇒ ϕ(L) = 0 (9)






u(0, t) = ϕ(0)G(t) = 0=⇒ ϕ(0) = 0 (8)

and

u(L, t) = ϕ(L)G(t) = 0

=⇒ ϕ(L) = 0 (9)






u(0, t) = ϕ(0)G(t) = 0=⇒ ϕ(0) = 0 (8)

and

u(L, t) = ϕ(L)G(t) = 0=⇒ ϕ(L) = 0 (9)





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

G′(t) = −λkG(t)

(since the IC provides spatial, x, information, while (7) is an ODE intime t).

Instead, (7) provides us only with

G(t) = ce−λkt

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




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

G′(t) = −λkG(t)

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

G(t) = ce−λkt

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




Solution of the Two-Point BVP

We now solve

ϕ′′(x) = −λϕ(x)ϕ(0) = ϕ(L) = 0.

This kind of problem is discussed in detail in MATH 252 (see, e.g.,Chapter 5 of [Zill]).The characteristic equation of this ODE is

r2 = −λ,

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





We now solve

ϕ′′(x) = −λϕ(x)ϕ(0) = ϕ(L) = 0.

This kind of problem is discussed in detail in MATH 252 (see, e.g.,Chapter 5 of [Zill]).

The characteristic equation of this ODE is

r2 = −λ,






We now solve

ϕ′′(x) = −λϕ(x)ϕ(0) = ϕ(L) = 0.


r2 = −λ,






We now solve

ϕ′′(x) = −λϕ(x)ϕ(0) = ϕ(L) = 0.


r2 = −λ,

which is obtained from another Ansatz, namely ϕ(x) = erx .

What are the roots r (and therefore the general solution ϕ)?





We now solve

ϕ′′(x) = −λϕ(x)ϕ(0) = ϕ(L) = 0.


r2 = −λ,





For a real separation constant λ there are three cases.

Case I, λ > 0: In this case, r2 = −λ gives us

r = ±i√λ

along with the general solution

ϕ(x) = c1 cos(√

λx)

+ c2 sin(√

λx).

Now we use the BCs:

ϕ(0) = 0 = c1 cos 0︸︷︷︸=1

+c2 sin 0︸︷︷︸=0

=⇒ c1 = 0

ϕ(L) = 0c1=0= c2 sin

(√λL)

=⇒ c2 = 0 or sin√λL = 0

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

c1 = 0 and sin√λL = 0.




For a real separation constant λ there are three cases.Case I, λ > 0: In this case, r2 = −λ gives us

r = ±i√λ


ϕ(x) = c1 cos(√

λx)

+ c2 sin(√

λx).

Now we use the BCs:

ϕ(0) = 0 = c1 cos 0︸︷︷︸=1

+c2 sin 0︸︷︷︸=0

=⇒ c1 = 0

ϕ(L) = 0c1=0= c2 sin

(√λL)

=⇒ c2 = 0 or sin√λL = 0







r = ±i√λ


ϕ(x) = c1 cos(√

λx)

+ c2 sin(√

λx).

Now we use the BCs:

ϕ(0) = 0 = c1 cos 0︸︷︷︸=1

+c2 sin 0︸︷︷︸=0

=⇒ c1 = 0

ϕ(L) = 0c1=0= c2 sin

(√λL)

=⇒ c2 = 0 or sin√λL = 0

The solution c1 = c2 = 0 is not desirable (since it leads to the trivialsolution ϕ ≡ 0).

Therefore, at this point we conclude






r = ±i√λ


ϕ(x) = c1 cos(√

λx)

+ c2 sin(√

λx).

Now we use the BCs:

ϕ(0) = 0 = c1 cos 0︸︷︷︸=1

+c2 sin 0︸︷︷︸=0

=⇒ c1 = 0

ϕ(L) = 0c1=0= c2 sin

(√λL)

=⇒ c2 = 0 or sin√λL = 0






Our conclusionsc1 = 0 and sin

√λL = 0

do not yet specify the solution ϕ, so we still have work to do.

Note that the equation sin√λL = 0 is true whenever

√λL = nπ for any

positive integer n.In other words, we get

λn =(nπ

L

)2, n = 1,2,3, . . .

Each eigenvalue λn =(nπ

L

)2 gives us an eigenfunctionϕn(x) = c2 sin

nπL

x , n = 1,2,3, . . .

— each one of which is a solution to the BVP.





√λL = 0

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

√λL = 0 is true whenever


positive integer n.

In other words, we get

λn =(nπ

L

)2, n = 1,2,3, . . .


L


nπL

x , n = 1,2,3, . . .






√λL = 0

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

√λL = 0 is true whenever


positive integer n.In other words, we get

λn =(nπ

L

)2, n = 1,2,3, . . .


L


nπL

x , n = 1,2,3, . . .





Case II, λ = 0: In this case, r2 = −λ = 0 implies r = 0 or

ϕ(x) = c1 + c2x .

The BCs lead to:ϕ(0) = 0 = c1

ϕ(L) = 0c1=0= c2L =⇒ c2 = 0

Now we have only the solution c1 = c2 = 0, i.e., the trivial solutionϕ ≡ 0.Since by definition ϕ ≡ 0 cannot be an eigenfunction, this implies thatλ = 0 is not an eigenvalue for our BVP.In other words, this case does not contribute to the solution.





ϕ(x) = c1 + c2x .


ϕ(L) = 0c1=0= c2L

=⇒ c2 = 0






ϕ(x) = c1 + c2x .


ϕ(L) = 0c1=0= c2L =⇒ c2 = 0






ϕ(x) = c1 + c2x .


ϕ(L) = 0c1=0= c2L =⇒ c2 = 0

Now we have only the solution c1 = c2 = 0, i.e., the trivial solutionϕ ≡ 0.

Since by definition ϕ ≡ 0 cannot be an eigenfunction, this implies thatλ = 0 is not an eigenvalue for our BVP.In other words, this case does not contribute to the solution.





ϕ(x) = c1 + c2x .


ϕ(L) = 0c1=0= c2L =⇒ c2 = 0

Now we have only the solution c1 = c2 = 0, i.e., the trivial solutionϕ ≡ 0.Since by definition ϕ ≡ 0 cannot be an eigenfunction, this implies thatλ = 0 is not an eigenvalue for our BVP.

In other words, this case does not contribute to the solution.





ϕ(x) = c1 + c2x .


ϕ(L) = 0c1=0= c2L =⇒ c2 = 0





Case III, λ < 0: Now r2 = −λ︸︷︷︸>0

implies r = ±√−λ or

ϕ(x) = c1e√−λx + c2e−

√−λx .

The BCs lead to:

ϕ(0) = 0 = c1 + c2 =⇒ c2 = −c1

ϕ(L) = 0c2=−c1= c1e

√−λL − c1e−

√−λL

or c1e√−λL = c1e−

√−λL

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 ≡ 0).

RemarkInstead of ϕ(x) = c1e

√−λx + c2e−

√−λx we could have used the

alternate formulation ϕ(x) = c1 cosh(√−λx

)+ c2 sinh

(√−λx

)— to

the same effect.




Case III, λ < 0: Now r2 = −λ︸︷︷︸>0


ϕ(x) = c1e√−λx + c2e−

√−λx .

The BCs lead to:

ϕ(0) = 0 = c1 + c2

=⇒ c2 = −c1

ϕ(L) = 0c2=−c1= c1e

√−λL − c1e−

√−λL


√−λL



√−λx + c2e−



)+ c2 sinh

(√−λx

)— to

the same effect.




Case III, λ < 0: Now r2 = −λ︸︷︷︸>0


ϕ(x) = c1e√−λx + c2e−

√−λx .

The BCs lead to:

ϕ(0) = 0 = c1 + c2 =⇒ c2 = −c1

ϕ(L) = 0c2=−c1= c1e

√−λL − c1e−

√−λL


√−λL



√−λx + c2e−



)+ c2 sinh

(√−λx

)— to

the same effect.




Case III, λ < 0: Now r2 = −λ︸︷︷︸>0


ϕ(x) = c1e√−λx + c2e−

√−λx .

The BCs lead to:

ϕ(0) = 0 = c1 + c2 =⇒ c2 = −c1

ϕ(L) = 0c2=−c1= c1e

√−λL − c1e−

√−λL


√−λL



√−λx + c2e−



)+ c2 sinh

(√−λx

)— to

the same [email protected] MATH 461 – Chapter 2 21



Summary (so far)

The two-point BVP

ϕ′′(x) = −λϕ(x)ϕ(0) = ϕ(L) = 0

has eigenvalues

λn =(nπ

L

)2, n = 1,2,3, . . .

and eigenfunctions

ϕn(x) = sinnπx

L, n = 1,2,3, . . .

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




Summary (cont.)

The PDE-BVP

∂

∂tu(x , t) = k

∂2

∂x2u(x , t), for 0 < x < L, t > 0

u(0, t) = u(L, t) = 0 for t > 0

has solutions

un(x , t) = ϕn(x)Gn(t)

= sinnπx

Le−λnkt

= sinnπx

Le−k(

nπL )

2t , n = 1,2,3, . . .




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, i.e.,

limt→∞

u(x , t) = 0.

We see that each

un(x , t) = sinnπx

Le−k(

nπL )

2t

satisfies this property.




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, i.e.,

limt→∞

u(x , t) = 0.

We see that each

un(x , t) = sinnπx

Le−k(

nπL )

2t

satisfies this property.




By the principle of superposition any linear combination of un,n = 1,2,3, . . ., will also be a solution, i.e.,

u(x , t) =∑

n

Bn sinnπx

Le−k(

nπL )

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) =∑

n

Bn sinnπx

L

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




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 MATLABscript Soundwaves.m.

If 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.




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 MATLABscript Soundwaves.m.If 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.




On March 27, 2008, researchers announced that they had found asound recording made by Édouard-Léon Scott de Martinville on April9, 1860 — 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 [email protected] MATH 461 – Chapter 2 27

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

Phonautogram by Édouard-Léon Scott de Martinville

Recorded on April 9, 1860: 1861 Deposit, Académie des Sciences (No. 324), Number 5

2008

Other

10840.803

eng - www.firstsounds.org

AuClairdelaLune.mp3Media File (audio/x-mp3)




Fourier analysis can be used to take apart and analyze complexsounds.

The MATLAB GUI touchtone lets us analyze which buttons werepressed on a touch-tone phone.Wave-like phenomena also play a fundamental role in

heat flow and other diffusion problems (e.g, the spreading ofpollutants),vibration problems,sound and image file compression (e.g., MP3 or JPEG files),filtering of noisy audio or video.




Fourier analysis can be used to take apart and analyze complexsounds.The MATLAB GUI touchtone lets us analyze which buttons werepressed on a touch-tone phone.

Wave-like phenomena also play a fundamental role inheat flow and other diffusion problems (e.g, the spreading ofpollutants),vibration problems,sound and image file compression (e.g., MP3 or JPEG files),filtering of noisy audio or video.




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

heat flow and other diffusion problems (e.g, the spreading ofpollutants),vibration problems,sound and image file compression (e.g., MP3 or JPEG files),filtering of noisy audio or video.



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

Le−k(

mπL )

2t

will satisfy the entire heat equation problem, i.e., the series solution

(10) collapses to just one term, so Bn =

{0 if n 6= m1 if n = m

.




ExampleIf the initial temperature distribution f is of the form

f (x) =M∑

n=1

Bn sinnπx

L,

then

u(x , t) =M∑

n=1

Bnun(x , t) =M∑

n=1

Bn sinnπx

Le−k(

nπL )

2t

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




What about other initial temperature distributions f?

The general idea will be to use an infinite series (i.e., a Fourier series)to represent an arbitrary f ,

i.e., we will show that any (with some mildrestrictions) function f can be written as

f (x) =∞∑

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) =∞∑

n=1

Bn sinnπx

Le−k(

nπL )

2t .

Remaining question: How do the coefficients Bn depend on f?




What about other initial temperature distributions f?

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

f (x) =∞∑

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) =∞∑

n=1

Bn sinnπx

Le−k(

nπL )

2t .

Remaining question: How do the coefficients Bn depend on f?




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 · b‖a‖‖b‖

,

and therefore the vectors a and b are orthogonal, a ⊥ b (orperpendicular, i.e., θ = π2 ), if and only if a · b = 0.

In terms of the vector components this becomes

a ⊥ b ⇐⇒ a · b =n∑

i=1

aibi = 0.

If A,B are two sets of vectors then A is orthogonal to B if a · b = 0for every a ∈ A and every b ∈ B.A is an orthogonal set (or simply orthogonal) if a · b = 0 for everya,b ∈ A with a 6= b.




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 · b‖a‖‖b‖

,

and therefore the vectors a and b are orthogonal, a ⊥ b (orperpendicular, i.e., θ = π2 ), if and only if a · b = 0.In terms of the vector components this becomes

a ⊥ b ⇐⇒ a · b =n∑

i=1

aibi = 0.

If A,B are two sets of vectors then A is orthogonal to B if a · b = 0for every a ∈ A and every b ∈ B.A is an orthogonal set (or simply orthogonal) if a · b = 0 for everya,b ∈ A with a 6= b.




Orthogonality (of functions)We can let our “vectors” be functions, f and g, defined on some interval[a,b]. Then f and g are orthogonal on [a,b] with respect to the weightfunction ω if and only if 〈f ,g〉 = 0,

where the inner product is defined by

〈f ,g〉 =∫ b

af (x)g(x)ω(x) dx .

RemarkOrthogonality of vectors is usually discussed in linear algebra,while orthogonality of functions is a topic that belongs to functionalanalysis.Note that orthogonality of functions always is specified relative toan interval and a weight function.There are many different classes of orthogonal functions such as,e.g., orthogonal polynomials, trigonometric functions, or wavelets.Orthogonality is one of the most fundamental (and useful)concepts in mathematics.




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

〈f ,g〉 =∫ b







〈f ,g〉 =∫ b


RemarkOrthogonality of vectors is usually discussed in linear algebra,while orthogonality of functions is a topic that belongs to functionalanalysis.

Note that orthogonality of functions always is specified relative toan interval and a weight function.There are many different classes of orthogonal functions such as,e.g., orthogonal polynomials, trigonometric functions, or wavelets.Orthogonality is one of the most fundamental (and useful)concepts in mathematics.





〈f ,g〉 =∫ b


RemarkOrthogonality of vectors is usually discussed in linear algebra,while orthogonality of functions is a topic that belongs to functionalanalysis.Note that orthogonality of functions always is specified relative toan interval and a weight function.

There are many different classes of orthogonal functions such as,e.g., orthogonal polynomials, trigonometric functions, or wavelets.Orthogonality is one of the most fundamental (and useful)concepts in mathematics.





〈f ,g〉 =∫ b


RemarkOrthogonality of vectors is usually discussed in linear algebra,while orthogonality of functions is a topic that belongs to functionalanalysis.Note that orthogonality of functions always is specified relative toan interval and a weight function.There are many different classes of orthogonal functions such as,e.g., orthogonal polynomials, trigonometric functions, or wavelets.

Orthogonality is one of the most fundamental (and useful)concepts in mathematics.





〈f ,g〉 =∫ b






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

orthogonal on the interval [−1,1] with respect to the weightfunction ω(x) ≡ 1.

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

p3(x) = αx2 + βx − 1

is orthogonal to both p1 and p2.

The polynomials p1,p2,p3 are known as the first three Legendrepolynomials.

SolutionAltogether, we need to show that∫ 1

−1pj(x)pk (x)ω(x) dx = 0, whenever j 6= k = 1,2,3




Solution (cont.)1 p1(x) = 1 and p2(x) = x are orthogonal since∫ 1

−1p1(x)︸︷︷︸=1

p2(x)︸︷︷︸=x

ω(x)︸︷︷︸=1

dx

=

∫ 1−1

x dx =x2

2

∣∣∣∣1−1

= 0.

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





−1p1(x)︸︷︷︸=1

p2(x)︸︷︷︸=x

ω(x)︸︷︷︸=1

dx =∫ 1−1

x dx

=x2

2

∣∣∣∣1−1

= 0.






−1p1(x)︸︷︷︸=1

p2(x)︸︷︷︸=x

ω(x)︸︷︷︸=1

dx =∫ 1−1

x dx =x2

2

∣∣∣∣1−1

= 0.





Solution (cont.)2 We need to find α and β such that both∫ 1

−1p1(x)p3(x)ω(x) dx =

∫ 1−1

p2(x)p3(x)ω(x) dx = 0.

This leads to∫ 1−1

(αx2 + βx − 1

)dx =

[α

x3

3+ β

x2

2− x

]1−1

=23α− 2 != 0

and∫ 1−1

x(αx2 + βx − 1

)dx =

[α

x4

4+ β

x3

3− x

2

2

]1−1

=12β

!= 0,

so that we have α = 3, β = 0 and

p3(x) = 3x2 − 1.





−1p1(x)p3(x)ω(x) dx =

∫ 1−1

p2(x)p3(x)ω(x) dx = 0.


(αx2 + βx − 1

)dx

=

[α

x3

3+ β

x2

2− x

]1−1

=23α− 2 != 0

and∫ 1−1

x(αx2 + βx − 1

)dx =

[α

x4

4+ β

x3

3− x

2

2

]1−1

=12β

!= 0,


p3(x) = 3x2 − 1.





−1p1(x)p3(x)ω(x) dx =

∫ 1−1

p2(x)p3(x)ω(x) dx = 0.


(αx2 + βx − 1

)dx =

[α

x3

3+ β

x2

2− x

]1−1

=23α− 2 != 0

and∫ 1−1

x(αx2 + βx − 1

)dx =

[α

x4

4+ β

x3

3− x

2

2

]1−1

=12β

!= 0,


p3(x) = 3x2 − 1.





−1p1(x)p3(x)ω(x) dx =

∫ 1−1

p2(x)p3(x)ω(x) dx = 0.


(αx2 + βx − 1

)dx =

[α

x3

3+ β

x2

2− x

]1−1

=23α− 2

!= 0

and∫ 1−1

x(αx2 + βx − 1

)dx =

[α

x4

4+ β

x3

3− x

2

2

]1−1

=12β

!= 0,


p3(x) = 3x2 − 1.





−1p1(x)p3(x)ω(x) dx =

∫ 1−1

p2(x)p3(x)ω(x) dx = 0.


(αx2 + βx − 1

)dx =

[α

x3

3+ β

x2

2− x

]1−1

=23α− 2 != 0

and∫ 1−1

x(αx2 + βx − 1

)dx =

[α

x4

4+ β

x3

3− x

2

2

]1−1

=12β

!= 0,


p3(x) = 3x2 − 1.





−1p1(x)p3(x)ω(x) dx =

∫ 1−1

p2(x)p3(x)ω(x) dx = 0.


(αx2 + βx − 1

)dx =

[α

x3

3+ β

x2

2− x

]1−1

=23α− 2 != 0

and∫ 1−1

x(αx2 + βx − 1

)dx

=

[α

x4

4+ β

x3

3− x

2

2

]1−1

=12β

!= 0,


p3(x) = 3x2 − 1.





−1p1(x)p3(x)ω(x) dx =

∫ 1−1

p2(x)p3(x)ω(x) dx = 0.


(αx2 + βx − 1

)dx =

[α

x3

3+ β

x2

2− x

]1−1

=23α− 2 != 0

and∫ 1−1

x(αx2 + βx − 1

)dx =

[α

x4

4+ β

x3

3− x

2

2

]1−1

=12β

!= 0,


p3(x) = 3x2 − 1.




Orthogonality of Sines

We now show that the functions{sin

πxL, sin

2πxL, sin

3πxL, . . .

}are orthogonal on [0,L] with respect to the weight ω ≡ 1.

To this end we need to evaluate∫ L0

sinnπx

Lsin

mπxL

dx

for different combinations of integers n and m.

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




Case I, m 6= n: Using the trigonometric identity

sin A sin B =12

(cos(A− B)− cos(A + B))

we get∫ L0

sinnπx

Lsin

mπxL

dx =12

∫ L0

[cos

((n − m)πx

L

)− cos

((n + m)

πxL

)]dx





sin A sin B =12


we get∫ L0

sinnπx

Lsin

mπxL

dx =12

∫ L0

[cos

((n − m)πx

L

)− cos

((n + m)

πxL

)]dx

=12

[L

(n − m)π sin((n − m)πx

L

)− L

(n + m)πsin((n + m)

πxL

)]L0





sin A sin B =12


we get∫ L0

sinnπx

Lsin

mπxL

dx =12

∫ L0

[cos

((n − m)πx

L

)− cos

((n + m)

πxL

)]dx

=12

[L


L

)− L


πxL

)]L0

=12

[L

(n − m)π (sin(n − m)π − sin 0)−L

(n + m)π(sin(n + m)π − sin 0)

]





sin A sin B =12


we get∫ L0

sinnπx

Lsin

mπxL

dx =12

∫ L0

[cos

((n − m)πx

L

)− cos

((n + m)

πxL

)]dx

=12

[L


L

)− L


πxL

)]L0

=12

[L

(n − m)π(

sin (n − m)︸︷︷︸integer

π − sin 0)− L

(n + m)π(

sin (n + m)︸︷︷︸integer

π − sin 0)]





sin A sin B =12


we get∫ L0

sinnπx

Lsin

mπxL

dx =12

∫ L0

[cos

((n − m)πx

L

)− cos

((n + m)

πxL

)]dx

=12

[L


L

)− L


πxL

)]L0

=12

[L

(n − m)π(

sin (n − m)︸︷︷︸integer

π

︸︷︷︸=0

− sin 0︸︷︷︸=0

)− L

(n + m)π(

sin (n + m)︸︷︷︸integer

π

︸︷︷︸=0

− sin 0︸︷︷︸=0

)]= 0.




Case II, m = n: Using the trigonometric identity

sin2 A =12

(1− cos 2A)

we get ∫ L0

sin2nπx

Ldx =

12

∫ L0

(1− cos 2nπx

L

)dx


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math 461: fourier series and boundary value problemsfass/notes461_ch2.pdf · 2015-09-24 · math...

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