Chapter 4

1D Linear Second-Order Equations

In this chapter, our primary focus is on introducing the series solution approach for 1D heat

and wave diﬀerential equations. Building upon the foundations laid in the preceding chapters,

where we explored the derivation of these equations in higher dimensions and analyzed

their overarching behavior, we now delve into the speciﬁc context of solving heat and wave

equations within b o unded media extending along the x-axis, deﬁned as the interval (x

; x

As we navigate this chapter, we encounter a crucial factor that heavily inﬂuences the

nature of the solution: the conditions at the boundary points x=x

and x=x

. This prompts

the eme rgence of boundary value problems a s a pivo tal concept in th is context. To enhance

our understanding, we introduce a renowne d theorem, the Sturm-Liouville theorem, w hich

assumes a critical role in our exploration of the e igenfunction expan sion method. This

method becomes instrumental in representing the solutions of 1D partial diﬀerential equa-

tions.

4.1 From PDE to e igenvalue problem

This section serves as a bridge between partial diﬀerential equations (PDEs) and eigenvalue

problems.

4.1.1 Outline of the method

We'll focus on the context of an open interval Ω=(x

; x

), where we encounter a PDE that

governs the function u=u(x; t):

= L[u]

where L takes the f o rm of a 1D diﬀerential o perator:

L := a(x)

+ b(x)

+ c(x): (4.1)

In our scenarios, a; b, and c are continuou s functions deﬁned within Ω. This equation ﬁnds

practical applications, like modeling temperature distribution along an extended conductive

rod within Ω. However, to establish a complete formulation, we need to determine conditions

at both endp o ints, x=x

and x=x

. This is because the diﬀerential equation is conﬁned to

the open interval Ω.

We proceed by assuming that the function u adhere s to a particular condition at th ese

endpoints:

αu + βu

= 0:

Additionally, the initial temperature distribution, u(x; 0), along the rod must be s peciﬁed.

The interplay between the thermal energy distribution along the rod and heat diﬀusivity

generates the dynamic heat behavior of the system. By combining these elements, we present

a comprehensive initial-boundary value problem:

= L[u] on Ω

αu + βu

= 0 on bnd(Ω)

u(x; 0) = f(x) initial condition

This synthesis of the diﬀerential operator L, boundary conditions, and initial condition lays

the foundation fo r solving various heat and wave equations, gradually unraveling the intricate

dynamics of these systems.

As L is a linear operator that operates solely on the spatial variable x, we can treat the

solution of the equation u

= L[u] as an ordinary diﬀerential equation for u and express it as:

= L[u], as an ordinary diﬀerential equation fo r u and write it as: u(x; t) = φ(x) e

¡λt

where

φ(x) is an unknown function and λ is a constant. To verify its validity, substituting this into

the equation gives:

¡λφ(x) e

¡λt

= e

¡λt

L[φ];

which leads to: L[φ] = ¡λφ. To satisfy the prescribed boundary conditions for u, we arrive

at the subsequent boundary value problem, often termed the eigenvalue problem:



L[φ] = ¡λφ on Ω

αφ + βφ

= 0 ob bnd(Ω)

Remark 4.1. The use of terminology ca n be justiﬁed by drawing parallels with the e igen-

value problem in linear algebra. Remember that a vector v in R

is an eigenvector of a linear

transformation T :R

if T [v]+λv=0. In the context of our problem, the operator L acts

linearly on the space of smooth functions deﬁned on Ω. Consequently, a function φ: Ω!R

is termed an eigenfunction if L[φ](x) + λφ(x)=0 holds for all x in Ω. Following the analogy

with the eigenvalue problem in linear algebra, the va lue λ is referred to as the associated

eigenvalue of the eigenfunction φ. It's important to note that if φ is an eigenfunction of

the problem, then c φ(x) for any non-zero constant c is also an eigenfunction. Just like

eigenvectors in linear algebra, eigenfunctions are typically considered as non-trivial functions.

Solving an eigenvalue problem involves determining the eigenfunctions φ(x) and eigen-

values λ. When the problem can be solved, the solution of the equation u

= L[u] while

adhering to the boundary condition αu + β u

= 0 can be expressed as follows: u(x; t) =

φ(x) e

¡λt

Example 4.1. Consider the interval (0; 1) and let Ω represent this interval. Let's explore

the eigenvalue problem φ

= ¡λφ, where φ(x) is deﬁned on Ω. The boundary conditions

are set as φ(0) = 0 and φ(1) = 0. We want to ﬁnd values of λ and corresponding non-trivial

functions φ(x) that satisfy this equation.

2 1D Linear Second-Order Equations

First, let's show that the problem h as a solution when λ is greater th a n zero. To do this,

we multiply the equation by φ(x) and integrate it over the interval (0; 1). This gives us an

equation involving integrals and the boundary term:

jφ

(x)j

dx + φ

(x) φ(x)



= ¡λ

jφ(x)j

dx;

where we used the integration by parts formula for the integral at the le f t-hand side. The

boundary term φ

(x) φ(x)j

is equal zero due to the given boundary cond itions. This gives

the following formula for λ:

λ =

jφ

(x)j

jφ(x)j

≥0:

If we were to assume λ = 0, we'd ﬁnd that φ(x) turns out to be a constant function. But

considering the boundary conditions and continuity, we'd be left with the trivial function

φ(x) = 0. Therefore, λ must b e greater than zero.

With this understanding, we proceed to solve the equation φ

= ¡λφ as an ordinary

diﬀerential equation. The general soluti on for φ(x) can be expressed a s:

φ(x) = A cos( λ

x) + B sin( λ

x):

Plugging in the boundary condition φ(0) = 0, we ﬁnd that A = 0. Applying the second

boundary condition, we determine that the eigenvalues λ are given by λ = n

for n = 1; 2; ···.

For each eigenvalue λ

, we obtain an associated eigenfunction φ

(x) = B

sin(nπx),

where B

is a non-zero constant. This set of eigenvalues and eigenfunctions forms an inﬁnite

collection: φ

(x) =sin(πx), λ

= π

, φ

(x) =sin(2πx), λ

= 4π

, φ

(x) =sin(3πx), λ

= 9π

and

so on. These eigenfunctions provide the building blocks for s o lv ing more complex diﬀerential

equations and understanding the behav ior of solutions on the interval (0; 1).

Exercise 4.1. Let's explore another approach to demonstrate that λ > 0 in the previously discussed

eigenvalue p roblem. Thi s will involve direct calculations. Begin by considering the algebraic character-

istic equation for the diﬀerential equation φ

= ¡λφ, which results in r = ± ¡λ

Task 1: Suppose λ < 0. Proceed by establishing the g ener al solution for this ordinary diﬀerential

equation (ODE). Afterward, apply the given boundary conditions and deduce that the only possible

outcome is φ = 0.

Task 2 : Assume that λ = 0. Derive the general solution for this ODE and arrive at the conclusion

that φ = 0 is again the only feasible solution.

Task 3: Now f ocus on the case where λ> 0. Determin e the non-trivial eigenfunctions that correspond

to these positive eigenvalues.

Exercise 4.2. Building upon the insights gained from the previous example, let's apply that knowledge

to draw conclusions about the heat problem given as:



= u

x 2(0; 1)

u(0; t) = u(1; t) = 0

Task: Utilize the outcomes from the previous example and deduce that this heat problem possesses an

inﬁnite number of solutions, denoted as u

(x; t) = e

¡n

sin(nπx) for any positive integer n = 1; 2; ···.

These solutions are often referred to as “s eparated solutions” due to their form.

4.1 From PDE to eigenvalue problem 3

Additionally, employing the sup erposition principle, you can express the general solution as:

u(x; t) =

n=1

(x; t);

Here, the constants c

can take any arbitrary values, provided that the inﬁnite series converges to a valid

function.

4.1.2 Separation of variables

The derivation of the eigenva lue problem discussed a bove can also be expressed in terms of

the separation o f variables technique. By taking u(x; t) as the separated function:

u(x; t) = φ(x) U(t);

and substituting this into the heat equation, we arrive at:

L[φ]

;

where L[φ] represents the operator applied to φ(x). This equation is possible only when the

ratios are equal to the same dimensionless constant. Let's denote this constant by ¡λ. The

minus sign is used for historical reasons and does not carry a physical meaning, but it will

become clear why this choice is more appropriate for subsequent calculations.

By introducing ¡λ as the eigenvalue, we obtain the following ordinary diﬀerential equa-

tion for U (t):

= ¡λU ;

which is solved as U(t)=Ce

¡λt

, where C is a c o nstant. The eigenvalue λ and eigenfunction

φ(x) are determined by solving the second ordinary diﬀerential equation:

L[φ] = ¡λφ;

subject to the boundary conditions: αφ + βφ

= 0 at the boundary points.

The derivation holds true for gene ral 1 D pa rtial diﬀerential equations. Consider the

following second-order PD E for u(x; t) on the interval x 2[x

; x

(t)u

+ a

(t) u

= L[u]

u(x

; t) + β

; t) = 0

u(x

; t) + β

; t) = 0

;

where a

(t) a nd a

(t) a re continuous functions. The equation is general enough to encompass

all types of 1D heat and wave equations. To ﬁnd a solution to this PDE, we consider the

solution in the separated fo rm as u(x; t)=U(t)φ(x). By substituting this separated function

into the PDE, we obtain the equality

(t)U

+ a

(t)U

L[φ]

4 1D Linear Second-Order Equations

Obviously, the equality holds only if the ratio is a dimensionless constant since x and t are

indep endent variables. Therefore, we can write

(t)U

+ a

(t)U

L[φ]

= ¡λ;

where λ is a constant. Here, the negative sign does not carry any signiﬁcant physical meaning;

it is merely a historical convention. Therefore, the separated form of the solution leads to

the following ordinary diﬀerential equations:

1. For U(t), we have the equation:

(t) U

+ a

(t) U

= ¡λU ;

2. For φ(x), we have the eigenvalue problem:

L[φ] = ¡λφ on Ω

φ(x

) + β

) = 0

φ(x

) + β

) = 0

4.1.3 Series solution

Superposition solutions hold value when the coeﬃcients c

are carefully chosen to ensure

that the inﬁnite series converges into a smooth function u(x; t). The convergence of inﬁnite

function s eries is a complex topic beyond this book's scope. Nevertheless, we provide illus-

trative examples to convey the concept of f unction series convergence.

Example 4.2. Consider the interval (0; 1), w here we delve into the followin g inﬁnite series:

n=1

sin(nπx):

Upon observation, it's evident that this series doesn't converge to a continuous function

within the interval (0; 1). A clear instance a rises when x =

, res ulting in the series becoming

a numerical series:

n=1

(¡1)

n¡1

2n ¡1

This numerical series exhibits conditional convergence, lacking absolute convergence. Now

explore the s eries:

n=1

sin(nπx):

We ascertain that this series demonstrates a bso lute convergence. This is evident from the

inequality:



n=1

sin(nπx)



≤

n=1

< 1:

The ﬁgure below depicts the truncated series of the above inﬁnite series up to 100 terms :

4.1 From PDE to eigenvalue problem 5

0 0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1.2

These examples underscore the intricate convergence behaviors of inﬁnite series and their

implications on the continuity of functions within a speciﬁed interval.

In the follow ing sections, we will delve into other ty pes of convergence. However, the

discussed form of convergence remains fundamentally crucial for comprehending initial value

problems. Consider th e subsequent heat problem as an example:



= u

x 2(0; 1)

u(0; t) = u(1; t) = 0

;

alongside the initial condition u(x; 0) = f(x). Given the general series solution for this heat

problem as:

u(x; t) =

n=1

¡λ

sin(nπx):

The signiﬁcance of the convergenc e of this inﬁnite series as t approaches 0 becomes evident.

This convergence property implies:

lim

t!0

u(x; t) = lim

t!0

n=1

¡λ

sin(nπx) =

n=1

sin(nπx) = f (x):

Consequently, the coeﬃcients c_n must be chosen carefully, ensuring that:

lim

N !1

n=1

sin(nπx) = f (x):

As an illustration, consider the case where f (x) = x(1 ¡ x). For this scenario, appropriate

values for c

are determined as follows:

(

n: odd

0 n: even

With this speciﬁc selection of c

, the series:

n=1

¡λ

sin(nπx);

6 1D Linear Second-Order Equations

successfully converges to the initial condition as t approaches 0, aligning with the prescribed

values of c

Exercise 4.3. Let Ω be the interval Ω: (0; 1):

a) Solve the following eigenvalue problem



= ¡λφ on Ω

(0) = φ

(1) = 0 on bnd(Ω)

b) Use this eigenvalue problem and solve the following heat problem



= ku

on Ω

(0; t) = u

(1; t) = 0

;

for k > 0.

Exercise 4.4. Let Ω be the interval Ω: (0; 1):

a) Solve the following eigenvalue problem



= ¡λφ on Ω

φ(0) = φ

(1) = 0 on bnd(Ω)

b) Use this eigenvalue problem and solve the following heat problem



= ku

on Ω

u(0; t) = u

(1; t) = 0

;

for k > 0.

c) Find the steady state solution

lim

t!1

u(x; t):

Exercise 4.5. Let Ω be the interval Ω: (0; 1):

a) Solve the following eigenvalue problem



= ¡λφ on Ω

(0) = φ(1) = 0 on bnd(Ω)

b) Use this eigenvalue problem and solve the following heat problem



= ku

on Ω

(0; t) = u(1; t) = 0

;

for k > 0.

c) Find the steady state solution

lim

t!1

u(x; t):

Exercise 4.6. Let Ω be the interval Ω: (0; 1):

a) Fin d eigenvalues and eigenfunctions of the following eigenvalue problem

= ¡λφ



= 0



= 0

b) Use these eigenfunct ions and solve the following wave equation

= c

x 2

;



; t



= u

; t



= 0

u(x; 0) = cos(x) + sin(2x)

(x; 0) = 0

4.1 From PDE to eigenvalue problem 7

4.1.4 Problems

Problem 4.1. Find the eigenvalues an d eigenfunctions of the following eigenvalue pr oblem

(

= ¡λφ



= φ



= 0

Problem 4.2. Find the eigenvalues an d eigenfunctions of the following eigenvalue pr oblem

= ¡λφ

φ(¡1) = φ(1)

(¡1) = φ

(1)

Problem 4.3. Find the eigenvalues an d eigenfunctions of the following eigenvalue pr oblem

= ¡λφ

¡1



= 0



= 0

Problem 4.4. Find the general ser ies solution of the following wave equation

(

= c

x 2(0; 1)

u(0; t) = u(1; t) = 0

Problem 4.5. Consider the following damped wave equation

(

+ 2ξu

= c

x 2(0; 1)

u(0; t) = u(1; t) = 0

a) For what value of damping factor ξ, the system is in the underdamped state? The underdamped

state is the state where the t-compo nent of the solution is in the form

U(t) = e

σt

[A cos(!t) + sin(!t)];

for some real parameters σ and !.

b) Find the general series solution of the problem when the system is under damped.

Problem 4.6. Consider the following eigenvalue problem on x 2(1; e)

L[φ] = ¡λφ

φ(1) = 0

φ(e) = 0

;

where L stands for the diﬀerential operator

L := x

+ x

a) Use the transformation x = e

and convert the given eigenvalue problem to the the following one

(s) = ¡λ'(s)

'(0) = 0

'(1) = 0

Solve t he transformed problem and ﬁnd eigenvalues and eigenfunctions of the oper ator L.

b) Use this results and solve the following problem

= x

+ xu

u(1; t) = 0

u(e; t) = 0

8 1D Linear Second-Order Equations

Problem 4.7. Consider the following eigenvalue problem on x 2(0; 1)

= ¡λφ

φ(0) = 0

φ(1) + φ

(1) = 0

a) Prove that the problem is solvable for a non-trivial function φ only if λ > 0.

b) Verify that there are inﬁnitely many eigenvalues λ

< λ

< ···. Find the smallest eigenvalue λ

a numerical method and the correspond eigenfunction φ

(x).

Problem 4.8. Consider the following eigenvalue problem on x 2(0; 1)

= ¡λφ

φ(0) = 0

φ(1) ¡φ

(1) = 0

a) Prove that the problem is solvable for a non-trivial function φ only if λ ≥0.

b) Verify that there are inﬁnitely many eigenvalues 0 = λ

< λ

< ···. What is the ﬁrst eigenfunction

(x).

Problem 4.9. Consider the following eigenvalue problem on x 2(0; 1):

+ 2φ

= ¡λφ

φ(0) = 0

φ(1) = 0

Show that the eigenvalues of the problem is strictly greater than 1. The root of the characteristic equation

of the ordinary diﬀerential equation φ + 2φ + λφ = 0 are

1;2

= ¡1 ± 1 ¡λ

Give argument why λ > 0 is the condition for the solvability of the eigenvalue problem for a non-trivial

eigenfunction φ(x).

Problem 4.10. We Consider the eigenvalue problem

+ 2φ

= ¡λφ

φ(0) = 0

φ(1) = 0

a) Multiply the equation by φ and integrate the equation in [0; 1]. Use the boundary condition and

show

(x) φ(x) dx = 0:

b) The part a) gives the result

λ =

jφ

(x)j

jφ(x)j

Prove the inequality

jφ(x)j

dx <

jφ

(x)j

dx;

if φ(0) = φ(1) = 0. Hint: write

φ(x) =

(x) dx;

and conclude

jφ(x)j≤

jφ

(x)jdx:

4.1 From PDE to eigenvalue problem 9

Use the Cau chy-Schwarz inequality

jφ

(x)jdx <



jφ

(x)j



;

and prove the claim.

4.2 Sturm-L iouville problem

As we observed above, the solution of a general linear 1D partial diﬀerential equation strongly

relies on th e solvability of the following eigenvalue problem

L[φ] = ¡λφ

φ(x

) + β

) = 0

φ(x

) + β

) = 0

;

where L is the diﬀerential operator (4.1).

While there is no general method to solve this problem, the Sturm-Liouville theorem

provides valuable insights into its solutions. This theorem is a powerful tool in mathematical

physics and plays a signiﬁcant role in u nderstanding the behavior of diﬀerential equations.

To ga in a rigorous understanding of the assertions of the theorem, it is essential to

familiarize ourselves with some advanced mathematical concepts. These include function

spaces, the notion of inner product, and the concepts of convergence of function series or

sequences.

4.2.1 Inner product

Recall the concept of the dot product between two vectors in R

. Extending this operation

to functions, we deﬁne the inner product as a natural generalization. Let f(x) and g(x) be

two real-va lued functions deﬁne d on the interval [x

; x

]. The inner pro duct hf ; gi is de ﬁned

as the integral

hf ; gi=

f(x) g(x) dx: (4.2)

The deﬁned inner product exhibits the following prop erties:

1. Positivity: hf ; f i≥0, and hf ; f i= 0 only if f ≡0.

2. Commutativity: hf ; g i=hg; f i

3. Homogeneity: hλf ; gi=λhf ; g i for any λ 2R

4. Additivity: hf ; g + hi=hf ; gi+hf ; hi, as long as the associated integrals exist.

To grasp the natural extension of the dot product, let us recall its deﬁnition between two

arbitrary vectors u~ = (u

; :::; u

) and v~ =(v

; :::; v

), which is given by:

u~ ·v~ =

j=1

10 1D Linear Second-Order Equations

It is worth noting that the Riemann sum of the integral (4.2) can be expressed a s:

hf ; g i= lim

n!1

j=1

f(x

) g(x

) δx

Exercise 4.7. Another approach to deﬁning the inner product, which we will utilize in our subsequent

discussions, is known as the weighted inner product. In this case, we introduce a positive function σ(x).

The weighted product, denoted as hf ; gi

, is deﬁned as:

hf ; gi

f(x) g(x) σ(x) dx

Verify that this weighted inner product also sa tisﬁes th e properties of positi vity, commutativity, homo-

geneity, an d additivity, just like the previously deﬁned inner products.

Now, we can deﬁne the notion of orthogonality between two functions using the inner

product.

Deﬁnition 4.1. Two non-trivial real-valued functions f and g deﬁned on the interval [x

; x

]

are called orthogonal if their inner product satisﬁes hf ; gi=0.

Example 4 .3. Consider the set of functions



sin

nπ



; x 2[0; L]



for n=1; 2; 3; :::; which

are deﬁned on the interval [0; L]. It is evident that these functions a re mutually orthogonal,

meaning that th eir inner p roduct s a tis ﬁe s:

sin



nπ



; sin



mπ

E

sin



nπ



sin



mπ



dx = 0;

for n =/ m. However, when n=m, we have:



sin



nπ





sin



nπ



dx =

Similarly, the functions in the set



cos

nπ



; x 2[0; L]



for n=0; 1; 2; 3; ::: are also mutually

orthogonal. Additionally, the inner product satisﬁes:

sin



nπ



; cos



mπ

E

= 0;

for all integer values of n and m.

In add ition to the orthogonality, we use the inner production notion and deﬁne the mag-

nitude of a function. Let f(x) be a function deﬁned on the interval [x

; x

]: The magnitude

or norm of f , compatible with the inner product h;i, is deﬁned as kf k= hf ; f i

. In general,

the norm o f a function should satisfy the follow ing properties

1. kf k≥0, and if kf k= 0, then f ≡0.

2. kλf k= jλjkf k for any λ 2R.

3. kf + gk≤kf k+ kgk.

4.2 S turm-Liouville problem 11

Exercise 4.8. Prove that functions in the set

f1; cos(nx); sin(nx)g

deﬁned in the interval [¡π; π] are mutually orthogonal:

hcos(nx); cos(mx)i= 0; n =/ m

hcos(nx); sin(mx)i= 0; 8n; m

Find the norm of each function.

Exercise 4.9. Show that the functions in the set



sin



2n ¡1

πx



;

are mutually orthogonal. What are the norms of these functions?

4.2.2 Approximating a function by orthogonal functions

The signiﬁcance of the orthogonality condition in our subsequent discussions lies in the

following f a ct: Assume that f (x) is a function deﬁned on the interval [x

; x

]. Furthermore,

assume that the functions in the set fφ

(x); x 2[x

; x

]g are mutually orthogonal with respect

to the inner product h; i, and that f(x) is approximated in terms of these functions as:

f(x) ≈c

(x) + c

(x) + ···;

for some undetermined co eﬃcients c

. T he orth o g o nality condition enables u s to determine

these parameters as:

hf ; φ

kφ

Note that for any ﬁxed j, we have

hf ; φ

i= c

hφ

; φ

i+ ···+ c

hφ

; φ

i+ ···+ c

hφ

; φ

i+ ···= c

hφ

; φ

and thus

hf ; φ

kφ

;

for any ﬁxed j, where kφ

= hφ

; φ

Example 4.4. The functions in the set fsin(nπx)g

n=1

deﬁned on the domain x 2[0; 1] are

mutually orthogonal. We can now ﬁnd the best approximation of the function f (x) = x,

x 2[0; 1] in terms of the functions in the set fsin(nπx)g

n=1

for so me number N ≥1. The best

approximation function f

(x) is given by

(x) =

n=1

hf ; sin(nπx)i

ksin(nπx)k

sin(nπx):

We have

hf ; sin(nπx)i=

x sin(nπx) dx =

¡cos(nπ)

nπ

and

ksin(nπx)k

:= hsin(nπx); sin(nπx)i=

jsin(nπx)j

dx =

;

12 1D Linear Second-Order Equations

and then, the best approximation is obtained as

(x) =

n=1

¡2 cos(nπ)

nπ

sin(nπx):

The ﬁgures below show the g raph of f(x) and f

(x) for some values of N:

0 0.5 1

0.2

0.4

0.6

0.8

0 0.5 1

0.2

0.4

0.6

0.8

0 0.5 1

0.2

0.4

0.6

0.8

0 0.5 1

0.2

0.4

0.6

0.8

Remark 4.2. (Fourier sine series) In the a bove e xample, we used the set of eigenfunctions

= fsin(nπx)g of the eigenvalue problem



= ¡λφ

φ(0) = φ(1) = 0

The series in terms of these eigenfunctions is known as Fourier sine series. The ﬁgure below

illustrate some of these orthogonal functions:

0 0.2 0.4 0.6 0.8 1

-1

-0.5

0.5

4.2 S turm-Liouville problem 13

Example 4.5. The functions in the set fcos(nπx)g

n=0

are mutually orthogonal. The best

approximation of the function f (x) = x in the set fcos(nπx)g

n=0

is given by

(x) =

n=0

cos(nπx):

For n = 0, we have

x dx =

;

and for n = 1; 2; ···, the coeﬃcients c

are

= 2

x c o s(nπx) =

2(1 ¡cos(nπ))

The ﬁgures below show the graph of f(x) = x and its best approximation for some values of

0 0.5 1

0.5

0 0.5 1

0.5

0 0.5 1

0.5

0 0.5 1

0.5

Remark 4.3. The functions φ

= cos(nπx) that we have used to approximate the function

f(x) are the eigenfunctions of the eigenvalue problem



= ¡λφ

(0) = φ

(1) = 0

14 1D Linear Second-Order Equations

The series in terms of these eigenfunctions i s known as Fourier cosine series. The ﬁgure below

illustrate some of these orthogonal functions:

0 0.2 0.4 0.6 0.8 1

-1

-0.5

0.5

Exercise 4.10. Express the function f (x) = x deﬁned on x 2 [0; 1] in terms of mutually orthogonal

functions φ

= sin

2n ¡ 1

πx



Exercise 4.11. Express the function f (x) = x deﬁned on x 2 [0; 1] in terms of mutually orthogonal

functions φ

= cos

2n ¡ 1

πx



Exercise 4.12. Express the function f(x) = 1 + x deﬁned on x 2[¡1; 1] in terms of mutually orthogonal

functions φ

= f1; cos(nπx); sin(nπx)g. These functions are the eigenfunction of the eigenvalue problem

= ¡λφ

φ(¡1) = φ(1)

(¡1) = φ

(1)

The series in terms of these functions are known as Fourier series.

4.2.3 Sturm-Liouville theorem

The theorem for the eigenvalue problem can be stated as follows:

Theorem 4.1. Consider the eigenvalue problem deﬁned by

a(x)φ

+ b(x)φ

+ c(x)φ = ¡λφ

φ(x

) + β

) = 0

φ(x

) + β

) = 0

; (4.3)

where a; b, and c are continuous functions with a(x)>0 in the interval [x

; x

The problem has infinitely many real eigenvalues arranged in increasing order:

<λ

<···. Furthermore, the eigenvalues diverge to inﬁnity as n tends to inﬁnity:

lim

n!1

=1:

4.2 S turm-Liouville problem 15

2. For each eigenvalue λ

, there exists a unique eigenfunction φ

(x) (up to a constant

multiplication).

3. The eigenfunctions φ

(x) for n=1; 2; ··· are mutually orthogonal with respect to the

weight function σ(x) deﬁned as

σ(x) =

a(x)

b(x)

a(x)

This orthogonality is represented by the equation hφ

; φ

=0 for n=/ m.

4. The set of functions fφ

(x)g for n=1; 2; ··· forms a basis for piecewise continuously

diﬀerentiable functions deﬁned in the interval [x ¡0; x

]. This implies that any piece-

wise continuously diﬀerentiable function f (x) deﬁned in [x

; x

] can be approximated

(x) ≈

n=1

hf ; φ

kφ

(x)

in the following sense:

lim

N !1

jf (x) ¡ f

(x)j= 0;

as long as x is a continuity point for f.

This theorem establishes the existence of eigenvalues and eigenfunctions, their orthogo-

nality, and thei r role in approximating piecewise continuously diﬀerentiable functions.

The result of this theorem is crucial in solving linear second-order PDEs. As an example,

let's consider the equation:

= L[u]

φ(x

) + β

) = 0

φ(x

) + β

) = 0

;

where L is the diﬀerential o perator (4.1). By assuming a separable solution o f the form

u(x; t)= φ(x)e

¡λt

, we arrive at the eigenvalue problem (4.3). For each pair (λ

; φ

), we obtain

a solution u

(x; t)=e

¡λ

(x), which is referred to as the separated solution.

Example 4.6. Consider the function f (x) deﬁned as:

f(x) =

(

≤x ≤

0 otherwise

deﬁned for x 2(0; 1). We can expand this f unction using diﬀerent basis sets generated from

various eigenvalue problems. The ﬁgure below illustrates the representation of this function

using four sets of eigenfunctions, truncated to N = 10 terms.

16 1D Linear Second-Order Equations

0 0.5 1

0.2

0.4

0.6

0.8

0 0.5 1

0.2

0.4

0.6

0.8

0 0.5 1

0.2

0.4

0.6

0.8

0 0.5 1

0.2

0.4

0.6

0.8

1. The ﬁrst ﬁgure at the left is generated using the set of functions fsin(nπx)g

n=1

which

are eigenfunctions of the following problems:



= ¡λφ

φ(0) = φ(1) = 0

The weight function for this set is σ(x) = 1.

2. The second ﬁgure at the right is generated using the set of functions fcos(nπx)g

n=1

which are eigenfunctions of the following problems:



= ¡λφ

(0) = φ

(1) = 0

The weight function for this set is σ(x) = 1.

3. The third ﬁgure is generated using the set of functions fsin(z

x)g

n=0

where z

are

the roots of the equation

sin(z) + z cos(z) = 0:

These functions are eigenfunction o f the eigenvalue problem

= ¡λφ

φ(0) = 0

φ(1) + φ

(1) = 0

The weight function for this set is σ(x) = 1.

4.2 S turm-Liouville problem 17

4. The last ﬁ g ure is generated using the set of functions fe

¡x

sin(nπx)g

n=1

which are

eigenfunctions of the following problems:



+ 2φ

= ¡λφ

φ(0) = φ(1) = 0

The weight function for this set is σ(x) = e

It is worth noting that in all series representations of the f unction f(x), the series co nverge

to the average of the right and left limits of f (x) at its di scontinuity points located at x =

and x =

. This convergence behavior is a characteristic feature of series expansions for

functions with discontinuities.

Exercise 4.13. Consider the following eigenvalue problem on x 2(1; e)

(

+ xφ

= ¡λφ

φ(1) = φ(e) = 0

a) Determine the weight function σ(x) that makes the eigenfunctions of the problem orthogonal.

b) The eigenfunctions o f the problem are φ

(x) = sin(nπ ln x). Calculate directly that these eigen-

functions are orthogonal with respect to σ(x).

Exercise 4.14. Consider the following eigenvalue problem

+ 2φ

= ¡λφ

φ(0) = 0

(1) = 0

Determines the eigenfunctions of the problem and the weight function σ(x) for the orthogonality of the

these e igenfunctions. Note that the ﬁrst eigenvalue is λ = 1, and for λ > 1, you derive an equation of the

form:

z cos(z) ¡sin(z) = 0;

which has inﬁnitely many roots.

4.2.4 Problems

Problem 4.11. There are alternative forms of the inner product beyond the one deﬁned in equation

(4.2). Let's consider the set of continuously diﬀerentiable functions deﬁned on the interval [x

; x

denoted by C

; x

). We can explore a diﬀerent deﬁnition of the inner product, denoted as hf ; g i,

which is given by:

hf ; gi=

f(x) g(x) dx +

(x) g

(x) dx:

As an exercise, we can show that this alternative deﬁnition satisﬁes the fundamental properties of

po sitivity, commutativity, homogeneity, and additivity, just like the inner product deﬁned in equation

(4.2).

Problem 4.12. Let T

be the space of all 2 ×2 matrices.

i. Show that the following operation is an inner product in T

hA; Bi= tr A

ii. Generalize this f act for T

, the space of all n ×n matrices.

iii. Is the o peration hA; Bi= tr AB an inner product in T

18 1D Linear Second-Order Equations

Problem 4.13. Several theorems in plane geometry can be proved by the concept of dot product. Here

are two of such problems:

a) Prove the cosine law in triangles, i.e.,

= a

+ b

¡2ab cos(θ);

where θ is the angle between sid es a and b.

b) Consider the triangle shown in the ﬁgure where the side d bisects the side c. Use vector operations

to prove the Apollonius law a

+ b

+ 2d

Problem 4.14. Propose a norm for T

, the set of 2 ×2 matrices.

Exercise 4.15. (Cauchy inequality) Cauchy's inequality sta tes the following inequality:

hf ; gi≤kf kkgk:

To prove this inequality, we start by expanding the right-hand side of the following inequality for any

complex number λ:

0 ≤hf + λg; f + λgi:

Using this inequality, we can then show the Cauchy inequality.

kf + g k≤kf k+ kgk:

Use this inequality and prove the triangle inequality

kf + gk≤kf k+ kgk

Problem 4.15. Prove the following version of the triangle inequality for arbitrary positive number " > 0.

kf + g k≤

kf k+ " kgk:

Problem 4.16. If x

; :::; x

are positive numbers, prove the following identities:

a) n

≤(x

+ ···+ x

)



+ ···+



b) (x

+ ···+ x

)

≤n(x

+ ···+ x

Problem 4.17. If the re lation kλ

f + λ

gk = kλ

f + λ

gk holds for all real numbers λ

; λ

, show

kf k= kgk.

Exercise 4.16. Let C([x

; x

]) be the set of all continuous functions deﬁned on [x

; x

a) Show that the operation kf k

deﬁnes as

kf k

= max

x2[x

]

jf(x)j;

is a norm for f 2C([x

; x

]). For this, you n eed to show that this operation satisﬁes all properties

of a norm.

b) For any inner product, show that the following equality holds

kf + g k

¡kf ¡ gk

= 4hf ; gi:

4.2 S turm-Liouville problem 19

Use this to prove that there is no inner product hf ; f i

= kf k

Problem 4.18. Assume that fφ

(x)g f or n = 1; 2; ··· is a set of mutually orthogonal functions. The best

approximation of a functions f(x) in the set fφ

(x)g

n=1

is deﬁned as

(x) =

n=1

(x);

such that

kf ¡ f

k≤kf ¡ gk;

for any other g(x) constructed by a linear combination of functions in the set fφ

(x)g

n=1

. If the norm

is compatible with the inner product h; i, prove that f

is the best approximation of f if c

are

hf ; φ

hφ

; φ

Hint: Deﬁne h = g ¡ f

, and use the inequality kf ¡ g k≥0.

Problem 4.19. Let f

is the best approximation of function f in terms of the mutually orthogonal

functions in the set fφ

n=1

a) s how tha t f ¡ f

is orthogonal to all functions φ

b) For any function g as

g =

n=1

;

for a rbitrary b

, show that hf ¡ f

; gi= 0.

Problem 4.20. As we observed in previous sections, the set of orthogonal functions



sin

nπ



for

n = 1; 2; ···, and x 2[0; L] are eigenfunctions of the eigenvalue problems



= ¡λφ

φ(0) = φ(L) = 0

Similarly, the set of orthogonal functions



cos

nπ



for n = 0; 1; 2; ··· are generated by the eigenvalue

problem



= ¡λφ

(0) = φ

(L) = 0

Consider t he following eigenvalue problem:

= ¡λφ

φ(0) = 0

φ(1) + φ

(1) = 0

a) Use a numerical method and ﬁnd ﬁrst six eigenvalues o f the problem λ

; :::; λ

. You can verify

that the eigenfunctions φ

(x) = sin( λ

x) are mutually orthogonal for the values of λ

; :::; λ

b) Find the bets approximation of the function f(x) = sin(πx) for x 2[0; 1] in terms of fφ

(x)g

n=1

You can use the following code in Matlab:

g=@(s) sin(s)+s.*cos(s);

z=zeros(1,6);

for n=1:6

z(n)=fzero(g,3*n);

end

C=integral(@(x) sin(pi*x).*sin(z(1:6)'.*x),0,1,'arrayvalue',true)./

integral(@(x) sin(z(1:6)'.*x).^2,0,1,'arrayvalue',true);

x=0:0.01:1;

fhat=sum(C.*sin(z(1:6)'.*x));

plot(x,sin(pi*x),x,fhat)

20 1D Linear Second-Order Equations

In the above code, the set z are equal to λ

Problem 4.21. In the book of ODEs, we saw that Legendre polynomials P

(x) are orthogonal functions

in [ ¡1; 1]. Find the best approximation of f (x) = sin(πx), x 2 [¡1; 1] in the set fP

(x)g

n=1

. You can

use the following code in Matlab to draw f(x) and its best approximation in the same coordinate. The

cod e provides also the error:

kf ¡ f

k:=

¡1

jsin(πx) ¡ f

(x)j

N=4;

C=integral(@(x) sin(pi*x).*legendreP(1:N,x),-1,1,'arrayvalued',tr ue).*(2*(1:N)+1)/

x=-1:0.01:1;

fhat=0;

for i=1:N

fhat=fhat+C(i).*legendreP(i,x);

end

plot(x,sin(pi*x),x,sum);

err=sqrt(trapz((sin(pi*x)-fhat).^2));

title('error=',err)

Problem 4.22. Consider the set of functions f'

= 1; '

= x; '

= x

g deﬁned in the domain [¡1; 1].

a) Fin d the best approximation of function f(x) = sin(πx) in terms of f'

n=0

. It is worth noting

that these functions are orthogonal. To solve this, let's write f

= a

+ a

x + a

, and minimize

the norm kf ¡ f

k by assuming that f ¡ f

is orthogonal to functions in the set f'

g, i.e.,

¡1

(sin(x) ¡a

¡b

x ¡c

= 0;

for n = 0; 1; 2.

b) Determine the square error deﬁned by th e following formula:

¡1

jf(x) ¡ f

(x)j

dx:

c) Repeat parts a) and b) for the set f'

= 1; '

= x; '

= x

; '

= x

g. As you can see, we need to

recalculate all the coeﬃcients. However, this is not the case for orthogonal functions. This is one

of the reasons why orthogonality is highly valued in app lied sciences.

4.3 From Strum-L iouville problem to PDEs

4.3.1 Homogeneous 1D problems

To demonstrate the application of the Sturm-Liouville problem in solving partial diﬀerential

equations, we consider the homogeneous heat equation given by:

= L[u] x 2(x

; x

)

u(x

; t) + β

; t) = 0

u(x

; t) + β

; t) = 0

u(x; 0) = f(x)

;

where L represents a linear second-order partial diﬀerential operator in the variable x:

L[u] = a(x) u

x x

+ b(x) u

+ c(x) u:

4.3 From Strum-Liouville p roblem to PDEs 21

The associated eigenvalue problem for the spatial variable x is:

L[φ(x)] = ¡λφ(x);

φ(x

) + β

) = 0

φ(x

) + β

) = 0

According to the Sturm-Liouville theorem, there exists an inﬁnite set of eigenvalues λ

₁

<λ

₂

<···

with corresponding orthogonal eigenfunctions φ₁(x); φ₂(x); ···, which are orthogonal with

respect to the weight function σ(x). Furthermore, the s et fφₙ(x)g forms a basis f o r piece-

wise continuously diﬀerentiable functions deﬁned on the interval (x₀; x₁). Therefore, the

desired solution u(x; t) can be expressed as a series in terms of this set:

u(x; t) =

n=1

(t) φ

(x);

where Uₙ(t) are coeﬃcient functions that need to be determined appropriately for the series

to be a valid solution to the problem. Substituting this series into the equation yields:

n=1

(t) φ

(x) =

n=1

(t) L [φ

(x)] =

n=1

¡λ

(t) φ

(x):

This equality leads to the following ordinary diﬀerential equation for U ₙ(t):

+ λ

= 0;

which can be solved to obtain the function

(t) = C

¡λ

Finally, we arrive at the solution:

u(x; t) =

n=1

¡λ

(x):

To ensure that this series satisﬁes the initial condition at t=0, we require:

lim

t!0

u(x; t) = u(x; 0) =

n=1

(x):

By the g iven initial condition, we obtain the following equation:

f(x) =

n=1

(x);

22 1D Linear Second-Order Equations

where the orthogonality of eigenfunctions φₙ(x) determines Cₙ as:

hf ; φ

kφ

f(x) φ

(x) σ(x) dx

Finally, the true solution of the given problem is:

u(x; t) =

n=1

hf ; φ

kφ

¡λ

(x):

This series solution known as the eigenfunction series solution provides converges to a

smooth so lution for t > 0 even if the initial condition is not continuous. This method

can be employed to solve a wide range of 1D second-order PDEs in t and x for a two-

variable function u(x; t).

Example. We will solve the standard heat problem with the given conditions:

= ku

x x

x 2(0; 1)

u(0; t) = 0

u(1; t) = 0

u(x; 0) = f(x)

The associated eigenvalue problem is



= ¡λφ

φ(0) = φ(1) = 0

;

Solving the eigenvalue problem, we obtain the eigenpairs λ

and φ

= sin(nπx). Using

these eigenfunctions, the superposition solution can be expressed a s:

u(x; t) =

n=1

¡kn

sin(nπx):

The coeﬃcients c

can be determined by solving the equation:

f(x) =

n=1

sin(nπx):

The orthogonality of the eigenfunctions φ

with respect to the weight function σ = 1 leads

to the coeﬃcients:

= 2

f(x) sin(nπx) dx:

Let's consider the speciﬁc initial condition given as:

f(x) =

(

≤x ≤

0 otherwise

4.3 From Strum-Liouville p roblem to PDEs 23

The coeﬃcients c

can be calculated as:

nπ



cos



nπ



¡cos



3nπ



The ﬁgures below illustrate the initial condition and the solution at diﬀerent instances of

time for k = 0.1.

0 0.5 1

0.5

0 0.5 1

0.5

0 0.5 1

0.5

0 0.5 1

0.5

Example 4.7. Let us solve the following parabolic equation

= u

x x

+ 2u

u(0; t) = 0

u(1; t) = 0

u(x; 0) = f(x)

;

for k > 0 a constant. The associated eigenvalue problem for the given PDE is

+ 2φ

= ¡λφ

φ(0) = 0

φ(1) = 0

The characteristic equation of the ODE is

+ 2r + λ = 0;

24 1D Linear Second-Order Equations

which has roots r

1;2

= ¡1 ± 1 ¡λ

. It is observed that the eigenvalue problem is s o lvable

only for λ > 1. In th is case, the solution φ(x) is given by:

φ(x) = c

¡x

cos( λ ¡1

x) + c

¡x

sin( λ ¡1

x):

Applying the given boundary conditions, we ﬁnd c

= 0, and λ

= 1 + n

with associated

eigenfunctions φ

(x)=e

¡x

sin(nπx), which are orthogonal with respect to the weight function

σ(x) = e

. The superposition solution can be expressed as:

u(x; t) =

n=1

¡(1+n

)kt

¡x

sin(nπx):

To determine the coeﬃcients c

, we need to satisfy the initial condition f (x). Thus, the series

solution should satisf y the equation:

f(x) =

n=1

¡x

sin(nπx);

and the coeﬃcients can be computed as:

= 2

f(x) e

¡x

sin(nπx) e

dx;

where the f a ctor e

represents the weight function σ(x). The ﬁgures below depict the graph

of the initial heat proﬁle f (x) = xe

¡x

and the solution u(x; t) at diﬀerent time values for

k = 0.1.

0 0.5 1

0.1

0.2

0.3

0.4

0 0.5 1

0.1

0.2

0.3

0.4

0 0.5 1

0.1

0.2

0.3

0.4

0 0.5 1

0.1

0.2

0.3

0.4

4.3 From Strum-Liouville p roblem to PDEs 25

Example 4.8. Let's solve the damped wave equation

+ 2ξu

= u

x x

u(0; t) = 0

(π; t) = 0

;

where ξ > 0 is the damping factor. This equation represents an elastic string ﬁxed at x=0

and free at x=1.

To solve the problem, we ﬁrst form the associated eigenva lue problem:

= ¡λφ

φ(0) = 0

(π) = 0

The s o luti o ns of the eigenvalue problem are given by the pair s λ

(2n ¡ )

, and φ

sin

2n ¡ 1



for n = 1; 2; ···. Since the set fφ

(x)g forms a basis for fun ctions deﬁned on x2[0;

π], we can write the solution of the given PDE as:

u(x; t) =

n=1

(t) φ

(x);

where U

(t) are undetermined coeﬃcients. This series is known as the eigenfunction expan-

sion of the unknown solution u(x; t). To determine U

(t), we substitute the eigenfunction

series into the PDE:

n=1

(t) + 2ξU

(t)] φ

(x) =

n=1

(t) φ

(x);

and using the relation φ

= ¡λ

, we obtain the following ODE for U

(t):

(t) + 2ξU

(t) + λ

(t) = 0:

Assuming ξ < λ

, the solution of the above ODE is given by:

(t) = e

¡ξt

cos(!

t) + B

sin(!

t)];

where !

= λ

¡ξ

. The superposition solution of the given PDE is:

u(x; t) =

n=1

¡ξt

cos(!

t) + B

sin(!

t)] φ

(x):

The coeﬃcients A

and B

are determined using the initial conditions u(x; 0) and u

(x; 0) for

the equation. The ﬁgures below show the wave for u(x; 0) = 0, u

(x; 0) = sin

3π



and ξ = 0.1

26 1D Linear Second-Order Equations

0 0.5 1

-1

-0.5

0.5

0 0.5 1

-1

-0.5

0.5

0 0.5 1

-1

-0.5

0.5

0 0.5 1

-1

-0.5

0.5

Exercise 4.17. Find the series solution of the following parabolic type PDE on the domain [1; e]

= x

+ xu

u(1; t) = u(e; 0) = 0

u(x; 0) = ln(x)

Exercise 4.18. We aim to solve th e following wave equation

= c

u(0; t) = 0

u(1; t) + u

(1; t) = 0

u(x; 0) = sin(πx)

(x; 0) = 0

a) Determine ﬁrst 6 eigenvalues and eigenfunctions of the following eigenvalue prob lem

= ¡λφ

φ(0) = 0

φ(1) + φ

(1) = 0

b) Use the d erived eigenfunctions to ﬁnd a truncated series solution of the wave equation.

Exercise 4.19. Find the eigenva lues and eigenfunctions of the eigenvalue problem:

+ 2φ

= ¡λφ x 2(0; 1)

φ(0) = 0

(1) = 0

4.3 From Strum-Liouville p roblem to PDEs 27

Use the result and write the solution of the following heat equation

= u

+ 2u

x 2(0; 1)

u(0; t) = 0

(1; t) = 0

u(x; 0) = xe

¡x

4.3.2 Non-homogeneous eq u ations

In this section, we apply the eigenfunction expansion method to solve non-homogeneous

linear second-order partial diﬀerential equati o ns. To illustrate the method, let us consider

the following heat problem:

= L[u] + h(x; t)

u(0; t) = u(1; t) = 0

u(x; 0) = 0

The associated eigenvalue problem for the g iven equation is



L[φ] = ¡λφ

φ(0) = φ(1) = 0

Since the set of eigenfunctions fφ

(x)g forms a basis for functions deﬁned on the interval

[0; 1], we express the desired solution as a series:

u(x; t) =

n=1

(t) φ

(x):

Substituting this series into the heat equation results in:

n=1

(t) + λ

] φ

(x) = h(x; t): (4.4)

To proceed, we can represent the source term h as the series in terms of fφ

(x)g as:

h(x; t) =

n=1

(t) φ

(x);

where H

(t)=

hh; φ

kφ

. Substituting the series representation of h into the series equation (4.4),

we obtain th e following ordinary diﬀerential equation for U

(t):

(t) + λ

= H

(t):

For example, let's consider h(x; t)=e

¡t

(x). Expanding h in terms of fφ

g yields: H

(t) = 0

for n =/ 1 and H

(t) = e

¡t

. Therefore, we obtain the following initial value problems:



+ λ

= e

¡t

(0) = 0

;



+ λ

= 0

(0) = 0

; n = 2; 3; ···:

28 1D Linear Second-Order Equations

The initial conditions for Uₙ(t) are chosen to satis f y the zero initial condition u(x; 0)=0.

Solving this system of equations, we ﬁnd U

(t) = 0 fo r n = 2; 3; ···, and U

(t) =

¡ 1

¡t

¡λ

), and ﬁnally the solution is given by:

u(x; t) =

¡1

¡t

¡e

¡λ

) φ

(x):

Remark 4.4. (System interpretation) Let's examine the a bove example from a system

perspective. We can view the heat equation as a dynamic system that responds to diﬀerent

input sources. In this case, we have a speciﬁc input source represente d by h=H₁ (t)sin(πx).

The heat system can be redeﬁned as:

u = (@

¡L)

¡1

(t) φ

(x)]:

Since φ

is an eigenfunction of the operator L, the response of the heat system to this input

is of the general form u=U (t)φ₁(x). T his can be represented using a block diagram as:

u = U(t) φ

(x)

Heat system

(t)φ

(x)

¡L)

¡1

Now, let's consider a more general source term h(x; t) = h

(t) φ

(x) + h

(t) φ

(x). By

applying the superposition principle, we can express the desired solution as:

u = U

(t) φ

(x) + U

(t) φ

(x):

This is depicted schematically in the ﬁgure below:

Heat system

u = U

(t) φ

(x) + U

(t) φ

(x)

(t) φ

(x)

(t) φ

(x)

¡L)

¡1

This argument extends for arbitrary summations h =

(t) φ

(x) as shown below:

u =

(t) φ

(x)

¡L)

¡1

Heat system

(t)φ

(x)

Here each coeﬃcient function U

(t) is determined by solving the associated ordinary

diﬀerential equation:

+ λ

= H

(t):

Remark 4.5. (Geometric i nterpretation) Another perspective to consider is the ge o -

metric interpretation of the given partial diﬀerential equation. Each eigenfunction φₙ(x) can

be seen as a direction in an inﬁnite-dimensional vector space spanned by the functions in

the set fφₙ(x)g. Alo ng each direction, the given partial diﬀerential equation reduces to an

ordinary diﬀerential equation of the form:

+ λ

= H

(t):

4.3 From Strum-Liouville p roblem to PDEs 29

In this geometric interpretation, we can view the partial diﬀerential equation as an inﬁnite

system of ordinary diﬀerential equations, where each equation is deﬁned along the direction

of φ

. Drawing a parallel to our approach for solving ﬁrst-order partial diﬀerential equations,

we can observe similarities in the method. In both ca ses, we identify characteristic curves

along which the given partial diﬀerential equation transforms into an ordinary diﬀerential

equation. This observation highlights the underlying sim ilarity between the two methods

and reinforces th e understanding of the eigenfunction expansion method.

Exercise 4.20. Find the series solution of the following heat prob lem

= u

+ sin(2πx)

u(0; t) = u(1; t) = 0

u(x; 0) = 0

Exercise 4.21. Find the solution of the following wave equation

= u

+ sin(t) sin



u(0; t) = u

(1; t) = 0

u(x; 0) = 0

(x; 0) = 0

Exercise 4.22. Consider the following heat problem

= u

+ e

¡t

f(x)

u(0; t) = u(1; t) = 0

u(x; 0) = 0

;

where f(x) is the following function

f(x) =

n=1

¡2(¡1)

nπ

sin(nπx):

a) Write the solution to the problem.

b) Supp ose f(x) = x. Write down the solution to the equation. Hint: write down f(x) as

f(x) =

n=1

¡2(¡1)

nπ

sin(nπx);

employ the same method you used for the part a).

30 1D Linear Second-Order Equations

Exercise 4.23. Consider the following heat equation

= u

u(0; t) = u(1; t) = 0

u(x; 0) = f (x)

;

where f(x) = φ

(x), and φ

(x) = sin(nπx) for n = 1; 2; ···.

a) Since the heat system is triggered only by the function φ

(x), write the solution u(x; t) as

u(x; t) = U (t) φ

(x):

Determine U (t).

b) Now, let f(x) be the following function:

f(x) =

n=1

¡2(¡1)

nπ

(x):

Find the solution of the equation.

c) Now, let f(x) = x + sin(πx). Determine the solution to the equation.

4.3.3 Non-homogeneous boundary con d itions

So far, our discussion has focused on problems with homogeneous boundary conditions. Now,

let's consider a case where the boundary conditions are non-homogeneous. The following

example will illustr ate this scenario.

Example 4.9. Let's continue with the given equation:

= u

+ 2u

¡e

¡2x

u(0; t) = 1

u(1; t) = 1 +

¡2

We consider the solution as u(x; t)=V (x)+w(x; t), where V (x) reﬂects the contribution of

the time-independent sources and w(x; t) satisﬁes the homogeneous equation. Substituting

this solution into the heat equation gives:

= V

+ 2V

+ w

x x

+ 2w

+ e

¡2x

To keep w as a homogeneous e quation, we take V (x) to satis f y the equation:

+ 2V

= e

¡2x

We transfer the non-zero boundary conditions to V (x), resulting in the following boundary

value problem for V :

+ 2V

= e

¡2x

V (0) = 1

V (1) = 1 +

¡2

4.3 From Strum-Liouville p roblem to PDEs 31

Solving this equation gives us the speciﬁc solution for V (x):

V (x) = 1 ¡

¡2x

For w(x; t), we have the following homogeneous equation and boundary conditions:

= w

x x

+ 2w

w(0; t) = 0

w(1; t) = 0

The general series solution for w is given by:

w(x; t) =

n=1

¡λ

(x);

where φ

¡x

sin(nπx), and λ

=1+n

. Finally, the general solution for the given problem

is obtained as:

u(x; t) = 1 ¡

¡2x

n=1

¡λ

(x):

The parameters C

can be determ ined by the initial condition of the p roblem. For example,

if u(x; 0)=0, then we have:

¡2x

¡1 =

n=1

(x):

Using the orthogonality of eigenfunctions with respect to the weight function σ =e

, the

coeﬃcients C

are determined as:

h¡V ; φ

kφ

Finally, we obtain the solution a s:

u(x; t) = 1 ¡

¡2x

n=1

hV ; φ

kφ

¡λ

(x):

In general, when the equation or its boundary conditions are functions independent of

time, it can be beneﬁcial to split the solution into two terms: 1) a pure function of x denoted

as V (x), and 2) a function of both x an d t denoted as w(x; t). The d esire d solution can then

be written as the sum of these two terms: u(x; t)=V (x)+w(x; t).

The function V (x) incorporates all the non-homogeneous terms that are independent of

time, while w(x; t) satisﬁes the zero boundary conditions. This decomposition allows us to

separate the time-independent part f rom the time-dependent part of the solution, making

the problem more manageable.

Example 4.10. Let's solve the following equation

= u

¡6x + e

¡t

sin(πx)

u(t; 0) = 0; u(t; 1) = 1

u(0; x) = x

32 1D Linear Second-Order Equations

The boundary condition is non-homogeneous. We take u(x; t) as the sum of two terms:

u(t; x) = V (x) + w(t; x):

By substituting u into the diﬀerential equation, we obtain:

= V

+ w

¡6x + e

¡t

sin(πx);

and thus the equation for V (x) is:



= 6x

V (0) = 0; V (1) = 1

The diﬀerential equation for V (x) is solved and we ﬁnd V (x)=x

. The equation fo r w(t; x)

becomes:

= w

+ e

¡t

sin(πx)

w(t; 0) = w(t; 1) = 0

w(x; 0) = 0

We observe that w(x; t) is triggered onl y by the term h(x; t) = e

¡t

( x), wher e

(x)= sin(nπx). Therefore, we can write the solution as:

w(t; x) = W (t) sin(πx);

where W (t) is an u ndetermined function. Substituting this series into the equation for

w(t; x), we obtain the following ﬁrst-order ODE for W (t):

+ π

= e

¡t

The solution to this ODE is:

W (t) = C e

¡π

¡1

¡t

The initial condition w(x; 0)=0 implies W (0)=0, which determines C = ¡

¡ 1

. Therefore,

the solution u(x; t) is given by:

u(t; x) = x

¡1

(¡e

¡π

+ e

¡t

) sin(πx):

It is observed that the derived solution satisﬁes the partial diﬀere ntial equation, as well as

the given boundary an d initial cond itions. Additionally, we note that:

lim

t!1

u(t; x) = x

= V (x);

and thus the function V (x) is also referred to as the steady-state solution in this case. The

function w(x; t) vanishes with tim e and is called the transient solution.

Exercise 4.24. Solve the following heat problem

= u

¡x

u(0; t) = ¡1; u

(π; t) = 0

u(x; 0) =

4.3 From Strum-Liouville p roblem to PDEs 33

Exercise 4.25. Solve the following wave problem

= u

u(0; t) = 1

(π; t) = 0

u(x; 0) = 0

(x; 0) = 0

4.3.4 Problem

Problem 4.23. Let Ω be the domain Ω:=(0; π). We aim to solve the following problem:

= u

+ 2u

+ sin(t) e

¡x

sin



on Ω

u(0; t) = 0

(π; t) + u(π; t) = 0

a) Fin d eigenvalues λ and their associated eigenfunctions φ of the following eigenvalue problem:



+ 2φ

= ¡λφ

φ(0) = 0; φ

(π) + φ(π) = 0

You can assume that λ > 1. Determine the weight function for the orthogonality of the eigenfunc-

tions too.

b) Use the results from part a) to ﬁnd the solution of the given wave problem when the initial

condition is given by: u(x; 0)=0, u

(x; 0) = 0.

Problem 4.24. Find the eigenvalues and eigenfunctions of the following eigenvalue problem

= ¡λφ x 2(¡1; 1)

φ(¡1) = 0

φ(1) = 0

Use the result and ﬁnd the series solution of the following heat problem

= u

u(¡1; t) = 0

u(1; t) = 0

u(x; 0) = x

Problem 4.25. Consider the following heat equation

= @

[(1 + x)

]

u(0; t) = 0

(1; t) = 0

Find the form of the eigenfunctions and use a numerical method to determine ﬁrst ﬁve eigenvalues and

associated eigenfunctions. Use these values and write downs of the series solution of the heat equation.

Find the ﬁrst ﬁve coeﬃcients o f the series solution if u(x; 0) is given as

u(x; 0) =

100

1 + x

Problem 4.26. Consider the following heat equation

= x

+ 3xu

+ u

u(1; t) = 0; u(e; t) + u

(e; t) = 0

u(x; 0) = x

a) Use a numerical method to ﬁnd the ﬁrst 3 eigenvalues of the associated eigenvalue problem

34 1D Linear Second-Order Equations

b) Use these eigenfunct ions and write downs an approximate series solution t o the heat equation.

Problem 4.27. Consider the wave equ ation:

= c

u(0; t) = u(1; t) = 0

u(x; 0) = f (x)

(x; 0) = 0

a) Write the series solution of the equation.

b) Use the trigonometric identity for terms cos(c nπt) sin(nπx) and conclude the solution can be

written as

u(x; t) =

odd

(x ¡ct) + f

odd

(x + ct)

Problem 4.28. (Fourier series) Conside r the following eigenvalue problem

(θ) = ¡λφ(θ) θ 2(¡π; π)

φ(¡π) = φ(π)

(¡π) = φ

(π)

The physics of the problem is a circle, and φ(θ) is the function deﬁned on this circle. The boundary

condition of the problem is not Robin condition and then we are not allowed to use the result of the

Sturm-Liouville theorem.

a) Show that the eigenvalues of the problem are λ

= n

, n = 0; 1; 2; ··· and eigenfunctions are

fcos(nθ)g

n=0

and fsin(nθ)g

n=1

. These functions are know and the Fourier eigenfunctions.

b) It is known that the eigenfunctions form an orthogonal basis for piecewise continuously diﬀeren-

tiable functions deﬁned on θ 2[¡π; π]. Use the orthogonality condition as ﬁnd an exp ansion for

the function

f(θ) =



1 ¡π < θ < 0

0 0 < θ < 0

You can use the f ollowing code in M atla b and see the graph of f(θ) and its approximation for a

truncated series.

%length

L=pi;

%functon

f=@(x) 1*(x>-L & x<0)+0*(x>0 & x<L);

%number of terms

n=10;

%the coefficients of Fou rier series

a0=integral(@(x) f(x),-L,L)/(2*L);

a=integral(@(x) f(x).*cos((1:n)'.*x*pi/L),-L,L,'arrayvalued',true)/L;

b=integral(@(x) f(x).*sin((1:n)'.*x*pi/L),-L,L,'arrayvalued',true)/L;

%making fhat

x=-L:0.01:L;

fhat=a0+sum(a.*cos((1:n)'*x*pi/L))+sum(b.*sin((1:n)'*x*pi/L));

%plofing functions

plot(x,f(x),x,fhat);

c) Use the above result and solve the following wave equation for u(θ; t) on the unit circle

= u

θθ

¡π < θ < π

u(¡π; t) = u(π; t)

(¡π; t) = u

(π; t)

4.3 From Strum-Liouville p roblem to PDEs 35

The ﬁgures below depicts the graph of the solution a t some instances of time for the initial

conditions

u(θ; 0) =

(

¡θ

¡

θ +



< θ <

0 otherwise

;

and u

(θ; 0) = 0.

0°

90°

180°

270°

0°

90°

180°

270°

0°

90°

180°

270°

0°

90°

180°

270°

Problem 4.29. Solve the following wave problem

= u

+ 2u

+ u + tx

u(0; t) = u(π; t) = 0

u(x; 0) = x; u

(x; 0) = 0

Problem 4.30. We know that functions φ

(x) = x sin(nπx) for n = 1; 2; ··· solve the following eigenvalue

problem:

(

¡2xφ

+ 2φ = ¡λx

φ(1) = φ(2 ) = 0

a) Determine the associated eigenvalue λ

b) Write down the series solution of the following problem

= u

u(1; t) = u(2; t) = 0

u(x; 0) = sin(πx)

Problem 4.31. Consider the following problem



= u

u(0; t) = 0; u(π; t) = t

The boundary condition at x = π is a function of time t.

a) Take u as: u(x; t) = V (x; t) + w(x; t). Substituting this into the equation results in:

+ V

= V

+ w

36 1D Linear Second-Order Equations

Assume that V satisﬁes the following boundary value problem

= 0

V (0; t) = 0

V (π; t) = t

Determine the function V (x; t).

b) Substituting this function into the equation for w yields:

= w

w(0; t) = 0

w(π; t) = 0

Solve t his equation for w and ﬁnd the solution to the problem if u(x; 0) = 0.

Problem 4.32. Consider the following wave equation

(

= u

u(0; t) = 0; u(π; t) = sin(t)

a) Take u as: u(x; t) = V (x; t) + w(x; t). Substituting this into the equation results in:

+ V

= V

V + w

Assume that V satisﬁes the following boundary value problem

V = 0

V (0; t) = 0

V (π; t) = sin(t)

Determine the function V (x; t).

b) Substituting this function into the equation for w yields:

¡sin(t) sin



= w

w(0; t) = 0

w(π; t) = 0

Solve t his equation for w and ﬁnd the solution to the problem if u(x; 0) = 0.

Problem 4.33. Solve the following heat problem

= x

+ 4u

u(1; t) = ¡1; u(e; t) = 1

u(x; 0) = x

Problem 4.34. Solve the following heat problem:

= u

+ 2u

u(0; t) = 0; u(1; t) = 1

u(x; 0) = x

Problem 4.35. Solve the following equation

= @

[(1 + x)

] ¡2

u(0; t) = 0; u

(1; t) = 1

u(x; 0) = 102ln (1 + x)

4.3 From Strum-Liouville p roblem to PDEs 37

Problem 4.36. Solve the following equation

= @

[(1 + x)

] +

1 + x

u(0; t) = 0; u

(1; t) = 1

u(x; 0) =

1 + x

Problem 4.37. Solve the following equation

= u

+ 2u

+ e

¡t

¡x

sin(πx)

u(0; t) = u(1; t) = 0

u(x; 0) = 0

Problem 4.38. Consider the following problem

= u

u(0; t) = u(1; t) = 0

u(x; 0) = f (x)

Show that the solution to the problem is equal to the solution to the following problem

= u

+ δ(t) f (x)

u(0; t) = u(1; t) = 0

u(x; 0) = 0

;

where δ(t) is the Dirac delta function.

Problem 4.39. Consider the following problem

= u

+ h(t) f (x)

u(0; t) = u(1; t) = 0

u(x; 0) = 0

;

where h = 0 f or t ≤0. Show that the series solution of the problem is

u(x; t) =

n=1

(h(t) ∗e

¡λ

) φ

(x);

where φ

= sin(nπx) are the eigenfunctions of the associated eigenvalue prob lem, λ

are the associated

eigenvalues, F

are the coeﬃcients of the expansion of f(x) in terms of the functions in fφ

g, and

h(t) ∗e

¡λ

is t he convolution of e

¡λ

and h(t) deﬁned a s

h(t) ∗e

¡λ

h(τ) e

¡λ

(t¡τ )

dτ:

Problem 4.40. Consider the following problem

= u

u(0; t) = u(1; t) = 0

u(x; 0) = 0; u

(x; 0) = g(x)

Show that the solution to the problem is equal to the solution to the following problem

= u

+ δ(t) g(x)

u(0; t) = u(1; t) = 0

u(x; 0) = 0; u

(x; 0) = 0

;

38 1D Linear Second-Order Equations

where δ(t) is the Dirac delta function.

Problem 4.41. Solve the following damped wave equation

+ 2u

= u

+ sin(x)

u(0; t) = u(1; t) = 0

u(x; 0) = 0; u

(x; 0) = 1

Problem 4.42. Consider the following equation

= u

+ 2u

u(0; t) = 0

(1; t) = 0

a) The associated eigenvalue problem is



+ 2φ

= ¡λφ

φ(0) = φ(1) = 0

Use a numerical method and ﬁnd ﬁrst ﬁve eigenvalues and the associated eig enfu nctions.

b) Use these set and approximate the solution of t he heat problem if u(x; 0) = 1.

Problem 4.43. Consider the following problem

= u

+ u

u(0; t) = 0

u(π; t) = 1

Let us take u(x; t) = V (x) + w(x; t), and deriving the following equation for V :



+ V = 0

V (0) = 0; V (π) = 1

Show that this equation is not solvable. What is your sug gestion to solve the given heat problem?

4.4 Theoretical aspects of the method

The method we introduced above is known as t he eigenfunction expansion method, which

is based on expanding the desired solution in terms o f the eigenfunctions of an eigenvalue

problem.

4.4.1 Function s p aces and convergence

In linear algebra, we learn that a set of vectors can form a vector space when equipped with

the operations of vector addition (+) and scalar multiplication. If the set is closed under these

operations, it creates a vector space that can serve as the domain or range of linear mappings.

Similarly, in the context of diﬀerential operators, the domain and range are vector func-

tion spaces. Consider two functions, f and g, deﬁned on the same domain (x

; x

). The

vector addition of f and g is deﬁned as follows:

(f + g) (x) = f (x) + g(x);

4.4 Theoretical aspects of the method 39

for all x2(x

; x

). Similarly, scalar multiplication λf is deﬁned as:

(λf)(x) = λf (x):

These operations maintain closure within the function space.

One important class of function spaces is the C

spaces. Recall that a function f(x)

is continuously diﬀerentiable in the interval (x

; x

) if its derivative f

(x) is a co ntinuous

function in the same interval. We denote this as f 2C

; x

). A function f is said to be

continuously diﬀerentiable of order r in the interval (x

; x

), denoted as C

; x

), if its k

derivative f

(k)

(x) is in t he vector space C(x

; x

) of continu o us functions, for all k up to r.

Exercise 4.26. Consider the function

f(x) =

(

x sin



x =/ 0

0 x = 0

Show that the function is not diﬀerentiable at x = 0. The functions

f(x) =

(

sin



x =/ 0

0 x = 0

is diﬀerentiable at x = 0 but not in C

(¡1; 1).

Indeed, function spaces, such as the space spa nned by fsin(nπx)g for n = 1; 2; :::, are n ot

of ﬁnite dimension. They contain linearly independent functions (which are orthogonal in

this case), making it tricky to deﬁne conve rgen ce in such spaces. The meaning of a s eries like:

f(x) =

n=1

(x);

One approach is to deﬁne the norm of a function f 2C

; x

) as:

kf k=

k=0

(k)

(x)j

;

where the norm measures the square roo t of the sum of squared derivatives. Convergence of

the series in this n o rm would imply:

lim

N !1



f(x) ¡

n=1

(x)



= 0:

Another notion of norm for a function f 2C

; x

) denoted by kf k

, can be deﬁned as:

kf k

k=0

sup

x2(x

)

(k)

(x)j;

which measures the supremum (or maximum) value of the derivative s. Consequently, the

notion of convergence changes based on this new deﬁnition.

40 1D Linear Second-Order Equations

In this book, the focus is mainly on po intwise convergence, which means that for any

ξ 2(x

; x

), and a ny " > 0, there exists an N = N(ξ,") such that:



f(ξ) ¡

n=1

(ξ)



< ":

This means that the series converges to the function f(x) at each point ξ w ithin the interval

; x

) with an arbitrary small error ".

Exercise 4.27. Sh ow that the sequence of functions f

(x) = x

for x 2[0; 1] and n = 1; 2; ··· converges

po intwise to the function

f(x) =



1 x = 1

0 0 ≤x < 1

However, if we deﬁne the norm

kuk

= sup

x2[0;1]

ju(x)j;

the sequence f

(x) does not converges to f(x) in this norm. Show that if we change the norm to

kuk=

ju(x)j

;

the sequence f

(x) converges to f(x) in this norm.

Exercise 4.28. Consider the sequence of functions

(x) =

0 < x <

≤x < 1

Show that f

(x) converges pointwise to the zero functions f (x) = 0 for x 2(0; 1). However, f

(x) does

not converges to the zero function in norm compatible with the inner product.

Exercise 4.29. Assume φ

(x) are mutually orthogonal functions and the sequence of functions

(x) =

k=1

hf ; φ

kφ

;

converges in norm to functions f(x), where the norm is compatible with the inner product h; i. Show

the equality

kf k

= lim

n!1

k=1

hf ; φ

kφ

Exercise 4.30. Consider the sequence of functions f

(x) =

sin(nx) o n 0 ≤x ≤π for n = 1; 2; ···.

i. Show that f

(x) !0 in the norm compatible with the inner product:

hf ; gi=

f(x)g(x)dx:

ii. If the inner product changes to the following one, show that the sequence f

(x) diverges:

hf ; gi=

f(x)g(x)dx +

(x)g

(x)dx:

Exercise 4.31. Consider the sequence of functions

(x) =

k=0

;

4.4 Theoretical aspects of the method 41

for n = 1; 2; ··· in the interval [0; 1]. Show that the sequence converges pointwise and in norm to the

function f(x) = e

In this book, we also consider the function spaces o f piecewise continuous functio ns

PC(x

; x

) an d piecewise continuously diﬀerentiable fu nc tions PC

; x

Deﬁnition 4.2. A function f(x), deﬁned on t he interval (x

; x

), is called a piecewise

continuous function if:

a) f is continuous every where in the interval, except possibly at some ﬁnite points.

b) If z is a discontinuous point for f, then the left and right limits of f at z exist.

c) The right limit of f exists at x

, and the left limit exists at x

Deﬁnition 4.3. A function f(x) belongs to the space PC

; x

) if f

(x) belongs to the

space PC(x

; x

), except possibly at some ﬁnit e points. In other words, f (x) is piecewise

continuously diﬀerentiable, and its derivative f

(x) is a piecewise continuous function.

These function spaces a llow us to work with functions that may have discontinuities at

certain points within the interval (x

; x

). Piecewise continuous functions are characterized

by their continuity properties and the existence of left and right limits at potential discon-

tinuities. Piecewise continuously diﬀerentiable functions, on the other hand, additionally

possess piecewise continuous derivatives, except at a ﬁnite number of points.

Example 4.11. Let's consider the function:

f(x) =



1 x 2Q

0 x 2Q

;

where Q is the set of rational numbers. This function is not piecewise continuous in any

interval (a; b) because it h a s inﬁnitely many discontinuities or jumps within the i nterval.

It is known as the Dirichlet function and serves as a classic example of a function that is

discontinuous almost everywhere.

Example 4.12. Now, let's examine the function:

f(x) =

(

x =/ 0

0 x = 0

;

This function is no t in the space PC(¡1; 1) because it exhibits an inﬁnite jump at x=0.

Similarly, consider the function:

f(x) =

(

sin



x =/ 0

0 x = 0

This function i s also not in the space PC(¡1; 1) since it oscillates inﬁnitely near x=0.

Example 4.13. Let's examine the function f(x)=x

. This function belongs to the space

PC(¡1; 1) since it is continuous in the interval (¡1; 1). However, it does not belong to the

space PC

(¡1; 1) because its derivative, f

(x), is not piecewise continuous within the interval.

42 1D Linear Second-Order Equations

Similarly, the function

f(x) =

(

sin



x =/ 0

0 x = 0

is continuous and diﬀerentiable in the interval (¡1; 1). However, its derivative , f

(x), is not

piecewise continuous at x=0 as it exhibits inﬁnite oscillation near this point.

Exercise 4.32. Which one o f the following functions belong to CP

or CP?

i. f (x) = x

2/3

, x 2[0; 1]

ii. f (x) = xjxj, x 2¡1 ; 1

iii. f (x) = jxj

, x 2[¡1; 1]

iv. f(x) =



1 0 < x < 1

0 otherwise

4.4.2 Proof of the orthogonality of eigenfunctions

The complete proof of the Sturm-Liouville theorem requires advanced tools that a re beyond

the scope of this book. However, we can establish two key aspects of the theorem through

an elementary approach. The crucial step in achieving the proof is to reframe the problem

in a symmetric form. To accomplish this, we need to introduce the concept of symmetry for

a diﬀerential operator.

The notion of inner product extends to another important concept known as symmetric

operators. This concept can be related to the concept of symmetric matrices in linear algebra.

In linear alge bra, a square matrix A = [a

] is considered s ymmetric if a

. It is known

that symmetric ma trices have real eigenvalues and orthogonal eigenvectors.

We can extend this concept to lin ear diﬀerential operators using the inner product.

Deﬁnition 4.4. A linear diﬀerential operator L is said to be symmetric with respect to the

inner product h; i if the following equality holds for all functions f and g in its domain:

hL[f ]; gi= hf ; L[g]i:

This deﬁnition establishes a symmetry relationship between the operator L and the inner

product, similar to the symmetry observed in symmetric matric es.

Example 4.14. Consider the diﬀerential operator L :=

deﬁned on the set of functions

φ(x) in C

(0; 1) which satisﬁes the boundary conditions φ(0) = φ(1) = 0. Let f and g be any

two functions in this set. We have

hL[f ]; gi=

(x) g(x) dx;

and through int egration by parts:

(x) g(x) dx = f

(x) g(x)j

(x) g

(x) dx = ¡

(x) g

(x) dx:

4.4 Theoretical aspects of the method 43

The last equality is obtained by using the boundary conditions for g(x). Another integration

by parts and use the boundary condition f o r f(x) yields

(x) g

(x) dx =

(x) g

(x) dx = hf ; L[g]i

and thus L is symmetric on the deﬁned set of functions.

Consider the diﬀerential equation:

a(x) φ

+ b(x) φ

+ c(x)φ = ¡λφ;

where x 2(x

; x

), a(x) > 0 in [x

; x

], and a(x); b(x) and c(x) are continuou s functions. By

dividing the e quation by the term

σ =

a(x)

b(x)

a(x)

;

the equation can be rewritten as



b(x)

a(x)



c(x)

a(x)

b(x)

a(x)

φ = ¡λ

a(x)

b(x)

a(x)

φ:

Taking p(x) = e

b(x)

a(x)

, q(x) =

c(x)

a(x)

b(x)

a(x)

, the equation can be expressed as:

[p(x) φ

] + q(x) φ = ¡λσ(x) φ:

Theorem 4.2. The operator

L[φ] :=

(p(x) φ

) + q(x) φ;

acting on functions in the vector set C

; x

) that satisfy the boundary condition:



φ(x

) + β

) = 0

φ(x

) + β

) = 0

;

is symmetric with respect to the inner product h; i.

Proof. For any functions φ; is th e deﬁned set of functions, we have

hL[φ]; i=

(p(x) φ

) +

q(x) φ :

Applying integration by parts to the ﬁrst integral yields:

(p(x) φ

) = p(x) φ

¡p(x) φ

(p(x)

) φ:

We need to show that the boundary terms are zero. The boundary term at x = x

is:

p(x

)[φ

) (x

) ¡φ(x

)

)] = p(x

) W (φ; )(x

);

44 1D Linear Second-Order Equations

where W (φ; ) represents the Wronskian of functions φ and . Due t o the boundary

conditions:



φ(x

) + β

) = 0

) + β

) = 0

;

we have W (φ; )(x

) =/ 0, otherwise α

= β

= 0, which is impossible. The same argument

applies to the boundary term at x = x

. Therefore, we can conclude that:

(p(x) φ

) =

(p(x)

) φ;

which leads to:

hL[φ]; i= hφ; L[ ]i;

completing the proof. 

Theorem 4.3. The eigenvalues of L in the deﬁned vector space of functions are real.

Proof. Assume L[φ] = λφ. If λ is a complex eigenvalue, then by taking complex conjugate

of the equation, we obtain L[φ

] = λ

, where λ

; φ

represents the complex conjugate of λ; φ

respectively. Now, by symmetricity, we have

¡λ

jφ(x)j

dx = hL[φ]; φ

i= hφ; L[φ

]i= ¡λ

jφ(x)j

dx;

and thus λ = λ

, which implies λ is a real value. 

Theorem 4.4. Suppose the eigenfunctions φ

; φ

are associated with distinct eigenvalues

and λ

, respectively, in the eigenvalue problem:

L[φ] = ¡λσφ;

Then φ

and φ

are orthogonal with respect to the weight function σ.

Proof. By assumption, we have:



L[φ

] = ¡λ

σ(x) φ

L[φ

] = ¡λ

σ(x) φ

;

where λ

=/ λ

. Using the property of symmetricity, we can write:

¡λ

hσφ

; φ

i= hL[φ

]; φ

i= hφ

; L[φ

]i= ¡λ

hφ

; σφ

From this, we have (λ

¡λ

)hφ

; φ

= 0 which implies that hφ

; φ

= 0, satisfying the

orthogonality condition. 

Exercise 4.33. Write down the following eigenvalue problems in the symmetric Sturm-Liouville

form. Use the energy method in each case and show that their eigenvalues are strictly positives



+ 2xφ

= ¡λφ

φ(0) = 0; φ

(1) = 0



(1 + x)φ

+ xφ

= ¡λφ

φ(0) = φ

(0); φ

(0) = φ

(1)

4.4 Theoretical aspects of the method 45

(

+ 2φ

= ¡λe

¡2x

φ(0) = 0; φ

(1) = 0

Exercise 4.34. Consider the opera tor

L[φ] =

(p(x)φ

) + q(x)φ;

with t he domain of functions φ in the vector space

U = fφ 2C

; x

); φ(x

) = φ(x

); φ

) = φ

)g:

Prove that if p(x

) = p(x

), then L is symmetric with respect to the inner product h; i and conclude the

that eigenvalues of the following problem a re real, and their eigenfunctions are orthogonal with respect

to the weight functions σ > 0

(p(x)φ

) + q(x)φ = ¡λσφ

φ(x

) = φ(x

)

) = φ

)

Prove that if p(x

) = p(x

), then the operator which is deﬁned in the equation is symmetric and therefore

its eigenvalues are real and eigenfunctions are orthogonal with respect to weight function σ.

Exercise 4.35. It can be shown that there exists only one eigenfunction (up to a multiplication by

constants) associated to each eigenvalue for the Strum-Liouville problem. The scheme of proof is a s

follows: Let φ

and φ

be the eig enfunctions of the Sturm-Liouville eigenvalue pr oblem associated to the

same eigenvalue λ. Prove that the Wronskian W (φ

; φ

) = 0 and conclude that φ

and φ

are linearly

dependent.

Exercise 4.36. The Sturm-Liouville theorem states that the eigenvalues of the problem forms an

increasing set of value λ

!1. This exercise s et a lower bound for the sequence. Show that λ satisﬁ es

the inequality

λ ≥

q(x)jφj

σ jφj

;

and conclude that if q < 0 then λ > 0.

Exercise 4.37. Show that linear operator L[φ] = φ

+ φ

is not symmetric in the following vector space

U = fφ 2C

(0; 1 ); φ(0) = φ(1) = 0g:

Exercise 4.38. Show that linear operator L[φ] = e

+ e

is symmetric in the following space

U = fφ 2C

(0; 1 ); φ(0) = φ(1) = 0g:

Exercise 4.39. Show that linear operator L[φ] = φ

+ e

φ is symmetric in the following space

U = fφ 2C

(0; 1); φ(0) = φ(1); φ

(0) = φ

(1)g:

Exercise 4.40. Show that linear operator L[φ] = φ

is anti-symmetric in the following space

U = fφ 2C

(0; 1 ); φ(0) = φ

(0); φ(1) = φ

(1)g

An o perator L is anti-symmetric if hL[φ]; i= ¡hφ; L[ ]i.

Exercise 4.41. Suppose G(x) is a continuous function in [0 ; 1], and G(¡x) = G(x). Show that the

op erator

L[φ](x) =

G(x ¡t)φ(t) dt;

is symmetric.

46 1D Linear Second-Order Equations

Exercise 4.42. Consider the following eigenvalue problem:

(

+ xφ

+ 2φ = ¡λφ

(1) = φ

(e) = 0

a) Fin d the eigenvalues and eigenfunctions of the problem.

b) Find an expansion of the function f (x) = 1 , x 2 [1; e] in terms of the ﬁrst four eigenfunctions of

the problem.

Exercise 4.43. Consider the linear opera tor L[φ] =

¡x

) with the domain of functions φ in the

vector space

U = fφ 2C

(0; ln2); φ(0) = φ(ln2) = 0g:

a) Show that L is symmetric.

b) Find λ

and σ(x) such that the eigenfunctions of the eigenvalue problem

L[φ

] = ¡λ

σ(x) φ

;

are φ

(x) = sin(nπe

c) The set fφ

n=1

is a basis for PC

(0; ln2). Find the best approximation of the function f(x) = 1

in the space spanfφ

n=1

. That is, ﬁnd f

; :::; f

such that

k1 ¡f

¡···¡ f

k≤kf ¡ gk;

for a ny g 2spanfφ

n=1

. Draw this approximation and compare it with the function f.

d) Consider the following heat problem

= L[u]

u(0; 0) = u(ln2; t) = 0

u(x; 0) = 1

Since fφ

n=1

is a basis, we can approximate the solution u as

u(x; 0) ≈

n=1

(t) φ

(x):

Find the ordinary diﬀerential equation o f U

(t) along φ

(x). This diﬀerential equation describes

the change of the temperature along ea ch φ

(x) for n = 1; ···5 .

4.4.3 Operation on inﬁni te series

The type of convergence of inﬁnite series imposes certain limitations when wo rking with

them. For instance, consider the function f (x) = 1, x 2 (0; 1) which belongs to the set

(0; 1) and can be represented in terms of the basis fsin(nπx)g for n = 1; 2; ··· as

1 =

n=1

2 ¡2 co s(nπ)

nπ

sin(nπx):

Taking the derivative of this eq uality yields:

0 =

n=1

[2 ¡2 cos(nπ)] cos(nπx);

However, this result is clearly incorrect since the function g(x)=0 is represented as a non-

trivial series in terms of the orthogonal set fcos(nπx)g. The ﬁgure below shows the function

f(x) in x 2(0; 1) and its series representation in a wider domain.

4.4 Theoretical aspects of the method 47

0 1

-1.5

-1

-0.5

0.5

1.5

-1 0 1 2

-100

-50

100

Where does this error come f rom? On e possible explanation for this discrepancy lies in

the behavior of the series representation at the endpoints of the interval (0; 1). Although

the function f (x) is continuous on (0; 1), its derivative is not continuous at x = 0 and x = 1.

This leads to a discontinuity in the derivative series representation, which can result in

inconsistencies in the equality above.

Furthermore, when extending the series representation of the function f (x) to a wider

interval beyond (0; 1), it exhibits an odd extension of f (x). This means that a jump is

formed at the endp o ints x = 0 and x = 1, due to the discontinuity of the original function.

Consequently, the derivative of th ese jumps generates a Dirac delta function at these points.

It is important to note that the Dirac delta function is a distribution that is concentrated

at a speciﬁc point and has the property of being zero everywhere except at that point,

where it behaves like an impulse. The presence of the Dirac delta function-like behavior in

the derivative seri es further contributes to the discrepancies observed in the equality, as it

introduces additional irregularities near the endpoints.

Now, let's consider the function f (x)=x(1 ¡ x). The series representation o f f (x) in

terms of the basis fsin(nπx)g is given by:

x(1 ¡x) =

n=1

4 ¡4 cos(nπ)

sin(nπx):

Taking the derivative of this eq uation yields the equality:

1 ¡2x =

n=1

4 ¡4 cos(nπ)

cos(nπx):

It can be veriﬁed that the latter series is a true series representation of the function f

(x)=1 ¡

2x in terms of the orthogonal basis fcos(nπx)g. The ﬁgure below shows the original series

of the function f (x) and its derivative, both truncated to N = 50 terms.

It is important to note that since the odd extension of the function f (x) is continuous

at the endpoints, the derivative of the series of f (x) converges pointwise to the derivative

(x) in the interval (0; 1). This behavior ensures that the series accurately represents the

derivative of the function within the given interval.

48 1D Linear Second-Order Equations

0 1

-0.3

-0.2

-0.1

0.1

0.2

0.3

-1 0 1 2

-1

-0.5

0.5

Theorem 4.5. Let f be a function belonging to C

; x

), and let fφ

(x)g for n=1; 2; ::: be

an orthogonal basis generated by a Sturm-Liouville eigenvalue problem. Su ppose the series

representation of f (x) in the interval (x

; x

) is given by:

f(x) =

n=1

(x):

If f(x

)= f (x

), then the derivative of f(x) can be represented as:

(x) =

n=1

(x):

There are no restrictions on other operations such as integration, addition, subtraction,

or multiplication when working with series representations. These operations can be applied

to series in a straightforward manner.

Exercise 4.44. Let f (x) and g(x) be functions with series representations as follows:

f(x) =

n=1

(x); g(x) =

n=1

(x);

where fφ

(x)g is an orthogonal basis generated by a Sturm-Liouville eigenvalue problem.

a) Integration: The integral of the sum of two series can be obtained by integrating each term of the

series individually:

[f(t) + g(t)] dt =

n=1

+ b

]

(t) dt;

for a ny x 2(x

; x

b) Addition and subtraction: The sum or diﬀerence of two series can be obtained by adding or

subtracting the corresponding terms of the series:

f(x) ± g(x) =

n=1

±b

] φ

(x)

c) Multiplication: The product of two serie s can be obtained by multiplying each term of one series

with each term of the other ser ies

f(x) g(x) =

n=1

m=1

(x) φ

(x)

4.4 Theoretical aspects of the method 49

In each case, the resulting series is still a representation of the corresponding operation applied to the

functions f(x) and g(x).

50 1D Linear Second-Order Equations

Appendix A

Fourier series

A. 1 Trigonometric and complex Fourier series

Consider the eigenvalue problem deﬁned on a unit circle ¡π ≤θ ≤π:

(θ) = ¡λφ(θ)

φ(¡π) = φ(π)

(¡π) = φ

(π)

: (A.1)

The periodic boundary condition given in the problem is natural, as φ(θ) should be periodic

with a period of 2π. Notice that φ(θ), the eigenfunction of this problem, is invariant up

to multiplication by a constant; tha t is, if φ(θ) solves the equation, so does cφ(θ) for any

c2R. Although this problem is not in the standard form of the Sturm-Liouville p roblem,

its eigenfunctions and eigenvalues share resemblances with the properties of the standard

Sturm-Liouville problem.

Utilizing the energy method, one can show that the eigenvalues are non-negative:

¡π

(θ) φ(θ) dθ = ¡λ

¡π

jφ(θ)j

dθ:

Integration by parts yields:

¡π

jφ

(θ)j

dθ = λ

¡π

jφ(θ)j

dθ;

Thus, we have λ ≥0. Furthermore, for λ =0, the eigenfunction is a constant fu nc tion, and

we ca n assume, without loss of generality, that φ

(θ)=1. For λ>0, the eigenfunctions are

given by:

(θ) 2fco s(nθ); sin(nθ)g; n 2N:

Furthermore, the set of eigenfunctions f1; cos(nx); sin(nx)g, n 2N forms a basis for piecewise

continuously diﬀerentiable functions deﬁne don the unit circle. That is, for any f (θ); ¡π ≤

θ ≤π, there are constants a

and b

such that:

f(θ) =

n=0

cos(nθ) +

n=1

sin(nθ):

Since the set f1; cos(nθ); sin(nθ)g is an orthogonal basis with respect to the inner product

h; i deﬁned as:

¡π

(θ) φ

(θ) dθ = 0;

the parameters a

and b

are determined by the inner product of f and cos(nθ) and sin(nθ)

respectively.

The eigenvalue problem (A.1) can be modiﬁed to represent functions on any symmetric

domain (¡L; L) as follows:

= ¡λφ

φ(¡L) = φ(L)

(¡L) = φ

(L)

The set of solutions to this modiﬁed eigenvalue problem, with the normalizing factor c = 1,

is given by:

(x) 2

1; cos



nπ



; sin



nπ

o

; n 2N:

The associated eigenvalues for this problem are λ

Remark A.1. The set



1; cos

nπ



; sin

nπ



is commonly known as the Fourier trigono-

metric basis. By using the Euler formula e

iθ

=cos(θ) + i sin(θ), we can express this bas is in the

complex form as



nπ



; n 2Z. The refore, a piecewise c o ntinuously diﬀerentiable function

deﬁned on [¡L; L] can be represented as:

f(x) =

n=¡1

in!x

;

where ! =

, and F

are constants that can be determined using the complex form of the

inner product hh; ii deﬁned as:

hhf ; gii:=

¡L

f(x) g¯(x) dx;

where g¯ represents the complex conjugate of g. Utilizing this deﬁnition for the orthogonal

eigenfunctions φ

(x) = e

in!x

, the parameters F

are determined as:

¡L

f(x) e

¡in!x

dx:

While the trigonometric basis



1; cos

nπ



; sin

nπ



and the complex basis



nπ



both

represent the same set of f unctions and are mathematically equivalent, the complex repre-

sentation is of ten preferred in theoretical works due to its simplicity and elegance. Thus, for

theoretical analysis and calculations, the complex basis oﬀers sig niﬁcant advantages over the

trigonometric basis.

Example A.1. Let's ﬁnd the complex Fo urier series of the function f(x)=x in the interval

(¡1; 1). Using the c o mplex Fourier series representation, we have:

x =

n=¡1;n =/ 0

i cos(nπ)

nπ

inπx

: (A.2)

52 Fourier se ries

Now, let's determine the complex Fourier series of the function f(x) =x

in the interval

(¡1; 1). Using the c o mplex Fourier series representation, we have:

n=¡1;n =/ 0

2 cos(nπ)

inπx

: (A.3)

Problem A.1. If f (x) is an odd function, show that F

are pure imaginary and F

¡n

= ¡F

(and thus

= 0). If f(x) is even function, show that F

are real and F

¡n

= F

Exercise A.1. Show that the set of functions



sin

nπ



; cos

2n ¡ 1

πx



is a basis for the piecewise

continuously diﬀerentiable functions deﬁned o n (¡L; L). Hint: Find an Sturm-Liouville problem that

generate the set.

Exercise A.2. Use either the Fourier sine or cosine series of the function f (x) = x on (0; 1) to prove

the following identity:

+ ···

Exercise A.3. Use either the Fourier sine or cosine series of the function f(x) = x

on x 2(0; 1) to prove

the following identity:

= 1 ¡

+ ···

Exercise A.4. Use either the Fourier sine or cosine series of the function f(x) = x on x 2(0; 2 ) to prove

the following identity:

= 1 ¡

+ ···:

Exercise A.5. Use either the Fourier sine or cosine series of the function f(x) = x

on x 2(0; 2) to prove

the following identity:

= 1 ¡

+ ···:

Exercise A.6. Let f( x) = e

¡x

on (¡1; 1).

a) Write down the Fourier series of f and draw S

(x); S

(x).

b) Write down the Fourier sine and cosine series of the function f (x) = e

¡x

deﬁned on x 2 (0 ; 1).

Draw the partial sums S

and S

for each series.

You can us e the following code in Matlab to generate the plot of partial sum S

of a function f(x)

deﬁned in x 2(¡L; L)

%The length L

L=1;

%Determine the functon

f=@(x) exp(-x);

% the number of terms

n=10;

%the coefficients of Fourier sine series

a0=integral(@(x) f(x),-L,L)/(2*L);

a=integral(@(x) f(x).*cos((1:n)'*x*pi/L),-L,L,'arrayvalued',true)/L;

b=integral(@(x) f(x).*sin((1:n)'*x*pi/L),-L,L,'arrayvalued',true)/L;

x=-L:0.01:L;

fhat=a0+a'*cos((1:n)'*x*pi/L)+b'*sin((1:n)'*x*pi/L);

plot(x,fhat,x,f(x);

For the Fourier sine series, you can use the following code:

%The length L

L=1;

%Determine the functon

A.1 Trigonometric and complex Fourier series 53

f=@(x) exp(-x);

% the number of terms

n=10;

%the coefficients of Fourier sine series

b=2*integral(@(x) f(x).*sin((1:n)'*x*pi/L),0,L,'arrayvalued',true)/L;

x=0:0.01:L;

fhat=b'*sin((1:n)'*x*pi/L);

plot(x,fhat,x,f(x);

For the Fourier cosine series, you can use the following code:

%The length L

L=1;

%Determine the functon

f=@(x) exp(-x);

% the number of terms

n=10;

%the coefficients of Fourier sine series

a0=integral(@(x) f(x),0,L)/L;

a=2*integral(@(x) f(x).*cos((1:n)'*x*pi/L),0,L,'arrayvalued',true)/L;

x=0:0.01:L;

fhat=a0+a'*cos((1:n)'*x*pi/L);

plot(x,fhat,x,f(x);

Exercise A.7. Consider the function f(x) = cos(x).

a) Write down the Fourier series of f(x) on x 2

;



b) Write down the Fourier sine and cosine s eries on x 2



Exercise A.8. In this exercise we use the Fourier series to ﬁn d the series solut ion of and boundary

value ODE equation. Consider the following problem:



+ y = 1

y(0) = y(1) = 0

a) Assum e that the solution to the equation is written as

y(x) =

n=1

sin(nπx):

Substitute y(x) into the equation and ﬁnd coeﬃcients Y

b) Show that the obtained series is absolutely convergent.

Theorem A.1. (Parseval's Theorem) Let f (x) be a piecewise continuously diﬀerentiable

function in (¡L; L). The Fourier series of f (x) converges to f(x) in the norm sense, which

can be expressed as:

lim

N !1

kf ¡S

= lim

N !1

jf (x) ¡S

(x)j

dx = 0;

where S

(x) is the partial sum of its Fourier series given by:

(x) =

n=¡N

in!x

;

for ! =

Problem A.2. Use the above theorem and prove the following equality

n=¡1

¡L

jf(x)j

dx;

54 Fourier se ries

for piecewise continuously diﬀerentiable functions deﬁned on (¡L; L).

Hint: Expand the square norm



f ¡

n=¡N

in!x



f ¡

n=¡N

in!x

; f ¡

n=¡N

in!x

;

and conclude the following inequality known as the Bessel inequality:

k=¡N

≤

¡L

jf(x)j

dx:

Then use the Parseval theorem and conclude the equality.

A. 2 O perations on Fourier series

Proposition A.1. If f and g are piecewise continuously diﬀerentiable functions, then

f(x) + g(x) =

n=¡1

+ G

) e

in!x

: (A.4)

Proposition A.2. If f and g are piecewise continuously diﬀerentiable functions, then

f(x)g(x) =

n=¡1

k=¡1

n¡k

in!x

: (A.5)

Proof. Multiply f(x) by g(x) and write:

f(x)g(x) =

k=¡1

g(x) e

ik!x

: (A.6)

Insert the serie s of g(x) into the above series and obtain

k=¡1

l=¡1

il!x

ik!x

k=¡1

l=¡1

i(k+l)!x

: (A.7)

If we take k + l = n, then

f(x)g(x) =

n=¡1

k=¡1

n¡k

in!x

; (A.8)

and this complete the proof. 

Proposition A.3. Let f be a twice continuously diﬀerentiable function in (¡L; L), and

f(¡L) = f(L), t hen

(x) =

n=¡1

(in!) F

in!x

: (A.9)

Proof. We c an write

(x) =

n=¡1

in!x

; (A.10)

A.2 Operations on Fourier series 55

where by the continuity of f , we can wr ite

¡L

(x) e

¡in!x

dx=

f(x)e

¡in!x



in!

hhf ; e

in!x

ii: (A.11)

The condition f (¡L) = f (L) implies f(x)e

¡in!x

= 0, and then

in!

hhf ; e

in!x

ii=in!F

;

and this completes the proof. 

The continuity of f, and condition f(¡L) = f (L) are crucial for the proof that can

not be relaxed. For example, function f(x) = x deﬁned on [¡1; 1] has the series in basis

f1; cos(nπx); sin(nπx)g as

x =

n=1

¡2(¡1)

nπ

sin(nπx);

while f

(x) = 1, and it has the trivial series expansion.

Proposition A.4. Let f be a piecewise continuously diﬀerentiable function in (¡L; L), then

¡L

f(t) dt =

n=¡1

¡L

in!t

dt: (A.12)

Proof. Let F (x) be an anti-derivative of f, that is,

F (x) =

¡L

f(t) dt:

First we prove the proposition if F (L) = 0, that is,

F (L) =

¡L

f(t) dt = 0:

Obviously F is a continuously diﬀerentiable f unction , and thus

F (x) =

n=¡1

in!x

; (A.13)

where F

hhF ; e

in!x

ii. On the other hand, the condition F (L) = 0 implies F (¡L) = F (L),

and thus,

f(x) = F

(x) =

n=¡1

(in!) F

in!x

: (A.14)

This implies F

= (in!) F

, and for n =/ 0, we can write F

in!

. Therefore,

F (x) = F

n=¡1;n =/ 0

in!

in!x

: (A.15)

56 Fourier se ries

Since F is continuous and F (¡L) = 0, then

= ¡

n=¡1;n=/ 0

in!

¡inπ

: (A.16)

On the oth er hand, we have

n=¡1

¡L

in!t

dt =

n=¡1;n=/ 0

in!

in!x

¡e

¡inπ

) =

n=¡1;n =/ 0

in!

¡inπ

n=¡1;n =/ 0

in!

in!x

n=¡1;n=/ 0

in!

in!x

= F (x):

Hence, we have

F (x) =

n=¡1

¡L

in!t

dt:

Now, we relax the condition F (L) = 0. Deﬁne function g as g(x) = f(x) ¡F

. Since

¡L

g(x) dx =

¡L

f(x) dx ¡F

= 0; (A.17)

we write G(x) :=

¡L

g(t) dt as the f o llowing series

G(x) = G

n=¡1;n =/ 0

in!

in!x

;

where G

are:

hhg; e

in!x

ii=

hhf ; e

in!x

ii¡

hhF

; e

in!x

ii= F

hhF

; e

in!x

ii:

For n =/ 0, we have hhF

; e

in!x

ii= 0, a nd hence G

= F

for n =/ 0. On the other ha nd, we have:

= ¡

n=¡1;n=/ 0

in!

¡inπ

= F

;

and thus

G(x) = F

n=¡1;n=/ 0

in!

in!x

Note that

G(x) =

¡L

g(t) dt =

¡L

f(t) dt ¡F

(x + L);

which implies

¡L

f(t) dt = F

(x + L) + F

n=¡1;n =/ 0

in!

in!x

n=¡1

¡L

in!t

dt;

A.2 Operations on Fourier series 57

and this completes the proof. 

A. 3 A proof of the Four ier and Parseva l theorem

We present a proof of the Fourier theorem for continuously diﬀerentiable functi ons deﬁned

on the domain (¡π; π).

Theorem A.2. (Fourier) Assume that f 2 C

(¡π; π) and continuous on [¡π; π]. Then

the partial series

(x) =

k=¡n

ikx

;

converges pointwise to f(x), where F

2π

hhf ; e

ikx

ii.

Substituting F

into the partial sum, gives

(x) =

2π

k=¡n



¡π

f(ξ) e

¡ikξ

dξ



ikx

2π

¡π

f(ξ)

k=¡n

ik(x¡ξ)

: (A.18)

Exercise A.9. Show the following identity

k=¡n

ikx

sin

¡¡

n +



sin



; (A.19)

and conclude

¡π

sin

¡¡

n +



sin



dx = 2π: (A.20)

The 2π-periodic function

(x) =

sin

n +



sin



; (A.21)

is called the Dirichlet kernel. The ﬁgure below depicts this kernel for values n = 1; 3; 9.

−

= 9

= 3

= 1

Therefore, we can rewrite S

(x) as the following integral

(x) =

2π

¡π

f(ξ) D

(x ¡ ξ): (A.22)

58 Fourier se ries

The kernel D

(x) is a Dirac delta se quence for n !1.

Proposition A.5. Let f 2C

(¡π; π), then for all x 2(¡π; π), we have:

lim

n!1

2π

¡π

f(ξ) D

(x ¡ ξ)dξ = f (x): (A.23)

In order to prove the proposition, we need the following lemma.

Lemma A.1. Assume that f(x) is a continuous function in [¡π; π], then

lim

n!1

¡π

f(x) sin(nx) dx = 0: (A.24)

Now using the above lemma, we have

(x) = f (x) +

2π

¡π

(f (ξ) ¡f(x))D

(x ¡ ξ) dξ = f(x) +

2π

¡π

f(ξ) ¡f(x)

ξ ¡x

sin

x ¡ ξ



sin((n + 1/2)(x ¡ ξ))dξ:

Since f is a C

function, the function

f (ξ) ¡ f (x)

ξ ¡ x

is continuous. Similarly, the function

ξ ¡ x

sin



x ¡ ξ



is continuous as well, and thus

lim

n!1

¡π

f(ξ) ¡f(x)

ξ ¡x

sin

x ¡ ξ



sin((n + 1/2)(x ¡ ξ))dξ = 0:

Finally, we obtain

lim

n!1

(x) = f (x):

Theorem A.3. (Parseval) If f 2PC

(¡π; π), then partial sum S

(x) converges to f (x) in

norm, that is,

lim

n!1

kf ¡S

k= 0:

The proof is based on a fact from analysis that if f 2PC

(T ), then for every " > 0, there

is a function g 2C

(T ) such that

kf ¡ g k< ":

By this fact, we can write:

kf ¡S

k≤kf ¡gk+ kg ¡S

k< " + kg ¡S

On the oth er hand, we have:

kg ¡S

= 2π

k=¡n

¡G

) ≤2π

k=¡n

1/2

k=¡n

¡G

1/2

A.3 A proof of the Fourier and Parseval theorem 59

The last inequal ity is the Cauchy-Schwarz inequality. On the other hand, using the Bessel

inequality, we have

2π

k=¡n

¡G

≤

¡π

jf (x) ¡g(x)j

< "

Similarly, we have

k=¡n

≤

2π

¡π

jf (x)j

≤M

;

for some M > 0. This implies that kg ¡S

≤

and ﬁnally

kf ¡S

k< " +

Since " is arbitrary, we let " !0 and thus

lim

n!1

kf ¡S

k= 0:

A.3.1 Gibbs phenomena

In this chapter, we have observed the overshoot jump of partial series near discontinuity

points, which is commonly known as the Gibbs phenomenon, named after the American

physicist J. W. Gibbs. Now, we aim to ﬁnd an estimate for the maximum va lue of thi s jump.

For simplicity, we assume th a t the function f is deﬁned on the interval [¡π; π] and has a

discontinuity at x = 0. From the formula of the Dirichlet kernel D

(x), we know that D

(x)

vanishes at z = ±

2π

2n + 1

. The ﬁgure below illustrates D

(x) and its shift D

x ¡

2π



−

(

)

(

−

π/

We observe that the maximum of D

x ¡

2π



o ccur at the ﬁrst zero point of D

(x) which

is z =

2π

. We expect that the jump occu rs at the zero points of D

. For this, we need to

calculate S

2π

2n + 1





2π

2n + 1



2π

¡π

f(ξ)D



2π

2n + 1

¡ξ



dξ =

2π

¡π

f(ξ) D



2π

2n + 1

¡ξ



dξ +

2π

f(ξ)D



2π

2n + 1

¡ξ



dξ:

60 Fourier se ries

According to the shape of the Dirichlet kernel for suﬃciently large n, we have:

¡π

f(ξ) D



2π

2n + 1

¡ξ



dξ

∼

f(0

)

¡π



2π

2n + 1

¡ξ



dξ: (A.25)

Similarly, we have

f(ξ) D



2π

2n + 1

¡ξ



dξ

∼

f(0

)



2π

2n + 1

¡ξ



dξ: (A.26)

Direct calculation yields the fo llowing values for large n:

¡π



2π

2n + 1

¡ξ



dξ

∼

¡0.0897; (A.27)

and



2π

2n + 1

¡ξ



dξ = 1.0897: (A.28)

Consequently„ we obtain the following estimate for the jump at the right side of the jump

point x = 0:



2π

2n + 1



≈1.09f

(0) ¡0.09 f

(0): (A.29)

A similar estimate holds for the jump at the left side of x = 0, that is,



2π

2n + 1



≈1.09f

(0) ¡0.09f

(0): (A.30)

A.3 A proof of the Fourier and Parseval theorem 61