Class Notes 335

8/13/2019 Class Notes 335

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Integral Transforms and Their Applications

Claremont Graduate University, Mathematics M335Qidi Peng

Fall 2013

1 Introduction and BackgroundMotivation: ”Integral transforms” is a powerful operational method forsolving differential and integral equations.Main goal: To catch the main interests of integral transforms; to be able toregard transforms as operators; to have basic skills of complex analysis andfunctional analysis.

1.1 All Begin From Fourier TransformsThe concept of an integral transform is originated from the Fourier inte-gral theorem rst appeared in his seminal work ”La théorie analytique de lachaleur (1822)”. This theorem is stated as: if f satises Dirichlet’s conditionsin R , and absolutely integrable on R , then for any x ∈R ,

12(f (x + 0) + f (x −0) =

12π R R eik (x−y)f (y) dy dk, (1.1)

wheref (x + 0) = lim

u>x,u →xf (u), f (x −0) = limu


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where Cauchy derived: if we dene, for f under appropriate condition,

F : f −→ 1√ 2π R e−i (·)x f (x) dx,

then

1. F transforms a function to another function. This is later called inte-gral operator which has nice properties.2. F is invertible. In fact,

F −1

(f ) = 1

√ 2π R ei (

·)y

f (y) dy.

Finally it was O. Heaviside who drove the Fourier transform (as well asLaplace transform) worldwidely famous. In his 2 papers ”On operationalmethods in physical mathematics - Part 1 and part 2”, 1892, he developedopperational methods for the solutions of telegraph equation and the sec-ond order hyperbolic partial differential equations with constant coefficients.Inspired by Fourier and Cauchy, Heaviside developed his new operationalmathematics, however, most of his works are lack of high level mathematicalsupport, namely, the rigor, even though the formulas worked well in practice.He rarely cared the functional space on which the operator dened. In ourcourse, we study the integral transforms from a mathematical point of viewrather than a physical or electric engineering point of view.

1.2 Denition of Integral Transforms and Their Appli-cations

Denition 1.1 (Absolutely integrable) A function f is called absolutely integrable on I ⊂R if and only if

I |f (x)|dx < + ∞.We can write f ∈ L1(I ) if f is absolutely integrable.We notice in the passage that this also means the Lebesgue integral ∫ I f (x) dxfor the measurable function f is well dened.

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Denition 1.2 (Banach space) The functional space

B is called a Banach

space if and only if (a) B is a normed space: there exists an application ∥ ·∥B verifying:

1. ∀ x ∈ B , ∥x∥B ≥ 0;2. ∀ x ∈ B and ∀ λ ∈R (or C ), ∥λx∥B = |λ|∥x∥B;3. ∀ x, y ∈ B ,

∥x + y∥B ≤ ∥x∥B + ∥y∥B.(b) B is complete: all Cauchy sequence of B converges in B .

Denition 1.3 (Operator on Banach space) T is called an operator if there exist 2 Banach spaces B 1, B 2 such that T is an application from B 1 toB 2: for all f ∈ B 1, T (f ) ∈ B 2.Denition 1.4 (Integral transforms) Let f ∈ L1([a, b]) and K ∈ L1([a, b]2)verify K (·, x)f (·) ∈ L1([a, b]), the integral transform T of f with kernel K is dened as: for x ∈ [a, b],

T (f )(x) := ba K (x, s )f (s) ds.Remark that this denition could be extended to the case where [ a, b] isreplaced by a subset of C .Examples of Banach spaces: R n , C ([a, b]), L p([a, b]).Examples of integral transforms:

1. When K (x, y) = χ [a,b ]×[a,b ](x, y), T (f )(x) = ∫ ba f (u) du. T is the usualintegral operator.2. When K (x, y) = 1√ 2π e−

ixy , T is Fourier transform.3. When K (x, y) = e−xy , for x > 0, Re(y) > 0, T is Laplace transform.

Properties of general integral transforms:

1. T is linear: ∀f, g ∈ L1([a, b]) and ∀λ ∈R (or C ),T (λf + g) = λT (f ) + T (g).

2. T is not necessarily invertible, it depends on the choice of the kernelK .3


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1.3 Some Interests of Integral TransformsHere we introduce several interests of integral transforms both in mathemat-ics and physics.

(1) Expansion of functions. For example, let f be a function whose Fouriertransform is well dened. For a continuous point x of f , by Fourierintegral theorem,

f (x) = 12π R ( R e−iky f (y) dy

:= g(k )

eixk dk

= 1√ 2π limn →+ ∞n

j = −n j +1

jg(k)eixk dk

≈ 1√ 2π

K

j = −K g( j )eixj .

The last approximation shows in fact an algorithm to evaluate the valueof f (x), starting from {g( j )} j∈[−K,...,K ], for K big enough.

(2) Selection among different kernels in order that the integral transform iswell dened. For example, for a period function sin x, x

≥ 0, its Fourier

transform does not exist. However, its Laplace transform exists.

(3) Solving ordinary, partial differential equations and integral equations.

(4) In mathematical statistics, we use Fourier transform and Laplace trans-form to describe probability distributions.

1.4 Exercises1. (Comples analysis) Describe geometrically the sets of points z in the

complex plane dened by the following relations.

• |z −z 1| = |z −z 2|, z 1, z 2 ∈C ;• 1/z = z ;• |z | = Re(z ) + 1;• Im(z ) = c, c ∈R .

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2. (Banach space) Prove C ([a, b]) is a Banach space.

3. (Integral transform) Determine the Fourier transform of the rectangularpulse 1 [−a,a ](t) (indicator function), a > 0.

2 Fourier TransformMotivation: In physics and electric engineering, Fourier transform has beenwidely used to solve initial value and boundary value problems. in mathe-matical probability and statistics, Fourier transform is used to describe andcompute probability distributions.Main goal: Review measure theory, determine Fourier transform, useproperties of Fourier transform, convolution theorem.

2.1 Fourier Transform under Lebesgue MeasureDenition 2.1 (Fourier transform) For f ∈ L1(R ), the Fourier transform of f is dened as: for all x ∈R ,

F (f )(x) := 1√ 2π

R

e−ixs f (s) ds.

And the inverse Fourier transform of f is: for all x ∈R ,

F −1(f )(x) = 1√ 2π R eixs f (s) ds.

We notice that the denition of Fourier transform is not unique, for thereexist at least two other versions:

(1) In electric engineering,

F (f )(x) := R e−2πixs f (s) ds,and

F −1(f )(x) := R e2πixs f (s) ds.5


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(2) In mathematical probability and statistics,

F (f )(x) := R eixs f (s) ds,and

F −1(f )(x) := 12π R e−ixs f (s) ds.

These 3 couples are not equal but there Fourier integral theorem’s presenta-tion are the same: they all verify

F −1(F (f ))( x) = 12π R R ei (x−s )k f (s) ds dk.Examples of Fourier transform:

(1) f (x) = e−ax 2 , x ∈R , a > 0.We haveF (f )(x) =

1√ 2a e−

x 24 a .

(2) f (x) = H (a−|x|), a > 0, x ∈R

(gate function), where H is the Heavisidestep function:

H (x) = { 1 if x ≥ 00 if x < 0.We have

F (f )(x) = 2π sin(ax )x .Question: Why are F (f ) in (1) and (2) real-valued?Answer: Since f is even over R .

The condition f ∈ L1(R ) is sufficient but not necessary for the existence of its Fourier transform. In fact it is a bit too strong as a condition of existencesince many simple functions such as constant function, periodic function donot have Fourier integral. In the next subsection we will extend the Fouriertransform to more generalized functions, under another measure.

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2.2 Fourier Transform under Dirac MeasureIn order to dene Fourier transform under Dirac measure, we need to rstrecall the following denitions.

Denition 2.2 ( σ- algebra) Let Ω be some set of elements. F is called a σ-algebra on Ω if and only if:1. All element of F is a subset of Ω and F ≠ ∅;2. For all A ∈ F , Ac∈ F ;

3. For {An }n∈I ⊂ F , where I is a at most countable set,∪

n∈

I An ∈ F .Examples of σ-algebra: Let Ω = {1, 2}, we can dene

F 1 = {{1}, {2}, {1, 2},∅}or

F 2 = {Ω,∅}.In measure theory, we need σ-algebra to construct measurable subsets of Ω.

Denition 2.3 (Measure) Let Ω ⊂ R

and F is a σ-algebra dened on Ω.Dene, for x ∈ Ω,δ x : F → {0, 1},

as

δ x (A) = { +∞ if x ∈ A0 if else .First we can see

δ x (s) = δ 0(x −s) = δ 0(s −x).Another more signicant property of δ x is: for any function f dened on Ω

and any s ∈ Ω, f (s)δ x (s) = f (x)δ x (s).This yields the following important remark:

Ω f (s)δ x (s) ds = f (x).7


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Notice that The Dirac measure is one approach to express discrete sum by

integral: for the at most countable set of indices I , for example:

k∈I

f (xk ) = R f (u)(k∈I δ x k (u) du.The following proposition is the principle result of this subsection:

Proposition 2.1 The Fourier transform can be also well dened under Dirac measure (of course if the function f is absolutely integrable in sense of Dirac).

Examples:

1. Determine the Fourier transform of δ 0.Solution: By the Remark (in red),

F (δ 0)(x) = 1√ 2π R e−isx δ 0(s) ds = e−ix 0√ 2π = 1√ 2π .

Notice that if we make an inverse Fourier transform to the above equa-tion, we get

δ 0(x) = 12π R eisx ds.

This fact implies the following elegant expressions:

• ∫ R eisx ds = ∫ R e−

isx ds = 2πδ 0(x);

• for f ≡ c ∈R (or C ), F (f )(x) = √ 2πcδ 0(x).2. Show that for x ∈R , H ′(x) = δ 0(x).Proof: Since

x−∞δ 0(s) ds = { 1 if x ≥ 00 if x < 0 = H (x),then H ′(x) = δ 0(x).

The most interesting point we can get from this example is that, from nowon we can write; for all Ω ⊂R ,

Ω f (s)δ 0(s) ds = Ω f (s) dH (s).This shows δ 0 is the density function and H is the distribution function of Dirac measure.

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2.3 Properties of Fourier Transform and ConvolutionTheorem

We make a summary of the basic properties of Fourier transform. Noticethat all these equations hold both under Lebesgue and Dirac measure. Leta ∈R be xed,(a) (shifting) F (f (· −a))( x) = e−iax F (f )(x);(b) (scaling) for a ̸= 0, F (a·)(x) = 1|a |F (f )(

xa );

(c) (conjugate)

F (f (

−·))( x) =

F (f )(x);

(d) (translation) F (eia ·f (·))( x) = F (f )(x −a);(e) (duality) F (F (f ))( x) = f (−x).The following theorems show the property of continuity of Fourier transform.

Theorem 2.2 If f ∈ L1(R ), then F (f ) is bounded and continuous on R .Lemma 2.3 (Riemann Lebesgue Lemma) If f ∈ L1(R ) and f is piece-wise continuous,

lim|x |→+ ∞|F (f )(x)| = 0.Proof: Use Lebesgue dominated convergence theorem. The following theo-rem tells a differentiation property of Fourier transform.

Theorem 2.4 If f ∈ L1(R ) and f is continuously differentiable (it is contin-uous and its differentiation is continuous function), and lim|x |→+ ∞ f (x) = 0 ,then

F (f ′)(x) = ( ix)F (f )(x).Proof: Use integral by parts and Riemann Lebesgue Lemma.The following corollary generalizes the above theorem.

Corollary 2.5 If f ∈ L1(R ) and f is n-times continuously differentiable,and lim|x |→+ ∞f (k )(x) = 0 for k = 0, 1, . . . , n −1, then

F (f (n ))(x) = ( ix )n F (f )(x).9


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Denition 2.4 (Convolution) Let f, g

∈ L1(R )

∩L2(R ), the convolution

of f, g is dened as

f ∗g(x) := 1√ 2π R f (x −s)g(s) ds = 1√ 2π R g(x −s)f (s) ds.

The convolution has its great interests mainly thanks to the following con-volution theorem.

Theorem 2.6 (Convolution theorem) For f, g ∈ L1(R ) ∩L2(R ),

F (f ∗g)(x) = F (f )(x)F (g)(x),or

f ∗g(x) = F −1(F (f )F (g) (x),or R f (x −s)g(s) ds = R eisx F (f )(s)F (g)(s) ds.

A straightforward application is the following Parseval’s relations.

Corollary 2.7 (Parseval’s relation) For f ∈ L1(R ) ∩L2(R ),

R |f (x)|2 dx = R |F (f )(x)|

2 dx.

Corollary 2.8 (General Parseval’s relation) For f, g ∈ L1(R ) ∩L2(R ),

R f (x)g(x) dx = R F (f )(x)F (g)(x) dx.2.4 ExercisesExercise 1: Find the Fourier transform of

f (x) = 11 + x2

.

Solution: Observe that

f (x) = π2F (e−|·|)(x),10


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then by the duality property of Fourier transform,

F (f )(x) = π2F (F (e−|·|))( x) = π2 e−|−x | = π2 e−|x |.Exercise 2: Find the Fourier transform of

f (x) = δ 0x −ct + δ 0x + ct.Solution:

F (f )(x) = 2π cos(ctx).Exercise 3: Show

i(F (f )(x) ′ = F (·f (·))( x).Proof: we have(F (f )(x) ′ =

1√ 2π( R e−isx f (s) ds ′.

Here we use a the Lebesgue dominated theorem for differentiation: if f (x, y)veries

1. f (x,

·)

∈ L1(R ) for x

∈R ;

2. ∂f (x,y )∂x exists and |∂f (x,y )∂x | ≤ g(y) with g ∈ L1(R ).Then

( R f (x, y) dy ′ = R∂f (x, y)

∂x dy.

By using this Lebesgue theorem we get

(F (f )(x) ′ = 1√ 2π R ∂e−isx∂x f (s) ds = −iF (·f (·))( x).

This yields the proof.

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3 Applications of Fourier TransformMotivation: Fourier transform is a powerful tool for solving ordinary dif-ferential equations, integral equations and partial differential equations whenthe solution is dened over R .Main goal: To know how to solve ordinary differential equations, integralequations and partial differential equations by using Fourier transform; Toknow how to evaluate some particular integrals by Cauchy residue theorem,convolution theorem and Parseval’s relation.

3.1 Applications of Fourier Transform to Ordinary D-ifferential Equations (ODE)

Fourier transforms can be used to solve linear ODE with constant coefficients.We rst introduce the key idea to solve differential equations and integralequations by using integral transforms. There are only three steps to take:

Step 1: Take integral transform to both sides of equation (as well as itsboundary or initial conditions);

Step 2: Solve the integral transform of the wanted solution;

Step 3: Determine the solution by taking inverse integral transform.

We will pursue this part by taking examples.

Example 3.1 (General linear ODE) We consider the nth order linear ODE with constant coefficients:

an y(n )(x) + an −1y(n −1) (x) + . . . + a1y′(x) + a0y(x) = f (x), (3.1)

where an , . . . , a 0 ∈ R are given and an ̸= 0 ; f is given, not identically van-ishing and its Fourier transform exists.Solution:

Step 1: Take Fourier transform to both sides.

n

k=0

akF (y(k ))(x) = F (f )(x).

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Using the fact that for all k

∈ {1, . . . , n

} (here we should suppose that

y(k )(x) → 0 as |x| → + ∞ for all k = 0, . . . , n −1),F (y(k ))(x) = ( ix )kF (y)(x),

we get

(n

k=0

ak (ix )k F (y)(x) = F (f )(x). (3.2)

Step 2: Solve F (y)(x).By using (3.2), we have

F (y)(x) = F (f )(x)∑nk=0 ak (ix)k. (3.3)

Step 3: Solve y(x) by taking inverse Fourier transform.It follows from (3.3) that

y(x) = F −1(F (y))( x) = 1√ 2π R eisx F (f )(x)∑nk=0 ak (is )k

ds.

Before introducing the second example, we would introduce Cauchy residuetheorem.

Denition 3.1 (Residue) The coefficient a−1 of the Laurent series of the complex valued function f on z 0:

f (z ) =+ ∞

k= −∞ak (z −z 0)k

is called the residue of f on z 0. It is denoted by Res(f, z 0).

Question: How to determine the residues of f on z 0?When f (z ) = h(z )g(z ) , with h being analytic on z 0 and z 0 is the zero of g of orderm (m ∈ N∗), then z 0 is called the pole of f of order m. The residue of f onz 0 is given as

Res (f, z 0) = limz→z0

1(m −1)!(

∂ (f (z )(z −z 0)m )∂ m z

.

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For example, if f (z ) = ez

z

−3i , then 3i is a pole of f of order 1, hence

Res (f, 3i) = limz→3i

ez = e3i .

Now we are ready to introduce Cauchy residue theorem.

Theorem 3.1 (Cauchy residue theorem) Let D be a domain containing a simple loop γ (a closed single line going round anticlockwise). Suppose that f is meromorphic on D with nitely many isolated poles at z 1, . . . , z n inside γ . Then

γ f (z ) dz = 2πin

k=1Res (f, z k ).

Here f is meromorphic at z 0 is equivalent to the fact that f (z ) can be writtenas

f (z ) =n

k= −mak (z −z 0)k ,

with m ∈N and n ∈N ∪{+ ∞}.Example of using Cauchy residue theorem: Calculate

∫ R

11+ x 2 dx.

Solution: Let us denote by iR + = (

−∞, +

∞) the closed loop in the half

plane. The meromorphic function f (x) = 11+ x 2 = 1(x + i )( x−i ) has one pole x = iinside the loop. Moreover,

Res (f, i ) = limx→i

1x + i

= −i2

.

Then by Cauchy residue theorem,

R 11 + x2 dx = 2πiRes (f, i ) = π.Notice that this result agrees with the classical approach:

R 11 + x2 dx = arctan( x) + ∞−∞ = π.Now we are ready to introduce the second example.

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Example 3.2 (Electric current in a simple circuit) The current I (t) in

a simple circuit containing the resistance R and inductance L satises the equation

LI ′(t) + RI (t) = E (t), (3.4)where L,R > 0 are given, E (t) = E 0e−a |t |, with a > 0 is the applied electro-magnetic force.Find I (t).

Solution: First we remark that this is the rst example with n = 1,(a0, a1) = ( R, L ), f (x) = E (x).

Step 1: Take Fourier transform to both sides of (3.4).We have

L(it )F (I )( t) + RF (I )( t) = F (E )(t).By using the fact that

F (E )(t) = E 0 2π aa2 + t2 ,we get

(iLt + R)F (I )( t) = E 0 2π aa2 + t2 .Step 2: Solve F (I )( t).It follows from Step 2 that

F (I )( t) = E 0 2π a(a2 + t2)( iLt + R) .Step 3: Solve I (t).

By taking the inverse Fourier transform,

I (t) = F −1(F (I ))( t) = aE 0iπL R e

ist

(s − iRL )(s + ai)(s −ai ) ds.

Now we are going to evaluate I (t). Observe that I (t) can be calculated by using Cauchy residue theorem. Notice that integral I (t) is ab-solutely convergent in iR + only when t > 0. If else supposethat t < 0 and Im(s) > 0, then Re(eist ) is increasing exponen-tially, yielding the integral I (t) is not absolutely convergent.Thus we need discuss of the distinct cases with respect to the sign of t:

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• When t > 0, the function f (s) = e

ist

(s

−iRL )( s+ ai )( s

−ai ) has two poles

iRL

and ai inside (−∞, + ∞). Hence by Cauchy residue theorem,

I (t) = aE 0iπL

2πi(Res (f, iRL

)+ Res (f,ai ) = E 0( e−atR −aL −

2aLe−RtLR2 −a2L2

.

(Here we should suppose R −aL ̸= 0 .)• When t < 0, set u = −s, we have

I (t) = −aE 0iπL

+ ∞

−∞

e−iut(u + iRL )(u + ai)(u

−ai )

du.

The function g(u) = e− iut

(u + iRL )( u + ai )( u−ai ) has only one pole ai inside

the loop R , hence by Cauchy residue theorem,

I (t) = −aE 0iπL

2πiRes (g,ai ) = E 0eat

aL + R.

• When t = 0, if we suppose I is continuous on 0, then I (0) =limt→0 I (t) = E 0R + aL ; if I is not continuous on 0, then no informa-tion on I (0) could be derived from this system of equations. We

disregard the value of I (0). It is worth remarking that by Fourier integral theorem,

I (t) = F −1(F (I ))( t)only when I is continuous on t.

Example 3.3 Find the solution of ODE:

−u(2) (x) + a2u(x) = f (x), x ∈R , a > 0. (3.5)Solution: This is the rst example with n = 2, (a0, a1, a2) = ( a2, 0, −1).Step 1: Take Fourier transform to both sides of (3.5). Here we should sup-

pose that the Fourier transforms of u, u ′′, f exist.

−(ix )2F (u)(x) + a2F (u)(x) = F (f )(x).16


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Step 2: Solve

F (u)(x).

F (u)(x) = F (f )(x)a2 + x2 .Step 3: Solve u(x).

By taking inverse Fourier transform, we have

u(x) = F −1(F (u))( x) = F −1( F (f )

a2 + ( ·)2)(x).

If we want to go further, we can use convolution theorem. Observe that

1a2 + x2 = F (g)(x),with g(x) = 1a√ π2 exp(−a|x|). Then u(x) = F −1(F (f )F (g))( x) = f ∗g(x) =

12a R f (s) exp(−a|x −s|) ds.

3.2 Application of Fourier Transform to Integral Equa-tion

We still take the 3 steps. In this subsection, we only take examples.

Example 3.4 (Fredholm integral) We consider

R f (s)g(x −s) ds + λf (x) = u(x), (3.6)where g,u, λ are given and g, u have Fourier transforms. Solve f (x).Solution:

Step 1: Take Fourier transform to both sides of (3.6). By convolution the-orem,

F (√ 2πf ∗g)(x) + λF (f )(x) = F (u)(x).This is equivalent to

√ 2πF (f )(x)F (g)(x) + λF (f )(x) = F (u)(x).17


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Step 2: Solve

F (f )(x).

F (f )(x) = F (u)(x)

√ 2 piF (g)(x) + λ.

Step 3: Solve f (x).By using inverse Cauchy transform,

f (x) = F −1(F (f ))( x) = 1√ 2π R eisx F (u)(s)√ 2πF (g)(s) + λ ds.

Remark: if the above integral is NOT convergent, think of other approaches.

Example 3.5 Find the solution f (x) of the integral equation

R f (x −s)f (s) ds = 1a2 + x2 , a > 0. (3.7)Remark: this is the above example with g = f, λ = 0, u(x) = 1a 2 + x 2 .Solution:

Step 1: Take Fourier transform to both sides of (3.7). By convolution the-orem,

√ 2π(F (f )(x)2 = F (

1(·)2 + a2

)(x) = π2 e−a |x |a .Step 2: Solve F (f )(x).

F (f )(x) = +− 12a e−a | x |2 .Step 3: Solve f (x).

By using inverse Cauchy transform,

f (x) = F −1(F (f ))( x) = +−F −1( 12a

e−a |·|

2 (x) = +− aπ 24x2 + a2 .18


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3.3 Applications of Fourier Transform to Partial Dif-ferential Equations (PDE)

We still illustrate the method by taking examples.

Example 3.6 (Dirichlet’s problem in the half-plane) Consider the fol-lowing solution of the Laplace equation in the half-plane:

uxx + uyy = 0, x ∈R , y ≥ 0u(x, 0) = f (x) x ∈R ,u(x, y) → 0 |x| → + ∞, y → + ∞Solution:

Step 1: Take Fourier transform to both sides of the system with respect tox (since the boundary conditions are with respect to x). The system becomes

(ix )2U (x, y) + ∂ 2 U (x,y )

∂y 2 = 0, x, y ∈RU (x, 0) = F (x) x ∈R ,U (x, y) → 0 |x| → + ∞, y → + ∞,

where U is the Fourier transform of u with respect to x and F is the Fourier transform of f .

Step 2: Solve U (x, y).Observe that

∂ 2U (x, y)∂y2 −x

2U (x, y) = 0

is a homogeneous linear ODE with respect to y. Then by using the method of characteristic polynomial:

θ2 −x2 = 0 .We get the solution vector space of U (x, y) has basis exy , e−xy , or saying

U (x, y) = C 1(x)exy + C 2(x)e−xy ,

where C 1(x), C 2(x) are two ’constants’ depending on x. The boundary conditions show that

C 1(x) = 0 x > 0,C 2(x) = 0 x ≤ 0,C 1(x) + C 2(x) = F (x),

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This yields

U (x, y) = { F (x)e−xy x > 0F (x)exy x ≤ 0. = F (x)e−|x |y .Step 3: Solve u(x, y).

By using inverse Fourier transform and the fact that (recall that y ≥ 0),

e−|x |y = F ( 2π y(·)2 + y2 )(x),we get

u(x, y) = F −1(U (·, y))( x, y) = F −1(F (f )(·)e−|·|y )(x, y) = f ∗g(x, y),where g(x) = 2π yx 2 + y2 .

When f (x) = T 0H (a − |x|), T 0, a > 0, we have u(x, y) =

T 0π

tan−1( 2ay

x2 + y2 −a2.

3.4 Evaluation of Denite IntegralsWe introduce 2 new approaches to evaluate some denite integrals: Convo-lution theorem and Parseval’s relation. Notice that Cauchy residue theoremis another approach that is often used. The key idea of this subsection is touse Fourier transforms’ table.

Example 3.7 (Convolution theorem) Evaluate

I (a, b) = R ds(s2 + a2)(s2 + b2) , a, b > 0.Solution: Observe that if we denote by

f (s) = e−a |s |, g(s) = e−b|s |,

then I (a, b) =

π2ab R F (f )(s)F (g)(s) ds.

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By using the following convolution theorem on x = 0:

R eisx F (f )(s)F (g)(s) ds = R f (s)F (g)(x −s) ds,we get

I (a, b) = π2ab R f (s)F (g)(−s) ds = π2ab R e−|s |(a + b) = πab(a + b) .

Example 3.8 (Parseval’s theorem) Evaluate the following denite inte-gral

I = + ∞0

s2 ds(s2 + a2) , a > 0.

Solution: Set f (s) = 12(s 2 + a 2 ) , then f ′(s) = − s(s 2 + a 2 ) and

F (f )(s) = π2 ( 12a )e−a |s |.By using Parseval’s relation,

I = R |f ′(s)|2 ds = R |F (f ′)(s)|2 ds = R s2f 2(s) ds = 2π(2a)5 .3.5 ExercisesExercise 1: (The Cauchy problem for the wave equation) Obtain the D’Alembertsolution of the initial value problem for the wave equation:

u tt = c2uxx , xR , t > 0, c > 0u(x, 0) = f (x) x ∈R ,u t (x, 0) = g(x) x ∈R .

Exercise 2: [One-Dimensional wave equation] Obtain the solution of the fol-

lowing wave equationu tt = c2uxx , xR , t > 0, c > 0u(x, 0) = 0 x ∈R ,u t (x, 0) = δ 0(x) x ∈R .

Remark this is a particular case of Exercise 1.

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Example: f (x) = 1x veries the rst condition but not the second one,

then its Laplace transform does not exist. The proof is:

+ ∞0 e−sxs ds = 10 e−sxs ds+ + ∞1 e−sxs ds = C

≥ e−x 10 1s ds+ C = e−x [logs]10+ C = + ∞.This computation can be used to show the Laplace transform of f (x) =1

x a exists only when a > 1.

• When f tends to 0 exactly with exponential rate, then the domain of x becomes more narrow than Re(x) > 0. For example, when f (x) =ce−

ax

with a > 0, then L(f )(x) exists only for Re(x) > a . This can beproved by the denition of absolutely integrable functions.Examples of basic Laplace transforms : The following list can be

obtained be direct computations. For other functions, please check the table.

L(c)(x) = cx

, Re(x) > 0.

L(ea (·))(x) = 1x −a

, Re(x) > a ∈R +

L(sin( a(·)))( x) = a

a2 + x2, a ∈R

L(cos(a(·)))( x) = xx2 + a2 , a ∈R

L((·)n )(x) = n!xn +1

, n ∈N∗

L((·)a )(x) = Γ(a + 1)

xa +1 , a > −1,

where Γ is the gamma function dened as, for x > 0,

Γ(x) = + ∞0 sx−1e−s ds.Examples: Γ(1) = 1, Γ( 12 ) = √ π, Γ(n) = ( n −1)! for n ∈N∗.4.2 Properties of Laplace TransformWe summarize the following basic properties of Laplace transforms, all of them can be obtained by the denition. Notice that most formulas in the

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4.3 Convolution TheoremWe should pay attention that the convolution for Laplace transform is dif-ferent from that of Fourier transform (no coefficient 1√ 2π ).

Denition 4.2 (Convolution) The convolution of f, g is given as (if it exists), for x ≥ 0,

f ∗g(x) = x0 f (x −s)g(s) ds.We still have convolution theorem for Laplace transforms:

Theorem 4.1 (Convolution theorem)

L(f ∗g)(x) = L(f )(x)L(g)(x),where x ≥ 0 and f, g are dened on [0, + ∞).

4.4 Inverse Laplace TransformThe inverse Laplace transform can be obtained by using Fourier integral

theorem, it is given as:

L−1(f )(x) = 12πi c+ i∞c−i∞ esx f (s) ds,

where c > 0 can be any constant and x ≥ 0.Some remarks should be made according to the above formula.Remarks:

1. L−1(f )(x) does not depend on c. In fact (c −i∞, c + i∞) denotes aloop strictly containing the half-plane {

z

∈C : Re(z )

≤ 0

}.

2. This inverse transform can be obtained directly by Fourier integraltheorem and change of variables.

Now we are looking for determining inverse laplace transforms. Here weintroduce 4 methods, they have both advantages and shortcomings.

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Method 1: (Partial fractional decomposition method + Laplace trans-

forms’ table)Key idea: Write

L(f )(x) = p(x)q (x)

,

where p, q are polynomials with deg( p) < deg(q ).

Example 4.1 Find L−1( 1(·)(( ·)−a ) ), a > 0.Solution: For x ≥ 0,L−1(

1(·)(( ·) −a)

)(x) = L−1

(1a

( 1

(·) −a − 1(·)

) (x)

= 1a(L−

1( 1(·) −a

)(x)

= eax−L−1( 1(·)

)(x)

=1= 1a

(eax −1).We remark that by using this result and Lebesgue dominated conver-gence, we can get

L−1( 1(·)2

)(x) = lima→0+ L−

1( 1

(·)(( ·) −a))(x) = lim

a→0+eax −1

a = x.

Method 2: (Convolution theorem)Key idea: Use the relation

L−1(L(f )L(g))( x) = f ∗g(x).Example 4.2 We take the above example and use convolution theo-rem.

L−1( 1

(·)(·) −a)(x) = L−1(L(1)L(ea (·)))( x) = 1∗ea (·)(x) = x0 eas ds = eax −1a .

Method 3: (Cauchy residue theorem) Key idea: Use the following rela-

tion:

L−1(f )(x) = 12πi c+ i∞c−i∞ esx f (s) ds =

n

k=1

Res (ex (·)f (·), z k ),

where {z 1, . . . , z n } are all the n holes of ex (·)f (·) inside the loop (c −i∞, c + i∞).26


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Example 4.3 Determine

L−1( (·)((

·)2 + a 2 )2 ), a

∈R .

Solution:

• When a ̸= 0 ,

L−1( (·)

((·)2 + a2)2)(x) =

12πi c+ i∞c−i∞ se

sx

(s2 + a2)2 ds.

The function f (s) = sesx

(s 2 + a 2 )2 has two poles {i|a|, −i|a|} of order 2inside the loop. Therefore

Res (f, i |a|) = lims→i|a |1

(2 −1)!dds((s −i|a|)

2

f (s) = xe|a |x

4i|a| .Similarly,

Res (f, −i|a|) = −xe|a |x4i|a|

.

Finally by Cauchy residue theorem,

L−1( (·)

((·)2 + a2)2)(x) = Res (f, i |a|) + Res (f, −i|a|) =

x sin(|a|x)2|a|

.

• When a = 0. The inverse Laplace transform reduces toL−1(

1(·)3

)(x) = x2

2 .

This can be seen from the table, or by Cauchy residue theorem, or by using Lesbegue dominated convergence:

L−1( 1(·)3

)(x) = lima→0 L−

1( (·)

((·)2 + a2)2)(x) = lim

a→0x sin(|a|x)

2|a| =

x2

2 .

Method 4: (Heaviside expansion theorem)Key idea: If f has a Taylor expansion to the innity around 0 (alsocalled Maclaurin series):

f (x) =+ ∞

k=0

f (k )(0)k!

xk ,

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then by taking inverse laplace transform,

L−1(f )(x) =+ ∞

k=0

f (k )(0)k! L−

1((·)k )(x) =+ ∞

k=0

f (k )(0)k!

k!xk+1

=+ ∞

k=0

f (k )(0)xk+1

.

The Heaviside expansion theorem is a way to determine all the coeffi-cients in the above series.

Theorem 4.2 (Heaviside expansion theorem) If f (x) = p(x )q(x ) , where p, q are two polynomials with deg( p) < deg(g). Then

L−1(f )(x) =

n

k=1

p(αk )

q ′(α k )exα k ,

where α1, . . . , α n are all the distinct roots of

q (x) = 0 .

Remark: This method can be used only for f having innite orderderivatives around 0. And it can determine the inverse Laplace trans-form (or Laplace transform) L−1(f )(x) only for a narrow range of xsince here f is not dened over R + but in the neighborhood of 0. How-ever, it is enough in practice to guess the form of inverse Laplace trans-form for all possible x such that Re(x) > 0.

Example 4.4 Find L−1( (·)(·)2 −3(·)+2 ).Solution: Here we have p(s) = s, q (s) = s2 −3s + 2, deg( p) = 1 0.Solution: Recall that cos(as ) has Taylor expansion around 0:

cos(as ) =+ ∞

k=0

(−1)k (as )2k(2k)!

.

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By taking Laplace transform, we get

L(cos(a(·)))( x) =+ ∞

k=0

(−1)k a2kL((·)2k )(x)(2k)!

=+ ∞

k=0

(−1)k a2k(2k)!

(2k)!x2k+1

= 1x

+ ∞

k=0 (−|ax |

2 k .

• When |x| > a , the geometric series converges and is summable.Using the fact ∑+ ∞k=0 xk = 11−x , we have L(cos(a(·)))( x) =

1x

( 1

1 + |ax |2) =

xa2 + x2

.

• When |x| ≤ a, the series diverges. By this method we can not derive any information on L(cos(a(·)))( x) when |x| < a even we know very well L(cos(a(·)))( x) =

xa2 + x2

for all Re(x) > 0.

4.5 ExercisesExercise 1: Using heaviside power series to nd L−1(f )(x) for

f (s) = 1s exp(−1s ).

Solution: Observe that

f (s) =+ ∞

k=0

(−1)kk!

(1s

)k+1 .

Taking the inverse Laplace transform,

L−1(f )(x) =

+ ∞

k=0

(−1)kk! L

−1(( 1(·)

)k+1 )(x) =+ ∞

k=0

(−1)kk!k!

xk .

Notice that this series converges for all x ∈C since+ ∞

k=0|(−1)kk!k!

xk | ≤+ ∞

k=0

|x|kk!

= e|x | < + ∞.

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Thus for any x such that Re(x) > 0,

L−1(f )(x) =+ ∞

k=0

(−1)kk!k!

xk .

Exercise 2: Use Maclaurin series to nd L(f )(x), wheref (s) =

1s

(e−as −e−bs ),for a,b > 0.Solution: f has Maclaurin series

f (s) =+ ∞

k=1

(−a)k −(−b)kk!

sk−1.

By taking Laplace transform,

L(f )(x) =+ ∞

k=1

(−a)k −(−b)kk! L((·)

k−1)(x) =+ ∞

k=1

(−a)k −(−b)kkxk

.

Notice the above series converges only when |x| > a and |x| > b. Thereforewhen |x| > max{a, b},

L(f )(x) =+ ∞

k=1

(−a)k −(−b)kkx k

= −log(1 + ax

) + log(1 + bx

) = log(x + bx + a

).

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5 Applications of Laplace TransformsMotivation: Laplace transform is an effective tool for solving large numberof linear ODE, PDE and IE with initial or boundary conditions, where thesolution is dened on R + . It is also quite useful to evaluate denite integrals.Main goal: Know how to solve ODE, PDE, IE by using Laplace transforms;know how to evaluate denite integrals.

5.1 Applications of Laplace Transforms to ODEWe still take the three steps:

Step 1: Take Laplace transform to both sides of equation.

Step 2: Solve L(f )(x), where f is the solution of equation.Step 3: Solve f (t) by using the inverse Laplace transform.

Most useful formula in this section:

• Laplace transforms of derivatives: suppose that f is nth order differ-entiable over R+ and its nth order derivative has Laplace transform,then for Re(x) > 0,

L(f (n ))(x) = xn L(f )(x) −n −1

k=0

xn −1−k f (k )(0).

Examples: we often use this result up to fth order derivative.

L(f ′)(x) = xL(f )(x) −f (0);L(f (2) )(x) = x2L(f )(x) −xf (0) −f ′(0);L(f (3) )(x) = x3L(f )(x) −x2f (0) −xf ′(0) −f (2) (0).

That is why Laplace transform can be used to solve initial or boundaryvalue problems.

• Laplace transforms of exponentials: for a ∈R and Re(x) > max{0, a},

L(ea (·))(x) = 1x −a

.

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This relations is often used to determine the inverse Laplace transforms

(after partial fraction decomposition):

L−1( 1

(·) −a)(t) = eat .

Example 5.1 (First order ODE with initial conditions) Solve x(t) from the following ODE:

{ x′(t) + px(t) = f (t), p ∈R , t > 0x(0) = a a ∈R .Solution:


{ sL(x)(s) −x(0) + pL(x)(s) = L(f )(s), Re(s) > 0x(0) = a a ∈R .This yields

(s + p)L(x)(s) −a − L(f )(s) = 0 .

Step 2: Solve L(x)(s).L(x)(s) =

a + L(f )(s)s + p

= as + p

+ L(f )(s)s + p

.

Step 3: Solve x(t).By using inverse Laplace transform and convolution theorem,

x(t) = aL−1( 1

(·) + p)(t) + L−1(L(f )L(e− p(·)) (t) = ae− pt + f ∗e− p(·)(t).

Example 5.2 ( 3nd order ODE) Denote by Df (t) = df (t )d t . Solve x(t) from { (D3 + D2 −6D)x(t) = 0 , t > 0(x(0), x′(0), x (2) (0)) = (1 , 0, 5).

Solution:

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For Re(s) > 0,

L(D3x)(s) + L(D2x)(s) −6L(Dx)(s) = 0⇐⇒(s3L(x)(s) −s2x(0) −sx ′(0) −x(2) (0)+(s2L(x)(s) −sx (0) −x′(0) −6(sL(x)(s) −x(0) = 0⇐⇒ (s3 + s2 −6s)L(x)(s) −s2 −s + 1 = 0 .

Step 2: Solve L(x)(s).

L(x)(s) = s2 + s

−1

s(s2 + s −6) = s2 + s

−1

s(s + 3)( s −2) .Step 3: Solve x(t).

We use partial fraction decomposition method. Assume there exist a, b, csuch that

f (s) = s2 + s −1s(s + 3)( s −2)

= as

+ bs + 3

+ cs −2

.

Then,

a = lims→0

sf (s) = 16

;

b = lims→−3

(s + 3) f (s) = 13

;

c = lims→2

(s −2)f (s) = 12

.

Finally,

L(x)(s) = f (s) = 16s

+ 1

3(s + 3) +

12(s −2)

.

By taking inverse Laplace transform, for t > 0,

x(t) = 16L−

1( 1(·)

)( t)+13L−

1( 1

(·) + 3)(t)+

12L−

1( 1

(·) −2)(t) =

16

+e−3t

3 +

e2t

2 .

Laplace transforms can also be used to solve linear system of ODE or PDE.

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Example 5.3 (System of rst order ODE) Solve the following ODE:

dX (t )d t = AX (t), t > 0, A =

0−2

13 , X = x 1x 2

X (0) = 01 .

Solution:

Step 1: Take Laplace transform to both sides of system (multivariate Laplace transform).For Re(s) > 0,

s

L(X )(s)

−X (0) = A

L(X )(s).

Step 2: Solve L(X )(s).

L(X )(s) = ( sI 2×2 −A)−1X (0) = s0 0s − 0−2 13−1 0

1 = 1s − 2 − 1s − 1

2s − 2 − 1s − 1

.

Step 3: Solve X (t), t > 0.By taking inverse Laplace transform,

X (t) = e2 t −e t2e2 t −e t .

5.2 Application of Laplace Transform to PDE with Ini-tial and Boundary Conditions

We still illustrate the approaches by taking examples. The main difficulty inthis part is to choose the ”right” variable to take Laplace transform.

Example 5.4 (First order initial boundary problem) Solve the follow-ing equation:

ut + xux = x, t > 0, x > 0u(x, 0) = 0 , x > 0

u(0, t ) = 0 , t > 0.Solution:

Step 1: Take Laplace transform to both sides of equation.Since both t,x > 0, we can take Laplace transform either to t or to x.Observe it is easy to compute when taking Laplace transform to t. Thus

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we denote by U (x, s ) the Laplace transform of u(x, t ) with respect to t.

Then,

{ sU (x, s ) + xU x = xs , Re(s) > 0, x > 0U (0, s) = 0 .⇐⇒ { U x + sx U (x, s ) = 1s , Re(s) > 0, x > 0U (0, s) = 0 .

Step 2: Solve U (x, s ).We solve the rst order ODE with respect to x by using integrating factor method:

U (x, s ) = U (0, s) + e−∫ x0

sy dy

x

01s e

∫ q0

sy dy dq = xs(s + 1) .

Step 3: Solve u(x, t ).By using inverse Laplace transform with respect to t. For t > 0,

u(x, t ) = L−1(U (x, ·))( t) = xL−1( 1(·) −

1(·) + 1

(t) = x(1 −e−t ).

Example 5.5 Find the solution of the following PDE:

xu t + ux = x, t > 0, x > 0u(x, 0) = 0 , x > 0u(0, t ) = 0 , t > 0.

Unlike the above example, here we had better take Laplace transform withrespect to x since it is easier to solve the remaining ODE after transform.We omit the detail of solution here since it is very similar to the above one.Solution:

u(x, t ) = t −(t − 12

x2)H (t − x2

2 ),

where H is the Heaviside unit step function.

5.3 Application of Laplace Transform to IEWe are going to solve the following type of IE:

f (t) = h(t) + λ t0 g(t −s)f (s) ds,35


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where h, g,λ are given.

The key idea is to use convolution theorem and to take the following threesteps:

Step 1: Take Laplace transform. using convolution theorem, we get

L(f )(s) = L(h)(s) + λL(g)(s)L(f )(s).Step 2: Solve L(f )(s).

L(f )(s) = L(h)(s)

1 −λL(g)(s).

Step 3: Solve f (t), t > 0.

f (t) = L−1 L(h)(·)

1 −λL(g)(·)(t).

Example 5.6 Solve the following IE:

f (t) = a + λ t0 f (s) ds.Solution:

Step 1: Take Laplace transform. For Re(s) > 0,

L(f )(s) = as

+ λL(f )(s)L(1)(s)⇐⇒

L(f )(s) = as

+ λL(f )(s)s

.

Step 2: Solve L(f )(s).

L(f )(s) =as

1

− λs

= as

−λ

.

Step 3: Solve f (t), t > 0.By taking Laplace transform,

f (t) = aL−1( 1

(·) −λ)(t) = aeλt .

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Example 5.7 Solve the IE.

{ f (t) = a sin t + 2 ∫ t0 f ′(s) sin(t −s) ds, t > 0f (0) = 0 .Solution:

Step 1: Take Laplace transform.

L(f )(s) = a1 + s2

+ 2L(f ′)(s)L(sin(·))( s)=

a1 + s2

+ 2(sL(f )(s) −f (0))

1 + s2

= a1 + s2

+ 2sL(f )(s)

1 + s2 .

Step 2: Solve L(f )(s).

L(f )(s) =a

1+ s 2

1 − 2s1+ s 2=

a(1 −s)2

.

Step 3: Solve f (t), t > 0.By taking inverse Laplace transform, for t > 0,

f (t) = L−1( a

(1 −(·)2)(t) = ate t .

5.4 Evaluate Denite IntegralsThe Laplace transform can be used to determine the integrals whose inte-grand is two variate function. The key idea is due to the following versionof Fubini theorem: if f has Laplace transform and its Laplace transform isabsolutely integrable over ( a, b), then

L b

af (·, x) dx (t) =

b

a L(f (·, x))( t) dx.Example 5.8 Evaluate

I (t) = + ∞0 sin(tx )x(1 + x2) dx.37


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Solution: By using Fubini theorem, for s > 0 (here we only consider real

valued function),

L(I )(s) = + ∞0 L(sin(( ·)x))( s)x(1 + x2) dx= + ∞0 dx(x2 + s2)(1 + x2)=

11 −s2

+ ∞0

1x2 + s2 −

1x2 + 1

dx

= 11

−s2

tan −1( xs )s −tan −

1(x)+ ∞0

= π2(

1s −

11 + s

.

Finally by taking inverse Laplace transform, we get

I (t) = π2

(1 −e−t ).Notice that

+ ∞0 1x2 + s2 − 1x2 + 1 dxcan also be calculated through Cauchy residue theorem, only needing to ob-serve that the integrand function is even with respect to x:

+ ∞0 1x2 + s2 − 1x2 + 1 dx = 12 + ∞−∞ 1x2 + s2 − 1x2 + 1 dx.5.5 ExercisesExercise 1: Solve the following linear system of ODE:

x′(t) = x(t) −2y(t),y′(t) = y(t) −2x(t), t > 0x(0) = 1 ,y(0) = 0 .

Solution:

{ x(t) = et cos(2t),y(t) = −et sin(2t), t > 0.38


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Exercise 2: Solve the following IE:

f (t) = sin(2 t) + t0 f (t −s) sin(s) ds.Solution:

f (t) = 34

sin(2t) + t2

.

Exercise 3: Evaluate

I (t) = + ∞0 e−tx 2 dx, t > 0.Solution:

I (t) =√ π2√ t .

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6 Finite Fourier Cosine and Sine TransformsMotivation: Finite Fourier transforms are methods for solving ODE, PDEand IE with initial and boundary conditions over (0 , a).Main goal: Know the denition and basic properties of nite Fourier trans-forms; know the convolution theorem; know how to use the nite Fouriertransforms to solve PDE.

6.1 Denitions of Finite Fourier TransformsDenition 6.1 (Finite Fourier sine transform) If f is almost everywhere continuous on a nite interval (0, a) (a > 0), the nite Fourier sine transform (if exists) is dened by: for n ∈N∗,

F s (f )(n) = a0 f (x)sin(nπx

adx.

Remark:

• The sufficient and necessary condition for the existence of nite Fouriersine transform of f is:

a

0

f (x)sin

(nπx

adx < +

∞.

A simple sufficient condition could be

a0 |f (x)|dx < + ∞.• F s (f )(n) ∈R . Therefore in this chapter, all is real-valued.• This integral can be used to solve ODE, PDE and IE where the solutionis dened on a nite interval as (0 , a).

Question: Where does this transform come?Answer: It is due to the decomposition of Fourier sine series: if f ∈ L2([0, a])and f is continuous on x, then

f (x) = 2a

+ ∞

n =1F s (f )(n)sin(

nπxa

,

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where

{F s (f )(n)

}n∈

N∗ are the Fourier wavelet coefficients. This equation can

also be viewed as a discrete version of the Fourier integral theorem, hencethe inverse nite Fourier sine transform is given as (if f is continuous on x)

F −1s (f )(x) = 2a

+ ∞

n =1

f (n)sin(nπx

a.

Denition 6.2 (Finite Fourier cosine transform) If f is a.s. continu-ous over (0, a), the nite Fourier cosine transform is dened as: for n ∈N ,

F c(f )(n) = a0 f (x)cos(

nπxa

dx.

The inverse nite Fourier cosine transform is given as

F −1c (f )(x) = f (0)

a +

2a

+ ∞

n =1

f (n)cos(nπx

a.

Notice that here n = 0 is not vanishing mainly because cos(0) = 1 ̸= 0.Example 6.1 Find the nite Fourier sine and cosine transforms of

(a) f (x) = 1; (b) f (x) = x.

Solution:(a) For n ∈N∗,

F s (1)(n) = a0 sin(nπx

adx

= − anπ

cos nπx

aa0

= anπ

(1 −cos(nπ ))=

anπ

(1 −(−1)n ).For n

∈N ,

F c(1)( n) = a0 cos(nπx

adx

= a sin(nπ )

nπ .

The results could be different via n:

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• When n

̸= 0, sin( nπ ) = 0 implies

F c(1)(n) = 0 .• When n = 0,

F c(1)(0) = lims→0a sin(s)

s = a sin′(0) = a cos(0) = a.

Finally,

F c(1)(n) = { a, n = 00, n ≥ 1.(b) For n ∈N∗,

F s ((·))( n) = a

0x sin(nπxa dx

= −x cos( nπxa )

a0 −∫ a0 cos(nπxa ) dxnπ

a

= a2(−1)n +1

nπ .

For n ∈N ,F c((·))( n) = a0 x cos

(nπx

adx

= x sin(nπxa ) a0 −∫

a0 sin( nπxa ) dxnπa

= a2 sin(nπ )

nπ − a2(1 −cos(nπ ))

(nπ )2 .

• When n ̸= 0,F c((·))( n) = −

a2(1 −cos(nπ ))(nπ )2

= a2

(nπ )2((−1)n −1 .• When n = 0, we use L’Hopital’s rule to calculate the limit:

F c((·))(0) = lims→0 a2

sin ss − a2

(1 −cos s)s2 = a2 −lims→0 a2

sin s2s = a2

2 .

Finally we conclude

F c((·))( n) = a2

2 , n = 0a 2

(nπ )2 ((−1)n −1 , n ≥ 1.42


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6.2 Basic Properties of Finite Fourier TransformsProperties:

1. The transforms of derivatives of f (if exist!):

• F s (f ′)(n) = −( nπa )F c(f )(n);• F s (f (2) )(n) = −( nπa )2F s (f )(n) + ( nπa )(f (0) + ( −1)n +1 f (a));• F c(f ′)(n) = ( nπa )F s (f )(n) + ( −1)n f (a) −f (0);• F c(f (2) )(n) = −( nπa )2F c(f )(n) + ( −1)n f ′(a) −f ′(0).

Remark: F s (f (2) ), F c(f ′) can be used to solve initial value problems;F c(f

(2)

) can be used to solve boundary value problems; F s (f ′) does notdepend on initial nor boundary value but it is not very tractable to useit for solving equations since the inverse nite sine transform of F c(f )is hard to tell.

2. The nite Fourier transforms can be extended to transforms of functionover (−a, a ).

Denition 6.3 (Odd periodic extension and even periodic extension)A function f 1(x) (a.s. continuous)is said to be the odd periodic extension of the function f (x) with period 2π if f 1(−x) = −f 1(x) and f 1(x) = f (x) for x > 0 or equivalently:

f 1(x) = { f (x), 0 < x < π−f (−x), −π < x < 0.Notice that usually we don’t care the value of f 1(0).A function f 2(x) (a.e. continuous)is said to be the even periodic extension of the function f (x) with period 2π if f 1(−x) = f 1(x) and f 1(x) = f (x) for x > 0 or equivalently:

f 2(x) = { f (x), 0 < x < πf (−x), −π < x < 0.Notice that f 1 and f 2 can be furthermore extended to R by the period 2π:for any x ∈R ,

f 1(x + 2π) = f 1(x), f 2(x + 2π) = f 2(x).

These extensions have the following properties, here we only consider oneperiod a ∈ (−π, π ):

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6.4 Applications of Finite Fourier Transforms to PDEThe key of solving PDE is to choose the right transform between sine andcosine transforms and to choose the right variable to take transforms. Letillustrate by examples.

Example 6.2 Suppose that the temperature distribution u(x, t ) satises the following equation:

ut = Ku xx , 0 ≤ x ≤ a, t > 0, K > 0u(0, t ) = u(a, t ) = 0 , t > 0u(x, 0) = f (x), 0

≤ x

≤ a.

Solution:

Step 1: Take nite Fourier transform.Here we should take nite Fourier sine transform with respect to x since x ∈ [0, a] (nite Fourier transform) and u(0, t ) = u(a, t ) = 0 are initial conditions (sine transform). Denote by U (n, t ) the nite Fourier sine transform of u(x, t ) with respect to x. Thus,

{ U t = −K ( nπa )2U (n, t ), n ∈N∗, t > 0, K > 0U (n, 0) = F s (f )(n), n ∈

N

∗.

Step 2: Solve U (n, t ).By using integrating factor method we get

U (n, t ) = F s (f )(n) exp(−K (nπa

)2t).

Step 3: Solve u(x, t ).By taking inverse nite Fourier sine transform,

u(x, t ) = 2a

+ ∞n =1

F s (f )(n) exp(−K (nπa

)2t)sin(nπx

a ).

Notice that this series is convergent and can be approximated by numerical method.

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Example 6.3 Now we consider a similar model for temperature distribution:

u t = Ku xx , 0 ≤ x ≤ a, t > 0, K > 0ux (0, t ) = ux (a, t ) = 0 , t > 0u(x, 0) = f (x), 0 ≤ x ≤ a.

Here we should take nite Fourier cosine transform with respect to x sinceux (0, t ) = ux (a, t ) = 0 are boundary conditions and the rest is similar to theabove example. We omit the detail of solution.Solution:

u(x, t ) =

∫ a

0 f (s) ds

a +

2

a

+ ∞

n =1 F c(f )(n) exp(

−K (

nπ

a )2t)cos(

nπx

a ).

The following example is quite interesting. It shows that we can use a com-bine transforms to solve PDE.

Example 6.4 (Applications of the joint Laplace and nite Fourier transform)We consider the following PDE and would solve u(x, t ):

utt = cuxx , 0 ≤ x ≤ l,t > 0, c > 0u(x, 0) = f (x), 0 ≤ x ≤ lut (x, 0) = g(x), 0 ≤ x ≤ lu(0, t ) = u(l, t ) = 0 , t > 0.

There are both initial or boundary conditions on x and t. Then it is con-venient to use integral transforms to with respect to both variables x and t.For t, it is proper to take Laplace transform since t > 0 and u(x, 0) = f (x); for x, it is proper to take nite Fourier sine transform since 0 ≤ x ≤ l and u(0, t ) = u(l, t ) = 0 .Solution:

Step 1: Take integral transforms. Let us rst take Laplace transform with respect to t. We denote by U (x, s ) the Laplace transform of u(x, t ) with

respect to t. Hence we have

{ s2U (x, s ) −sf (x) −g(x) = cU xx , 0 ≤ x ≤ lU (0, s) = U (l, s ) = 0 .In order to solve the above U (x, s ), we had better then take nite Fourier sine transform with respect to x. Denote by V (n, s ) the nite Fourier

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transform of U (x, s ) with respect to x. Hence

s2V (n, s ) −sF s (f )(n) −F s (g)(n) = −c(nπl

2V (n, s ).

V (n, t ) is in fact the joint Laplace and nite Fourier sine transform of u(x, t ) on (n, s ).

Step 2: Solve V (n, t ).

V (n, t ) = sF s (f )(n) + F s (g)(n)

s2 + c( nπl )2 =

sF s (f )(n)s2 + a2

+ F s (g)(n)s2 + a2

,

where a = c( nπl )2.

Step 3: Solve u(x, t ).By taking the inverse nite Fourier sine transform with respect to nand the inverse Laplace transform with respect to s, we get nally

G(n, t ) = L−1(V )(n, t ) = F s (f )(n) cos(at ) + F s (g)(n) sin(at )

a ,

and

u(x, t ) = F −1s (G)(x, t ) = 2a+ ∞

n =1 (F s (f )(n)cos(at )+ F s (g)(n) sin(at )a sin(nπxl .

Remark: we recall that the joint Laplace and nite Fourier sine transformof f (x, t ), x ∈ (0, a), t > 0 is given as

V (n, t ) = + ∞0 e−st( a

0f (x, t ) sin(

nπxl

) dx ds.

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7 Finite Laplace TransformsMotivation: The Laplace transform of f (t) = eat

2, a > 0 does not exist, the

reason is that the function f tends to innity faster than exponential as t →.Hence we introduce nite Laplace transform to remedy this inconvenience.Finite Laplace transform can be viewed as some generalization of Laplacetransform, since all results by Laplace transforms can be obtained from niteLaplace transform by taking T → + ∞.Main goal: Know the denition and basic operational properties of niteLaplace transform; know how to solve ODE by using nite Laplace transform.

7.1 IntroductionThe Laplace transform of

f (t) = eat2

, a > 0

does not exist since

+ ∞0 eas 2 e−st ds = + ∞0 es (as −t ) ds ≥ + ∞t +1a es ds = + ∞.However, the following transform of f exists: for T > 0,

T

0eas 2 e−st ds < + ∞.

Note that in physics we often deal with systems over a nite time interval(0, T ), thus the above transform has interests of its own right.

Denition 7.1 (Finite Laplace transform) The nite laplace transform of f almost everywhere continuous over (0, T ) is dened for T > 0, t ∈C ,

LT (f )( t) = T 0 f (s)e−st ds.By Fourier integral theorem, the inverse Laplace transform of f is given as: for T > 0 and all x ∈ (0, T ),

L−1T (f )(x) = 12πi c+ i∞c−i∞ esx ds,

where c > 0 is a xed constant.

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Remarks:

• The sufficient and necessary condition for the existence of the niteLaplace transform is

T 0 |f (s)|e−st ds < + ∞.• One sufficient condition of the existence of the nite Laplace transformof f is

|f (s)||s| s→0−−→ 0,

or equivalently:f (s) = o(

1s

), as s → 0.

• By Fourier integral theorem (this is true for all integral transforms),the exact relation between nite Laplace transform and its inverse isfor each x ∈ (0, T ),

L−1T (LT (f ))( x) = 12(f (x+ ) + f (x−) .

Thus

L−1T (

LT (f ))( x) = f (x) only when f is continuous on x.

Examples of nite Laplace transforms:

• For f (x) ≡ 1,

LT (1)( t) = T 0 e−st ds = e−st−t T s=0 = 1−e−tT

t .

• For f (x) = eax ,

LT (eax )( t) =

T

0e−(t−a )s ds = 1−e−(t−a )T

T −a .

• For f (x) = x,

LT (x)(t) = T 0 se−st ds = [se−st ]T 0 −∫ T

0 e−st ds

−t =

1t2 −

e−tT t

(1t + T ).

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• For f (x) = sin( ax ) and f (x) = cos( ax ).

LT (sin( ax ))( t) = a

t2 + a2 − e−tT t2 + a2(t sin(aT ) + a cos(aT ) ;

LT (cos(ax ))( t) = t

t2 + a2 +

e−tT t2 + a2(a sin(aT ) −t cos(aT ) .

• LT (H (x −a))( t) = 1t (e−at −e−tT ).

7.2 Basic properties of nite laplace transformsWe enumerate several important properties of nite Laplace transforms thatwe might need for solving ODE, PDE or IE. Since nite Laplace transformis very similar to Laplace transform, then nearly all properties of Laplacetransform could be generalized to nite Laplace transforms.

(1) (Shifting) For a ∈R ,Lt (e−ax f (x))( t) = LT (f )( t + a).

(2) (Scaling) For a > 0,

LT (f (ax ))( t) = 1aLaT (f )(

ta

).

(3) (Finite Laplace transform of derivatives)For n ∈N∗,

LT (f (n ))( t) = tn LT (f )( t) +n

k=1

e−st tn −k f (k−1) (s) T s=0 .

(4) (Finite Laplace transform of integrals)

LT

( (·)0 f (u) du (t) = 1t

(LT (f )( t) −e−sT T 0 f (u) du .

(5) (Derivatives of nite Laplace transform)For n ∈N∗, dn

dtn LT (f )( t) = ( −1)n LT ((·)n f (·))( t).

The most important property that one may use to solve differentialequations is the following:

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(6)

LT (f ) is an entire function. We recall that a complex valued function g is

an entire function has at least the following three equivalent denitions:

• g is holomorphic over the complex plane C ;• g is differentiable all over C ;• There exists a sequence (an )n ≥0 such that the following series con-verges for all possible t:

g(t) =+ ∞

n =0

an tn .

Proof: We only use the last denition to show that LT (f ) is entirefunction. We have

LT (f )( t) = T 0 f (s)e−st ds= T 0 f (s)

+ ∞

n =0

(−st )nn!

ds.

Since

1n! T 0 f (s)(−st )n ds ≤ 1n! T 0 |f (s)|(|t|T )n ds = (|t|T )nn! T 0 |f (s)|ds < + ∞,and

+ ∞

n =0

(|t|T )nn! T 0 |f (s)|ds = e|t |T T 0 |f (s)|ds < + ∞,

hence by Lebesgue dominated convergence theorem (we can change thepositions of integral and sum),

LT (f )( t) =+ ∞

n =0 T

0f (s)

(−st )nn!

ds =+ ∞

n =0 T

0f (s)

(−s)nn!

ds

= a ntn .

Finally by denition LT (f ) is an entire function.51


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7.3 Applications of nite Laplace trnasforms to solveordinary differential equations

The nite laplace transforms allow to solve the ODE, PDE with the initialand boundary conditions as the following type:

u(0) = a, u′(0) = b, . . . ;

oru(T ) = a, u′(T ) = b . . . .

We can show that the above two conditions are equivalent provided that

LT (u) exists. Thus the main difficulty is to determine either u(T ) starting

from u(0), u ′(0) or u(0) starting from u(T ), u ′(T ).

Example 7.1 Solve x(t) from the following equation:

{ x′(t) + ux(t) = At, t ∈ [0, T ], u ,A ∈R , u ̸= 0x(0) = a, a ∈R .Solution: The range of t is nite ( [0, T ]) and the initial condition is x(0) =a, then we could consider taking nite Laplace transform.

Step 1: Take nite laplace transform to both sides of the equation and we

get: for s ∈C ,sLT (x)(s) −x(0)

= a+ e−sT x(T ) + uLT (x)(s) = A( 1s 2 − 1s e−sT ( 1s + T ) ,

x(0) = a.

Step 2: (The most difficult) Solve LT (x)(s).From the above equation we obtain

LT (x)(s) = as + u −

e−sT x(T )s + u

+ As + u

( 1s2 −

1s

e−sT ( 1s

+ T ) .

Question: How to determine Lt (x)(t)? Strategy: Take two steps:(1) Let x(T ) = 0 in the expression of LT (x)(t) and nd the poles of the remaining part.

Assume x(T ) = 0 , we nd only one simple pole s = −u.52


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(2) Establish equation by letting the residue of

LT (x) on

−u be van-

ishing, and solve x(T ):lim

s→−u(s + u)LT (x)(s) = 0 .

Equivalently,

x(T ) = AT

u − Au2

+ Au2

e−uT + ae−uT .

Hence, by inserting x(T ) into the expression of LT (x)(s),

LT (x)(s) = ( a+

A

u2 )

(1 −e−(s+ u )T

s + u+

1

u(A

s2 (1

−e−sT )

−

AT

s e−sT

−

A

u2

(1

s(1

−e−sT ) .

Step 3: Solve x(t).Here we propose two approaches to solve x(t):

• One classical method is to take inverse nite Laplace transform of LT (x)(s) and get: for t ∈ (0, T ],x(t) =

Atu −

Au2

+ Au2

e−ut + ae−ut .

• Another method is to remark the formula of x(T ) is in fact true for any T > 0, then we can replace T by t and get for t ∈ (0, T ],

x(t) = At

u − Au2

+ Au2

e−ut + ae−ut .

Note that by using this method the nding LT (x) is no longer needed.Finally we conclude

{ x(t) = Atu − Au 2 + Au 2 e−ut + ae−ut , t ∈ (0, T ]x(0) = a.Example 7.2 Solve the simple harmonic oscilator governed by

{ x′′(t) + w2x(t) = F, t ∈ [0, T ], w,F ∈R , w ̸= 0x(0) = a, x′(0) = u, a,u ∈RSolution: With similar reason we can consider nite Laplace transform.Here we use the second method to solve x(t).

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Step 1: Take nite Laplace transform to both sides of equation. For s

∈C ,

s2LT (x)(s)−s x (0)

= a−x′(0)

= u+ sx(T )e−sT + x′(T )e−sT + w2LT (x)(s) =

F s (1−e−sT ).

Step 2: Solve x(T ).Recall that LT (x) is an entire function.(1) Assume x(T ) = 0 in the above equation, the remaining part has

simple poles s = iw,−iw.(2) Establish the equation by letting the residues of LT (x) be vanishing:

lims→iw

(s −iw)LT (x)(s) = 0 ,lim

s→−iw(s + iw)LT (x)(s) = 0 .

Equivalently,

{ 12iw(aiw + u −iwe−iwT x(T ) −e−iwT x(T ) + F iw (1 −e−iwT ) = 0,− 12iw(−aiw + u + iweiwT x(T ) −eiwT x(T ) + F −iw (1 −eiwT ) = 0.From this linear system of equations on (x(T ), x′(T )) , we can solve x(T ) = ( a −

F w2 ) cos(wT ) +

uw sin(wT ) +

F w2 .

Step 3: Solve x(t).Since the expression of x(T ) is valid for all T > 0, then we have for all t ∈ (0, T ],

x(t) = ( a − F w2

)cos(wt) + uw

sin(wt) + F w2

.

Notice that limt→0+

x(t) = a, then we conclude for all t ∈ [0, T ],

x(t) = ( a − F w2 )cos(wt) + uw sin(wt) + F w2 .Example 7.3 Solve the signal pulse equation:

{ y′′(x) = δ 0(x −a), 0 < a < T, x ∈ [0, T ]y(0) = y(T ) = 0 .54


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Solution: Notice that we can also use nite Fourier sine transform to solve

this equation. As a practice, we still consider nite laplace transform here.We also remark that we can’t apply Method 2 to solve this problem, simply because by using the fact that LT (y) is an entire function, we can’t solve y(T )Step 1: Take nite Laplace transform to both sides of equation: for s ∈C ,

s2LT (y)(s) −s y(0)

=0−y′(0) + s y(T )

=0e−sT + y′(T )e−sT = e−as .

Step 2: Solve LT (y)(s).Observe that

LT (y)(s) = 1s2(e−

as

+ y′(0) −e−sT

y′(T )When removing y(T ) in the expression of LT (y), the remaining part has a pole s = 0 of order 2. Then we establish the following linear system of equations on (y′(0), y′(T )):

lims→0

sLT (y)(s) = 0 ,lims→0

s2 dds LT (x)(s) = 0 .This yields,

{ y′(0) = aT −1,y

′(T ) = a

T .

Hence,

LT (y)(s) = y′(0)

s2 (1−e−sT −sT e−sT +e−ass2 (1−e−s (T −a )+ s(a−T )e−s (T −a ) .

Step 3: Solve y(x).By taking inverse nite laplace transform, we get nally for all x ∈[0, T ],

y(x) = x(aT −1 + ( x −a)H (x −a).

Remark: It is wrong to use the relation y′(T ) = aT to solve y(t) as y(t) =log(t), since there is some hidden restriction of the variable T behind this equation as T > a. In fact the complete equation without considering the restriction t > a is: for all t ∈ (0, T ),

y′(t) = a −tt

+ H (t −a) + ( t −a)δ 0(t −a).

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8 Hankel TransformsMotivation: The Hankel transforms are very useful in solving PDE in cylin-drical polar coordinates.Main goal: Know the denition of the Bessel functions; know the deni-tion and basic properties of the Hankel transforms; Use Hankel transformsto solve PDE in physical problems.

8.1 IntroductionIn this chapter we are interested in solving PDE in cylindrical coordinates,namely, Laplace equation in cylindrical coordinates. For example: considerequation

∇2(x1, x2, z ) = f (x1, x2, z ),

where ∇2 denotes the Laplace operator dened as

∇2 = (

∂ 2

∂x 21+

∂ 2

∂x 22+

∂ 2

∂z 2.

If we set (x1, x2, z ) = ( r cos(θ), r sin(θ), z ) (cylindrical coordinate), the equa-tion becomes

( ∂ 2

∂r2 +

1

r

∂

∂r +

∂ 2

r2∂θ

2 + ∂ 2

∂z 2 u(r cos(θ), r sin(θ), z ) = f (r cos(θ), r sin(θ), z ).

In particular, when θ = z = 0, we have

∇2u(r, 0, 0) = ∇2u(r ) = (

∂ 2

∂r 2 +

1r

∂ ∂r

u(r ).

Now we are ready to introduce the Hankel function. The denition of theHankel function relies on the Bessel functions, thus we would rst introduceBessel functions.

Denition 8.1 Bessel functions are the solutions of Bessel’s differential e-quation:

t2 d2f (t)dt2

+ t df (t)dt

+ ( t2 −a2)f (t) = 0 ,where a ∈C .Now we only consider the case when a ∈ Z . When a ∈ Z , the solution of Bessel’s differential equation is a linear combination of two functionals: J a (t)(no singularity on 0) and Y a (t) ( 0 is a singularity).

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• J a is called Bessel function of rst kind of order a: for a

∈Z , it can be

dened as, for t ∈R ,J a (t) =

12π 2π0 ei (as −t sin( s )) ds.

• Y a (t) is called Bessel function of second kind of order a: for a ∈ Z , it can be dened as, for t ∈R ,Y a (t) = lim

k→aJ k (t) cos(kπ ) −J −k (t)

sin(kπ ) ,

where J k (t) is denotes the Bessel function of the rst kind of real-valued k.

Properties of J a (t):

(1) For a ∈Z , b ∈R ,

J a (t) = 12π 2π + bb ei (as −t sin s ) ds.

Proof: It suffices to show

2π + b

bei (as −t sin s ) ds =

2π

0ei (as −t sin( s )) ds.

We use the periodic property of function f (t) = ei (as −t sin s ) . Notice thatf has period 2π. Then, for any b ∈R , if we let k be the unique integersuch that b−2kπ ∈ [0, 2π), we have

2π + bb f (s) ds = 2π + b−2kπb−2kπ f (s) ds=

2π

b−2kπ + 2π + b−2kπ

2π f (s) ds

= 2πb−2kπ + b−2kπ

0f (s) ds

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(2) For n

∈Z , t

∈R ,

J n (t) = 1π π0 cos(ns −t sin(s)) d s.

Now we are ready to dene Hankel transform.

Denition 8.2 The Hankel transform of f : R + → R of order n ∈ Z is dened as: for t ∈R ,

Hn (f )( t) = + ∞0 rJ n (rt )f (r ) dr.Here we denote the integrand’s variable by r in order to the meaning of cylindrical coordinate. Its inverse transform is given as H −1n = H n : for t ≥ 0,

H−1n (f )( t) = + ∞0 rJ n (rt )f (r ) dr.Remarks:

(1) The sufficient and necessary condition of the existence of Hankel trans-form:

+ ∞0 |

rJ n(rt )f (r )

|dr < +

∞.

One simple sufficient condition of existence is:

−0+ ∞|rf (r )|dr < + ∞.(2) The Hankel transform comes from the 2- D Fourier transform where we

take ( x, y) = r (cos(θ), sin(θ)).

8.2 Properties of Hankel transformsWe summarize the basic operational properties of Hankel transforms as fol-lows:

(1) (Scaling)

Hn (f (a(·)))( t) = 1a2Hn (f )(

ta

), a > 0.

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(2) (Parseval’s relation)

+ ∞0 rf (r )g(r ) dr = + ∞0 tHn (f )( t)Hn (g)(t) dt,where f, g are such that the both sides of the above equation exist.

(3) (Hankel transform of derivatives)

• If n ∈N∗ and [rf (r )]+ ∞0 = 0, then

Hn (f ′)( t) = t2n((n −1)H

n +1 (t) −(n + 1)Hn −1(f )( t) .In particular, when n = 1,

H1(f ′)(t) = −tH0(f )( t).

• (Important property for solving differential equations) For n ∈Z ,if [rf (r )]+ ∞0 = [rf ′(r )]+ ∞0 = 0,

then

Hn((∇2− n2

(·)2)f (·) (t) := Hn(

df dr

+1r

df dr −

n2

r 2f (r ) (t) = −t2Hn (f )( t).

In particular, when n = 0,

H0( df dr

+ 1r

df dr

(t) = −t2H0(f )( t).And when n = 1,

H1( df dr

+ 1r

df dr −

1r 2

f (r ) (t) = −t2H1(f )( t).

The last two particular cases are widely used for nding solutions of PDE in cylindrical polar coordinates.

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8.3 Applications of Hankel transformsExample 8.1 (Free vibration of a large circle membrane) Find the so-lution u(r, t ) of the following PDE:

c2(∂ 2 u

∂r 2 + 1r

∂u∂r =

∂ 2 u∂t 2 , c > 0, r, t > 0

u(r, 0) = f (r ), r > 0ut (r, 0) = g(r ), r > 0.

Solution: Observe that the PDE can be written as

c2

(∇

2r u(r, t ) =

∂ 2u∂t 2

,

then it is convenient to take Hankel transform of order 0.

Step 1: Take Hankel transform of order 0 to both sides of equation. Denote by U (x, t ) the Hankel transform of u(r, t ) with respect to r . Then we get

∂ 2 U (x,t )∂t 2 + c

2x2U (x, t ) = 0 ,U (x, 0) = H0(f )(x),U t (x, 0) = H0(g)(x).0

Step 2: Solve U (x, t ).

By the method of characteristic polynomial,θ2 + c2x2 = 0 ,

this yields θ = +−icx.

Hence U (x, t ) = c1(x)eicxt + c2(x)e−icxt .

Notice that the constants c1(x), c2(x) should verify the following linear system due to the initial conditions:

{ c1(x) + c2(x) = H0(f )(x),c1(x) −c2(x) = H0 (g)( x )icx .⇐⇒

c1(x) = 12(H0(f )(x) + H0 (g)( x )icx ,c2(x) = 12(H0(f )(x) − H0 (g)( x )icx .60


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Finally,

U (x, t ) = H0(f )(x)cos(cxt) + 1cxH0(g)(x) sin(cxt ).

Step 3: Solve u(r, t ).By using the inverse Hankel transform, we get

u(r, t ) = + ∞0 sJ 0(sr )U (s, t ) ds = + ∞0 sJ 0(sr )(H0(f )(s)cos(cst )+ 1csH0(g)(s) sin(cst ) ds.

Example 8.2 (Steady temperature distribution in a semi-innite solid)Find the solution u(r, z ) from the following equation:

{ ∂ 2 u

∂r 2 + 1r

∂u∂r +

∂ 2 u∂z 2 = −Qq (r ), Q > 0, r, z > 0u(r, 0) = 0 . r > 0

Solution: We still consider Hankel transform of order 0.

Step 1: Take Hankel transform of order 0 to both sides of equation. Denote by U (x, z ) the Hankel transform of order 0 of u(r, z ) with respect to r.

{ ∂ 2 U (x,z )

∂z 2

−x2U (x, z ) =

−Q

H0(q )(x),

U (x, 0) = 0 .

Step 2: Solve U (x, z ).Observe that this is a second degree non homogeneous differential equa-tion.Recall that if for t ≥ t0, X (t) veries

a0X ′′(t) + a1X ′(t) + a2X (t) = g(t), a0 ̸= 0 .

Then the general solution X (t) is: for t ≥ t0,

X (t) = t

t 0

ϕ2(t −s)a0

g(s) ds + X (t0)ϕ1(t) + X ′(t0)ϕ2(t),

where ϕ1,ϕ2 are two unique solutions of the following system

a0ϕ′′(t) + a1ϕ′(t) + a2ϕ(t) = 0

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with initial conditions

{ ϕ1(t0) = 1 , ϕ′1(t0) = 0;ϕ2(t0) = 0 , ϕ′2(t0) = 1 .Now we are going to solve U (x, z ). Recall that we have

{ ∂ 2 U (x,z )

∂z 2 −x2U (x, z ) = −QH0(q )(x),U (x, 0) = 0 .First solve

ϕ′′(z ) −x2ϕ(z ) = 0with initial conditions

{ ϕ1(0) = 1 , ϕ′1(0) = 0;ϕ2(0) = 0 , ϕ′2(0) = 1 .By using characteristic polynomial method, we get for z > 0,

{ ϕ1(z ) = 12 (exz + e−xz ),ϕ2(z ) = 12 (exz −e−xz ).Finally, by inserting ϕ1,ϕ2 into the expression of U (x, z ), we get

U (x, z ) = z0 12x (ex (z−x ) −e−x (z−x )) ds(−QH0(q )(x) + U z (x, 0) 12x

(exz −e−xz )=

Qx2

(1 −cosh(xz ))H0(q )(x) + Asinh(xz )

x ,

where A = U z (x, 0) is a real valued function only depending on x.

Step 3: Solve u(r, z ).By taking inverse Hankel transform, we get

u(r, z ) = + ∞0

sJ 0(sr )U (s, z ) ds

= + ∞0 sJ 0(sr )(Qs2

(1 −cosh(sz ))H0(q )(s) + A sinh( sz )

sds.

Class Notes 335

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