LAPLACE TRANSFORM, FOURIER TRANSFORM AND … · 2018-08-15 · More Laplace transforms 3 2. Fourier analysis 9 2.1. Complex and real Fourier series (Morten will probably teach this

LAPLACE TRANSFORM, FOURIER TRANSFORM AND DIFFERENTIAL EQUATIONS

XU WANG

These notes for TMA4135 (first seven weeks) are based on Erwin Kreyszig’s book [2], Dag Wessel-Berg’s video: http://video.adm.ntnu.no/serier/4fe2d4d3dbe03, and references [1, 3, 4] (among them[1] is very short and readable).

CONTENTS

1. Laplace transform 11.1. Basic facts 11.2. Laplace transform of derivatives, ODEs 21.3. More Laplace transforms 32. Fourier analysis 92.1. Complex and real Fourier series (Morten will probably teach this part) 92.2. Fourier Sine and Cosine series 132.3. Parseval’s identity 142.4. Fourier transform 152.5. Fourier inversion formula 162.6. Fourier transform of derivative and convolution 183. Partial differential equations 193.1. Functions of several variables 193.2. Solve wave equation by Fourier series 213.3. Solve heat equation by Fourier series 233.4. Solve heat equation by Fourier transform 244. Appendix: Definition of e, π and Euler’s formula 254.1. Where does e come from ? 254.2. Definition of the exponential function 264.3. Definition of π and trigonometric functions 27References 28

1. LAPLACE TRANSFORM

1.1. Basic facts.

Definition 1.1. Let f(t), t ≥ 0 be a given function. We call

F (s) :=

∫ ∞0

e−stf(t) dt,

the Laplace transform of f(t) and write

F = L(f), f = L−1F.

Date: August 16, 2018.1

2 XU WANG

Remark: One can prove that the Laplace transform L is injective (see page 9 in [1]), that is thereason whyL−1 is well defined (for a precise formula ofL−1, see page 10 in [1]). To compute Laplacetransforms, we need:

(1) d(fg) = fdg + gdf,

∫ b

adf = f(b)− f(a),

where df := f ′(t)dt.

Example 1:

(2) L(1) =1

s, L−1(

1

s) = 1, s > 0.

Example 2:

(3) L(ekt) =1

s− k, L−1(

1

s− k) = ekt, s > k.

Example 3: L(et2) does not exist, ∫ ∞

0e−stet

2dt =∞,

for all real number s.

Remark: Laplace transform is linear: By linearity, we mean for all real numbers a, b,

(4) L(af + bg) = aL(f) + bL(g).

Application 1:

L(3 + 2e5t) = 3L(1) + 2L(e5t) =5(s− 3)

s− 5, s > 5.

Application 2: Since1

s2 − 3s+ 2=

1

(s− 1)(s− 2)=

1

s− 2− 1

s− 1.

linearity gives

L−1(1

s2 − 3s+ 2) = L−1(

1

s− 2)− L−1(

1

s− 1) = e2t − et.

Proposition 1.2. L(tn) = n!sn+1 , for n = 1, 2, · · · , and s > 0.

1.2. Laplace transform of derivatives, ODEs.

Theorem 1.3 (Laplace transform of the derivative).

L(f ′) = sL(f)− f(0).

Example: Solve:y′ = y, y(0) = 1.

The answer isy(t) = et.

Remark: Apply the theorem to f ′, we get

L(f ′′) = sL(f ′)− f ′(0) = s(sL(f)− f(0))− f ′(0) = s2L(f)− sf(0)− f ′(0).

LAPLACE TRANSFORM AND FOURIER TRANSFORM 3

Remark: Apply the Laplace transform to a differential equation

y′′ + ay′ + by = c(t), a, b ∈ R,then we get

s2Y − sy(0)− y′(0) + a(sY − y(0)) + bY = C,

i.e(s2 + as+ b)Y = (s+ a)y(0) + y′(0) + C,

thus the inverse transform gives the solution

y = L−1(Y ) = L−1

((s+ a)y(0) + y′(0) + C

s2 + as+ b

).

Example: Considery′′ + 4y′ + 4y = 0, y(0) = 0, y′(0) = 1.

then the above formula gives

y = L−1

(1

s2 + 4s+ 4

)= L−1

(1

(s+ 2)2

).

How to compute the inverse Laplace transform of 1(s+2)2

? Is it related to L−1( 1s2

) = t ? We willanswer them in the next section.

1.3. More Laplace transforms.

1.3.1. s-Shifting. The following formula∫ ∞0

e−steatf(t) dt = F (s− a), F (s),

gives

Theorem 1.4 (s-Shifting theorem).

L(eatf(t)) = F (s− a).

Example:

L−1

(1

(s+ 2)2

)= e−2tt.

Example:

L(ekt) =1

s− k.

Take k = iw,

L(eiwt) =1

s− iw=

s+ iw

s2 + w2.

Euler’s formula (see the appendix) gives

(5) L(coswt) =s

s2 + w2, L(sinwt) =

w

s2 + w2.

Example:

L−1(s+ 2

s2 + 4) = L−1(

s

s2 + 4) + L−1(

2

s2 + 4) = cos 2t+ sin 2t.

Exercises:

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1. Find the inverse Laplace transform of ss2+2s+2

. The answer is

L−1(s

s2 + 2s+ 2) = e−t(cos t− sin t).

2. Solve y′′ − y = t, y(0) = y′(0) = 1. The answer is

y = et +1

2(et − e−t)− t.

Definition 1.5 (sinh t and cosh t).

sinh t :=et − e−t

2, cosh t :=

et + e−t

2.

Exercises:

3. Compute Laplace transform of sinh t and cosh t. The answer is

L(sinh t) =1

s2 − 1, L(cosh t) =

s

s2 − 1.

4. Compute Laplace transform of

f(t) = 1 if 3 < t < 4; f(t) = 0 otherwise.

The answer is

L(f) =e−3s − e−4s

s.

1.3.2. Laplace transform of integrals. Put

g(t) =

∫ t

0f(τ) dτ,

theng′ = f, g(0) = 0.

ThusF = L(f) = L(g′) = sG− g(0) = sG

gives G = Fs , i.e.

L(∫ t

0f(τ) dτ

)=L(f)

s.

Exercises:

1. Show that

L−1

(1

s(s2 + 1)

)= 1− cos t.

2. Show that

L−1

(1

s· 1

s(s− 1)

)= et − 1− t.


1.3.3. Using step functions.

Definition 1.6 (Step function). Let a ≥ 0, the step function u(t− a) is defined as follows

u(t− a) = 0, for 0 ≤ t < a; u(t− a) = 1, for t ≥ a.In case a = 0 we call u(t) the Heaviside function.

Exercise:

1. Draw the graphs off(t) = u(t− 1)− u(t− 3),

f(t) = (u(t)− u(t− π)) sin t,

f(t) = u(t) + u(t− 1) + · · ·+ u(t− n) + · · · .

2. Show that

L(u(t− a)) =e−as

s.

3. Compare the graphs of u(t− a)f(t− a) with that of f(t).

Theorem 1.7 (t-Shifting theorem).

L(u(t− a)f(t− a)) = e−asL(f).

Example: Since1

s− 2= L(e2t),

we haveL−1(e−s

1

s− 2) = u(t− 1)e2(t−1).

Exercise:

1. Compute Laplace transform of

f(t) =

∞∑n=0

u(t− n).

(Hint:∑∞

n=0 rn = 1

1−r for |r| < 1) Answer

L

( ∞∑n=0

u(t− n)

)=

1

s(1− e−s).

RC-Circuit equation (see page 29 section 1.5 and page 93 section 2.9 of Kreyszig’s book): R,Cpostive constants, i(t), e(t) functions:

Ri(t) +1

C

∫ t

0i(τ) dτ = e(t).

Apply the Laplace transform, we get

RI(s) +1

C· I(s)

s= E(s),

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i.e.

I(s) =E(s)

R+ 1Cs

=s

s+ 1RC

E(s)

R.

Exercise:

2. Compute i(t) when

e(t) = u(t− 1)− u(t− 2), R = C = 1.

Answeri(s) = u(t− 1)e−(t−1) − u(t− 2)e−(t−2).

1.3.4. Dirac delta function. The delta function δ(t− a) is defined by∫ ∞0

f(t)δ(t− a) dt = f(a).

In case f(t) = e−st, we have ∫ ∞0

e−stδ(t− a) dt = e−as.

ThusL(δ(t− a)) = e−as.

Exercise: solvey′′ + y = δ(t− 1), y(0) = y′(0) = 0.

Answery = L−1(e−sL(sin t)) = u(t− 1) sin(t− 1).

1.3.5. Convolution. Let f(t), g(t) be two functions for t ≥ 0.

Definition 1.8 (Convolution of f and g).

(f ? g)(t) :=

∫ t

0f(τ)g(t− τ) dτ, t ≥ 0.

Examples:

1 ? t =t2

2,

et ? et = tet,

f(t) ? 1 =

∫ t

0f(τ) dτ.

Theorem 1.9 (Laplace transform of convolution).

L(f ? g) = L(f) · L(g).


Exercises:

1. Compute tm ? tn. Hint: use tm ? tn = L−1L(tm ? tn), answer

tm ? tn =m!n!

(m+ n+ 1)!tm+n+1, m, n = 0, 1, · · ·

2. Compute L−1( 1(s2+1)2

). Hint: use L−1( 1(s2+1)2

) = L−1(L(sin t) · L(sin t)). Answer

L−1(1

(s2 + 1)2) =

sin t− t cos t

2.

3. Solve y′′ + y = sin t, y(0) = 0, y′(0) = 1. Hint: use Exercise 2. Answer

y = sin t+sin t− t cos t

2=

3 sin t− t cos t

2.

4. Solve: y −∫ t

0 (t− τ)y(τ) dτ = 1. Hint: use∫ t

0 (t− τ)y(τ) dτ = y ? t. Answer

y =et + e−t

2= cosh t.

1.3.6. Non-homogeneous linear ODEs. Consider

y′′ + by′ + cy = r(t),

given y(0) and y′(0), we have

s2Y − sy(0)− y′(0) + b(sY − y(0)) + cY = R(s).

Thus

Y =1

s2 + bs+ c·R(s) +

sy(0) + y′(0) + by(0)

s2 + bs+ c:= K(s) ·R(s) +G(s),

we get

y = k ? r + g.

Example: Consider

y′′ + y = r(t), y(0) = y′(0) = 0.

Apply the Laplcae transform, we have

s2Y + Y = L(r).

Thus

Y =1

s2 + 1· L(r),

which gives

y(t) = sin t ? r.

8 XU WANG

1.3.7. Derivative of the Laplace transform. Apply differential to

F (s) =

∫ ∞0

e−stf(t) dt,

we get

F ′(s) =

∫ ∞0

d(e−st)

dsf(t) dt =

∫ ∞0

e−st · (−tf(t)) dt = L(−tf(t)).

Example 1:

L(t sin t) = −(1

s2 + 1)′ =

2s

(s2 + 1)2.

Example 2: Let F (s) = ln(1 + s−2). Then

F ′ = (ln(1 + s2)− ln(s2))′ =2s

1 + s2− 2

s.

ThusL−1(F ′) = 2 cos t− 2.

By the above theorem, we haveL−1(F ′) = −tf(t).

thus

f(t) = L−1(ln(1 + s−2)) =2− 2 cos t

t.

1.3.8. System of differential equations. Look at this example:

y′1 = −y1 + y2; y′2 = −y1 − y2 + f(t), y1(0) = y2(0) = 0.

Apply the Laplace transform, we get

sY1 = −Y1 + Y2; sY2 = −Y1 − Y2 + F (s).

Thus(s+ 1)Y1 − Y2 = 0;

Y1 + (s+ 1)Y2 = F (s).

The first equation gives Y2 = (s+ 1)Y1, together with the second, we have

Y1 = F (s)(1 + (s+ 1)2)−1.

Use the first equation again,

Y2 = F (s)(s+ 1)(1 + (s+ 1)2)−1.

Thusy1 = f(t) ? (e−t sin t), y2 = f(t) ? (e−t cos t).

when f(t) = e−t, we get

y1 =

∫ t

0e−(t−τ)e−τ sin τ dτ = e−t(1− cos t), y2 = e−t sin t.


1.3.9. Homework for Laplace transform. Please compute Laplace transform of

1. f(t) = t, if 0 ≤ t ≤ a, f(t) = 0, if t > a. Answer

F (s) =1

s2− e−as

s2− ae

−as

s.

2. f(t) = u(t− π) sin t. Answer F = − e−πs

s2+1.

3. Solve the following equation

i′(t) + 2 i(t) +

∫ t

0i(τ) dτ = δ(t− 1), i(0) = 0.

Answeri(t) = u(t− 1)(e−(t−1) − e−(t−1)(t− 1)).

2. FOURIER ANALYSIS

2.1. Complex and real Fourier series (Morten will probably teach this part).

2.1.1. Complex Fourier series. Fix p > 0, if

f(x+ p) = f(x), ∀ x ∈ R,then call f a periodic function with period p.

Example: periodic function:

1. A polynomial is periodic if and only if it is a constant;2. eλx has period 2π if and only if λ = in, n ∈ Z.

The main theorem in Fourier analysis is the following:

Theorem 2.1 (Fourier 1807). If f has period 2π and is smooth enough then we have

f(x) =∑n∈Z

cn einx, ∀ x ∈ R.

——The proof (see Page 63 in [3]) is not assumed in this course.

What does "smooth enough" mean? It means that f is piecewise smooth and

f(x0) =f(x0+) + f(x0−)

2,

if f is not smooth at x0.

How to compute cn ? We shall prove that

cn =1

2π

∫ π

−πf(x)e−inx dx.

In fact , by the above theorem, we have1

2π

∫ π

−πf(x)e−inx dx =

∑m∈Z

cm2π

∫ π

−πeimxe−inx dx.

If m = n then ∫ π

−πeimxe−inx dx =

∫ π

−π1 dx = 2π.

10 XU WANG

If m 6= n then ∫ π

−πei(m−n)x dx =

∫ π

−πd(ei(m−n)x

i(m− n)) = 0.

Thus ∑m∈Z

cm2π

∫ π

−πeimxe−inx dx = cn.

Definition 2.2. We call

f(x) =∑n∈Z

cneinx

the complex Fourier series of f and

cn :=1

2π

∫ π

−πf(x)e−inx dx, n ∈ Z,

the complex Fourier coefficients of f .

Example: Consider

f(x) = 1, 0 < x < π; f(x) = −1, −π < x < 0,

and

f(0) = f(π) = f(−π) = 0.

Then we know f is smooth enough and

2πcn =

∫ π

−πf(x)e−inx dx =

∫ π

0e−inx dx−

∫ 0

−πe−inx dx.

Since ∫ π

0e−inx dx =

∫ π

0d(e−inx

−in) =

(−1)n − 1

−in,

and ∫ 0

−πe−inx dx =

∫ 0

−πd(e−inx

−in) =

1− (−1)n

−in,

we have

2πcn =2(1− (−1)n)

in,

i.e.

cn =2

inπ, n odd; cn = 0, n even.

Thus the complex fourier series of f is

f(x) =∑m∈Z

2

i(2m+ 1)πei(2m+1)x.


2.1.2. (Real) Fourier series. In the previous example, we have

f(x) =2

iπ

(eix +

e3ix

3+ · · ·

)+

2

iπ

(e−ix

−1+e−3ix

−3+ · · ·

),

thus

f(x) =2

iπ

((eix − e−ix) +

e3ix − e−3ix

3+ · · ·

).

Euler’s formula giveseinx − e−inx = 2i sinnx,

thus

f(x) =4

π

(sinx+

sin 3x

3+ · · ·

),

In particular, it gives

1 = f(π

2) =

4

π

(1− 1

3+

1

5− 1

7+ · · ·

).

Thus

1− 1

3+

1

5− 1

7+ · · · = π

4,

which is a famous formula obtained by Leibniz in 1673 from geometric considerations.

For a general function f , by the Euler formula, we have

f(x) =∑

cneeinx =

∑cn(cosnx+ i sinnx),

which gives

f(x) = c0 +∞∑n=1

cn(cosnx+ i sinnx) +∞∑n=1

c−n(cosnx− i sinnx).

Thus we have

f(x) = c0 +∞∑n=1

((cn + c−n) cosnx+ i(cn − c−n) sinnx) ,

Recall that

(cn + c−n) =1

2π

∫ π

−πf(x)(e−inx + einx) dx =

1

π

∫ π

−πf(x) cosnx dx,

and

i(cn − c−n) =i

2π

∫ π

−πf(x)(e−inx − einx) dx =

1

π

∫ π

−πf(x) sinnx dx,

thus we get

Theorem 2.3. If f has period 2π and is smooth enough then it has the following Fourier seriesexpansion

f(x) = a0 +∞∑n=1

(an cosnx+ bn sinnx),

where a0, an, bn are the Fourier coefficients of f such that

a0 = c0 =1

2π

∫ π

−πf(x) dx,

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and for n = 1, 2 · · · , we have

an =1

π

∫ π

−πf(x) cosnx dx,

and

bn =1

π

∫ π

−πf(x) sinnx dx.

Example: Consider

f(x) = 0, −π < x < 0; f(x) = x, 0 ≤ x < π.

Then

2πa0 =

∫ π

−πf(x) dx =

∫ π

0x dx =

π2

2,

and

πan =

∫ π

−πf(x) cosnx dx =

∫ π

0x cosnx dx =

∫ π

0x d(

sinnx

n) = −

∫ π

0

sinnx

ndx.

Since

−∫ π

0

sinnx

ndx =

∫ π

0d(

cosnx

n2) =

(−1)n − 1

n2,

we get

a0 =π

4, a2m = 0, a2m−1 =

−2

(2m− 1)2π, m = 1, 2 · · · .

Moreover, we have

πbn =

∫ π

−πf(x) sinnx dx =

∫ π

0x sinnx dx =

∫ π

0x d(− cosnx

n) =

π(−1)n+1

n+

∫ π

0

cosnx

ndx,

Notice that ∫ π

0

cosnx

ndx =

∫ π

0d(

sinnx

n2) = 0,

thus

bn =(−1)n+1

n.

Thus

f(x) =π

4− 2

π

(cosx+

cos 3x

32+ · · ·

)+

(sinx− sin 2x

2+

sin 3x

3+ · · ·

).

Take x = 0 then we get

0 =π

4− 2

π(1 +

1

32+ · · · ),

i.e.

1 +1

32+

1

52+ · · · = π2

8.

Exercise: Use 1 + 132

+ 152

+ · · · = π2

8 to prove that

1 +1

22+

1

32+ · · · = π2

6.


Proposition 2.4. Put δmn = 1 if m = n and δmn = 0 if m 6= n then∫ π

−πcosnx cosmxdx =

∫ π

−πsinnx sinmxdx = πδmn, m, n = 1, 2, · · · ,

and ∫ π

−πcosnx dx = 2πδn0,

∫ π

−πcosnx sinmx = 0, m, n = 0, 1, 2, · · · .

Proof. Follows from the Euler formula

cosnx =einx + e−inx

2, sinnx =

einx − e−inx

2i,

and ∫ π

−πeinxe−imx dx = 2πδmn.

Try to give the details yourself. �

Exercise: Try to use the above proposition to prove the formulas for an, bn in Theorem 2.3.

2.2. Fourier Sine and Cosine series.

Definition 2.5. We say that f is odd if f(−x) = −f(x); f is even if f(−x) = f(x).

Example: For every positive integer n, we know that cosnx is even and sinnx is odd.

Application: If f is even then ∫ π

πf(x) dx = 2

∫ π

0f(x) dx.

If f is odd then ∫ π

πf(x) dx = 0.

In particular, if f is odd then all

an =1

π

∫ π

−πf(x) cosnx dx = 0;

if f is even then all

bn =1

π

∫ π

−πf(x) sinnx dx = 0.

Thus we get:

Theorem 2.6. Assume that f has period 2π and is smooth enough. If f is odd then it can be writtenas a Fourier sine series

f(x) =∞∑n=1

bn sinnx, bn =2

π

∫ π

0f(x) sinnx dx.

If f is even then it can be written as a Fourier cosine series

f(x) =1

π

∫ π

0f(x) dx+

∞∑n=1

an cosnx, an =2

π

∫ π

0f(x) cosnx dx.

14 XU WANG

Odd or Even extension: Let f be a function in (0, π). Then we can extend f to an odd function,say fo such that

fo(−x) = −f(x), x ∈ (0, π);

we can also extend f to an even function, say fe such that

fe(−x) = f(x), x ∈ (0, π).

Exercise:

1. Draw the graph of the odd extension fo and the even extension fe of the following function

f(x) = x, 0 < x <π

2; f(x) =

π

2,π

2< x < π.

2. Find the Fourier cosine series of fe. Answer

fe(x) =3π

8+

2

π

(− cosx− 2 cos 2x

22− cos 3x

32− cos 5x

52− · · ·

).

3. Find the Fourier sine series of fo. Answer

fo(x) =

(2

π+ 1

)sinx+

(0− 1

2

)sin 2x+

(−2

32π+

1

3

)sin 3x

+

(0 +

1

4

)sin 4x+

(2

52π+

1

5

)sin 5x+ · · · .

2.3. Parseval’s identity. Let f be a smooth enough function with period 2π. Consider the complexFourier series expansion of f

f =∑n∈Z

cneinx.

We have ∫ π

−π|f(x)|2 dx =

∫ π

−π

∑n,m∈Z

cncmeinx−imx dx.

Now use that ∫ π

−πeinx−imx dx = 0,

if m 6= n and ∫ π

−πeinx−imx dx = 2π,

if m = n. We get the following Parseval identity∫ π

−π|f(x)|2 dx = 2π

∑n∈Z|cn|2.

Example: Consider the last example in section 2.1.1:

f(x) = 1, 0 < x < π; f(x) = −1, −π < x < 0,

andf(0) = f(π) = f(−π) = 0.

We know that f has the following complex Fourier series expansion:

f(x) =∑m∈Z

2

i(2m+ 1)πei(2m+1)x.


Thus the Parseval identity gives

1

2π

∫ π

−π|f(x)|2 dx = 1 =

∑m∈Z

4

(2m+ 1)2π2=

8

π2

(1 +

1

32+

1

52+ · · ·

),

which gives another proof of

1 +1

32+

1

52+ · · · = π2

8.

2.4. Fourier transform.


f(w) :=1√2π

∫ ∞−∞

f(x)e−iwx dx,

the Fourier transform of f and write f = F(f).

Example: F1: Fourier transform of f(x) = 1 if |x| < 1 and f(x) = 0 otherwise:

f(w) :=1√2π

∫ ∞−∞

f(x)e−iwx dx =1√2π

∫ 1

−1e−iwx dx

if w 6= 0 then ∫ 1

−1e−iwx dx =

∫ 1

−1d(e−iwx

−iw) =

e−iw

−iw− eiw

−iw=

2 sinw

w.

Notice that

limw→0

2 sinw

w= 2 =

∫ 1

−1dx = f(0).

Thus we can write

f(w) =1√2π

(2 sinw

w

)=

√2

π

sinw

w.

Example: F2: Fourier transform of f(x) = e−x if x > 0 and f(x) = 0 otherwise:

f(w) :=1√2π

∫ ∞−∞

f(x)e−iwx dx =1√2π

∫ ∞0

e−xe−iwx dx =1√2πL(e−iwt)(1).

Recall that

L(e−iwt)(s) =1

s+ iw,

thus

f(w) =1√2π· 1

1 + iw.

16 XU WANG

2.4.1. From Complex Fourier series to inverse Fourier transform. : Assume that f is smooth enoughin −N < x < N and f = 0 when |x| > N . For each L > N , let us define a periodic function fLsuch that

fL(x) = f(x), |x| < L; fL(x+ 2L) = fL(x).

Then we know that

gL(x) = fL

(Lx

π

),

has period p and is smooth enough. Thus

gL(x) =∑

cneinx, cn =

1

2π

∫ π

−πgL(x)e−inx dx.

ThusfL(x) = gL

(πxL

)=∑

cneinπx

L .

Consider v = Lxπ , we can write

cn =1

2π

∫ L

−LfL(v)e−in

πvL d(

πv

L) =

1

2L

∫ ∞−∞

f(v)e−inπvL dv =

√2π

2Lf(nπ

L).

which gives

f(x) =

√π

2

∑n∈Z

f(nπL ) · einπxL

L.

Put∆w =

π

L,

then we havef(x) =

1√2π

∑n∈Z

f(n∆w) · eix·n∆w∆w.

Assume that f(w)eixw is integrable in −∞ < x < ∞. Let L goes to infty, the above formula givesthe following Fourier inversion formula:

f(x) =1√2π

∫ ∞−∞

f(w)eixw dw.

We say that f(x) is the inverse Fourier transform of f(w) and write f = F−1(f).

2.5. Fourier inversion formula. When do we have the Fourier inversion formula ? It is known that(see Page 141 Theorem 1.9 in [3]) the Fourier inversion formula is true if f is smooth and rapidlydecreaing, in the sense that

supx∈R|x|k|f (l)(x)| <∞, for every k, l ≥ 0,

where f (l) denotes the l-th derivative of f .

Example: Let f(x) = e−x2

2 . We shall use Fourier inversion formula to prove

(6) f(w) :=1√2π

∫ ∞−∞

e−x2

2 e−iwxdx = e−w2

2 = f(w).

Step 1: Look at the derivative of f(w):

(f)′(w) =1√2π

∫ ∞−∞

e−x2

2 (−ix)e−iwxdx = F(−ixf(x)).


Notice that (e−x2

2 )′ = e−x2

2 (−x), thus

f ′(w) =i√2π

∫ ∞−∞

e−iwxd(e−x2

2 ) =−i√2π

∫ ∞−∞

e−x2

2 d(e−iwx) = −wf(w).

Now we have (f(w)e

w2

2

)′= (−w + w)

(f(w)e

w2

2

)= 0.

Thus f(w)ew2

2 is a constant, i.e.

f(w)ew2

2 ≡ f(0)e0 = f(0).

Now we have

f(w) = f(0)e−w2

2 = f(0)f(w)

Step 2: The Fourier inversion formula implies (notce that f is smooth and rapidly decreasing)

f(x) = F−1(f) = f(0)F−1(f) = f(0)f(−x) = (f(0))2f(x).

Thus f(0) = ±1. Since

f(0) =1√2π

∫ ∞−∞

e−x2

2 dx > 0,

we get

f(0) = 1, f = f.

2.5.1. Normal distribution. f(0) = 1 gives

(7)1√2π

∫ ∞−∞

e−x2

2 dx = 1,

(one may also use integration on R2 to compute the following integral directly, see page 138 formula(6) in [3]), consider

u =√tx+ µ, t > 0, µ ∈ R,

then (7) becomes the following classical formula in Gauss’s normal distribution theory

(8)∫ ∞−∞

1√2πt

e−(u−µ)2

2t du = 1,

where

(9) f(u |µ, t) :=1√2πt

e−(u−µ)2

2t ,

is the probability density of the normal distribution with expectation µ and variance t.

18 XU WANG

2.6. Fourier transform of derivative and convolution. In this section, we only consider functionsthat are smooth and rapidly decreasing. The main result is the following:

Theorem 2.8. Let F(f)(w) = 1√2π

∫∞−∞ f(x)e−iwxdx be the Fourier transform of f . Then

1) F(af + bg) = aF(f) + bF(g);2) F(f ′) = iwF(f);3) (F(f))′ = −iF(xf(x)).

Proof. We only prove 2). Notice that∫ ∞−∞

f ′(x)e−iwx dx =

∫ ∞−∞

e−iwx df =

∫ ∞−∞

d(e−iwxf)−∫ ∞−∞

f(x) d(e−iwx).

Since f is smooth and repadly decreasing, we have∫ ∞−∞

d(e−iwxf) = limx→∞

e−iwxf(x)− limx→−∞

e−iwxf(x) = 0.

Thus ∫ ∞−∞

f ′(x)e−iwx dx = −∫ ∞−∞

f(x) d(e−iwx) = iw

∫ ∞−∞

f(x)e−iwx dx,

which implies 2). �

Example: F(xe−x2

2 ) = −iwe−x2

2 : By 3), we have

F(xe−x2

2 ) = i(F(e−x2

2 ))′ = i(e−w2

2 )′ = −iwe−x2

2 .

Convolution: Let us first recall the definition of convolution for functions f, g defined on [0,∞):

(f ? g)(t) :=

∫ t

0f(τ)g(t− τ) dτ.

Notice that if we extend f, g to functions on R such that

f = g = 0, when x ≤ 0.

Then we can write(f ? g)(x) :=

∫ ∞−∞

f(u)g(x− u) du.

Definition 2.9. The convolution of two functions on R is defined by

(f ? g)(x) :=

∫ ∞−∞

f(u)g(x− u) du.

Similar as the Laplace transform, we have

(10) F(f ? g) =√

2πF(f) · F(g).

Proof. [Need integration on R2]. We have√

2πF(f ? g) =

∫ ∞−∞

(f ? g)(x)e−iwx dx =

∫ ∞−∞

(∫ ∞−∞

f(u)g(x− u) du

)e−iwx dx.

Change the order of integration, we get√

2πF(f ? g) =

∫ ∞−∞

f(u)

(∫ ∞−∞

g(x− u)e−iwx dx

)du =

√2π

∫ ∞−∞

f(u)F(g)(w)e−iwu du,

which gives√

2πF(f ? g) = 2πF(f) · F(g). �


3. PARTIAL DIFFERENTIAL EQUATIONS

3.1. Functions of several variables. In Laplace transform and Fourier transform, the functionse−st, e−iwx depend on two variables. We can look at e−st as a map, say

f : (s, t) 7→ f(s, t) := e−st,

from R2 to R. We say that the map f defines a function on R2. It is clear that

f(x, y) = x2 + y2,

is a function on R2;f(x, y, z) = x+ y + z,

is a function on R3 and

f(t, x) =1√2πt

e−x2

2t ,

is a function on (0,∞)× R.

3.1.1. Graph. If f is a function on U ⊂ Rn then we call the following set

Gf,U := {(x, f(x)) ∈ Rn+1 : x ∈ U},

in Rn+1 the graph of f over U .

Exercise: Draw the graph of f(x, y) = x2 + y2 over the unit disc.

3.1.2. Partial derivatives. x-partial derivative of f(x, y) means derivative of f with y fixed:

∂f

∂x(x, y) := lim

h→0

f(x+ h, y)− f(x, y)

h,

we write

fx :=∂f

∂x, fxy :=

∂fx∂y

.

Example: If f(x, y) = x2 + y2 + xy then

fx = 2x+ y, fy = 2y + x, fxy = 1 = fyx, fxx = 2, fyy = 2.

Example: If

f(t, x) =1√2πt

e−x2

2t ,

then

ft =1√2π· −1

2· t

−32 · e−

x2

2t +1√2πt

e−x2

2t · −x2

2· −1

t2=x2 − t

2t2f,

and

fx = −xtf, fxx = −1

tf +

x2

t2f =

x2 − tt2

f.

Thus we get that

(11) ft =1

2fxx.

20 XU WANG

3.1.3. Directional derivative and gradient. Let n in the unit sphere be a given direction, then we call

fn(p) := limh→0

f(p+ hn)− f(p)

h,

the derivative of f along direction n at p. If we write

n = (a, b, c), p = (x, y, z),

then

fn(p) = limh→0

f(x+ ah, y + bh, z + ch)− f(x, y, z)

h.

Example: If

f(x, y, z) = c0 + c1x+ c2y + c3z,

then

f(hn) = f(ah, bh, ch) = c0 + (c1a+ c2b+ c3c)h,

which gives

(12) fn(0) = c1a+ c2b+ c3c = (fx(0), fy(0), fz(0)) · n.


∇f(p) := (fx1(p), · · · , fxm(p)),

the gradient of f(x1, · · · , xm) at p.

For a general smooth function, we have the following generalization of (12):

(13) fn(p) = ∇f(p) · n.

Since |n| = 1, the above formula gives

fn(p) = ∇f(p) cos θ,

where θ denotes the angle from∇f(p) to n.

3.1.4. PDEs. We shall study the following partial differential equations

1. Wave equation

utt = uxx.

2. Heat equation

ut =1

2uxx,

Homework: Find the background of the above two equations in [2] or wikipedia.

Example: By (11), we know that

f(t, x) :=1√2πt

e−x2

2t ,

satisfies the above heat equation. We call f(t, x) the heat kernel.


3.2. Solve wave equation by Fourier series. Let us solve the wave equation

utt = uxx,

with boundary conditionsu(t, 0) = u(t, π) = 0, ∀ t ≥ 0;

and initial conditions

u(0, x) = f(x), ut(0, x) = g(x), ∀ 0 ≤ x ≤ π.Step 1: Separating variables: Find solutions of the form

u(t, x) = G(t)F (x),

Sinceutt = G′′F, uxx = GF ′′,

our equation becomesG′′F = GF ′′,

thusG′′

G=F ′′

F≡ k,

where k is constant (notice that k does not depend on t and x).

Step 2: Fit boundary conditions: Notice that the boundary conditions

G(t)F (0) = G(t)F (π) = 0,

is equivalent toF (0) = F (π) = 0.

In case k = 0, then F ′′ ≡ 0, i.e F (x) = ax+ b. The boundary conditions give F ≡ 0.In case k = µ2 > 0, then the general solution for

F ′′ = µ2F,

is F = Aeµx +Be−µx, then the boundary conditions give

A+B = 0, Aeµπ +Be−µπ = 0,

thus A = B = 0.Thus the only possible case is k = −p2 < 0, then the general solution for

F ′′ = −p2F,

is F = A cos px+B sin px, F (0) = 0 gives A = 0. Thus F = B sin px, B 6= 0, but F (π) = 0 givessin pπ = 0, i.e.

p = n, n = 1, 2, · · · ,(notice that sin−px = − sin px, thus up to a constant they give the same solution).

Summary: The boundary condition implies that p = n2, n = 1, 2, · · · , and

F = Fn(x) = sinnx.

Now let us solveG′′ = −n2G,

the general solution isGn(t) = Bn cosnt+ Cn sinnt.

22 XU WANG

Now we know that each

un(t, x) = Gn(t)Fn(t) = (Bn cosnt+ Cn sinnt) sinnx,

satisfies the wave equation and the boundary conditions, so does

u(t, x) =

∞∑n=1

(Bn cosnt+ Cn sinnt) sinnx.

Step 3: Fit the initial conditions: Choose Bn and Cn such that

u(0, x) = f(x), ut(0, x) = g(x).

i.e.∞∑n=1

Bn sinnx = f(x);

and∞∑n=1

nCn sinnx = g(x);

Consider odd extension fo, go of f, g, by Theorem 2.6, if fo and go is smooth enough, then it is enoughto choose

Bn =2

π

∫ π

0f(x) sinnx dx,

and

nCn =2

π

∫ π

0g(x) sinnx dx.

Now we know that if f, g have smooth enough odd extension then

u(t, x) =∞∑n=1

(Bn cosnt+ Cn sinnt) sinnx,

solves the wave equation (of course we have it check that it converges).

Example: When g = 0 we have

u(t, x) =∞∑n=1

Bn cosnt sinnx.

Thus we can write

u(t, x) =1

2

∞∑n=1

(Bn sinn(x− t) +Bn sinn(x+ t)) =1

2(fo(x− t) + fo(x+ t)),

i.e. u(t, x) is the superposition of two travelings of the initial wave.

Exercise: In case

f(x) = x, 0 ≤ x ≤ π

2, f(x) = π − x, π

2< x ≤ π,

try to draw the graph of u(t, x) for t = 0, π8 ,π4 ,

π2 , π.


3.3. Solve heat equation by Fourier series. Consider the heat equation

ut =1

2uxx,

with boundary conditionsu(t, 0) = u(t, π) = 0, ∀ t ≥ 0;

and initial conditionsu(0, x) = f(x), ∀ 0 ≤ x ≤ π.

Step 1: Separating variables: Find solutions of the form

u(t, x) = G(t)F (x),

Sinceut = G′F, uxx = GF ′′,

our equation becomes

G′F =1

2GF ′′,

thus2G′

G=F ′′

F≡ k,

Step 2: Fit the boundary conditions: Same as the wave equation, we have

k = −n2, n = 1, 2 · · · , nand

Fn(x) = sinnx.

Then

G′ =n2

2G,

gives

Gn(t) = Bne−n

2

2t,

Thus the general solution is

u(t, x) =

∞∑n=1

Bne−n

2

2t sinnx.

Step 3: Fit the initial conditions: We hope

u(0, x) =∞∑n=1

Bn sinnx = f(x).

By Theorem 2.6, we get

Bn =2

π

∫ π

0f(x) sinnx dx,

Example: f(x) = 2 sinx then

B1 = 2, B2 = · · · = Bn = 0.

Thusu(t, x) = 2e−

t2

2 sinx,

Notice that u goes to zero as t goes to infinity.

24 XU WANG

Exercise: Find u with

f(x) = x, 0 ≤ x ≤ π

2, f(x) = π − x, π

2≤ x ≤ π.

3.4. Solve heat equation by Fourier transform. Still the heat equation

ut =1

2uxx.

But this time we consider the initial condition for all x in R (thus no boundary conditions):

u(0, x) = f(x), ∀ −∞ < x <∞.Step 1: Reduce to ODE by Fourier transform: Consider Fourier transform of ut with respect to the

x variable

F(ut) = ut(w) =1√2π

∫ ∞−∞

ut(t, x)e−ixw dx.

Then we have

F(ut) = F(1

2uxx) =

1√2π

∫ ∞−∞

1

2uxx(t, x)e−ixw dx.

Recall that if ux is smooth and rapidly decreasing then∫ ∞−∞

uxxe−ixw dx =

∫ ∞−∞

e−ixw d(ux) = −∫ ∞−∞

ux d(e−ixw) = iw

∫ ∞−∞

uxe−ixw dx,

the same computation for u gives∫ ∞−∞

uxe−ixw dx = iw

∫ ∞−∞

ue−ixw dx,

thus we have

F(ut) =−w2

2F(u), F(u) :=

1√2π

∫ ∞−∞

u(t, x)e−ixw dx

Notice that we also haveF(ut) = (F(u))t.

Thus F(u) satisfies the following ODE.

(F(u))t =−w2

2F(u).

Step 2: Solve the ODE and fit the initial condition: The general solution is

F(u)(t, w) = c(w)e−w2

2t.

Notice that our initial condition implies

F(u)(0, w) =1√2π

∫ ∞−∞

f(x)e−ixw dx = F(f).

Thusc(w) = F(f).

Now we haveF(u)(t, w) = F(f) · e

−w2

2t.

Step 3: use Fourier convolution formula: Recall that (see (6))

e−u22 =

1√2π

∫ ∞−∞

e−y22 e−iyu dy.


Takeu = w

√t, y =

x√t,

we get

e−w2

2t =

1√2πt

∫ ∞−∞

e−x22t e−ixw dy =

1√tF(e

−x22t ).

Thus

F(u)(t, w) = F(f) ·(

1√tF(e

−x22t )

)Now the Fourier convolution formula gives

u(t, x) =1√2πt

(f ? e−x22t ) =

∫ ∞−∞

f(p)1√2πt

e−(x−p)2

2t dp,

Summary: the solution u(t, x) is given by convolution of the initial temperature distribution withthe heat kernel.

4. APPENDIX: DEFINITION OF e, π AND EULER’S FORMULA

4.1. Where does e come from ? Recall that: Let A : Cn → Cn be a linear map (here linear mapmeans A(au + bv) = aA(u) + bA(v) for all a, b in C and all u, v in Cn). We call u 6= 0 in Cn aneigenvector of A if

(14) Au = λu,

where λ is a constant in C.

What is an eigenvector of the derivative ?

By (14), we want to find function u : R→ C such that

u′ = λu.

Power series method: Assume that

u(x) = a0 + a1x+ a2x2 + · · ·+ anx

n + · · · .

The following lemma gives:

u′(x) = a1 + 2a2x+ · · ·+ nanxn−1 + (n+ 1)an+1x

n + · · · .

Lemma 4.1. (xn)′ = nxn−1, n = 1, 2, · · · .

Proof. If n = 1 then

x′(x) = lim4x→0

(x+4x)− x4x

= 1.

Assume the Lemma for n = 1, · · · , N − 1. Then (fg)′ = f ′g + fg′ gives

(xN )′ = (xN−1)′ · x+ xN−1 · x′ = (N − 1)xN−2 · x+ xN−1 = NxN−1.

The proof is complete. �

26 XU WANG

Exercise: Why we have (fg)′ = f ′g + fg′ ?

Nowu′ = λu⇔ λan = (n+ 1)an+1, n = 0, 1, · · · .

Thus

an+1 =λan

(n+ 1)=

λ2an−1

(n+ 1)n= · · · = λn+1a0

(n+ 1)n · · · 1=

λn+1a0

(n+ 1)!,

where we definen! = 1 · 2 · · ·n.

Then we have

u(x) = u0 · (1 + λx+ · · ·+ (λx)n

n!+ · · · ).

PutE(x) := 1 + x+ · · ·+ xn

n!+ · · · .

Since for every C > 0,

limn→∞

Cn

n!= 0,

we know that E(x) converges for all x in C.

Theorem 4.2. E(λx) is a unique solution of the eigenvalue equation

u′ = λu,

with initail condition u(0) = 1.

Definition 4.3. We shall define

e := E(1) = 1 + 1 +1

2+ · · ·+ 1

n!+ · · · .

4.2. Definition of the exponential function. Let us write

e2 = e · e, e3 = e2 · e,and define em inductively by

en+1 = en · e.Since e is positive, we can take the q − th root of em, we write it as e

mq . Thus for every x ∈ Q, ex is

well defined. The following lemma tells us that E(x) is an extension of ex from Q to C.

Lemma 4.4. For every x ∈ Q, we have ex = E(x).

Proof. Since E(1) = e, it suffices to prove

(15) E(λ1)E(λ2) = E(λ1 + λ2),

for every λ1, λ2 in C. Notice that

(E(λ1x)E(λ2x))′ = E(λ1x)′E(λ2x) + E(λ2x)′E(λ1x).

PutG(x) = E(λ1x)E(λ2x).

Apply E(λx)′ = λE(λx), we getG′ = (λ1 + λ2)G.

Notice that G(0) = 1. Thus Theorem 4.2 implies that

G(x) = E((λ1 + λ2)x).


Take x = 1, we get E(λ1)E(λ2) = E(λ1 + λ2). �

Exercise: Find a direct proof of E(λ1)E(λ2) = E(λ1 + λ2) without using Theorem 4.2.

Definition 4.5. We shall use the same symbol ex to denote E(x) for all x in C and call ex theexponential function. If x > 0 then we define lnx as the unique real solution of elnx = x.

By Theorem 4.2, we know that ex is fully determined by

(ex)′ = ex, e0 = 1.

4.3. Definition of π and trigonometric functions. : Fix P0 = (1, 0) in the unit circle

S1 := {(x, y) ∈ R2 : x2 + y2 = 1}.

A counterclockwise rotation of P0 gives a arc P0P . The length, say θ(P ), of the arc P0P is a functionof P . It is clear that the circumference diameter ratio is equal to θ(−1, 0).

Definition 4.6 (Definition of π). We shall write the circumference diameter ratio as π.

Denote byF : θ(P ) 7→ P,

the inverse function of 0 ≤ θ(P ) ≤ 2π.

Definition 4.7. We shall write F (θ) = (cos θ, sin θ).

Notice thatF (0) = (1, 0) = F (2π), F (π) = (−1, 0), |F (θ)| ≡ 1.

In particular, it givessin(0) = sin(2π) = 0, cos(0) = cos(2π) = 1.

By definition of θ, we have ∫ θ

0|F ′(θ)| dθ = θ, 0 ≤ θ ≤ 2π,

which gives|F ′(θ)| ≡ 1.

Now F (θ) · F (θ) ≡ 1 implies

F ′ · F + F · F ′ = 2F · F ′ ≡ 0.

Hence F ′⊥F , thus we know that

F ′(θ) = (− sin θ, cos θ), or F ′(θ) = (sin θ,− cos θ).

But notice that F ′(0) = (0, 1), thus we must have

F ′(θ) = (− sin θ, cos θ),

which is equivalent to(cos θ + i sin θ)′ = i(cos θ + i sin θ).

Notice that cos 0 + i sin 0 = 1, thus Theorem 4.2 gives

Theorem 4.8 (Euler’s formula). eiθ = cos θ + i sin θ.

28 XU WANG

Take θ = π, we get the following Euler’s identity

eiπ = −1.

Moreover, apply (15), we geteiθ1eiθ2 = ei(θ1+θ2),

thus by Euler’s formula, we have

(cos θ1 + i sin θ1)(cos θ2 + i sin θ2) = cos(θ1 + θ2) + i sin(θ1 + θ2),

i.e.

(16) cos(θ1 + θ2) = cos θ1 cos θ2 − sin θ1 sin θ2,

and

(17) sin(θ1 + θ2) = sin θ1 cos θ2 + cos θ1 sin θ2.

REFERENCES

[1] B. Berndtsson, Fourier and Laplace transforms, can be found in:www.math.chalmers.se/Math/Grundutb/CTH/mve025/1718/Dokument/F-analysny.pdf.

[2] E. Kreyszig, Advanced Engineering Mathematics, 10 th edition, internation student version.[3] E. Stein and R. Shakarchi, Fourier analysis, Princeton lectures in analysis.[4] A. Vretblad, Fourier Analysis and Its Applications, GTM 233.

DEPARTMENT OF MATHEMATICAL SCIENCES, NORWEGIAN UNIVERSITY OF SCIENCE AND TECHNOLOGY, NO-7491 TRONDHEIM, NORWAY

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LAPLACE TRANSFORM, FOURIER TRANSFORM AND … · 2018-08-15 · More Laplace transforms 3 2. Fourier analysis 9 2.1. Complex and real Fourier series (Morten will probably teach this