APPLIED MATHEMATICS

Part 5:

Partial Differential Equations

Wu-ting Tsai
Contents

11 Partial Differential Equations 2

11.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . 3
11.2 Modeling: Vibrating String. Wave Equation . . . . . . . . . 5
11.3 Separation of Variables. Use of Fourier Series . . . . . . . . 7
11.4 Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . 15
11.5 Laplace Equation in a Rectangular Domain . . . . . . . . . 23
11.6 Two-Dimensional Wave Equation. Use of Double Fourier Series 27
11.7 Heat Equation: Use of Fourier Integral . . . . . . . . . . . . 34
11.8 Heat Equation: Use of Fourier Transform . . . . . . . . . . 38
11.9 Heat Equation. Use of Fourier Cosine and Sine Transforms . 41
11.10 Wave Equation. Use of Fourier Transform . . . . . . . . . . 43

1
Chapter 11

Partial Differential Equations

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.1 Basic Concepts

Partial differential equation ⇐⇒ Ordinary differential equation








order of differential equation


linear ⇐⇒ nonlinear


 homogeneous ⇐⇒ nonromogeneous


For examples:

∂ 2u 2
2∂ u
=c one-dimensional wave equation
∂t2 ∂x2
2
∂u 2∂ u
=c one-dimensional heat equation
∂t ∂x2

∂ 2u ∂ 2u
+ =0 two-dimensional Laplace equation
∂x2 ∂y 2

∂ 2u ∂ 2u
+ = f (x, y) two-dimensional Poisson’s equation
∂x2 ∂y 2

Note:
u(x, y) = x2 − y 2, ex cos y and ln(x2 + y) all satisfy the two-dimensional
Laplace equation
⇒ “boundary condition” makes the solution unique
(or “initial condition” when t is one of the variable).

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Theorem : Superposition of solutions

If u1 and u2 are any solutions of a linear and homogeneous partial differ-
ential equation in R
⇒ u = c1u1 + c2u2, where c1 and c2 are constant, is also solution of the
equation in R ✷

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.2 Modeling: Vibrating String. Wave Equation

Consider a vibrating, elastic string with length L and deformation u(x, t).

Assumptions:
1. homogeneous string (i.e. constant density)
2. perfectly elastic and cannot sustain bending
3. neglect gravitation force (i.e. tension  gravitational force)
4. small deformation (i.e. du/dx is small)
5. u is in a plane only

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Since du/dx (slope) is small

⇒ no horizontal motion
⇒ constant horizontal tension

T1 cos α = T2 cos β = T

in vertical direction:
 
2
 ∂ u
 T2 sin β − T1 sin α = ρ∆x 2
÷T
∂t

T2 sin β T1 sin α ρ∆x ∂ 2u

⇒ − = · 2
T2 cos β T1 cos α T ∂t

ρ∆x ∂ 2u
⇒ tan β − tan α = · 2
T ∂t
    
∂u  ∂u
 + ∆x −    ρ ∂ 2u
⇒ ∂x x ∂x x = · 2
∆x T ∂t
∆x → 0

∂ 2u 1 ∂ 2u T
⇒ = where c2 ≡ ✷
∂x2 c2 ∂t2 ρ

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.3 Separation of Variables. Use of Fourier Series

One-dimensional wave equation:

∂ 2u 2
2∂ u
=c ——— (1)
∂t2 ∂x2
Boundary conditions:

u(0, t) = 0 ——— (2)

u(L, t) = 0 ——— (3)

Initial conditions:

u(x, 0) = f (x) ——— (4)


∂u 
 = g(x) ——— (5)
∂t t=0

Method of separation of variables (product method):

u(x, t) = F (x)G(t)




 ∂ 2u d2 G



 = F 2 ≡ F G̈


 ∂t2 dt
⇒
 



 ∂ 2 u d2 F



 = G≡F G
∂x2 dx2

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

(1) ⇒ F G̈ = c2F G

G̈ F
⇒ 2G
= =k
c
   F
  
function of t function of x
where k is constant to be determined.






F − kF = 0 ——— (6)

⇒
 


 G̈ − c2kG = 0 ——— (7)

F (x)

F − kF = 0 F (0) = F (L) = 0

• If k = 0 ⇒ F (x) = ax + b


 F (0) = b = 0

 F (L) = aL = 0

⇒a=b=0

⇒ F (x) = 0 ⇒ u(x, t) = 0 ×

• If k = µ2 > 0 ⇒ F (x) = Aeµx + Be−µx



 F (0) = A + B = 0
 F (L) = AeµL + Be−µL = 0


⇒A=B=0

⇒ F (x) = 0 ⇒ u(x, t) = 0 ×

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

•
• • k = −p2 < 0 ⇒ F (x) = A cos px + B sin px


 F (0) = A = 0

 F (L) = B sin pL = 0

nπ
⇒ pL = nπ ⇒p= , n = 1, 2, 3 . . .
L
nπ
⇒ F (x) ≡ Fn(x) = sin x, n = 1, 2, 3 . . .
L

nπ 2
2
and k = −p = −
L

G(t)
  
2 nπ 2
G̈ − c − G=0
L
cnπ
⇒ G̈ + λ2nG = 0, λn =
L
G(t) ≡ Gn(t) = Bn cos λnt + Bn∗ sin λn t

•
• • un(x, t) = Fn (x)Gn(t)
nπ
= (Bn cos λnt + Bn∗ sin λnt) sin x (n = 1, 2, 3 . . .)
L
So far, un(x, t) satisfies the differential equation (1), and the boundary con-
ditions (2) and (3).
un (x, t) is called eigenfunction or characteristic function.
λn is called eigenvalue or characteristic value.

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

A single solution un (x, t) does not satisfy initial condition.

Since the differential equation (1) is linear and homogeneous, we therefore
assume the solution of (1) is:
∞

u(x, t) = un (x, t)
n=1

∞
 nπ
= (Bn cos λnt + Bn∗ sin λnt) sin x
n=1 L

Initial conditions:






u(x, 0) = f (x) ——— (4)


∂u 


  = g(x) ——— (5)
∂t t=0
 nπ
(4) ⇒ u(x, 0) = Bn sin x = f (x) (Fourier sine series of f (x))
n=1 L

2 L nπx
⇒ Bn = f (x) sin dx n = 1, 2, 3 . . .
L 0 L

∂u  ∞
 ∗ nπx
(5) ⇒  = (B λ n cos λ t)
n t=0 sin
∂t t=0 n=1 n L
∞
 nπx
= Bn∗ λn sin = g(x) (Fourier sine series of g(x))
n=1 L

2 L nπx
⇒ Bn∗ λn = g(x) sin dx
L 0 L

2 L nπx
⇒ Bn∗ = g(x) sin dx n = 1, 2, 3 . . . ✷
λn L 0 L

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Discussion of eigenfunctions and eigenvalues:

(1) The eigenfunction

nπ
un (x, t) = (Bn cos λnt + Bn∗ sin λnt) sin x
L
λn cn
represents a harmonic motion having frequency (cycles/unit time) = .
2π 2L
This motion is called the normal mode.

(2) The frequency (eigenvalue)

cn T
wn = , c2 =
2L ρ
where T = tension, ρ = density
⇒ increase tension or reduce density will increase the frequency.

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Discussion of the solution:

∞

u(x, t) = un(x, t)
n=1
∞
 nπ
= (Bn cos λnt + Bn∗ sin λnt) sin x
n=1 L
cnπ
λn =
L




2 L nπx


 Bn = f (x) sin dx, u(x, 0) = f (x)



 L 0 L

 




 ∗ 2 L nπx ∂u 

 Bn = g(x) sin dx,  = g(x)

λn L 0 L ∂t t=0

Consider the special case in which the string starts from rest, i.e. g(x) = 0,
and Bn∗ = 0, then
∞
 cnπ nπ
u(x, t) = Bn cos t sin x
n=1 L L
∞  
1  nπ nπ
= Bn sin (x − ct) + sin (x + ct)
2 n=1 L L
1  ∞ nπ 1  ∞ nπ
= Bn sin (x − ct) + Bn sin (x + ct)
2 n=1

L
 
2 n=1

L
 
Fourier sine series of f ∗ (x−ct) Fourier sine series of f ∗ (x+ct)
1 ∗
= [f (x − ct) + f ∗(x + ct)]
2
where f ∗ (•) is an odd periodic function, and f (•) = f ∗ (•) for • ∈ [0, 2L].

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

1
Physical meaning of u(x, t) = [f ∗ (x − ct) + f ∗ (x + ct)] :
2

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Ex :

g(x) = 0





2k L




x, 0<x<

 L 2
f (x) = 
 



 2k L


 (L − x), <x<L
L 2
Bn∗ = 0

2 L nπx
Bn = f (x) sin dx
L 0 L
 
n+1
∞
8k   (−1) (2n − 1)π (2n − 1)πc 
= 2  sin x · cos t
π n=1 (2n − 1)2 L L

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.4 Heat Equation

Consider the heat conduction along a heat-conducting homogeneous rod with

temperature distribution u(x, t).

Assumptions:
1. homogeneous rod with uniform cross section A and constant density ρ,
specific heat c, thermal conductivity κ
2. insulated laterally so heat flows only in x-direction only
3. temperature is constant at all points of a cross section
4. rate of heat conduction is proportional to −∂u/∂x

We are going to derive the differential equation governing the temperature

distribution u(x, t) along the rod by conservation of energy.

The amount of internal energy within ∆x segment is:

 x+∆x
Q(x, t) = x
σρAu(x̃, t)dx̃

where σ = specific heat (amount of energy required to raise the temperature

of a unit mass by one unit).

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

• Rate of change of internal energy within ∆x

∂Q  x+∆x ∂u
= = x σρA (x̃, t)dx̃
∂t ∂t
• Rate at which heat flows into ∆x
∂u
= −κA (x, t)
∂x
• Rate at which heat flows out of ∆x
∂u
= −κA (x + ∆x, t)
∂x
•
• • conservation of energy requires that
∂Q  x+∆x ∂u
= x σρA (x̃, t)dx̃
∂t ∂t
∂u
= σρA (ξ, t)∆x (x < ξ < x + ∆x) [by mean value theorem]
∂t
∂u ∂u
= −κA (x, t) + κA (x + ∆x, t)
∂x ∂x
 
∂u ∂u ∂u
⇒ σρA (ξ, t)∆x = κA  (x + ∆x, t) − (x, t)
∂t ∂x ∂x
 
∂u ∂u
∂u

κ ∂x (x + ∆x, t) − (x, t)
⇒ (ξ, t) − ∂x =0
∂t σρ ∆x

As ∆x → 0
2
∂u 2∂ u k
(x, t) − c (x, t) = 0 c2 =
∂t ∂x2 σρ

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Initial condition:

u(x, 0) = f (x) (0 ≤ x ≤ L)

Boundary condition:

u(0, t) = 0, u(L, t) = 0 (t ≥ 0)

Note:
For 3 − D heat equation (see §9.8):
 
2 2 2
∂u ∂ u ∂ u ∂ u
= c2 
 + + 
∂t ∂x2 ∂y 2 ∂z 2

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Fourier series solution of the heat equation:

2
∂u 2∂ u
−c = 0 for t ≥ 0, 0 ≤ x ≤ L ——— (1)
∂t ∂x2

u(0, t) = 0 for t ≥ 0, x = 0 

——— (2)
u(L, t) = 0 for t ≥ 0, x = L 


u(x, 0) = f (x) for t = 0, 0 ≤ x ≤ L ——— (3)

By method of separation of variables:

u(x, t) = F (x) · G(t)

(1) ⇒ F Ġ = c2F G
 2 p2 t





p2 ⇒ Ġ = c2p2G ⇒ G(t) = Cec ×






Ġ F 

⇒ = = 0 ⇒ Ġ = 0 ⇒ G(t) = C ×
c2G F 



 





 −p2




 F + p2F = 0 ——— (6)
⇒
 
 Ġ + c2p2G = 0 ——— (7)

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

F (x):

(6) ⇒ F (x) = A cos px + B sin px

(2) ⇒ F (0)G(t) = 0, F (L)G(t) = 0

Since G(t) = 0 ⇒ F (0) = F (L) = 0

•
• • A=0 B sin pL = 0
Since B = 0 ⇒ sin pL = 0, i.e. pL = nπ, n = 1, 2, 3 . . .

nπ
⇒ p= n = 1, 2, 3 . . .
L

nπx
Fn (x) = sin px = sin n = 1, 2, 3 . . .
L

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

G(t):
cnπ
(3) ⇒ Ġ + λ2nG = 0 λn =
L
2
⇒ Gn(t) = Bne−λn t n = 1, 2, 3 . . .

•
• • eigenf unction un(x, t) :
nπx −λ2nt
un (x, t) = Fn (x) · Gn(t) = Bn sin ·e
L
with corresponding eigenvalue λn :
cnπ
λn =
L

In order to satisfy initial conduction (3) at t = 0 :

∞
 ∞
 nπx −λ2nt
u(x, t) = un (x, t) = Bn sin e
n=1 n=1 L
∞
 nπx
⇒ u(x, 0) = f (x) = Bn sin (Fourier sine series of f (x))
n=1 L

2 L
⇒ Bn = f (x) sin nπxdx ✷
L 0

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Note:
∞
 nπx −λ2n t
u(x, t) = Bn sin ·e
n=1 L
⇒ the temperature alway decays.

Ex :

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Physical meanings of the boundary conditions

u(0, t) = u(L, t) = constant

⇒ Temperatures at x = 0 and L remain constant, but heat can still transfer
across x = 0 and L

∂u ∂u
(0, t) = (L, t) = 0
∂x ∂x
⇒ No heat transfer across x = 0 and L, i.e. these two ends are insulated,
but the temperature at these two boundary may vary.

(See Page 654, Problem Set 11.5, Problem 13)

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.5 Laplace Equation in a Rectangular Domain

Consider the steady-state two-dimensional heat flow:

 
2 2
∂u ∂ u ∂ u
= c2 

2
+ 2  = 0 (••• steady state)
∂t ∂x ∂y

∂ 2u ∂ 2u
2
⇒∇ u= + =0 ← two-dimensional Laplace equation
∂x2 ∂y 2

• Boundary value problem (bvp) :

If the boundary conditions are given as







u given ⇒ Dirichlet problem











 ∂u



 given ⇒ Neumann problem
∂n












∂u




u given on some boundary , given on the rest boundary



∂n

 ⇒ mixed problem

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Ex :

Method of separation of variables :

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

d2 F d2 G
⇒ G+F 2 =0
dx2 dy

1 d2 F 1 d2 G
⇒− = ≡k
F dx2 G dy 2





d2 F


 + kF = 0 ——— (1)



 dx2
⇒




 d2 G



 − kG = 0 ——— (2)
dy 2

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

If k < 0 :
√ √
kx − kx
(1) ⇒ F (x) = Ae + Be

Boundary conditions: u(0, y) = 0 and u(a, y) = 0

⇒ F (0) = 0, F (a) = 0


 A +√ B = 0 √
⇒
 Ae ka + Be− ka = 0
√ √
ka − ka
⇒ A(e −e )=0

•
• • A = 0, B=0 ×

If k > 0 :
√ √
(1) ⇒ F (x) = A cos kx + B sin kx

Boundary conditions:

F (0) = 0 ⇒A=0
√ √ nπ
F (a) = 0 ⇒ F (a) = B sin ka = 0 ⇒ k=
a

nπ 2
⇒ kn =
a

nπ
and Fn (x) = sin x
a

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

d2 G
(2) ⇒ − kn G = 0
dy 2
√ √
kn y − kn y
⇒ G(y) = Gn(y) = Ane + Bn e

Boundary condition:
u(x, 0) = 0 ⇒ G(0) = 0

⇒ A n + Bn = 0
√ √
kn y − kn y
√
kn y
√
− kn y (e −e )
⇒ Gn (y) = An(e −e ) = 2An
2
•
√
• • Gn(y) = A∗n sinh kn y A∗n ≡ 2An

•
• • un (x, y) = Fn (x) · Gn(y)
nπx nπy
= A∗n sin sinh
a a
∞
 ∞
 nπx nπy
⇒ u(x, y) = un (x, y) = A∗n sin sinh
n=1 n=1 a a
Since u(x, b) = f (x) :
∞
 nπb nπx
⇒ u(x, b) = A∗n sinh sin = f (x)
n=1   a a
≡ bn
nπb 2  a nπx
⇒ bn = A∗n sinh = 0 f (x) sin dx
a a a
2  a nπx
• ∗
• • An = f (x) sin dx ✷
a sinh( nπb
a
) 0 a

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.6 Two-Dimensional Wave Equation. Use of Double Fourier

Series

Consider the motion of a stretched elastic membrane, such as a drumhead.

⇒ Two dimensional wave equation:

∂ 2u 2
2 ∂ u ∂ 2u T
= c ( 2 + 2 ), c2 =
∂t2 ∂x ∂y ρ

u(x, y, t) = vertical deformation

T = tension force per unit length
ρ = mass of the membrane per unit area

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Consider the problem of a vibrating rectangular membrane:

 
2 2 2
∂ u 2 ∂ u ∂ u
= c  + , 0 ≤ x ≤ a, 0 ≤ y ≤ b ——— (1)
∂t2 ∂x2 ∂y 2
Boundary conditions:

u(x, y, t) = 0 on the boundary for t ≥ 0 ——— (2)

Initial condition:






u(x, y, 0) = f (x, y) ——— (3)









∂u

 (x, y, 0) = g(x, y) ——— (4)
∂t

By method of separation of variables we first determine the solutions of (1)

that satisfy the boundary condition (2) :

u(x, y, t) = F (x, y) · G(t)

(1) ⇒ F G̈ = c2(Fxx G + Fyy G) ÷ (c2F G)

G̈ 1 2
⇒ = (Fxx + Fyy ) ≡ −ν (cannot be zero or posititice)
c2G F




 G̈ + λ2G = 0 (λ ≡ cν) ——— (6)
⇒
 
 Fxx + Fyy + ν 2F = 0 ——— (7)

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai





 G̈ + λ2G = 0 ——— (6)



 Fxx + Fyy + ν 2F = 0 ——— (7)

(7) is called two-dimensional Helmholtz equation, if ν = 0 it become Laplace

equation.

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Applied Mathematics — Partial Differential Equation Wu-ting Tsai

• Solution of Helmholtz equation (7) Fxx + Fyy + ν 2F = 0

Again, using method of separation of variables :

F (x, y) = H(x) Q(y)

d2 H d2 Q
(7) ⇒ 2
Q + H 2 + ν 2HQ = 0 ÷ (HQ)
dx dy

1 d2 H 1 d2 Q 2 2
⇒ = − ( + ν Q ) ≡ −k (must be negative)
H dx2 Q dy 2


 d2 H





 2
+ k 2H = 0


 dx
⇒
 


 d2 Q




 2
+ (ν 2 − k 2)Q = 0 (ν 2 − k 2 ≡ p2)
dy


 H(x) = A cos kx + B sin kx

⇒
  Q(y) = C cos py + D sin py

Since u(x, y, t) = 0 on x = 0, x = a, y = 0, y = b,






H(0) = 0 ⇒A=0





 mπ





H(a) = 0 ⇒ sin ka = 0 ⇒k= , m = 1, 2, 3 . . .
 a
⇒
 



Q(0) = 0 ⇒C=0



 nπ






Q(b) = 0 ⇒ sin pb = 0 ⇒p= , n = 1, 2, 3 . . .
 b

30
Applied Mathematics — Partial Differential Equation Wu-ting Tsai





mπx




H(x) = Hm (x) = sin
• a
• • 
 nπy



 Q(y) = Q (y) = sin
 n
b

⇒ F (x, y) = Fmn (x, y) = Hm(x) Qn (y)

mπx nπy
= sin sin (m, n = 1, 2, 3 . . .)
a b

• Solution of (6) G̈ + λ2G = 0

λ = cν p2 = ν 2 − k 2
√  !
! mπ nπ
⇒ λ = c ν = c p + k = c"(
2 2 2 )2 + ( )2
a b
!
m2 n2
!
!
⇒ λ = λmn "
= cπ 2 + 2
a b

(6) ⇒ G̈ + λ2mnG = 0
∗
⇒ G(t) = Gmn(t) = Bmn cos λmnt + Bmn sin λmnt

•
• • umn (x, y, t) = Fmn (x, y) · Gmn(t)
∗ mπx nπy
= ( Bmn cos λmnt + Bmn sin λmn t ) · sin · sin
a b

umn (x, y, t) is the eigenfunction with the corresponding eigenvalue λmn .

31
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Since every umn(x, y, t) satisfies (1) and (2), so is the summation:

∞
 ∞

u(x, y, t) = umn (x, y, t)
m=1 n=1

∞
 ∞
 mπx nπy
∗
= ( Bmn cos λmnt + Bmn sin λmnt ) · sin · sin
m=1 n=1 a b

Now apply the initial conditions (3) and (4):

 ∞ ∞
   mπx nπy





(3) ⇒ u(x, y, 0) = Bmn sin sin = f (x, y) ——— (8)

 m=1 n=1 a b





 ∞  ∞

 ∂u  ∗ mπx nπy



 (4) ⇒ (x, y, 0) = Bmn λmn sin sin = g(x, y) − (9)
∂t m=1 n=1 a b

(8) and (9) are called double Fourier series of f (x, y) and g(x, y).

We can generalized the Euler formula in one dimension to two dimension:

 ba m πx n πy
0 0
(8) sin sin dxdy
a b
 ∞
a b  ∞
 mπx m πx nπy n πy
⇒ 0 0 m=1 n=1
Bmn sin · sin · sin · sin dxdy
a a b b
 ab m πx n πy
= 0 0
f (x, y) sin · sin dxdy
a b
∞
 ∞
 #
a$ mπx m πx %  b$ nπy n πy %&
⇒ Bmn 0
sin · sin dx × 0 sin · sin dy
m=1 n=1 a a b b
 ab m πx n πy
= 0 0
f (x, y) sin · sin dxdy
a b

32
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

 a mπx m πx
0
sin · sin dx
a a
1  a# πx nπx &
= 0 cos(m − m ) − cos(m + m ) dx
2 a   a 
=0
1a πx
= 0 cos(m − m ) dx
2 a


 0 if m = m
=
 1
2 a if m = m

Similarly,

 b nπy n πy 
 0 if n = n
0
sin · sin dy = 
 1 b if
b b 2 n=n
  
a  b  a  b mπx nπy
⇒ Bmn = 0 0 f (x) sin · sin dxdy
2 2 a b

• 4  a b mπx nπy
• • Bmn = f (x, y) sin · sin dxdy (m, n = 1, 2, 3 . . .)
ab 0 0 a b
Similarly, from (9) we have:

∗ 4 ab mπx nπy

Bmn = g(x, y) sin · sin dxdy ✷
abλmn 0 0 a b

33
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.7 Heat Equation: Use of Fourier Integral

Consider the heat equation along a heat conducting rod extending to ±∞:

 2



∂u 2∂ u

 =c (−∞ < x < +∞)


∂t ∂x2







 u(x, 0) = f (x) (−∞ < x < +∞)

Again by method of separation of variables:

u(x, t) = F (x) · G(t)







F + p2F = 0 ⇒ F (x) = A cos px + B sin px




 2 p2 t

 Ġ + c2p2G = 0 ⇒ G(t) = e−c

• 2 p2 t
• • u(x, t; p) = F (x) · G(t) = (A cos px + B sin px)e−c

Note that there are no boundary conditions to determine p

This means any values of p is fine.
•
• • We therefor write
 p
u(x, t) = 0
u(x, t; p)dp
 p 2 p2 t
−c
= 0
[A cos px + B sin px]e dp

34
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

$ %
∞
Recall the Fourier integral for f (x) −∞
|f (x)| dx < ∞
 ∞
f (x) = 0
[A(w) cos wx + B(w) sin wx]dw
 
 1 ∞




A(w) = π −∞
f (x̃) cos wx̃dx̃


 



 1 ∞
B(w) = π −∞
f (x̃) sin wx̃dx̃

Now apply the initial conditions:

 ∞
u(x, 0) = 0
[A cos px + B sin px]dp = f (x)
 
 1 ∞




A(p) = π −∞ f (x̃) cos px̃dx̃

⇒
 



 1 ∞
B(p) = π −∞ f (x̃) sin px̃dx̃

• 1∞ ∞
• • u(x, t) = [ ( −∞ f (x̃) cos px̃dx̃ ) cos px
π 0
 ∞ 2 p2 t
−c
+( −∞
f (x̃ ) sin px̃dx̃ ) sin px ] e dp
1∞ ∞ −c2 p2 t
= [ f (x̃)(cos px̃ cos px + sin px̃ sin px)dx̃ ] e dp
π 0 −∞
1∞ ∞ −c2 p2 t
= [ f (x̃) cos p(x̃ − x)dx̃ ] e dp
π 0 −∞
1∞  ∞
−c2 p2 t
= f (x̃) [ cos p(x̃ − x) · e dp ] dx̃
π −∞  0  

35
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

s ds
Let p ≡ √ , dp = √
c t c t
 ∞ 2 p2 t
0
cos p(x̃ − x) · e−c dp
 ∞ s 2 ds (x̃ − x)
= 0
cos √ (x̃ − x) · e−s √ b≡ √
c t c t 2c t
1 $ ∞ −s2
= √ cos 2bs· e ds )
c t 0
√ 
1  π −b2 
= √  e 
c t 2
√ 2
π − (x̃−x)
= √ e 4c2t
2c t


√ 
2

1 ∞ π (x̃−x)
•
• • u(x, t) = f (x̃) 
 √ e− 4c2t  dx̃
π −∞ 2c t
1 ∞ (x̃−x)2
− 2
= √ f (x̃) · e 4c t dx̃
2c πt −∞

36
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Ex :
If f (x) =

2
u0  1 − (x̃−x)
u(x, t) = √ −1
e 4c2 t dx̃
2c πt

If U0 = 100 0C, c2 = 1 cm2/sec.

37
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.8 Heat Equation: Use of Fourier Transform


 2

 ∂u 2∂ u



 =c (−∞ < x < ∞) ——— (1)

∂t ∂x2







 u(x, 0) = f (x) (−∞ < x < ∞) ——— (2)

Recall that:
1 ∞
û(w) = F{u(x)} = √ −∞
u(x)e−iwx dx
2π

F{u (x)} = iwF{u(x)}

F{u (x)} = −w2F{u(x)}

Take Fourier Transform of (1),

   
∂u    ∂ 2u 
 
2 2 2
⇒F =c F ∂x2  = −c w F{u}
∂t

Also,
 
∂u 
 1  ∞ ∂u −iwx 1 ∂ # ∞ −iwx
& ∂
F = √ e dx = √ ue dx = F{u}
∂t 2π −∞ ∂t 2π ∂t −∞ ∂t

38
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

∂
⇒ F{u} = −c2w2F(u)
∂t
∂ û(w, t)
i.e. = −c2w2û(w, t) ——— (3)
∂t
•
• • we have transformed the partial differential equation (1) into an ordinary
differential equation (3) !!

Similarly, taking Fourier transform of the initial condition (2), gives:

û(w, 0) = fˆ(w) ——— (4)

2 w2t
(3) ⇒ û(w, t) = Ce−c

(4) ⇒ û(w, 0) = C = fˆ(w)

• 2 2
• • û(w, t) = fˆ(w) · e−c w t

39
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Take inverse Fourier transform of û(w, t):

  ∞

 √1 −iwx




û(w) = 2π −∞
u(x)e dx


  ∞

û(w)eiwxdw



 u(x) = √1
2π −∞

2 2
û(w, t) = fˆ(w) · e−c w t

1 ∞ ˆ −c2 w 2 t iwx
⇒ u(x, t) = √ f (w) · e · e dw
2π −∞

1 ∞ 1 ∞ −iwx̃ −c2 w 2 t iwx

= √ [ √ f (x̃)e dx̃ ] e · e dw
2π −∞ 2π −∞

1 ∞ ∞ −iw(x̃−x) −c2 w 2 t
= [ f (x̃)e dx̃ ] · e dw
2π −∞ −∞
1 ∞  ∞
−iw(x̃−x) −c2 w 2 t
= f (x̃) [ e · e dw ] dx̃
2π −∞ −∞

1∞  ∞
−c2 w 2 t
= f (x̃) [ cos w(x̃ − x) · e dw ] dx̃
π −∞ 0
√  
π −(x̃−x)
= √ e 4c2t
2c t

Same as using Fourier integral !! ✷

40
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.9 Heat Equation. Use of Fourier Cosine and Sine Trans-

forms

Consider the heat equation along a semi-infinite rod:

2
∂u 2∂ u
=c (0 ≤ x < ∞)
∂t ∂x2
Recall Fourier cosine and sine transforms of derivatives :
 
2



 Fc {f } = wFs{f } − π
f (0)



 Fs {f } = −wFc{f }
 
 2 2



 Fc {f } = −w Fc{f } − π
f (0)
 

 2 2

 Fs {f } = −w Fs{f } + π
wf (0)
 
2 ∞
where Fc {f } = fˆc(w) = π 0
f (x) cos wxdx
 
2 ∞
Fs {f } = fˆs(w) = π 0 f (x) sin wxdx
⇒
If the boundary condition at x = 0 is :
u(0.t) = g(t) we use Fourier sine transform
∂u
(0, t) = g(t) we use Fourier cosine transform
∂x

41
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

Ex :
∂u ∂u
= c2 2 0 ≤ x < ∞ ——— (1)
∂t ∂x
u(x, 0) = f (x) 0 ≤ x < ∞ ——— (2)

u(0, t) = 0 t ≥ 0 ——— (3)

Take Fourier sine transform of (1) :

    !
∂u  ∂ ûs
  ∂ 2u 
  !2
!
2 2 2
⇒ Fs   = = c Fs 
 2 

= −c w Fs {u} + " wu(0, t)
∂t ∂t ∂x π

= −c2w2ûs(w, t)

∂ ûs
i.e. = −c2w2ûs
∂t
2w2t
⇒ ûs (w, t) = C(w)e−c

Take Fourier sine transform of the initial condition (2): u(x, 0) = f (x)

⇒ ûs (w, 0) = fˆs (w) = C(w)

 
!
! 2 
• 2 2 ! ∞  2 w2t
• • ûs (w, t) = fˆs (w)e−c w t = "

0
f (x̃) sin wx̃dx̃ e−c
π

Take inverse Fourier sine Transform:

• 2 ∞∞ −c2 w 2 t
• • u(x, t) = f (x̃) sin w x̃ · e · sin wx dx̃dw ✷
π 0 0

42
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

11.10 Wave Equation. Use of Fourier Transform

Consider wave equation along an infinitely long string :




 ∂ 2u 2
2∂ u



 =c (−∞ < x < ∞) ——— (1)






∂t2 ∂x2





 u(x, 0) = f (x) ——— (2)











ut (x, 0) = 0 ——— (3)







 u → 0, ux → 0 as |x| → ∞ ——— (4)

Take Fourier transform of (1):

   
2
 ∂u  ∂ 
 ∂ 2u 

F 2  = 2
F{u} = c2F  2
 ∂x  = −c2 2
w F{u}
∂t ∂t

⇒ ûtt + c2w2û = 0, û = û(w, t)

•
• • û(w, t) = A(w) cos cwt + B(w) sin cwt

Take Fourier transform of the initial conditions (2) and (3):

û(w, 0) = fˆ(w) = A(w)

ût (w, 0) = 0 = cwB(w)

⇒ û(w, t) = fˆ(w) cos cwt

1ˆ # &
icwt −icwt
= f (w) e +e
2

43
Applied Mathematics — Partial Differential Equation Wu-ting Tsai

1 ∞ −iwx
F{f (x − a)} = √ f (x − a)e dx x − a ≡ p, dx = dp
2π −∞

1 ∞
= √
−∞
f (p)e−iw(p+a)dp
2π

1 −iwa  ∞ −iwp
=√ e −∞
f (p)e dp
2π

i.e. F{f (x − a)} = e−iwaF{f (x)}

•
• • F −1 {F{f (x − a)}} = F −1 {e−iwaF{f (x)}} = f (x − a)

i.e. the inverse Fourier Transform of e−iwaF{f (x)} is f (x − a)

Similarly, the inverse Fourier transform of e+iwaF{f (x)} is f (x + a)

1ˆ # &
icwt −icwt
Since û(w, t) = f (w) e +e
2

• 1
• • u(x, t) = [f (x − ct) + f (x + ct)] ✷
2

44