ABSTRACT OF CHAP. 12

ENGINEERING MATHEMATICS II

010.141

Byeng Dong Youn (윤병동윤병동윤병동)윤병동

CHAP. 12

Partial Differential Equations (PDEs)

ABSTRACT OF CHAP. 12

PDEs in Chap. 12 are models of various physical and geometrical problems (the solutions) depend on two or

more variables, usually on time t and one or several space variables.

 We concentrate on the most important PDEs of applied mathematics, the wave equations governing the vibrating string (Sec. 12.2) and vibrating membrane (Sec. 12.7), the heat equation (12.5), and Laplace equation (Secs. 12.5 and 12.10).

 We derive these PDEs from physics and consider methods for solving initial and boundary value problems, that is, methods of obtaining solutions satisfying conditions that are given by the physical situation.

CHAP. 12.1

BASIC CONCEPTS

Classes of the PDEs.

EXAMPLES OF PDE

One-D Wave equation

One-D Heat equation

Two-D Laplace equation

Two-D Wave Equation

Two-D Poisson equation

Three-D Laplace Equation

2 2

2

2 2

u u

t c x

∂ ∂

∂ = ∂

2 2

2

u u

t c x

∂ ∂

∂ = ∂

2 2

2 2 0

u u

x y

∂ ∂

+ =

∂ ∂

( )

2 2

2 2 ,

u u

f x y

x y

∂ ∂

+ =

∂ ∂

2 2 2

2

2 2 2

u u u

t c x y

 

∂ ∂ ∂

=  + 

∂ ∂ ∂ 

2 2 2

2 2 2 0

u u u

x y z

∂ ∂ ∂

+ + =

∂ ∂ ∂

A PDE is an equation involving one or more partial derivatives of a function.

Fundamental Theorem

EXAMPLE OF WAVE EQUATION

Problem, as a solution to the wave equation

( ) ( )

( )

( )

( ) ( )

1 4

1 4 2

1 2 4

1 1

4 4

2

1 1

16 4

2

1 2 1 1

4 16 4

( , ) sin 9 sin 9 cos 9 sin

81 sin 9 sin sin 9 cos

sin 9 cos

81 sin 9 sin sin 9 sin is identically true,

u t x t x

u t x

t

u t x

t

u t x

x

u t x

x

t x c t x

= ⋅

∂ =

∂

∂ = − ⋅

∂

∂ = ⋅

∂

∂ = ⋅ −

∂

− ⋅ = − ⋅

( )

2 2

2

2 2

u u

t c x

∂ ∂

∂ = ∂ u t x( , )

EXAMPLE OF HEAT EQUATION

2 2

2 4

4

4

2

4 2

4 4 2

2

cos 3

4 cos 3 3 sin 3

9 cos 3

4 cos 3 9 cos 3 Then 4

9

t

t

t

t

t t

u u

t c x

u e x

u e x

t

u e x

x

u e x

x

e x e x c

c

−

−

−

−

− −

∂ ∂

∂ = ∂

=

∂ = −

∂

∂ = −

∂

∂ = −

∂

− = − ⋅

=

EXAMPLE OF LAPLACE EQUATION

2 2

2 2

2 2

2 2

0

0

0 0

x

x

x

x

x

x x

u u

x y

u e sin y u e sin y x

u e sin y x

u e cos y y

u e sin y

y

e sin y e sin y

∂ ∂

+ =

∂ ∂

=

∂ =

∂

∂ =

∂

∂ =

∂

∂ = −

∂

− =

=

HOMEWORK IN 12.1

HW1. Problems 1, 3, 5

HW2. Problems 14, 15

HW3. Problems 18, 19

HW4. Problem 23

CHAP. 12.2

MODELING: VIBRATING STRING, WAVE EQUATION

Modeling the wave equation.

VIBRATING STRING AND THE WAVE EQUATION

General assumptions for vibrating string problem:

mass per unit length is constant; string is perfectly elastic and no resistance to bending.

tension in string is sufficiently large so that gravitational forces may be neglected.

string oscillates in a plane; every particle of string moves vertically only;

and, deflection and slope at every point are small in terms of absolute value.

Deflected string at fixed time t

DERIVATION OF WAVE EQUATION

Deflected string at fixed time t

T₁, T₂ = tension in string at point P and Q

T₁ cos α = T₂ cos β = T, a constant (as string does not move in horizontal dir.)

Vertical components of tension:

– T₁ sin α and T₂ sin β

DERIVATION OF WAVE EQUATION (Cont.)

2 2

2 1

Let = length PQ, = mass unit length , and has mass Newton's Law:

If is the vertical position,

sin si

x x x

F mass acceleration

u u acceleration

t

T T

∆ ∆ ∆

= ×

∂ =

∂

−

ρ ρ

β

( )

( )

( )

2 2

2

2 1

2

2 1

n sin sin

tan tan equation 2

cos cos

At P is distance from origin , tan is slope Likewise at Q, tan

x

x x

x u

t

T T x u

T T T t

x u

x u

x _{+ ∆}

= ∆ ∂

∂

∆ ∂

− = − =

∂

= ∂

∂

= ∂

∂

α ρ

β α β α ρ

β α

α

β

DERIVATION OF WAVE EQUATION (Cont.)

2 2

2 2

2 2

2 2

2 2

2

Substituting and 1

as 0 L.H.S. becomes

Let

This is the 1-D Wave equation If or

x x x

u u u

x : x x x T t

x , u

x

T u u

c , so that c

t x

T c

+∆

 ∂ ∂  ρ ∂

÷ ∆  −  =

∆  ∂ ∂  ∂

∆ → ∂

∂

∂ ∂

= =

ρ ∂ ∂

↑ ρ ↓ ↑

CHAP. 12.3

SOLUTION BY SEPARATING

VARIABLES. USE OF FOURIER SERIES

Solving the PDEs using separating variables and

Fourier series and interpreting the solution.

Solution by Separating Variables

Solution by Separating Variables

Solution by Separating Variables

Solution by Separating Variables

Solution by Separating Variables

Solution by Separating Variables

Physical Interpretation

Let us consider when the initial velocity g(x) be identically zero.

( ) ( )

1

1 1

( , ) cos sin ,

1 1

sin sin

2 2

n n n

n

n n

n n

n x cn

u t x B

L L

n n

B x ct B x ct

L L

∞

=

∞ ∞

= =

= =

   

=  − +  + 

   

∑

∑ ∑

π π

λ λ

π π

( ) ( )

* *

( , ) 1

u t x = 2  f x ct− + f x+ct 

The first term represents a wave that is traveling to the right as t increases.

The second is a wave that is traveling to the left as t increases.

EXAMPLE 1

EXAMPLE 1

HOMEWORK IN 12.3

HW1. Problem 1

HW2. Problems 11, 12

HW3. Problems 15, 16

CHAP. 12.5

HEAT EQUATION: SOLUTION BY FOURIER SERIES

Also called the diffusion equation.

From prior work the heat equation is:

In one dimension (laterally insulated):

Some boundary conditions at each end:

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

Initial Condition:

u(x, 0) = f(x), specified 0 ≤ x ≤ L

HEAT EQUATION

2 2

2

u u

t c x

∂ ∂

∂ = ∂

2 2 2

u k

c u, c

t

∂ = ∇ =

∂ σρ

HEAT EQUATION

HEAT EQUATION

HEAT EQUATION

EXAMPLE 1

EXAMPLE 2

EXAMPLE 3

EXAMPLE 4

HOMEWORK IN 12.5

HW1. Problems 5,7

CHAP. 12.6

HEAT EQUATION: SOLUTION BY FOURIER INTEGRALS AND

TRANSFORMS

Extension to bars of infinite length.

INTRODUCTION

USE OF FOURIER INTEGRALS

USE OF FOURIER INTEGRALS

Using the inner integral and variable transformation

EXAMPLE1

EXAMPLE1

HOMEWORK IN 12.6

HW1. Problems 3,5

CHAP. 12.7

MODELING: MEMBRANE, TWO- DIMENSIONAL WAVE EQUATION

Extension to bars of infinite length.

VIBRATING MEMBRANE AND

THE TWO-DIMENSIONAL WAVE EQUATION

Three Assumptions:

The mass of the membrane is constant, the membrane is perfectly flexible, and offers no resistance to bending;

The membrane is stretched and then fixed along its entire boundary in the x-y plane. Tension per unit length (T) which is caused by stretching is the same at all points and in the plane and does not change during the motion;

The deflection of the membrane u(x,y,t) during vibratory motion is small compared to the size of the membrane, and all angles of

inclination are small

GEOMETRY OF VIBRATING "DRUM"

Vibrating Membrane

NOW FOR NEWTON'S LAW

Divide by ρ∆x∆y:

Take limit as ∆x → 0, ∆y → 0

This is the two-dimensional wave equation

(^{ρ∆x y∆} ) ^∂_∂t^{u = T y u∆}

[ (

)

(

) ]

[ ⁽

⁾

⁽

⁾ ]

2

x 1 x 2 y 1 y

2 − − 2

( ) ( ) ( ) ( )

∂

∂ ρ

2 x 1 x 2 y 1 y

= T u + x, y u , y

+ u x , y + y u , y y

u t

x x

x

x

2

∆ 2

∆

∆

∆

− −













∂² ₂ ∂² ∂²

= c +

u  u u

 

 T = c²

ρ

WAVE EQUATION

In Laplacian form:

where

c² = T/ρ

 Some boundary conditions: u = 0 along all edges of the boundary.

 Initial conditions could be the initial position and the initial velocity of the membrane

 As before, the solution will be broken into separate functions of x,y, and t.

 Subscripts will indicate variable for which derivatives are taken.

∂

∂

2 2

u 2

t = c² ∇ u

SOLUTION OF 2-D WAVE EQUATION

Let

u(x, y, t) = F(x, y)G(t) substitute into wave equation:

divide by c²FG, to get:

This gives two equations, one in time and one in space. For time,

where λ = cν, and, what is called the amplitude function:

also known as the Helmholtz equation.

( )

&&

G

G F_xx

c = 1

F + F

2 yy = −ν²

&&

G + λ²G = 0

( )

FG&& = c² F G _xx + F_yyG ;

F_xx + F + _yy ν²F = 0

SEPARATION OF THE HELMHOLTZ EQUATIONS

Let F(x, y)=H(x)Q(y) and, substituting into the Helmholtz:

Here the variables may be separated by dividing by HQ:

Note: p² = ν² – k²

As usual, set each side equal to a constant, -k² . This leads to two ordinary linear differential equations:

2 2

2

2 Q = 2 +

d H d Q

H HQ

dx − ^ dy ν ^

 

1 ² 1

2

2

H 2

d H

dx Q

d Q

dy Q

= −  + ² = k²

 

 − ν

d H

dx H d Q

dy Q

2 2

2

0 2 0

+ k² = , + p² =