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Solution to a Wave Equation in One Dimension, Slides of Engineering Analysis

The solution to a wave equation in one dimension using various methods, including the use of trigonometric identities and the d'alembert solution. The document also discusses the behavior of wave phenomena and the importance of boundary conditions in numerical solutions.

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Classification/multidimensional PDEs March 2, 2009
ME 501B Engineering Analysis 1
Classification of
Classification of PDEs
PDEs and
and
Multidimensional
Multidimensional PDEs
PDEs
Larry Caretto
Mechanical Engineering 501B
Seminar in Engineering
Seminar in Engineering
Analysis
Analysis
March 2, 2009
2
Overview
Review last class
Wave equation solutions by separation of
variables and D’Alembert approach
Characteristics and classification of
partial differential equations
General analysis
Parabolic equations
Elliptic equations
Hyperbolic equations
Solving a wave equation problem
3
Midterm Exam
Wednesday, March 11
Covers material on diffusion and
Laplace equations
Includes material up to and including
tonight’s February 23 lecture and
homework for Monday, March 2
Open book and notes, including
homework solutions
Focus on working with existing solutions
4
Review Wave Equation
Usual assumption u(x,t) = X(x)T(t)
2
2
2
2
2
2
)(
)(
1)(
)(
11 λ=
=
x
xX
xX
t
tT
tT
c
Result is
function of t
equal to
function of x
[][]
)cos()sin()cos()sin(
)()(),(
xDxCctBctA
xXtTtxu
λλλλ
++
=
=
Use above solution as starting point
Boundary conditions at x = 0 and x = L
Initial conditions on u and u/tat x = 0
5
Review General Solution
=
+
=
1
sincossin),(
nnn L
xn
L
ctn
B
L
ctn
Atxu
πππ
π
π
=
L
mdx
L
xm
xg
m
A
0
sin)(
2
=
L
mdx
L
xm
xf
L
B
0
sin)(
2
π
Solution for u(x,t) with
initial and boundary
conditions
u(x,0) = f(x); u/x|0= g(x)
u(0,t) = u(L,t) = 0 speedwaveisc
tLx x
u
c
t
u
0,0
2
2
2
2
2
=
6
Review Use of Trig Identities
2 sin z cos y = sin(z + y) + sin(z y)
2 sin z sin y = cos(z y) – cos(z + y)
=
+π
π
=
1
)(
cos
)(
cos
2
1
),(
n
nL
ctxn
L
ctxn
Atxu
=
+
=
1
sincossin),(
nnn L
xn
L
ctn
B
L
ctn
Atxu
πππ
π
+
+π
+L
ctxn
L
ctxn
Bn)(
sin
)(
sin
sin zcos ysin y
pf3
pf4
pf5

Partial preview of the text

Download Solution to a Wave Equation in One Dimension and more Slides Engineering Analysis in PDF only on Docsity!

Classification of PDEsClassification ofPDEs andand

Multidimensional PDEsMultidimensionalPDEs

Larry Caretto Mechanical Engineering 501B Seminar in EngineeringSeminar in Engineering AnalysisAnalysis March 2, 2009

2

Overview

  • Review last class
    • Wave equation solutions by separation of variables and D’Alembert approach
  • Characteristics and classification of partial differential equations - General analysis - Parabolic equations - Elliptic equations - Hyperbolic equations
  • Solving a wave equation problem

3

Midterm Exam

  • Wednesday, March 11
  • Covers material on diffusion and Laplace equations
  • Includes material up to and including tonight’s February 23 lecture and homework for Monday, March 2
  • Open book and notes, including homework solutions
  • Focus on working with existing solutions 4

Review Wave Equation

  • Usual assumption u(x,t) = X(x)T(t)

2 2

2 2

2 2

=−λ ∂

x

X x t X x

Tt c Tt

Result is function of t equal to function of x [ sin( ) cos( )][ sin( ) cos( )]

A ct B ct C x D x

uxt TtX x

  • Use above solution as starting point
    • Boundary conditions at x = 0 and x = L
    • Initial conditions on u and ∂u/∂t at x = 0

5

Review General Solution

=

1

( ,) sin cos sin n

n n L

nx L

nct B L

nct uxt A π π π

∫ ⎟⎠

⎞ ⎜ ⎝

⎛ π π

=

L m (^) L dx gx mx m

A 0

2 ()sin ∫ ⎟

⎞ ⎜ ⎝

=^ Lm (^) L dx f x mx L

B 0

(^2) ()sin π

  • Solution for u(x,t) with initial and boundary conditions - u(x,0) = f(x); ∂u/∂x| 0 = g(x) - u(0,t) = u(L,t) = 0 ciswavespeed

x Lt

x

u c t

u

0 , 0

2

2 2 2

2

≤ ≤ ≥

6

Review Use of Trig Identities

  • 2 sin z cos y = sin(z + y) + sin(z – y)
  • 2 sin z sin y = cos(z – y) – cos(z + y)

= ⎪⎩

⎪ ⎨

⎧ ⎥ ⎦

⎤ ⎢ ⎣

⎡ ⎟ ⎠

⎜ ⎞ ⎝

⎟− ⎛^ π + ⎠

⎜ ⎞ ⎝

= ⎛^ π − 1

( ) cos ( ) cos 2

1 (,) n

n (^) L

n x ct L

uxt A n x ct

=

1

( ,) sin cos sin n

n n L

nx L

nct B L

nct uxt A π π π

⎪⎭

⎪ ⎬

⎫ ⎥ ⎦

⎤ ⎢ ⎣

⎡ ⎟ ⎠

⎞ ⎜ ⎝

  • ⎛^ π − ⎟ ⎠

⎞ ⎜ ⎝

  • ⎛^ π + L

n x ct L

B n x ct n sin ( ) sin ( )

sin y cos y sin z

7

Review D’Alembert Solution

  • Wave phenomena: u(x,t) is wave amplitude varying with space, x, and time, t
  • c is wave speed
  • Over any x region and t ≥ 0

2

2 2 2

2

x

u c t

u

  • D’Alembert solution using ξ = x + ct and η = x – ct, u(x,0) = f(x) and ∂u/∂x|t=0 = g(x)

[ ] ∫

= + + − +

x ct

xct

g d c

uxt f x ct f x ct (ν ) ν 2

( ) ( )^1 2

(,)^1

8

Review D’Alambert II

  • We see that the solution obtained by separation of variables agrees with the D’Alambert solution for one case
  • Solution shows propagation of wave shapes without damping - Look at meaning of f(x + ct) and f(x – ct) - As time increases f(x + ct) retains the shape of the initial condition and moves to the left - Similarly f(x – ct) retains its shape and moves to the right

9

Review Wave Propagation

-5 -4 -3 -2 -1 0 1 2 3 4 5 distance, x

time, t

t = 0 ct=1ct = 2 ct = 3 x+ct x+ct x-ctx-ct

10

Review Boundaries

  • With g(x) = 0 solution is Fourier sine series which is periodic, odd function

=

1

sin

n

n (^) L

n x B

ux f x

-

-0.

0

0.

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 x

f(x) Actual solution Periodic extensions

⎥ ⎦

⎤ ⎢ ⎣

⎡ ⎟ ⎠

⎜ ⎞ ⎝

⎟− ⎛ ⎠

⎜ ⎞ ⎝

⎟− ⎛ ⎠

⎜ ⎞ ⎝

= ⎛ 5

3 sin 5

2 sin 2 2 sin 20 2 2

π π π π

m m m m Bm

11

Review Time Evolution

  • Look at evolution when ct = 0.4 (^2)

f x ct f x ct u xt

-

-0.

0

0.

1

-2 -1.5 -1 -0.5 (^0) x 0.5 1 1.5 2

f(x)

  • For larger values of x ± ct, periodic extensions move into 0 ≤ x ≤ L = 1

Initial Condition

f(x+ct)/2 = f(x+0.4)/2 f(x-ct)/2 = f(x-0.4)/

-0.

0

0.

1

0 0.2 0.4 0.6 0.8 1 x

f(x+ct)

ct = 0.

-0.

0

0.

1

0 0.2 0.4 0.6 0.8 1 x

f(x-ct) ct = 0.

-0.

0

0.

1

0 0.2 0.4 0.6 0.8 1 x

u(x,t) ct = 0.

Phase behavior of sine function causes initial wave form to be reflected at bound- aries

19

Behavior of Equation Types

  • Domain of dependence for u(x 1 ,y 1 )
    • The area (in x-y space) whose u values affect the value of u(x 1 ,y 1 )
  • Region of influence of u(x 1 ,y 1 )
    • The area (in x-y space) whose u values are affected by the value of u(x 1 ,y 1 )
  • Importance for specifying boundary conditions and for numerical solutions

Right-running charac- teristic (dy/dx > 0 )

Left-running charac- teristic (dy/dx < 0 )

x 1 , y 1

Initial Condition Curve

Domain of dependence

20

Hyperbolic Equations

  • Domain of dependence shown on previous chart
  • Region of influence is region of characteristics leaving x 1 , y 1
  • Conditions outside domain of dependence should not affect solution
  • Important point for numerical algorithms which can violate this principle for inappropriate choices of step sizes

21

Elliptic PDEs

  • Imaginary characteristics for elliptic equations like Laplace and Poisson’s
  • Entire solution region is both domain of dependence and region of influence - Must have closed boundaries
  • This means that any change in any boundary condition can affect the solution at any point in the region - Effects may be small far from boundary, but will be present 22

Parabolic PDEs

  • Parabolic equations typically involve time and space as coordinates
  • Consider region 0 ≤ x ≤ L and t > 0
    • Domain of dependence at any point x 1 , t 1 is entire domain at previous times: 0 ≤ x ≤ L and 0 ≤ t < t 1 - Any change in initial conditions or boundary conditions for t < t 1 will change solution here
    • Region of influence at x 1 , t 1 is entire region for future times 0 ≤ x ≤ L and t > t (^1)

23

Example Question

  • You are solving the diffusion equation in the region 0 ≤ x ≤ L and t > 0 with an initial condition u(x,0) = f(x) and the following boundary conditions - t < 12 s: u(0,t) = u(L,t) = 0 - t ≥ 12 s: u(0,t) = u(L,t) = a = 1
  • If you have a solution to this problem for a = 1, how does the solution change for t < 12 s, if you set a = 2?

24

Work on Homework Problem

  • Page 547, problem 12 – solve the wave equation for 0 ≤ x ≤ L = π and t ≥ 0 with boundary and initial conditions shown

⎩⎨^ (^ )

⎧ π− π ≤ ≤π

≤ ≤π

∂ = = =

= = ∂

∂ = x x

x x t

u u t uLt ux

c x

u c t

u t^0.^012

  1. 01 0 2 ( 0 ,) (,) 0 (, 0 ) 0

0 1 0

2

2 2 2

2

[ sin( ) cos( )][ sin( ) cos( )]

A ct B ct C x D x

uxt TtX x λ + λ λ + λ

  • Start with separation of variables result

25

Work on Homework Problem

⎩⎨^ (^ )

⎧ π− π ≤ ≤π

≤ ≤π

∂ = = =

= = ∂

= ∂ ∂

∂ = x x

x x t

u

u t uLt ux

c x

c u t

u t^0.^012

  1. 01 0 2 ( 0 ,) (,) 0 (, 0 ) 0

0 1 0

2

2 2 2

2

[ sin( ) cos( )][ sin( ) cos( )]

A ct B ct C x D x

uxt TtX x λ + λ λ + λ

  • For u(0,t) = 0 we must have D = 0
  • For u(L,t) = 0 we must have λL = nπ (n an integer)

26

Work on Homework Problem

u ( x , t )= [ An sin( λ n ct )+ Bn cos( λ nct )] sin( λ nx )

  • From solution for u = 0 at x = 0 and x = L, we have for u(x,0) = f(x), ut(x,0) = g(x)

⎞ ⎜ ⎝

=^ Lm (^) L dx gx mx mc

A 0

(^2) ()sin π

π ∫^

⎟ ⎠

⎞ ⎜ ⎝

=^ Lm (^) L dx f x mx L

B 0

(^2) ()sin π

  • In this problem f(x) = u(x,0) = 0 so we have all B (^) m = 0

27

Work on Homework Problem

=

1

( ,) sin( )sin( ) n

uxt An λ nct λ nx

⎞ ⎜ ⎝

=^ Lm (^) L dx

mx gx mc

A 0

()sin 2 π π

  • Use function given for g(x) to get A (^) m

∫ ∫^ (^ ) ⎟

⎞ ⎜ ⎝

  • − ⎛ ⎟ ⎠

⎞ ⎜ ⎝

= ⎛ π π

π π π π

π π 2

2

0

(^2 0). 01 sin (^20). 01 sin dx L

x mx m c

dx L

x mx mc

Am

  • Set L = π and c = 1 28

Multidimensional Equations

  • Can have equations in three space dimensions and time
  • Classification as elliptic, parabolic, or hyperbolic does not apply to equations with more than two dimensions
  • Coordinates can have elliptic-like, parabolic-like, and hyperbolic-like behavior in multidimensional equations - E. g., time is a parabolic coordinate

29

Multidimensional Laplace

  • General Laplace equation for three dimensions

0

2 ∇ u =

φ

φ φ θ φ

θ

∂ ⎟+ ⎠

⎜ ⎞ ⎝

⎛ ∂

∂ ∂

∂ ∇ =

∂ ⎟+ ⎠

⎜ ⎞ ⎝

⎛ ∂

∂ ∂

∂ ∇ =

+∂ ∂

+∂ ∂

∇ =∂

u r

u r

u r r

u r r r Sphere u

z

u u r r

u r r r Cylindrical u

z

u y

u x

Cartesian u u

2 2

2 2 2

2 2 2 2 2 2

2

2 2

2 2 2

2

2 2

2 2

2 2

1 cot sin

1 1

1 1

30

Multidimensional Diffusion

  • General diffusion equation for three dimensions

u t

u (^) 2 = ∇ ∂

φ

φ φ θ φ

θ

∂ ⎟+ ⎠

⎜ ⎞ ⎝

⎛ ∂

∂ ∂

∂ ∇ =

∂ ⎟+ ⎠

⎜ ⎞ ⎝

⎛ ∂

∂ ∂

∂ ∇ =

+∂ ∂

+∂ ∂

∇ =∂

u r

u r

u r r

u r r r Sphere u

z

u u r r

u r r r Cylindrical u

z

u y

u x

Cartesian u u

2 2

2 2 2

2 2 2 2 2 2

2

2 2

2 2 2

2

2 2

2 2

2 2

1 cot sin

1 1

1 1

37

Initial Condition

  • Get general solution as sum of all eigenfunctions and get initial condition

⎟ ⎠

⎞ ⎜ ⎝

⎛ ⎟ ⎠

⎞ ⎜ ⎝

= ⎛

⎥⎥⎦

⎤ ⎢⎢⎣

⎡ (^) ⎟ ⎠ ⎜ ⎞ ⎝ ⎟+⎛ ⎠ ⎜ ⎞ ⎝ −⎛ H

my L

u xyt C e nx n m

nL mH t nm

α π π π π ( , ,) sin sin 1 1

2 2

⎟ ⎠

⎞ ⎜ ⎝

⎛ ⎟ ⎠

⎞ ⎜ ⎝

= = ⎛

∞ = H

my L

u xy f xy C nx n m nm ( , , 0 ) (, ) sin π^ sin^ π 1 1

  • Multiply by [sin(pπx/L) sin(qπy/H)]dxdy and integrate for 0 ≤ x ≤ L and 0 ≤ y ≤ H 38

Initial Condition Gives Cmn

22

sin sin

sin sin sin sin

(, )sin sin

0

2 0

2

0 0 1 1

0 0

dx C H^ L L

dy px H

C qy

dxdy H

qy L

px H

my L

nx C

dxdy H

qy L

px f xy

pq

H L pq

H L n m nm

H L

⎟^ = ⎠

⎞ ⎜ ⎝

⎛ ⎥ ⎦

⎤ ⎢ ⎣

⎡ ⎟ ⎠

⎞ ⎜ ⎝

= ⎛

⎟ ⎠

⎜ ⎞ ⎝

⎟ ⎛ ⎠

⎜ ⎞ ⎝

⎟ ⎛ ⎠

⎜ ⎞ ⎝

⎟ ⎛ ⎠

⎜ ⎞ ⎝

⎟ = ⎠

⎜ ⎞ ⎝

⎟ ⎛ ⎠

⎜ ⎞ ⎝

π π

π π π π

π π

  • Use orthogonality of sine and previous results for sin^2 integrals to get

∫∫ ⎟⎠ ⎜ ⎞ ⎝ ⎟ ⎛^ π ⎠ ⎜ ⎞ ⎝ = ⎛^ π

H L pq (^) H dxdy qy L fxy px C HL 0 0

(^4) (,)sin sin