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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 ∫ ⎟
⎠
⎞ ⎜ ⎝
=^ L ⎛ m (^) 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
⎪⎭
⎪ ⎬
⎫ ⎥ ⎦
⎤ ⎢ ⎣
⎡ ⎟ ⎠
⎞ ⎜ ⎝
⎞ ⎜ ⎝
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
- 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
- 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)
⎞ ⎜ ⎝
=^ L ⎛ m (^) L dx gx mx mc
A 0
(^2) ()sin π
π ∫^
⎟ ⎠
⎞ ⎜ ⎝
=^ L ⎛ m (^) 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
⎞ ⎜ ⎝
=^ L ⎛ m (^) 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
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
