Uploaded by love love

PDE

advertisement
Partial Differential Equations
Assoc. Prof. Dr. Siridech Boonsang
Electrical Engineering
Partial Differential Equations Background
• Physical problems are governed by
many PDEs
• Some are governed by first order PDEs
• Numerous problems are governed by
second order PDEs
• A few problems are governed by
fourth-order PDEs.
Examples
2
T
 T
Heat Conduction Equation (1  D) :
C
2
t
x
  2T  2T 
T

Heat Conduction Equation ( 2  D) :
 C

 x 2 y 2 
t


2  D Laplace Equation :  2  
2
2  D Poisson Equation :   
 2
x
2
 
x
2
2


 2
y
2
 
y
2
2
0
 f ( x, y )
• Laplace’s Equation: 2u = uxx +uyy +uzz = 0
• unknown: u(x,y,z)
• gravitational / electrostatic potential
Heat Equation: ut = a22u
unknown: u(t,x,y,z)
heat conduction
Wave Equation: utt = a22u
unknown: u(t,x,y,z)
wave propagation
Examples (contd.)
1  D Wave Equation :
2
 u
t
2
C
2
 u
x
2
2  D Wave Equation ( vibrating membrane) :
2
  2u  2u 
 u

 C

2
 x 2 y 2 
t


4thOrder Equations :
2
4
 u
 u
C
0
2
4
t
x
4
4 
2
  4u

u

u

u


D
2

 h
0
2 2
4
2
 x 4
x y
y 
t

( Vibrating
beam)
( Vibrating
plate)
• Schrödinger Wave Equation
• quantum mechanics
• (electron probability densities)
• Navier-Stokes Equation
• fluid flow (fluid velocity & pressure)
Classification of Partial Differential
Equations (PDEs)
There are 6 basic classifications:
(1) Order of PDE
(2) Number of independent variables
(3) Linearity
(4) Homogeneity
(5) Types of coefficients
(6) Canonical forms for 2nd order PDEs
(1)
Order of PDEs
The order of a PDE is the order of the highest
partial derivative in the equation.
Examples:
u  2 u
 2
t x
(2nd order)
u u
(1st order)

t x
u
 3u
 u 3  sin x
(3rd order)
t
x
(2) Number of Independent Variables
Examples:
(2 variables: x and t)
(3 variables: r, q, and t)
u  2 u
 2
t x
u  2 u 1 u 1  2 u
 2
 2 2
t r
r r r q
(3)
Linearity
PDEs can be linear or non-linear. A PDE is linear if
the dependent variable and all its derivatives appear
in a linear fashion (i.e. they are not multiplied
together or squared for example.
 2u
Examples: (Linear)
t
(Non-linear) u
(Linear)
(Non-linear)
(Non-linear)
(Linear)
 2u

u
0
t
 2u
 2u
x 2
x
2
y
y
2
e
 2u
x
2
 sin t
0
2
u
u
x  y  u2  0
x
y
 2 u  u 
y


sin
u

e
2  x 
 
x
 2u
 2u  2u
2
 2  sin x
2
 x y  y
x
2
(Non-linear)
2
t
u
 u 
1
  u
y
 x 
(4)
Homogeneity
A PDE is called homogenous if after writing the
terms in order, the right hand side is zero.
Examples:
 2u  2u
 2  f ( x, y )
(Non-homogeneous)
2
x
(Homogeneous)
(Homogenous)
y
 2u
u

0
2
t
x
 2u
t
2

 2u
x
2
u
Examples
(Non-homogeneous)
(Homogeneous)
u u

 u5
x  t
 ( u  5)  ( u  5)

 u5
x
t
(5)
Types of Coefficients
If the coefficients in front of each term involving the dependent
variable and its derivatives are independent of the variables
(dependent or independent), then that PDE is one with constant
coefficients.
Examples
(Variable coefficients)
 2u
x
2
(C constant; constant coefficients)
x
2

u
2
y
 2u
x
2
2
0
C
 2u
t
2
0
(6) Canonical forms for 2nd order PDEs
(Linear)
 2u
 2u
 2u
u
u
A 2 B
 C 2  D  E  Fu  G
 x y
x
y
x
y
(Standard Form)
where A, B, C, D, E, F, and G are either real
constants or real-valued functions of x and/or y.
B 2  4AC  0 
PDE is Elliptic
B 2  4AC  0 
PDE is Parabolic
B 2  4AC  0 
PDE is Hyperbolic
Parabolic PDE  solution “propagates” or diffuses
Hyperbolic PDE  solution propagates as a wave
Elliptic PDE  equilibrium
This terminology of elliptic, parabolic, and hyperbolic, reflect the analogy
between the standard form for the linear, 2nd order PDE and conic sections
encountered in analytical geometry:
Ax 2  Bxy  Cy 2  Dx  Ey  F  0
2
B
 4AC  0 one obtains the equation for an ellipse,
for which when
when B 2  4AC  0 one obtains the equation for a parabola, and when
B 2  4AC  0
one gets the equation for a hyperbola.
Examples
 2u
2
 2u
0
x
y
(a)
Here, A=1, B=0, C=2, D=E=F=G=0  B24AC = 0 - 4(1)(2) = -8 < 0  this equation is elliptic.
2
 2u
2
2
 2u
0
x
y
(b)
Here, A=1, B=0, C=-2, D=E=F=G=0  B2-4AC = 0 - 4(1)(-2) = 8 > 0
 this equation is hyperbolic.
2
2
 2u
u
2 0
2
y
x
(c)
Here, A=1, B=0, E=-2, C=D=F=G=0  B2-4AC = 0 - 4(1)(0) = 0 
this equation is parabolic.
Examples
 2u
 2u  2u
4
 2 0
2
 x y  y
x
(d)
Here, A=1, B=-4, C=1, D=E=F=G=0  B2-4AC =
16 - 4(1)(1) = 12 > 0  this equation is hyperbolic.
 2u
 2u
 2u
3 2 4
5 2 0
 x y
x
y
(e)
Here, A=3, B=-4, C=-5, D=E=F=G=0  B2-4AC = 16 - 4(3)(-5) = 76 > 0  this
equation is hyperbolic.
 2u
 2u
 2u
u
u
3 2 4
 5 2  8  9  6u  27e xy
 x y
x
y
y
(f) x
Here, A=3, B=-4, C=-5, D=8, E=-
9, F=6, G=27exy  B2-4AC = 16 - 4(3)(-5) > 0  this equation is hyperbolic.
Examples
y
 2u

 2u
0
x
y
(g)
Here, A=y, B=0, C=-1, D=E=F=G=0  B2-4AC = 0 - 4(y)(-1) = 4y 
for y>0, this equation is hyperbolic; for y=0, this equation is
parabolic; for y<0, this equation is elliptic.
2
2
y
Hyperbolic
x
Elliptic
Parabolic
Examples
 2u
2
 2u
2  u
 2x
 (1  y ) 2  0
2
 x y
x
y
(h)
Here, A=1, B=2x, C=1-y2, D=E=F=G=0  B2-4AC = 4x2 - 4(1)(1-y2)
= 4x2+4y2-4 or x2+y2 >,=,< 0
y
Hyperbolic
Elliptic
x
Parabolic on
surface of circle
Examples
 2u
 2u
u
u

y

x

y
u0
2
 x y
x
y
x
(i)
Here, A=1, B=-y, C=0, D=E=F=G=0  B2-4AC = y2  for y=0, this
equation is parabolic; for y0, this equation is hyperbolic.
y
Hyperbolic
x
Hyperbolic
Parabolic
Example
(j)
 2u
 2u
2
(sin x ) 2  (sin 2x )
 (cos x ) 2  x
 x y
x
y
2
 2u
Here, A=sin2x, B=sin2x, C=cos2x, D=E=F=G=0  B2-4AC =
sin22x-4sin2xcos2x = 4sin2xcos2x-4sin2xcos2x = 0  this
equation is parabolic everywhere.
Solution Technique
• Elliptic equations in engineering are typically used to
characterize steady-state, boundary value problems.
• For numerical solution of elliptic PDEs, the PDE is
transformed into an algebraic difference equation.
• Because of its simplicity and general relevance to
most areas of engineering, we will use a heated plate
as an example for solving elliptic PDEs.
The Laplacian Difference Equations/
 2T  2T
 2 0
2
Laplace Equation
x
y
 2T Ti 1, j  2Ti , j  Ti 1, j
O[(x)2]

x 2
x 2
 2T Ti , j 1  2Ti , j  Ti , j 1

O[(y)2]
2
2
y
y
Ti 1, j  2Ti , j  Ti 1, j Ti , j 1  2Ti , j  Ti , j 1

0
2
2
x
y
x  y
Ti 1, j  Ti 1, j  Ti , j 1  Ti , j 1  4Ti , j  0
Laplacian difference
equation.
Holds for all interior points
• In addition, boundary conditions along the edges must be
specified to obtain a unique solution.
• The simplest case is where the temperature at the boundary is
set at a fixed value, Dirichlet boundary condition.
• A balance for node (1,1) is:
T21  T01  T12  T10  4T11  0
T01  75
T10  0
 4T11  T12  T21  0
• Similar equations can be developed for other interior points to
result a set of simultaneous equations.
• The result is a set of nine simultaneous equations with nine
unknowns:
4T11
 T21
 T12
 T11  4T21  T13
 T21
 75
 T22
 4T31
 T11
0
 T32
 4T12  T22
 T21
 T12
 T31
 50
 T13
 4T22  T32
 T22
 75
 T23
 4T32
 T12
 T22
 T32
0
 T33
 50
 4T13  T23
 175
 T13
 100
 4T23  T33
 T23
 4T33  150