Contributed by:
SECTIONS:
1. Partial Differential Equations (PDEs).
2. What is a PDE?
3. Examples of Important PDEs.
4. Classification of PDEs.
1.
Methods
Topic 9
Partial Differential Equations
(PDEs)
Lectures 37-39
KFUPM
(Term 101)
Section 04
Read 29.1-29.2 & 30.1-30.4
CISE301_Topic9 1
2.
Lecture 37
Partial Differential
Equations
Partial Differential Equations (PDEs).
What is a PDE?
Examples of Important PDEs.
Classification of PDEs.
CISE301_Topic9 2
3.
Partial Differential
A partial differential equation (PDE) is an
equation that involves an unknown function
and its partial derivatives.
Example :
2 u ( x, t ) u ( x , t )
2
x t
PDE involves two or more independent variables
(in the example x and t are independent variables)
CISE301_Topic9 3
4.
2 u ( x, t )
u xx
x 2
2 u ( x, t )
u xt
x t
Order of the PDE order of the highest order derivative.
CISE301_Topic9 4
5.
Linear PDE
A PDE is linear if it is linear in the unknown
function and its derivatives
Example of linear PDE :
2 u xx 1 u xt 3 utt 4 u x cos(2t ) 0
2 u xx 3 ut 4 u x 0
Examples of Nonlinear PDE
2 u xx u xt 2 3 utt 0
u xx 2 u xt 3 ut 0
2 u xx 2 u xt ut 3 ut 0
CISE301_Topic9 5
6.
Representing the Solution of a
PDE
(Two Independent Variables)
Three main ways to represent the solution
T ( x1 , t1 ) T=5.2
t1 T=3.5
x1
Different curves are Three dimensional The axis represent
used for different plot of the function the independent
values of one of the T(x,t) variables. The value
independent variable
of the function is
displayed at grid
points
CISE301_Topic9 6
7.
Heat Equation Different curve is
used for each value
of t
ice ice Temperature Temperature at
x different x at t=0
Thin metal rod insulated
everywhere except at the
edges. At t =0 the rod is
placed in ice
2 T ( x, t ) T ( x, t ) Position x
2
0
x t Temperature at
T (0, t ) T (1, t ) 0 different x at t=h
T ( x,0) sin( x)
CISE301_Topic9 7
8.
Examples of PDEs
PDEs are used to model many systems in
many different fields of science and
engineering.
Important Examples:
Laplace Equation
Heat Equation
Wave Equation
CISE301_Topic9 8
9.
Laplace Equation
2u ( x , y , z ) 2u ( x , y , z ) 2u ( x , y , z )
2
2
2
0
x y z
Used to describe the steady state distribution of
heat in a body.
Also used to describe the steady state
distribution of electrical charge in a body.
CISE301_Topic9 9
10.
Heat Equation
u( x, y , z, t ) 2u 2 u 2u
2 2 2
t x y z
The function u(x,y,z,t) is used to represent
the temperature at time t in a physical body
at a point with coordinates (x,y,z)
is the thermal diffusivity. It is sufficient to
consider the case = 1.
CISE301_Topic9 10
11.
Simpler Heat Equation
2
T ( x, t ) T ( x, t )
x
t x 2
T(x,t) is used to represent the temperature
at time t at the point x of the thin rod.
CISE301_Topic9 11
12.
Wave Equation
2u ( x , y , z , t ) 2 2
u 2
u 2
u
2
c 2 2 2
t x y z
The function u(x,y,z,t) is used to represent the
displacement at time t of a particle whose
position at rest is (x,y,z) .
The constant c represents the propagation
speed of the wave.
CISE301_Topic9 12
13.
Classification of PDEs
Linear Second order PDEs are important
sets of equations that are used to model
many systems in many different fields of
science and engineering.
Classification is important because:
Each category relates to specific engineering
problems.
Different approaches are used to solve these
categories.
CISE301_Topic9 13
14.
Linear Second Order PDEs
A second order linear PDE (2 - independent variables)
A u xx B u xy C u yy D 0,
A, B, and C are functions of x and y
D is a function of x, y , u, u x , and u y
is classified based on (B 2 4 AC) as follows :
B 2 4 AC 0 Elliptic
B 2 4 AC 0 Parabolic
B 2 4 AC 0 Hyperbolic
CISE301_Topic9 14
15.
Linear Second Order PDE
Examples (Classification)
2u ( x , y ) 2u ( x , y )
Laplace Equation 2
2
0
x y
A 1, B 0, C 1 B 2 4 AC 0
Laplace Equation is Elliptic
One possible solution : u( x, y ) e x sin y
u x e x sin y , u xx e x sin y
u y e x cos y , u yy e x sin y
u xx u yy 0
CISE301_Topic9 15
16.
Linear Second Order PDE
Examples (Classification)
2u ( x , t ) u ( x , t )
Heat Equation 2
0
x t
A , B 0, C 0 B 2 4 AC 0
Heat Equation is Parabolic
______________________________________
2 2
2 u ( x , t ) u( x, t )
Wave Equation c 2
2
0
x t
A c 2 0, B 0, C 1 B 2 4 AC 0
Wave Equation is Hyperbolic
CISE301_Topic9 16
17.
Boundary Conditions for
To uniquely specify a solution to the PDE,
a set of boundary conditions are needed.
Both regular and irregular boundaries are
possible. t
2u( x, t ) u ( x, t )
Heat Equation : 2
0
x t region of
interest
u(0, t ) 0
u(1, t ) 0
1 x
u( x,0) sin( x )
CISE301_Topic9 17
18.
The Solution Methods for
Analytic solutions are possible for simple
and special (idealized) cases only.
To make use of the nature of the
equations, different methods are used to
solve different classes of PDEs.
The methods discussed here are based on
the finite difference technique.
CISE301_Topic9 18
19.
Lecture 38
Parabolic
Equations
Parabolic Equations
Heat Conduction Equation
Explicit Method
Implicit Method
Cranks Nicolson Method
CISE301_Topic9 19
20.
Parabolic Equations
A second order linear PDE (2 - independent variables x , y )
A u xx B u xy C u yy D 0,
A, B, and C are functions of x and y
D is a function of x, y, u, u x , and u y
is parabolic if B 2 4 AC 0
CISE301_Topic9 20
21.
Parabolic Problems
T ( x, t ) 2 T ( x, t )
Heat Equation :
t x 2
T (0, t ) T (1, t ) 0
T ( x,0) sin( x ) ice ice
x
* Parabolic problem ( B 2 4 AC 0)
* Boundary conditions are needed to uniquely specify a solution.
CISE301_Topic9 21
22.
Finite Difference
Divide the interval x into sub-intervals,
each of width h
Divide the interval t into sub-intervals,
each of width k
t
A grid of points is used for
the finite difference solution
Ti,j represents T(xi, tj)
Replace the derivates by
x
finite-difference formulas
CISE301_Topic9 22
23.
Finite Difference Methods
Replace the derivatives by finite difference formulas
2T
Central Difference Formula for 2 :
x
2T ( x, t ) Ti 1, j 2Ti , j Ti 1, j Ti 1, j 2Ti , j Ti 1, j
2
2
x ( x ) h2
T
Forward Difference Formula for :
t
T ( x, t ) Ti , j 1 Ti , j Ti , j 1 Ti , j
t t k
CISE301_Topic9 23
24.
Solution of the Heat
• Two solutions to the Parabolic Equation
(Heat Equation) will be presented:
1. Explicit Method:
Simple, Stability Problems.
2. Crank-Nicolson Method:
Involves the solution of a Tridiagonal system
of equations, Stable.
CISE301_Topic9 24
25.
Explicit Method
T ( x, t ) 2T ( x, t )
t x 2
T ( x, t k ) T ( x, t ) T ( x h, t ) 2T ( x, t ) T ( x h, t )
k h2
k
T ( x, t k ) T ( x, t ) 2 T ( x h, t ) 2T ( x, t ) T ( x h, t )
h
k
Define 2
h
T ( x, t k ) T ( x h, t ) (1 2 ) T ( x, t ) T ( x h, t )
CISE301_Topic9 25
26.
Explicit Method
How Do We Compute?
T ( x, t k ) T ( x h, t ) (1 2 ) T ( x, t ) T ( x h, t )
T(x,t+k)
T(x-h,t) T(x,t) T(x+h,t)
CISE301_Topic9 26
27.
Convergence and
T ( x, t k ) can be computed directly using :
T ( x, t k ) T ( x h, t ) (1 2 ) T ( x, t ) T ( x h, t )
Can be unstable errors are magnified
1 h2
To guarantee stability, (1 2 ) 0 k
2 2
This means that k is much smaller than h
This makes it slow.
CISE301_Topic9 27
28.
Convergence and Stability of the
Convergence
The solutions converge means that the
solution obtained using the finite difference
method approaches the true solution as the
steps x and t approach zero.
Stability:
An algorithm is stable if the errors at each
stage of the computation are not magnified
as the computation progresses.
CISE301_Topic9 28
29.
Example 1: Heat
Solve the PDE :
2u(x,t) u(x,t)
2
0
x t
u(0, t ) u(1, t ) 0
ice ice
u( x,0) sin( x )
x
Use h 0.25, k 0.25 to find u( x, t ) for x [0,1], t [0,1]
k
2 4
h
CISE301_Topic9 29
30.
Example 1
2 u ( x , t ) u ( x , t )
2
0
x t
u ( x h , t ) 2u ( x , t ) u ( x h , t ) u ( x , t k ) u ( x , t )
2
0
h k
16 u( x h, t ) 2u ( x, t ) u( x h, t ) 4 u( x, t k ) u( x, t ) 0
u( x, t k ) 4 u( x h, t ) 7 u ( x, t ) 4 u ( x h, t )
CISE301_Topic9 30
31.
Example 1
u( x, t k ) 4 u( x h, t ) 7 u( x, t ) 4 u( x h, t )
t=1.0 0 0
t=0.75 0 0
t=0.5 0 0
t=0.25 0 0
t=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0 x=0.25 x=0.5 x=0.75 x=1.0
CISE301_Topic9 31
32.
Example 1
u(0.25,0.25) 4 u(0,0) 7 u(0.25,0) 4 u(0.5,0)
0 7 sin( / 4) 4 sin( / 2) 0.9497
t=1.0 0 0
t=0.75 0 0
t=0.5 0 0
t=0.25 0 0
t=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0 x=0.25 x=0.5 x=0.75 x=1.0
CISE301_Topic9 32
33.
Example 1
u(0.5,0.25) 4 u (0.25,0) 7 u(0.5,0) 4 u(0.75,0)
4 sin( / 4) 7 sin( / 2) 4 sin(3 / 4) 0.1716
t=1.0 0 0
t=0.75 0 0
t=0.5 0 0
t=0.25 0 0
t=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0 x=0.25 x=0.5 x=0.75 x=1.0
CISE301_Topic9 33
34.
Remarks on Example 1
The obtained results are probably not accurate
because : 1 2 7
For accurate results : 1 2 0
h 2 (0.25)2
One needs to select k 0.03125
2 2
k
For example, choose k 0.025, then 2 0.4
h
CISE301_Topic9 34
35.
Example 1 – cont’d
u( x, t k ) 0.4 u( x h, t ) 0.2 u( x, t ) 0.4 u( x h, t )
t=0.10 0 0
t=0.075 0 0
t=0.05 0 0
t=0.025 0 0
t=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0 x=0.25 x=0.5 x=0.75 x=1.0
CISE301_Topic9 35
36.
Example 1 – cont’d
u(0.25,0.025) 0.4 u(0,0) 0.2 u(0.25,0) 0.4 u(0.5,0)
0 0.2 sin( / 4) 0.4 sin( / 2) 0.5414
t=0.10 0 0
t=0.075 0 0
t=0.05 0 0
t=0.025 0 0
t=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0 x=0.25 x=0.5 x=0.75 x=1.0
CISE301_Topic9 36
37.
Example 1 – cont’d
u(0.5,0.025) 0.4 u(0.25,0) 0.2 u(0.5,0) 0.4 u(0.75,0)
0.4 sin( / 4) 0.2 sin( / 2) 0.4 sin( 3 / 4) 0.7657
t=0.10 0 0
t=0.075 0 0
t=0.05 0 0
t=0.025 0 0
t=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0 x=0.25 x=0.5 x=0.75 x=1.0
CISE301_Topic9 37
38.
Crank-Nicolson Method
The method involves solving a Tridiagona l system of linear equations.
The method is stable (No magnification of error).
We can use larger h, k (compared to the Explicit Method).
CISE301_Topic9 38
39.
Crank-Nicolson Method
Based on the finite difference method
1. Divide the interval x into subintervals of width h
2. Divide the interval t into subintervals of width k
3. Replace the first and second partial derivatives with their
backward and central difference formulas respectively :
u( x, t ) u( x, t ) u ( x, t k )
t k
2 u ( x , t ) u ( x h , t ) 2u ( x , t ) u ( x h , t )
2
x h2
CISE301_Topic9 39
40.
Crank-Nicolson Method
2 u ( x , t ) u ( x , t )
Heat Equation : 2
becomes
x t
u ( x h , t ) 2u ( x , t ) u ( x h , t ) u ( x , t ) u ( x , t k )
2
h k
k
2
u( x h, t ) 2u( x, t ) u( x h, t ) u( x, t ) u( x, t k )
h
k k k
2 u ( x h, t ) (1 2 2 ) u ( x, t ) 2 u( x h, t ) u ( x, t k )
h h h
CISE301_Topic9 40
41.
Crank-Nicolson Method
k
Define 2 then Heat equation becomes :
h
u( x h, t ) (1 2 ) u( x, t ) u ( x h, t ) u( x, t k )
u(x-h,t) u(x,t) u(x+h,t)
u(x,t - k)
CISE301_Topic9 41
42.
Crank-Nicolson Method
The equation :
u( x h, t ) (1 2 ) u( x, t ) u( x h, t ) u( x, t k )
can be rewritten as :
ui 1, j (1 2 ) ui , j ui 1, j ui , j 1
and can be expanded as a system of equations (fix j 1) :
u0,1 (1 2 ) u1,1 u2,1 u1,0
u1,1 (1 2 ) u2,1 u3,1 u2,0
u2,1 (1 2 ) u3,1 u4,1 u3,0
u3,1 (1 2 ) u4,1 u5,1 u4,0
CISE301_Topic9 42
43.
Crank-Nicolson Method
u( x h, t ) (1 2 ) u( x, t ) u( x h, t ) u( x, t k )
can be expressed as a Tridiagonal system of equations :
1 2 u1,1 u1,0 u0,1
1 2 u u
2,1 2, 0
1 2 u3,1 u3,0
u u u
1 2 4,1 4,0 5,1
where u1,0 , u2,0 , u3,0 , and u4,0 are the initial temperature values
at x x0 h, x0 2h, x0 3h, and x0 4h
u0,1 and u5,1 are the boundary values at x x0 and x0 5h
CISE301_Topic9 43
44.
Crank-Nicolson Method
The solution of the tridiagonal system produces :
The temperature values u1,1, u2,1, u3,1 , and u4,1 at t t0 k
To compute the temperature values at t t0 2k
Solve a second tridiagonal system of equations ( j 2)
1 2 u1,2 u1,1 u0,2
1 2 u u
2, 2 2 ,1
1 2 u3,2 u3,1
1 2 u u
4,2 4,1 u5, 2
To compute u1,2 , u2,2 , u3,2 , and u4,2
Repeat the above step to compute temperature values at t0 3k , etc.
CISE301_Topic9 44
45.
Example 2
Solve the PDE :
2u( x, t ) u( x, t )
2
0
x t
u(0, t ) u(1, t ) 0
u( x,0) sin( x )
Solve using Crank - Nicolson method
Use h 0.25, k 0.25 to find u( x, t ) for x [0,1], t [0,1]
CISE301_Topic9 45
46.
Example 2
Crank-Nicolson Method
2 u ( x , t ) u ( x , t )
2
0
x t
u ( x h , t ) 2u ( x , t ) u ( x h , t ) u ( x , t ) u ( x , t k )
2
h k
16 u ( x h, t ) 2u( x, t ) u( x h, t ) 4 u( x, t ) u( x, t k ) 0
k
Define 2 4
h
4 u ( x h, t ) 9 u ( x, t ) 4 u( x h, t ) u ( x, t k )
4 ui 1, j 9 ui , j 4 ui 1, j ui , j 1
CISE301_Topic9 46
47.
Example 2
4u0,1 9u1,1 4u2,1 u1,0 9u1,1 4u2,1 sin( / 4)
4u1,1 9u2,1 4u3,1 u2,0 4u1,1 9u2,1 4u3,1 sin( / 2)
4u2,1 9u3,1 4u4,1 u3,0 4u2,1 9u3,1 sin(3 / 4)
u1,4 u2,4 u3,4
t4=1.0 0 0
u1,3 u2,3 u3,3
t3=0.75 0 0
u1,2 u2,2 u3,2
t2=0.5 0 0
u1,1 u2,1 u3,1
t1=0.25 0 0
t0=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x0=0.0 x1=0.25 x2=0.5 x3=0.75 x4=1.0
CISE301_Topic9 47
48.
Example 2
Solution of Row 1 at t1=0.25 sec
The Solution of the PDE at t1 0.25 sec is the solution
of the following tridiagonal system of equations :
9 4 u1,1 sin(0.25 )
4 9 4 u sin(0.5 )
2,1
4 9 u3,1 sin(0.75 )
u1,1 0.21151
u2,1 0.29912
u3,1 0.21151
CISE301_Topic9 48
49.
Example 2:
Second Row at t2=0.5 sec
4u0,2 9u1,2 4u2,2 u1,1 9u1,2 4u2,2 0.21151
4u1,2 9u2,2 4u3,2 u2,1 4u1,2 9u2,2 4u3,2 0.29912
4u2,2 9u3,2 4u4,2 u3,1 4u2,2 9u3,2 0.21151
u1,4 u2,4 u3,4
t4=1.0 0 0
u1,3 u2,3 u3,3
t3=0.75 0 0
u1,2 u2,2 u3,2
t2=0.5 0 0
u1,1 u2,1 u3,1
t1=0.25 0 0
t0=0 0 0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x0=0.0 x1=0.25 x2=0.5 x3=0.75 x4=1.0
CISE301_Topic9 49
50.
Example 2
Solution of Row 2 at t2=0.5 sec
The Solution of the PDE at t2 0.5 sec is the solution
of the following tridiagonal system of equations :
9 4 u1, 2 u1,1 0.21151
4 9 4 u u 0.29912
2, 2 2,1
4 9 u3,2 u3,1 0.21151
u1, 2 0.063267
u2, 2 0.089473
u3,2 0.063267
CISE301_Topic9 50
51.
Example 2
Solution of Row 3 at t3=0.75 sec
The Solution of the PDE at t3 0.75 sec is the solution
of the following tridiagonal system of equations :
9 4 u1,3 u1, 2 0.063267
4 9 4 u u 0.089473
2,3 2, 2
4 9 u3,3 u3,2 0.063267
u1,3 0.018924
u2,3 0.026763
u3,3 0.018924
CISE301_Topic9 51
52.
Example 2
Solution of Row 4 at t4=1 sec
The Solution of the PDE at t4 1 sec is the solution
of the following tridiagonal system of equations :
9 4 u1, 4 u1,3 0.018924
4 9 4 u u 0.026763
2, 4 2, 3
4 9 u3,4 u3,3 0.018924
u1, 4 0.0056606
u2,4 0.0080053
u3, 4 0.0056606
CISE301_Topic9 52
53.
The Explicit Method:
•One needs to select small k to ensure stability.
•Computation per point is very simple but many
points are needed.
Cranks Nicolson:
• Requires the solution of a Tridiagonal system.
• Stable (Larger k can be used).
CISE301_Topic9 53
54.
Lecture 39
Elliptic
Equations
Elliptic Equations
Laplace Equation
Solution
CISE301_Topic9 54
55.
Elliptic Equations
A second order linear PDE (2 - independent variables x , y )
A u xx B u xy C u yy D 0,
A, B, and C are functions of x and y
D is a function of x, y , u, ux , and u y
is Elliptic if B 2 4 AC 0
CISE301_Topic9 55
56.
Laplace Equation
Laplace equation appears in several
engineering problems such as:
Studying the steady state distribution of heat in a
body.
Studying the steady state distribution of electrical
charge in a body.
2 2
T ( x, y ) T ( x, y )
2
2
f ( x, y )
x y
T : steady state temperatu re at point (x, y)
f ( x, y ) : heat source (or heat sink)
CISE301_Topic9 56
57.
Laplace Equation
2 T ( x, y ) 2 T ( x , y )
2
2
f ( x, y )
x y
A 1, B 0, C 1
B 2 4 AC 4 0 Elliptic
Temperature is a function of the position (x and y)
When no heat source is available f(x,y)=0
CISE301_Topic9 57
58.
Solution Technique
A grid is used to divide the region of
interest.
Since the PDE is satisfied at each point in
the area, it must be satisfied at each point
of the grid.
A finite difference approximation is
obtained at each grid point.
2 T ( x, y ) Ti 1, j 2Ti , j Ti 1, j 2 T ( x, y ) Ti , j 1 2Ti , j Ti , j 1
2
,
x 2
x y 2
y 2
CISE301_Topic9 58
59.
Solution Technique
2 T ( x, y ) Ti 1, j 2Ti , j Ti 1, j
2
2
,
x x
2 T ( x, y ) Ti , j 1 2Ti , j Ti , j 1
2
2
y y
2 T ( x, y ) 2 T ( x, y )
2
2
0
x y
is approximated by :
Ti 1, j 2Ti , j Ti 1, j Ti , j 1 2Ti , j Ti , j 1
2
2
0
x y
CISE301_Topic9 59
60.
Solution Technique
Ti 1, j 2Ti , j Ti 1, j Ti , j 1 2Ti , j Ti , j 1
2
2
0
x y
( Laplacian Difference Equation)
Assume : x y h
Ti 1, j Ti 1, j Ti , j 1 Ti , j 1 4Ti , j 0
CISE301_Topic9 60
61.
Solution Technique
Ti , j 1
Ti 1, j Ti , j Ti 1, j
Ti , j 1
Ti 1, j Ti 1, j Ti , j 1 Ti , j 1 4Ti , j 0
CISE301_Topic9 61
62.
It is required to determine the steady
state temperature at all points of a heated
sheet of metal. The edges of the sheet are
kept at a constant temperature: 100, 50,
0, and 75 degrees. 100
75 50
The sheet is divided
to 5X5 grids. 0
CISE301_Topic9 62
63.
Known
Example To be determined
T1, 4 100 T2, 4 100 T3, 4 100
T1,3 T2,3
T0,3 75 T3,3
T4,3 50
T1, 2 T2, 2 T3, 2
T0, 2 75 T4, 2 50
T1,1 T2,1 T3,1
T0,1 75 T4,1 50
T1, 0 0 T2, 0 0 T3, 0 0
CISE301_Topic9 63
64.
Known
First Equation To be determined
T1, 4 100 T2, 4 100
T1,3 T2,3
T0,3 75
T1, 2 T2, 2
T0, 2 75
T0,3 T1, 4 T1, 2 T2,3 4T1,3 0
75 100 T1, 2 T2,3 4T1,3 0
CISE301_Topic9 64
65.
Known
Another Equation To be determined
T1, 4 100 T2, 4 100 T3, 4 100
T1,3 T2,3 T3,3
T1, 2 T2, 2 T3, 2
T1,3 T2, 4 T3,3 T2,2 4T2,3 0
T1,3 100 T3,3 T2,2 4T2,3 0
CISE301_Topic9 65
66.
The Rest of the Equations
4 1 0 1 T1,1 75
1 4 1 0 1 T2,1 0
0 1 4 0 0 1 T 50
3,1
1 0 0 4 1 0 1 T1, 2 75
1 0 1 4 1 0 1 T 0
2, 2
1 0 1 4 0 0 1 T3,2 50
1 0 0 4 1 0 T 175
1,3
1 0 1 4 1 T2,3 100
1 0 1 4 T
3,3 150
CISE301_Topic9 66
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