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ENGRD 241 Lecture Notes              Section 8: Partial Differential Equations         Page 8-1 of 8-18

PARTIAL DIFFERENTIAL EQUATIONS (PDE's) (C&C PART 8)
Mathematics is the language of science. PDEs are the analytical expression of processes that occur
across space and/or time. (See Chapra and Canale, Chap. 32)

• A PDE is an equation w/ derivatives of an unknown function with respect to 2
or more independent variables
• Describe the behavior of critical engineering phenomenon
– Wave propagation
– Fluid flow (air or liquid)
– Vibration
– Mechanics of solids (e.g. stress or strain in a machine part or a structure)
– Heat flow and distribution
– Electric fields and potentials
– Diffusion of chemical in air or water
– Electromagnetism and quantum mechanics
– Weather Prediction

Learning Objectives
(1) Be able to distinguish between the 3 classes of 2nd-order linear PDEs.
Know physical problems each class represents and physical/mathematical
characteristics of each.
(2) Be able to describe differences between finite-difference and finite-element
methods for solving PDEs.
(3) Be able to solve Elliptic (Laplace/Poisson) PDEs using finite differences.
(4) Be able to solve Parabolic (Heat/Diffusion) PDEs using finite differences.

ENGRD 241 Focus: Linear 2nd-Order PDE's
u is the dependent variable and x and y are the independent variables
2 u        2 u   2 u
A          B        C 2 D 0
x 2        xy   y
u u
u(x,y), A(x,y), B(x,y), C(x,y), and D(x,y,u,     , )
x y
The PDE is nonlinear if A, B, or C include u, u/x, or u/y,
or if D is nonlinear in u and/or its first derivatives.

Classification (C&C, pp. 813-816)
Similar to the technique used to obtain an analytical solution,
B2 – 4AC < 0 ––––> Elliptic               (e.g. Laplace Eq.)
2
B – 4AC = 0 ––––> Parabolic               (e.g. Heat Eq.)
B2 – 4AC > 0 ––––> Hyperbolic             (e.g. Wave Eq.)
Each category describes different phenomena.
Mathematical properties correspond to those phenomena.
ENGRD 241 Lecture Notes                                                    Section 8: Partial Differential Equations                                     Page 8-2 of 8-18

2
Elliptic (B – 4AC < 0) [steady-state in time] (C&C Chap. 29, p. 820)
• typically characterize steady-state systems (no time derivative)
– temperature           – torsion
– pressure              – membrane displacement
– electrical potential

• closed domain with boundary conditions given in terms of:
u                u     u
u(x,y)                     (in terms of    and    )
                x     y
Typical examples:
         0
 2u  2u                          Laplace Eq.
 u
2
 2               u u
x 2 y     D(x, y, u, x , y ) Poisson Eq.


A = 1, B = 0, C = 1                                          ––> B2 – 4AC = – 4 < 0

Boundary Conditions (BC’s) for Elliptic Equations:
Dirichlet:     u provided along all of the edge
u
Neumann:                                                     (normal derivative) provided along all of the edge

Mixed:                                                u provided for some of the edge
u
for the remainder of the edge

Elliptical PDE’s are analogous to Boundary Value ODE’s
u(x,y=1) given on boundary, or u/y
u(x=0,y) given on the boundary, or u/x

u(x=1,y) given on the boundary, or u/x

u(x,y)
estimated from
Laplace Equation
y

u(x,y=0) given on boundary, or u/y
x
ENGRD 241 Lecture Notes              Section 8: Partial Differential Equations         Page 8-3 of 8-18

2
Parabolic (B – 4AC = 0) [first derivative in time] (C&C Chap. 30, p. 840)
• variation in both space (x,y) and time, t
• typically provided
– initial values:          u(x,y,t = 0)
– boundary conditions:     u(x = xo,y = yo, t) for all t (or Neumann)
u(x = xf, y = yf, t) for all t (or Neumann)
• all changes are propagated forward in time, i.e., nothing goes backward in time;
changes propagated across space at decreasing amplitude.
• Typical example: Heat Conduction or Diffusion Equation
u     2u        u
1D:           k 2  D(x, u, )
t    x          x
2
A = k, B = 0, C = 0  B – 4AC = 0

u       2u  2u          u u
2D:           k  2  2   D(x, y, u, , )
t      x
      y           x y
= k2u + D

u(x=0,t)            Estimate u(x,t)            u(x=1,t)
given                   from                   given
t      on boundary              the Heat              on boundary
for all t             Equation                 for all t

u(x,t =0) given on
boundary
as initial conditions for or
x
u/t

c(x,t) = concentration
at time, t, and distance,   x.
• An elongated reactor with a single entry and exit point and uniform cross sectional area A.
• A mass balance is developed for a finite segment Δx along the tank's longitudinal axis to
derive a differential equation for concentration. (V = A Δx):

c                     c(x)           c(x)      c(x)  c(x) 
V       Q c(x)  Q c(x)       x   D A         DA                x  –            kVc(x)
t                      x              x        x     x x       
Flow in        Flow out        Dispersion in        Dispersion out             Decay reaction
c       c       c Q c
2
As t and  x  0,               D 2         kc
t       t     x   A x
ENGRD 241 Lecture Notes                                                   Section 8: Partial Differential Equations                                               Page 8-4 of 8-18

2
Hyperbolic (B – 4AC > 0) [second derivative in time] (C&C, pp. 814 and 816)
• variation in both space (x,y) and time, t
• requires:
– two initial values:                                                u(x,y,t = 0)
u
(x,y,t = 0) "initial velocity"
t
– boundary conditions:      u(x = xo,y = yo, for all t) (or Neumann)
u(x = xf,y = yf, for all t) (or Neumann)
• All changes are propagated forward in time, i.e., nothing goes back in time
• Typical example:Wave Equation

 2u       1   2u             u u 
1D:                                     2   D  x, y, u, ,   0
x 2       c2   t 
                 x t 

1                 4
A = 1, B = 0, C =                               –> B2 – 4AC =    > 0
c2               c2
NOW the PDE is 2nd ORDER IN TIME
• Models
– vibrating string
– water waves
– voltage change in a wire
u(1,t) given on the boundary for all t
u(0,t) given on the boundary for all t

Estimate u(x,t) from
Wave Equation                                                                  Specifying
u(1,t) provides
u(1,t)/t

x
u(x,t=0) & u/t given on
boundary as init. conds. for 2u/t2
ENGRD 241 Lecture Notes                          Section 8: Partial Differential Equations   Page 8-5 of 8-18

Numerical Methods for Solving PDE's:
Numerical methods for solving different types of PDE's reflect the different
character of the problems.

• Laplace -- solve all at once for steady state conditions.
• Parabolic (heat) and Hyperbolic (wave) equations --
Integrate initial conditions forward through time.
• Finite Difference (FD) Approaches (C&C Chs. 29 & 30)
Based on approximating solution at a finite # of points and using
finite divide differences to replace derivatives at each point.

• Finite Element (FE) Method (C&C Ch. 31, p. 857)
Based on approximating solution on an assemblage of simply shaped
(triangular, quadrilateral) finite pieces or "elements" which together
make up the (perhaps complexly shaped) domain.
In this course, we concentrate on FD applied to elliptic & parabolic eqns.

Solving Elliptic PDE's by FD Method (C&C Chap. 29, p. 820)
• Solve all at once.
• Liebmann Method:
– Based on BC’s and finite diff. approx., formulate system of equations
– Use Gauss-Seidel or other methods to solve system of equations
         0
 2u
         2u                                Laplace Eq.
 u
2
              u u
x 2 y2 D(x, y, u, , )                             Poisson Eq.
           x y
1. Discretize domain into grid of evenly spaced points (nodes)
2. For nodes where u is unknown, use FDD of O (h2)

2u          u i1, j  2u i, j  u i 1, j
                                     O(x 2 )
x   2
(x)   2

2u          u i, j1  2u i, j  u i, j1
                                     O(y 2 )
y   2
(y)   2

w/ x=y=h, substitute into main equation

 2u    2 u ui 1, j  u i 1, j  u i, j1  u i, j1  4ui, j
                                                           O(h 2 )
x 2
y  2
h  2

3. Using BC’s write n*m equations for u(xi=1:m,yj=1:n) or n*m unknowns

4. Solve this banded system with an efficient scheme. Using Gauss-Siedel
iteratively yields the Liebmann Method.
ENGRD 241 Lecture Notes                       Section 8: Partial Differential Equations                     Page 8-6 of 8-18

2               2
       u              u                       j+1
For the Laplace Equation:              2
           2
0
x              y
Then the Laplace “molecule” is, if x = y:                                        j

Ti+1,j + Ti-1,j + Ti,j+1 + Ti,j-1 + 4Ti,j = 0
j-1
i-1          i       i+1
• Primary (solve for first): u(x,y) = T(x,y) = temperature distribution
• Secondary (solve for second derivative):
T                 T
heat flux : q x   k     and q y   k     obtain by employing:
x                 y

T Ti1, j  Ti1, j                         T Ti,j1  T1,j1
                                            
x      2x                                  y      2y
then obtain resultant flux and direction θ (+ clockwise from +x axis):
 qy 
qn =    q2  q2 ;
x    y               = tan-1                            qx > 0
 qx 

 qy 
 = tan-1   +                       qx < 0 with  in radians
 qx 

C&C Example 29.2, consider point (1,1), i.e., lower left interior point:

T = 100o C
T      Ti1, j  Ti1, j
q x  k       k
x           2x
78.6 76.1 69.7
T = 75o C                                               T = 50o C              T      Ti, j1  Ti, j1
63.2 56.1 52.3                                  q y  k       k
y            2y
43.0 33.3 33.9
k  = 0.49 cal/s cm oC
T = 0o C

qx ~ -0.49 (33.3-75)/(210cm) = 1.02 cal/(cm2s)
qy ~ -0.49 (63.2-0)/(210cm) = -1.55 cal/(cm2s)

q n  q x 2  q y 2  1.022  1.552  1.86 cal /(cm 2  s)

 qy         1  1.55 
  tan 1         tan            56.7
 qx             1.02 
ENGRD 241 Lecture Notes           Section 8: Partial Differential Equations             Page 8-7 of 8-18

• Neumann B.C.'s (i.e., derivatives at edges provided)
– employ imaginary (“phantom”) points outside of domain (as with BV ODEs)
– use FD to obtain information at phantom point, i.e.,
T1,j + T-1,j + T0,j+1 + T0,j-1 – 4T0.j = 0            [*]
T                T     T1,j  T1,j
if given       then use          
x                x         2x
T
to obtain        T-1,j = T1,j – 2 x
x
Substituting into FD approximation of 2T  0 on the boundary [*]:
T
2T1,j – 2 x    + T0,j+1 + T0,j-1 – 4T0,j = 0
x

T T3,1 - T5,1
Example: Given a derivative (Neumann) BC at (4,1):          =
y    2Δy

T = 100o C                       The Laplace molecule for the point
(4,1) is:

T4,2 + T4,0 + T3,1 + T5,1  4T4,1 = 0
1,1    1,2    1,3
with the boundary condition:
T = 75o C                               T = 50   o
2,1    2,2    2,3       C                             T
T5,1  T3,1  2y
y
3,1    3,2    3,3                 = T3,1  0
Thus, the Laplace molecule becomes:
Insulated ==> T/ y = 0         T4,2 + T4,0 + 2T3,1  4T4,1 = 0

• Irregular boundaries
– use unevenly spaced molecules close to edge
– use finer mesh
– use Finite Element Method (C&C Chap. 31)
ENGRD 241 Lecture Notes             Section 8: Partial Differential Equations        Page 8-8 of 8-18

Solution of Parabolic PDE's by FD Method (C&C Chap. 30, p. 840)
• using BC's and finite difference approx. to formulate model
• integrate IC's forward through time
• we'll investigate, for parabolic systems only:
– explicit schemes: stability criterion (C&C, p. 841)
– implicit schemes
- Simple Implicit (C&C, p. 845)
- Crank-Nicolson (C&C, p. 849)
- Alternating Direction (A.D.I), 2D-space (C&C, p. 852)

Prototype problem, Heat Equation:
T    2T
1D           k                       Find T(x,t)
t    x 2
T     2T 2T 
2D            k                   Find T(x,y,t)
t     x

2
y 2 

Given initial temperature distribution as well as boundary temps. (or rate of
change of temp.) with
k
k =         = Coefficient of thermal diffusivity
C
k' = coef. of thermal conductivity
C = heat capacity
 = density

General Procedure for Solution of Parabolic PDE’s

1. Discretize domain into a grid of evenly spaced points (nodes)
2. Express derivatives in terms of Finite Difference Approximation of O(x2) in x
and O(t) in t
2T 2T           T
,       and      ––––> Finite Differences
 x 2  y2        t
3. Choose h = x = y & t and use IC's & BC's to solve problem by systematically

• Explicit Schemes (C&C 30.2, p. 841)
Express all future (t+t) values, T(x,t+t),
in terms of current (t) and previous (t–t)
information, which is known.

• Implicit Schemes (C&C 30.3 – 30.4, p. 845)
Express all future (t+t) values, T(x,t+t),
in terms of future (t+t) and current (t) information.
ENGRD 241 Lecture Notes              Section 8: Partial Differential Equations    Page 8-9 of 8-18

Notation:
Use subscript(s) to indicate spatial points.
Use superscript to indicate time: Tim1  T(xi , t m1 )

NOTE the notation change: C&C uses l (lower case L) instead of m, but m is less
likely to be confused with 1 (one).

Explicit Schemes for Solution of Parabolic PDEs (C&C Chap. 30.2, p. 841)

Express a future state, Tim1 only in terms of the present state Tim .
T    2T
For the Heat Eq.       k      , substitute for the derivatives:
t    x 2

2T   T m  2Tim  Tim1
 i1           
 O(x 2 ) Centered Difference Eqn.
x 2
(x) 2

T Tim1  Tim
            O(t)                   Forward Difference Eqn.
t    (t)

t         k  t
and solve for Tim1 with   k                           to obtain:
(x) 2       C ( x) 2

Tim1  Tim  (Tim1  2Tim  Tim1 ) = (1  2)Tim  (Tim1  Tim1)
                                           
ENGRD 241 Lecture Notes                   Section 8: Partial Differential Equations          Page 8-10 of 8-18

T    2T
C&C Example 30.1: Heat equation:                         k
t    x 2
t                          t
k = 0.82 cal/ s·cm·oC, 10-cm long rod,
t = 2 secs, x = 2.5 cm (# segs. = 4)
Heat Eqn.
I.C.'s:           T(0<x<10, t=0) = 0° C

100° C
Example

50° C
B.C.'s:           T(x=0, all t) = 100° C
T(x=10, all t) = 50° C
t
= k              = 0.262
(x)2

Tim1  Tim  (Tim1  2Tim  Tim1 )
                                              X=0            0° C     X = 10 cm

t=6
Left                                           Right
t=4
Bndry              26.2°     0°                 Bndry
t=2         100°C                                           50°C
t=0
0 °C
Initial temperature

From t = 0 secs. (m = 0), find results at t = 2 secs. (m = 1):

T11 = T10 + l(T00 +T10 +T20 ) = 0 + 0.262[100–2(0)+0] = 26.2°

T21 = T20 + l(T10 +T20 +T30 ) = 0 + 0.262[0–2(0)+0] = 0°

T31 = T30 + l(T20 +T30 +T40 ) = 0 + 0.262[0–2(0)+50] = 13.1°

t=6
Left              38.7°     10.3°     19.3°    Right
t=4
Bndry              26.2°     0°        13.1°    Bndry
t=2          100°C                                           50°C
t=0
0 °C
Initial temperature

From t = 2 secs. (m = 1), find results at t = 4 secs. (m = 2):

T12 = T11 + l(T01 +T11 +T21 ) = 26.2 + 0.262[100–2(26.2)+0] = 38.7°

T22 = T21 + l(T11 +T21 +T31 ) = 0 + 0.262[26.2–2(0)+13.1] = 10.3°

T32 = T31 + l(T21 +T31 +T41 ) = 13.1 + 0.262[0–2(13.1)+50] = 19.3°
ENGRD 241 Lecture Notes                           Section 8: Partial Differential Equations        Page 8-11 of 8-18

Solution of Parabolic PDE's by Explicit Schemes
• Disadvantage: low accuracy, O (t) accurate in time.
Subject to instability; must use "small" t 's ––> requires many steps !

Implicit Schemes for Parabolic PDEs

Express Tim1 in terms of Tjm 1 , Tim , and possibly also Tjm (in which j = i–1 and i+1 )
Represents spatial and time domain. For each new time, write m (# of interior
nodes) equations and simultaneously solve for m unknown values (banded
system).

T    2T
The 1-D Heat Equation:                   k
t    x 2
Simple Implicit Method (C&C Chap. 30.3, p. 845) substituting:

2T                               
Tim1 1  2Tim1  Tim1 1
                 
                                 O(x 2 )             Centered FDD
x   2
(x)   2

T Tim1  Tim
            O(t)                                           Backward FDD
t     t
results in:
t
                         
Tim1 1  (1  2)Tim1  Tim1 1  Tim
                                                 with  = k
(x)2
Requires IC's for case where m = 0, i.e., Ti0 = is given for all i
Requires BC's to provide values for 1st & last nodes (i = 0 and n+1) for all m.

Implicit Method
Tim  Tim1 1  1  2  Tim1  Tim1 1





t=6
m +1
t=4            Left                                          Right
m                                                Bndry                                          Bndry
i-1    i     i+1              t=2           100°C                                          50°C
t=0
0 °C
Initial temperature

- T2       = T1 + l + T0
m +1        m +1        m              m+1
At the Left boundary:                 (1+2)T1
- Ti+1
m +1               m +1         m +1      m
Away from boundary:                   -Ti-1      + (1+2)Ti                      = Ti
- Ti-1      = Ti + l + Ti+1
m +1         m +1     m             m +1
At the Right boundary:                (1+2)Ti
ENGRD 241 Lecture Notes           Section 8: Partial Differential Equations       Page 8-12 of 8-18

m = 3; t = 6

m = 2; t = 4      Left                                                  Right
Bndry                                                  Bndry
100°C                                                  50°C
m = 1; t = 2
T11       T21      T31      T41
m = 0; t = 0
0 °C
Initial temperature
With  = 0.4

At the Left boundary: (1+2)T11 - T21 = T10 + T01
1.8 T11 – 0.4 T21 = 0 + 0.4*100 =          40
Away from boundary:        -Ti-11 + (1+2)Ti1 – Ti+11 = Ti0
-0.4 T11 + 1.8 T21 – 0.4 T31 = 0 =     0
-0.4 T21 + 1.8 T31 – 0.4 T41 = 0 =      0
At the Right boundary:   (1+2)T31 – T21 = T30 + T41
1.8 Tim+1 – 0.4 Ti-1m+1 = 0 + 0.4*50) =      20

 1                     T11 
 1.8 -0.4 0   0   T1  40                27.6 
-0.4 1.8 -0.4 0  T 1   0                          
                  2  =  
T21  6.14 
 
                    =        
 0 -0.4 1.8 -0.4  T 1   0                 1
 T3   4.03
                  3
                           
0 -0.4 1.8   1  20               1  12.0 
          
0
T4 
                      T4 

m = 3; t = 6

m = 2; t = 4      Left                                                 Right
Bndry            T12       T22      T32      T42      Bndry
100°C                                                 50°C
m = 1; t = 2
23.6     6.14     4.03     12.0
m = 0; t = 0
0 °C
Initial temperature

At the Left boundary: (1+2)T12 - T22 = T11 + T02
1.8 T12 – 0.4 T22 = 23.6+ 0.4*100 =              78.5
Away from boundary:        – Ti-12 + (1+2)Ti2 – Ti+12 = Ti1
– 0.4*T12+ 1.8*T22 – 0.4*T32 = 6.14 =         6.14
– 0.4*T22 + 1.8*T32 – 0.4*T42 = 4.03 =        4.03
ENGRD 241 Lecture Notes                     Section 8: Partial Differential Equations   Page 8-13 of 8-18

At the Right boundary:         (1+2)T32 – T42 = T31 + T42
1.8*T32 – 0.4*T42 = 12.0 + 0.4*50 =              32.0

 2                                 T12 
 1.8 -0.4 0   0   T1  78.5                          38.5 
-0.4 1.8 -0.4 0  T 2  6.14                                    
                  2  =                            T22  14.1 
 
 2                              2=       
 0 -0.4 1.8 -0.4   T   4.03                        T3  9.83 
                     3
 0                            
0 -0.4 1.8   2  32.0                         2  20.0 
T4 
                                  T4  
 


Crank-Nicolson Method Implicit Method (C&C Chap. 30.4, p. 849)

Provides scheme which has 2nd-order accuracy in both space and time.
Averages 2nd-derivatives in space for Tm+1 and Tm.

2T                                                    
1  Tim1  2Tim  Tim1 Tim1 1  2Tim1  Tim1 1 
             
                                           
  O(x)
2
x 2       2       (x) 2
( x) 2



T Tim 1  Tim
             O(t 2 )              (central difference in time now)
t      t
t
resulting in, with  = k            :
(x )2
                         
Tim1 1  2(1  )Tim1  Tim1 1  Tim1  2(1  )Tim  Tim1
                                                       

Requires I.C.'s for case where m = 0, Ti0 = given.
Requires B.C.'s in order to write expression for T0 1 and Tn 1 .
m         m
1

xi-    xi      xi+
1               1
m+1
t         Left                                     Right
tm+1/    Bndry                                     Bndry
tm
2     100°C                                     50°C

0 °C
Initial temperature

                         
Tim1 1  2(1  )Tim1  Tim1 1  Tim1  2(1  )Tim  Tim1
                                                       
ENGRD 241 Lecture Notes                  Section 8: Partial Differential Equations    Page 8-14 of 8-18

Summary of Solution of Parabolic PDE's by Implicit Schemes
 Unconditionally stable,
 t choice governed only by accuracy. [ CN Error O(t2) ]
 May be able to take larger t  fewer steps

 More difficult calculations, especially for 2D and 3D spatially.
 For 1D spatially, effort  same as explicit because system is
tridiagonal.

Stability Analysis of Numerical Solution to the Heat Equation
T    2T
Consider the classical solution of the Heat Equation:    k
t    x 2
To find the form of the solution, try: T(x,t) = e-at sin(x)
Substituting this into the Heat Equation yields:
 a T(x,t) =  k 2 T(x,t)
OR        a = k 2
 T(x,t) = e-k t sin(x)
2

Each sin component of the initial temperature distribution decays as exp{-k2t).

Now consider FD schemes as advancing one step with a "transition equation":

{Tm+1 } = [A] {Tm }          w/ [A] a function of 

w/ {Tm } = T1m , T2 ,
m
, Tim ,      m
, Tn   
w/ zero boundary conditions
First step can be written:       {T1} = [A] {T0}       w/ {T0} = initial conditions
Second step as:                  {T2} = [A] {T1} = [A]2{T0}
and mth step as:                 {Tm } = [A] {Tm-1} = [A]m{T0}
(Here "m" is an exponent on [A])
   For the influence of the initial conditions and any errors to decay with time,
it must be the case that A  1.0
   If || A || > 1.0 , some eigenvectors of the matrix [A] can grow without bound
generating ridiculous results. In such cases the method is said to be
unstable.
   Taking r = A  A 2 = max. eigenvalue of [A] for symmetric A ("spectral
norm"), the maximum eigenvalue describes the stability of the method.
ENGRD 241 Lecture Notes                               Section 8: Partial Differential Equations              Page 8-15 of 8-18

Illustration of Instability of Explicit Method (for a simple case)
Consider Explicit Method for 1D spatial:                             Tim1  Tim1  (1  2)Tim  Tim1
                     

Worst case solution:         Tim  r m (1)i (high frequency x-oscillations in index i)
Substitution of this solution into the difference equation yields:
i1                                           i1
r m1  1  r m  1                 1  2  r m  1  r m  1
i                                            i

1                                    1
r    1               1  2     1
or                 r=1–4

If initial conditions are to decay and nothing “explodes,” we need:
–1 < r < 1 or         0 <  < 1/2.

For no oscillations we want:                            0< r <1               or    0 <  < 1/4.

Stability of Simple Implicit Method
Consider 1D spatial case:                                                 
Tim1 1  (1  2)Tim1  Tim1 1  Tim
                         

Tim  r m  1
i
Worst case solution:
Substitution of this solution into difference equation yields:
i1                                                    i1
r m1  1                 1  2  r m1  1  r m1  1              r m  1
i                                  i

r [– (–1)–1 + (1 + 2) –  (–1)+1] = 1
or           r = 1/[1 + 4 ]
i.e.,        0 < r < 1 for all  > 0

Stability of Crank-Nicolson Implicit Method
                         
Consider 1D spatial case: Tim1 1  2(1  )Tim1  Tim1 1  Tim1  2(1  )Tim  Tim1
                                                       

Tim  r m  1
i
Worst case solution:
Substitution of this solution into difference eqn. yields:
i1                                                     i1
r m1  1           2 1    r m1  1  r m1  1
i

i1                                            i1
 2 1    r m  1  r m  1
r m  1
i

r [– (–1)–1 + 2(1 + ) –  (–1)+1] =  (–1)–1 + 2(1 – ) +  (–1)+1
or       r = [1 – 2  ] / [1 + 2 ]
i.e.,    |r| < 1 for all  > 0
ENGRD 241 Lecture Notes                     Section 8: Partial Differential Equations               Page 8-16 of 8-18

Roots for Stability Analysis of Parabolic Heat Eq.

1.00
r(explicit)
r(implicit)
0.50                                                                                  r(C-N)
analytical

r   0.00
0              0.25                       0.5                  0.75               1           1.25

-0.50

-1.00

The Implicit Methods are Unconditionally Stable :
Magnitude of all eigenvalues of [A] is < 1 for all values of .
Therefore, x and t are selected based solely upon overall accuracy.

The Explicit Method is Conditionally Stable:
kt            1                      1 (x) 2
Explicit, 1D Spatial *                               or    t 
(x) 2          2                      2 k
1
         can still yield oscillation 1D 
2
1
         ensures no oscillation
4
1
         tends to minimize truncation error
6

kt         1                         1 h2
Explicit, 2D Spatial                            or        t         w/ h = x = y
(h) 2       4                         4 k
ENGRD 241 Lecture Notes                       Section 8: Partial Differential Equations         Page 8-17 of 8-18

Parabolic PDE's in two spatial dimensions (C&C 30.5, p. 852)
T       2T 2T 
2D:           k 2                  Find T(x,y,t)
t      x      y2 
            

Explicit solutions :
(x) 2  (y) 2
Stability criterion:               t 
8k
h2
if h  x  y  t 
4k

Implicit solutions : No longer tridiagonal

Alternating-Direction Implicit (ADI) Method (C&C 30.5.2, p. 852)

• Provides a method for using tridiagonal matrices for solving parabolic
equations in 2 spatial dimensions.
• Each time increment is implemented in two steps:

first direction                             second direction

Explicit                                                           tg+1
Implicit
yi+1
yi                             tg+1/
yi-1                                    2
xi-           x        xi+
yi+1                                    1             i        1
yi                                                                          tg
yi-1
xi-          xi         xi+
1                       1
first half-step                                   second half-step
13.1°

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