1. No category

Chapter 9
NUMERICAL SOLUTION OF PARTIAL
DIFFERENTIAL EQUATIONS
1
CLASSIFICATION
 A general second order partial differential equation may be
written as AUXX + BUXY + CUYY + DUX + EUY + FU = 0 where
A, B, C, D, E, F are in general functions of x and y.
 The above equation is classified as follows:
●
●
●
Elliptic if
Parabolic if
Hyperbolic if
B2 – 4AC < 0
B2 – 4AC = 0
B2 – 4AC > 0
2
Standard Examples
Elliptic equations
Laplace equation
 2u
x 2
Poisson equation
 2u
x 2
+
 2u
y 2
=0
+
 2u
y 2
= f (x, y)
Parabolic equations
One dimensional heat equation
 2u
x 2
=
1
2
u
t
Hyperbolic equations
One dimensional wave equation
 2u
x 2
=
1
2
 2u
t 2
3
ELLIPTIC EQUATIONS
 Finite Difference Method—In solving elliptic equations, we approximate
the derivatives using finite differences.
u ( x 0  h, y 0 )  u ( x 0 , y 0 )
h
ux (x0, y0) =
Also ux (x0, y0) =
uxx (x0, y0) =
uy(x0, y0) =
[Forward difference]
u (x 0 , y 0 )  u (x 0  h, y 0 )
h
[Backward difference]
u( x 0  h, y 0 )  2u( x 0 , y 0 )  u(x 0  h, y 0 )
h2
u(x 0 , y 0  k)  u(x 0 , y 0 )
k
[Forward difference]
Also ux (x0, y0) =
u(x 0 , y 0 )  u(x 0 , y 0  k)
k
and uyy (x0, y0) =
u( x 0 , y 0  k)  2u( x 0 , y 0 )  u(x 0 , y 0  k)
k2
[Backward difference]
4
Graphical Representation
I = 0
1
i=m
j=n
2
xij+1
xi-1, j
xij
xi+1, j
xi, j+1
J=2
J=1
J=0
5
Representation & Approximation
Then
(i, j+1)
(i -1, j)
(i, j)
(i, j- 1)
(i +1, j)
ux =
=
uy =
=
uxx =
uyy =
u i 1, j  u ij
[Forward difference]
h
u i , j  u i 1, j
h
u i ,1 j  u i , j
h
u i , j  u i , j1
h
[Backward difference]
[Forward difference]
[Backward difference]
u i 1, j  2u ij  u i 1, j
h
u i , j1  2u i , j  u ij1
k2
6
Solution of Elliptic Equations
 The most important type of elliptic equation is Laplace equation
uxx + uyy = 0.

Approximating the derivatives by difference expressions, we get
uij = 14 (ui-1, j + ui+1,j + ui,j-1 + ui,j +1) when h = k
u(i, j+1)
u(i -1, j)
u(i, j)
u(i +1, j)
u(i, j- 1)
This is called diagonal averaging.
7
Diagonal averaging
(ui-1, j+1)
(ui-1,j-1)
(ui+1,j+1)
(ui+1,j-1)
Also we can use the values of u at the diagonal
points.
ui,j = 14 (ui-1, j-1 + ui+1,j+1 + ui+1, j-1 + ui-1,j +1)
8
Liebmann’s Method
 To solve uxx + uyy = 0, in a square region R whose boundary is
C. The square region is sub-divided into small squares. The
values on the boundary points are given. We have to find the
values for the function u (x,y) in the interior mesh points.
 For this we use the approximation using cross-averaging
wherever possible and diagonal averaging otherwise.
 This iterative process is continued until the values at each mesh
point converge.
 While applying this method, symmetry about the horizontal,
vertical and diagonal lines should be taken into consideration.
9
Example
Find by Liebmann’s Method the values at the interior lattice
points of a square plate of the harmonic function u whose
boundary values are given in the figure.
0
0
0
0
0
1
4
9
U1
U2
U4
U5
U6
U7
U8
U9
3
U3
6
16
15
9
14
13
12
10
Solution
Using cross averaging and diagonal averaging, the successive iterations
yield the following values:
u1
u2
2.5000 5.625
u3
u4
u5
u6
u7
u8
u9
10.000 3.1250 6.0000 9.8750 3.0000 6.1250 9.5000
2.4375 5.6094 9.8711 2.8594 6.1172 9.8721 2.9948 6.1530 9.5063
2.3672 5.5888 9.8652 2.8698 6.1209 9.8731 3.0057 6.1582 9.5078
2.3647 5.5877 9.8652 2.8728 6.1230 9.8740 3.0078 6.1597 9.5084
2.3651 5.5883 9.8656 2.8740 6.1240 9.8745 3.0084 6.1602 9.5087
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PARABOLIC EQUATION
Most important parabolic equation
One dimensional heat equation
2
u

u
2
=
t
x 2
where
2
k
=
c
and c is the specific heat,  is the density and k is the
thermal conductivity of the material.
The above equation can be written as:
uxx = aut
1
where a = 2

12
Bender – Schmidt Method
Consider the equation uxx = aut with boundary
conditions
u(0, t) = T0 and u(1, t) = T1
The initial condition is u (x, 0) = f(x). Let h be the spacing
for x and k be the spacing for t.
Using finite difference approximation for derivatives,
and applying boundary conditions and considering the
k
special case  =
= ½ , we get
2
h a
13
Bender – Schmidt Method
ui,j+1 = ½ (ui-1,j + ui+1,j)
This formula means that the value of u at x = xi and
t = tj+1 is the arithmetic mean of the values of u at the
surrounding points xi-1 and xi+1 at the previous time tj.
B
C
ui-1,j
ui,j
ui,j+1
ui+1,j
Value at A = ½ (Value
at B + Value at C)
A
14
Example
Find the value of the function u (x,t) satisfying
the equation
u
 2u
4 2
t
x
The boundary conditions are u (0, t)  0  u (8, t)
1 2
and the initial condition is u ( x,0)  4 x  x
2
at the points x = i, i = 0, 1, 2, 3, 4 and t = (1/8) j,
j = 0, 1, 2, 3, 4, 5.
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Solution
1
k
1
1
Given a = , h = 1,  = 2 = 4k. If  = then k = .
4
2
8
h a
i
0
1
2
3
4
5
6
7
8
0
0
3.5
6
7.5
8
7.5
6
3.5
0
1
0
3
5.5
7
7.5
7
5.5
3
0
2
0
2.75
5
6.5
7
6.5
5
2.75
0
3
0
2.5
4.625
6
6.5
6
4.625
2.5
0
4
0
2.3125
4.25
5.5525
6
5.5625
4.25
2.125
0
5
0
2.125
3.9875
5.125
5.5625
5.125
3.9875
2.125
0
j
16
HYPERBOLIC EQUATION
Most important hyperbolic equation
One – dimensional wave equation
 u
2  u
a
2
t
x
2
2
17
Solution of One–Dimensional Wave
Equation
a2 uxx – utt = 0
The boundary conditions are u
The initial conditions are
(0, t) = 0 = u (1, t)
u (x, 0) = f (x) and ut (x, 0) = 0.
Let h be the spacing for x and k be the spacing for t.
Using finite difference approximation for derivatives,
and applying boundary conditions and considering the
k
1
special case  =
= , we get
h
a
18
Solution of One–Dimensional Wave
Equation
 The boundary conditions u(0, t) = 0, u(1, t) = 0 can be written in
the from u 0, j = 0 and un,j = 0 when l = nh.
 The initial condition u (x, 0) = f (x) can be expressed as:
ui,0 = f (ih) i = 1, 2,…,
 That is ui,0 = fi. For the initial condition ut(x, 0) = 0, we use
Central Difference Approximation for the derivative and write
u i ,1
f i 1  f i 1

2
19
Solution of One – Dimensional Wave
Equation
ui, j+1 = ui+1, j + ui-1,j – ui, j-1
ui-1, j-1
ui-1, j
ui, j-1
ui, j
ui+1, j
ui+1, j
ui, j+1
Note:
u(xi,tj+1)=(sum of surrounding values at previous time) – u(xi,tj-1)
20
Example
Tabulate the pivotal values for the equation
 2u  2u
16 2  2
x
t
Given that
u (0, t) = 0, u (5, t) = 0, u (x, 0) = x2 (5 – x) and ut (x, 0) = 0
Assume h = 1.
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Solution
i
0
1
2
3
4
5
0
0
4
12
18
16
0
1
0
6
11
14
9
0
2
0
7
8
2
-2
0
3
0
2
-2
-8
-7
0
4
0
-9
-14
-11
-6
0
5
0
-16
-18
-12
-4
0
6
0
-9
-14
-11
-6
0
7
0
2
-2
-8
-7
0
8
0
7
8
2
-2
0
9
0
6
11
14
9
0
10
0
4
12
18
16
0
j
ui,j+1 = ui-1, j + ui+1, j – ui, j-1
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The End……
Wish you all the best…
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