MODELLING PROBLEMS #2
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Consider the “arrested topographic wave” problem:
y 
r
 xx  0
fs
x 
Ff
rg
at x  0
  0 at x  
where r = 0.1 cm/sec, s =0.001, f = 10-4 sec.
Solve two problems:
10 cm 0  x  50 km
a) F = 0 ,  ( x, 0)  
;
x  50 km
 0
b) F = -1 cm2/sec2,  ( x,0)  0
Solve numerically for  at -500 km < y < 0 in each problem using both the forward
time/centered space scheme and the fully implicit scheme.
Compare your results to the appropriate analytical solutions. Explain the behavior in
physical terms.
Ans:
Looking into the basic equation for the “arrested topographic wave” problem
r
(1)
 y   xx  0
fs
Equation (1) resembles the archetype of this kind of equation: heat diffusion equation,
with y takes the role of time and no source term is considered. So the simple way to solve
above problem numerically is to let y = -t. then, equation is transformed into:
r
(1)
 t   xx  0
fs
1) Applying the explicit forward time/ centered space scheme to equation (1):
 n 1   n
n
n
n
r  i1  2 i   i1
0
t
fs
(x) 2
Where n denotes the nth computing node in y direction and i means the ith one in x
direction. To solve above problem, we just need compute out the value of  in 1 in
i
i

equation (2) for each (i, n) node. That is:
r t
 in1   in 
( in1  2 in   in1 )
2
fs (x)
Then equation (3) can be solved under specific boundary and initial conditions.
(2)
(3)
10 cm 0  x  50 km
For case a) F = 0 ,  ( x, 0)  
x  50 km
 0
Ff
 0 at x  0
rg
Which means along y direction at x = 0 boundary,  Bn   Bn1 . For the boundary of
x   , the value of  always be the constant of 0.
10 cm 0  x  50 km
The initial condition for case a) is like  ( x, 0)  
x  50 km
 0
The computing result for case a) using method 1) is presented in fig.1.a
From F  0   x 
For case b) F = -1 cm2/sec2,  ( x,0)  0
 Bn1   Bn
f
xf
  Bn   Bn1 
at x  0
x
rg
rg
Which is the boundary condition at x = 0. Similarly, for the boundary of x   , the
value of  always be the constant of 0.
The initial condition for case a) is like  ( x, 0)  0
The computing result for case a) using method 1) is presented in fig.2.a
From F  1   x 

2) Applying the implicit forward time/ centered space scheme to equation (1):
 n 1   n
i
i
t


n 1
n 1
n 1
r  i1  2 i   i1

0
fs
(x) 2
r t
2r t
r t
 in11  (1 
) in 1 
 in11   in
2
2
2
fs (x)
fs (x)
fs (x)
Equation (5)can only be solved using simultaneous equations for the entire grids
accompanied by boundary conditions.
10 cm 0  x  50 km
For case a) F = 0 ,  ( x, 0)  
x  50 km
 0
(4)
(5)
Ff
 0 at x  0
rg
Which means along y direction at x = 0 boundary,  Bn 1   Bn11 . For the boundary of
x   , the value of  always be the constant of 0.
10 cm 0  x  50 km
The initial condition for case a) is like  ( x, 0)  
x  50 km
 0
From F  0   x 
The computing result for case a) using method 2) is presented in fig.1.b
For case b) F = -1 cm2/sec2,  ( x,0)  0
 Bn11   Bn1
f
xf
  Bn1   Bn11 
at x  0
x
rg
rg
Which is the boundary condition at x = 0. Similarly, for the boundary of x   , the
value of  always be the constant of 0.
The initial condition for case a) is like  ( x, 0)  0
The computing result for case a) using method 1) is presented in fig.2.b
From F  1   x 

3) Analytical solutions
For case a)
For case a, here, the analytical solution is referred from Walter A. Strauss’s book, as
below:
Let k 
r
fs

2
2
1
 ( x, t ) 
[e  ( x  y ) / 4 kt  e ( x  y ) / 4 kt ] ( y )dy

4 kt 0
10 cm 0  x  50 km
Where  ( y ) is the initial value of  , that is  ( x, 0)  
x  50 km
 0
The boundary condition is:  x ( x, t )  0 at x  0
The image of analytical solution is shown in fig1.c.
(6)
(7)
For case b)
For case b, we still want to use the analytical solution listed in case a. but, in this case,
the boundary condition is a little different from case a in that
Ff
(8)
 x ( x, t ) 
at x  0
rg
The initial condition is  ( x,0)  0
(10)
Ff
x
rg
Taking equation (11) into basic equation (1), boundary condition(8) and initial
condition (10), we then get a new one:
r
St  S xx  0
fs
S x  0 at x  0
Then let us define a function S ( x, t )   ( x, t ) 
(11)
(12)
Ff
x
rg
According to analytical solution (6), we can write out the solution for the new function
S(x,t).
S ( x, 0)  
Let k 
r
fs

Ff
 ( x  y )2 / 4 kt
 ( x  y ) 2 / 4 kt
S ( x, t )  
[
e

e
]ydy

rg 4 kt 0

(13)

2
2
Ff
Ff
S ( x, t )  
e  ( x  y ) / 4 kt ydy 
e  ( x  y ) / 4 kt ydy


rg 4 kt 0
rg 4 kt 0

(14)

For the first part in right side of equation (14): 
2
Ff
e  ( x  y ) / 4 kt ydy

rg 4 kt 0
(15)
Let y  x  m and take it into equation (15):
Equation (15) is transformed into: 

Ff 4kt

e x
rg 2 4 kt
Ff
rg 4 kt

e
 m2 / 4 kt
(m  x)dm
x

x
2
/ 4 kt
Ffx
Ffx
 m2 / 4 kt
 m2 / 4 kt

e
dm

e
dm


rg 4 kt 0
rg 4 kt 0
(16)

2
Ff
e  ( x  y ) / 4 kt ydy
For the second part in right side of equation (14): 

rg 4 kt 0
Let y  x  m and take it into equation (17):
(17)

Equation (17) is transformed into: 
2
Ff
e  m / 4 kt (m  x)dm

rg 4 kt x

x
2
2
2
Ff 4kt
Ffx
Ffx
e  x / 4 kt 
e  m / 4 kt dm 
e  m / 4 kt dm


rg 2 4 kt
rg 4 kt 0
rg 4 kt 0
Finally, the analytical solution for problem b is equal to:
x
2
Ff 4kt  x2 / 4 kt
2 Ffx
S ( x, t )  
e

e  m / 4 kt dm

rg 4 kt
rg 4 kt 0



 ( x, t )  
x
2
Ff 4kt  x2 / 4 kt
2 Ffx
Ff
e

e  m / 4 kt dm 
x

rg
rg 4 kt
rg 4 kt 0
(18)
The image of analytical solution is shown in fig2.c.
Analysis:
Both Problem A and B describe the diffusion process for a substance for example heat.
In Problem A, we can image a case:
At the initial position, that is 0  x  50 km ; t  0 , there have a constant, even amount
of heat at any place. For x  50 km ; t  0 , no heat exists. And at x=0, there has a wall
which is insulted from heat transfer. As time pass by, due to the heat difference at
x  50 km , heat begin conducts from high heat to the low heat area, which then lead to
the temperature difference in area originally have constant, even heat. Finally, heat
gradually move into originally on heat area, and tend to average in the whole domain.
Contrasting fig1.a and fig1.b with analytically solution fig 1.c, it can be seen that the
result with explicit numerical method is more close to the analytically solution. In
comparison, the result with implicit numerical method is less precise. It may be the
reason that implicit method has stronger effect of numerical damping.
In Problem B, it shows just like there has a heat source at boundary x = 0 and input heat
at constant speed. As time pass by, heat gradually propagate into the computation domain.
Due to the diffusion process can not balance the process of heat input, therefore, as we
can see, temperature increase at boundary x = 0.
Contrasting fig2.a and fig2.b with analytically solution fig2.c, it can be seen that the
result with explicit numerical method is also more close to the analytically solution. In
comparison, the result with implicit numerical method is less precise. It may be the
reason that implicit method has stronger effect of numerical damping.
Method1 Problem A
0
0.09
0.08
2
0.0
0.07
0.07
0.06
1
0.0
0.0
5
-150
0.06
0.03
-200
-250
0.05
-300
0.04
1
0.0
0.04
0.05
0.02
y=km
0.
01
0.04
-100
3
0.0
-50
0.06
0.1
9
0.0
08
0.
-350
0.03
04
0.
0
50
0.01
0.01
-500
0.02
-450
0.02
0.03
-400
100
x=km
150
200
fig 1.a explicit numerical method for problem a
Method2 Problem A
0.00.
2 01
0.
07
3
0.0
0.04
08
0.
-50
0.06
.09
00.1
0
0.09
0.0
5
0.06
2
0.0
-150
0.08
01
0.
-100
0.07
-250
0.05
0.02
0.05
1
0.0
0.04
0.06
0.03
-300
0.04
-350
0.03
-450
-500
0.02
0.03
-400
0.01
0.
04
y=km
-200
0
50
0.01
100
x=km
150
fig 1.b implicit numerical method for problem a
200
00.05
0.04
0.07
0.08
0.01
0.09
.0.06
0
2
03
0.
0
0.1
Analytical Solution A
0.02
0.03
0.05
0.04
0.07
0.06
0.08
0.01
0.09
0. 08
0.0
5
0.04
0.
06
0.07
1
0.0
0.03
-150
0.08
2
0.0
7
0.0
0.09
0.
01
-50
-100
-200
0.06
-250
0.05
0. 05
0.02
0.03
-350
0.04
1
0.0
0.0
4
-300
0.03
-400
0.02
0
50
100
x=km
0.01
0.01
-500
0.02
4
0.0
-450
150
200
fig 1.c analytical solution for problem a
Method1 Problem B
0
02
0.
-50
01
0.
-150
0.07
3
0.0 .04
0
-100
0.06
2
0.0
0.05
01
0.
-200
-250
-350
0.03
1
0.0
2
0.0
4
0.0
0.06
-300
0.04
3
0.0
0.05
y=km
y=km
0.02
0.03
0.05
0.04
0.07
0.06
0.08
0.01
0.09
0.02
1
0.0
50
0.01
2
0.0
0.04
0
0.06
-500
0.07
-450
3
0.0
0.05
-400
100
x=km
150
fig 2.a explicit numerical method for problem b
200
Method2 Problem B
0
0.0
1
0.06
01
0.
-150
0.07
3
0.0 4
0.0
-100
02
0.
-50
-350
0.03
1
0.0
0.01
0
2
0.0
0.04
0.05
0.06
0.07
-450
0.02
3
0.0
-400
-500
0.04
2
0.0
4
0.0
0.05
0.06
-300
1
0.0
-250
3
0.0
y=km
0.05
2
0.0
-200
50
100
x=km
150
200
fig 2.b implicit numerical method for problem b
Analytical Solution B
0
01
0.
-150
3
0.0 0.04
-100
0.07
02
0.
-50
0.06
1
0.0
-350
0.03
1
0.0
2
0.0
4
0.0
-300
0.04
3
0.0
-250
0.05
0.06
y=km
0.05
2
0.0
-200
0.02
0.01
100
x=km
1
0.0
50
2
0.0
0.04
0
0.05
0.06
-500
0.07
-450
3
0.0
-400
150
fig 2.c analytical solution for problem b
200