solution of PDEs using MATLAB

1. 1
Computational Method to Solve
the Partial Differential
Equations (PDEs)
By
Dr. Khurram Mehboob
1

2. Mehboob, K
NE: 402 computation Method in Nuclear
Engineering
Department of Nuclear Engineering,
KAU, Jeddah, KSA
2017. 5. 1
How to solve 1st Order PDEs
using MATLAB
2

3. Contents
• Solving PDEs using MATLAB (Explicit method)
- FTCS method
- Lax method
- Crank Nicolson method
- Jacobi’s method
- Simultaneous-over-relaxation (SOR) method
• Solving PDEs using MATLAB
- Examples of PDEs
3

4. PDE (Partial Differential Equation)
PDEs are used as mathematical models for phenomena in
all branches of engineering and science.
2 2 2
2 2
2 ( , , , , )
u u u u u
A B C D x t u
t xt x t x
    
  
    
u u
and
t x
 
 
linear in
- Three Types of PDEs:
1) Elliptic:
 Steady heat transfer, flow and diffusion
2) Parabolic:
 Transient heat transfer, flow and diffusion
3) Hyperbolic:
 Transient wave equation
2
0B AC 
2 2
2 2 0u u
x y
  
 
Laplace equations
2
0B AC 
2
2
u uD
t x
 
 
Diffusion Equation
2
0B AC  2 2
2 2 2
1u u
x c t
 
 
Wave Equation
4

5. Similarity method to solve PDE
often the solution of a PDE together wit its initial and boundary problem depends upon its
variables. In such cases the problem can be reduce to the combination ODEs with its
independent variables
The key to the similarity solution is the diffusing equation is that both the equations and
initial and boundary conditions are under the scaling x→lx and t →l2x.
Therefore, x/√t is independent transformation. In general the function should be of
form
The u(x,t)=U(z)
2
2
u u
t x
 
 
, 0t x 
( , ) 1 0
( , ) 0 0
u x t t
u x t x
 
 
( / )u t f x t 

x
t
 
2
2
1 0
2
U U
 
  
 
(0) 1
( ) 0
U
U

 
2
/4
0
( ) s
U C e ds D

 
 
2
/41( ) s
U e ds





 
2
/4
/
1( , ) s
x t
u x t e ds



 
5

6. FTCS method for Advection Eq
Advection equation
The Forward time and central space (FTCS)
method implementation
,
0,1........
0,1........
( , )
n o n
j o n
n j n j
x x n x n N
t t j x j t
u u x t
   
   

, 1 , 1, 1,
2
n j n j n j n j
u u u u
c
t x
  
 
 
 
0u uc
t x
  
 
FTCS is explicit scheme
, 1 , 1, 1,
( )
2n j n j n j n j
c tu u u u
x  
  

6

7. Implementation in MATLAB
FTCS is explicit scheme
Square-wave Test for the Explicit Method to solve the Advection Equation
clear;
% Parameters to define the advection equation and the range in space and time
Lmax = 1.0; % Maximum length
Tmax = 1.0; % Maximum time
c = 1.0; % Advection velocity
% Parameters needed to solve the equation within the explicit method
maxt = 3000; % Number of time steps
dt = Tmax/maxt;
n = 30; % Number of space steps
nint=15; % The wave-front: intermediate point from which u=0
dx = Lmax/n;
b = c*dt/(2.*dx);
% Initial value of the function u (amplitude of the wave)
for i = 1:(n+1)
if i < nint
u(i,1)=1.;
else
u(i,1)=0.;
end
x(i) =(i-1)*dx;
end
% Value of the amplitude at the boundary
for k=1:maxt+1
u(1,k) = 1.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the explicit method
for k=1:maxt % Time loop
for i=2:n % Space loop
u(i,k+1) =u(i,k)-b*(u(i+1,k)-u(i-1,k));
end
end
% Graphical representation of the wave at different selected times
plot(x,u(:,1),'-',x,u(:,10),'-',x,u(:,50),'-',x,u(:,100),'-')
title('Square-wave test within the Explicit Method I')
xlabel('X')
ylabel('Amplitude(X)')
0u uc
t x
  
 
( , ) 1 0
( , ) 0 0
u x t t
u x t x
 
 
7
0.025
2
c t
x
 


8. FTSC Lax method for Advection Eq
Advection equation
To increase the stability of FTCS method simply
talke the average of term u(n,j) ,
0,1........
0,1........
( , )
n o n
j o n
n j n j
x x n x n N
t t j x j t
u u x t
   
   

0u uc
t x
  
 
FTCS is explicit scheme, 1 , 1, 1,
( )
2n j n j n j n j
c tu u u u
x  
  

1, 1,
,
( )
2
n j n j
n j
u u
u  


1, 1,
, 1 1, 1,
( )
( )
2 2
n j n j
n j n j n j
u u c tu u u
x
 
  
   

LEX method (explicit)
8

9. Implementation in MATLAB
LEX method is explicit scheme
0u uc
t x
  
 
( , ) 0 0 ( , ) 1 0u x t t u x t x   
wave Test for the Lax method to solve the Advection Equation
clear;
% Parameters to define the advection equation and the range in space and time
Lmax = 1.0; % Maximum length
Tmax = 1.; % Maximum time
c = 1.0; % Advection velocity
% Parameters needed to solve the equation within the Lax method
maxt = 350; % Number of time steps
dt = Tmax/maxt;
n = 300; % Number of space steps
nint=50; % The wave-front: intermediate point from which u=0 (nint<n)!!
dx = Lmax/n;
b = c*dt/(2.*dx);
% The Lax method is stable for abs(b)=< 1/2 but it gets difussed unless abs(b)= 1/2
% Initial value of the function u (amplitude of the wave)
for i = 1:(n+1)
if i < nint
u(i,1)=1.;
else
u(i,1)=0.;
end
x(i) =(i-1)*dx;
end
% Value of the amplitude at the boundary at any time
for k=1:maxt+1
u(1,k) = 1.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the Lax method
for k=1:maxt % Time loop
for i=2:n % Space loop
u(i,k+1) =0.5*(u(i+1,k)+u(i-1,k))-b*(u(i+1,k)-u(i-1,k));
end
end
% Graphical representations of the evolution of the wave
figure(1)
mesh(x,time,u')
title('Square-wave test within the Lax Method'); xlabel('X'); ylabel('T')
figure(2)
plot(x,u(:,1),'-',x,u(:,20),'-',x,u(:,50),'-',x,u(:,100),'-')
title('Square-wave test within the Lax Method')
xlabel('X')
ylabel('Amplitude(X)')
9
1.01
2
c t
x
 

1.0
2
c t
x
 

0.5
2
c t
x
 


10. CTCS method for Advection
Advection equation
The Forward time and central space (FTCS)
method implementation
,
0,1........
0,1........
( , )
n o n
j o n
n j n j
x x n x n N
t t j x j t
u u x t
   
   

0u uc
t x
  
 
Center point in time
2 2, 1 , 1 1, 1,
( ) ( )
2 2
n j n j n j n j
u u u uu uo t o x
t t x x
   
       
   
Center point in Space
, 1 , 1 1, 1,
2 2
n j n j n j n ju u
t
u u
c
x
   


 

, 1 , 1 1, 1,
2 ( )
2n j n j n j n j
c tu u u u
x   
  

, 1 , 1 1, 1,
( )n j n j n j n j
c tu u u u
x   
  

CTCS is also an explicit scheme
10

11. CTCS Lax method for Advection Eq
Advection equation
To increase the stability of CTCS method simply
take the average of term u(n,j-1)
,
0,1........
0,1........
( , )
n o n
j o n
n j n j
x x n x n N
t t j x j t
u u x t
   
   

0u uc
t x
  
 
CTCS explicit scheme, 1 , 1 1, 1,
( )n j n j n j n j
c tu u u u
x   
  

1, 1 1, 1
, 1
( )
2
n j n j
n j
u u
u    



1, 1 1, 1
, 1 1, 1,
( )
( )
2
n j n j
n j n j n j
u u c tu u u
x
   
  
   

LEX CTCS method (explicit)
, 1 , 1 1, 1,
( )n j n j n j n j
c tu u u u
x   
  

11

12. Implementation in MATLAB (CTCS)
0u uc
t x
  
 
( , ) 1 0
( , ) 0 0
u x t t
u x t x
 
 
% Matlab Program : Square-wave Test for the Explicit Method to solve the advection Equation
clear;
% Parameters to define the advection equation and the range in space and % time
Lmax = 1.0; % Maximum length
Tmax = 1.; % Maximum time
c = 1.0; % Advection velocity
% Parameters needed to solve the equation within the explicit method
maxt = 100; % Number of time steps
dt = Tmax/maxt;
n = 100; % Number of space steps
nint=15; % The wave-front: intermediate point from which u=0
dx = Lmax/n;
b = c*dt/(dx)
% Initial value of the function u (amplitude of the wave)
for i = 1:(n+1)
if i < nint
u(i,1)=1.;else
u(i,1)=0.;
end
x(i) =(i-1)*dx;
end
% Value of the amplitude at the boundary
for k=2:maxt+1
u(1,k) = 1.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the explicit method
for k=2:maxt % Time loop
for i=2:n % Space loop
u(i,k+1) =u(i,k-1)-b*(u(i+1,k)-u(i-1,k));
end
end
% Graphical representation of the wave at different selected times
plot(x,u(:,10),'-',x,u(:,50),'-',x,u(:,100),'-',x,u(:,maxt),'-')
title('Square-wave test within the Explicit Method')
xlabel('X')
ylabel('Amplitude(X)')
1c t
x
 

0.5c t
x
 

12

13. Implementation in MATLAB (CTCS)
0u uc
t x
  
 
( , ) 1 0
( , ) 0 0
u x t t
u x t x
 
 
% Matlab Program : Square-wave Test for the Explicit Method to solve the advection Equation
clear;
% Parameters to define the advection equation and the range in space and % time
Lmax = 1.0; % Maximum length
Tmax = 1.; % Maximum time
c = 1.0; % Advection velocity
% Parameters needed to solve the equation within the explicit method
maxt = 100; % Number of time steps
dt = Tmax/maxt;
n = 100; % Number of space steps
nint=15; % The wave-front: intermediate point from which u=0
dx = Lmax/n;
b = c*dt/(dx)
% Initial value of the function u (amplitude of the wave)
for i = 1:(n+1)
if i < nint
u(i,1)=1.;else
u(i,1)=0.;
end
x(i) =(i-1)*dx;
end
% Value of the amplitude at the boundary
for k=2:maxt+1
u(1,k) = 1.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the explicit method
for k=2:maxt % Time loop
for i=2:n % Space loop
u(i,k+1) =0.5*(u(i+1,k-1)+u(i-1,k-1))-b*(u(i+1,k)-u(i-1,k));
end
end
% Graphical representation of the wave at different selected times
plot(x,u(:,10),'-',x,u(:,50),'-',x,u(:,100),'-',x,u(:,maxt),'-')
title('Square-wave test within the Explicit Method')
xlabel('X')
ylabel('Amplitude(X)')
0.5c t
x
 

1c t
x
 

13

14. Mehboob, K
NE: 402 computation Method in Nuclear
Engineering
Department of Nuclear Engineering,
KAU, Jeddah, KSA
2017. 5. 3
How to solve 2nd Order PDEs
using MATLAB
14

15. Diffusion Problem
),(
),(),(
2
2
txS
x
txu
D
t
txu






L
x
xu
conditionsinitial
tutLututu
conditionsBoundry
2
cos100)0,(
),(),(),(),0( 21



ydiffusivitThermalD :
SourceHeattxS :),(
),(
),(),(
txS
x
txu
D
xt
txu













15

16. FTCS method for the heat equation
FTCS ( Forward Euler in Time and Central difference in Space )
):),(),(),(),0((
),(),(),(
21
2
2
ydiffusivitthermalDtutLututu
txStxu
x
Dtxu
t







Heat equation in a slab
),( 1
t
uu
t
txu n
j
n
jnj




 
2
11
2
2
2),(
x
uuu
x
txu n
j
n
j
n
jnj




 
n
jnj utxu ),(Let
Forward Euler in Time
2nd order Central difference in Space
1
2 1 1
( 2 )n n n n n n
j j j j j j
D tu u tS u u u
x

 
    

1
1 1
2
2
( , )
n n n n n
j j j j j
u u u u u
D s x t
t x

 
  
 
 
FTCS explicit scheme for diffusion Equation
16

17. Implementation in MATLAB (FTCS heat equation)
% Heat Diffusion in one dimensional wire within the % Explicit Method
clear;
% Parameters to define the heat equation and the range in space and time
L = 1.; % Length of the wire
T =1.; % Final time
% Parameters needed to solve the equation within the explicit method
maxk = 2500; % Number of time steps
dt = T/maxk;
n = 50; % Number of space steps
dx = L/n;
cond = 1/4; % Conductivity
b = cond*dt/(dx*dx); % Stability parameter (b=<1)
% Initial temperature of the wire: a sinus.
for i = 1:n+1
x(i) =(i-1)*dx;
u(i,1) =sin(pi*x(i));
end
% Temperature at the boundary (T=0)
for k=1:maxk+1
u(1,k) = 0.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the explicit method
for k=1:maxk % Time Loop
for i=2:n; % Space Loop
u(i,k+1) =u(i,k) + b*(u(i-1,k)+u(i+1,k)-2.*u(i,k));
end
end
% Graphical representation of the temperature at different selected times
figure(1)
plot(x,u(:,1),'-',x,u(:,100),'-',x,u(:,300),'-',x,u(:,600),'-')
title('Temperature within the explicit method')
xlabel('X')
ylabel('T')
figure(2)
mesh(x,time,u')
title('Temperature within the explicit method')
xlabel('X')
ylabel('Temperature')
),(
),(),(
2
2
txS
x
txu
D
t
txu






)sin()0,(
0),(,0),(,0),0(
xxu
conditionsinitial
txstLutu
conditionsBoundry


17

18. Implementation in MATLAB (FTCS heat equation)
Courant condition for
FTCS
1
2
2



x
tD
2
2 1.01D t
x
 
2
2 1
2
D t
x
 

18

19. CTCS method for the heat equation
FTCS ( Forward Euler in Time and Central difference in Space )
):),(),(),(),0((
),(),(),(
21
2
2
ydiffusivitthermalDtutLututu
txStxu
x
Dtxu
t







Heat equation in a slab
2
),( 11
t
uu
t
txu n
j
n
jnj




 
2
11
2
2
2),(
x
uuu
x
txu n
j
n
j
n
jnj




 
n
jnj utxu ),(Let
Center difference in Time
2nd order Central difference in Space
1 1
2 1 1
22 ( 2 )n n n n n n
j j j j j j
D tu u tS u u u
x
 
 
     

1 1
1 1
2
2
( , )
2
n n n n n
j j j j j
u u u u u
D s x t
t x
 
 
  
 
 
FTCS explicit scheme for diffusion Equation
19

20. CTCS method for the heat equation% Heat Diffusion in one dimensional wire within the % Explicit Method
clear; clc;
% Parameters to define the heat equation and the range in space and time
L = 1.; % Length of the wire
T =1.; % Final time
% Parameters needed to solve the equation within the explicit method
maxk = 600; % Number of time steps
dt = T/maxk;
n = 20; % Number of space steps
dx = L/n;
cond = 1/132; % Conductivity
b =2.*cond*dt/(dx*dx) % Stability parameter (b=<0.01)
% Initial temperature of the wire: a sinus.
for i = 1:n+1
x(i) =(i-1)*dx;
u(i,1) =sin(pi*x(i));
end
% Temperature at the boundary (T=0)
for k=1:maxk+1
u(1,k) = 0.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the explicit method
for k=2:maxk % Time Loop
for i=2:n; % Space Loop
u(i,k+1) =u(i,k-1) + 0.5*2*b*(u(i-1,k)+u(i+1,k)-2.*u(i,k));
end
end
% Graphical representation of the temperature at different selected times
figure(1)
plot(x,u(:,1),'-',x,u(:,10),'-',x,u(:,30),'-',x,u(:,60),'-')
title('Temperature within the explicit method')
xlabel('X')
ylabel('T')
figure(2)
mesh(x,time,u')
title('Temperature within the explicit method')
xlabel('X')
ylabel('Temperature')
),(
),(),(
2
2
txS
x
txu
D
t
txu






)sin()0,(
0),(,0),(,0),0(
xxu
conditionsinitial
txstLutu
conditionsBoundry


0126.0
2
2



x
tD
20

21. Stability of FTCS and CTCS
FTCS is first-order accuracy in time and second-order
accuracy in space.
So small time steps are required to achieve reasonable
accuracy.
Courant condition for FTCS
2
1
2



x
tD
)2(
2
2 112
11 n
j
n
j
n
j
n
j
n
j
n
j uuu
x
tD
tSuu 





CTCS method for heat equation
(Both the time and space derivatives are
center-differenced.)
However, CTCS method is unstable for
any time step size.
2
2 1
2
D t
x
 

2
2 0.0126D t
x
 

CTCS (unstable)
FTCS (Stable)
21

22. Lax method (FTCS)
Simple modification to the FTCS method
In the differenced time derivative,
)2()( 112112
11 n
j
n
j
n
j
n
j
n
j
n
j
n
j uuu
x
tD
tSuuu 





t
uuu
t
uu n
j
n
j
n
j
n
j
n
j




 

)]([)( 112
111
Replacement by average value from surrounding grid points
The resulting difference equation is
Courant condition for Lax method
4
1
2



x
tD
( Second-order accuracy in both time and space )
22

23. Lax method(CTCS)
Simple modification to the CTCS method
In the differenced time derivative,
)2(
2
2)( 112
1
1
1
12
11 n
j
n
j
n
j
n
j
n
j
n
j
n
j uuu
x
tD
tSuuu 









t
uuu
t
uu n
j
n
j
n
j
n
j
n
j




 




2
)]([
2
)( 1
1
1
12
1111
Replacement by average value from surrounding grid points
The resulting difference equation is
Courant condition for Lax method
4
1
2



x
tD
( Second-order accuracy in both time and space )
23

24. Implementation (FTCS-Lax) Method
% Initial temperature of the wire: a
sinus.
for i = 1:n+1
x(i) =(i-1)*dx;
u(i,1) =sin(pi*x(i));
end
% Temperature at the boundary (T=0)
for k=1:maxk+1
u(1,k) = 0.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the explicit
method
for k=2:maxk % Time Loop
for i=2:n; % Space Loop
u(i,k+1) =0.5*(u(i+1,k)+u(i-1,k)) +
0.5*b*(u(i-1,k)+u(i+1,k)-2.*u(i,k));
end
end
2
2 0.1786D t
x
 

2
2 1
2
D t
x
 

Stability condition
2
2 0.0047D t
x
 

24

25. Implementation Lax method(CTCS)
2
2 0.3571D t
x
 

2
2 1
2
D t
x
 

Stability condition
% Initial temperature of the wire:
a sinus.
for i = 1:n+1
x(i) =(i-1)*dx;
u(i,1) =sin(pi*x(i));
end
% Temperature at the boundary (T=0)
for k=1:maxk+1
u(1,k) = 0.;
u(n+1,k) = 0.;
time(k) = (k-1)*dt;
end
% Implementation of the explicit
method
for k=2:maxk % Time Loop
for i=2:n; % Space Loop
u(i,k+1) =0.5*(u(i+1,k-1)+u(i-1,k-
1))+ 0.5*2*b*(u(i-1,k)+u(i+1,k)-
2.*u(i,k));
end
end
25

26. Implementation Lax method(CTCS)
26

27. Crank Nicolson Algorithm ( Implicit Method )
BTCS ( Backward time, centered space ) method for heat equation
)2( 11
1
2
n
j
n
j
n
jx
tn
j
n
j
n
j TTTtSTT 

 
( This is stable for any choice of time steps,
however it is first-order accurate in time. )
Crank-Nicolson scheme for heat equation
taking the average between time steps n-1 and n,

))22(
)(
1
1
11
111
1
2
11
2








n
j
n
j
n
j
n
j
n
j
n
jx
t
n
j
n
j
n
j
n
j
TTTTTT
SStTT

( This is stable for any choice of time steps and second-order accurate in time. )
a set of coupled linear equations for n
jT
27

28. Crank Nicolson Algorithm
Initial
conditions
Plot
Crank-
Nicolson
scheme
)/sin(),(
22
/
LxetxT Lt


Exact solution
28

29. Crank Nicolson Algorithm
29

30. Multiple Spatial Dimensions
),,(),,(),,( 2
2
2
2
tyxStyxT
yx
tyxT
t















FTCS for 2D heat equation
)2()2( 112112
1 n
jk
n
jk
n
jk
n
kj
n
jk
n
kj
n
jk
n
jk
n
jk TTT
y
t
TTT
x
t
tSTT 









Courant condition for this scheme
22
22
2
1
yx
yx
t



( Other schemes such as CTCS and Lax can be easily extended to multiple dimensions. )
30

31. Wave equation with nonuniform wave speed
),,(),(),,( 2
2
2
2
2
2
2
tyxz
yx
yxctyxz
t 













2D wave equation
0
),()0,,(),,()0,,(
00
00


vz
yxvyxzyxzyxz 
txxtxz
txztyztyz
2sin)1(),0,(
,0),1,(),,1(),,0(


Initial condition :
Boundary
condition :
 ])5.0(5)5.0(2exp[7.01),( 22
10
1
 yxyxcWave speed :
)2()2(2 112
22
112
22
11 n
jk
n
jk
n
jk
jkn
kj
n
jk
n
kj
jkn
jk
n
jk
n
jk TTz
y
tc
zzz
x
tc
zzz 








CTCS method for the wave equation :
)(
2
1
2
22



yxfor
tc
Courant condition :
31

32. Wave equation with nonuniform wave speed
Since evaluation of the nth timestep
refers back to the n-2nd step,
for the first step, a trick is employed.
jk
t
jkjkjk atvzz 20
01 2


)2()2( 112
2
0
1
00
12
2
n
jk
n
jk
n
jk
jk
kjjkkj
jk
jk TTz
y
c
zzz
x
c
a  




Since initial velocity and value,
0
0
1
0
0


jk
jkjk
z
zv
32

33. Wave equation with nonuniform wave speed
33

34. Wave equation with nonuniform wave speed
34

35. 2D Poisson’s equation
),(),()( 2
2
2
2
yxyxyx
  



Poisson’s equation
Direct Solution for Poisson’s equation
jk
jkjkjkkjjkkj
yx







 
2
11
2
11 22
Centered-difference the spatial derivatives
35

36. Jacobi’s method ( Relaxation method )
Direct solution can be difficult to program efficiently.
Relaxation methods are relatively simple to code,
however, they are not as fast as the direct methods.
Idea :
I. Poisson’s equation can be thought of as the equilibrium solution to the heat
equation with source.
II. Starting with any initial condition, the heat equation solution will eventually
relax to a solution of Poisson’s equation.













 

)2(
1
)2(
1
2
1
11211222
22
1 n
jk
n
jk
n
jk
n
kj
n
jk
n
kjjk
n
jk
n
jk
yxyx
yx














  )(
1
)(
1
2
1
11211222
22
n
jk
n
jk
n
kj
n
kjjk
yxyx
yx

),(),(),,(
t
),(),( 22
yxyxtyxyxyx  


 )
2
1
( 22
22
yx
yx
t



FTCS
(Maximum time step satisfying Courant condition)
36

37. Jacobi method
37

38. Simultaneous OverRelaxation (SOR)
The convergence of the Jacobi method is quite slow.
Furthermore, the larger the system, the slower the convergence.
Simultaneous OverRelaxation (SOR) :
the Jacobi method is modified in two ways,



















)(
1
)(
1
2
)1(
1
112
1
112
22
22
1
n
jk
n
jk
n
kj
n
kjjk
n
jk
n
jk
yx
yx
yx



1. Improved values are used as soon as they become available.
2. Relaxation parameter ω tries to overshoot for going to the final result.
( 1<ω<2)
38

39. Simultaneous OverRelaxation (SOR)
39

40. MATLAB The Language of Technical Computing MATLAB PDE
Run: dftcs.m >> dftcs
dftcs - Program to solve the diffusion equation using the Forward Time Centered Space scheme.
Enter time step: 0.0001
Enter the number of grid points: 51
Solution is expected to be stable
2
2
),(),(
x
txT
t
txT






40

41. MATLAB The Language of Technical Computing MATLAB PDE
Run: dftcs.m >> dftcs
dftcs - Program to solve the diffusion equation using the Forward Time Centered Space scheme.
Enter time step: 0.00015
Enter the number of grid points: 61
WARNING: Solution is expected to be unstable
41

42. MATLAB The Language of Technical Computing MATLAB PDE
Run: neutrn.m >> neutrn Program to solve the neutron diffusion equation using the FTCS.
Enter time step: 0.0005
Enter the number of grid points: 61
Enter system length: 2 => System length is subcritical
Solution is expected to be stable
Enter number of time steps: 12000
),(
),(),(
2
2
txn
x
txn
t
txn
 





42

43. MATLAB The Language of Technical Computing MATLAB PDE
Run: neutrn.m >> neutrn Program to solve the neutron diffusion equation using the FTCS.
Enter time step: 0.0005
Enter the number of grid points: 61
Enter system length: 4 => System length is supercritical
Solution is expected to be stable
Enter number of time steps: 12000
43

44. MATLAB The Language of Technical Computing MATLAB PDE
Run: advect.m >> advect
advect - Program to solve the advection equation using the various hyperbolic PDE schemes:
FTCS, Lax, Lax-Wendorf
Enter number of grid points: 50
Time for wave to move one grid spacing is 0.02
Enter time step: 0.002
Wave circles system in 500 steps
Enter number of steps: 500
FTCS FTCS
0
),(),(






x
txu
t
txu

44

45. MATLAB The Language of Technical Computing MATLAB PDE
Run: advect.m >> advect
advect - Program to solve the advection equation using the various hyperbolic PDE schemes:
FTCS, Lax, Lax-Wendorf
Enter number of grid points: 50
Time for wave to move one grid spacing is 0.02
Enter time step: 0.02
Wave circles system in 50 steps
Enter number of steps: 50
Lax Lax
45

46. MATLAB The Language of Technical Computing MATLAB PDE
Run: relax.m >> relax
relax - Program to solve the Laplace equation using Jacobi, Gauss-Seidel and SOR methods on a
square grid
Enter number of grid points on a side: 50
Theoretical optimum omega = 1.88184
Enter desired omega: 1.8
Potential at y=L equals 1
Potential is zero on all other boundaries
Desired fractional change = 0.0001
0
),(),(
2
2
2
2






y
yx
x
yx 
46

47. 47
The END
47