coverting fortran to matlab

[dpb]

Show us your work on the problem to date...where did you get stuck?

HINT: Turn the GOTO into a WHILE structure.

[hasan damaj]

clc
clear
% Laplace equation
% Define the length of the grid
L  = 5 ;
% Define the Width of the grid
B = 5 ; 
nx = 4 ;
ny = 4 ;
dx = 100;
dy =  100;
h=zeros(nx,ny)
amax = 0
% boundry conditions
h(1,:)  = [ 8.04 , 7.68 , 7.19 , 6.82 ]; 
h(nx,:) = [ 8.53 , 8.41 , 8.33 , 8.29 ];
h(:,1)  = [ 8.04 , 8.18 , 8.36 , 8.53 ];
h(:,ny) = [ 6.82 , 7.56 , 7.99 , 8.29 ];
while amax > 0.01
for j = 2 : nx-1
    for i = 2 : ny-1
    oldval = h(i,j);
    h(i,j) = ( h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1)/4) ;
    e = abs(h(i,j)) - oldval;
    if e > amax
        amax = e
    end
    end
end
h

can u help to continuing the code?

i know i need to use a while loop but i dont know were and how?

[dpb]

Get rid of the the clc and clear and incorporate as a function, first.

You'll have to set AMAX to a value greater than the error tolerance (0.01) in order for the while loop to execute.

You can, of course, write the whole expression in MATLAB w/o looping over the H array using vector operations. filter is useful for this.

[hasan damaj]

ok no problem.

line 22 to line 28.

but i need ur help to continue the code and thank u

[hasan damaj]

please continue this code program for me..knowing i will attach the picture for the results and thank you

Answer 1

[J Chen]

The following statements are wronng

    h(i,j) = ( h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1)/4) ;
    ..
    e = abs(h(i,j)) - oldval;

It should be

    h(i,j) = ( h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1) ) /4 ;
    ..
    e = abs( h(i,j) - oldval );

Change while amax > 0.01 to

while 1
    ..
if ( amax > 0.01 )
    amax = 0;
else
    break
end

end

One end statement has been missing in your code.

[hasan damaj]

can u just write ur suggestion in my code and send it to me please.

were do i put my while loop??

[dpb]

Read the original code...where do you think it should go?

[hasan damaj]

ur code is not clear

while 1 .... ???????

.. (what is this)??????

if ( amax > 0.01 )

amax = 0;

else

break

end

end

[J Chen]

Replace "while amax > 0.01" with "while 1".

Add

if ( amax > 0.01 )
    amax = 0;
else
    break
end

just before the corresponding end for the while, i.e., at the end of the while loop. Try your code and figure out the rest!

[dpb]

Where is the looping structure in the original? That's what you need to emulate.

I don't quite agree with the other poster's suggestion (not that it won't work but it isn't my "cup of tea" in how I'd write it).

I pointed out above the modifications needed to write the while as

amax=1;
while amax>E  % set E to desired tolerance value
  ...
end

[Walter Roberson]

while 1 ....       ???????
 ..  (what is this)??????

In MATLAB, if and while are considered true provided that all of the values in the condition are non-zero. The constant 1 there will always be non-zero, so this code is expressing an infinite loop.

MATLAB happens to use the numeric value 1 for the logical value true so a directly equivalent way of writing while 1 is while true -- which is a form I am more likely to write.

Answer 2

[hasan damaj]

h=zeros(nx,ny);
amax = 0;
h(1,:)  = [ 8.04 , 7.68 , 7.19 , 8.53];
h(end,:) = [6.82 , 7.56 , 7.99 , 8.29 ];
h(2:3,1)  = [7.68;7.19];
h(2:3,end)  = [8.41; 8.33];


while amax >= 0.01
for j = 2 : nx-1
    for i = 2 : ny-1
        oldval = h(i,j);
        h(i,j) =  h(i-1,j) + h(i+1,j) + h(i,j-1) + h(i,j+1)/4 ;
        e = abs(h(i,j)) - oldval;
        if e > amax
            amax = e;
        end
        end
    end
end
h

its giving me zeros in h(2,3) h(3,2) h(3,3) h(2,2)

how can i get to the results same as above