MATLAB

$\%$ Step $1$: Create a matrix psiold with vertical values spanning from $0$ to $16$ for every column

rows $=17$; $\%$ Number of rows

cols $=30$; $\%$ Number of columns

psiold $=$ repmat$((0$:$16)\text{'},1,$ cols$)$;

$\%$ Step $2$: Create a zeros matrix psinew that is the same size as psiold

psinew $=$ zeros$($rows$,$ cols$)$;

$\%$ Step $3$: Create a tolerance variable tol

tol $=0\mathrm{.}001$;

$\%$ Step $4$: Create a counter variable

counter $=0$;

$\%$ Step $5$: While loop condition

while any$($any$($abs$($psinew $-$ psiold$)>$ tol$))$

$\%$ Step $6$: Nested for loop to update psinew excluding the boundaries

for i $=2$:rows$-1$

for j $=2$:cols$-1$

psinew$($i$,$j$)=($psiold$($i$+1,$j$)+$ psiold$($i$-1,$j$)+$ psiold$($i$,$j$+1)+$ psiold$($i$,$j$-1))/4$;

end

end

$\%$ Step $7$: Define the first column of psinew as values spanning from $0$ to $16$

psinew$($:$,1)=(0$:$16)\text{'}$;

$\%$ Step $8$: Define the last column of psinew as the same as the second to last column

psinew$($:$,$cols$)=$ psinew$($:$,$cols$-1)$;

$\%$ Step $9$: Define the first row of psinew as equal to $0$

psinew$(1,$:$)=0$;

$\%$ Step $10$: Define the last row of psinew as equal to $16$

psinew$($rows$,$:$)=16$;

$\%$ Step $11$: Define rows $1$ through $9$ and columns $21$ through $29$ of psinew as equal to zero

psinew$(1$:$10,21$:cols$-1)=0$;

$\%$ Step $12$: If statement to check tolerance and update psiold and psinew

if any$($any$($abs$($psinew $-$ psiold$)>$ tol$))$

psiold $=$ psinew;

psinew $=$ zeros$($rows$,$ cols$)$;

end

$\%$ Step $13$: Increment the counter

counter $=$ counter $+1$;

end

$\%$ Step $14$: Display the counter

disp$([\text{'}$Number of iterations: $\text{'},$ num$2$str$($counter$)])$

$\%$ Step $15$: Create a contour plot of psinew

contour$($psinew$,16)$

$\%$ Step $16$: Set axis equal, label, and title the contour plot

axis equal

xlabel$(\text{'}$Column Index'$)$

ylabel$(\text{'}$Row Index'$)$

title$(\text{'}$Contour Plot of psinew'$)$

$\%$ Step $17$: Discussion of tolerance changes

$\%$ As the tolerance is increased, the counter value $($number of iterations$)$ decreases

$\%$ because the stopping criterion is met more quickly. Conversely, as the tolerance

$\%$ is decreased, the counter value increases because it takes longer to satisfy

$\%$ the stopping condition with a stricter tolerance.