How to change the values and do the functions Newton-Raphson - arrays

Question

I need to iterate Newton-Raphson in MATLAB. It seems easy but I cannot figure out where I am wrong. The problem is:
For mmm=1:
1) If m=1 take c1=c1b and c2=1-c1 and do the loop for u1,2(i) and p1,2(i)
2)If m=2 take c1=c1+dc and c2=1-c1, and this time do the loop with new c1 and c2 for u1,2(i) and p1,2(i)
3) If m=3 take c1=(c1*st(1)-(c1-dc)*st(2))/(st(1)-st(2)) and do the loop for new c1 and c2.
Then increase the iteration number: mmm=2 ;
mmm keeps count of the number of N-R iterations. The first iteration has mmm=1, the second mmm=2, etc. (This particular run only do 2 iterations).
sumint are inside of the integrals.
I need to plot these figures in the code but MATLAB gives errors below. Please help me.
Relevant part of the code:
ii=101;
u = cell(2, 1);
ini_cond = [0,0];
for i = 1:2;
u{i} = zeros(1,ii);
u{i}(:, ii) = ini_cond(i) * rand(1, 1);
end
for i=1:ii;
fikness=fik*sin(pi.*x);
u{1}(i)=(c1-H1D*(x-0.5)+AD/2.*(x-0.5).^2)./(H1-0.5*fikness-A*(x-0.5));
u{2}(i)=(c2+H1D*(x-0.5)-AD/2.*(x-0.5).^2)./(1.-H1+0.5*fikness+A*(x-0.5));
end
p = cell(2, 1);
q = cell(2, 1);
for i = 1:2;
p{i} = zeros(1,ii);
q{i} = zeros(1,ii);
end
p{1}(1)=0.5*(1.-u{1}(1).^2);
q{1}(1)=0;
p{2}(1)=0.5*(1.-u{2}(1).^2);
q{2}(1)=0;
for i=2:101
q{1}(i)=q{1}(i-1)-dx*(u{1}(i-1)-ub{1}(i-1))./dt;
p{1}(i)=0.5*(1.-u{1}(i).^2)+q{1}(i);
q{2}(i)=q{2}(i-1)-dx*(u{2}(i-1)-ub{2}(i-1))./dt;
p{2}(i)=0.5*(1.-u{2}(i).^2)+q{2}(i);
end
st = zeros(2, length(t));
st(1,:)=p{1}(100)-p{2}(100);
m=m+1;
if m==3;
c1=(c1*st(1)-(c1-dc)*st(2))/(st(1)-st(2));
c2=1-c1;
end
for i = 1:2;
sumint{i} = zeros(1,length(t));
end
sumint = cell(2, 1);
sumint{1}(1)=0.5*(p{2}(1)-p{1}(1));
sumint{2}(1)=0.5*(p{2}(1)-p{1}(1)).*(-1/2);
for i=2:ii-1;
x=(i-1)*dx;
sumint{1}(i)=sumint{1}(i-1)+(p{2}(i)-p{1}(i));
sumint{2}(i)=sumint{2}(i-1)+(p{2}(i)-p{1}(i))*(x-1/2);
end
H1DDOT=-sumint{1}.*dx./rmass;
H1D=H1D+dt*H1DDOT;
H1=H1+dt*H1D;
ADDOT=sumint{2}*dx./rmomi;
AD=AD+dt*ADDOT;
A=A+dt*AD;
H1L=H1+A.*0.5;
H1R=H1-A.*0.5;
H2=1.-H1;
rat1=AD./ADinit;
rat2=ADDOT./AD;
u are the velocities p are the pressures c1,c2 are the camber effects H1DDOT and ADDOT are the second derivation of H1 and A. sum1 and sum2 are the inside of the integrals to define the values of H1DDOT and ADDOT. H1DDOT and ADDOT are functions of time.

Answer 1

As you can see from the message, the error is with this line:
sumint{2}(i)=sumint{2}(i-1)+(p{2}(i)-p{1}(i)).*(x-1/2);
Now, let's find out why:
sumint{2}(i) = ...
This part means you want to insert whatever is on the right side to the ith position of the array in cell sumint{2}. That means, the right side must be a scalar.
Is it?
Well, sumint{2}(i-1)+(p{2}(i)-p{1}(i)) is certainly a scalar, since you use a single value as index to all the vectors/arrays. The problem is the multiplication .*(x-1/2);.
From the code above, it is clear that x is a vector / array, (since you use length(x) etc). Multiplying the scalar sumint{2}(i-1)+(p{2}(i)-p{1}(i)), by the vector x, will give you back a vector, which as mentioned, will not work.
Maybe you want the ith value of x?
There are several other strange things going on in your code, for instance:
for i=1:101;
fikness=fik*sin(pi.*x);
u{1}=(c1-H1D*(x-0.5)+AD/2.*(x-0.5).^2)./(H1-0.5*fikness-A*(x-0.5));
u{2}=(c2+H1D*(x-0.5)-AD/2.*(x-0.5).^2)./(1.-H1+0.5*fikness+A*(x-0.5));
end
Why do you have a loop here? You are not looping anything, you are doing the same calculations 100 times. Again, I guess it should be x(i).
Update:
Your new error is introduced because you changed x into a scalar. Then u = zeros(1,length(x)) will only be a scalar, meaning u{1}(i) will fail for i ~= 1.