Function optimization with some conditions conditions

I want to find the h values that meet the following conditions, h(j)<Hmax, h(j+1)-h(j)<lvasc(j) and h(j+1)-h(j)>lsa(j), the values of lvasc and lsa are calculated in teh function. About the input values ht is a array with a size ht(n,m), Lc and s are scalar and vmax size is vmax (1,m), i have done the function below but it returns me the error:

Error using vertcat

Dimensions of arrays being concatenated are not consistent.

Error in hp2 (line 38)

b = [b1;b2;b3];

I dont know how to solve this, thanks in davance

Mathematically, I need to find the values of h that minimize the function I, which is the summation of the subtraction between ht(n,m) and h(m), and h has to satisfy several conditions. The first condition is that h(end) < 2.5 * Lc. The next condition is that h(j+1) - h(j) < (1/36 * 0.5 / vmax(j+1)) or (0.75 * s / 9.81) / (Lc * 2), whichever is more restrictive. Finally, h(j+1) - h(j) > (0.5 * s * 0.4 * (ht(j+1) - ht(j))) / (vmax(j+1) * 9.81).

n = 10;
m = 50;
ht = rand(n,m);
Lc = 2;
s = 15;
vmax = rand(1,m);
[h,fval,exitflag] = hp2(ht,Lc,s,vmax)

ans = 1×2

50 1

ans = 1×2

1 49

ans = 1×2

1 49

Error using vertcat
Dimensions of arrays being concatenated are not consistent.

Error in solution>hp2 (line 49)
b = [b1;b2;b3];

function [h,fval,exitflag] = hp2(ht,Lc,s,vmax)

s1=s/1000;
Hmax=2.5*Lc;
Imax=0.75*s1/9.81;
lvasc=1/36*0.5./vmax;

[n,m]=size(ht);

for j=1:m-1
    lvasc(j)=1/36*0.5./vmax(j+1);
    lv(j)=min(Imax,lvasc(j));
    lsa(j) = ((ht(j+1)-ht(j)) * (0.5 * s1 * 0.4)) / (vmax(j+1) * 9.81);
end

I=@(h) sum(sum((ht-h).^2));

h0 = zeros(1,m);

v1 = ones(m,1);
w1 = -ones(m-1,1);
A1 = diag(v1) + diag(w1,1);
b1 = [zeros(m-1,1);Hmax];

v2 = -ones(m,1);
w2 = ones(m-1,1);
A2 = diag(v2) + diag(w2,1);
A2(end,:) = [];
b2 = lv;

v3 = -ones(m-1,1);
w3 = ones(m,1);
A3 = diag(v3,1) + diag(w3);
A3(end,:) = [];
b3 = -lsa;

size(b1)
size(b2)
size(b3)
A = [A1;A2;A3];
b = [b1;b2;b3];

[h,fval,exitflag] = fmincon(I,h0,A,b);

end

Answer 1

[Torsten]

The dimensions would be correct if you use

b = [b1;b2.';b3.'];

instead of

b = [b1;b2;b3];

but you'd better post the problem you are trying to solve in a mathematical form.

[Jon Bilbao]

I wil´l try to explain it in the next message but before that could you help me add the condition h(1) =0, to get this I added what is below but matlab returns me: h(1)=-4.814824860968090e-35.

Aeq = [1,zeros(1, m-1)];
beq = 0;
[h,fval,exitflag] = fmincon(I,h0,A,b,Aeq,beq);

[Torsten]

I'd call it a good numerical approximate result for h(1) = 0.

[Jon Bilbao]

Yeah, but in the problem that i am trying to solve thats physically imposible so if it can be made i would like to change it to 0.

[Torsten]

function [h,fval,exitflag] = hp2(ht,Lc,s,vmax)
  ...
  [h,fval,exitflag] = fmincon(I,h0,A,b);
  h(1) = 0;
end

Now you've set it to 0.