# function call and getting array error

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Thin Rupar Win on 9 Nov 2021
Commented: Steven Lord on 10 Jan 2024
I recently learn matlab for my education purpose. I would like to know how to solve the Index error in my code as I call function in for loop and getting error. I have already check array boundary and can't find the answer for the case. I deeply request you how to solve this case.
clear
% nx and ny must be the same size with t, n_part of the DEM loop
nx=4;ny=9;
% f_ equation for collisions
f=zeros(nx,ny,9);
feq=zeros(nx,ny,9);feq_f=zeros(nx,ny,9);
% velocities for x and y direction
u=zeros(nx,ny);v=zeros(nx,ny);
% velocities on solid particle
% for x direction
U_s_x=zeros(nx,ny,9);U_px=[0,0];Omega_px=[0,0];Xpx=[0,0];
% for y direction
U_s_y=zeros(nx,ny,9);U_py=[0,0];Omega_py=[0,0];Xpy=[0,0];
% position of particle and omega_s
omega_s=zeros(nx,ny,9);
% force term Ff,Tf on solid particle
Ffx=zeros(nx,ny,9);Ffy=zeros(nx,ny,9);
Tfx=zeros(nx,ny,9);Tfy=zeros(nx,ny,9);
rho=ones(nx,ny);x=zeros(nx);y=zeros(ny);
w(9)=zeros;
% initialization of the system
w=[1/9 1/9 1/9 1/9 1/36 1/36 1/36 1/36 4/9];
cx=[1 0 -1 0 1 -1 -1 1 0];
cy=[0 1 0 -1 1 1 -1 -1 0];
c2=1./3.;
% derivative rate
dx=1.0;dy=1.0;
x1=(nx-1)/(ny-1);y1=1.0;
dx=x1/(nx-1);
dy=y1/(ny-1);
x=(0:dx:x1);
y=(0:dy:y1);
u0=0.2;
% relaxiation time
alpha=0.02;
Re=u0*(ny-1)/alpha;
% omega = 1/tau
omega=1./(3.*alpha+0.5);
% toleriant and error
%count=0;tol=1.0e-4;error=10.;erso=0.0;
% setting velocity
for j=2:ny-1
u(1,j)=u0;
end
t_lbm=50;
for t=1:t_lbm
[f,Ffx,Ffy,Tfx,Tfy]=up_collition(nx,ny,u,v,cx,cy,U_s_x,U_px,Omega_px,Xpx,U_s_y,U_py,Omega_py,Xpy,feq,rho,w,feq_f,f,omega_s,Ffx,Ffy,Tfx,Tfy);
end
Index in position 1 exceeds array bounds. Index must not exceed 1.

Error in solution>up_collition (line 76)
U_s_x(i,j,k)=U_px(i,j)+Omega_px(i,j)*(x(i)+0.5*(cx(k)))-Xpx(i,j);
function[f,Ffx,Ffy,Tfx,Tfy]=up_collition(nx,ny,u,v,cx,cy,U_s_x,U_px,Omega_px,Xpx,U_s_y,U_py,Omega_py,Xpy,feq,rho,w,feq_f,f,omega_s,Ffx,Ffy,Tfx,Tfy)
% the weighting function of solid particles k Bk,
% the local solid ratio e_k divided
e_k=[1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 0];
e_total=1;
for j=1:ny
for i=1:nx
t1=u(i,j)*u(i,j)+v(i,j)*v(i,j);
for k=1:9
x=zeros(length(i));% later add with dx term, let dt =1;
t2=u(i,j)*cx(k)+v(i,j)*cy(k);
Bk(k)=e_k(k).*(0.56-0.5)/(1-e_total+(0.56-0.5));
% velocity on solid particle
% the equaiton is Us=UP+omega_px *(x +0.5 *c(k)*dt)-Xp
U_s_x(i,j,k)=U_px(i,j)+Omega_px(i,j)*(x(i)+0.5*(cx(k)))-Xpx(i,j);
U_s_y(i,j,k)=U_py(i,j)+Omega_py(i,j)*(x(i)+0.5*(cy(k)))-Xpy(i,j);
t22=U_s_x(i,j,k)*cx(k)+U_s_y(i,j,k)*cy(k);
t11=U_s_x(i,j,k)*U_s_x(i,j,k)+U_s_y(i,j,k)*U_s_y(i,j,k);
feq(i,j,k)=rho(i,j)*w(k)*(1.0+3.0*t2+4.5*t2*t2-1.5*t1);
feq_f(i,j,k)=rho(i,j)*w(k)*(1.0+3.0*t22+4.5*t22*t22-1.5*t11);
f(i,j,k)=f(i,j,k)-(0.9091*(f(i,j,k)-feq(i,j,k)))+(Bk(k)*(feq_f(i,j,k)-feq(i,j,k)));
omega_s(i,j,k)=feq_f(i,j,k)-f(i,j,k)+((f(i,j,k)-feq(i,j,k)).*0.42);
% force term and torque term for DEM
Ffx(i,j,k)=sum(Bk(k)).*sum(omega_s(i,j,k))*cx(k);
Ffy(i,j,k)=sum(Bk(k)).*sum(omega_s(i,j,k))*cy(k);
Tfx(i,j,k)=sum((x(i)-Xpx)*Bk(k)).*sum(omega_s(i,j,k)).*cx(k);
Tfy(i,j,k)=sum((x(i)-Xpy)*Bk(k)).*sum(omega_s(i,j,k)).*cy(k);
end
end
end
end
Rik on 9 Nov 2021
You're indexing several variables. Did you check each one? One of them is resulting in an error. You best chance to solve the problem is to figure out which variable is causing the problem.
Thin Rupar Win on 9 Nov 2021
Sir, I tried checking the size and array in my file. I still cant find the reason for error.

Sudharsana Iyengar on 9 Nov 2021
U_s_x is a 4x9x9 matrix but U_px is 1x2 matrix. So there is no U_px(3,4) did you check on that.
U_s_x(i,j,k)=U_px(i,j)+Omega_px(i,j)*(x(i)+0.5*(cx(k)))-Xpx(i,j);
U_s_y(i,j,k)=U_py(i,j)+Omega_py(i,j)*(x(i)+0.5*(cy(k)))-Xpy(i,j);
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Thin Rupar Win on 9 Nov 2021
Sir, thank you very much for your point. Now I can solve them with giving to initial zeros(x,y). Have a nice day.
Steven Lord on 10 Jan 2024
FYI, one way to help debug this type of issue is to set an error breakpoint, run the code to reproduce the error, and examine the variables used on the line where MATLAB enters debug mode to make sure you're not trying to retrieve an element that doesn't exist of one of those variables.