# function call and getting array error

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Thin Rupar Win on 9 Nov 2021
Commented: Thin Rupar Win on 9 Nov 2021
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
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); ##### 2 CommentsShow 1 older commentHide 1 older comment
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.