Hi,

I thought that for bounce back boundary conditions yo simply assign the opposite distribution function at that boundary. But in this code i have looked at, it assigns the opposite forcing function rather than the distribution function as shown below? Can somebody explain to me why this is so?

f4(M,1:N)=F2(M,1:N);

f7(M,1:N)=F5(M,1:N);

f8(M,1:N)=F6(M,1:N);

% bottom: bounce back, no-slip

f2(1,1:N)=F4(1,1:N);

f5(1,1:N)=F7(1,1:N);

f6(1,1:N)=F8(1,1:N);

I am also having similar confusion with neumann boundary condition…

% right: Neumann

f3(1:M,N)=F1(1:M,N)-2*fstar(4)/cs2*(cx(4)*u(1:M,N-1)+cy(4) v(1:M,N-1));*(cx(8)

f7(1:M,N)=F5(1:M,N)-2fstar(8)/cs2

*u(1:M,N-1)+cy(8)*(cx(7)*u(1:M,N-1)+cy(7)*v(1:M,N-1));

*v(1:M,N-1));*

f6(1:M,N)=F8(1:M,N)-2fstar(7)/cs2f6(1:M,N)=F8(1:M,N)-2

Thanks. James.