1. Implementing our backward order final equation
Jennifer Tanyer
March 27, 2002

2. Goals: Balance the discretized equation from Bob’s presentation
Develop a system of linear equations from the given equation
Modify previous code to solve the implicit function and get values for our unknown concentration at time t+1

3. Our backward order discretized equation:
From Bob’s presentation:

4. Problems with this equation:
The system is not balanced i.e.(# of variables on the LHS  # of variables on the RHS)
LHS accounts for interior nodes only
RHS accounts for interior and boundary nodes
# of Unknowns > #of Equations because of this. The difference in the number is due to the lack of representation for the boundary points

5. Solutions:
Add the boundary condition for each boundary node onto the LHS of the equation: c(x,y,t) = 0.
Since we can’t solve this system explicitly, we must generate a system of linear equations that establish an algorithm for solving for our unknowns
Our unknown : concentration at time t + 1

6. Step 1: Developing the new linear system
Place all of our unknowns: t + 1 onto the LHS of our equation
The RHS will be equal to our concentration at time t

7. Step 1:

8. Step 2: Replace equations with their proper node labels

9. Step 3: Combine like terms
Refer to poisson.m
These new combined equations will represent w(k,k), w(k,E), w(k,W), w(k,N), and w(k,S) respectively.

10. Step 4: Implement the new equation into ‘concentrate.m’
%Modified concentrate.m
deltat=0.5;
D=0.1;
rho=1;
time=10;
ntimes=round(time/deltat);
NRPOINTS=max(size(a10));
for k=1:NRPOINTS
Position=a10(k,1:2);
Concentration(k,1)=g(Position);
end

11. Modified concentration.m continued…
for m=1:ntimes m
w = zeros(1528, 1528);
rhs = zeros(1528,1);
deltaXsq = 1/deltaX^2;
deltaYsq = 1/deltaY^2;
for k = 1:1528
N = nb(k,find((nbor(k,:) == 1)));
W = nb(k,find((nbor(k,:) == 3)));
S = nb(k,find((nbor(k,:) == 5)));
E = nb(k,fgind((nbor(k,:) == 7)));

12. gAnd finally… uy = scaledvelocity(k,2); ux = scaledvelocity(k,1);
if(~g(N) & ~isempty(W) & ~isempty(S) & ~isempty(E))
ux = scaledvelocity(k,1);
uy = scaledvelocity(k,2);
w(k,N) = ((D/rho) * (1/deltaYsq) – uy * (1/(deltaY)*2));
w(k,S) = ((D/rho) * (1/deltaYsq) + uy * (1/(deltaY)*2));
w(k,E) = ((D/rho) * (1/deltaXsq) - ux * (1/(deltaX)*2));
w(k,W) = ((D/rho) * (1/deltaXsq) + ux * (1/(deltaX)*2));
w(k,k) = (-1/deltat – (D/rho)*(2/(deltaXsq)) – (D/rho) *(2/(deltaYsq));
rhs(k,1) = concentration(k,m)/deltat;
else
w(k,k) = 1;
rhs(k,1) = 0;
end
xjen=w\rhs;
concentration(:,m+1)=xjen;

13. In the End…
You now have an algorithm of linear equations that you can solve to get a 1532 x 1532 matrix of concentrations at each node.
You can use MATLAB to solve Ax = b (w * x) = rhs by using either The inverse of A >>x=inv(A)*b OR Row Reduction >>x1=A\b
We no longer have approximate values for our concentrations…we now have actual values