# Solving system of equations

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EldaEbrithil on 27 May 2020
Commented: darova on 30 May 2020
Hi all
i have a question about solving this system of equations. Tt, Pt and M are related to space and time due to i and j; i want to solve the system maintaining that dependence, so the result will be a matrix respectively for Tt, Pt and M. When i try to solve, i obtain "Out of range subscript." error. gamma, deltax and deltat are constant
Thanks to all
Tt=zeros(length(x),length(t));
Pt=zeros(length(x),length(t));
M=zeros(length(x),length(t));
Tt(1,1)=3.000555630247608e+02;
Pt(1,1)=2.201018491400215e+05;
M(1,1)=0.023565919700319;
for j=1:length(t)-1
for i=2:length(x)-1
Alla = cell(length(x),length(t));
Allb = cell(length(x),length(t));
Allc = cell(length(x),length(t));
syms Tt Pt M
[sola,solb,solc]=vpasolve(Tt(i,j+1)==0.5*(Tt(i+1,j)-Tt(i-1,j))+((1+((gamma-1)/2)*M(i,j)^2)^(gamma/(gamma-1)))*((Tt(i+1,j)-Tt(i-1,j))*deltat/(2*deltax))+((1+((gamma-1)/2)*M(i,j)^2))*((Pt(i+1,j)-Pt(i-1,j))*deltat/(2*deltax)),...
Pt(i,j+1)==0.5*(Pt(i+1,j)-Pt(i-1,j))+2*((1+((gamma-1)/2)*M(i,j)^2)^(gamma/(gamma-1)))*((Tt(i+1,j)-Tt(i-1,j))*deltat/(2*deltax))+3*((1+((gamma-1)/2)*M(i,j)^2))*((Pt(i+1,j)-Pt(i-1,j))*deltat/(2*deltax)),...
M(i,j+1)==0.5*(M(i+1,j)-M(i-1,j))+2*((1+((gamma-1)/2)*M(i,j)^2)^(gamma/(gamma-1)))*((Tt(i+1,j)-Tt(i-1,j))*deltat/(2*deltax))+3*((1+((gamma-1)/2)*M(i,j)^2))*((Pt(i+1,j)-Pt(i-1,j))*deltat/(2*deltax)));
Alla{i,j} = sola;
Allb{i,j} = solb;
Allc{i,j} = solc;
end
end
EldaEbrithil on 28 May 2020
Yes i understand, but i think it is what similar to what i have done in my code, the only difference is related to the typology of discretization: you have used a forward discretiation in space and time, i have used a Forward Time Centered Space, FTCS discretization. Thi is the only difference, but the problem i have is easier than you think: i do not understand how to write the code for solving the system of equations practically.

darova on 28 May 2020
Here is a simple example. I hope it's clear enough. TR, TL, TD - boundary conditions (right, left and down boundaries) ##### 2 CommentsShowHide 1 older comment
darova on 30 May 2020
I can't check it. It's too complicated, sorry

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