Solving system of equations

조회 수: 2 (최근 30일)
EldaEbrithil
EldaEbrithil 2020년 5월 27일
댓글: darova 2020년 5월 30일
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
  댓글 수: 17
darova
darova 2020년 5월 27일
I can't explain it here
can be re-written as (P(i,j+1)-P(i,j))/dt
can be re-written as (P(i+1,j)-P(i,j))/dx
you what i mean?
Read about this method. Read about "Method of lines"
EldaEbrithil
EldaEbrithil 2020년 5월 28일
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.

댓글을 달려면 로그인하십시오.

답변 (1개)

darova
darova 2020년 5월 28일
Here is a simple example. I hope it's clear enough. TR, TL, TD - boundary conditions (right, left and down boundaries)
  댓글 수: 2
EldaEbrithil
EldaEbrithil 2020년 5월 30일
I have tried to implement the method for the equation tht you give me in the.m file but i am not very confident about the results
clc,clear
% problem definition and discretization
dx = 0.01;
dt = 0.008;
xdomain = [0 1];
tdomain = [0 1];
nx = round((xdomain(2)-xdomain(1))/dx);
nt = round((tdomain(2)-tdomain(1))/dt);
x = linspace(xdomain(1),xdomain(2),nx);
t = linspace(tdomain(1),tdomain(2),nt);
u = zeros(nt,nx);
% du/dt - 2*t*du/dx = 0
u(1,:) = sin(2*pi*x);
for k = 1:nt-1
for i = 1:nx-1
% Predictor step
u(k+1,i) = 2*t(k)*dt/dx*(u(k,i+1)-u(k,i)) + u(k,i);
end
end
figure(1);set(gcf,'Visible', 'off')
plot(x,u(85,:))
figure(4);set(gcf,'Visible', 'off')
surf(x,t,u)
%%%%%LAX WENDROFF%%%%%
dx2 = 0.01;
dt2 = 8e-4;
xdomain2 = [0 1];
tdomain2 = [0 1];
nx2 = round((xdomain2(2)-xdomain2(1))/dx2);
nt2 = round((tdomain2(2)-tdomain2(1))/dt2);
x2= linspace(xdomain2(1),xdomain2(2),nx2);
t2 = linspace(tdomain2(1),tdomain2(2),nt2);
u2 = zeros(nx2,nt2);
u2(:,1) = sin(2*pi*x2);%initial condition
for i=2:nx2-1
for j=1:nt2-1
u2(i,j+1)=u2(i,j)+(2*t2(j)*dt2/(2*dx2))*(u2(i+1,j)-u2(i-1,j))+((dt2^2)/(2*dx2))*(u2(i+1,j)-u2(i-1,j))+2*((dt2^2)/(dx2^2))*(t2(j)^2)*(u2(i+1,j)-2*u2(i,j)+u2(i-1,j));
stab(j)=2*t2(j)*dt2/(2*dx2);%always less then one
end
end
figure(2);set(gcf,'Visible', 'off')
plot(x2,u2(:,85))
figure(3);set(gcf,'Visible', 'off')
surf(t2,x2,u2)
darova
darova 2020년 5월 30일
I can't check it. It's too complicated, sorry

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Fluid Dynamics에 대해 자세히 알아보기

제품


릴리스

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by