First order partial differential equations system - Numerical solution
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Hello everyone!
I would like to solve a first order partial differential equations (2 coupled equations) system numerically. I just want to make sure that my thoughts are correct.
So firstly, I will start by doing a discretization to each of the two equations and then I will use ode15s to solve the ordinary differential equations that I got from the first step. Am I right? Which method of discretization do you recommend me to use?
Note: My equations are similar to the linear advection equation but with a source term.
Thank you very much.
Malak
2 件のコメント
採用された回答
Torsten
2019 年 1 月 18 日
編集済み: Torsten
2019 年 1 月 18 日
%program solves
% dy1/dt + v1*dy1/dx = s1*y1*y2
% dy2/dt + v2*dy2/dx = s2*y1*y2
% y1(0,x) = y2(0,x) = 0
% y1(t,1) = y2(t,0) = 1
% for 0 <= t <= 400
function main
nx = 500;
y1 = zeros(nx,1);
y1(end) = 1.0;
y2 = zeros(nx,1);
y2(1) = 1.0;
ystart = [y1;y2];
tstart = 0.0;
tend = 400;
nt = 41;
tspan = linspace(tstart,tend,nt);
xstart = 0.0;
xend = 1.0;
x = linspace(xstart,xend,nx);
x = x.';
v1 = -0.005;
v2 = 0.0025;
v = [v1 v2];
s1 = 3.0e-3;
s2 = -4.0e-3;
s = [s1 s2];
[T,Y] = ode15s(@(t,y)fun(t,y,nx,x,v,s),tspan,ystart);
plot(x,Y(20,1:nx), x,Y(20,nx+1:2*nx))
end
function dy = fun(t,y,nx,x,v,s)
y1 = y(1:nx);
y2 = y(nx+1:2*nx);
dy1 = zeros(nx,1);
dy2 = zeros(nx,1);
dy1(1:nx-1) = -v(1)*(y1(2:nx)-y1(1:nx-1))./(x(2:nx)-x(1:nx-1)) + s(1)*y1(1:nx-1).*y2(1:nx-1);
dy1(nx) = 0.0;
dy2(1) = 0.0;
dy2(2:nx) = -v(2)*(y2(2:nx)-y2(1:nx-1))./(x(2:nx)-x(1:nx-1)) + s(2)*y1(2:nx).*y2(2:nx) ;
dy = [dy1;dy2];
end
12 件のコメント
Sreejath S
2020 年 10 月 26 日
Hi Torsten,
Sorry for commenting on this post, rather than asking my own question. I have a very similar system of coupled first order hyperbolic PDES. But my system has a major diference from the example code you posted.
The first equation has dy1/dt and dy2/dx, and the second equation has dy2/dt and dy1/ds. Kindly note the change (marked bold) in the following lines from the original problem you posted:
% dy1/dt + v1*dy2/dx = s1*y1*y2
% dy2/dt + v2*dy1/dx = s2*y1*y2
% y1(0,x) = y2(0,x) = 0
% y1(t,1) = y2(t,0) = 1
% for 0 <= t <= 400
Can this type of coupled problem be solved using MOL?
I ran the code with this change, but I could not get a solution. The solution is not converging.Kindly help in this regard.
Sincerly,
Sreejath
その他の回答 (2 件)
Malak Galal
2019 年 2 月 22 日
1 件のコメント
Torsten
2019 年 2 月 22 日
Hello Malak,
nx and nt can be chosen completely independent from each other.
Best wishes
Torsten.
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