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First order partial differential equations system - Numerical solution

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Malak Galal
Malak Galal on 16 Jan 2019
Commented: Torsten on 22 Feb 2019
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 Galal
Malak Galal on 18 Jan 2019
Thank you very much for your reply.
Is there any document or website that could be of use to me?
I am a bit lost because I have been trying to understand how to use the upwind method to solve my system, but I couldn't really find anything similar to my system. I saw some examples about the advection equation, but they all deal with one equation and with no source term.
My problem is that the source terms in my system couple the two equations.
I would appreciate your help very much! Thanks in advance.

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Accepted Answer

Torsten on 18 Jan 2019
Edited: Torsten on 18 Jan 2019
%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))
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];


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Malak Galal
Malak Galal on 31 Jan 2019
Okay! So, basically it has nothing to do with ODE45, it all depends on the discretization method; am I correct?

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More Answers (2)

Malak Galal
Malak Galal on 22 Feb 2019
Hi Torsten,
I have a question concerning the number of points in space and time (nx and nt). Using a proper discretization method, if I want to use for instance nx=1e5, does nt have to be the same as nx or close to it? It seems to me that when I increase the number of points in space, the number of points in time have to be increased as well. Could you please help me understand the relationship between nx and nt?
Thank you very much!

  1 Comment

Torsten on 22 Feb 2019
Hello Malak,
nx and nt can be chosen completely independent from each other.
Best wishes

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