MATLAB Answers


First order partial differential equations system - Numerical solution

Asked by Malak Galal on 16 Jan 2019
Latest activity Commented on by 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.


Start with first-order upwind.
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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3 Answers

Answer by Torsten
on 18 Jan 2019
Edited by Torsten
on 18 Jan 2019
 Accepted Answer

%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];


Okay! So, basically it has nothing to do with ODE45, it all depends on the discretization method; am I correct?
Okay. Thank you very much for your help!

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Answer by Malak Galal on 21 Jan 2019

Hi Torsten,
Thank you very much for your help!


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Answer by 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

Hello Malak,
nx and nt can be chosen completely independent from each other.
Best wishes

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