Asked by Malak Galal
on 16 Jan 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

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))

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

Malak Galal
on 31 Jan 2019

Torsten
on 31 Jan 2019

Correct.

Malak Galal
on 31 Jan 2019

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!

Malak

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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!

Malak

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## 2 Comments

## Torsten (view profile)

Direct link to this comment:https://kr.mathworks.com/matlabcentral/answers/440015-first-order-partial-differential-equations-system-numerical-solution#comment_661000

## Malak Galal (view profile)

Direct link to this comment:https://kr.mathworks.com/matlabcentral/answers/440015-first-order-partial-differential-equations-system-numerical-solution#comment_661779

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