Zuhair
Posts:
9
Registered:
10/18/13


Soving PDE using pdepe
Posted:
Oct 18, 2013 3:25 AM


Hi
I got problem with solving pde using pdepe. Here is my code:
% Neumann boundary conditions (bc2.m) function [pl,ql,pr,qr] = bc2(xl,ul,xr,ur,t) pl = [0; 0]; ql = [1; 1]; pr = [0; 0]; qr = [1; 1];  % Initial conditions (initial2.m) function value = initial2(x); value = [0.95*x; 0.05*0.5];  % Equations PDE (eqn2.m) function [c,b,s] = eqn2(x,t,u,DuDx) c = [1; 1]; % coefficiet of time derivatives b = [0.001; 0.001].* DuDx; % diffusion terms s = [u(1)*(1(u(1)/x)(1.5)*(u(2)/x));u(2)*(1(u(2)/0.5)(1.5)*(u(1)/0.5))]; % LotkaVolterra 2species competition model  %Solving PDE stored in eqn2.m, bc2.m, and initial2.m close clear clc m = 0; x = linspace(0,1,1000); t = linspace(0,1,10000); sol = pdepe(m,@eqn2,@initial2,@bc2,x,t); u1 = sol(:,:,1); u2 = sol(:,:,2);
plot(x,u1(end,:)); hold all plot(x,u2(end,:)); xlabel('x') ylabel('u(x)')
This is the LotkaVolterra competition model with diffusion term (pde). When I try to solve this model with Neumann boundary conditions, I do not get the correct figure i.e. the correct solution of u1 should look like sigmoid curve which start at 0 (when x=0) and finish at 1 (when x = 1). And the correct solution of u2 also looks like sigmoid curve (but in the opposite direction to u1) and it start at 0.5 (when x = 0) and end at 0 (when x = 1).
The initial conditian that should be used is [0.95*x; 0.05*0.5]. And diffusion constant of 0.001.
Thanking you in advance for your help.

