diffusion with difference finite metod

조회 수: 2 (최근 30일)
bianchi fiammetta
bianchi fiammetta 2019년 2월 6일
댓글: Torsten 2019년 2월 6일
Good morning.
I do not understand the error in this script:
param.Cafeed=0.014158;% concentration [mol/dm3]
param.v=0.1;
param.L=1.1245; % led leight [dm]
param.D=0.001;
nx = 100; % to define the dimension of the initial condition
x = linspace(0,param.L,nx);
h = x(2)-x(1);
y0 = zeros(nx,1);
y0(1) = param.Cafeed;% Boundary condition
tspan = linspace(0,30,nx);
[t,y] = ode15s(@(t,y)fun(t,y,param,h),tspan,y0);%To solve the time dependence
function dy=fun(t,y,param,h)
M = length(y)+1;
dy = zeros(M+1,1);
dy(1)= 0;
for i=2:M
dy(i) = -param.v*(y(i)-y(i-1))/h+(param.D/h^2*(-2*y(i)+y(i-1)+y(i+1)));
end
thank you
  댓글 수: 1
Torsten
Torsten 2019년 2월 6일
편집: Torsten 2019년 2월 6일
  1. You have M-1 unknowns y, but you try to provide M time derivatives dy.
  2. You reference y(M) and y(M+1) in your loop which do not exist.
  3. You have to prescribe a boundary condition at x=L. Open boundaries are not allowed for second-order PDEs.
Best wishes
Torsten.

댓글을 달려면 로그인하십시오.

답변 (1개)

bianchi fiammetta
bianchi fiammetta 2019년 2월 6일
Thank you.
  댓글 수: 1
Torsten
Torsten 2019년 2월 6일
By the way:
This kind of equation can be solved using PDEPE.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Eigenvalue Problems에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by