pdepe boundary condition help!!

2 views (last 30 days)
KANCHETI
KANCHETI on 10 Mar 2025
Commented: KANCHETI on 11 Mar 2025
H0 = 1;
% Spatial and time mesh
y = linspace(0, H0, 20);
t = linspace(0, 1, 200);
% Solve the PDE
sol = pdepe(0, @pdefun, @icfun, @bcfun, y, t);
% Extract the solution
lambda = sol(:,:,1).';
function [c, f, s] = pdefun(y, t, lambda, DlambdaDx)
nu = 1.7e-28;
X = 0.1;
G0 =1e7;
kB = 1.3806488e-23;
theta = 298;
m_d = 0.5/(kB*nu*theta);
% Define the PDE coefficients
c = 1 / nu;
f = m_d/lambda * (G0*nu*(1+1/lambda^2))*DlambdaDx + kB*theta*(1/(lambda*(lambda-0.99))-1/lambda^2-2*X/lambda^3)*DlambdaDx;
s = 0;
end
function lambda0 = icfun(y)
% Initial condition
if y < 1 %H0
lambda0 = 1;
else
lambda0 = 1.498;
end
end
function [pl, ql, pr, qr] = bcfun(yl, lambda_l, yr, lambda_r, t)
% Boundary conditions
pl = 0;
ql = 1;
pr = lambda_r-1.498 ;
qr = 0;
end
I am trying to solve this problem and confused about the boundary condition where dLambda/dX = 0 at x = 0. I am not getting the results. Can someone explain how to apply initial and boundary condtions? Thank you very much.

Accepted Answer

Torsten
Torsten on 10 Mar 2025
Edited: Torsten on 10 Mar 2025
Note that the f you define in "pdefun" doesn't fit the formula from the paper. You forgot to multiply both terms with m_d/lambda, not only the first term.
H0 = 1;
% Spatial and time mesh
y = linspace(0, H0, 40);
t = linspace(0, 1, 20);
% Solve the PDE
sol = pdepe(0, @pdefun, @icfun, @bcfun, y, t);
% Extract the solution
lambda = sol(:,:,1);
plot(y,[lambda(1,:);lambda(5,:);lambda(10,:);lambda(15,:);lambda(end,:)])
function [c, f, s] = pdefun(y, t, lambda, DlambdaDx)
nu = 1.7e-28;
X = 0.1;
G0 =1e7;
kB = 1.3806488e-23;
theta = 298;
m_d = 0.5/(kB*nu*theta);
% Define the PDE coefficients
c = 1 / nu;
f = m_d/lambda * (G0*nu*(1+1/lambda^2) + kB*theta*(1/(lambda*(lambda-0.99))-1/lambda^2-2*X/lambda^3))*DlambdaDx;
s = 0;
end
function lambda0 = icfun(y)
% Initial condition
if y < 1 %H0
lambda0 = 1;
else
lambda0 = 1.498;
end
end
function [pl, ql, pr, qr] = bcfun(yl, lambda_l, yr, lambda_r, t)
% Boundary conditions
pl = 0;
ql = 1;
pr = lambda_r-1.498 ;
qr = 0;
end
  3 Comments
Torsten
Torsten on 11 Mar 2025
Edited: Torsten on 11 Mar 2025
Also could you tell me how can i calculate μ from equation 44 for every value of λ.
μ = nu*(G0*λ-G0/λ+ kB*theta/nu*(log(1-1/λ)+1/λ+X/λ^2))
Also if λ = 1, i need to replace log(1-1/λ) by log(1-0.999/λ)
I don't know about the physical background of your problem. So I can't give advice how the case λ = 1 is to be handled in the equations.
KANCHETI
KANCHETI on 11 Mar 2025
Thank you for your time.

Sign in to comment.

More Answers (0)

Community Treasure Hunt

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

Start Hunting!