pdepe boundary condition help!!

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KANCHETI
KANCHETI il 10 Mar 2025
Commentato: KANCHETI il 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.

Risposta accettata

Torsten
Torsten il 10 Mar 2025
Modificato: Torsten il 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 Commenti
Torsten
Torsten il 11 Mar 2025
Modificato: Torsten il 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 il 11 Mar 2025
Thank you for your time.

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