清空筛选条件
清空筛选条件

problem onn numerical solution

5 次查看（过去 30 天）
ABDO
ABDO 2024-4-26
回答： Vinay 2024-9-2
= 0.3; % Rayon de lissage
dx=0.3;
N =floor(1/dx);
x =0:dx:(pi/sqrt(3)); % Positions des particules
dt = 1/4;
t = 0:dt:1;
M = length(t) - 1;
% Initialisation de la solution approchée et de la matrice A
A = zeros(N, N);
b=zeros(N,1);
uapp=zeros(N,1);
%sol et cond exacte
u_exact = @(x) cos(sqrt(3)*x);
% Construction de la matrice A et du vecteur b
for i = 1:N
for j =N
r_ij = abs(x(i) - x(j));
if r_ij <= h
d2W_dx2 = second_derivative_monaghan_kernel(r_ij, h);
W_value = monaghan_kernel(r_ij, h);
A(i, j) = (u_exact(x(j))- u_exact(x(i))) * d2W_dx2;
b(i) = -3 * u_exact(x(j)) * W_value ;
end
end
end
u_app(1)=1;
u_app(N)=u_exact(pi/sqrt(3));
uapp = b\A;
uapp = [0; uapp];
% Affichage des résultats
plot(x , uapp, 'r');
hold on;
plot(x, u_exact(x), 'b--');
xlabel('x');
ylabel('u');
legend('Solution approchée (SPH)', 'Solution exacte');
title('Solution approchée et exacte par SPH');
grid on;
% Fonctions du noyau de Monaghan
function W_value =monaghan_kernel(r_ij, h)
C= (1/ h);
if r_ij < h
W_value =C*((2 / 3) - (r_ij / h)^2 +(1/2 *(r_ij / h)^3));
elseif r_ij<2*h
W_value =C*((1/6)*(2-(r_ij / h))^3);
else
W_value=0;
end
end
function d2W_dx2 = second_derivative_monaghan_kernel(r_ij, h)
C = (1 / h); % Corrected the denominator
if r_ij < h
d2W_dx2 = -2 * C * (1 - (3 * (r_ij / h)));
elseif r_ij < 2*h
d2W_dx2 = 2 * C * (2 - (r_ij / h)); % Removed extra characters
else
d2W_dx2 = 0;
end
end
I have a problem simulataing a numerical solution for this program
Ineed some help urgent
  5 个评论
显示 3更早的评论
Torsten
Torsten 2024-4-26
It would help if you explained the problem you are trying to solve.
Further - as @Walter Roberson mentionned : the first line of your code is incomplete. Did you mean
h = 0.3; % Rayon de lissage
?
ABDO
ABDO 2024-4-26
Smoothing length h is a length that allows the support of a cut-off core to be fixed W

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Vinay
Vinay 2024-9-2
Hii ABDO,
The Smoothed Particle Hydrodynamics (SPH) method calculates the approximate values by solving the linear equation. The error arises in the code due to the following issues.
  • Loop Mistake: In the nested loop, the variable iterates for ‘j’ = ‘N’, which should be for ‘j’ = 1:N. This means the loop is only iterating over the last element of ‘x’, causing an error.
% Construction of matrix A and vector b
for i = 1:N
for j = 1:N % Corrected loop range
r_ij = abs(x(i) - x(j));
if r_ij <= h
d2W_dx2 = second_derivative_monaghan_kernel(r_ij, h);
W_value = monaghan_kernel(r_ij, h);
% Update A(i, j) based on the interaction
A(i, j) = d2W_dx2;
% Accumulate b(i) with contributions from j
b(i) = b(i) - 3 * u_exact(x(j)) * W_value;
end
end
end
  • Matrix Solver: The calculation ‘uapp’ = b\A is incorrect because ‘b’ is a column vector and ‘A’ is a square matrix. The solver solves the ‘A \ b’ to find ‘uapp’.
% Solve system
uapp = A \ b; % Solve the linear system
  • Incorrect dimension: The dimension of the ‘x’ and the ‘uapp’ are not compatible, the dimension of ‘uapp’ should be same as that of ‘x’.
x = 0:dx:(pi/sqrt(3)); % Positions of particles
N = length(x); % Update N to match the length of x

类别

在 Help Center 和 File Exchange 中查找有关 Numerical Integration and Differentiation 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by