Thomas algorithm - tridiagonal matrix
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Is there any other way to code and solve the tridiagonal matrix? the idea would be to try to get the plot shown. Matlab beginner, so, no sure how to do it. Any help will be greatly appreciated. Thanks
clear
cm=1/100;
delta = 1*cm;
nu=1e-6;
Uinf=1;
H=2*delta;
y=linspace(0,H,40);
u=Uinf*erf(3*y/delta);
dy=mean(diff(y));
dx=100*dy;
Q=nu*dx/Uinf/dy^2;
a=-Q;
c=-Q;
b=1+2*Q;
N=length(y)-2;
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);
%Constant Ue
Ue = @(x) Uinf;
u = u(2:end-1);
x=0;
uall=[0,u,Ue(x)];
采纳的回答
clear
cm=1/100;
delta = 1*cm;
nu=1e-6;
Uinf=1;
H=2*delta;
y=linspace(0,H,40);
u=Uinf*erf(3*y/delta);
dy=mean(diff(y));
dx=100*dy;
Q=nu*dx/Uinf/dy^2;
a=-Q;
c=-Q;
b=1+2*Q;
N = length(y);
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);
M(1,:) = [1,zeros(1,N-1)];
M(end,:) = [zeros(1,N-1),1];
%Constant Ue
Ue = @(x) Uinf;
u = u(2:end-1);
x=0;
uall=[0,u,Ue(x)];
sol = M\uall.';
plot(y,sol)
grid on
5 个评论
Cesar Cardenas
2023-3-2
Torsten
2023-3-2
These are the changes I made:
N = length(y);
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);
M(1,:) = [1,zeros(1,N-1)];
M(end,:) = [zeros(1,N-1),1];
instead of
N=length(y)-2;
M = diag(b*ones(1,N)) + diag(c*ones(1,N-1),1) + diag(a*ones(1,N-1),-1);
Think about what they mean.
Cesar Cardenas
2023-3-2
Torsten
2023-3-2
You can assign values to certain elements in a matrix by using a loop. But if the above line to define M is correct, it's elegant, isn't it ? Why do you want to define it differently ?
Cesar Cardenas
2023-3-2
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