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2D cylindrical boundriondition problem

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suketu vaidya
suketu vaidya 2021-1-7
回答： Alan Stevens 2021-1-8
function cyl3
close all;
L = 1; % total height of the coaxial cable
hr = 0.01;
hz = 0.01; % these should be the r coordinates of grid
vr = 0.8:hr:0.2;
vz = 0:hr:10; % these should be the z coordinates of grid
nr = length(vr);
nz = length(vz);
n = nr*nz;
A = zeros(n);
b = zeros(n,1);
for jz = 1:nz
for ir = 1:nr
i = (jz-1) * nr + ir; % the global index of the ir,jz vertex
r = vr(ir); % the geometrical r coordinate
z = vz(jz); % the geometrical z coordinate
if (r == 0.8) || (r == 0.2) || (z == 0) ||(z == 10)
% implement boundary condition for r_a ?
A(i,i) = 1;
b(i) = 0;
else
% implement the main equation here
A(i,i) = -2/hr^2 - 2/hz^2;
A(i,i+1) =1/hr^2 + (1/(2*r)*hr) ;
A(i,i-1) = 1/hr^2 - (1/(2*r)*hr);
A(i,i+nr) = 1/hz^2;
A(i,i-nr) = 1/hz^2;
end
end
end
% solve the problem
u = A\b;
s = reshape(u,nr,nz);
% ss = zeros(nz,nr);
% for jz = 1:nz
% for ir = 1:nr
% i = (jz-1) * nr + ir;
% ss(jz,i) = u(i);
% end
%end
surf(s)
end

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回答（1 个）

Alan Stevens
Alan Stevens 2021-1-8
It looks like
vr = 0.8:hr:0.2;
is the cause of the problem. This leads to vr being empty!

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