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clc
disp("the soln of eg 8.2-->");
for i=1:100, x(i)=0
end
iter=0, e1=1, f=1
k=.1,D=10^-9,r=.01,delta_r=r/10,t1=k*delta_r^2/D
while e1>1e-6 & f>1e-6 do iter=iter+1,for i=1:100, x1(i)=x(i),
end, for i=2:100, a(i)=1-(1/(i-1))
end, b(1)=-6-.t1*x1(1), for i=2:100, b(i)=-2-t1*x1(i)
end, c(1)=6, for i=2:99,c(i)=1+(1/(i-1))
end, for i=1:99,d(i)=0, end, d(100)=-100/99,
i=1, n=100, Beta(i)=b(i),
Gamma(i)=d(i)/Beta(i)
i1=i+1,
for j=i1:n, Beta(j)=b(j)-a(j)*c(j-1)/Beta(j-1),
Gamma(j)=(d(j)-a(j)*Gamma(j-1))/Beta(j),
end
x(n)=Gamma(n)
n1=n-i,
for k=1:n1, j=n-k,x(j)=Gamma(j)-c(j)*x(j+1)/Beta(j)
end
e1=abs(x(1)-x1(1)),
f=abs(x(100)-x1(100)),
end
disp("the solution by TDMA of node 77 to 99 by 2nd order rxn. is");
for i=77:100,
disp(x(i));
end
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