MATLAB Answers

Any mathematical mistake in my script ?

1 view (last 30 days)
Hi, I am trying to plot the self-derived analytical solution using matlab. However, I get different answer from what I obtained using excel sheet.
I suspect the mistake is in somewhere within this script, where I could have miss out.
Could anyone spare a help here ?
Self-derived Analytical Solution :
syms n y real
assume([n y] >= 0);
Y = 0:0.02:1;
b =1; t=1; br=1; Pr=1;
%t = t*
K1=(1-exp(-((2.*n)+1).^2.*pi.^2.*t./(4.*Pr)))./(((2.*n)+1).^2.*pi.^2);
K2=((((2.*n)+1).^2.*pi.^2.*exp(-((2.*n)+1).^2.*pi.^2.*t./(4.*Pr)))-(((2.*n)+1).^2.*pi.^2.*cos(2.*b.*t))-(8.*b.*Pr.*sin(2.*b.*t)))./((64.*b.^2.*Pr.^2)+(((2.*n)+1).^4.*pi.^4));
T = subs( sum( subs( 8.*br./(((2.*n)+1).*pi).*(K1+K2).*sin((((2.*n)+1)./2).*pi.*y), n, 1:100 )), y, Y );
Tn = double(T);
disp(Tn);
plot(Tn,Y);
Thanks,
Andy

Accepted Answer

Alan Stevens
Alan Stevens on 10 Aug 2021
Easier to check like this
b=1; t=1; br=1; Pr=1;
d = @(n) (2*n + 1)*pi;
c = @(n) d(n).^2;
k1 = @(n) ( 1 - exp( -c(n)*t/(4*Pr) ) )./c(n);
k2 = @(n) c(n).*exp( -c(n)*t/(4*Pr) ) - c(n)*cos(2*n*t) - 8*b*Pr*sin(2*b*t);
k3 = @(n) c(n).^2 + 64*b^2*Pr^2;
K = @(n) k1(n) + k2(n)./k3(n);
term = @(y,n) 8*br./d(n).*K(n).*sin(d(n)*y/2);
y = 0:0.02:1;
N = numel(y);
T = zeros(1,N);
for i=1:N
for n = 0:100
T(i) = term(y(i),n) + T(i);
end
end
figure
plot(y,T),grid
xlabel('y'), ylabel('T')
  3 Comments
Chee Hao Hor
Chee Hao Hor on 12 Aug 2021
Alright, for loop method, it is substituting the equation every sum set of n, giving numerical value for each discretized y. That's why no need used element wise.
Whereas my original script only substitute the sum up y at the last step, so need element wise.
Thanks @Alan Stevens !

Sign in to comment.

More Answers (0)

Products


Release

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by