Any mathematical mistake in my script ?

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

Answer 1

[Alan Stevens]

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')

[Chee Hao Hor]

Hi @Alan Stevens, thanks for helping me, sincerely.

I see you use a different approach (for loop method) from what i am using (system sum method) for a summation solution.

Yours look simpler and clean.

I have 3 questions from here:

1. Shouldn't it be all using .* and ./ since we are summing up the "n" term by term (consider as element wise sum up right) ? I tried using your current script, it gives same results with I added all with .* and ./. I could not imagine how they sum up using mixed of *; .*; / and ./.
2. In your case, you first sum up the T for a particular of y location (here is up to 100 n terms, increment 1 each time). Then do the same for rest of the discretized y which defined by array of zeros N elements (from 0 to 1, increment 0.02 each time), am i right ?
3. If i have a similar analytical solution that consists of triple summation terms, it involved piecewise function, using for loop method should work also right ?

Thanks,

Andy :)

[Alan Stevens]

"1. Shouldn't it be all using .* and ./ since we are summing up the "n" term by term (consider as element wise sum up right) ? I tried using your current script, it gives same results with I added all with .* and ./. I could not imagine how they sum up using mixed of *; .*; / and ./."

Yes, you can replace the .* etc by just . I put .* initially, as I hadn't decided if I would use a loop or vector indexing.

"2. In your case, you first sum up the T for a particular of y location (here is up to 100 n terms, increment 1 each time). Then do the same for rest of the discretized y which defined by array of zeros N elements (from 0 to 1, increment 0.02 each time), am i right ?"

Yes, that's right.

"3. If i have a similar analytical solution that consists of triple summation terms, it involved piecewise function, using for loop method should work also right ?"

Yes, it should.

[Chee Hao Hor]

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 !