I am trying to generate these 3 plots

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

So... you are asking the volunteers to do a search to try to find the paper that those plots appear in, and to read and understand the paper, and to write the appropriate code for you?

[DGM]

It sounds to me like you're asking to replicate an entire modeling study -- based on three graphs. The difference is unsubtle and important.

[Ive J]

First start by reading a heat transfer textbook (Holman maybe). Then try to formulate the temp profiles of those 3 objects, and ask a clear question.

[Mr.DDWW]

clc;clear all;

n=100000;
alphat=1;
y_b=linspace(0,1);
% y_b=.1
for j=1:length(y_b)
    sum_D=0;
    for i=1:n
        sum_D=sum_D+(((-1)^(i-1))/(((i-1)+0.5)*pi))*exp(-((i-1)+0.5)^2*pi^2*alphat)*cos((i-1)+0.5)*pi*y_b(j);
    end
    F(j)=2*sum_D;
end
plot(y_b,F);grid on;

I tried but not correct

[Walter Roberson]

Regardless of the theoretical shape, you can see from the below with very reduced n that after n = 2, the terms are too small to make any difference in the summation.

n=5.0000;
alphat=1;
y_b=linspace(0,1,10);
% y_b=.1
Pi = sym(pi);
for j=1:length(y_b)
    sum_D = sym(zeros(1,n));
    for i=1:n
        sum_D(i)=(((-1)^(i-1))/(((i-1)+0.5)*Pi))*exp(-((i-1)+0.5)^2*Pi^2*alphat)*cos((i-1)+0.5)*Pi*y_b(j);
    end
    vpa(sum_D)
    F(j)=2*sum(sum_D);
end

ans = 

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F(1:min(end,10)).'

ans = 

vpa(ans)

ans = 

plot(y_b,F);grid on;

simplify(F ./ F(2))

ans = 

vpa(ans)

ans = 

Answer 1

[Walter Roberson]

        sum_D(i)=(((-1)^(i-1))/(((i-1)+0.5)*Pi))*exp(-((i-1)+0.5)^2*Pi^2*alphat)*cos((i-1)+0.5)*Pi*y_b(j);

Look at that. How does the change of j affect it ?

j does not show up until y_b(j) which is a multiplier. And it is constant for all the different I values. So the result is going to be same as if you calculated for all the i values without including the y_b(j) term, and then multiplied by the y_b(j) term afterwards.

Which means that the result is going to be exactly the same each time, except multiplied by y_b(j) . And since the y_b is linear, that means you get a linear result.