1-d heat transfer equation

now i am wirte the heat equation code. it tisn't work ,,

L=10cm, dx=1cm n=10 T(11,200)

the initial condition is T(*,1)= 293K

the boundary condition is T(1,*)=373k , T(11,*)=T(10,*)

and this is my code

clear
clc
L=10;
n=11;
dx=L/n;
dt=0.005;
t=1;
nt=200;
alpha=0.001;
beta=alpha*dt*(1/(dx)^2);
T0=293;
T1=373;

for j=1:nt
    for i=2:n-1
        T1(i)=((1-2*beta)*T0(i)+beta*T0(i-1)+beta*T0(i+1));
    end
end
plot(T1)

[KSSV]

Your T0 is a scalar, and you are using it as a vector.

Answer 1

[Walter Roberson]

T0=293;

That is a scalar.

    for i=2:n-1        

i starts at 2

        T1(i)=((1-2*beta)*T0(i)+beta*T0(i-1)+beta*T0(i+1));

You use T0(i) and T0(i+1) . On the first iteration that would be trying to access T0(2) and T0(2+1) . But T0 is a scalar.

for j=1:nt

Your for i loop does not use j at all, and has no feedback -- the calculation of T1(i) has no reliance on T1(i) calculated from an earlier for j value. Therefore in your code, your for j is a waste of time: every j iteration is going to produce the same T1 vector.