- #1
wumple
- 60
- 0
Hi,
So if I start with the boundary conditions
[tex]U(0,t) = T1 [/tex] and [tex] U(L,t) = T2 [/tex]
and T1 does not equal T2, it seems that you are supposed to look at the 'steady state solution' (solution as t goes to infinity)?
which satisfies
[tex] T''(x) = 0 [/tex]
so the solutions are
[tex]T(x) = Ax + B[/tex]
and then you fit the BCs to this 'steady-state solution'
Why is this the 'steady state solution'? And why does the second derivative have to be 0? After a long period of time, wouldn't the entire bar (minus the ends) be the same temperature, so then just T'(x) = 0?
Thanks
So if I start with the boundary conditions
[tex]U(0,t) = T1 [/tex] and [tex] U(L,t) = T2 [/tex]
and T1 does not equal T2, it seems that you are supposed to look at the 'steady state solution' (solution as t goes to infinity)?
which satisfies
[tex] T''(x) = 0 [/tex]
so the solutions are
[tex]T(x) = Ax + B[/tex]
and then you fit the BCs to this 'steady-state solution'
Why is this the 'steady state solution'? And why does the second derivative have to be 0? After a long period of time, wouldn't the entire bar (minus the ends) be the same temperature, so then just T'(x) = 0?
Thanks