wumple
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Hi,
So if I start with the boundary conditions
U(0,t) = T1 and U(L,t) = T2
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
T''(x) = 0
so the solutions are
T(x) = Ax + B
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
U(0,t) = T1 and U(L,t) = T2
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
T''(x) = 0
so the solutions are
T(x) = Ax + B
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