# Question about heat equation

1. Apr 7, 2012

### wumple

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