Why is T''(x) Zero in the Steady State Solution of the Heat Equation?

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In the steady state solution of the heat equation, T''(x) equals zero because the temperature distribution along the bar becomes constant over time, leading to a linear temperature profile represented by T(x) = Ax + B. This condition arises when the system has reached equilibrium, where the temperature at each point no longer changes, despite the boundary conditions at the ends being different. The steady state reflects a balance between heat conduction and the constant temperatures maintained at the boundaries. Although the entire bar does not reach the same temperature, the steady state indicates that the temperature gradient is constant, resulting in no further changes in temperature over time. Thus, T'(x) can be non-zero, but T''(x) must be zero to signify that the temperature distribution is stable.
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
 
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If the two ends of the bar are always held at different temperatures, how can the entire bar reach the same temperature? The equation T"=0 comes from a differential conductive heat balance on a small section of the bar. At steady state, the temperature at all points along the bar are not changing with time.
 
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