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

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SUMMARY

The discussion centers on the steady state solution of the heat equation, specifically why T''(x) equals zero. The boundary conditions U(0,t) = T1 and U(L,t) = T2 indicate that the ends of the bar are held at different temperatures. In steady state, the temperature distribution T(x) is linear, represented by T(x) = Ax + B, because the second derivative T''(x) must be zero, indicating no curvature in the temperature profile. This reflects a condition where the temperature does not change over time, despite the ends being at different temperatures.

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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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