What are the physical implications of these BCs to the 1D heat equation?

Question

The heat equation in 1 dimension states that:
$$frac{partial T(x,t)}{partial t} = alpha frac{partial^2 T(x,t)}{partial x^2}, , quad alpha > 0$$
And my boundaries on a finite rod of length $L$ are:
begin{align}
partial_xT(x,t) = partial_xT(L,t) = 0
end{align}
What do they mean physically. Also what are the differences between the BC's above and below:
begin{align}
partial_xT(0,0) = partial_xT(L,0) = 0
end{align}

Gert · Accepted Answer

Your first BC doesn't make any sense:
$$begin{align}
T'(x,t) = T'(L,t) = 0
end{align}$$
because $T'(x,t)=0$ means there's no heat flux at all, for any $x$. That's nonsensical. Note also that your BCs must be in partials because you're looking for a function $T(x,t)$
It should be:
$$begin{align}
partial_xT(0,t) = partial_xT(L,t) = 0
end{align}$$
Then it means that at all times $t$ there is no heat flow in or out of the points $x=0$ and $x=L$. I.o.w. the rod is perfectly insulated at both its ends.
Note that this is so because heat flux in $text{1D}$ is given by:
$$q=-kfrac{partial T(x,t)}{partial x}$$
As regards:
$$begin{align}
partial_xT(x,0) = partial_xT(L,0) = 0
end{align}$$
It's an initial condition but it makes little sense. A useful IC would be simply:
$$T(x,0)=f(x)$$
which describes the initial temperature distribution ($t=0$)

What are the physical implications of these BCs to the 1D heat equation?

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