Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
$$ k \frac{dT^2}{d^2 x} +heat generation =\rho c \frac{dT} {dt} $$
I find it counter-intuitive that if temperature gradient is linear, temperature won't change with time at a point.
I mean if there is a difference in temperature with space, shouldn't heat flow, and flow till there is a difference no more?
So first off, the case $T(x)=x$ on the whole line is nonphysical, so you should first forget about that. A temperature distribution on an unbounded domain will go to some constant at infinity in a reasonable physical situation. Intuitively speaking, you can't exchange heat with infinity.
On a bounded domain, you have boundary conditions. For example, if there's no bulk heat input but you have boundary conditions $T(0)=0,T(1)=1$ then indeed $T(x)=x$ is a steady state. But the point is that there actually is heat flow here: heat is flowing into the rod on the right at a certain rate and leaving the rod on the left at exactly the same rate. The temperature profile isn't changing despite this flow.
On the other hand, if you require that no heat be conducted on the edges either, then the suitable boundary condition is $T_x=0$ on the boundary. A nonconstant linear profile does not satisfy this boundary condition. If you try to prepare a linear profile in this situation, intuitively your profile will really be something which looks linear and then rapidly flattens out near the boundaries. The little regions where the transitions from $T'=1$ to $T'=0$ occur will have nonzero $T''$, so heat will start to flow and you'll get a flat temperature profile in the long run.
It will probably never get more intuitive as to understanding how energy conservation is stated to obtain this equation and Fourier's law relating the heat flux density vector $\mathbf{q}$ to the temperature $T$. Energy conservation states that the variation of temperature in time in a given volume stems from the difference of outgoing and ingoing heat flux, which is quantified by the divergence of $\mathbf{q}$. So no temperature variation actually means $\mathbf{\nabla}.\mathbf{q}=0$. In 1D, this is just $\frac{\partial q}{\partial x}=0$, the heat flux density remains constant along $x$. Because of Fourier law $\mathbf{q}=-k\nabla T $, the temperature is then linear.
