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$$ k \frac{dT^2}{d^2 x} +heat generation =\rho c \frac{dT} {dt} $$

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?

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  • $\begingroup$ Rohit, keep in mind that heat is not temperature. $\endgroup$ Sep 26, 2019 at 16:31

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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.

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  • $\begingroup$ "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" why isn't it heating up the medium though? $\endgroup$
    – Rohit
    Sep 26, 2019 at 14:53
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    $\begingroup$ Everything is balanced by the boundary conditions. $\endgroup$
    – user196418
    Sep 26, 2019 at 14:55
  • $\begingroup$ @Rohit The boundary conditions actually are heating the domain, but in such a way that the fluxes exactly cancel out. Assuming $k=1$ for definiteness, you have a flux of $1$ into $x=1$ from outside the domain and a flux of $-1$ from $x=1$ going out into the domain. At $x=0$ instead you have a flux of $-1$ out of the domain and a flux of $1$ coming from inside the domain. The picture is the same at each interior point: on some small interval, the flux through the right endpoint into the interval is $1$ and the flux through the left endpoint into the interval is $-1$. $\endgroup$
    – Ian
    Sep 26, 2019 at 14:55
  • $\begingroup$ So you mean to say that if the heat is flowing out of the edges slower than it is flowing through the medium, then temperature gradient cannot be linear? $\endgroup$
    – Rohit
    Sep 26, 2019 at 15:02
  • $\begingroup$ @Rohit Basically yes, but I am a bit hesitant about that because of some mathematical subtleties about this "mismatched boundary condition" problem. $\endgroup$
    – Ian
    Sep 26, 2019 at 15:14
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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.

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