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When I compute the heat equation, I found that gross of heat do not change,it is same as our real world.But as I know , the heat equation is found according to the way of heat conduction. Why the mode of heat spread contains heat conservation?

Below is my calculate, I assume $\Omega$ is compact manifold without boundary,it means $\partial\Omega=\varnothing$ $$ \left\{ \begin{aligned} & u_t=\Delta u \text{ in $\Omega\times (0,T]$} \\ & u(x,0)=u_0 \end{aligned} \right. $$ Then the gross of heat is $$ G_{heat}=\int_\Omega u dV $$ Then, show $G_{heat}$ don't change alone time $$ \begin{align} \frac{dG_{heat}}{dt} &=\int_\Omega \frac{du}{dt}dV \\ &=\int_\Omega\Delta u dV \\ &=\int_{\partial\Omega} \frac{\partial u}{\partial\overrightarrow n}dV \\ &=0 \end{align} $$ So, $G_{heat}$ don't change alone time.It means heat conservation.Why the mode of heat spread contains heat conservation?I want a physic explanation,how does local behave decides global behave?

I just a master of math,and my English is poor, so, I don't know whether rightly I describe my question.If not , welcome any edit or counsel.

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  • $\begingroup$ I do not understand you last identity, shouldn't it be $\int_\Omega\Delta u dV=\int_{\partial\Omega} \nabla u . \vec{n} dS$ ? $\endgroup$ – user83548 Dec 23 '15 at 14:55
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    $\begingroup$ Looking at the Wikipedia article would have told you that "The heat equation is derived from Fourier's law and conservation of energy". So...you're asking why an equation derived from conservation of energy conserves energy? $\endgroup$ – ACuriousMind Dec 23 '15 at 14:56
  • $\begingroup$ @ACuriousMind but you can have have heat sources and sinks. Are not these represented in the boundary conditions? for instance, if you have a hot spot at the center (I might remember this wrong, and I could not figure it it out doing a quick search) $\endgroup$ – user83548 Dec 23 '15 at 15:12
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    $\begingroup$ @brucesmitherson: Note that the question assumes $\Omega$ is a "compact manifold without boundary", so $\partial \Omega$ is empty, and so all the sources and sinks balance within $\Omega$, since it can't have net flow out of it. $\endgroup$ – ACuriousMind Dec 23 '15 at 15:16
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Quick note on terminology: in thermodynamics, heat is associated with energy transfers; the related property associated with the state of a system is called energy. A better name for the "gross of heat" would be the "total energy" of the system.

As commenters have pointed out, the heat equation is derived from the First Law of Thermodynamics. A more direct way of proving that the total energy is conserved is to apply the First Law of Thermodynamics to the system defined by the manifold $\Omega$:

$$ \frac{\text{d}}{\text{d}t} U_\Omega = \dot{Q} - P \frac{\text{d}}{\text{d}t} V + \dot{W}_\text{other} $$ Where

  • $\frac{\text{d}}{\text{d}t} U_\Omega$ is the rate of change in the total internal energy of the system
  • $\dot{Q}$ is the rate of heat transfer across the boundary, which is zero since there is no boundary
  • $-P\frac{\text{d}}{\text{d}t}V$ is the boundary work associated with expanding/contracting the system
  • $\dot{W}$ is the rate at which work crosses the boundary due to other effects (e.g. electricity), which is again zero because there is no boundary

If the volume of $\Omega$ is constant, then the boundary work is zero and thus the total internal energy of the system is constant.

To get back to your original question, if you apply the First Law of Thermodynamics to a system where only conduction occurs, substitute Fourier's Law of Conduction, and also assume that $u$ is linearly related to $T$, then the result takes the form of the Heat Equation. This is why the heat equation is called "the heat equation." As was mentioned in the comments, the heat equation conserves energy because the equations that it is derived from conserve energy.

If you prefer a more physical line of reasoning, the heat equation only represents the effects of conduction. In conduction, energy flows from high temperature to low temperature, but the total is conserved. Conduction conserves energy, so it follows that the heat equation conserves energy.

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  • $\begingroup$ Thanks very much.Are there any thermodynamics book which is interesting and suitable for student of mathematics? $\endgroup$ – lanse7pty Dec 24 '15 at 6:20

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