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A solid, homogeneous object occupies the region $D\subseteq \mathbb{R}^3$, and is completely insulated. Its initial temperature is given by $u(x,0) = \phi(x)$, for some function $\phi$. So $u$ satisfies the heat equation $u_t = k\Delta u$ with boundary condition $\frac{\partial u}{\partial n} = 0$ on $\partial D$. After a long time, the object reaches a steady, uniform temperature. In terms of $f$, what is this temperature?

I normally don't post a question without at least some attempt but I have no idea how to start this. Any suggestions are greatly appreciated.

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  • $\begingroup$ What is $f$? Do you mean $\phi$? $\endgroup$ – Robert Lewis May 28 '17 at 17:16
  • $\begingroup$ The region D isn't defined anywhere. Is it just everything above the x-axis? $\endgroup$ – Kaynex May 28 '17 at 17:17
  • $\begingroup$ @Kaynex, The OP seems to suggest that it's in $\mathbb R^3$. $\endgroup$ – Sharat V Chandrasekhar May 28 '17 at 18:59
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From the boundary condition, we have $$ \oint_{\partial D} \boldsymbol\nabla u \cdot d\mathbf{A} = \oint_{\partial D}\frac{\partial u}{\partial n} dA = 0. $$ From the divergence theorem, we have $$ \oint_{\partial D} \boldsymbol\nabla u \cdot d\mathbf{A} = \int_D\nabla^2u \,dV = \int_D\frac{1}{k}\frac{\partial u}{\partial t} dV = \frac{d}{dt}\int_D\frac{u}{k}\,dV $$ assuming $k$ is constant in time (if it is also constant in space, you can pull it out of the integral entirely). In other words $$ \frac{d}{dt}\int_D \frac{u}{k}\, dV = 0 $$ This should not be surprising, as heat conduction conserves energy and the boundary condition says there's no energy outflow from $D$. From here I think you can get the answer.

EDIT: So continuing from here, we have that the integral in the final state must equal the integral in the initial state. Thus, $$ \int\frac{u(\mathbf{x},0)}{k}dV = \int_D\frac{\phi(\mathbf{x})}{k}dV = \int_D\frac{u(\mathbf{x},\infty)}{k}dV = \int_D\frac{T}{k}dV = T\int_D\frac{dV}{k}. $$ So $$ T = \frac{\int_D \frac{\phi(\mathbf{x})}{k}dV}{\int_D\frac{dV}{k}} $$ If $k$ is constant in space, we can multiply the top and bottom by $k$ to get $$ T = \frac{\int_D \phi(\mathbf{x}) dV}{\int_D dV} = \frac{1}{V}\int_D\phi(\mathbf{x})dV $$ where $V$ is the volume of $D$.

This is as far as you can go without knowing the actual function $\phi(\mathbf{x})$.

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  • $\begingroup$ I am not sure where to go from where you started, I understand your work. But what exactly to I need to do from here? $\endgroup$ – Scooby May 28 '17 at 17:26
  • $\begingroup$ If the time derivative of a quantity is zero, that means its initial value is equal to its final value. So calculate the value of the integral in the initial state where $u = \phi(\mathbf{x})$ and in the final state where $u = T$, then set them equal to each other and solve for T. $\endgroup$ – eyeballfrog May 28 '17 at 17:30
  • $\begingroup$ I am sorry I am a bit slow at the moment. Could you re-edit to show me how to calculate the integral in the initial state where $u = \phi(x)$? $\endgroup$ – Scooby May 28 '17 at 17:33
  • $\begingroup$ See my answer below $\endgroup$ – Sharat V Chandrasekhar May 28 '17 at 17:34
  • $\begingroup$ @SharatVChandrasekhar I am not sure how that shows me how to calculate the value of the integral in the inital state where $u = \phi(x)$? $\endgroup$ – Scooby May 28 '17 at 17:45
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If the object is completely insulated, then there is no loss of thermal energy in the system and so if the thermal capacity ($\rho c_p$) of the object is spatially invariant, then the equilibrium temperature will be

$$ u_{\text{eq}} = \frac{1}{V}\int_V\phi({\bf x})\text dv $$

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  • $\begingroup$ If you're trying to perform an actual set of calculations, then you would need to pay careful attention to a possible singularity at $t=0$ if $\phi({\bf x}), {\bf x} \in \partial D$ does not satisfy the $\frac{\partial u}{\partial n}=0$ condition. $\endgroup$ – Sharat V Chandrasekhar May 28 '17 at 17:31
  • $\begingroup$ Could you post a detailed solution I am still having a hard time with this one $\endgroup$ – Scooby May 28 '17 at 17:55

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