# Application of the heat equation, PDE

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.

• What is $f$? Do you mean $\phi$? – Robert Lewis May 28 '17 at 17:16
• The region D isn't defined anywhere. Is it just everything above the x-axis? – Kaynex May 28 '17 at 17:17
• @Kaynex, The OP seems to suggest that it's in $\mathbb R^3$. – Sharat V Chandrasekhar May 28 '17 at 18:59

## 2 Answers

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 $\phi(\mathbf{x})$.

• 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? – Scooby May 28 '17 at 17:26
• 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. – eyeballfrog May 28 '17 at 17:30
• 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 $u = \phi(x)$? – Scooby May 28 '17 at 17:33
• See my answer below – Sharat V Chandrasekhar May 28 '17 at 17:34
• @SharatVChandrasekhar I am not sure how that shows me how to calculate the value of the integral in the inital state where $u = \phi(x)$? – Scooby May 28 '17 at 17:45

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

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

• 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. – Sharat V Chandrasekhar May 28 '17 at 17:31
• Could you post a detailed solution I am still having a hard time with this one – Scooby May 28 '17 at 17:55