# Heat equation

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 1.62.

The concepts of thermal conductivity and specific heat capacity can be combined to derive the heat equation, which governs how heat spreads through an object with a non-uniform temperature distribution. We’ll derive the one-dimensional version of the heat equation.

Suppose we have a bar of material with some initial temperature distribution ${T\left(x,0\right)}$, where ${T}$ is a function of position ${x}$ along the bar and time ${t}$, so ${T\left(x,0\right)}$ is the initial state of the bar at ${t=0}$. Consider two adjacent slices in the bar, each of width ${\Delta x}$. The first slice is bounded by ${x_{1}}$ and ${x_{2}}$ and the second slice by ${x_{2}}$ and ${x_{3}}$. According to heat conduction equation, the amount of heat ${Q_{2}}$ flowing into the second slice from the first slice in time interval ${\Delta t}$ is

$\displaystyle \frac{Q_{2}}{\Delta t}=-k_{t}A\frac{T_{2}-T_{1}}{\Delta x} \ \ \ \ \ (1)$

where ${k_{t}}$ is the thermal conductivity and ${A}$ is the cross-sectional area of the bar.

Similarly, the amount of heat flowing out of the second slice on the other side is

$\displaystyle \frac{Q_{1}}{\Delta t}=-k_{t}A\frac{T_{3}-T_{2}}{\Delta x} \ \ \ \ \ (2)$

The difference ${Q_{2}-Q_{1}}$ is stored in the second slice and will cause a change ${\Delta T}$ in temperature within the slice. If the heat capacity of the material is ${c}$ and its mass density is ${\rho}$ then

$\displaystyle \frac{Q_{2}-Q_{1}}{\left(A\Delta x\right)c\rho}=\Delta T \ \ \ \ \ (3)$

Plugging in the values for ${Q_{1}}$ and ${Q_{2}}$ we get

$\displaystyle \frac{\Delta T}{\Delta t}=\frac{k_{t}}{c\rho}\left[\frac{T_{1}-2T_{2}+T_{3}}{\left(\Delta x\right)^{2}}\right] \ \ \ \ \ (4)$

In the limit ${\Delta x\rightarrow0}$, the quantity in brackets goes to ${\partial^{2}T/\partial x^{2}}$.

If you haven’t seen this form before, the argument goes like this. For some function ${f\left(x\right)}$, the second derivative is defined as

 $\displaystyle \frac{d^{2}f}{dx^{2}}$ $\displaystyle \equiv$ $\displaystyle \lim_{\Delta x\rightarrow0}\frac{f^{\prime}\left(x+\Delta x\right)-f^{\prime}\left(x\right)}{\Delta x}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \lim_{\Delta x\rightarrow0}\frac{1}{\Delta x}\frac{f\left(x+2\Delta x\right)-f\left(x+\Delta x\right)-\left[f\left(x+\Delta x\right)-f\left(x\right)\right]}{\Delta x}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \lim_{\Delta x\rightarrow0}\frac{f\left(x+2\Delta x\right)-2f\left(x+\Delta x\right)+f\left(x\right)}{\left(\Delta x\right)^{2}} \ \ \ \ \ (7)$

This has the same form as 4. Thus in the limit, we get the heat equation

$\displaystyle \frac{\partial T}{\partial t}=\frac{k_{t}}{c\rho}\frac{\partial^{2}T}{\partial x^{2}} \ \ \ \ \ (8)$

One solution of this equation is

$\displaystyle T\left(x,t\right)=T_{0}+\frac{A}{\sqrt{t}}e^{-x^{2}/4Kt} \ \ \ \ \ (9)$

where

$\displaystyle K\equiv\frac{k_{t}}{c\rho} \ \ \ \ \ (10)$

This can be verified by taking the derivatives, and we find that

 $\displaystyle \frac{\partial^{2}T}{\partial x^{2}}$ $\displaystyle =$ $\displaystyle \frac{A\left(x^{2}-2Kt\right)}{4K^{2}t^{5/2}}e^{-x^{2}/4Kt}\ \ \ \ \ (11)$ $\displaystyle \frac{\partial T}{\partial t}$ $\displaystyle =$ $\displaystyle \frac{A\left(x^{2}-2Kt\right)}{4Kt^{5/2}}e^{-x^{2}/4Kt}=K\frac{\partial^{2}T}{\partial x^{2}} \ \ \ \ \ (12)$

The function 9 with ${T_{0}=0}$ and ${A=1/2\sqrt{\pi K}}$ is actually a delta function in the limit ${t\rightarrow0}$. Using Maple, we find that

$\displaystyle \frac{1}{2\sqrt{\pi K}}\int_{-\infty}^{\infty}\frac{e^{-x^{2}/4Kt}}{\sqrt{t}}dx=1 \ \ \ \ \ (13)$

For any ${x\ne0}$, ${\lim_{t\rightarrow0}\frac{e^{-x^{2}/4Kt}}{\sqrt{t}}=0}$ so in this limit, the function has an infinitely high spike at ${x=0}$ and an integral of 1, which are the conditions of a delta function. As time increases, the function becomes a standard Gaussian curve which gradually spreads out until as ${t\rightarrow\infty}$, it becomes zero, so ${\lim_{t\rightarrow\infty}T\left(x,t\right)=T_{0}}$. This is what we’d expect since the heat will gradually diffuse over the bar until everywhere is at the same background temperature. Here are some plots of ${T\left(x,t\right)}$ for various values of ${t}$, with ${A=1/2\sqrt{\pi K}}$, ${K=1}$ and ${T_{0}=0}$:

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