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)


\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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