Chapter 5, Problem 67RE

To determine

To find: the temperature distributed that results from an initial hot spot concentrated at the origin, need to compute $\underset{a\to 0}{\lim }T(x,t)$ using L’Hospital Rule.

Answer to Problem 67RE

  $\underset{a\to 0}{\lim }T(x,t)=\frac{C{e}^{-\frac{x^2}{(4kt)}}}{\sqrt{4\pi kt}}.$

Explanation of Solution

Given information: Suppose that the temperature in a long. Thin rod placed along the

x − axis is initially $\frac{C}{2a}\text{if}\left|x\right|\le a\text{and}\text{if}\left|x\right|>a$ . It can be shown that if the heat diffusivity

of the rode is k ,then the temperature of the rod at the point x at time t is:

  $T(x,t)=\frac{C}{a\sqrt{4\pi kt}}{\displaystyle \underset{0}{\overset{a}{\int }}{e}^{-\frac{{(x-u)}^2}{4kt}}}du$ .

Formula Used:

2^nd Fundamental Theorem of calculus:

  $\frac{d}{dt}{\displaystyle \underset{a}{\overset{x}{\int }}f(t)dt=f(x)}$ .

L-Hospital Rule:

  $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f\text{'}(x)}{g\text{'}(x)}.$

Calculation:

  $\begin{array}{c}\underset{a\to 0}{\lim }T(x,t)=\underset{a\to 0}{\lim }\frac{C}{a\sqrt{4\pi kt}}{\displaystyle \underset{0}{\overset{a}{\int }}{e}^{-\frac{{(x-u)}^2}{4kt}}}du\\ =\frac{C}{\sqrt{4\pi kt}}\underset{a\to 0}{\lim }\frac{{\displaystyle \underset{0}{\overset{a}{\int }}{e}^{-\frac{{(x-u)}^2}{4kt}}}du}{a}\end{array}$

Since the limit is of the form $\frac{0}{0}$ , can use the L-Hospital Rule.

Differentiate the numerator using the 2^nd Fundamental Theorem of calculus

  $\begin{array}{c}=\frac{C}{\sqrt{4\pi kt}}\underset{a\to 0}{\lim }\frac{e^{-\frac{{(x-a)}^2}{(4kt)}}}{1}\\ \underset{a\to 0}{\lim }T(x,t)=\frac{C{e}^{-\frac{x^2}{(4kt)}}}{\sqrt{4\pi kt}}.\end{array}$