202 Chapter 2 The Heat Equation
(a)(b)Figure 11 (a) Graph of exp(−y^2 )and (b) graph of sgn(x+y
√
4 kt)exp(−y^2 ).
The tails beyond±x/
√
4 kthave the same areas, with opposite signs.Figure 12 Graphs of the solution of the problem in Eqs. (11) and (12),
u(x,t)=erf(x/
√
4 kt),forxin the range−2to2andforkt= 0 .01, 0.1, 1, and 10.
Asktincreases, the graph ofu(x,t)collapses toward thex-axis.
A simple modification leads to the conclusion that the complementary error
function,u(x,t)=erfc(x/
√
4 kt)is the solution of this problem with zero ini-
tial condition and constant boundary condition,
∂^2 u
∂x^2 =1
k∂u
∂t,^0 <x,^0 <t,
u( 0 ,t)= 1 , 0 <t,
u(x, 0 )= 0 , 0 <x.