2.12 The Error Function 201
We are interested in the error function because of its role in solving the heat
equation. First we shall show thatthe solution of the problem
∂^2 u
∂x^2=^1
k∂u
∂t, −∞<x<∞, 0 <t, (11)
u(x, 0 )=sgn(x), −∞<x<∞ (12)is u(x,t)=erf(x/
√
4 kt).(Recallthatsgn(x)has the value−1ifxis negative or
+1ifxis positive.) The easy way to prove this statement is to verify it directly.
(See Exercises 4 and 5.) Here, we shall arrive at the same conclusion, starting
from Eq. (3),
u(x,t)=√^1
4 πkt∫∞
−∞sgn(x′)e−(x−x′)^2 /^4 ktdx′. (13)First, change the variable of integration toy=(x′−x)/√
4 kt.Thendy=
dx′/
√
4 kt,andu(x,t)=1
√π∫∞
−∞sgn(
x+y√
4 kt)
e−y^2 dy. (14)Now the functione−y^2 is even, and the sgn function changes from−1to+1at
y=−x/
√
4 kt. Thus, the integrand of Eq. (14) is as shown in Fig. 11. The tail
to the left of−x/
√
4 kthas the same area as the tail to the right ofx/√
4 ktbut
opposite sign. These two areas cancel, leaving
u(x,t)=√^1
π∫x/√ 4 kt−x/√ 4 kte−y^2 dy.Finally, use the symmetry of the integrand to halve the interval of integra-
tion and double the result:
u(x,t)=√^2
π∫x/√ 4 kt0e−y^2 dy=erf(
x/√
4 kt)
.
This is the result we wanted to arrive at. Figure 12 shows graphs ofu(x,t)=
erf(x/
√
4 kt)as a function ofxfor several values ofkt.
Because erf( 0 )=0, the functionu(x,t)=erf(x/√
4 kt)must also be the so-
lution of the problem
∂^2 u
∂x^2=^1
k∂u
∂t, 0 <x, 0 <t,u( 0 ,t)= 0 , 0 <t,u(x, 0 )= 1 , 0 <x.