Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 142

Its solution is, by Eq. (3.27) in Section 3.4,

0 0

N 0
2 √

e − ( x − ξ ) 2

N ( x , t )= N 0 V ( x , ξ , t ) d ξ =

4 Dt d

ξ

π Dt − ∞ − ∞

+ ∞ + ∞

N 0
2 √ π

N 0
√ π e − ( x + ξ ) 2

2

e − ξ =

4 Dt d

ξ = d ξ

Dt

x
2 √ Dt

0

+ ∞

x
2 √ Dt

N 0
√ π

2 2

e − ξ e − ξ

=

ξ − d d ξ

0 0

1

2 erfc

x
2 √ Dt

N 0
2

2
√ π

N 0 x

2 √ Dt

2

e − ξ

= − d ξ = .

0

N 0

At x = 0, N ( 0 , t )=

2 . Therefore, the density of the mixed material is constant at

the interface x = 0 and is equal to the half of the initial density N 0 .