´ ρ η ´ ρ y,t 2 D t η t = y D t π ∫ ´ ρ y,t = 1 − e dξ 2 D t ´ ρ y→∞,t≥ 0 = 0 ´ ρ y = 0, t≥ 0 = 1 ´ ρ y,t ρ − ρ ρ y,t = ´ ρ y,t − ρ ´ ρ y,t

Appendix

Exact solution of one-dimensional diffusion equation for liquid water penetrating wood

The exact solution of Eq. (2), which is the unsteady one-dimensional diffusion equation, in the text is derived here.

As an initial condition, it is assumed that the test log contains uniform water concentration,

H2O (y, t < 0) = H2O, init, in the whole region at a time before water supply (t < 0). As a boundary condition, it is assumed that the maximum water concentration, H2O (y = 0, t) = H2O, max, can be obtained at the bottom surface of the test log (y = 0), and that the initial concentration, H2O (y→∞, t)

= H2O, init, is sustained at the top of the sufficiently long test log (y→∞). The length of the used test log is L [m].

The water concentration, H2O (y, t), is represented by the normalized concentration,

´ ρ

_H2O

( y ,t )

[-], standardized by the following equation.

´ ρ

_H2O

( y ,t )= ρ

_H2O

( y , t )−ρ

_{H2O, init}

ρ

_{H2O ,max}

− ρ

_{H2O , init} ^(A1)

Further assuming that the effective diffusion coefficient, Deff, is always a constant value, Eq. (2), the initial condition, and the boundary conditions are given by the following equations using

´ ρ

_H2O

( y ,t )

_.

∂ ρ ´

_H2O

( y , t )

∂ t = D

_eff

∂

²

´ ρ

_H2O

( y , t )

∂ y

^{2 (A2)}

´ ρ

_H2O

( y ,t < 0 )=0

´ ρ

_H2O

( y=0,t ≥ 0 )=1

´ ρ

_H2O

( y →∞ , t ≥ 0)=0

When Eq. (A2) is transformed by

√ ^D

^eff

^t

, which is called the Boltzmann transformation, the exact solution can be obtained from the following equation (Bergman et al. 2017).

´ ρ

_H2O

( y ,t )=1− 2

√ ^π ∫

0 η

e

^−ξ2

dξ

_(A3)

where  is a dimensionless variable obtained by dividing the position y by the Boltzmann transformation

√ ^D

^eff

^t

^.

η (t )= y

2 √ ^D

^eff

^t

^(A4)

When the normalized concentration distribution,

´ ρ

_H2O

( y ,t )

, as the exact solution (5) is expressed by using  instead of the position y, the transformed concentration distribution,

´ ρ

_H2O

(η)

, is expressed by just one distribution that does not depend on time, t.

As shown below, the water uptake mass, dMH2O(t), is obtained by dividing Eq. (A3) by the area, A [m²], of the bottom surface of the test log and integrating from the bottom surface position (y

= 0) to the top position of the test log (y = L) against position y.

dM

_H2O

(t )= A ρ

_H2O

( y=0 ) ∫

0

L

( ¹⁻ √ ^{2 π} ∫

0 η

e

^−ξ2

dξ ) ^dy

^(A5)

Here, assuming that the length, L, is infinite and converting it to the dimensionless variable , Eq.

(A5) becomes the following:

dM

_H2O

(t )= A ρ

_H2O

( y=0 ) × 2 √ ^D

eff

t ∫

0

∞

( ¹⁻ √ ^{2 π} ∫

0 η

e

^−ξ2

dξ ) ^dη

^(A6)

The integral term in Eq. (A6) can be calculated and has the following values:

∫

0

∞

( ¹⁻ √ ^{2 π} ∫

0 η

e

^−ξ2

dξ ) ^dη= √ ^{1 π}

^(A7)

By substituting the integral value of Eq. (A7) into Eq. (A6) and rearranging the equation, the relationship between the water uptake mass, dMH2O(t), and cumulative water supply time, t, can be obtained as shown in Eq. (5).