4 Analytical Solution of the Segregation Problem for One Dimension

In order to assess the numerical scheme for the problem described in Section 3.5.3 it is useful to construct an analytical solution for a special one-dimensional case. For the sake of simplicity, we consider in both segments intrinsic diffusion (Section 3.3.1). As segment $S_0$ the region $x>0$ is assumed and as segment $S_1$ region $x<0$. The one-dimensional segregation problem fullfils the following initial and interface conditions,

$$C_0(x,0) = C_{init} x>0 \quad C_1(x,0) = 0 x<0,$$ (180)

$$D_0 \frac{\partial C_0}{\partial x} = - h \Bigl(C_1-\frac{C_0}{m}\Bigr)$$ (181)

$$D_0 \frac{\partial C_0}{\partial x} = D_1 \frac{\partial C_1}{\partial x}, x=0.$$ (182)

Note that condition (3.117) also means that the media from both sides of the interface has an infinite length. We are searching for the solution of the problem given by (3.30), (3.117), (3.118) and (3.119) in the form,

$$x > 0, \quad C_0(x,t) = A_0 + B_0 C(x,t,\alpha_0,D_0)$$ (183)

$$x < 0, \quad C_1(x,t) = A_1 + B_1 C(-x,t,\alpha_1,D_1)$$ (184)

where $A_0, A_1, B_0, B_1, \alpha_0, \alpha_1$ are constants to be determined and

$$C(x,t,\alpha,D)= \Bigl (\frac{x}{2 \sqrt{D t}}\Bigr) - \exp\Bigl(\frac{h x \alpha + h^2 t \alpha^{2}}{D}\Bigr) \Bigl (\frac{x}{2 \sqrt{D t}} + \frac{h\sqrt{D t} \alpha}{D}\Bigr )$$ (185)

is a solution of the diffusion equation ( $\ref{eq:intrinsic-diffusion:1}$) for the case of the surface evaporation condition already studied in [38].
We determine constants $A_0, A_1, B_0, B_1, \alpha_0, \alpha_1$ from the initial and interface conditions (3.117), (3.118), (3.119) as follows. From the initial conditions we have

$$A_0 = C_{init} \quad A_1 = 0.$$ (186)

The interface condition (3.119) yields

$$\frac{D_0}{D_1}\frac{\frac{\partial C_0}{\partial x}}{\frac{\part... ...{D_0}\Bigr )}{\text{erfc} \Bigl (\frac{h\sqrt{D_1 t} \alpha_1}{D_1}\Bigr )}=-1.$$ (187)

This equation is fullfiled if

$$\frac{\alpha_0}{\sqrt{D_0}}=\frac{\alpha_1}{\sqrt{D_1}} \quad B_0\alpha_0 = -B_1\alpha_1.$$ (188)

From (3.118) and (3.123) follows,

$$\frac{h}{D_0}\frac{\Bigl(C_{1} - \frac{C_{0}} {m}\Bigr)}{\frac{\p... ...}{D_1}) \text{erfc} \Bigl (\frac{h\sqrt{D_1 t} \alpha_1}{D_1}\Bigr)\Bigr )- \$$

$$-\frac{B_0}{m}\Bigl(1-\exp(\frac{h^2 t \alpha_0^{2}}{D_0}) \Bigl(\frac{h\sqrt{D_0 t} \alpha_0}{D_0}\Bigr )\Bigr)-\frac{C_{init}}{m}\Bigr)=1.$$ (189)

The last equality is ensured for the condition,

$$B_1-\frac{B_0}{m} = \frac{C_{init}}{m} \quad -B_1+\frac{B_0}{m} = B_0\alpha_0.$$ (190)

By solving the equation system given by (3.125) and (3.127) we have,

$$B_0 = -\frac{C_{init}}{1+m\sqrt{\frac{D_0}{D_1}}},\quad B_1 = \fr... ...+ \sqrt{\frac{D_1}{D_0}}},\quad\alpha_0 = \frac{1}{m} + \sqrt{\frac{D_0}{D_1}},$$

$$\alpha_1 = 1 + \frac{1}{m}\sqrt{\frac{D_1}{D_0}}.$$ (191)

So we can write a solution for the problem posed by (3.30), (3.117), (3.118) and (3.119). For $x>0$,

$$C_0(x,t) = C_{init} \Bigl ( 1- \Bigl ( \frac{1}{1+m\sqrt{\frac{D_0}{D_1}}}\Bigr)\Bigl ( \Bigl (\frac{x}{2 \sqrt{D_0 t}}\Bigr) -$$

$$- \exp\Bigl(\frac{h x \alpha_0 + h^2 t \alpha_{0}^{2}}{D_0}\Bigr) \Bigl (\frac{x}{2 \sqrt{D_0 t}} + \frac{h\sqrt{D_0 t} \alpha_1}{D_0}\Bigr ) \Bigr ) \Bigr),$$ (192)

and for $x<0$,

$$C_1(x,t) = \frac{C_{init}}{ m + \sqrt{\frac{D_1}{D_0}}}\Bigl ( \Bigl (-\frac{x}{2 \sqrt{D_1 t}} \Bigr) -$$

$$-\exp\Bigl(\frac{-h x \alpha_1 + h^2 t \alpha_{1}^{2}}{D_1} \Bigr ) \Bigl (-\frac{x}{2 \sqrt{D_1 t}} + \frac{h\sqrt{D_1 t} \alpha_1}{D_1}\Bigr ) \Bigr ).$$ (193)