Episode 04.
Linear Diffusion in Plate Solids - Unsteady State
Summary
In this episode, we describe the diffusion of plate solids within a finite range. How would we calculate the concentration If the diffusion occurs in an unsteady state? To describe this, first, we find a solution for a PDE under specific conditions. Then, we utilize the solution to examine diffusion scenarios for various cases.
1. Introduction
Continuing from the previous episode, we explain the diffusion of a plate solid in an unsteady state. In this episode, we will discuss the scenario where the solid diffuses in an unsteady state within a finite range. Before we delve into the specifics, let’s solve the following PDE.
\( {\small\displaystyle\begin{align} \frac{\partial u}{\partial t} \quad&=\quad k\frac{\partial^2 u}{\partial x^2},\qquad (0\lt x \lt L) \\[15pt] C(x,t)\quad&=\quad \begin{cases} 0,&\text{for }x\,=\,0 \\[15pt] 0,&\text{for }x\,=\,L \end{cases} \\[15pt] C(x,0) \quad&=\quad f(x) \end{align} \tag*{} } \)
When solving the above PDE, we utilize the separation of variables method. Below is the process of finding the solution to this PDE, assuming the function u consists of X(x) for position and T(t) for time.
Step 1: Separation of variables | ||||
|---|---|---|---|---|
| Suppose u | \( {\small\displaystyle u(x,t) \quad=\quad X(x)T(t) } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle \frac{\partial u}{\partial t} \,=\, XT’,\qquad \frac{\partial^2 u}{\partial x^2} \,=\, X^{\prime\prime}T } \) | |
| \( {\small\displaystyle \frac{1}{k}\cdot\frac{T’}{T} \,=\, \frac{X^{\prime\prime}}{X} \,=\, -\,\sigma^2 } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle \mathrm{i})\quad T’\,+\,k\sigma^2T \quad=\quad 0 } \) | ||
| \( {\small\displaystyle \mathrm{ii})\quad X^{\prime\prime}\,+\,\sigma^2X \quad=\quad 0 } \) | ||||
| \( {\small\displaystyle \mathrm{i}) } \) | \( {\small\displaystyle T \quad=\quad qe^{-k\sigma^2t} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle e^{-k\sigma^2t}\,[Asin(\sigma x)\,+\,Bcos(\sigma x)] } \) | |
| \( {\small\displaystyle \mathrm{ii}) } \) | \( {\small\displaystyle X \quad=\quad psin(\sigma x)\,+\,rcos(\sigma x) } \) | |||
| Step 2: Using BCs | ||||
| \( {\small\displaystyle\begin{align} e^{-k\sigma^2t}\,(Asin(\sigma L)\,+\,Bcos(\sigma L)) \quad&=\quad 0 \\[15pt] Be^{-k\sigma^2t} \quad&=\quad 0 \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} \sigma \quad&=\quad \frac{n\pi}{L},\qquad n\,=\,1,\,2,\,3,\,\cdots \\[15pt] B \quad&=\quad 0 \end{align} } \) | \( {\small\displaystyle\begin{align} u(0,t)\,&=\,0\\[10pt] u(L,t)\,&=\,0 \end{align} } \) | |
| so, | \( {\small\displaystyle\begin{align} u_n(x,t) \quad=\quad A_n\exp\left[-k\left(\frac{n\pi}{L}\right)^2\right]sin\left(\frac{n\pi}{L}x\right), \\[10pt] n\,=\,1,\,2,\,3,\cdots \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle u(x,t) \quad=\quad \sum_{n=1}^{\infty} \, A_n\exp\left[-k\left(\frac{n\pi}{L}\right)^2\right]sin\left(\frac{n\pi}{L}x\right) } \) | |
| Step 3: Using ICs | ||||
| \( {\small\displaystyle u(x,0) \quad=\quad \sum_{n=1}^{\infty}\,A_n\,sin\left(\frac{n\pi}{L}x\right) } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle f(x) \quad=\quad A_1sin\left(\frac{\pi}{L}x\right)\,+\, A_2sin\left(\frac{2\pi}{L}x\right)\,+\, A_3sin\left(\frac{3\pi}{L}x\right)\,+\cdots } \) | ||
| \( {\small\displaystyle \int_{0}^{L} \,f(x)sin\left(\frac{n\pi}{L}x\right)dx \quad=\quad A_n\int_{0}^{L}\,sin^2\left(\frac{n\pi}{L}x\right)dx } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle A_n \quad=\quad \frac{2}{L} \int_{0}^{L}\, f(x)sin\left(\frac{n\pi}{L}x\right)dx } \) | \( {\small\displaystyle \int_{0}^{L}\,sin^2\left(\frac{n\pi}{L}x\right)dx \,=\, \frac{L}{2} } \) | |
\( {\small\displaystyle\begin{align} u(x,t) \quad=\quad \sum_{n=1}^{\infty}\,A_n\exp\left[-k\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right) \tag*{} \\[15pt] \text{where } A_n \,=\, \frac{2}{L}\int_{0}^{L}\,f(x)sin\left(\frac{n\pi}{L}x\right)dx \end{align} } \)
Now, let’s discuss the diffusion of a plate solid within a finite range when it occurs in an unsteady state.
2. Various cases of linear diffusion in plate solids
Let’s take a closer look at how the concentration changes in each case when diffusion occurs only in the x-axis direction. As before, we utilize Fick’s 2nd Law and the diffuison coefficient is constant.
Case 1: when the boundary conditions are 0
\( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad&=\quad D\frac{\partial^2 C}{\partial x^2}\qquad(0\lt x \lt L) \\[15pt] C(x,t)\quad&=\quad \begin{cases} 0,&\text{for }x\,=\,0 \\[15pt] 0,&\text{for }x\,=\,L \end{cases} \\[15pt] C(x,0) \quad&=\quad f(x) \end{align} \tag*{} } \)
The conditions of case 1 are the same as those mentioned in the introduction. Therefore, the concentration under these conditions is as follows.
\( {\small\displaystyle\begin{align} C(x,t) \quad=\quad \frac{2}{L}\sum_{n=1}^{\infty}\,\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right)\int_0^L\,f(x')sin\left(\frac{n\pi}{L}x'\right)dx' \tag{Eq 4.1} \end{align} } \)
If we represent Case 1 to graph based on the given conditions, it would appear as shown in (Figure 4.1).
\( {\small\displaystyle\begin{align} C(x,0) \quad=\quad 10,\qquad L\quad=\quad 1,\qquad D\quad=\quad 0.1 \tag*{} \end{align} } \)
(Figure 4.1)
Case 2: when the boundary conditions are constant values
In this case, we can set the PDE based on Fick’s 2nd law and conditions as follows.
\( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad&=\quad D\frac{\partial^2 C}{\partial x^2}\qquad(0\lt x \lt L) \\[15pt] C(x,t)\quad&=\quad \begin{cases} C_1,&\text{for }x\,=\,0 \\[15pt] C_2,&\text{for }x\,=\,L \end{cases} \\[15pt] C(x,0) \quad&=\quad f(x) \end{align} \tag*{} } \)
Considering the concentration C as the sum of new variables u and v, we can express the equation and conditions for the variables as shown in (Table 4.1).
\( {\small\displaystyle\begin{align} C \quad=\quad u\,+\,v \end{align} \tag*{} } \)
| Variable | C | u | v |
|---|---|---|---|
| Equation | \( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad=\quad D\frac{\partial^2 C}{\partial x^2} \end{align} } \) | \( {\small\displaystyle\begin{align} \frac{\partial u}{\partial t} \quad=\quad D\frac{\partial^2 u}{\partial x^2} \end{align} } \) | \( {\small\displaystyle\begin{align} \frac{\partial^2 v}{\partial x^2} \quad=\quad 0 \end{align} } \) |
| Boundary Condtions | \( {\small\displaystyle\begin{align} C(0,t) \quad&=\quad C_1 \\[10pt] C(L,t) \quad&=\quad C_2 \end{align} } \) | \( {\small\displaystyle\begin{align} u(0,t) \quad&=\quad 0 \\[10pt] u(L,t) \quad&=\quad 0 \end{align} } \) | \( {\small\displaystyle\begin{align} v(0,t) \quad&=\quad C_1 \\[10pt] v(L,t) \quad&=\quad C_2 \end{align} } \) |
| Initial Condtions | \( {\small\displaystyle C(x,0) \quad=\quad f(x) } \) | \( {\small\displaystyle u(x,0) \quad=\quad f(x)\,-\,v } \) | - |
In the introduction section, we’ve already derived solutions for a case such as u. Using (Eq 4.1), we can arrange the concentration C as follows.
\( {\small\displaystyle\begin{align} C(x,t) \quad=\quad \sum_{n=1}^{\infty}A_n\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right)\,+\,v \\ \tag{Eq 4.2} \\ \text{where } A_n \,=\, \frac{2}{L}\int_{0}^{L}\,\{f(x)\,-\,v\}sin\left(\frac{n\pi}{L}x\right)dx \end{align} } \)
Based on the given BCs, we can obtain the function for v.
\( {\small\displaystyle\begin{align} v(x,t) \quad=\quad \frac{C_2\,-\,C_1}{L}x\,+\,C_1 \tag*{} \end{align} } \)
And substituting v into (Eq 4.2), An and C become:
\( {\small\displaystyle\begin{align} A_n \quad&=\quad \frac{2}{\pi}\cdot\frac{C_2cos(n\pi)\,-\,C_1}{n}\,+\,\frac{2}{L}\int_{0}^{L}\,f(x')sin\left(\frac{n\pi}{L}x\right)dx' \tag{Eq 4.3} \\[15pt] C(x,t) \quad&=\quad \frac{C_2\,-\,C_1}{L}x\,+\,C_1 \,+\, \frac{2}{\pi}\sum_{n=1}^{\infty}\frac{C_2cos(n\pi)\,-\,C_1}{n}\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right) \\ \tag{Eq 4.4} \\ \quad&\qquad \,+\, \frac{2}{L}\sum_{n=1}^{\infty}\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right)\int_{0}^{L}\,f(x')sin\left(\frac{n\pi}{L}x'\right)dx' \end{align} } \)
Case 3: when the boundary conditions vary with time.
\( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad&=\quad D\frac{\partial^2 C}{\partial x^2}\qquad(0\lt x \lt L) \\[15pt] C(x,t)\quad&=\quad \begin{cases} \phi_1(t),&\text{for }x\,=\,0 \\[15pt] \phi_2(t),&\text{for }x\,=\,L \end{cases} \\[15pt] C(x,0) \quad&=\quad f(x) \tag*{} \end{align} } \)
Just like in Case 2, we can approach the problem by considering the concentration C as a combination of two variables, u and v. You can find the equations and conditions for u and v in (Table 4.2).
| Variable | C | u | v |
|---|---|---|---|
| Equation | \( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad=\quad D\frac{\partial^2 C}{\partial x^2} \end{align} } \) | \( {\small\displaystyle\begin{align} \frac{\partial u}{\partial t} \quad=\quad D\frac{\partial^2 u}{\partial x^2} \end{align} } \) | \( {\small\displaystyle\begin{align} \frac{\partial v}{\partial t} \quad=\quad D\frac{\partial^2 v}{\partial x^2} \end{align} } \) |
| Boundary Condtions | \( {\small\displaystyle\begin{align} C(0,t) \quad&=\quad \phi_1(t) \\[10pt] C(L,t) \quad&=\quad \phi_2(t) \end{align} } \) | \( {\small\displaystyle\begin{align} u(0,t) \quad&=\quad 0 \\[10pt] u(L,t) \quad&=\quad 0 \end{align} } \) | \( {\small\displaystyle\begin{align} v(0,t) \quad&=\quad \phi_1(t) \\[10pt] v(L,t) \quad&=\quad \phi_2(t) \end{align} } \) |
| Initial Condtions | \( {\small\displaystyle C(x,0) \quad=\quad f(x) } \) | \( {\small\displaystyle u(x,0) \quad=\quad f(x) } \) | \( {\small\displaystyle v(x,0) \quad=\quad 0 } \) |
For u, since the conditions are the same as those we covered earlier, we can apply (Eq 4.1). For v, we will determine the solution through Laplace transform. Let’s take the Laplace transform to both sides of the PDE.
\( {\small\displaystyle\begin{align} \mathscr{L}\left[\frac{\partial v}{\partial t}\right] \quad=\quad D\mathscr{L}\left[\frac{\partial^2 v}{\partial x^2}\right] \tag{Eq 4.5} \end{align} } \)
| Step 1: Laplace transform | ||||
|---|---|---|---|---|
| From (Eq 4.5) | \( {\small\displaystyle\begin{align} sV\,-\,v(x,0) \quad=\quad D\frac{\partial^2 V}{\partial x^2} \end{align} } \) | \( {\small\displaystyle\begin{align} \longrightarrow \end{align} } \) | \( {\small\displaystyle\begin{align} V^{”}\,-\,\frac{s}{D}V \quad=\quad 0 \end{align} } \) | \( {\small\displaystyle\begin{align} L[v] \,&=\,V \\[10pt] v(x,0) \,&=\,0 \end{align} } \) |
| Step 2: Solve ODE | ||||
| Suppose V | \( {\small\displaystyle\begin{align} \lambda^2e^{\lambda x}\,-\,\frac{s}{D}e^{\lambda x} \quad=\quad 0 \end{align} } \) | \( {\small\displaystyle\begin{align} \longrightarrow \end{align} } \) | \( {\small\displaystyle\begin{align} \lambda \,=\, \pm\sqrt{\frac{s}{D}} \end{align} } \) | \( {\small\displaystyle\begin{align} V \,=\, e^{\lambda x} \end{align} } \) |
| \( {\small\displaystyle\begin{align} \text{Thus }\quad V \quad=\quad A\exp\left(\sqrt{\frac{s}{D}}x\right)\,+\,B\exp\left(-\sqrt{\frac{s}{D}}x\right) \end{align} } \) | ||||
| Using BCs | \( {\small\displaystyle\begin{align} A \quad&=\quad \frac {\Phi_2(s) \,-\, \Phi_2(s)\exp\left(-L\sqrt{\frac{s}{D}}\right)} {2sinh\left(L\sqrt{\frac{s}{D}}\right)} \end{align} } \) | \( {\small\displaystyle\begin{align} \text{and} \end{align} } \) | \( {\small\displaystyle\begin{align} B \quad&=\quad \frac {\Phi_1(s)\exp\left(L\sqrt{\frac{s}{D}}\right) \,-\, \Phi_2(s)} {2sinh\left(L\sqrt{\frac{s}{D}}\right)} \end{align} } \) | \( {\small\displaystyle\begin{align} L[\phi_1(t)]\,&=\,\Phi_1(s), \\[10pt] L[\phi_2(t)]\,&=\,\Phi_2(s) \end{align} } \) |
| \( {\small\displaystyle\begin{align} \text{Thus }\quad V \quad&=\quad \dfrac{\Phi_1sinh\left[\sqrt{\frac{s}{D}}(L\,-\,x)\right]}{sinh\left(L\sqrt{\frac{s}{D}}\right)} \,+\, \dfrac{\Phi_2sinh\left[\sqrt{\frac{s}{D}}x\right]}{sinh\left(L\sqrt{\frac{s}{D}}\right)} \\[15pt] &=\quad \Phi_1(s)\cdot s\cdot\underbrace{\frac{sinh\left[\sqrt{\frac{s}{D}}(L\,-\,x)\right]}{s\cdot sinh\left(L\sqrt{\frac{s}{D}}\right)}}_{V_1} \,+\, \Phi_2(s)\cdot s\cdot\underbrace{\frac{sinh\left[\sqrt{\frac{s}{D}}x\right]}{s\cdot sinh\left(L\sqrt{\frac{s}{D}}\right)}}_{V_2} \end{align} } \) | ||||
| Step 3: Inverse laplace Transform | ||||
| \( {\small\displaystyle \mathscr{L}^{-1}[V] \quad=\quad \mathscr{L}^{-1}\left[\Phi_1(s)\cdot s V_1(s)\right] \,+\, \mathscr{L}^{-1}\left[\Phi_2(s)\cdot s V_2(s)\right] } \) | \( {\small\displaystyle\begin{align} \longrightarrow \end{align} } \) | \( {\small\displaystyle\begin{align} v \quad=\quad \mathscr{L}^{-1}\left[\Phi_1(s)\cdot L\left[v_1’\right]\right] \,+\, \mathscr{L}^{-1}\left[\Phi_2(s)\cdot L\left[v_2’\right]\right] \end{align} } \) | \( {\small\displaystyle \mathscr{L}\left[\frac{\partial f}{\partial t}\right] \,=\, sF\,-\,f_0 } \) | |
| Convolution | \( {\small\displaystyle v \quad=\quad \mathscr{L}^{-1}\left[\Phi_1\right]\,*\,\left[v’_1\right] \,+\, \mathscr{L}^{-1}\left[\Phi_2\right]\,*\,\left[v’_2\right] } \) | \( {\small\displaystyle\begin{align} \longrightarrow \end{align} } \) | \( {\small\displaystyle v \quad=\quad \int_0^{t}\,\left[ \phi_1(\lambda)v’_1(x,t\,-\,\lambda) \,+\, \phi_2(\lambda)v’_2(x,t\,-\,\lambda)\right] \,d\lambda } \) | |
| * For V1 | ||||
| Conditions | \( {\small\displaystyle\begin{align} \phi_1(t) \quad=\quad 1 \\[10pt] \phi_2(t) \quad=\quad 0 \end{align}} \) | \( {\small\displaystyle\begin{align} \longrightarrow \end{align} } \) | \( {\small\displaystyle\begin{align} V_1 \quad=\quad \frac{sinh\left[\sqrt{\frac{s}{D}}(L\,-\,x)\right]}{s\cdot sinh\left(L\sqrt{\frac{s}{D}}\right)} \end{align}} \) | \( {\small\displaystyle\begin{align} \Phi_1(s) \,&=\, \frac{1}{s} \\[10pt] \Phi_2(s) \,&=\, 0 \end{align}} \) |
| By (Eq 4.4) | \( {\small\displaystyle\begin{align} \mathscr{L}^{-1}[V_1] \quad=\quad v_1 \quad&=\quad 1\,-\,\frac{x}{L}\,-\,\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{1}{n}\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right) \\[15pt] \longrightarrow\quad \frac{dv_1}{dt} \quad&=\quad \frac{2D}{\pi}\sum_{n=1}^{\infty}\left(\frac{n\pi}{L}\right)^2\frac{1}{n}\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right) \end{align} } \) | |||
| ** For V2 | ||||
| Conditions | \( {\small\displaystyle\begin{align} \phi_1(t) \quad=\quad 0 \\[10pt] \phi_2(t) \quad=\quad 1 \end{align}} \) | \( {\small\displaystyle\begin{align} \longrightarrow \end{align} } \) | \( {\small\displaystyle\begin{align} V_2 \quad=\quad \frac{sinh\left[\sqrt{\frac{s}{D}}x\right]}{s\cdot sinh\left(L\sqrt{\frac{s}{D}}\right)} \end{align}} \) | \( {\small\displaystyle\begin{align} \Phi_1(s) \,&=\, 0 \\[10pt] \Phi_2(s) \,&=\, \frac{1}{s} \end{align}} \) |
| By (Eq 4.4) | \( {\small\displaystyle\begin{align} \mathscr{L}^{-1}[V_2] \quad=\quad v_2 \quad&=\quad \frac{x}{L}\,+\,\frac{2}{\pi} \sum_{n=1}^{\infty}\frac{cos(n\pi)}{n}\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right) \\[15pt] \longrightarrow\quad \frac{dv_2}{dt} \quad&=\quad \frac{-2D}{\pi}\sum_{n=1}^{\infty}\left(\frac{n\pi}{L}\right)^2\frac{cos(n\pi)}{n}\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right) \end{align} } \) | |||
\( {\small\displaystyle\begin{align} v \quad=\quad \frac{2D\pi}{L^2}\sum_{n=1}^{\infty} n\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right]sin\left(\frac{n\pi}{L}x\right)\int_{0}^{t}\,\exp\left[D\left(\frac{n\pi}{L}\right)^2\lambda\right]\left\{\phi_1(\lambda)\,-\,(-1)^n\phi_2(\lambda)\right\}\,d\lambda \tag{Eq 4.6} \end{align} } \)
As C is the sum of u and v, we can represent the concentration C using (Eq 4.1) and (Eq 4.6) as follows.
\( {\small\displaystyle\begin{align} C \quad=\quad \frac{2}{L}\sum_{n=1}^{\infty} &\exp\left[-D\left(\frac{n\pi}{L}\right)^2t\right] sin\left(\frac{n\pi}{L}x\right) \\ \tag{Eq 4.7} \\ &\times\left[\int_{0}^{L}\,f(x^{'})sin\left(\frac{n\pi}{L}x^{'}\right)dx^{'}\,+\,\frac{n\pi D}{L}\int_0^{t}\,\exp\left[D\left(\frac{n\pi}{L}\right)^2\lambda\right]\left(\phi_1(\lambda)\,-\,(-1)^n\phi_2(\lambda)\right)d\lambda\right] \end{align} } \)
Nomenclature
BCs : boundary conditions
\({\small\displaystyle C}\) : concentration
\({\small\displaystyle D}\) : diffusion coefficient
\({\small\displaystyle \text{ICs}}\) : initial conditions
\({\small\displaystyle L}\) : length
\({\small\displaystyle \text{PDE}}\) : partial differential equation
\({\small\displaystyle t}\) : time
\({\small\displaystyle u,\,v,\,x}\) : variables
\(\small\mathscr{L}\) : laplace transform