Episode 07.
Three-Dimensional Diffusion in Cylindrical Solids - Unsteady State
Summary
This episode covers three-dimensional diffusion in a cylindrical solid under an unsteady state. Unlike previous episodes, we discuss how to calculate the concentration when the boundary conditions vary not with time but with radius, azimuth, or height.
1. Introduction
In this episode, we discuss three–dimensional diffusion within a cylindrical solid under an unsteady state. When dealing with the three–dimensional diffusion in a cylindrical solid, we can classify whether the height of the solid is infinite or finite. Let’s explore how to derive solutions to (Eq 2.3) based on various scenarios and conditions.
\( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad&=\quad \frac{D}{r}\left[ \frac{\partial}{\partial r}\left(r\frac{\partial C}{\partial r}\right)\,+\, \frac{\partial}{\partial\theta}\left(\frac{1}{r}\frac{\partial C}{\partial\theta}\right)\,+\, \frac{\partial}{\partial z}\left(r\frac{\partial C}{\partial z}\right) \right] \tag{Eq 2.3} \end{align} } \)
(Figure 7.1)
2. 3D - diffusion in infinite cylindrical solids
Let’s discuss that three-dimensional diffusion in a cylindrical solid with an infinite height. In this scenario, the initial condition is a function of the radius, azimuthal angle, and height.
Case 1: when boundary condition is 0.
\( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad&=\quad \frac{D}{r}\left[ \frac{\partial}{\partial r}\left(r\frac{\partial C}{\partial r}\right)\,+\, \frac{\partial}{\partial\theta}\left(\frac{1}{r}\frac{\partial C}{\partial\theta}\right)\,+\, \frac{\partial}{\partial z}\left(r\frac{\partial C}{\partial z}\right) \right],\qquad (0\lt r\lt a) \\[15pt] C(a,\theta,z,t) \quad&=\quad 0 \\[10pt] C(r,\theta,z,0) \quad&=\quad f(r,\theta,z) \tag*{} \end{align} } \)
So far, we’ve obtained the solution to the PDE under these conditions using the separation of variables method. Based on Episode 06, we can infer that the forms of each variable are as follows.
\( {\small\displaystyle C \,=\, R\Theta HT \quad\longrightarrow\quad \begin{align} R\,&=\,\alpha J_{\lambda}(\sigma r) \\[10pt] \Theta\,&=\,acos(\lambda\theta)\,+\,bsin(\lambda\theta) \\[10pt] H\,&=\,pcos(\mu z)\,+\,qsin(\mu z) \\[10pt] T\,&=\,\exp\left[-D\left(\sigma^2\,+\,\mu^2\right)t\right] \end{align} \tag*{} } \)
Let’s solve the PDE using the forms of each variable. In the case where the cylindrical solid has an infinite height, the solution to this PDE can be obtained through the following steps.
| Step 1: Fourier series | ||||
|---|---|---|---|---|
| Initial condition | \( {\small\displaystyle\begin{align} f(r,\theta,z) \quad=\quad \sum_{n=0}^{\infty} A_ncos(n\theta)\,+\,B_nsin(n\theta) \end{align} } \) | where | \( {\small\displaystyle\begin{align} A_n(r,z) \,&=\, \sum_{\sigma} \alpha_n(z)J_n(\sigma r)& \\[10pt] B_n(r,z) \,&=\, \sum_{\sigma} \beta_n(z)J_n(\sigma r)& \end{align} } \) | \( {\small\displaystyle\begin{align} &A_n,\,B_n \\[8pt] &\text{ are functions of } r,z \end{align} } \) |
| Step 2: Fourier Integral | ||||
| \( {\small\displaystyle \mathrm{i}. } \) | \( {\small\displaystyle\begin{align} \alpha_n(z) \quad=\quad \int_0^{\infty}\,a(\epsilon)cos(\epsilon z)\,+\,b(\epsilon)sin(\epsilon z)\,d\epsilon \end{align} } \) | where | \( {\small\displaystyle\begin{align} a(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,\alpha_n(\omega)cos(\omega\epsilon)\,d\omega \\[10pt] b(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,\alpha_n(\omega)sin(\omega\epsilon)\,d\omega \end{align} } \) | |
| \( {\small\displaystyle\begin{align} \alpha_n(z) \quad&=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\, \alpha_n(\omega) \left[cos(\omega\epsilon)cos(\epsilon z)\,+\,sin(\omega\epsilon)sin(\epsilon z)\right]\,d\omega d\epsilon \\[10pt] &=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\, \alpha_n(\omega) \left[\frac{cos\epsilon(\omega\,+\,z)\,+\,cos\epsilon(\omega\,-\,z)}{2}\,-\,\frac{cos\epsilon(\omega\,+\,z)\,-\,cos\epsilon(\omega\,-\,z)}{2}\right]\,d\omega d\epsilon \\[10pt] &=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\,\alpha_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon \end{align} } \) | ||||
| \( {\small\displaystyle \mathrm{ii}. } \) | \( {\small\displaystyle\begin{align} \beta_n(z) \quad&=\quad \int_0^{\infty}\,c(\epsilon)cos(\epsilon z)\,+\,d(\epsilon)sin(\epsilon z)\,d\epsilon \\[10pt] &=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\,\beta_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon \end{align} } \) | where | \( {\small\displaystyle\begin{align} c_n(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,\beta_n(\omega)cos(\omega\epsilon)\,d\omega \\[10pt] d_n(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,\beta_n(\omega)sin(\omega\epsilon)\,d\omega \end{align} } \) | |
| \( {\small\displaystyle\begin{align} f(r,\theta,z) \quad&=\quad \frac{1}{\pi}\sum_{\sigma}\sum_{n=0}^{\infty} J_n(\sigma r) \left[\int_0^{\infty}\int_{-\infty}^{\infty}\,\alpha_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon\, cos(n\theta)\,+\,\int_0^{\infty}\int_{-\infty}^{\infty}\,\beta_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon\, sin(n\theta)\right] \\[10pt] &=\quad \frac{1}{\pi}\sum_{\sigma}\sum_{n=0}^{\infty}J_n(\sigma r)\left[\int_0^{\infty}\int_{-\infty}^{\infty}\,\{\alpha_n(\omega)cos(n\theta)\,+\,\beta_n(\omega)sin(n\theta)\}\times cos\epsilon(\omega\,-\,z)\,d\omega d\epsilon\right] \end{align} } \) | ||||
| \( {\small\displaystyle\begin{align} \text{Thus},\qquad& \\[10pt] C \quad&=\quad f(r,\theta,z)\cdot T \\[10pt] &=\quad \frac{1}{\pi}\sum_{\sigma}\sum_{n=0}^{\infty}\left[\int_0^{\infty}\int_{-\infty}^{\infty}\,\exp\left[-D\left(\sigma^2\,+\,\epsilon^2\right)t\right]J_n(\sigma r)\{\alpha_n(\omega)cos(n\theta)\,+\,\beta_n(\omega)sin(n\theta)\}cos\epsilon(\omega\,-\,z)\,d\omega d\epsilon\right] \end{align} } \) | ||||
\( {\small\displaystyle\begin{align} C \quad=\quad \frac{1}{\pi}\sum_{\sigma}\sum_{n=0}^{\infty}\int_0^{\infty}\int_{-\infty}^{\infty}\,\exp\left[-D\left(\sigma^2\,+\,\epsilon^2\right)t\right]J_n(\sigma r)\{\alpha_n(\omega)cos(n\theta)\,+\,\beta_n(\omega)sin(n\theta)\} \\ \tag{Eq 7.1} \\ \times cos\epsilon(\omega\,-\,z)\,d\omega d\epsilon \end{align} } \)
Case 2: when boundary condition is variable with azimuth and height
\( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad&=\quad \frac{D}{r}\left[ \frac{\partial}{\partial r}\left(r\frac{\partial C}{\partial r}\right)\,+\, \frac{\partial}{\partial\theta}\left(\frac{1}{r}\frac{\partial C}{\partial\theta}\right)\,+\, \frac{\partial}{\partial z}\left(r\frac{\partial C}{\partial z}\right) \right],\qquad (0\lt r\lt a) \\[15pt] C(a,\theta,z,t) \quad&=\quad F(\theta,z) \\[10pt] C(r,\theta,z,0) \quad&=\quad f(r,\theta,z) \tag*{} \end{align} } \)
Let’s assume that \({\small\displaystyle C}\) is composed of the sum of \({\small\displaystyle u}\) and \({\small\displaystyle v}\). The conditions for each variable are detailed in (Table 7.1).
| Variables | \( {\small\displaystyle C } \) | \( {\small\displaystyle u } \) | \( {\small\displaystyle v } \) |
|---|---|---|---|
| Equation | \( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad=\quad D\nabla^2C \end{align} } \) | \( {\small\displaystyle\begin{align} \nabla^2u \quad=\quad 0 \end{align} } \) | \( {\small\displaystyle\begin{align} \frac{\partial v}{\partial t} \quad=\quad D\nabla^2v \end{align} } \) |
| Boundary condition | \( {\small\displaystyle\begin{align} C(a,\theta,z,t) \quad=\quad F(\theta,z) \end{align} } \) | \( {\small\displaystyle\begin{align} u(a,\theta,z,t) \quad=\quad F(\theta,z) \end{align} } \) | \( {\small\displaystyle\begin{align} v(a,\theta,z,t) \quad=\quad 0 \end{align} } \) |
| Initial condition | \( {\small\displaystyle\begin{align} C(r,\theta,z,0) \quad=\quad 0 \end{align} } \) | - | \( {\small\displaystyle\begin{align} v(r,\theta,z,0) \quad=\quad f(r,\theta,z,0)\,-\,u \end{align} } \) |
| \( {\small\displaystyle\begin{align} ^*\nabla^2C \,=\, \frac{\partial^2C}{\partial r^2}\,+\, \frac{1}{r}\frac{\partial C}{\partial r}\,+\, \frac{1}{r^2}\frac{\partial^2C}{\partial\theta^2}\,+\, \frac{\partial^2C}{\partial z^2} \end{align} } \) | |||
We can derive a solution to the PDE for variable \({\small\displaystyle u}\) through the following steps.
| Step 1: Fourier series | ||||
|---|---|---|---|---|
| Boundary condition | \( {\small\displaystyle\begin{align} F(\theta,z) \quad=\quad \sum_{n=0}^{\infty} A_n(z)cos(n\theta)\,+\,B_n(z)sin(n\theta) \end{align} } \) | \( {\small\displaystyle\begin{align} &A_n,\,B_n \\[8pt] &\text{ are functions of } z \end{align} } \) | ||
| Step 2: Fourier Integral | ||||
| \( {\small\displaystyle \mathrm{i}. } \) | \( {\small\displaystyle\begin{align} A_n(z) \quad=\quad \int_0^{\infty}\,a(\epsilon)cos(\epsilon z)\,+\,b(\epsilon)sin(\epsilon z)\,d\epsilon \end{align} } \) | where | \( {\small\displaystyle\begin{align} a(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,A_n(\omega)cos(\omega\epsilon)\,d\omega \\[10pt] b(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,A_n(\omega)sin(\omega\epsilon)\,d\omega \end{align} } \) | |
| \( {\small\displaystyle\begin{align} A_n(z) \quad&=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\, A_n(\omega) \left[cos(\omega\epsilon)cos(\epsilon z)\,+\,sin(\omega\epsilon)sin(\epsilon z)\right]\,d\omega d\epsilon \\[10pt] &=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\, A_n(\omega) \left[\frac{cos\epsilon(\omega\,+\,z)\,+\,cos\epsilon(\omega\,-\,z)}{2}\,-\,\frac{cos\epsilon(\omega\,+\,z)\,-\,cos\epsilon(\omega\,-\,z)}{2}\right]\,d\omega d\epsilon \\[10pt] &=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\,A_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon \end{align} } \) | ||||
| \( {\small\displaystyle \mathrm{ii}. } \) | \( {\small\displaystyle\begin{align} B_n(z) \quad&=\quad \int_0^{\infty}\,c(\epsilon)cos(\epsilon z)\,+\,d(\epsilon)sin(\epsilon z)\,d\epsilon \\[10pt] &=\quad \frac{1}{\pi} \int_0^{\infty}\int_{-\infty}^{\infty}\,B_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon \end{align} } \) | where | \( {\small\displaystyle\begin{align} c_n(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,B_n(\omega)cos(\omega\epsilon)\,d\omega \\[10pt] d_n(\epsilon) \,&=\, \frac{1}{\pi}\int_{-\infty}^{\infty}\,B_n(\omega)sin(\omega\epsilon)\,d\omega \end{align} } \) | |
| \( {\small\displaystyle\begin{align} \text{Thus,}\qquad& \\[10pt] u \quad&=\quad \frac{1}{\pi}\sum_{\sigma}\sum_{n=0}^{\infty} \frac{J_n(i\sigma r)}{J_n(i\sigma a)} \left[\int_0^{\infty}\int_{-\infty}^{\infty}\,\alpha_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon\, cos(n\theta)\,+\,\int_0^{\infty}\int_{-\infty}^{\infty}\,\beta_n(\omega)[cos\epsilon(\omega\,-\,z)]\,d\omega d\epsilon\, sin(n\theta)\right] \\[10pt] &=\quad \frac{1}{\pi}\sum_{\sigma}\sum_{n=0}^{\infty} \frac{J_n(i\sigma r)}{J_n(i\sigma a)} \left[\int_0^{\infty}\int_{-\infty}^{\infty}\,\{\alpha_n(\omega)cos(n\theta)\,+\,\beta_n(\omega)sin(n\theta)\}\times cos\epsilon(\omega\,-\,z)\,d\omega d\epsilon\right] \end{align} } \) | ||||
\( {\small\displaystyle\begin{align} u \quad=\quad \frac{1}{\pi}\sum_{\sigma}\sum_{n=0}^{\infty}\int_0^{\infty}\int_{-\infty}^{\infty}\,\frac{J_n(i\sigma r)}{J_n(i\sigma a)}\{\alpha_n(\omega)cos(n\theta)\,+\,\beta_n(\omega)sin(n\theta)\} cos\epsilon(\omega\,-\,z)\,d\omega d\epsilon \tag{Eq 7.2} \end{align} } \)
Now, using (Eq 7.1) and (Eq 7.2), we can obtain the variable \({\small\displaystyle v}\) under these conditions and calculate the concentration \({\small\displaystyle C}\).
3. 3D - diffusion in finite cylindrical solids
Now, let’s discuss three-dimensional diffusion in a cylindrical solid with a fixed height.
Case 1: when boundary conditions are 0.
\( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad&=\quad \frac{D}{r}\left[ \frac{\partial}{\partial r}\left(r\frac{\partial C}{\partial r}\right)\,+\, \frac{\partial}{\partial\theta}\left(\frac{1}{r}\frac{\partial C}{\partial\theta}\right)\,+\, \frac{\partial}{\partial z}\left(r\frac{\partial C}{\partial z}\right) \right],\qquad \begin{pmatrix} 0\lt r\lt a\\ 0\lt z\lt L \end{pmatrix} \\[15pt] C(a,\theta,z,t) \quad&=\quad 0\qquad C(a,\theta,0,t) \quad=\quad 0\qquad C(a,\theta,L,t) \quad=\quad 0 \\[15pt] C(r,\theta,z,0) \quad&=\quad f(r,\theta,z) \tag*{} \end{align} } \)
Using the separation of variables method, the concentration \({\small\displaystyle C}\) under these conditions can be determined as follows.
\( {\small\displaystyle\begin{align} C \quad&=\quad \sum_{\sigma}\sum_{m=1}^{\infty}\sum_{n=0}^{\infty} \exp\left[-D\left(\sigma^2\,+\,\frac{m^2\pi^2}{L^2}\right)t\right]J_n(\sigma r)sin\left(\frac{m\pi}{L}z\right)(A_{\sigma,m,n}cos(n\theta)\,+\,B_{\sigma,m,n}sin(n\theta)) \tag{Eq 7.3} \end{align} } \)
\( {\small\displaystyle\begin{align} \text{where}\qquad \\ A_{\sigma,m,0} \quad&=\quad \frac{2\int_0^a\,rJ_0(\sigma r)\,dr\int_0^L\,sin\left(\frac{m\pi}{L}z\right)\,dz\int_{-\pi}^{\pi}\,f(r,\theta,z)\,d\theta}{\pi a^2L\left[J_0^{\prime}(\sigma a)\right]^2} \\[10pt] A_{\sigma,m,n} \quad&=\quad \frac{4\int_0^a\,rJ_n(\sigma r)\,dr\int_0^L\,sin\left(\frac{m\pi}{L}z\right)\,dz\int_{-\pi}^{\pi}\,f(r,\theta,z)cos(n\theta)\,d\theta}{\pi a^2L\left[J_n^{\prime}(\sigma a)\right]^2} \\[10pt] B_{\sigma,m,n} \quad&=\quad \frac{4\int_0^a\,rJ_n(\sigma r)\,dr\int_0^L\,sin\left(\frac{m\pi}{L}z\right)\,dz\int_{-\pi}^{\pi}\,f(r,\theta,z)sin(n\theta)\,d\theta}{\pi a^2L\left[J_n^{\prime}(\sigma a)\right]^2} \tag*{} \end{align} } \)
Case 2: when boundary conditions are variable with radius, azimuth, and height.
| Variables | \( {\small\displaystyle C } \) | \( {\small\displaystyle u } \) | \( {\small\displaystyle v } \) |
|---|---|---|---|
| Equation | \( {\small\displaystyle\begin{align} \frac{\partial C}{\partial t} \quad=\quad D\nabla^2C \end{align} } \) | \( {\small\displaystyle\begin{align} \nabla^2u \quad=\quad 0 \end{align} } \) | \( {\small\displaystyle\begin{align} \frac{\partial v}{\partial t} \quad=\quad D\nabla^2v \end{align} } \) |
| Boundary condition | \( {\small\displaystyle\begin{align} C(a,\theta,z,t) \quad&=\quad F_1(\theta,z) \\[10pt] C(r,\theta,0,t) \quad&=\quad F_2(r,\theta) \\[10pt] C(r,\theta,L,t) \quad&=\quad F_3(r,\theta) \end{align} } \) | \( {\small\displaystyle\begin{align} u(a,\theta,z,t) \quad&=\quad F_1(\theta,z) \\[10pt] u(r,\theta,0,t) \quad&=\quad F_2(r,\theta) \\[10pt] u(r,\theta,L,t) \quad&=\quad F_3(r,\theta) \end{align} } \) | \( {\small\displaystyle\begin{align} v(a,\theta,z,t) \quad&=\quad 0 \\[10pt] v(r,\theta,0,t) \quad&=\quad 0 \\[10pt] v(r,\theta,L,t) \quad&=\quad 0 \end{align} } \) |
| Initial condition | \( {\small\displaystyle\begin{align} C(r,\theta,z,0) \quad=\quad 0 \end{align} } \) | - | \( {\small\displaystyle\begin{align} v(r,\theta,z,0) \quad=\quad f(r,\theta,z,0)\,-\,u \end{align} } \) |
| \( {\small\displaystyle\begin{align} ^*\nabla^2C \,=\, \frac{\partial^2C}{\partial r^2}\,+\, \frac{1}{r}\frac{\partial C}{\partial r}\,+\, \frac{1}{r^2}\frac{\partial^2C}{\partial\theta^2}\,+\, \frac{\partial^2C}{\partial z^2} \end{align} } \) | |||
We can assume that variable \({\small\displaystyle u}\) is composed once again as the sum of \({\small\displaystyle u_1}\), \({\small\displaystyle u_2}\) and \({\small\displaystyle u_3}\). Let’s solve the PDEs based on the conditions of each variable.
\( {\small\displaystyle\begin{align} u \quad=\quad u_1\,+\,u_2\,+\,u_3 \tag*{} \end{align} } \)
First, \({\small\displaystyle u_1}\) satisfies the following PDE along with its boundary conditions. Solving for this PDE, we can obtain (Eq 7.4).
\( {\small\displaystyle\begin{align} \text{PDE }&:\quad\nabla^2u_1 \,=\,0 \\[15pt] \text{BCs }&:\quad u_1(a,\theta,z) \,=\,F_1(\theta,z),\qquad u_1(r,\theta,0) \,=\,0,\qquad u_1(r,\theta,L) \,=\,0 \tag*{} \end{align} } \)
| Step 1: Separation of variable | ||||
|---|---|---|---|---|
| \( {\small\displaystyle\begin{align} u_1 \quad=\quad R\Theta H \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} R(r) \quad&=\quad \alpha J_n(i\sigma r) \\[10pt] \Theta(\theta) \quad&=\quad acos(\lambda\theta)\,+\,bsin(\lambda\theta) \\[10pt] H(z) \quad&=\quad p_1cos(\sigma z)\,+\,q_1sin(\sigma z) \end{align} } \) | ||
| Step 2: Using BCs | ||||
| \( {\small\displaystyle \text{For z} } \) | \( {\small\displaystyle\begin{align} H(0) \quad&=\quad p_1 \\[10pt] H(L) \quad&=\quad p_1cos(\sigma L)\,+\,q_1sin(\sigma L) \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} p_1 \quad&=\quad 0 \\[10pt] \sigma \quad&=\quad \frac{m\pi}{L},\quad m\,=\,1,\,2,\,3,\,\cdots \end{align} } \) | |
| \( {\small\displaystyle \text{For }\theta } \) | \( {\small\displaystyle\begin{align} 2bsin(\lambda\pi) \quad=\quad 0 \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} \lambda \quad=\quad n,\quad n\,=\,1,\,2,\,3,\,\cdots \end{align} } \) | \( {\small\displaystyle \Theta(\pi)\,=\,\Theta(-\pi) } \) |
| \( {\small\displaystyle \text{For }r } \) | \( {\small\displaystyle\begin{align} R(r) \quad=\quad \alpha_mJ_m\left(i\frac{m\pi}{L}r\right) \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} R(r) \quad=\quad \beta_m\frac{J_n\left(i\frac{m\pi}{L}r\right)}{J_n\left(i\frac{m\pi}{L}a\right)},\quad m=1,\,2,\,3,\,\cdots \end{align} } \) | |
| \( {\small\displaystyle\begin{align} \text{Thus,}\quad& \\[10pt] &u_1 \quad=\quad \sum_{m=1}^{\infty}\sum_{n=0}^{\infty} \frac{J_n\left(i\dfrac{m\pi}{L}r\right)}{J_n\left(i\dfrac{m\pi}{L}a\right)}sin\left(\frac{m\pi}{L}z\right) \left[A_{m,n}cos(n\theta)\,+\,B_{m,n}sin(n\theta)\right] \end{align} } \) | ||||
| Step 3: Fourier Series | ||||
| \( {\small\displaystyle r\,=\,a } \) | \( {\small\displaystyle\begin{align} u_1 \quad&=\quad \sum_{m=1}^{\infty}\sum_{n=0}^{\infty}sin\left(\frac{m\pi}{L}z\right)\left[A_{m,n}cos(n\theta)\,+\,B_{m,n}sin(n\theta)\right] \end{align} } \) | = | \( {\small\displaystyle\begin{align} F_1(\theta,z) \end{align} } \) | |
| \( {\small\displaystyle\begin{align} F_1(\theta,z) \quad&=\quad \frac{a_0}{2}\,+\, \sum_{n=1}^{\infty}a_ncos(n\theta)\,+\,b_nsin(n\theta) \end{align} } \) | where | \( {\small\displaystyle\begin{align} a_n \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\, F_1(\theta,z)cos(n\theta)\,d\theta \\[10pt] b_n \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\, F_1(\theta,z)cos(n\theta)\,d\theta \end{align} } \) | ||
| \( {\small\displaystyle\begin{align} \sum_{m=1}^{\infty}A_{m,0}sin\left(\frac{m\pi}{L}z\right) \quad&=\quad \frac{1}{2\pi}\int_{-\pi}^{\pi}\,F_1(\theta,z)\,d\theta \\[10pt] \sum_{m=1}^{\infty}A_{m,n}sin\left(\frac{m\pi}{L}z\right) \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\,F_1(\theta,z)cos(n\theta)\,d\theta \\[10pt] \sum_{m=1}^{\infty}B_{m,n}sin\left(\frac{m\pi}{L}z\right) \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\,F_1(\theta,z)sin(n\theta)\,d\theta \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} A_{m,0} \quad&=\quad \frac{1}{\pi L} \int_0^{L}\int_{-\pi}^{\pi}\,F_1(\theta,z)sin\left(\frac{m\pi}{L}z\right)\,d\theta dz \\[10pt] A_{m,n} \quad&=\quad \frac{2}{\pi L} \int_0^{L}\int_{-\pi}^{\pi}\,F_1(\theta,z)cos(n\theta)sin\left(\frac{m\pi}{L}z\right)\,d\theta dz \\[10pt] B_{m,n} \quad&=\quad \frac{2}{\pi L} \int_0^{L}\int_{-\pi}^{\pi}\,F_1(\theta,z)sin(n\theta)sin\left(\frac{m\pi}{L}z\right)\,d\theta dz \end{align} } \) | ||
\( {\small\displaystyle\begin{align} u_1 \quad&=\quad \sum_{m=1}^{\infty}\sum_{n=0}^{\infty} \frac{J_n\left(i\dfrac{m\pi}{L}r\right)}{J_n\left(i\dfrac{m\pi}{L}a\right)}sin\left(\frac{m\pi}{L}z\right) \left[A_{m,n}cos(n\theta)\,+\,B_{m,n}sin(n\theta)\right] \tag{Eq 7.4} \\[15pt] \text{where}\qquad \\[10pt] A_{m,0} \quad&=\quad \frac{1}{\pi L} \int_0^{L}\int_{-\pi}^{\pi}\,F(\theta,z)sin\left(\frac{m\pi}{L}z\right)\,d\theta dz \\[10pt] A_{m,n} \quad&=\quad \frac{2}{\pi L} \int_0^{L}\int_{-\pi}^{\pi}\,F(\theta,z)cos(n\theta)sin\left(\frac{m\pi}{L}z\right)\,d\theta dz \\[10pt] B_{m,n} \quad&=\quad \frac{2}{\pi L} \int_0^{L}\int_{-\pi}^{\pi}\,F(\theta,z)sin(n\theta)sin\left(\frac{m\pi}{L}z\right)\,d\theta dz \end{align} } \)
Now, let’s solve about \({\small\displaystyle u_2}\), which satisfies the following conditions.
\( {\small\displaystyle\begin{align} \text{PDE }&:\quad\nabla^2u_2 \,=\,0 \\[15pt] \text{BCs }&:\quad u_2(a,\theta,z) \,=\,0,\qquad u_2(r,\theta,0) \,=\,F_2(r,\theta),\qquad u_2(r,\theta,L) \,=\,0 \tag*{} \end{align} } \)
| Step 1:Separation of variables | ||||
|---|---|---|---|---|
| \( {\small\displaystyle\begin{align} u_2 \quad=\quad R\Theta H \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} R(r) \quad&=\quad \alpha J_{\lambda}(i\sigma r) \\[10pt] \Theta(\theta) \quad&=\quad acos(\lambda\theta)\,+\,bsin(\lambda\theta) \\[10pt] H(z) \quad&=\quad c_1\exp(i\sigma z)\,+\,c_2\exp(-i\sigma z) \end{align} } \) | ||
| Step 2: Using BCs | ||||
| \( {\small\displaystyle \text{For }z } \) | \( {\small\displaystyle\begin{align} H(z) \quad=\quad c_1\exp(\mu z)\,+\,c_2\exp(-\mu z) \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} c_3\exp[\mu(L\,-\,z)]\,+\,c_4\exp[-\mu(L\,-\,z)] \end{align} } \) | \( {\small\displaystyle \mu \,=\, i\sigma } \) |
| \( {\small\displaystyle\begin{align} =\quad &c_3\left[\frac{\exp[\mu(L\,-\,z)]\,+\,\exp[-\mu(L\,-\,z)]}{2}\right]\,+\, c_4\left[\frac{\exp[\mu(L\,-\,z)]\,+\,\exp[-\mu(L\,-\,z)]}{2}\right] \\[10pt] &\,+\, c_3\left[\frac{\exp[\mu(L\,-\,z)]\,-\,\exp[-\mu(L\,-\,z)]}{2}\right]\,-\, c_4\left[\frac{\exp[\mu(L\,-\,z)]\,-\,\exp[-\mu(L\,-\,z)]}{2}\right] \\[10pt] =\quad &p_2cosh[\mu(L\,-\,z)]\,+\,q_2sinh[\mu(L\,-\,z)] \end{align} } \) | ||||
| \( {\small\displaystyle\begin{align} H(L) \quad=\quad p_2 \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} p_2 \quad=\quad 0 \end{align} } \) | ||
| \( {\small\displaystyle\begin{align} H(z) \quad=\quad q_2sinh[\mu(L\,-\,z)] \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} H(z) \quad=\quad q\frac{sinh[\mu(L\,-\,z)]}{sinh(\mu L)} \end{align} } \) | ||
| \( {\small\displaystyle \text{For }\theta } \) | \( {\small\displaystyle\begin{align} 2bsin(\lambda\pi) \quad=\quad 0 \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} \lambda \quad=\quad n,\quad n=0,\,1,\,2,\,\cdots \end{align} } \) | \( {\small\displaystyle \Theta(\pi) \,=\, \Theta(-\pi) } \) |
| \( {\small\displaystyle \text{For }r } \) | \( {\small\displaystyle\begin{align} R(r) \quad=\quad \alpha J_n(\mu r) \end{align} } \) | where | \( {\small\displaystyle\begin{align} \mu\,\text{is the roots of}& \\[10pt] &J_n(\mu a) \quad=\quad 0 \end{align} } \) | |
| \( {\small\displaystyle\begin{align} \text{Thus,}\quad& \\[10pt] &u_2 \quad=\quad \sum_{\mu}\sum_{n=0}^{\infty} \frac{sinh[\mu(L\,-\,z)]}{sinh(\mu L)} J_n(\mu r) \left[A_{\mu,n}cos(n\theta)\,+\,B_{\mu,n}sin(n\theta)\right] \end{align} } \) | ||||
| Step 3: Fourier series | ||||
| \( {\small\displaystyle z\,=\,0 } \) | \( {\small\displaystyle\begin{align} u_2 \quad=\quad \sum_{\mu}\sum_{n=0}^{\infty} J_n(\mu r) \left[A_{\mu,n}cos(n\theta)\,+\,B_{\mu,n}sin(n\theta)\right] \end{align} } \) | = | \( {\small\displaystyle F_2(r,\theta) } \) | |
| \( {\small\displaystyle\begin{align} F_2(r,\theta) \quad=\quad \frac{a_0}{2}\,+\, \sum_{n=1}^{\infty}a_ncos(n\theta)\,+\,b_nsin(n\theta) \end{align} } \) | where | \( {\small\displaystyle\begin{align} a_n \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\, F_2(r,\theta)cos(n\theta)\,d\theta \\[10pt] b_n \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\, F_2(r,\theta)sin(n\theta)\,d\theta \end{align} } \) | ||
| \( {\small\displaystyle\begin{align} \sum_{\mu}A_{\mu,0}J_0(\mu r) \quad&=\quad \frac{1}{2\pi}\int_{-\pi}^{\pi}\, F_2(r,\theta)\,d\theta \\[10pt] \sum_{\mu}A_{\mu,n}J_n(\mu r) \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\, F_2(r,\theta)cos(n\theta)\,d\theta,\quad n\ge 1 \\[10pt] \sum_{\mu}B_{\mu,n}J_n(\mu r) \quad&=\quad \frac{1}{\pi}\int_{-\pi}^{\pi}\, F_2(r,\theta)sin(n\theta)\,d\theta \end{align} } \) | \( {\small\displaystyle \longrightarrow } \) | \( {\small\displaystyle\begin{align} A_{\mu,0} \quad&=\quad \frac{\int_0^a\,rJ_0(\mu r)\,dr\int_{-\pi}^{\pi}\,F_2(r,\theta)\,d\theta}{\pi a^2\left[J_0^{\prime}(\mu a)\right]^2} \\[10pt] A_{\mu,n} \quad&=\quad \frac{2\int_0^a\,rJ_n(\mu r)\,dr\int_{-\pi}^{\pi}\,F_2(r,\theta)cos(n\theta)\,d\theta}{\pi a^2\left[J_n^{\prime}(\mu a)\right]^2} \\[10pt] B_{\mu,n} \quad&=\quad \frac{2\int_0^a\,rJ_n(\mu r)\,dr\int_{-\pi}^{\pi}\,F_2(r,\theta)sin(n\theta)\,d\theta}{\pi a^2\left[J_n^{\prime}(\mu a)\right]^2} \end{align} } \) | ||
\( {\small\displaystyle\begin{align} u_2 \quad&=\quad \sum_{\mu}\sum_{n=0}^{\infty} \frac{sinh[\mu(L\,-\,z)]}{sinh(\mu L)} J_n(\mu r) \left[A_{\mu,n}cos(n\theta)\,+\,B_{\mu,n}sin(n\theta)\right] \tag{Eq 7.5} \\[15pt] \text{where}\qquad \\[10pt] A_{\mu,0} \quad&=\quad \frac{\int_0^a\int_{-\pi}^{\pi}\,rF_2(r,\theta)J_0(\mu r)\,d\theta\,dr}{\pi a^2\left[J_0^{\prime}(\mu a)\right]^2} \\[10pt] A_{\mu,n} \quad&=\quad \frac{2\int_0^a\int_{-\pi}^{\pi}\,rF_2(r,\theta)cos(n\theta)J_n(\mu r)\,d\theta\,dr}{\pi a^2\left[J_n^{\prime}(\mu a)\right]^2},\quad n\ge 1 \\[10pt] B_{\mu,n} \quad&=\quad \frac{2\int_0^a\int_{-\pi}^{\pi}\,rF_2(r,\theta)sin(n\theta)J_n(\mu r)\,d\theta\,dr}{\pi a^2\left[J_n^{\prime}(\mu a)\right]^2} \end{align} } \)
Finally, solving the PDE for \({\small\displaystyle u_3}\) in a similar way as for \({\small\displaystyle u_2}\), we obtain (Eq 7.6). By comparing the equation when the height is L and the Fourier series of \({\small\displaystyle F_3}\), we can derive \({\small\displaystyle A_{m,n}}\) and \({\small\displaystyle B_{m,n}}\) in (Eq 7.6).
\( {\small\displaystyle\begin{align} \text{PDE }&:\quad\nabla^2u_3 \,=\,0 \\[15pt] \text{BCs }&:\quad u_3(a,\theta,z) \,=\,0,\qquad u_3(r,\theta,0) \,=\,0,\qquad u_3(r,\theta,L) \,=\,F_3(r,\theta) \tag*{} \end{align} } \)
\( {\small\displaystyle\begin{align} u_3 \quad&=\quad \sum_{\mu}\sum_{n=0}^{\infty} \frac{sinh(\mu z)}{sinh(\mu L)} J_n(\mu r) \left[A_{\mu,n}cos(n\theta)\,+\,B_{\mu,n}sin(n\theta)\right] \tag{Eq 7.6} \\[15pt] \text{where}\qquad \\[10pt] A_{\mu,0} \quad&=\quad \frac{\int_0^a\int_{-\pi}^{\pi}\,rF_3(r,\theta)J_0(\mu r)\,d\theta\,dr}{\pi a^2\left[J_0^{\prime}(\mu a)\right]^2} \\[10pt] A_{\mu,n} \quad&=\quad \frac{2\int_0^a\int_{-\pi}^{\pi}\,rF_3(r,\theta)cos(n\theta)J_n(\mu r)\,d\theta\,dr}{\pi a^2\left[J_n^{\prime}(\mu a)\right]^2},\quad n\ge 1 \\[10pt] B_{\mu,n} \quad&=\quad \frac{2\int_0^a\int_{-\pi}^{\pi}\,rF_3(r,\theta)sin(n\theta)J_n(\mu r)\,d\theta\,dr}{\pi a^2\left[J_n^{\prime}(\mu a)\right]^2} \end{align} } \)
Now, we can calculate the concentration under these specific conditions using (Eq 7.3) to (Eq 7.6).
Nomenclature
\({\small\displaystyle C}\) : concentration
\({\small\displaystyle D}\) : diffusion coefficient
\({\small\displaystyle J}\) : bessel function
\({\small\displaystyle r}\) : radius
\({\small\displaystyle t}\) : time
\({\small\displaystyle u,\,v}\) : variables
\({\small\displaystyle z}\) : height
\({\small\displaystyle \theta}\) : azimuth