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Mathematics LibreTexts

6.4.1: Fourier's Method

Separation of variables ansatz \(c(x,t)=v(x)w(t)\) leads to the eigenvalue problem, see the arguments of Section 4.5,

\begin{eqnarray}
\label{ewpar1}\tag{6.4.1.1}
-\triangle v&=&\lambda v\ \ \mbox{in}\ \Omega\\
\label{ewpar2} \tag{6.4.1.2}
\frac{\partial v}{\partial n}&=&0\ \ \mbox{on}\ \partial\Omega,
\end{eqnarray}

and to the ordinary differential equation

\begin{equation} \tag{6.4.1.3}
\label{ewpar3}
w'(t)+\lambda Dw(t)=0.
\end{equation}

Assume \(\Omega\) is bounded and \(\partial\Omega\) sufficiently regular, then the eigenvalues of (\ref{ewpar1}), (\ref{ewpar2}) are countable and

$$0=\lambda_0<\lambda_1\le\lambda_2\le\ldots\to\infty.$$

Let \(v_j(x)\) be a complete system of orthonormal (in \(L^2(\Omega)\)) eigenfunctions.
Solutions of (\ref{ewpar3})  are

$$w_j(t)=C_je^{-D\lambda_jt},$$

where \(C_j\) are arbitrary constants.

According to the superposition principle,

$$c_N(x,t):=\sum_{j=0}^N C_je^{-D\lambda_jt}v_j(x)$$

is a solution of the differential equation (\ref{ewpar1}) and

$$c(x,t):=\sum_{j=0}^\infty C_je^{-D\lambda_jt}v_j(x),$$

with

$$C_j=\int_\Omega\ c_0(x)v_j(x)\ dx,$$

is a formal solution of the initial-boundary value problem (6.4.1)-(6.4.3).

Diffusion in a tube

Consider a solution in a tube, see Figure 6.4.1.1.

Figure 6.4.1.1: Diffusion in a tube

Assume the initial concentration \(c_0(x_1,x_2,x_3)\) of the substrate in a solution  is constant if \(x_3=const.\) It follows from a uniqueness result below that the solution of the initial-boundary value problem \(c(x_1,x_2,x_3,t)\) is independent of \(x_1\) and \(x_2\).

Set \(z=x_3\), then the above initial-boundary value problem reduces to

\begin{eqnarray*}
c_t&=&Dc_{zz}\\
c(z,0)&=&c_0(z)\\
c_z&=&0,\ \ z=0,\ z=l.
\end{eqnarray*}

The (formal) solution is

$$c(z,t)=\sum_{n=0}^\infty C_ne^{-D\left(\frac{\pi}{l}n\right)^2 t}\cos\left(\frac{\pi}{l}nz\right),$$

where

\begin{eqnarray*}
C_0&=&\frac{1}{l}\int_0^l\ c_0(z)\ dz\\
C_n&=&\frac{2}{l}\int_0^l\ c_0(z)\cos\left(\frac{\pi}{l}nz\right)\ dz,\ \ n\ge1.
\end{eqnarray*}

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