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Mathematics LibreTexts

6.4.1: Fourier's Method

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

    -\triangle v&=&\lambda v\ \ \mbox{in}\ \Omega\\
    \label{ewpar2} \tag{}
    \frac{\partial v}{\partial n}&=&0\ \ \mbox{on}\ \partial\Omega,

    and to the ordinary differential equation

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

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


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


    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

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


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

    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

    Figure 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

    c_z&=&0,\ \ z=0,\ z=l.

    The (formal) solution is

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


    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.