6.4.1: Fourier's Method
( \newcommand{\kernel}{\mathrm{null}\,}\)
Separation of variables ansatz c(x,t)=v(x)w(t) leads to the eigenvalue problem, see the arguments of Section 4.5,
−△v=λv in Ω∂v∂n=0 on ∂Ω,
and to the ordinary differential equation
w′(t)+λDw(t)=0.
Assume Ω is bounded and ∂Ω sufficiently regular, then the eigenvalues of (6.4.1.1), (6.4.1.2) are countable and
$$0=\lambda_0<\lambda_1\le\lambda_2\le\ldots\to\infty.\]
Let vj(x) be a complete system of orthonormal (in L2(Ω)) eigenfunctions.
Solutions of (6.4.1.3) are
$$w_j(t)=C_je^{-D\lambda_jt},\]
where Cj 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 (6.4.1.1) 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 c0(x1,x2,x3) of the substrate in a solution is constant if x3=const. It follows from a uniqueness result below that the solution of the initial-boundary value problem c(x1,x2,x3,t) is independent of x1 and x2.
Set z=x3, then the above initial-boundary value problem reduces to
ct=Dczzc(z,0)=c0(z)cz=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),\]
where
C0=1l∫l0 c0(z) dzCn=2l∫l0 c0(z)cos(πlnz) dz, n≥1.
Contributors and Attributions
Integrated by Justin Marshall.