tdiffusion.tex - pism - [fork] customized build of PISM, the parallel ice sheet model (tillflux branch)
HTML git clone git://src.adamsgaard.dk/pism
DIR Log
DIR Files
DIR Refs
DIR LICENSE
---
tdiffusion.tex (3521B)
---
1 \documentclass{article}
2 \usepackage[margin=1in]{geometry}
3 \usepackage{amsmath}
4 \usepackage{hyperref}
5 \parindent=0in \parskip=0.5\baselineskip
6
7 \usepackage{fancybox}
8 \usepackage{xcolor}
9 \newcommand{\highlight}[1]{{\color{red!80!black} \fbox{$ \displaystyle #1 $} }}
10
11 \begin{document}
12
13 \title{Exact solution for a simple IBVP with Dirichlet and Neumann BC}
14 \maketitle
15
16 \section{Introduction}
17 \label{sec:introduction}
18
19 This exact solution can be used to test the implementation of the
20 enthalpy and temperature column solvers, both in the ice and in the
21 bedrock.
22
23 This is sufficient to test the bedrock thermal unit code, but only
24 goes half way towards checking the enthalpy code: in these notes I
25 consider a simple \emph{diffusion only} problem without advection or
26 reaction terms.
27
28 \section{The problem}
29 \label{sec:problem}
30
31 \begin{align}
32 \label{eq:1}
33 \text{PDE:} \qquad & u_{t} = \alpha^{2}\, u_{zz},\quad 0 < z < L,\\
34 \label{eq:2}
35 \text{BCs:} \qquad &\left\{
36 \begin{aligned}
37 u(0,t) &= U_{0},\\
38 u_{z}(L,t) &= Q_{L},
39 \end{aligned}\right.\\
40 \label{eq:3}
41 \text{IC:} \qquad & u(z,0) = \phi(z).
42 \end{align}
43
44 In the PISM context
45 \begin{equation}
46 \label{eq:14}
47 \alpha^{2} = \frac{k}{\rho c_{p}},
48 \end{equation}
49 where $k$ is the thermal conductivity, $\rho$ is the density, and
50 $c_{p}$ is the specific heat capacity of ice.
51
52 To transform this problem with \emph{non-homogeneous} boundary conditions
53 into one with homogeneous ones I define $v(z,t)$ by subtracting the
54 steady state solution from $u$:
55 \begin{equation}
56 \label{eq:4}
57 v(z,t) = u(z,t) - \left( U_{0} + Q_{L}\cdot z \right).
58 \end{equation}
59
60 It is easy to check that $v_{t} = u_{t}$, $v_{zz} = u_{zz}$, and $v$ satisfies
61
62 \begin{align}
63 \label{eq:15}
64 \text{PDE:} \qquad & v_{t} = \alpha^{2}\, v_{zz},\quad 0 < z < L,\\
65 \label{eq:5}
66 \text{BCs:} \qquad &\left\{
67 \begin{aligned}
68 v(0,t) &= 0,\\
69 v_{z}(L,t) &= 0,
70 \end{aligned}\right.\\
71 \label{eq:16}
72 \text{IC:} \qquad & v(z,0) = \phi(z) - [U_{0} + Q_{L}z].
73 \end{align}
74
75 This new problem (equations (\ref{eq:15})--(\ref{eq:16})) can be solved using separation of variables, as follows.
76
77 Assume that $v(z,t)$ can be written as
78 \begin{equation}
79 \label{eq:6}
80 v(z,t) = Z(z)\cdot T(t),
81 \end{equation}
82 then the PDE (\ref{eq:15}) becomes
83 \begin{equation}
84 \label{eq:7}
85 ZT' = \alpha^{2}Z''T,
86 \end{equation}
87 or
88 \begin{equation}
89 \label{eq:8}
90 \frac{T'}{\alpha^{2}T} = \frac{Z''}{Z} = -\lambda^{2},
91 \end{equation}
92 for some constant $\lambda$, and so
93 \begin{align}
94 \label{eq:9}
95 T' + \lambda^{2}\alpha^{2}T &= 0,\\
96 \label{eq:10}
97 Z'' + \lambda^{2}Z &= 0.
98 \end{align}
99
100 Solving equations (\ref{eq:9}) and (\ref{eq:10}) gives
101 \begin{align}
102 T(t) &= C_{1}e^{-(\lambda\alpha)^{2}t},\\
103 Z(z) &= C_{2}\sin(\lambda z) + C_{3}\cos(\lambda z),
104 \end{align}
105 or
106 \begin{equation}
107 \label{eq:11}
108 v(z,t) = e^{-(\lambda\alpha)^{2}t}\left( A\sin(\lambda z) + B\cos(\lambda z) \right)
109 \end{equation}
110 Combining (\ref{eq:11}) with (\ref{eq:5}) gives
111 \begin{align}
112 B &= 0,\\
113 \cos(\lambda\cdot L) &= 0,
114 \end{align}
115 and so fundamental solutions have the form
116 \begin{align}
117 \label{eq:12}
118 v_{n}(z,t) &= \highlight{a_{n}\,e^{-(\lambda_{n}\alpha)^{2}t}\sin\left( \lambda_{n} z \right)},\\
119 \lambda_{n} &= \highlight{\frac{1}{L} \left( \frac{\pi}{2} + n\,\pi \right),\quad n \in \mathbf{Z}}.
120 \end{align}
121
122 \end{document}