Important: Please read the installation page for details about how to install the toolboxes. $\newcommand{\dotp}[2]{\langle #1, #2 \rangle}$ $\newcommand{\enscond}[2]{\lbrace #1, #2 \rbrace}$ $\newcommand{\pd}[2]{ \frac{ \partial #1}{\partial #2} }$ $\newcommand{\umin}[1]{\underset{#1}{\min}\;}$ $\newcommand{\umax}[1]{\underset{#1}{\max}\;}$ $\newcommand{\umin}[1]{\underset{#1}{\min}\;}$ $\newcommand{\uargmin}[1]{\underset{#1}{argmin}\;}$ $\newcommand{\norm}[1]{\|#1\|}$ $\newcommand{\abs}[1]{\left|#1\right|}$ $\newcommand{\choice}[1]{ \left\{ \begin{array}{l} #1 \end{array} \right. }$ $\newcommand{\pa}[1]{\left(#1\right)}$ $\newcommand{\diag}[1]{{diag}\left( #1 \right)}$ $\newcommand{\qandq}{\quad\text{and}\quad}$ $\newcommand{\qwhereq}{\quad\text{where}\quad}$ $\newcommand{\qifq}{ \quad \text{if} \quad }$ $\newcommand{\qarrq}{ \quad \Longrightarrow \quad }$ $\newcommand{\ZZ}{\mathbb{Z}}$ $\newcommand{\CC}{\mathbb{C}}$ $\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\EE}{\mathbb{E}}$ $\newcommand{\Zz}{\mathcal{Z}}$ $\newcommand{\Ww}{\mathcal{W}}$ $\newcommand{\Vv}{\mathcal{V}}$ $\newcommand{\Nn}{\mathcal{N}}$ $\newcommand{\NN}{\mathcal{N}}$ $\newcommand{\Hh}{\mathcal{H}}$ $\newcommand{\Bb}{\mathcal{B}}$ $\newcommand{\Ee}{\mathcal{E}}$ $\newcommand{\Cc}{\mathcal{C}}$ $\newcommand{\Gg}{\mathcal{G}}$ $\newcommand{\Ss}{\mathcal{S}}$ $\newcommand{\Pp}{\mathcal{P}}$ $\newcommand{\Ff}{\mathcal{F}}$ $\newcommand{\Xx}{\mathcal{X}}$ $\newcommand{\Mm}{\mathcal{M}}$ $\newcommand{\Ii}{\mathcal{I}}$ $\newcommand{\Dd}{\mathcal{D}}$ $\newcommand{\Ll}{\mathcal{L}}$ $\newcommand{\Tt}{\mathcal{T}}$ $\newcommand{\si}{\sigma}$ $\newcommand{\al}{\alpha}$ $\newcommand{\la}{\lambda}$ $\newcommand{\ga}{\gamma}$ $\newcommand{\Ga}{\Gamma}$ $\newcommand{\La}{\Lambda}$ $\newcommand{\si}{\sigma}$ $\newcommand{\Si}{\Sigma}$ $\newcommand{\be}{\beta}$ $\newcommand{\de}{\delta}$ $\newcommand{\De}{\Delta}$ $\newcommand{\phi}{\varphi}$ $\newcommand{\th}{\theta}$ $\newcommand{\om}{\omega}$ $\newcommand{\Om}{\Omega}$
This tours studies linear diffusion PDEs, a.k.a. the heat equation. A good reference for diffusion flows in image processing is Weickert98.
from __future__ import division
import nt_toolbox as nt
from nt_solutions import pde_1_diffusion_linear as solutions
%matplotlib inline
%load_ext autoreload
%autoreload 2
The heat equation reads $$ \forall t>0, \quad \pd{f_t}{t} = \nabla f_t $$ for a function $f_t : \RR^2 \rightarrow \RR$ and where $f_0$ (the solution at initial time $t=0$) is given.
The Laplacian operator reads $$ \Delta f = \pdd{f}{x_1} + \pdd{f}{x_2}. $$
The flow is discretized in space by considering a discrete image of $N = n \times n$ pixels.
n = 256
Load an image $f_0 \in \RR^N$, that will be used to initialize the flow at time $t=0$.
name = 'hibiscus'
f0 = load_image(name, n)
f0 = rescale(sum(f0, 3))
Display it.
imageplot(f0)
The flow is discretized in time using an explicit time-stepping $$ f^{(\ell+1)} = f^{(\ell)} + \tau \Delta f^{(\ell)}. $$ We use finite difference Laplacian $$ (\Delta f)_i = \frac{1}{h^2}\pa{ f_{i_1+1,i_2}+f_{i_1-1,i_2}+f_{i_1,i_2+1}+f_{i_1,i_2-1}-4f_j }$$ where we assume periodic boundary conditions, and where $h = 1/N$ is the spacial step size.
h = 1/ n
delta = lambda f: 1/ h^2 * div(grad(f))
The step size $\tau$ should satisfy $$ \tau < \frac{h^2}{4} $$ for the discretized flow to be stable.
The discrete solution $f^{(\ell)}$ converges to the continuous solution $f_t$ at time $t = \tau \ell$ if both $\tau \rightarrow 0$ and $h \rightarrow 0$ under the condition $\tau/h^2 < 1/4$.
Select a small enough step size.
tau = .5 * h^2/ 4
Final time.
T = 1e-3
Number of iterations.
niter = ceil(T/ tau)
Initialize the diffusion at time $t=0$.
f = f0
One step of discrete diffusion.
f = f + tau * delta(f)
Exercise 1
Compute the solution to the heat equation.
solutions.exo1()
## Insert your code here.
The solution to the heat equation can be computed using a convolution $$ \forall t>0, \quad f_t = f_0 \star h_t $$ where $\star$ denotes the convolution of continuous functions $$ f \star h(x) = \int_{\RR^2} f(y) g(x-y) d y $$ and $h_t$ is a Gaussian kernel of width $\sqrt{t}$ $$ h_t(x) = \frac{1}{4 \pi t} e^{ -\frac{\norm{x}^2}{4t} } $$
One can thus approximate the solution using a discrete convolution. Convolutions can be computed in $O(N\log(N))$ operations using the FFT, since $$ g = f \star h \qarrq \forall \om, \quad \hat g(\om) = \hat f(\om) \hat h(\om). $$
cconv = lambda f, h: real(ifft2(fft2(f).*fft2(h)))
Define a discrete Gaussian blurring kernel of width $\sqrt{t}$.
t = [0: n/ 2 -n/ 2 + 1: -1]
[X2, X1] = meshgrid(t, t)
normalize = lambda h: h/ sum(h(: ))
h = lambda t: normalize(exp(-(X1.^2 + X2.^2)/ (4*t)))
Define blurring operator.
heat = lambda f, t: cconv(f, h(t))
Example of blurring.
imageplot(heat(f0, 2))
Exercise 2
Display the heat convolution for increasing values of $t$.
solutions.exo2()
## Insert your code here.