Linear Diffusion Flows

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This tours studies linear diffusion PDEs, a.k.a. the heat equation. A good reference for diffusion flows in image processing is Weickert98.

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Heat Diffusion

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.

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n = 256;

Load an image $f_0 \in \RR^N$, that will be used to initialize the flow at time $t=0$.

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name = 'hibiscus';
f0 = load_image(name,n);
f0 = rescale( sum(f0,3) );

Display it.

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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.

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h = 1/n;
delta = @(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.

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tau = .5 * h^2/4;

Final time.

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T = 1e-3;

Number of iterations.

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niter = ceil(T/tau);

Initialize the diffusion at time $t=0$.

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f = f0;

One step of discrete diffusion.

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f = f + tau * delta(f);

Exercise 1

Compute the solution to the heat equation.

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%% Insert your code here.

Explicit Solution using Convolution

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 $$ g = f \star h \qarrq \forall \om, \quad \hat g(\om) = \hat f(\om) \hat h(\om). $$

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cconv = @(f,h)real(ifft2(fft2(f).*fft2(h)));

Define a discrete Gaussian blurring kernel of width $\sqrt{t}$.

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t = [0:n/2 -n/2+1:-1];
[X2,X1] = meshgrid(t,t);
normalize = @(h)h/sum(h(:));
h = @(t)normalize( exp( -(X1.^2+X2.^2)/(4*t) ) );

Define blurring operator.

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heat = @(f, t)cconv(f,h(t));

Example of blurring.

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Exercise 2

Display the heat convolution for increasing values of $t$.

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%% Insert your code here.