## Conductivity matrix for isotropic diffusion problems in FED

In isotropic diffusion, the conductivity is adaptive to the local structure of the signal. But in comparison to anisotropic diffusion, the conductivity is still the same in all directions. Mathematically, the conductivity factor in the diffusion equation depends now on the signal $$u$$ instead of being a constant

$$\frac{\partial u(x,t)}{\partial t} = g(u(x, t)) \cdot \frac{\partial^2 u(x,t)}{\partial^2 x^2}. \label{eq:FEDIsotropic_PDE}$$

In case of images and isotropic diffusion, the conductivity function may be related to the image gradient

$$g(u(x, t)) = \frac{1}{\left(\frac{\left\| \nabla u(x,t) \right\|}{\lambda }\right)^2+1} \label{eq:FEDIsotropic_ConductivityFunction}$$

with already a concrete function being used (more on this later) and where $$\lambda$$ controls how strong the conductivity function reacts1. To use Fast Explicit Diffusion (FED) for isotropic problems the matrix $$A$$ in

$$\fvec{u}_{k+1} = \left( I + \tau \cdot A(\fvec{u}_k) \right) \cdot \fvec{u}_k \label{eq:FEDIsotropic_Discrete}$$

now depends on the discrete signal vector $$\fvec{u}_k$$2 evolved to the diffusion time $$k$$. To achieve isotropic diffusion in this scheme $$A(\fvec{u}_k)$$ must somehow encode the information from \eqref{eq:FEDIsotropic_ConductivityFunction}. This article deals with this questions and gives a definition of $$A(\fvec{u}_k)$$ like described in this paper.

### Conductivity function

Before diving into the matrix definition, I first want to analyse the conductivity function from \eqref{eq:FEDIsotropic_ConductivityFunction} a bit more. Even though only this special definition of $$g(u(x, t))$$ will be used here, let me note that others are possible, too (see e.g. Digital Image Processing by Burger and Burge, p. 437 for an overview) and which may behave slightly different.

So, what is the point of the conductivity function? Like in the homogeneous diffusion this parameter controls how fast the diffusion proceeds. But since $$g(u(x, t))$$ now depends on the signal value this means that the diffusion behaves differently depending on the current structure described by the signal. E.g. if $$\fvec{u}$$ describes an image diffusion means blurring an image and $$g(u(x, t))$$ would make this blurring adaptive to the current image structure. More precisely, since the image gradient is used in \eqref{eq:FEDIsotropic_ConductivityFunction} this means that the blurring behaves differently at edges (high gradient magnitude) and homogeneous areas (low gradient magnitude). Often it is desired to suppress blurring around edges and enhance it in homogeneous areas3.

To achieve this behaviour, $$g(u(x, t))$$ should return low values (resulting in low blurring) for higher gradient magnitudes and high values (resulting in high blurring) for lower gradient magnitudes. This is exactly the behaviour which is modelled by \eqref{eq:FEDIsotropic_ConductivityFunction}. The following animation shows this function with control over the parameter $$\lambda$$ which scales the magnitude value. Larger $$\lambda$$ means higher magnitude values conduct and get blurred as well; or in other words: more and more edges are included in the blurring process. One additional important note about this function is that it returns values in the range $$]0;1]$$.

The goal is to define $$A(\fvec{u}_k)$$ in a way that \eqref{eq:FEDIsotropic_Discrete} behaves as a good discretization of \eqref{eq:FEDIsotropic_PDE}. The first thing we do is that we apply the conductivity function in \eqref{eq:FEDIsotropic_ConductivityFunction} to all signal values

\begin{equation*} \fvec{g}_k = g(\fvec{u}_k). \end{equation*}

These conductivity values will then later be used in $$A(\fvec{u}_k)$$. Now, let me remind that $$A$$ from the homogeneous diffusion process is defined to be an approximation of the second order derivative and each row can basically be seen as a discrete filter in the form $$\left( 1, -2, 1 \right)$$, or equivalent: $$\left( 1, -(1+1), 1 \right)$$. You can think of this as a weighted process which compares the centre element with its left neighbour by weighting the left side positive and the centre negative and analogous on the right side. $$A(\fvec{u}_k)$$ should also encode this information somehow, but the weighting process will be a little bit more sophisticated. The idea is to define a filter like $$\left( a, -(a+b), b \right)$$ with4

\begin{align*} a &= \frac{1}{2} \cdot \left( g_{i-1} + g_{i} \right) \\ b &= \frac{1}{2} \cdot \left( g_{i} + g_{i+1} \right) \end{align*}

encoding the comparison process of the centre element $$i$$ with its neighbours in a similar way as it was the case in the homogeneous diffusion, but rather using not only the signal values but the conductivity information instead. Here, the conductivity information is directly encoded with $$a$$ being the average conductivity between the centre element and its left neighbour and $$b$$ being the average conductivity between the centre element and its right neighbour.

Let's see how this filter expands for an $$i$$ which is not the border element:

\begin{equation*} \begin{pmatrix} a \\ -(a+b) \\ b \\ \end{pmatrix} = \frac{1}{2} \cdot \begin{pmatrix} g_{i-1} + g_{i} \\ -(g_{i-1} + g_{i} + g_{i} + g_{i+1}) \\ g_{i} + g_{i+1} \\ \end{pmatrix} = \frac{1}{2} \cdot \begin{pmatrix} g_{i-1} + g_{i} \\ -g_{i-1} - 2\cdot g_{i} - g_{i+1} \\ g_{i} + g_{i+1} \\ \end{pmatrix} \end{equation*}

As you can see, the weighting process is well defined since in total the elements sum up to zero. The next step is to use this pattern in the definition of $$A(\fvec{u}_k)$$. Let's start with a one-dimensional signal vector.

### 1D example

I want to use a signal vector $$\fvec{u} \in \mathbb{R}^{9 \times 1}$$ containing nine discrete elements. I don't use concrete values here because I think it won't be helpful for understanding the general computation pattern. Providing an example on the index level is more informative in my opinion. The reason for a $$9 \times 1$$ vector lies in the fact that it is easily possible to use the same example later for the 2D case.

In this example, the matrix $$A(\fvec{u}_k)$$ has the dimension $$9 \times 9$$ and is defined as

\begin{equation*} A(\fvec{u}_k) = \frac{1}{2} \cdot \begin{pmatrix} -g_{0}-g_{1} & g_{0}+g_{1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ g_{0}+g_{1} & -g_{0}-2 g_{1}-g_{2} & g_{1}+g_{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & g_{1}+g_{2} & -g_{1}-2 g_{2}-g_{3} & g_{2}+g_{3} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & g_{2}+g_{3} & -g_{2}-2 g_{3}-g_{4} & g_{3}+g_{4} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & g_{3}+g_{4} & -g_{3}-2 g_{4}-g_{5} & g_{4}+g_{5} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & g_{4}+g_{5} & -g_{4}-2 g_{5}-g_{6} & g_{5}+g_{6} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & g_{5}+g_{6} & -g_{5}-2 g_{6}-g_{7} & g_{6}+g_{7} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & g_{6}+g_{7} & -g_{6}-2 g_{7}-g_{8} & g_{7}+g_{8} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & g_{7}+g_{8} & -g_{7}-g_{8} \\ \end{pmatrix} \end{equation*}

Only the first and last row is a bit special since these are the border cases (signal values outside the border are set to zero). The rest of the matrix encodes the discussed weighting process between the centre element (diagonal components of the matrix) and its neighbours (left and right column of the diagonal components).

Remember that in \eqref{eq:FEDIsotropic_Discrete} the actual computation is $$A(\fvec{u}_k) \cdot \fvec{u}_k$$, so the matrix must be multiplied with the signal vector. For the second row of $$A(\fvec{u}_k)$$ this looks like

$$\begin{split} \frac{1}{2} \cdot \begin{pmatrix} g_{0}+g_{1} & -g_{0}-2 g_{1}-g_{2} & g_{1}+g_{2} \\ \end{pmatrix} \cdot \begin{pmatrix} u_{0} \\ u_{1} \\ u_{2} \\ \end{pmatrix} = \\ \frac{1}{2} \cdot \left( (g_{0}+g_{1}) \cdot u_{0} + (-g_{0}-2 g_{1}-g_{2}) \cdot u_{1} + (g_{1}+g_{2}) \cdot u_{2} \right). \end{split} \label{eq:FEDIsotropic_1DComputationLine}$$

In total, this results in

\begin{equation*} A(\fvec{u}_k) \cdot \fvec{u}_k = \frac{1}{2} \cdot \begin{pmatrix} \left(-g_{0}-g_{1}\right) u_{0}+\left(g_{0}+g_{1}\right) u_{1} \\ \left(g_{0}+g_{1}\right) u_{0}+\left(-g_{0}-2 g_{1}-g_{2}\right) u_{1}+\left(g_{1}+g_{2}\right) u_{2} \\ \left(g_{1}+g_{2}\right) u_{1}+\left(-g_{1}-2 g_{2}-g_{3}\right) u_{2}+\left(g_{2}+g_{3}\right) u_{3} \\ \left(g_{2}+g_{3}\right) u_{2}+\left(-g_{2}-2 g_{3}-g_{4}\right) u_{3}+\left(g_{3}+g_{4}\right) u_{4} \\ \left(g_{3}+g_{4}\right) u_{3}+\left(-g_{3}-2 g_{4}-g_{5}\right) u_{4}+\left(g_{4}+g_{5}\right) u_{5} \\ \left(g_{4}+g_{5}\right) u_{4}+\left(-g_{4}-2 g_{5}-g_{6}\right) u_{5}+\left(g_{5}+g_{6}\right) u_{6} \\ \left(g_{5}+g_{6}\right) u_{5}+\left(-g_{5}-2 g_{6}-g_{7}\right) u_{6}+\left(g_{6}+g_{7}\right) u_{7} \\ \left(g_{6}+g_{7}\right) u_{6}+\left(-g_{6}-2 g_{7}-g_{8}\right) u_{7}+\left(g_{7}+g_{8}\right) u_{8} \\ \left(g_{7}+g_{8}\right) u_{7}+\left(-g_{7}-g_{8}\right) u_{8} \\ \end{pmatrix} \end{equation*}

If we would use this in real applications, we would not really set up the matrix $$A(\fvec{u}_k)$$ and perform matrix multiplication. The matrix is way to sparse (containing too many zero elements) in order to make this computational efficient. Instead, it is more efficient to focus on the computation in \eqref{eq:FEDIsotropic_1DComputationLine} and abstract from that. If you look close, you see that the computation consists of six additions/subtractions and three multiplications. With a little bit of restructuring

\begin{align*} &(g_{0}+g_{1}) \cdot u_{0} + (-g_{0}-2 g_{1}-g_{2}) \cdot u_{1} + (g_{1}+g_{2}) \cdot u_{2} \\ &= (g_{0}+g_{1}) \cdot u_{0} + (-g_{0}-g_{1}) \cdot u_{1} + (-g_{1}-g_{2}) \cdot u_{1} + (g_{1}+g_{2}) \cdot u_{2} \\ &= (g_{0}+g_{1}) \cdot u_{0} - (g_{0}+g_{1}) \cdot u_{1} - (g_{1}+g_{2}) \cdot u_{1} + (g_{1}+g_{2}) \cdot u_{2} \\ &= (g_{0}+g_{1}) \cdot (u_{0} - u_{1}) + (g_{1}+g_{2}) \cdot (u_{2} - u_{1}) \end{align*}

we only have 5 additions/subtractions and 2 multiplications left. The following figure depicts the general computation pattern.

Remember that in order to calculate $$\tau_i$$ in the FED scheme (\eqref{eq:FEDIsotropic_Discrete}) the maximum step size $$\tau_{\text{max}}$$ which is still stable in the explicit scheme is needed and can be approximated by the maximum eigenvalue of $$A(\fvec{u}_k)$$. Since the maximum return value of $$g(u(x,t))$$ is 1 we now know that a row of $$A(\fvec{u}_k)$$ has the values $$0.5 \cdot \left( 2, -4, 2 \right)$$ in the extreme case resulting in a maximum absolute eigenvalue of 4, hence $$\tau_{\text{max}} = 0.25$$ can be used5.

### 2D example

Adding an additional dimension to the diffusion process basically means that the second order derivative now needs to be approximated in two directions: $$A_x(\fvec{u}_k)$$ for the horizontal and $$A_y(\fvec{u}_k)$$ for the vertical direction. Both matrices are additively combined, meaning the FED scheme from \eqref{eq:FEDIsotropic_Discrete} changes to

$$\fvec{u}_{k+1} = \left( I + \tau_i \cdot \left( A_x(\fvec{u}_k) + A_y(\fvec{u}_k) \right) \right) \cdot \fvec{u}_k. \label{eq:FEDIsotropic_2DDiscrete}$$

Unfortunately, the second dimension brings also more effort for the border handling since more border accesses need to be considered. As an example, I want to analyse the $$3 \times 3$$ image

\begin{equation*} \fvec{u} = \begin{pmatrix} u_{0} & u_{1} & u_{2} \\ u_{3} & u_{4} & u_{5} \\ u_{6} & u_{7} & u_{8} \\ \end{pmatrix}. \end{equation*}

In the $$x$$-dimension only the vector components $$u_{1}, u_{4}$$ and $$u_{7}$$ are inside of the matrix, respectively the components $$u_{3}, u_{4}$$ and $$u_{5}$$ in the $$y$$-dimension. But this is only so drastic because the example is very small. If the matrix is larger (e.g. represents an image) the border cases may be negligible. To apply the matrices $$A_x(\fvec{u}_k)$$ and $$A_y(\fvec{u}_k)$$ in a similar way, as it was the case in the 1D example, we assume that the signal matrix is reshaped to a one-dimensional vector. This allows us to use similar computations as in the 1D example. First the definition of

\begin{equation*} A_x(\fvec{u}_k) = \frac{1}{2} \cdot \begin{pmatrix} -g_{0}-g_{1} &g_{0}+g_{1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ g_{0}+g_{1} & -g_{0}-2g_{1}-g_{2} &g_{1}+g_{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 &g_{1}+g_{2} & -g_{1}-g_{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -g_{3}-g_{4} &g_{3}+g_{4} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 &g_{3}+g_{4} & -g_{3}-2g_{4}-g_{5} &g_{4}+g_{5} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 &g_{4}+g_{5} & -g_{4}-g_{5} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -g_{6}-g_{7} &g_{6}+g_{7} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 &g_{6}+g_{7} & -g_{6}-2g_{7}-g_{8} &g_{7}+g_{8} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 &g_{7}+g_{8} & -g_{7}-g_{8} \\ \end{pmatrix} \end{equation*}

which now includes two more border cases (third and sixth row) but otherwise did not changed from the 1D case. The calculation in \eqref{eq:FEDIsotropic_2DDiscrete} is easiest done if $$A_x(\fvec{u}_k) \cdot \fvec{u}_k$$ and $$A_y(\fvec{u}_k) \cdot \fvec{u}_k$$ are applied separately. For the first matrix, this results in

\begin{equation*} A_x(\fvec{u}_k) \cdot \fvec{u}_k = \frac{1}{2} \cdot \begin{pmatrix} \left(-g_{0}-g_{1}\right) u_0+\left(g_{0}+g_{1}\right) u_1 \\ \left(g_{0}+g_{1}\right) u_0+\left(-g_{0}-2 g_{1}-g_{2}\right) u_1+\left(g_{1}+g_{2}\right) u_2 \\ \left(g_{1}+g_{2}\right) u_1+\left(-g_{1}-g_{2}\right) u_2 \\ \left(-g_{3}-g_{4}\right) u_3+\left(g_{3}+g_{4}\right) u_4 \\ \left(g_{3}+g_{4}\right) u_3+\left(-g_{3}-2 g_{4}-g_{5}\right) u_4+\left(g_{4}+g_{5}\right) u_5 \\ \left(g_{4}+g_{5}\right) u_4+\left(-g_{4}-g_{5}\right) u_5 \\ \left(-g_{6}-g_{7}\right) u_6+\left(g_{6}+g_{7}\right) u_7 \\ \left(g_{6}+g_{7}\right) u_6+\left(-g_{6}-2 g_{7}-g_{8}\right) u_7+\left(g_{7}+g_{8}\right) u_8 \\ \left(g_{7}+g_{8}\right) u_7+\left(-g_{7}-g_{8}\right) u_8 \\ \end{pmatrix} \end{equation*}

and reveals a very similar computation pattern as in the 1D case. For the $$y$$-dimension the matrix is defined as

\begin{equation*} A_y(\fvec{u}_k) = \frac{1}{2} \cdot \begin{pmatrix} -g_{0}-g_{3} & 0 & 0 & g_{0}+g_{3} & 0 & 0 & 0 & 0 & 0 \\ 0 & -g_{1}-g_{4} & 0 & 0 & g_{1}+g_{4} & 0 & 0 & 0 & 0 \\ 0 & 0 & -g_{2}-g_{5} & 0 & 0 & g_{2}+g_{5} & 0 & 0 & 0 \\ g_{0}+g_{3} & 0 & 0 & -g_{0}-2 g_{3}-g_{6} & 0 & 0 & g_{3}+g_{6} & 0 & 0 \\ 0 & g_{1}+g_{4} & 0 & 0 & -g_{1}-2 g_{4}-g_{7} & 0 & 0 & g_{4}+g_{7} & 0 \\ 0 & 0 & g_{2}+g_{5} & 0 & 0 & -g_{2}-2 g_{5}-g_{8} & 0 & 0 & g_{5}+g_{8} \\ 0 & 0 & 0 & g_{3}+g_{6} & 0 & 0 & -g_{3}-g_{6} & 0 & 0 \\ 0 & 0 & 0 & 0 & g_{4}+g_{7} & 0 & 0 & -g_{4}-g_{7} & 0 \\ 0 & 0 & 0 & 0 & 0 & g_{5}+g_{8} & 0 & 0 & -g_{5}-g_{8} \\ \end{pmatrix} \end{equation*}

which looks very different at the first glance. But the computation is actually the same, we only need to combine elements vertically now, e.g. $$u_{0}$$ (first row) and $$u_{3}$$ (fourth row), and the reshaped signal value offers no consecutive access here. For the combination with the signal vector as implied by \eqref{eq:FEDIsotropic_2DDiscrete} we get

\begin{equation*} A_y(\fvec{u}_k) \cdot \fvec{u}_k = \frac{1}{2} \cdot \begin{pmatrix} \left(-g_{0}-g_{3}\right) u_{0}+\left(g_{0}+g_{3}\right) u_{3} \\ \left(-g_{1}-g_{4}\right) u_{1}+\left(g_{1}+g_{4}\right) u_{4} \\ \left(-g_{2}-g_{5}\right) u_{2}+\left(g_{2}+g_{5}\right) u_{5} \\ \left(g_{0}+g_{3}\right) u_{0}+\left(-g_{0}-2 g_{3}-g_{6}\right) u_{3}+\left(g_{3}+g_{6}\right) u_{6} \\ \left(g_{1}+g_{4}\right) u_{1}+\left(-g_{1}-2 g_{4}-g_{7}\right) u_{4}+\left(g_{4}+g_{7}\right) u_{7} \\ \left(g_{2}+g_{5}\right) u_{2}+\left(-g_{2}-2 g_{5}-g_{8}\right) u_{5}+\left(g_{5}+g_{8}\right) u_{8} \\ \left(g_{3}+g_{6}\right) u_{3}+\left(-g_{3}-g_{6}\right) u_{6} \\ \left(g_{4}+g_{7}\right) u_{4}+\left(-g_{4}-g_{7}\right) u_{7} \\ \left(g_{5}+g_{8}\right) u_{5}+\left(-g_{5}-g_{8}\right) u_{8} \\ \end{pmatrix} \end{equation*}

Note how e.g. the middle line combines $$u_{1}$$ (top) with $$u_{4}$$ (middle) and $$u_{7}$$ (bottom) and how these elements are aligned vertically in the signal matrix $$\fvec{u}$$. Like in the 1D case, it is possible to optimize the computation for a matrix line, e.g. for the middle line in $$A_y(\fvec{u}_k) \cdot \fvec{u}_k$$

\begin{align*} & \left(g_{1}+g_{4}\right) \cdot u_{1} + \left(-g_{1}-2 g_{4}-g_{7}\right) \cdot u_{4} +\left(g_{4}+g_{7}\right) \cdot u_{7} \\ &= \left(g_{1}+g_{4}\right) \cdot u_{1} - \left(g_{1}+g_{4}\right) \cdot u_{4} - \left(g_{4} + g_{7}\right) \cdot u_{4} +\left(g_{4}+g_{7}\right) \cdot u_{7} \\ &= \left(g_{1}+g_{4}\right) \cdot \left( u_{1} - u_{4}\right) +\left(g_{4}+g_{7}\right) \cdot \left( u_{7} - u_{4} \right). \\ \end{align*}

The optimization alongside the $$x$$-direction can directly be copied from the 1D case. To sum up, the following figure illustrates the general computation pattern for the 2D case.

You can move this pattern around to any image location. If some of the computations touch the border, this part is set to zero. E.g. when centred around the element with index 3, the left subtraction/addition is neglected. This is in agreement with the definition of $$A(\fvec{u}_k)$$ where the computation is also reduced if out of border values are involved.

Given all this, it is relatively straightforward to implement a FED step (one iteration of a FED cycle). The following listing shows an example implementation in C++ using the OpenCV library6. The function takes the image from the previous iteration, the conductivity image for the current FED cycle (the conductivity image only changes after the FED cycle completes) and the current step size as input. The full source code is available on GitHub.


static cv::Mat FEDInnerStep(const cv::Mat& img, const cv::Mat& cond, const double stepsize)
{
// Copy input signal vector
cv::Mat imgCopy = img.clone();

// Apply the computation pattern to each image location
for (int row = 0; row < img.rows; row++)
{
for (int col = 0; col < img.cols; col++)
{
double xLeft = 0;
double xRight = 0;
double yTop = 0;
double yBottom = 0;

if (col > 0)
{
// 3 <--> 4
xLeft = (cond.at<double>(row, col - 1) + cond.at<double>(row, col)) * (img.at<double>(row, col - 1) - img.at<double>(row, col));
}
if (col < img.cols - 1)
{
// 4 <--> 5
xRight = (cond.at<double>(row, col) + cond.at<double>(row, col + 1)) * (img.at<double>(row, col + 1) - img.at<double>(row, col));
}
if (row > 0)
{
// 1 <--> 4
yTop = (cond.at<double>(row - 1, col) + cond.at<double>(row, col)) * (img.at<double>(row - 1, col) - img.at<double>(row, col));
}
if (row < img.rows - 1)
{
// 4 <--> 7
yBottom = (cond.at<double>(row, col) + cond.at<double>(row + 1, col)) * (img.at<double>(row + 1, col) - img.at<double>(row, col));
}

// Update the current pixel location with the conductivity based derivative information and the varying step size
imgCopy.at<double>(row, col) = 0.5 * stepsize * (xLeft + xRight + yTop + yBottom);
}
}

// Update old image
return img + imgCopy;
}



Note that this is basically the computation pattern from the figures with some additional logic for the border handling. If you have read my article about FED, you might have been a bit disappointed that no real application was shown. But without a definition of the conductivity matrix, it is hard to show something useful. Nevertheless, we have now all ingredients together so that it is possible to apply isotropic diffusion to an example image. The following animation lets you control this process starting from $$T=0$$ (original image) up to the diffusion time $$T=100$$. If you switch to the right tab, you can see the result of the corresponding homogeneous diffusion process (applied via Gaussian convolution).

As you can see, the isotropic diffusion results in blurred regions but the main edges stay intact. E.g. the fine left background structure is nearly completely lost at the end but the stripes of the clothes are still visible. This effect is especially noticeable when the result is compared with the Gaussian filter response, which blurs everything equally regardless of its content.

List of attached files:

1. Note that for anisotropic diffusion the conductivity function would depend on a tensor (e.g. the structure tensor) instead of the scalar valued image gradient resulting in a response which is selective for different orientations.
2. See the first footnote of my FED article for a description of the used notation.
3. I refer to Digital Image Processing by Burger and Burge, p. 433 for a detailed introduction to edge-preserving smoothing in the image processing domain.
4. Note that I omit the evolution index $$k$$ when accessing elements in the vector for simplicity reasons. So, $$g_i$$ refers to the $$i^{\text{th}}$$ component of the vector $$\fvec{g}_k$$ (with the current relevant evolution level $$k$$). But this is really only for notational simplicity, the conductivity still needs to be calculated for each evolution step $$k$$ and the signal vector still evolves over the diffusion time $$k$$ (cf. the explanation of the notation).
5. The Gershgorin circle theorem gives $$\left|\lambda_{\text{max}}\right|=4$$ based on the centre value $$\left|0.5 \cdot (-2)\right| = 2$$ with radius $$0.5 \cdot \left(2 + 2\right) = 2$$ and the relevant equation of the FED article states that in this case $$\tau_{\text{max}} = \frac{1}{\left|\lambda_{\text{max}}\right|} = \frac{1}{4}$$ is the maximum stable step size.
6. Note that this implementation is optimized for readability and not performance. E.g. you may want to change the matrix accesses to the pointer method before using the code in production environments.

## Introduction to Fast Explicit Diffusion (FED)

Diffusion, in general, is a concept which describes the propagation of particles over time within some substance. One good example is temperature diffusing in a closed body leading to the heat equation. The question hereby is: how does the temperature change over time at the considered spatial locations. Assuming for simplicity only 1D spatial locations, this can mathematically be described by a partial differential equation (PDE)

$$\frac{\partial u(x,t)}{\partial t} = g \cdot \frac{\partial^2 u(x,t)}{\partial^2 x^2}. \label{eq:FEDIntroduction_PDE}$$

with $$g$$ being a constant describing the conductivity of the homogeneous diffusion (e.g. how fast levels the temperature to equality). I am not going into details here why this equation looks like it looks (check the linked video for a great introduction). My concern is more about the discretization of this PDE. By using the Euler method, the discretization is achieved by

\begin{align*} t_{k+1} &= t_k + \tau \\ \fvec{u}_{k+1} &= \fvec{u}_k + \tau \cdot (g \cdot \Delta \fvec{u}_k) \\ \end{align*}

which proceeds over time in steps of $$\tau$$. The part in the brackets is basically the discrete version of \eqref{eq:FEDIntroduction_PDE} where $$\Delta$$ denotes an operator for the central finite difference approximation of the second order derivative $$\frac{\partial^2 u(x,t)}{\partial^2 x^2}$$ applied to the vector of signal values $$\fvec{u}_k$$1 at the time $$t_k$$ and the constant $$g$$ is re-used as is. Note that in this explicit scheme the time step $$\tau$$ is fixed over the complete process. But one step size might not be appropriate for every step; it may be better if some are larger and others are smaller. This problem is addressed by the Fast Explicit Diffusion (FED) algorithm. Basically, it introduces varying time steps $$\tau_i$$ resulting in more accuracy and higher performance (hence the “Fast” in the name). This article has the aim to introduce FED and to provide examples of how it can be used.

Before going into the details, let me note that FED can not only be applied to homogeneous diffusions (like the heat equation) but also to inhomogeneous, anisotropic or isotropic and even multi-dimensional diffusion problems. More precisely, FED can be applied to any finite scheme of the form [12-13]2

$$\fvec{u}_{k+1} = \left( I + \tau \cdot A \right) \cdot \fvec{u}_k \label{eq:FEDIntroduction_Arbitrary}$$

where $$A$$ is some symmetric and negative semidefinite matrix embedding the conductivity information. This includes the approximation of the second order derivative $$\Delta$$ and the constant $$g$$. I am coming to this point later, but first dive into what FED actually does.

The rest of this article is structured as follows: first, the concept behind FED is introduced by an example showing the basic rationale behind it. Next, the notation is simplified allowing for generalization of the concept behind FED to a wider range of problems. Then, a summary of the used parameters is given, and in the last section, some elaboration of the role of box filters in FED is provided.

### How does it work?

For simplicity, I stick to one-dimensional homogeneous diffusion problems and set the constant to $$g = 1$$. First, two important notes

• If a homogeneous diffusion is applied to a signal, it is equivalent to applying a Gaussian to that signal. More precisely, the total diffusion time $$T$$ maps to a Gaussian with standard deviation $$\sigma = \sqrt{2 \cdot T}$$ (e.g. Digital Image Processing by Burger and Burge, p. 435).
• If a filter whose coefficients are non-negative and sum up to 1 (i.e. $$\sum w_i = 1$$) is applied multiple times to a signal, it approximates a Gaussian convolution with that signal. This is known from the central limit theorem (CLT).

What has FED now to do with this? In essence, FED introduces an alternative way of applying such a filter by using iterations of explicit diffusion steps. The main finding is that filters (with the discussed properties) can be defined as

$$B_{2n+1} = \prod_{i=0}^{n-1} \left( I + \tau_i \Delta \right). \label{eq:FEDIntroduction_Cycle}$$

I directly try to be a bit more concrete here: $$B_{2n+1}$$ denotes a box filter of size $$2n+1$$ and $$I$$ the identity (e.g. 1). $$\Delta$$ is, again, the operator for an approximation of the second order derivative. If applied to a signal $$\fvec{u}$$, it may be defined as3

$$\Delta \fvec{u} = u(x-1, t) - 2 \cdot u(x, t) + u(x+1, t), \label{eq:FEDIntroduction_Delta}$$

so the spatial difference at some fixed time $$t$$. You can also think of this operation as a kernel $$\left( 1, -2, 1 \right)$$ convolved with the signal $$\fvec{u}$$. All iterations of \eqref{eq:FEDIntroduction_Cycle} together are called a FED cycle. Given this, there is only $$\tau_i$$ left. This is actually the heart of FED since it denotes the varying step sizes, and in this case, is defined as

$$\tau_i = \tau_{\text{max}} \cdot \frac{1}{2 \cos ^2\left(\frac{\pi (2 i+1)}{4 n+2}\right)} \label{eq:FEDIntroduction_TimeStepsBox}$$

with $$\tau_{\text{max}} = \frac{1}{2}$$ here (more on this parameter later). Unfortunately, I can't tell you why this works (the authors provide proves though [46 ff. (Appendix)]) but I can give an example.

We want to show that the FED cycle applied to the signal is indeed the same as convolving the signal with a box filter. Let's start by defining a small signal and a box filter of size $$n=1$$

\begin{equation*} \fvec{u} = \begin{pmatrix} 1 \\ 4 \\ 2 \\ 6 \\ \end{pmatrix} \quad \text{and} \quad \tilde{B}_3 = \frac{1}{3} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}. \end{equation*}

The convolution with reflected boundaries4 results in

$$\fvec{u} * \tilde{B}_3 = \left( 2.,2.33333,4.,4.66667 \right)^T. \label{eq:FEDIntroduction_ExampleConvolution}$$

For the FED approach, we first need to calculate the time step (only one in this case because of $$n=1$$)

\begin{equation*} \tau_0 = \frac{1}{2} \cdot \frac{1}{2 \cos ^2\left(\frac{\pi (2 \cdot 0+1)}{4 1+2}\right)} = \frac{1}{3} \end{equation*}

and then \eqref{eq:FEDIntroduction_Cycle} can be applied to $$u$$ by multiplication resulting in the following explicit diffusion step

\begin{equation*} B_3 \cdot \fvec{u} = \prod_{i=0}^{1-1} \left( I + \tau_i \Delta \right) \cdot \fvec{u} = \fvec{u} + \tau_0 \cdot \Delta \fvec{u}. \end{equation*}

How is $$\Delta \fvec{u}$$ defined here? It is basically \eqref{eq:FEDIntroduction_Delta} but extended to handle out of border accesses correctly

\begin{equation*} \Delta u_{j} = \begin{cases} &-& u_{j} &+& u_{j+1}, & j=1 \\ u_{j-1} &-& u_{j}, & & & j=N \\ u_{j-1} &-& 2 u_{j} &+& u_{j+1}, & \text{else} \\ \end{cases} \end{equation*}

and is applied to every vector component $$u_{j}$$. In this case, we have

\begin{equation*} \Delta \fvec{u} = \begin{pmatrix} -1+4 \\ 1-2\cdot 4+2 \\ 4-2\cdot 2 + 6 \\ 2-6 \\ \end{pmatrix} = \begin{pmatrix} 3 \\ -5 \\ 6 \\ -4 \\ \end{pmatrix} \end{equation*}

and then everything can be combined together

$$\fvec{u} + \tau_0 \cdot \Delta \fvec{u} = \begin{pmatrix} 1 \\ 4 \\ 2 \\ 6 \\ \end{pmatrix} + \frac{1}{3} \cdot \begin{pmatrix} 3 \\ -5 \\ 6 \\ -4 \\ \end{pmatrix} = \begin{pmatrix} 2 \\ 2.33333 \\ 4 \\ 4.66667 \\ \end{pmatrix}. \label{eq:FEDIntroduction_ExampleFED}$$

As you can see, \eqref{eq:FEDIntroduction_ExampleConvolution} and \eqref{eq:FEDIntroduction_ExampleFED} produce indeed identical results.

If larger filters and hence more iterations per FED cycle are needed, the same technique would be recursively applied. E.g. for $$n=2$$ we get

\begin{equation*} B_5 \cdot \fvec{u}= \prod_{i=0}^{2-1} \left( I + \tau_i \Delta \right) \cdot \fvec{u} = \fvec{u} + \tau_0 \cdot \Delta \fvec{u} + \tau_1 \cdot \Delta \fvec{u} + \tau_0 \cdot \tau_1 \cdot \Delta (\Delta \fvec{u}), \end{equation*}

where $$\Delta (\Delta \fvec{u})$$ effectively encodes an approximation for the fourth order derivative (it calculates the second order derivative of something which is already the second order derivative of the input signal). The concept stays the same for larger filters. The iterations of the FED cycle gradually reach the same state of the corresponding box filter response. To visualize this, the following animation shows the result of a filter with $$n=10$$ each time plotting the state up to the current iteration (i.e. the previous and current multiplications).

So far we only discussed exactly one FED cycle (which consists of iterations). It is also possible to apply multiple cycles just by repeating the procedure. This is then equal to iteratively applying multiple filters to the signal.

Recapitulate what we now have:

• multiple iterations (products in \eqref{eq:FEDIntroduction_Cycle}) form a FED cycle,
• a FED cycle is the same like convolving the signal with a (box) filter,
• multiple FED cycles are the same as the iterative convolution of the signal with (box) filters,
• applying multiple filters approximate a Gaussian filter response and
• the response of a Gaussian filter is the same as the result of the diffusion process.

Therefore, as a conclusion of this chain of logic: FED can be used as explicit diffusion scheme for homogeneous diffusion equations. But how is FED used for diffusion problems which are usually formulated like in \eqref{eq:FEDIntroduction_Arbitrary}? It basically includes \eqref{eq:FEDIntroduction_Cycle}:

$$\fvec{u}_{k+1} = \prod_{i=0}^{n-1} \left( I + \tau_i \cdot A \right) \cdot \fvec{u}_k. \label{eq:FEDIntroduction_DiscreteDiffusion}$$

This means that one diffusion step (from $$k$$ to $$k+1$$) applies a box filter with length $$n$$ to the signal. And with further diffusion steps, a Gaussian and hence the diffusion itself is approximated. The notion of $$A$$ is introduced in the next section.

You can also think of this as breaking apart the proceeding in diffusion time into multiple blocks (= FED cycles). Each FED cycle has a fixed step size of $$\theta = \sum_{i=0}^{n-1}\tau_i$$ and this size is the same for all cycles (like the usual step size in an explicit scheme). But each FED cycle is now divided into multiple inner steps (factors of \eqref{eq:FEDIntroduction_DiscreteDiffusion}) and each inner step has its own step size $$\tau_i$$. Additionally, all FED cycles share the same inner steps, e.g. each cycle starts with an inner step with step size $$\tau_0$$ and ends with $$\tau_{n-1}$$. The following figure visualizes this for the previous example.

We have now seen how FED generally works. The next part focuses on simplifying the notation.

### Matrix notation

Calculating FED like in the previous section may become unhandy for larger signals and/or larger filters. Luckily, there is an easier way to express the same logic using matrices. It turns out that this is also useful in generalising FED to arbitrary problems (as implied in the beginning).

The example so far is based on a PDE with some change of function value at each position over time. In \eqref{eq:FEDIntroduction_PDE} therefore the derivative definition depended on two variables. It is possible to transform the PDE to a system of ODEs by fixing space. The derivative can then be defined as

\begin{equation*} \frac{\mathrm{d} \fvec{u}(t)}{\mathrm{d} t} = A \cdot \fvec{u}(t). \end{equation*}

Where $$\fvec{u}(t)$$ is the vector containing the discrete set of spatial locations but which are still continuous in time $$t$$. You can think of this as each vector component fixing some $$x$$ position of the original continuous function $$u(x, t)$$ but the function value of this position (e.g. the temperature value) still varies continuously over time. The benefit of this approach is that the operator $$\Delta$$ is discarded and replaced by the matrix $$A \in \mathbb{R}^{N \times N}$$ which returns us to \eqref{eq:FEDIntroduction_Arbitrary}. $$A$$ is now responsible for approximating the second order derivative. To be consistent with $$\Delta$$ it may be defined for the previous example as

\begin{equation*} A = \begin{pmatrix} -1 & 1 & 0 & 0 \\ 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -1 \\ \end{pmatrix}. \end{equation*}

Note that the pattern encoded in this matrix is the same as in \eqref{eq:FEDIntroduction_Delta}: left and right neighbour weighted once and the centre value compensates for this. Hence, $$A \fvec{u}$$ is the same as $$\Delta \fvec{u}$$. By calculating

\begin{equation*} (I + \tau_1\cdot A) \fvec{u} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 4 \\ 2 \\ 6 \\ \end{pmatrix} + \tau_1 \cdot \begin{pmatrix} -1 & 1 & 0 & 0 \\ 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -1 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 4 \\ 2 \\ 6 \\ \end{pmatrix} = \begin{pmatrix} 2 \\ 2.33333 \\ 4 \\ 4.66667 \\ \end{pmatrix} \end{equation*}

we obtain exactly the same result as in \eqref{eq:FEDIntroduction_ExampleFED}. A FED cycle with multiple iterations is done by repeatedly multiplying the brackets (cf. \eqref{eq:FEDIntroduction_Cycle})

\begin{equation*} \fvec{u}_{k+1} = \left( I + \tau_0 \cdot A \right) \left( I + \tau_1 \cdot A \right) \cdots \left( I + \tau_{n-1} \cdot A \right) \fvec{u}_k. \end{equation*}

This is actually the basis to extend FED to arbitrary diffusion problems as denoted in \eqref{eq:FEDIntroduction_Arbitrary} [9-11]. The trick is to find an appropriate matrix $$A$$. If you, for example, encode the local conductivity information, i.e. let $$A$$ depend on the signal values: $$A(\fvec{u}_{k})$$, it is possible to use FED for inhomogeneous diffusion problems.

In the next section, I want to give an overview of the relevant parameters when using FED.

### Parameters

I already used some parameters, like $$\tau_{\text{max}}$$, without explaining what they do. The following list contains all previously discussed parameters plus introducing new ones which were there only implicitly.

• $$T$$: is the total diffusion time. A longer diffusion time means in general higher blurring of the signal. I already stated that the relation to the standard deviation is $$T = \frac{1}{2} \cdot \sigma$$. Usually, the desired amount of blurring $$\sigma$$ is given.
• $$M$$: a new parameter denoting the number of FED cycles used. So far exactly one cycle was used in the examples but usually multiple cycles are desired. Since multiple cycles correspond to multiple filters applied to the signal and more filters mean a better Gaussian approximation, this parameter controls the quality. Larger $$M$$ means better quality (better Gaussian approximation). This parameter must also be given.
• $$n$$: is the length of one FED cycle, i.e. the number of iteration it contains. It corresponds to the length of an equivalent kernel with size $$2\cdot n + 1$$. Usually, the total diffusion time $$T$$ should be accomplished by $$M$$ cycles of equal length. $$n$$ is therefore the same for all cycles and can be calculated as [11] \begin{equation*} n = \left\lceil -\frac{1}{2} + \frac{1}{2} \cdot \sqrt{1 + \frac{12 \cdot T}{M \cdot \tau_{\text{max}}}} \right\rceil. \end{equation*}
• $$\theta$$: a new parameter denoting the diffusion time for one FED cycle, i.e. how much further in time one cycle approaches. This is basically the sum of all steps $$\tau_i$$ \begin{equation*} \theta = \sum_{i=0}^{n-1} \tau_i \end{equation*} but it is also possible to calculate it explicitly for a given kernel, e.g. for a box filter [10] \begin{equation*} \theta_{\text{box}}(n) = \tau_{\text{max}} \cdot \frac{n^2 + n}{3}. \end{equation*}
• $$\tau_{\text{max}}$$: is the maximum possible time step from the Euler scheme which does not violate stability. It is defined by the stability region and is based on the eigenvalues of $$A$$. More precisely, the relation [12] $$\tau_{\text{max}} \leq \frac{1}{\left| \max_i{(\lambda_i)} \right|} \label{eq:FEDIntroduction_TauMax}$$ for the eigenvalues $$\lambda_i$$ must hold. It is possible to make a worst-case estimation of the eigenvalues with the Gershgorin circle theorem. This explains why $$\tau_{\text{max}} = \frac{1}{2}$$ is used here: the largest eigenvalue of $$A$$ is5 $$\lambda_{\text{max}} = 4$$ and $$\frac{2}{4} = \frac{1}{2}$$ is valid according to \eqref{eq:FEDIntroduction_TauMax}.

After giving an overview of the relevant parameters, I want to elaborate a little bit more on the relation between filters and FED.

### Why box filters?

You may have noticed that I kind of silently used \eqref{eq:FEDIntroduction_Cycle} as a representation of a box filter. Actually, this is even more general since any symmetric filter with its coefficients summing up to 1 can be used. So, why using a box filter? Is the box filter not the filter with such a destructive behaviour in the frequency domain? You could argue that, but let me remind you that in FED we usually use several iterations (i.e. $$M>1$$) and therefore applying multiple box filters which behave more like a Gaussian filter in the end. But there are even better reasons to use a box filter here. They are best understood when looking at how other filters behave [6-8].

In what do varying filters differ? Most importantly, the varying time steps $$\tau_i$$ used in \eqref{eq:FEDIntroduction_Cycle} depend on the concrete filter. The steps in \eqref{eq:FEDIntroduction_TimeStepsBox} e.g. correspond to a box filter. Other filters result in different step sizes [7]. If the $$\tau_i$$ are different, also their sum $$\theta(n)$$ is not the same for all filters. This means if we want to proceed with all filters a fixed amount of diffusion time per FED cycle, $$n$$ must also be defined differently. Therefore, different filters result in different long iterations per FED cycle.

Three filters will now be analysed under two basic criteria:

• Quality: the general goal is to approximate a Gaussian function since this is what the (homogeneous) diffusion equation implies. So the question is: how well do multiple FED cycles approximate a Gaussian?
• Performance: it is good to have high quality but it must also be achieved in a reasonable amount of time. In terms of FED, performance is defined by $$n$$, the number of iterations per cycle. As more iterations needed, as worse is the performance. So the question is: how many iterations are needed in total?

The test setup is now as follows: a total diffusion time of $$T=6$$ should be achieved in $$M=3$$ FED cycles to a signal of length 101 with a single peak in the middle

\begin{equation*} s(i) = \begin{cases} 1, & i = 51 \\ 0, & i \neq 51 \end{cases} \end{equation*}

The behaviour of three different filters will now be analysed: box, binomial and the so-called maximum variance (MV) kernel. The first column of the following table shows a graph for each kernel (for concrete definition see [6]).

Filter Cycle time $$\theta(n)$$ Iterations $$n$$ Performance Quality
$$\theta_{\text{box}}(n) = \frac{n^2+n}{6}$$ $$n_{\text{box}}=3$$ Good: $$\mathcal{O}_{\theta_{\text{box}}}(n^2)$$ Good
$$\theta_{\text{bin}}(n) = \frac{n}{4}$$ $$n_{\text{bin}}=8$$ Poor: $$\mathcal{O}_{\theta_{\text{bin}}}(n)$$ Very good
$$\theta_{\text{box}}(n) = \frac{n^2}{2}$$ $$n_{\text{MV}}=2$$ Very good: $$\mathcal{O}_{\theta_{\text{MV}}}(n^2)$$ Poor

The second column shows the cycle time $$\theta(n)$$ for each filter as a variable of $$n$$. As faster this function growths as better because then the diffusion proceeds faster. The following figure shows how the MV kernel is superior in this task. But note also that the box filter, even though growing slower, increases still in a quadratic magnitude of $$n$$ (hence $$\mathcal{O}_{\theta_{\text{box}}}(n^2)$$ in the 4th column of the table).

In the test scenario, each filter is supposed to make a total step of $$\frac{T}{M} = 2$$, so $$n$$ is adjusted for each filter in order to have proceeded the same diffusion time after each cycle. For the box filter, this leaves $$n_{\text{box}}=3$$ since $$\frac{3^2 + 3}{6} = 2$$. Therefore, over three iterations $$2 \cdot 3 = 6$$ FED multiplications are necessary. Similarly for the other values in the third column. It is also worth looking how the time steps $$\tau_i$$ accumulate over the multiplications (in only one FED cycle since the steps are equal among all cycles). The following figure shows how the $$\tau_i$$ accumulate to 2 for each filter revealing that the MV kernel wins this race.

So far for on the performance point of view, but what are the results regarding approximation quality of a Gaussian? To answer this question an error value is calculated. More concretely, the total of absolute differences between the filter response and the corresponding Gaussian after each FED cycle is used

$$E = \sum_{i=1}^{101} = \left| \operatorname{FED}_{T, \text{filter}}(s(i)) - G_{\sigma}(i) \right| \label{eq:FEDIntroduction_Error}$$

where $$\operatorname{FED}_{T, \text{filter}}(s(i))$$ denotes the result after applying the FED cycle for the specific filter up to the total diffusion time $$T$$ and $$G_{\sigma}(i)$$ is the Gaussian with the same standard deviation and mean as the filter response data.

Each of the filters is now applied three times to $$s(i)$$ by using three FED cycles. Since the signal is just defined as a single peak of energy 1, the first iteration will produce the filter itself in each case.

Even though it is visible that all tend to take the shape of a Gaussian, it is also clear that the binomial filter performs best on this task. The visual representation together with the error value now also explains the fifth column of the table.

Summarising, we can now see that no filter is perfect in both disciplines. But if we would need to choose one filter, we would probably select the box filter. It offers reasonably approximation quality and still good performance and this is exactly the reason why in FED the cycle times resulting from the box filter are used.

As a last note, let me show how the step sizes $$\tau_i$$ behave for larger $$n$$. The following figure shows the individual step sizes $$\tau_i$$ and their cumulative sum

\begin{equation*} \sum_{j=0}^{i} \tau_j \end{equation*}

for $$n=50$$. With higher cycle lengths the steps get even larger [12].

As you can see, in higher iterations the step sizes get larger and larger resulting in a great speedup (comparison between using a fixed time step of 0.5 and using the varying step sizes). What is more, there are many steps which actually exceed the stability limit from the explicit scheme. It can be shown though that at the end of a full cycle the result stays stable [10].

List of attached files:

1. A short remark on the notation I use here: \begin{align*} u(x,t) &: \text{continuous function in space and time.} \\ \fvec{u} &: \text{discrete signal vector at the start time $$t_0$$.} \\ \fvec{u}_k &: \text{discrete signal vector evolved to the diffusion step $$k$$.} \\ u_j &: \text{$$j^{\text{th}}$$ component of the vector $$\fvec{u}$$ (or $$\fvec{u}_k$$, I usually don't repeat the evolution index $$k$$} \\ &\phantom{:} \text{when noting individual vector elements) with $$j \in \left\{ 1, \ldots, N\right\}$$ and $$N$$ the number of} \\ &\phantom{:} \text{elements in the vector.} \\ \fvec{u}(t)&: \text{vector with discrete set of spatial locations which are still continuous in time.} \end{align*}
2. Numbers in square brackets are the page numbers of the journal version of the FED article: Cyclic schemes for PDE-based image analysis.
3. With $$h=1$$ here.
4. Meaning the signal values repeat at the boundaries: $$\left(\ldots, 4, 1, 1, 4, 2, 6, 6, 2, \ldots\right)$$.
5. $$\left| -2 \right| = 2$$ as the centre with radius $$1+1=2$$ leaving a total range of 4. The eigenvalues must also be real since $$A$$ is symmetric.

## Solving a homogeneous ODE by first-order substitution methods

Calculating analytically solutions to differential equations can be hard and sometimes even impossible. Methods exist for special types of ODEs. One method is to solve the ODE by separation of variables. The idea of substitution is to replace some variable so that the resulting equation has the form of such a special type where a solution exists. In this scenario, I want to look at homogeneous ODEs which have the form

$$y'(x) = F\left( \frac{y(x)}{x} \right). \label{eq:homogeneousODE}$$

They can be solved by replacing $$z=\frac{y}{x}$$ followed by separation of variables. Equations of this kind have the special property of being invariant against uniform scaling ($$y \rightarrow \alpha \cdot y_1, x \rightarrow \alpha \cdot x_1$$):

\begin{align*} \frac{\alpha \cdot \partial y_1}{\alpha \cdot \partial x_1} &= F\left( \frac{\alpha \cdot y_1}{\alpha \cdot x_1} \right) \\ \frac{\partial y_1}{\partial x_1} &= F\left( \frac{y_1}{x_1} \right) \end{align*}

Before analysing what this means, I want to introduce the example from the corresponding Lecture 4: First-order Substitution Methods (MIT OpenCourseWare), which is the source of this article. I derive the substitution process for this example later.

Imagine a small island with a lighthouse built on it. In the surrounding sea is a drug boat which tries to sail silently around the sea raising no attention. But the lighthouse spots the drug boat and targets the boat with its light ray. Panic-stricken the boat tries to escape the light ray. To introduce some mathematics to the situation, we assume that the boat always tries to escape the light ray in a 45° angle. Of course, the lighthouse reacts accordingly and traces the boat back. The following image depicts the situation.

We now want to know the boat's curve, when the light ray always follows the boat directly and the boat, in turn, evades in a 45° angle. We don't model the situation as a parametric curve where the position would depend on time (so no $$(x(t), y(t))$$ here). This also means that we don't set the velocity of the curve explicitly. Instead, the boat position just depends on the angle of the current light ray. Mathematically, this means that in the boat's curve along the x-y-plane the tangent of the curve is always enclosed in a 45° angle with the light ray crossing the boat's position. $$\alpha$$ is the angle of the light ray and when we assume that the lighthouse is placed at the origin so that the slope is just given by the fraction $$\frac{y}{x}$$, $$\alpha$$ is simply calculated as

\begin{equation*} \tan(\alpha) = \frac{y}{x}. \end{equation*}

Now we can define the tangent of the boat's curve, which is given by its slope value

$$y'(x) = f(x,y(x)) = \tan(\alpha + 45°) = \frac{\tan(\alpha) + \tan(45°)}{1 - \tan(\alpha) \cdot \tan(45°)} = \frac{\frac{y(x)}{x} + 1}{1 - \frac{y(x)}{x}} = \frac{x + y(x)}{x - y(x)}. \label{eq:slopeBoat}$$

In the first simplification step, a trigonometric addition formula is used. This again can be simplified so that the result fulfils the definition of \eqref{eq:homogeneousODE}. This means that the ODE can be solved by separation of variables if the substitution $$z(x) = \frac{y(x)}{x}$$ is made. We want to replace $$y'(x)$$, so we first need to calculate the derivative of the substitution equation

\begin{align*} y(x) &= z(x) \cdot x \\ y'(x) &= \frac{\partial y(x)}{\partial x} = z'(x) \cdot x + z(x). \end{align*}

Note that we calculate the derivative with respect to $$x$$ and not $$y(x)$$ (which is a function depending on $$x$$ itself). Therefore the product rule was used. Next we substitute and try to separate variables.

\begin{align*} y'(x) &= \frac{x + y(x)}{x - y(x)} \\ z'(x) \cdot x + z(x) &= \frac{x + z(x) \cdot x}{x - z(x) \cdot x} \\ \frac{\partial z(x)}{\partial x} \cdot x &= \frac{1 + z(x)}{1 - z(x)} - z(x) = \frac{1 + z(x)}{1 - z(x)} - \frac{\left( 1-z(x) \right) \cdot z(x)}{1-z(x)} = \frac{1 + z(x) - z(x) + z^2(x)}{1 - z(x)} \\ \frac{\partial z(x)}{\partial x} &= \frac{1 + z^2(x)}{1 - z(x)} \cdot \frac{1}{x} \\ \frac{1 - z(x)}{1 + z^2(x)} \partial z(x) &= \frac{1}{x} \cdot \partial x \\ \int \frac{1 - z(x)}{1 + z^2(x)} \, \mathrm{d} z(x) &= \int \frac{1}{x} \, \mathrm{d} x \\ \tan^{-1}(z(x)) - \frac{1}{2} \cdot \ln \left( z^2(x)+1 \right) &= \ln(x) + C \\ 0 &= -\tan^{-1}\left(z(x)\right) + \frac{1}{2} \cdot \ln \left( z^2(x)+1 \right) + \ln(x) + C \\ 0 &= -\tan^{-1}\left(z(x)\right) + \ln \left( \sqrt{z^2(x) + 1} \cdot x \right) + C \\ 0 &= -\tan^{-1}\left(z(x)\right) + \ln \left( \sqrt{z^2(x) \cdot x^2 + x^2} \right) + C \\ \end{align*}

(I used the computer for the integration step) We now have a solution, but we first need to substitute back to get rid of the $$z(x) = \frac{y(x)}{x}$$

\begin{align*} 0 &= -\tan^{-1}\left(\frac{y(x)}{x}\right) + \ln \left( \sqrt{\frac{y^2(x)}{x^2} \cdot x^2 + x^2} \right) + C \\ 0 &= -\tan^{-1}\left(\frac{y(x)}{x}\right) + \ln \left( \sqrt{y^2(x) + x^2} \right) + C \end{align*}

Next, I want to set $$C$$ to the starting position of the boat by replacing $$x = x_0$$ and $$y(x) = y_0$$

\begin{align*} C &= \tan^{-1}\left(\frac{y_0}{x_0}\right) - \ln \left( \sqrt{y_0^2 + x_0^2} \right) \end{align*}

The final result is then an implicit function

$$0 = -\tan^{-1}\left( \frac{y}{x} \right) + \ln\left( \sqrt{x^2 + y^2} \right) + \tan^{-1}\left( \frac{y_0}{x_0} \right) - \ln \left( \sqrt{x_0^2 + y_0^2} \right). \label{eq:curveBoat}$$

So, we now have a function where we plug in the starting point $$(x_0,y_0)^T$$ of the boat and then check every relevant value for $$y$$ and $$x$$ where the equation is fulfilled. In total, this results in the boat's curve. Since the boat's position always depends on the current light ray from the lighthouse, you can think of the curve being defined by the ray. To clarify this aspect, you can play around with the slope of the ray in the following animation.

As you can see, the boat's curve originates by rotating the light ray. Also, note that there are actually two curves. This is because we can enclose a 45° angle on both sides of the light ray. The right curve encloses the angle with the top right side of the line and the left curve encloses with the bottom right side of the line. Actually, one starting point defines already both curves, but you may want to depict the situation like two drug boats starting at symmetric positions.

I marked the position $$(x_c,y_c) = (1,2)$$ as “angle checkpoint” to see if the enclosed angle is indeed 45°. To check, we first need the angle of the light ray which is just the angle $$\alpha$$ defined above for the given coordinates

\begin{equation*} \phi_{ray} = \tan^{-1}\left( \frac{y_c}{x_c} \right) = \tan^{-1}\left( \frac{2}{1} \right) = 63.43°. \end{equation*}

For the angle of the boat, we need its tangent at that position which is given by its slope value. So we only need to plug in the coordinates in the ODE previously defined

\begin{equation*} f(x_c,y_c) = \frac{1 + 2}{1 - 2} = -3. \end{equation*}

Forwarding this to $$\tan^{-1}$$ results in the angle of the tangent

\begin{equation*} \phi_{tangent} = \tan^{-1}\left(-3\right) + 180° = -71.57° + 180° = 108.43°. \end{equation*}

I added 180° so that the resulting angle is positive (enclosed angles can only be in the range $$\left[0°;180°\right[$$). Calculating the difference $$\phi_{tangent} - \phi_{ray} = 108.43° - 63.43° = 45°$$ shows indeed the desired result. Of course, this is not only true at the marked point, but rather at any point instead, because that is the way we defined it in \eqref{eq:slopeBoat}.

Another way of visualising \eqref{eq:curveBoat} is to switch to the the polar coordinate system by using $$\theta = \tan^{-1}\left( \frac{y}{x} \right)$$ respectively $$r = \sqrt{x^2 + y^2}$$

\begin{align*} 0 &= -\theta + \ln\left( r \right) + \theta_0 - \ln \left( r_0 \right) \\ \ln \left( \frac{r_0}{r} \right) &= -\theta + \theta_0 \\ \frac{r_0}{r} &= e^{-\theta + \theta_0}, \end{align*}

and solve by $$r$$

$$r\left( \theta \right) = \frac{r_0}{e^{\theta_0 -\theta}}. \label{eq:curveBoatPolar}$$

We can now visualize this function using a polar plot where we move around a (unit) circle and adjust the radius accordingly to \eqref{eq:curveBoatPolar}. The result is a graph which looks like a spiral. Beginning from the starting point the light ray forces the boat to move counter-clockwise in a circle with increasing distance from the island. So, without considering physics (infinity light ray, ...) and realistic human behaviour (escaping strategy of the boat, ...), this cat-and-mouse game lasts forever.

Next, I want to analyse how the curve varies when the starting position of the boat changes. Again, each position of the curve is just given by the corresponding light ray crossing the same position. The curve in total is, therefore, the result of a complete rotation (or multiple like in the polar plot) of the light ray (like above, just with all possible slope values). In the next animation, you can change the starting position manually.

Do you remember the property of scale invariance for a homogeneous ODE introduced in the beginning? Let's have a lock what this means for the current problem. For this, it helps to analyse the different slope values which the equation $$f(x,y)$$ produces. This is usually done via a vector field. At sampled positions ($$x_s,y_s$$) in the grid, a vector $$(1, f(x_s,y_s))$$ is drawn which points in the direction of the slope value at that position (here, the vectors are normalized). So the vector is just a visualization technique to show the value of $$f(x,y)$$. Additionally, I added some isoclines where on all points on one line the slope value is identical. This means that all vectors along a line have the same direction (easily checked on the horizontal line).

You can check this if you move the boat along the line $$y=x$$. This will result in different curves, but the tangent of the boat's starting point is always the same (vertical). Actually, this is already the property of scale invariance: starting from one point, you can scale your point (= moving along an isocline) and always get the same slope value.

List of attached files: