# Laplacian of Gaussian Approximation and Gaussian Blur as the Solution of Heat Equation

While I was reading SIFT paper(Lowe, 2004), I came across the method that he apply "heat diffusion equation" to Gaussian function to derive that $$\frac{∂G}{∂σ} = σ∇^2G$$ I searched Wikipedia and found that original heat equation is

$$\frac{∂u}{∂t} = σ∇^2u$$ I wonder how can time derivative equation be applied to sigma derivative equation. Is there anybody to understand me?

• I don't understand what you're after. LoG usually used in Image Processing to apply a filter to approximate smooth Laplacian of the Image.
– Royi
Jul 19, 2018 at 6:43
• Lowe approximated LoG to DoG by using "heat diffusion equation". I don't understand why u function(in the heat equation) can be substituted with Gaussian function. Jul 19, 2018 at 7:05

I'm not sure I fully understood what's the issue you're having.
Yet I will show a simple property of the Gaussian filter which might make things clearer.

For simplicity, I will use 1D Signal.
Yet it is easy to extend it into 2D.

## The Problem

For $u \in \mathbb{R}$ (1D, unbounded domain), show that a Gaussian convolution with the initial condition solves the linear diffusion equation:

$${u}_{t} = {u}_{xx}, \quad u \left( t = 0; x \right) = f \left( x \right)$$

That is:

$$u \left( t; x \right) = f \left( x \right) \ast {g}_{ \sigma \left( t \right)} \left( x \right)$$

Where $\ast$ denotes convolution.

Find the function $F \left( \cdot \right)$ relating the evolution time $t$ to the Gaussian standard deviation $\sigma \: : \: \sigma = F \left( t \right)$.

## Solution

Looking at:

$$\frac{\partial u \left( x, t \right ) }{\partial t } = \frac{\partial^2 u \left( x, t \right ) }{\partial x^2} , \; u \left( x, t = 0 \right) = f \left( x \right)$$

Defining $U \left( s, t \right)$ as the Fourier Transform of $u \left( x, t \right)$ with respect to the parameter $x$.

\begin{align} \mathfrak{F} \left( \frac{\partial u }{\partial t} \right) & = \frac{\partial U \left( s, t \right ) }{\partial t} \\ \mathfrak{F} \left( \frac{\partial^2 u}{\partial x^2} \right) & = -4 {\pi}^{2} {s}^{2} U \left( s, t \right ) \\ \mathfrak{F} \left( f \left( x \right) \right) & = F \left( s \right) \end{align}

Hence the equation in the Fourier domain is given by:

$$\frac{\partial U \left( s, t \right ) }{\partial t} = -4 {\pi}^{2} {s}^{2} U \left( s, t \right ) , \; U \left( s, t = 0 \right) = F \left( s \right)$$

The solution of this Linear Ordinary Differential Equation is given by:

$$U \left( s, t \right) = F \left( s \right) {e}^{-4 {\pi}^{2} {S}^{2} t }$$

Hence the solution in the time domain is given by (Using the identity $\mathfrak{F} \left( {e}^{- \alpha {x}^{2}} \right) = \sqrt{\frac{\pi}{\alpha}} {e}^{- \frac{{\left( \pi f \right)}^{2}}{\alpha}}$):

$$u \left( x, t \right) = \frac{1}{\sqrt{2 \pi 2t}} {e}^{- \frac{{x}^{2}}{4 t}} = \frac{1}{\sqrt{2 \pi {\sigma \left( t \right)}^{2}}} {e}^{- \frac{{x}^{2}}{2 {\sigma \left( t \right)}^{2}}}$$

Comparing it to the classic Gaussian Function yields that $\sigma \left( t \right) = \sqrt{\left( 2 t \right)}$.

Remark
This is a solution of mine to a problem I was given in a course named Variational Methods in Image Processing.