# Diffusion/heat (in space)



My notes of the excellent lecture 29 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

The Diffusion equation governs motion of e.g. smoke in unmovable air, or dye in a solution.

## Definition

It is a partial differential equation. The unknown is a function, and the function relates the partial derivatives of that function to each other.


Let $$u$$ be the concentration of substance at a given point, $$u(x,y,z,t$$). Let $$\vec F$$ be the flow of substance.


For the heat equation $$u$$ would be the temperature.

## Understanding

Let $$\vec F$$ is the flow of smoke

1. Physics: smoke will flow from high concentration towards low concentration regions. So, in the direction where the concentration decreases the fastest. That is negative the gradient. So $$\vec F$$ is directed along $$-\nabla u$$. In fact, the flow is proportional to the gradient field $$\vec F = -k\,\nabla u \label{eq:understanding1}$$
2. Math: How does the flow affect the concentration? One place it will be decreasing, and another place it will be increasing. The divergence theorem relates $$\vec F$$ and $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \pdv{u}{t}$$.

Let’s try to understand that

Flux out of $$D$$ through $$S$$ measures the amount of smoke going through $$S$$ in unit time. $$\newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \oiint_S\vec F\cdot\hat n\,dS \nonumber$$


We can use the divergence theorem to compute this

$$\iiint_D\rm{div}(\vec F)\,dV =-\pdv{}{t}\left( \iiint_D u\,dV\right) \nonumber$$

Instead of taking the sum of all the amounts of smoke, and seeing how it changes over time, we can take all the little derivatives and then sum them. The derivative of the sum, is the sum of the derivatives

$$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \frac{d}{dt}\left(\sum_i u(x_i,y_i,z_i,t)\,\Delta V_i\right) = \sum_i\pdv{u}{t}(x_i,y_i,z_i,t)\,\Delta V_i \nonumber$$

Thus $$\iiint_D\underline{\rm{div}(\vec F)}\,dV = -\iiint_D \underline{\pdv{u}{t}}\,dV \nonumber$$

For any region $$D$$. $$\newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \tfrac{1}{\rm{Vol}(D)}\iiint_D \rm{div}(\vec F)\,dV = \tfrac{1}{\rm{Vol}(D)}\iiint_D-\pdv{u}{t}\,dV \\ \Rightarrow \text{avg. of }\rm{div}(\vec F)\text{ in }D = \text{avg. of }-\pdv{u}{t}\text{ in }D \nonumber$$

So, from the divergence theorem we derived $$\shaded{ \rm{div}(\vec F) = -\pdv{u}{t} } \nonumber$$

So, this is another way think about what divergence means: how much “stuff” is flowing out, how much you’re loosing.

If you combine that with equation $$\eqref{eq:understanding1}$$, you get the diffusion equation \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \left. \begin{align*} \vec F &= -k\,\nabla u \\ \pdv{u}{t} &= -\rm{div}(\vec F) \end{align*} \right\} \Rightarrow \shaded{ \pdv{u}{t} = +k\,\rm{div}(\nabla u)= k\,\nabla^2 u }