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I want to simulate the diffusion equation numerically.

$$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} $$

With the boundary condition

$$ \frac{\partial u}{\partial x} \bigg|_{x=R}=0 $$

I am having a conceptual problem with boundary conditions.

The book I'm reading seems to imply that this boundary condition prevents anything from leaving the system. i.e. the edge of my test tube.

To me it simply means that there is no change in concentration wrt position at the point $x=R$. Concentration could be changing with time etc but not with position. To me it would mean that the concentration is the same on one side of the test tube as the other.

Could someone help me understand how this boundary condition implies nothing is leaving the system?

Thanks

Rgkpdx
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RNs_Ghost
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3 Answers3

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{-\,\partiald[2]{{\rm u}\pars{x,t}}{x} + \partiald{{\rm u}\pars{x,t}}{t} = 0}$ leads to $\ds{\partiald{}{x}\bracks{\color{#f00}{-\partiald{{\rm u}\pars{x,t}}{x}}} + \partiald{{\rm u}\pars{x,t}}{t} = 0}$ which is the Continuity Equation which guarantees the particle conservation number. In particular, $\ds{{\rm J}_{x}\pars{x,t} = \color{#f00}{-\partiald{{\rm u}\pars{x,t}}{x}}}$ is the Current Particle $\ds{x}$-component.

$\ds{\color{#f00}{\left.-\,\partiald{{\rm u}\pars{x,t}}{x}\right\vert_{x\ =\ R}} = 0}$ guarantees that the particles do not cross the boundary at $\ds{x = R}$. In another words, the 'wall' at $\ds{x = R}$ confines the particles.

Felix Marin
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It basically means that the flux across the boundary R is zero. Which means that there is no concentration difference (if u is concentration for Diffusion Equation) across the wall. If there is no concentration change then there is nothing leaving/entering across this boundary. Because diffusion is concentration driven. Hence this is a barrier boundary. (Gradient of concentration is zero only at the boundary)

If u is temperature then it means there is no temperature difference across this boundary and the heat flow is zero. This can be thought as perfect insulation. (Gradient of temperature is zero across the boundary)

If u is pressure then this is a no-flow boundary and because there is no pressure difference across the boundary no fluid can enter/leave. This is because flow of fluids is dependent on pressure difference (gradient of Pressure).

Fahd Siddiqui
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Those boundary conditions are simulated by the random walk reflecting at the boundary. That is, the diffusing particles elastically collide with the boundary.

Stephen Montgomery-Smith
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