I have a question regarding an equation for changes in chemical concentration. The chemical concentration we are looking at changes due to diffusion, decay and production of the chemical. Since the diffusion happens so much faster than the other two, it is given that we can you a so-called quasi-steady assumption. This then gives the equation:

$$D \nabla^2c-a \cdot c+b\cdot c=0$$

where $a$ and $b$ are rates of decay and production, respectively. $c$ is the chemical concentration and D is the diffusion coefficient.

Is it correct that the reason we only have a derivative expression for the diffusion because it happens so fast relative to the other two, and therefore we can assume the chemical concentration for the two other processes to remain constant (using the quasi steady assumption), or is that not a correct reasoning for why the equation has the simple expression for decay and production, while we need a more complex expression for diffusion?


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  • $\begingroup$ Read about the diffusion equation (Fick's law). ▽² is a spatial derivative. $\endgroup$ – WYSIWYG Jan 24 '17 at 5:49
  • $\begingroup$ Diffusion by nature is associated with concentration by a second-derivate gradient. Decay and production are not necessarily. I think you're just asking if it's a fundamental thing. It is. $\endgroup$ – khaverim Jan 24 '17 at 6:18

This is an example of a reaction-diffusion system where the participating species both react as well as diffuse. Such systems can be described using partial differential equations.

The mathematical model describes how the system changes with time. So we have $$\frac{\mathrm dC}{\mathrm dt}=f(C)$$

where $C$ is a vector with each element denoting the concentration a species ($C_i$).

At quasi-steady state the rates of some of the species will remain unchanged over time.

$$\frac{\mathrm dC_i}{\mathrm dt}=0$$

The reaction part usually comes from mass action kinetic models (which gives you rates for zeroth/first/second order reactions).

The diffusion part comes from Fick's laws.

$$\frac{\partial C}{\partial t}=D\nabla^2C(x,y,z,t)$$

Here the $\nabla^2$ denotes spatial double derivatives $\large\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$

which kind of describe how the different species would be distributed in space. The reaction-diffusion equation is a sum of the reaction and the diffusion parts. At quasi-steady state, the concentration of a given species at any given point in the space would remain unchanged over time. This only sets the time derivative to zero, as I previously mentioned.

The time derivative denotes the net accumulation rate of a species. In the absence of diffusion, it will simply be the sum of rates of all reactions leading to formation of the species minus the sum of rates of all reactions causing removal of the species (principle of mass balance).

In the absence of reactions the diffusion terms will describe how the spatial pattern of the species will change over time.

At (quasi) steady state, the time derivative goes to zero which means that all the reaction and diffusion terms balance each other out such that there is no change in the concentration of the corresponding species over time.

  • $\begingroup$ Thank you very much for your reply! Does this mean that the expression for production and decay are defined as $a\cdot c$ and $b \cdot c$ also if we did not assume quasi-steady approximation, because the quasi-steady approximation just deals with time, and this equation for decay, production and diffusion is just over space? $\endgroup$ – David Jan 24 '17 at 9:09
  • $\begingroup$ @David At (quasi) steady state the time derivative of concentration becomes zero. In this case the time derivative is a sum of reaction rates described by the kinetics and spatial diffusion. In the absence of reactions i.e. when there are only diffusion terms, you can still solve for the steady state spatial distribution. When there is no diffusion then it would mean that the production and degradation reactions rates are equal thereby causing no net accumulation of product. (Note that the time derivative denotes accumulation rate). $\endgroup$ – WYSIWYG Jan 24 '17 at 10:47

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