In the Mass transport section, we applied conservation of a species to a control volume and found that:
\[\frac{\partial c}{\partial t} = -\frac{\partial J}{\partial x} - R ,\]
That section ended by pointing out the equation never says what the flux \(J\) is.
Fick’s first law supplies the missing piece if the flux is driven by diffusion. Putting the two together gives a single equation for \(c(x,t)\).
Start from the species balance. Conservation of species \(i\) in one dimension gives
\[\frac{\partial c}{\partial t} = -\frac{\partial J}{\partial x} - R .\]
Substitute Fick’s first law. The flux is \(J = -D\,\partial c/\partial x\), so
\[\frac{\partial c}{\partial t}
= -\frac{\partial}{\partial x}\!\left(-D\,\frac{\partial c}{\partial x}\right) - R .\]
Take \(D\) outside the derivative. For constant \(D\) the two minus signs cancel and \(D\) comes out in front, leaving a second derivative:
\[\frac{\partial c}{\partial t} = D\,\frac{\partial^2 c}{\partial x^2} - R .\]
The result is the reaction–diffusion equation:
\[\boxed{\;\frac{\partial c}{\partial t} = D\,\frac{\partial^2 c}{\partial x^2} - R\;}\]
Setting \(R = 0\) (no reaction) gives pure diffusion, known as Fick’s second law,
\[\frac{\partial c}{\partial t} = D\,\frac{\partial^2 c}{\partial x^2} .\]
The right-hand side is the curvature of the concentration profile. Where the profile curves upward (\(\partial^2 c/\partial x^2 > 0\)) the concentration rises, where it curves downward the concentration falls, and a straight-line profile does not change at all. Diffusion therefore smooths a profile out, and the larger \(D\) is, the faster it does so. The reaction term \(R\) then lowers the concentration everywhere the species is consumed.
The mass balance above was written for one dimension. But the full form contains all three dimensions (x,y,z).
\[\frac{\partial c}{\partial t}
= \nabla\cdot(D\,\nabla c) - R = D\,\nabla^2 c - R,\]
where \(\nabla^2 c = \partial^2 c/\partial x^2 + \partial^2 c/\partial y^2 + \partial^2 c/\partial z^2\) is the Laplacian. For a 1D problem (slab geometry, gradients only in \(x\)) this reduces to the 1D form above.
The characteristic diffusion time
How long does a profile take to respond to a change? The answer can be estimated using our knowledge of the random walk. A diffusing molecule wanders from where it started, and in one dimension its typical distance from the start, the root-mean-square displacement, is:
\(\sqrt{\langle x^2\rangle} = \sqrt{2Dt}\)
At what time, \(t = \tau\), have the molecules, on average, moved \(L\) distance?
\[\sqrt{2D\tau} = L
\qquad\Longrightarrow\qquad
\tau = \frac{L^2}{2D}.\]
This is the characteristic diffusion time.
The diffusion time grows as the square of the distance. For a graft of thickness \(L = 1\ \mathrm{mm}\) and an oxygen diffusion coefficient \(D = 2\times10^{-9}\ \mathrm{m^2\,s^{-1}}\) (Milo et al. 2010), the estimate gives
\[\tau \sim \frac{(10^{-3}\ \mathrm{m})^2}{2\,(2\times10^{-9}\ \mathrm{m^2\,s^{-1}})} = 250\ \mathrm{s} \approx 4\ \mathrm{min}.\]
A millimeter-thick graft reaches a steady-state oxygen profile over a few minutes: slow enough to matter during handling and medium changes, yet fast enough to assume that the concentration profile is constant during week-long culture.
A second graft is fabricated at twice the thickness. Using \(\tau \sim L^2/(2D)\), by what factor does the time to reach steady state change?
Because \(\tau \sim L^2/(2D)\) scales with the square of the thickness, doubling \(L\) raises \(\tau\) by a factor of \(2^2 = 4\). The thicker graft takes roughly four times as long to equilibrate, about sixteen minutes rather than four.
Exercises
Consider a one-dimensional control volume of dimensions \(\Delta x\,\Delta y\,\Delta z\) in which a species diffuses along \(x\) and is consumed at a volumetric rate \(R\) (moles per unit volume per unit time).
Write the mass balance for the volume, expressing accumulation as flux in, minus flux out, minus consumption.
Divide the balance through by the volume \(\Delta x\,\Delta y\,\Delta z\), then shrink the box, \(\Delta x \to 0\), to obtain \(\partial c/\partial t = -\partial J/\partial x - R\).
Substitute Fick’s first law \(J = -D\,\partial c/\partial x\) (constant \(D\)) to obtain the reaction–diffusion equation, and state the equation it reduces to at steady state.
- With each flux crossing a face of area \(\Delta y\,\Delta z\) and consumption acting over the volume \(\Delta x\,\Delta y\,\Delta z\),
\[\frac{\partial c}{\partial t}\,\Delta x\,\Delta y\,\Delta z = J_x\,\Delta y\,\Delta z - J_{x+\Delta x}\,\Delta y\,\Delta z - R\,\Delta x\,\Delta y\,\Delta z .\]
- Dividing every term by \(\Delta x\,\Delta y\,\Delta z\) cancels one face area from each flux term and leaves \(\partial c/\partial t = (J_x - J_{x+\Delta x})/\Delta x - R\). The quotient is a flux difference over the distance between the two faces, so as \(\Delta x\to 0\) it becomes a derivative,
\[\frac{\partial c}{\partial t} = -\frac{\partial J}{\partial x} - R .\]
- Substituting \(J = -D\,\partial c/\partial x\),
\[\frac{\partial c}{\partial t} = D\,\frac{\partial^2 c}{\partial x^2} - R .\]
At steady state \(\partial c/\partial t = 0\), leaving the ordinary differential equation \(D\,\mathrm{d}^2c/\mathrm{d}x^2 = R\).
A thin epidermal construct is \(L = 200\ \mu\mathrm{m}\) thick. Using the oxygen diffusion coefficient \(D = 2\times10^{-9}\ \mathrm{m^2\,s^{-1}}\) and the estimate \(\tau \sim L^2/(2D)\):
Estimate the time for the construct to approach a new steady state after a change in surface oxygen.
The graft of the worked example is \(L = 1\ \mathrm{mm}\) thick, five times thicker. How much longer is its diffusion time, and does the ratio match your two estimates?
\(\tau \sim L^2/(2D) = (200\times10^{-6})^2/(2\cdot2\times10^{-9}) = (4\times10^{-8})/(4\times10^{-9}) = 10\ \mathrm{s}\).
The diffusion time scales as \(L^2\), so a construct five times thicker has \(5^2 = 25\) times the diffusion time: \(25 \times 10\ \mathrm{s} = 250\ \mathrm{s} \approx 4\ \mathrm{min}\), matching the worked example. The thin construct equilibrates in seconds; the millimeter graft takes minutes.
Oxygen can reach cells either by diffusing through stagnant fluid or by being carried in flowing blood.
Which mechanism does Fick’s second law, as derived in this section, describe? What kind of term would a full mass balance need to add in order to describe the other?
Diffusion crosses a distance \(L\) in a time of order \(L^2/(2D)\), while bulk flow at speed \(v\) crosses it in a time of order \(L/v\). For oxygen with \(D = 2\times10^{-9}\ \mathrm{m^2\,s^{-1}}\), compute the diffusion time across \(L_1 = 10\ \mu\mathrm{m}\) (a single cell) and across \(L_2 = 1\ \mathrm{cm}\) (a tissue block).
Use the contrast to explain why single cells can rely on diffusion but centimeter-scale tissues require a blood supply.
Fick’s second law describes diffusion: the spreading of the species by random molecular motion, with no bulk movement of the medium. Describing transport by flowing fluid would require an advective (convective) term accounting for the solute the bulk flow carries.
\(\tau_1 \sim (10\times10^{-6})^2/(2\cdot2\times10^{-9}) = (10^{-10})/(4\times10^{-9}) = 0.025\ \mathrm{s}\), and \(\tau_2 \sim (10^{-2})^2/(2\cdot2\times10^{-9}) = (10^{-4})/(4\times10^{-9}) = 2.5\times10^{4}\ \mathrm{s} \approx 6.9\ \mathrm{h}\).
Because the diffusion time grows as the square of the distance, crossing a cell takes a fraction of a second, but crossing a centimeter takes about seven hours, far too slow to supply living tissue. Bulk flow, whose transport time grows only linearly with distance, delivers oxygen across large distances quickly, so tissues beyond roughly a millimeter in scale depend on perfusion rather than on diffusion from their surface.
A graft is required to re-equilibrate its oxygen profile within \(5\ \mathrm{min}\) of a change in the culture-medium oxygen level, using \(\tau \sim L^2/(2D)\) as the equilibration-time estimate, with \(D = 2\times10^{-9}\ \mathrm{m^2\,s^{-1}}\).
Solve \(\tau \sim L^2/(2D)\) for the maximum thickness \(L\) that meets the requirement.
Does the \(1\ \mathrm{mm}\) graft of the worked example satisfy the \(5\ \mathrm{min}\) requirement? Justify with its diffusion time.
- Setting \(\tau = 5\ \mathrm{min} = 300\ \mathrm{s}\) and solving \(L = \sqrt{2 D\,\tau}\),
\[L = \sqrt{2\,(2\times10^{-9})(300)} = \sqrt{1.2\times10^{-6}} \approx 1.1\times10^{-3}\ \mathrm{m} \approx 1.1\ \mathrm{mm}.\]
- The \(1\ \mathrm{mm}\) graft has \(\tau \sim (10^{-3})^2/(2\cdot2\times10^{-9}) = 250\ \mathrm{s} \approx 4.2\ \mathrm{min}\), which is under \(5\ \mathrm{min}\). It is thinner than the \(1.1\ \mathrm{mm}\) limit found in (a), so it does satisfy the requirement, though not with much room to spare: a graft built any thicker than about \(1.1\ \mathrm{mm}\) would fail to re-equilibrate quickly enough.
Write the reaction–diffusion equation for 1D diffusion with a zero-order reaction term \(R\). Identify what \(R > 0\) and \(R < 0\) represent physically.
\[\frac{\partial c}{\partial t} = D\,\frac{\partial^2 c}{\partial x^2} - R.\]
- \(R > 0\): net consumption of the diffusing species (e.g. cells consuming oxygen; the oxygen concentration falls).
- \(R < 0\): net production (e.g. cells producing a growth factor).
The reaction term adds or removes material independently of the spatial flux.
References
Milo, Ron, Paul Jorgensen, Uri Moran, Griffin Weber, and Michael Springer. 2010.
“BioNumbers—the Database of Key Numbers in Molecular and Cell Biology.” Nucleic Acids Research 38 (suppl_1): D750–53.
https://doi.org/10.1093/nar/gkp889.