20  Solving for Concentration

Steady-state

The previous section showed that at steady state the concentration at every point is constant, so the oxygen delivered by diffusion balances the oxygen consumed by the cells there. With zero-order consumption, that balance is the ordinary differential equation solved below.

NoteCore concept — the steady-state reaction-diffusion balance

For one-dimensional diffusion with a constant, uniform volumetric consumption rate, the steady-state oxygen concentration \(c(x)\) obeys

\[D\,\frac{\mathrm{d}^2 c}{\mathrm{d}x^2} = R \]

where \(c\) is concentration (\(\mathrm{mol\,m^{-3}}\)), \(x\) is depth (\(\mathrm{m}\)), \(D\) is the oxygen diffusion coefficient (\(\mathrm{m^2\,s^{-1}}\)), and \(R\) is the volumetric consumption rate (\(\mathrm{mol\,m^{-3}\,s^{-1}}\)).

The sign of \(R\). The reaction–diffusion equation of the previous section carries consumption as a sink, \(\partial c/\partial t = D\,\partial^2 c/\partial x^2 - R\), so \(R > 0\) is the magnitude of net consumption, as it is for cells consuming oxygen.

Boundary conditions

The boundary conditions of the previous section apply directly to the graft of Figure 19.2, a slab of thickness \(L\) resting on a Petri dish that oxygen cannot permeate. Let \(x\) measure position from that sealed base at \(x = 0\) out to the surface at \(x = L\), which is bathed in oxygenated medium.

No oxygen crosses the sealed base, so the flux, and with it the gradient \(\mathrm{d}c/\mathrm{d}x\), is zero. The well-mixed medium holds the surface at the fixed concentration \(c_0\). In symbols,

\[\left.\frac{\mathrm{d}c}{\mathrm{d}x}\right|_{x=0} = 0 \quad\text{(no flux through the sealed base),} \qquad c(L) = c_0 \quad\text{(fixed surface concentration),}\]

where \(L\) is the graft thickness (\(\mathrm{m}\)) and \(c_0\) is the oxygen concentration held at the surface by the culture medium.

Solving the equation

The reaction-diffusion equation is a second-order ordinary differential equation with a constant right-hand side, so it integrates directly. The result is a function that gives the oxygen concentration at any x in the graft.

NoteDerivation — the steady-state oxygen profile

Integrate the balance twice. Write the equation as \(\mathrm{d}^2 c/\mathrm{d}x^2 = R/D\), a constant. Integrating once gives the gradient,

\[\frac{\mathrm{d}c}{\mathrm{d}x} = \frac{R}{D}\,x + A ,\]

with \(A\) a constant of integration. Integrating a second time gives the concentration,

\[c(x) = \frac{R}{2D}\,x^2 + A\,x + B ,\]

with \(B\) a second constant. The two constants are fixed by the two boundary conditions.

Apply the no-flux base condition. The gradient at the sealed base is \(\mathrm{d}c/\mathrm{d}x|_{x=0} = A\), and the boundary condition sets it to zero, so \(A = 0\). The linear term drops out, leaving

\[c(x) = \frac{R}{2D}\,x^2 + B .\]

Apply the fixed-surface condition. At the surface \(c(L) = c_0\), so

\[c_0 = \frac{R}{2D}\,L^2 + B \qquad\Longrightarrow\qquad B = c_0 - \frac{R}{2D}\,L^2 .\]

The dimensional profile. Substituting \(B\) back gives the steady-state oxygen concentration at every depth,

\[c(x) = c_0 - \frac{R}{2D}\left(L^2 - x^2\right) .\]

The profile is a parabola with its minimum at the sealed base (\(x = 0\)), where \(c(0) = c_0 - RL^2/2D\), and its maximum at the surface (\(x = L\)), where \(c(L) = c_0\). The depth of the dip is set by the combination \(RL^2/2D\).

Non-dimensionalize the finished solution. Divide the profile by the surface concentration \(c_0\) and measure depth as the fraction \(x/L\), so both quantities are of order one across the graft:

\[\frac{c}{c_0} = 1 + \frac{R L^2}{2 D c_0}\left(\left(\frac{x}{L}\right)^2 - 1\right) .\]

Every dimension has canceled, and the four parameters \(R\), \(L\), \(D\), \(c_0\) enter only through the single dimensionless number. For the purpose of this class, we define this group as \(\phi\) and call it the Thiele modulus. Many texts define the modulus as \(L\sqrt{R/(D c_0)}\) instead, which is \(\sqrt{2\phi}\) in our notation:

\[\phi = \frac{R L^2}{2 D c_0} ,\]

This single number compares the rate of consumption to the rate of supply. The steady-state oxygen profile is then

\[\boxed{\; \frac{c}{c_0} = 1 + \phi\left(\left(\frac{x}{L}\right)^2 - 1\right) . \;}\]

Exercises

TipProblem 1

At steady state, Fick’s second law with zero-order consumption (\(R > 0\) the magnitude of the consumption rate) reduces to

\[D\,\frac{d^2c}{dx^2} = R.\]

  1. Integrate twice to obtain the general solution \(c(x)\).

  2. Apply the boundary conditions \(dc/dx\big|_{x=0} = 0\) (no flux through the sealed base) and \(c(L) = c_0\) (surface concentration maintained at \(c_0\)) to find the specific solution.

  3. Non-dimensionalize the result. Write it in terms of \(c/c_0\) and \(x/L\), and identify the single group the profile depends on.

  4. Sketch the concentration profile and state whether it is concave up or concave down for consumption (\(R > 0\)).

  1. \(d^2c/dx^2 = R/D\) (a positive constant, since \(R>0\) for consumption).

Integrate once: \(dc/dx = (R/D)\,x + A_1\).

Integrate again: \(c(x) = (R/D)\,x^2/2 + A_1\,x + A_2\).

  1. Boundary condition \(dc/dx|_{x=0} = 0\): \((R/D)\cdot0 + A_1 = 0 \Rightarrow A_1 = 0\).

Boundary condition \(c(L) = c_0\): \(c_0 = (R/D)\,L^2/2 + A_2 \Rightarrow A_2 = c_0 - RL^2/(2D)\).

Therefore: \[c(x) = c_0 - \frac{R}{2D}(L^2 - x^2) = c_0 - \frac{RL^2}{2D}\!\left(1-\frac{x^2}{L^2}\right).\]

  1. Divide through by \(c_0\) and measure depth as \(x/L\):

\[\frac{c}{c_0} = 1 + \frac{R L^2}{2 D c_0}\left(\frac{x^2}{L^2} - 1\right), \qquad\text{so}\qquad \frac{c}{c_0} = 1 + \phi\left(\left(\frac{x}{L}\right)^2 - 1\right),\]

with \(\phi = R L^2/(2 D c_0)\). Four parameters collapse into one: the shape of the profile depends on \(\phi\) alone, not on \(R\), \(D\), \(L\) and \(c_0\) separately.

  1. \(d^2c/dx^2 = R/D > 0\): the profile is concave up (curves upward). The concentration is lowest at \(x=0\) (the sealed base) and highest at \(x=L\) (the surface). This is the parabolic DermaGraft oxygen profile derived in full above.
TipProblem 2

The sealed-base assumption is now relaxed. Suppose the wound bed beneath the graft supplies oxygen across the base at a constant flux of magnitude \(J_0\) directed into the graft, toward larger \(x\), so the base condition becomes \(-D\,\dfrac{\mathrm{d}c}{\mathrm{d}x}\big|_{x=0} = J_0\), while the surface condition \(c(L) = c_0\) and the balance \(D\,\mathrm{d}^2 c/\mathrm{d}x^2 = R\) are unchanged.

  1. Integrate the equation and apply the two conditions to find the steady-state profile \(c(x)\).

  2. Verify that it reduces to the sealed-base result \(c(x) = c_0 - \tfrac{R}{2D}(L^2 - x^2)\) when \(J_0 = 0\).

  3. With the base flux switched on, the oxygen minimum is no longer at the base. Show that it now sits at the interior depth \(x^{*} = J_0/R\) (provided \(x^{*} < L\)), and explain physically why supplying oxygen at the base moves the most vulnerable point off the base.

  1. Integrating twice gives \(c(x) = \tfrac{R}{2D}x^2 + A x + B\). The base condition \(-D\,c'(0) = J_0\) fixes \(c'(0) = A = -J_0/D\), and the surface condition \(c(L) = c_0\) then fixes \(B = c_0 - \tfrac{R}{2D}L^2 + \tfrac{J_0}{D}L\). Hence

\[c(x) = c_0 - \frac{R}{2D}\left(L^2 - x^2\right) + \frac{J_0}{D}\left(L - x\right).\]

  1. Setting \(J_0 = 0\) deletes the final term and returns the sealed-base parabola.

  2. The minimum is where the gradient vanishes: \(c'(x) = \tfrac{R}{D}x - \tfrac{J_0}{D} = 0\), i.e. \(x^{*} = J_0/R\). Below \(x^{*}\) the gradient is negative (oxygen entering at the base diffuses toward the surface); above it the gradient is positive (oxygen entering at the surface diffuses toward the base); the two supplies meet at an interior minimum. Feeding oxygen through the base lifts the base concentration and relocates the point of greatest depletion into the interior, so the base is no longer the first place to starve.

TipProblem 3

Evaluate the profile for a graft of thickness \(L = 0.4\ \text{mm}\) using the representative dermal-graft parameters \(D = 2\times10^{-9}\ \text{m}^2/\text{s}\) (Milo et al. 2010), \(c_0 = 0.2\ \text{mol/m}^3\) (Place et al. 2017), and \(R = 1.19\times10^{-3}\ \text{mol/m}^3\text{/s}\) (Streeter and Cheema 2011).

  1. Compute the Thiele modulus \(\phi\).

  2. Find the oxygen concentration at the sealed base, both as a fraction \(c(0)/c_0\) of the surface value and in \(\mu\text{M}\) (note \(0.2\ \text{mol/m}^3 = 200\ \mu\text{M}\)).

  3. Find the concentration at mid-depth, \(x/L = 0.5\). Does oxygen still reach the base (\(c(0) > 0\))?

  1. \(\phi = \dfrac{R L^2}{2 D c_0} = \dfrac{(1.19\times10^{-3})(0.4\times10^{-3})^2}{2\,(2\times10^{-9})(0.2)} = \dfrac{1.90\times10^{-10}}{8\times10^{-10}} \approx 0.24.\)

  2. \(\frac{c(0)}{c_0} = 1 - \phi = 0.76\), so the base holds \(76\%\) of the surface value: \(c(0) = 0.76 \times 200\ \mu\text{M} = 152\ \mu\text{M}\) (\(= 0.152\ \text{mol/m}^3\)).

  3. At \(x/L = 0.5\), \(c/c_0 = 1 + \phi\,(0.5^2 - 1) = 1 + 0.24\,(-0.75) = 0.82\), i.e. \(c = 0.82 \times 200\ \mu\text{M} = 164\ \mu\text{M}\). Since \(\phi < 1\) the base stays oxygenated (at \(76\%\) of the surface concentration), so a \(0.4\ \text{mm}\) graft sits comfortably in the safe, reaction-limited regime.

TipProblem 4

Using the base concentration \(\frac{c(0)}{c_0} = 1 - \phi\), interpret the graft’s behavior in three regimes of the Thiele modulus.

  1. What does \(\phi \to 0\) mean physically, and what does the oxygen profile look like?

  2. What is special about \(\phi = 1\)?

  3. The linear model returns \(\frac{c(0)}{c_0} = 1 - \phi < 0\) for \(\phi > 1\). A concentration cannot be negative: what is actually happening in the graft?

  1. \(\phi \to 0\) means diffusion is fast compared with consumption (small \(R\) or \(L\), large \(D\) or \(c_0\)). Then \(c(0)/c_0 \to 1\): the base sits at essentially the surface concentration, the profile \(c/c_0 = 1 + \phi\left((x/L)^2-1\right)\) is nearly flat, and the cells are well oxygenated throughout. Consumption, not transport, is the bottleneck: this is the reaction-limited regime, and it is safe.

  2. At \(\phi = 1\) the base concentration \(\frac{c(0)}{c_0} = 1 - \phi\) reaches exactly zero: the deepest cells just consume the last of their oxygen. This marks the edge of viability and defines the critical thickness \(L^{*} = \sqrt{2 D c_0 / R}\), at which \(\phi = 1\).

  3. The negative value is unphysical and signals that the linear model has been pushed past its range of validity: consumption exhausts the oxygen at some finite depth, and below that depth an anoxic core (a region of oxygen-starved, dying cells) forms. Transport, not consumption, is now the bottleneck: this is the diffusion-limited regime, which the design must avoid by keeping \(\phi < 1\).

TipProblem 5

The critical thickness is the value of \(L\) at which the base oxygen just reaches zero.

  1. Set \(c(0)/c_0 = 0\) in \(\frac{c(0)}{c_0} = 1 - \phi\) and solve for the thickness \(L^{*}\) in terms of \(D\), \(c_0\), and \(R\).

  2. Show that the Thiele modulus can be written \(\phi = (L/L^{*})^2\).

  3. A graft is built at half the critical thickness, \(L = L^{*}/2\). What is its Thiele modulus, and what fraction of the surface oxygen concentration reaches the base?

  1. \(\frac{c(0)}{c_0} = 1 - \phi = 0\) gives \(\phi = 1\), i.e. \(\dfrac{R\,(L^{*})^2}{2 D c_0} = 1\), so

\[L^{*} = \sqrt{\frac{2 D c_0}{R}} .\]

  1. Since \(\phi = \dfrac{R L^2}{2 D c_0}\) and \((L^{*})^2 = \dfrac{2 D c_0}{R}\), dividing gives \(\phi = \dfrac{L^2}{(L^{*})^2} = \left(\dfrac{L}{L^{*}}\right)^2\). The Thiele modulus is the squared ratio of the graft thickness to the critical thickness.

  2. At \(L = L^{*}/2\), \(\phi = (1/2)^2 = 0.25\), so \(c(0)/c_0 = 1 - 0.25 = 0.75\): the base holds \(75\%\) of the surface oxygen concentration. Building at half the critical thickness leaves a comfortable margin. (For the representative parameters of Problem 3, \(L^{*} = \sqrt{2 D c_0/R} \approx 0.82\ \text{mm}\).)

TipProblem 6

Not every cell sits at the base; a useful summary of the whole graft is the depth-averaged oxygen fraction, \(\overline{c/c_0} = \displaystyle\int_0^1 \frac{c}{c_0}\,\mathrm{d}(x/L)\).

  1. Evaluate the integral for \(c/c_0 = 1 + \phi\left((x/L)^2-1\right)\) and show that \(\overline{c/c_0} = 1 - \tfrac{2}{3}\phi\).

  2. For a graft with \(\phi = 0.6\), compute both the depth-averaged fraction \(\overline{c/c_0}\) and the base fraction \(c(0)/c_0\). Which is the right quantity to design against, and why?

  1. \(\displaystyle \overline{c/c_0} = \int_0^1 \left[1 + \phi\left((x/L)^2-1\right)\right]\mathrm{d}(x/L) = 1 + \phi\int_0^1 \left((x/L)^2-1\right)\mathrm{d}(x/L) = 1 + \phi\left(\tfrac{1}{3} - 1\right) = 1 - \tfrac{2}{3}\phi .\)

  2. For \(\phi = 0.6\): \(\overline{c/c_0} = 1 - \tfrac{2}{3}(0.6) = 0.6\), while the base fraction is \(c(0)/c_0 = 1 - 0.6 = 0.4\). The base value is the smaller of the two and is the correct design target: viability is lost first at the least-oxygenated point, so designing to the average (\(60\%\)) would overstate the margin and could leave the base (\(40\%\)), or at larger \(\phi\) the deepest cells, starved. The average is a summary; the minimum is the constraint.

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.
Place, Trenton L., Frederick E. Domann, and Adam J. Case. 2017. “Limitations of Oxygen Delivery to Cells in Culture: An Underappreciated Problem in Basic and Translational Research.” Free Radical Biology and Medicine 113: 311–22. https://doi.org/10.1016/j.freeradbiomed.2017.10.003.
Streeter, Ian, and Umber Cheema. 2011. “Oxygen Consumption Rate of Cells in 3D Culture: The Use of Experiment and Simulation to Measure Kinetic Parameters and Optimise Culture Conditions.” Analyst 136 (19): 4013–19. https://doi.org/10.1039/c1an15249a.