Spatial Gradients from Localized Sources and Sinks

Why Localized Sources and Sinks Produce Distinctive Profiles

In tissue, a cell may release a mediator into its surroundings, while another cell or an interface removes it. Even at steady state, diffusion does not generally make concentration spatially uniform: a source supplies material locally, and a sink removes it locally. The resulting profile depends on the distance between them, the strength and kinetics of uptake, any loss throughout the tissue, and what happens at the domain boundaries.

We consider a concentration \(C(x)\) along a one-dimensional tissue coordinate \(x\), with constant diffusion coefficient \(D>0\). A localized source or sink is idealized as acting at a point. This approximation is useful when its physical extent is small compared with the spatial scale over which concentration changes. A point source has strength \(q\), measured as concentration × length / time in this one-dimensional model. A localized sink has an uptake rate with the same units.

Model: In a region without localized sources or sinks, steady diffusion with first-order bulk loss satisfies \(D C''(x)-kC(x)=0\), where \(k\geq 0\) is a loss rate per unit time. A point source of strength \(q\) at \(x=a\) adds \(q\delta(x-a)\) to the balance; a point sink with uptake \(v(C(b))\) at \(x=b\) subtracts \(v(C(b))\). Here \(\delta\) is the Dirac delta, representing a unit point contribution. Boundary conditions must also be specified. The concentration is continuous at an ideal point source or sink, but its derivative can jump there.

The steady assumption means that the concentration at each location is no longer changing in time. It does not mean that material stops moving: diffusion flux can continue from a source toward a sink. We use Fick's law, \(J=-D C'(x)\), where \(J\) is flux in the positive \(x\)-direction. In a steady model, local changes in that flux must balance local production and removal.

Jump Conditions at a Localized Source or Sink

The point idealization gives a precise condition for how the gradient changes at the source or sink. Consider the steady equation \(D C''+q\delta(x-a)=0\) near a source, with no bulk loss in that small region. Integrate across a small interval from \(a-\varepsilon\) to \(a+\varepsilon\). As \(\varepsilon\) tends to zero, the integral of the delta function is one, so

D\left[C'(a^+)-C'(a^-)\right]+q=0, \qquad C'(a^+)-C'(a^-)=-\frac{q}{D}.

The source therefore makes the slope decrease by \(q/D\) as \(x\) increases across it. For a sink with uptake \(v(C(b))\), the steady equation includes \(-v(C(b))\delta(x-b)\); integration gives \(D[C'(b^+)-C'(b^-)]=v(C(b))\). These jumps are consistent with flux balance: a source adds material to the diffusive flux leaving its location, while a sink removes it. Concentration remains continuous because the model adds a finite amount at a point rather than a concentration discontinuity.

Key result: For a point source at \(a\) and point sink at \(b\), the respective slope jumps are \(-q/D\) and \(v(C(b))/D\). These conditions, together with the differential equation between localized sites and the boundary conditions, determine the steady profile.

A Source with Distributed First-Order Loss

A useful reference case is a point source on an effectively infinite tissue domain, with first-order loss everywhere. The model is \(D C''-kC+q\delta(x-a)=0\), and the concentration must remain finite and tend to zero far from the source. Away from the source, the solutions of \(D C''-kC=0\) are exponential. Applying the decay condition on each side, continuity at \(a\), and the source slope jump gives

C(x)=\frac{q}{2\sqrt{Dk}}\exp\left(-\frac{|x-a|}{\lambda}\right), \qquad \lambda=\sqrt{\frac{D}{k}}.

The peak concentration is proportional to source strength \(q\), and the spatial scale \(\lambda\) is the diffusion length: larger \(D\) spreads the signal farther, while faster bulk loss \(k\) shortens its reach. The factor of two arises because a source on an infinite one-dimensional line sends material away on both sides. This solution relies on \(k>0\); with no bulk loss and no sink or boundary removal, a continuously active source cannot produce a steady profile on an infinite domain.

Worked Example: Concentration Around a Point Source

Suppose a hypothetical mediator has \(D=0.20\ \mathrm{mm^2/min}\), bulk loss \(k=0.050\ \mathrm{min^{-1}}\), and a point source at \(a=0\) with \(q=10\ \mathrm{nM\,mm/min}\). Assume a sufficiently large domain that the infinite-domain approximation is appropriate.

The diffusion length is \( \lambda=\sqrt{D/k}=\sqrt{0.20/0.050}\ \mathrm{mm}=2.00\ \mathrm{mm} \). The peak concentration is \( C(0)=q/(2\sqrt{Dk})=10/[2\sqrt{(0.20)(0.050)}]\ \mathrm{nM}=50.0\ \mathrm{nM} \).

At \(x=2.00\ \mathrm{mm}\), the distance from the source is one diffusion length, so \( C(2)=50.0\exp(-2/2)\ \mathrm{nM}=50.0e^{-1}\ \mathrm{nM}=18.4\ \mathrm{nM} \). The profile is symmetric around the source and declines exponentially on both sides. This conclusion assumes constant \(D\), linear bulk loss, and no nearby boundary that would alter the profile.

How Boundary Conditions Shape a Gradient

A finite tissue region requires boundary conditions as well as local source and sink rules. A no-flux boundary, \(C'=0\), represents an impermeable edge. A fixed-concentration boundary, \(C=C_{\mathrm{b}}\), represents contact with a maintained reservoir at concentration \(C_{\mathrm{b}}\). These conditions are not interchangeable: an impermeable edge retains material, whereas a maintained-concentration boundary can supply or remove it.

For example, a finite interval with a point source and first-order loss can be solved using a Green's function. With \(C(0)=C(L)=0\), define \(\kappa=\sqrt{k/D}\). The solution is

C(x)= \frac{q\,\sinh(\kappa x_<)\sinh\!\left(\kappa(L-x_>)\right)} {D\kappa\sinh(\kappa L)}, \qquad x_<=\min(x,a),\quad x_>=\max(x,a).

This profile satisfies the prescribed zero concentrations at both ends, solves \(D C''-kC=0\) on either side of the source, and has the required source slope jump at \(x=a\). The boundaries therefore lower the concentrations relative to an unbounded domain when they are close enough to matter. A no-flux boundary would require a different solution.

Worked Example: A Source Between Fixed-Concentration Boundaries

Consider a hypothetical interval \(0\leq x\leq 4\ \mathrm{mm}\) with \(C(0)=C(4)=0\). Let \(D=0.20\ \mathrm{mm^2/min}\), \(k=0.050\ \mathrm{min^{-1}}\), and \(q=10\ \mathrm{nM\,mm/min}\), with the source at \(a=2\ \mathrm{mm}\). The loss and diffusion parameters match the preceding example, but the boundaries are now only \(2\ \mathrm{mm}\) from the source.

Here \(\kappa=\sqrt{k/D}=0.50\ \mathrm{mm^{-1}}\). At the source, \(x_<=x_>=2\ \mathrm{mm}\), so \( C(2)=10\sinh(1)\sinh(1)/[0.20(0.50)\sinh(2)]\ \mathrm{nM} \). Using \(\sinh(1)=1.1752\) and \(\sinh(2)=3.6269\), this is \( 10(1.1752)^2/[0.10(3.6269)]=38.1\ \mathrm{nM} \).

At \(x=1\ \mathrm{mm}\), \(x_<=1\ \mathrm{mm}\) and \(x_>=2\ \mathrm{mm}\), giving \( C(1)=10\sinh(0.5)\sinh(1)/[0.10\sinh(2)]=16.9\ \mathrm{nM} \). The center concentration is below the \(50.0\ \mathrm{nM}\) infinite-domain peak because the maintained zero-concentration boundaries provide additional routes for material to leave. Symmetry of the source position and boundary values gives \(C(3)=C(1)\).

Localized Uptake and Steady-State Feasibility

A localized sink can have linear uptake, \(v(C)=k_s C\), or saturable uptake, \(v(C)=V_{\max}C/(K_m+C)\). In the saturable model, \(V_{\max}\) is the maximum uptake rate and \(K_m\) is the concentration at which uptake is half that maximum. Their units are respectively concentration × length / time and concentration in the one-dimensional point-sink formulation. Nonlinear uptake means the profile generally cannot be obtained by simply multiplying a fixed shape by source strength.

A particularly transparent case has a point source at \(a\) and a point sink at \(b>a\) on a finite interval with no-flux ends, no bulk loss, and no other sources or sinks. Integrating the steady balance over the whole interval shows that total production must equal total uptake. Thus \(v(C(b))=q\). For a linear sink this gives \(C(b)=q/k_s\); for a saturable sink, a finite steady concentration requires \(q<V_{\max}\), and solving \(q=V_{\max}C(b)/(K_m+C(b))\) gives \(C(b)=K_mq/(V_{\max}-q)\). If \(q\geq V_{\max}\), the sink cannot balance production, so no finite steady state exists under these assumptions.

Worked Example: A Saturable Localized Sink

Suppose a hypothetical source of strength \(q=1.00\ \mathrm{nM\,mm/min}\) is at \(a=1\ \mathrm{mm}\), and a saturable sink is at \(b=3\ \mathrm{mm}\). Let \(D=0.20\ \mathrm{mm^2/min}\), \(V_{\max}=1.50\ \mathrm{nM\,mm/min}\), and \(K_m=4.00\ \mathrm{nM}\). The interval is \(0\) to \(4\ \mathrm{mm}\), its ends are no-flux, and there is no bulk loss.

Steady mass balance requires sink uptake to equal the source strength. Since \(q<V_{\max}\), a finite sink concentration is possible: \( C(b)=K_mq/(V_{\max}-q)=(4.00)(1.00)/(1.50-1.00)\ \mathrm{nM}=8.00\ \mathrm{nM} \). Substitution verifies the balance: \( V_{\max}C(b)/(K_m+C(b))=(1.50)(8.00)/(4.00+8.00)=1.00\ \mathrm{nM\,mm/min} \).

Between the source and sink, the slope is \(-q/D=-1.00/0.20=-5.00\ \mathrm{nM/mm}\). Therefore the concentration drops by \(5.00(3-1)=10.0\ \mathrm{nM}\) from the source to the sink. It is \(18.0\ \mathrm{nM}\) on the source's left side, declines linearly to \(8.00\ \mathrm{nM}\) at the sink, and remains \(8.00\ \mathrm{nM}\) to the right boundary. The zero gradients outside the source-sink interval satisfy the no-flux conditions. If the source strength were raised to \(1.50\ \mathrm{nM\,mm/min}\) or more with all else unchanged, the sink could not remove material as fast as it is produced; this no-loss model would not have a finite steady state.

Assumptions, Diagnostics, and Model Choice

Before interpreting a calculated gradient, check whether the assumptions match the biological setting. Constant \(D\) assumes that the medium can be represented by a uniform effective diffusivity. First-order bulk loss assumes removal proportional to concentration over the range considered. A point source or sink assumes its physical extent is short relative to the gradient scale. A steady solution assumes sufficient time has elapsed and that source, sink, and boundary conditions remain stable.

Several checks help reveal a mistaken model or calculation:

Check global balance. With no-flux boundaries and no bulk loss, total uptake must equal total production. If it does not, a steady solution is impossible.
Check slope jumps. A source must produce a slope jump of \(-q/D\); a sink must produce one of \(v(C)/D\). A wrong sign often reverses the expected rise or fall.
Check concentration and units. A physically meaningful concentration profile should be nonnegative, and the units of each term in the steady balance must agree.
Check boundary compatibility. A solution for fixed boundary concentration cannot be used for an impermeable boundary. Boundary conditions may change both the concentration level and the gradient.
Check steady-state feasibility. A saturable sink with insufficient capacity cannot balance production unless another removal pathway, such as bulk loss or boundary flux, is present.

The infinite-domain exponential profile is convenient when bulk loss dominates the distant-boundary effects. The finite-domain Green's function is appropriate for fixed-concentration boundaries and distributed linear loss. For localized uptake with no bulk loss, a global balance plus piecewise slopes can be more direct; nonlinear uptake may require solving an algebraic balance for the sink concentration. More elaborate reaction-diffusion models are needed when uptake is distributed, diffusivity varies, or multiple interacting species materially alter the profile. These steady profiles describe spatial structure, not how quickly it is established.

Common pitfall: It is tempting to assume that a stronger source always produces a proportional increase in the entire profile. That is true for a linear model with fixed parameters and fixed boundary conditions, but not generally when a localized sink saturates. In a reflecting domain with no bulk loss, increasing production toward the sink's maximum capacity can drive the sink concentration sharply upward; at or above capacity, no finite steady state exists.

Check Your Understanding

Answer each question by identifying the model assumptions that matter.

In the infinite-domain point-source model, what happens to the diffusion length if \(D\) increases while \(k\) stays fixed?
A source of strength \(2\ \mathrm{nM\,mm/min}\) and a linear sink with \(k_s=0.50\ \mathrm{mm/min}\) are the only production and removal processes in a reflecting domain with no bulk loss. What must the steady concentration be at the sink?
For a saturable sink with \(V_{\max}=3\ \mathrm{nM\,mm/min}\), \(K_m=2\ \mathrm{nM}\), and source strength \(q=1\ \mathrm{nM\,mm/min}\), find the sink concentration required by steady balance.
Why can the finite-interval solution with zero concentration at each end not be used for two impermeable ends?

Solutions. (1) Since \(\lambda=\sqrt{D/k}\), increasing \(D\) increases the diffusion length. (2) Global balance gives \(k_sC=q\), so \(C=2/0.50=4.00\ \mathrm{nM}\). (3) Because \(q<V_{\max}\), \(C=K_mq/(V_{\max}-q)=(2)(1)/(3-1)=1.00\ \mathrm{nM}\). (4) Zero concentration prescribes exchange with fixed-concentration reservoirs; impermeable ends instead require zero flux, \(C'=0\), and therefore a different profile.