The Singular Limit

What setting $\ve = 0$ loses¶

The full problem is an initial value problem, one first-order equation and one condition to fix the one constant it produces:

Setting $\ve = 0$ deletes the derivative:

At $\tilde t = 0$ the reduced equation insists that $\tilde v(0) = \tilde F(0)$. If the cell started at any other speed, the reduced model does not approximate the initial condition badly; it contradicts it.

Our equation of motion is singular: the order went from 1 to 0. Yet the reduced equation came from a correct one by deleting a term with prefactor 10^-9, so it can only be wrong on a set small enough that nine orders of magnitude went unnoticed. Where that set is, and how small, follows from the two time scales in the equation.

Outer and inner problems¶

$\ve = \tau_i/T$ is a ratio of two times, and each gives a unit in which to measure time. $T = L/v_0$ belongs to the question, and writing the equation of motion in units of $T$, as in (1), is what compressed the layer to a point. $\tau_i = m/\gamma$ belongs to the cell. Measuring time in the cell’s own unit, $\sigma = t/\tau_i = \tilde t/\ve$, the chain rule gives $\dd/\dd\tilde t = \ve^{-1}\,\dd/\dd\sigma$, and (1) becomes

with no $\ve$ in front of the derivative. This is the unit the equation picks for itself.

(1) and (3) are the same equation in two units of time, and they differ only in what they resolve. On $\sigma$ the transient occupies an interval of order one and is resolved in full, while the migration has not begun. On $\tilde t$ the transient is compressed into a span $\ve$ of the axis and vanishes as $\ve \to 0$. The shape of the transient does not depend on $\ve$ at all; only the fraction of the $\tilde t$ axis it occupies does.

# The same solution on the two axes, with epsilon as a dial. Each frame is one
# value of epsilon; the slider below the picture selects it.
import base64, io
import numpy as np
import matplotlib.pyplot as plt
from IPython.display import display
plt.rcParams.update({"figure.dpi": 130, "font.size": 9, "axes.spines.top": False,
                     "axes.spines.right": False})
C_EXACT, C_OUTER, C_INNER, C_COMP, C_LAYER = "#111111", "#0072B2", "#D55E00", "#009E73", "#F0C808"
V0, FBAR = 0.0, 1.0            # start at rest; force balance value F/gamma = 1

def frame_png(eps):
    fig, ax = plt.subplots(1, 2, figsize=(6.6, 2.5))
    sig = np.linspace(0, 6, 600)
    ax[0].plot(sig, FBAR + (V0 - FBAR) * np.exp(-sig), color=C_EXACT, lw=1.8)
    ax[0].axhline(FBAR, color=C_OUTER, ls="--", lw=1.2)
    ax[0].set_xlabel(r"time of the cell  $\sigma = t/\tau_i = \tilde t/\varepsilon$")
    ax[0].set_ylabel(r"$\tilde v$")
    ax[0].set_title("the same solution: nothing moves")
    ax[0].set_xlim(0, 6); ax[0].set_ylim(-0.08, 1.18)
    t = np.linspace(0, 1, 4000)
    ax[1].plot(t, FBAR + (V0 - FBAR) * np.exp(-t / eps), color=C_EXACT, lw=1.8)
    ax[1].axhline(FBAR, color=C_OUTER, ls="--", lw=1.2)
    ax[1].axvspan(0, eps, color=C_LAYER, alpha=0.9, lw=0)
    ax[1].set_xlabel(r"time of the question  $\tilde t = t/T$")
    ax[1].set_title(rf"$\varepsilon = {eps:.3g}$: the layer is $\varepsilon$ wide")
    ax[1].set_xlim(0, 1); ax[1].set_ylim(-0.08, 1.18)
    fig.tight_layout()
    buf = io.BytesIO(); fig.savefig(buf, format="png", dpi=100); plt.close(fig)
    return base64.b64encode(buf.getvalue()).decode()

eps_values = np.logspace(0, -3, 25)
frames = [frame_png(e) for e in eps_values]
labels = [f"{e:.3g}" for e in eps_values]
k0 = int(np.argmin(np.abs(np.log10(eps_values) + 1)))   # open at epsilon = 0.1
page = f"""<!doctype html><html><body style="margin:0;font:13px sans-serif;text-align:center">
<img id="f" src="data:image/png;base64,{frames[k0]}" style="max-width:100%">
<div><label>&#949; = <b id="l">{labels[k0]}</b></label>
<input id="s" type="range" min="0" max="{len(frames)-1}" value="{k0}" style="width:60%"></div>
<script>
var F={frames!r}, L={labels!r};
document.getElementById("s").oninput=function(){{
  document.getElementById("f").src="data:image/png;base64,"+F[this.value];
  document.getElementById("l").textContent=L[this.value];}};
</script></body></html>"""
esc = page.replace("&", "&amp;").replace('"', "&quot;")
display({"text/html": f'<iframe srcdoc="{esc}" sandbox="allow-scripts" width="100%" '
                       'height="330" style="border:0"></iframe>'}, raw=True)

One solution, starting from rest with $\tilde F = 1$, on the two time axes. Left: on the cell’s time $\sigma$ the transient fills the frame, and does not change with $\ve$. Right: on the time of the question $\tilde t$ the same curve is squeezed into the yellow sliver of width $\ve$ at the origin. The slider (in the web version of these notes) starts at $\ve = 0.1$ and runs from 1 to 10^-3: the limit $\ve \to 0$ is the sliver closing. A cell’s $\ve$ is a million times smaller than 10^-3.

Letting $\ve \to 0$ in each of the two units of time gives a different, and simpler, problem.

For the cell the two problems are as follows.

On the time of the question. The outer problem is the reduced equation (2), so

It follows the force instantaneously, and it meets the initial condition only if $\tilde v_0 = \tilde F(0)$.

On the time of the cell. In (3) the force is $\tilde F(\ve\sigma) \to \tilde F(0)$ as $\ve \to 0$: over a few $\tau_i$ the force has no time to change. The inner problem and its solution are

It carries the initial condition, but knows only the initial value of the force.

Neither problem is wrong; each answers a different question. Ours is where the cell is after a body length, and on that time scale $\tilde v = \tilde F$ from the start. Whether the ignored microsecond matters for that question is settled in Error Analysis.

Checking against the exact solution¶

For constant $\tilde F$ the full problem can be solved exactly, so both reductions can be checked against it. By the integrating factor (Proposition 1; the steps are in Background: The Integrating Factor),

Letting $\ve \to 0$ with $\tilde t > 0$ fixed, $e^{-\tilde t/\ve} \to 0$ and the limit is the outer solution (4). Letting $\ve \to 0$ with $\sigma = \tilde t/\ve$ fixed, the exponential is $e^{-\sigma}$ and the limit is the inner solution (5). The outer and inner solutions are the two limits of one solution, depending on which unit of time is held fixed while $\ve$ shrinks.

Source

fig, ax = plt.subplots(figsize=(5.4, 2.7))
t = np.linspace(0, 1, 4000)
for eps, col in ((0.2, "#9ecae1"), (0.05, "#4292c6"), (0.01, C_OUTER)):
    ax.plot(t, np.exp(-t / eps), color=col, lw=1.9, label=rf"$\varepsilon = {eps}$")
ax.plot([0], [1], "o", color=C_INNER, ms=6, zorder=5)
ax.annotate("every curve starts at $1$,\nfor every $\\varepsilon$", xy=(0.0, 1.0),
            xytext=(0.17, 0.86), fontsize=8, color=C_INNER,
            arrowprops=dict(arrowstyle="->", color=C_INNER, lw=1.0))
ax.annotate("but for any fixed $\\tilde t > 0$\nit tends to $0$", xy=(0.45, 0.0),
            xytext=(0.45, 0.30), fontsize=8, color="#555555",
            arrowprops=dict(arrowstyle="->", color="#777777", lw=1.0))
ax.set_xlabel(r"time of the question  $\tilde t$"); ax.set_ylabel(r"$e^{-\tilde t/\varepsilon}$")
ax.set_xlim(-0.02, 1.0); ax.set_ylim(-0.03, 1.08)
ax.legend(loc="upper right", frameon=False)
ax.set_title("the term that carried the free constant");

The function $e^{-\tilde t/\ve}$ that multiplies the free constant $C$, for three values of $\ve$. Every curve equals 1 at $\tilde t = 0$, but at any fixed $\tilde t > 0$ it tends to 0 as $\ve \to 0$.

The interval $0 \le \tilde t \lesssim \ve$ on which the two limits differ is the initial layer. Its thickness $\ve$ is set by the equation, not chosen; in dimensional time it is one $\tau_i$, a microsecond.

Matching and the composite solution¶

Where each one is right:

In the overlap region both equal $\tilde F(0)$; if the two limits in (7) disagreed, the layer would have been scaled wrongly. Inside the layer $\tilde v_\mathrm{out} \approx \tilde v_\mathrm{common}$ and the composite reduces to $\tilde v_\mathrm{in}$; outside it $\tilde v_\mathrm{in} \approx \tilde v_\mathrm{common}$ and the composite reduces to $\tilde v_\mathrm{out}$.

For a constant force the composite is not an approximation: it reproduces the exact solution, because the outer solution is itself the common limit. When the force varies this is no longer true, and the next section measures the error.

A slowly varying force¶

With $\tilde F$ constant, the outer solution is the common limit and the composite collapses onto the inner solution, so the three curves coincide and there is nothing to see. A cell’s traction changes over minutes rather than over $\tau_i$, so the interesting case is a force that varies on the slow time scale. Take the simplest one.

Source

eps = 0.04
fig, ax = plt.subplots(figsize=(6.2, 3.1))
t = np.linspace(0, 0.6, 4000); sig = t / eps
exact = 1 + t - eps + (eps - 1) * np.exp(-sig)
outer = 1 + t
inner = 1 - np.exp(-sig)
comp = inner + outer - 1.0                          # common limit is F(0) = 1
ax.axvspan(4 * eps, 0.28, color=C_LAYER, alpha=0.28, lw=0)
ax.plot(t, exact, color=C_EXACT, lw=3.2, alpha=0.3, label="exact")
ax.plot(t, outer, color=C_OUTER, ls="--", lw=1.5, label=r"outer  $\tilde v_{\rm out} = \tilde F(\tilde t)$")
ax.plot(t, inner, color=C_INNER, ls=":", lw=2.0,
        label=r"inner  $\tilde v_{\rm in} = \tilde F(0) + (\tilde v_0 - \tilde F(0))e^{-\sigma}$")
ax.plot(t, comp, color=C_COMP, lw=1.3, label=r"composite  $\tilde v_{\rm in} + \tilde v_{\rm out} - \tilde v_{\rm common}$")
ax.set_xlabel(r"time of the question  $\tilde t = t/T$"); ax.set_ylabel(r"$\tilde v$")
ax.set_ylim(-0.08, 1.80)
ax.annotate("overlap region\n" + r"$\tau_i \ll t \ll T$", xy=(0.205, 0.45), fontsize=8,
            color="#8a6d00", ha="center")
ax.annotate("outer misses\nthe layer", xy=(0.004, 1.0), xytext=(0.045, 1.58), fontsize=8,
            color=C_OUTER, arrowprops=dict(arrowstyle="->", color=C_OUTER, lw=1.0))
ax.annotate("inner misses\nthe ramp", xy=(0.50, 1.0), xytext=(0.30, 1.12), fontsize=8,
            color=C_INNER, arrowprops=dict(arrowstyle="->", color=C_INNER, lw=1.0))
ax.legend(loc="lower right", frameon=False)
ax.set_title(r"a ramped force, $\tilde F(\tilde t) = 1 + \tilde t$,  with $\varepsilon = 0.04$");

The ramped force of Example 2 with $\ve = 0.04$. The outer solution has the late behavior right and the initial condition wrong; the inner solution has the initial condition right and never learns that the force is growing; both are correct in the overlap region (shaded); and the composite agrees with the exact solution to $O(\ve)$ everywhere.

The reason is (13). On the cell’s time scale $\tilde t = \ve\sigma$, so $\tilde F(\tilde t) = \tilde F(0) + \ve\sigma\tilde F'(0) + O(\ve^2)$: the variation of the force is itself $O(\ve)$ and invisible at leading order. The same bound holds for every force that varies on the slow time scale.