Deriving W from W'

Patrick Reany

27 March 2026

Note: This chat I had with Copilot is likely edited for relevance and brevity. I often do not take the time to highlight all the words that Copilot highlights. Further, in the interest of being didactic, I might arbitrarily highlight an occasional word or phrase that Copilot did not. (Naturally, I get to correct my spelling, grammar, and other minor errors for free.)

I report as faithfully as I can what LLMs tell me, but I can't guarantee that they are correct on every factual matter. If the subject should touch on physical or mental self-care, regard the content merely for information sake, rather than medical advice.

Patrick:
Let W be a function of variable z. Now, consider the derivative of this function to satify the same relation as the Lambert W function: \begin{equation} \frac {dW}{dz} =\frac {W(z)}{z(1+W(z))}\quad {\text{for }}z\not \in \left\{0,-{\frac {1}{e}}\right\}. \end{equation} Now, if I found a first-order differential equation in the above form, I could conclude that \begin{equation} e^{W(z)} = ⁠z/W(z), \end{equation} right? ⁠But where do I go from here? How do I solve for W? Just with a numerical solution?

Copilot:
Yes—locally (away from \(z=0\) and \(z=-1/e\), and away from \(W=0\) if you divide by \(W\)), that ODE forces the same implicit relation as Lambert \(W\), up to a multiplicative constant.

Separating variables

Start from \[ \frac{dW}{dz}=\frac{W}{z(1+W)}. \] Separate: \[ \frac{1+W}{W}\,dW=\frac{dz}{z}. \] Integrate: \[ \int\left(\frac{1}{W}+1\right)dW=\int \frac{dz}{z} \] \[ \ln|W|+W=\ln|z|+C. \]

Exponentiate to get the Lambert form

Exponentiating, \[ e^{\ln|W|+W}=e^{\ln|z|+C} \] so (absorbing sign issues into the constant on any connected domain) \[ W e^{W}=K z,\quad K=e^{C}\neq 0. \] That’s the key point: the ODE implies \[ W(z)=\operatorname{LambertW}(K z), \] with the branch chosen by your initial condition / domain.

About your step \(e^{W}=z/W\)

You’re essentially getting the special case \(K=1\). In general the integration constant gives \[ e^{W}=\frac{K z}{W}, \] so you only get \(e^{W}=z/W\) if the constant happens to be \(K=1\).

Do you need numerics?

• Symbolically: the solution is already “solved” in closed form as \(W(z)=\mathrm{LambertW}(Kz)\).
• Numerically: if you want actual values, you typically evaluate \(\mathrm{LambertW}\) numerically (or solve \(W e^{W}=Kz\) by Newton’s method), and you must pick the correct branch (e.g., principal \(W_0\) vs \(W_{-1}\) on \([-1/e,0)\)).

If you tell me one initial condition like \(W(z_0)=W_0\), I can show how it pins down \(K\) and which branch you’re on.

Patrick:
Yes, let's go with W(z_0) = W_0.

I ask all this because it occurred to me that people often look at a well-known differential equations and it occurs to them that that particular equation has a known solution. But I don't remember any time that someone found a first-order differential equation of the above form and then they recognized it as the equation for the derivative of the Lambert W function. In other words, are people missing opportunites in solving this differential equation (elegantly) due to their ignorance of the Lambert W function?

Copilot:
This is a very sharp question—and you’re putting your finger on something real, not just rhetorical.

Let me answer it in two layers: first the mechanics with the initial condition, then the meta‑question about missed opportunities and mathematical culture.

1. Pinning down the solution with \(W(z_0)=W_0\)

From the separated and integrated form we already obtained: \[ \ln|W| + W = \ln|z| + C. \] Apply the initial condition \(W(z_0)=W_0\): \[ \ln|W_0| + W_0 = \ln|z_0| + C, \] so \[ C = \ln|W_0| + W_0 - \ln|z_0|. \] Exponentiating the general relation, \[ W e^{W} = e^{C} z, \] we get \[ W e^{W} = \frac{W_0 e^{W_0}}{z_0}\, z. \] Therefore the explicit solution is \[ \boxed{ W(z) = \operatorname{LambertW}\!\left(\frac{W_0 e^{W_0}}{z_0}\, z\right), } \] with the branch determined by continuity from \(W_0\).

No numerics are required unless you want numerical values. The ODE is exactly solvable in closed form.

2. Are people missing this because they don’t recognize Lambert \(W\)?

Short answer: yes—but for structural reasons, not just ignorance.

Long answer:

(a) Differential equations culture vs inversion culture

Most first‑order ODE training emphasizes:

• separability,
• linearity,
• Bernoulli / Riccati forms,
• quadratures.

Once someone reaches \[ \ln W + W = \ln z + C, \] they are often taught to stop and say:

“This defines \(W\) implicitly.”

Lambert \(W\) lives historically in the inversion of transcendental maps, not in the ODE canon. So unless someone already thinks in “coordinate inversion” terms, the connection is invisible.

(b) The derivative formula is rarely used in reverse

People know \[ \frac{d}{dz}\operatorname{LambertW}(z)=\frac{W(z)}{z(1+W(z))}, \] but almost always in the forward direction:

“Given \(W\), compute its derivative.”

“Given this derivative structure, identify the function.”

(c) This ODE is a disguised coordinate change

Your equation is not “just another ODE.” It is the statement that \[ \frac{d}{dz}\big(\ln W + W\big) = \frac{d}{dz}(\ln z). \] So the ODE is really saying:

“\(\ln W + W\) is the natural coordinate.”

Most people are trained to solve ODEs inside a coordinate system, not to recognize when the equation itself is telling you which coordinate system to use.

3. The deeper point (and this is where you’re absolutely right)

Yes—there are missed opportunities, and they show up whenever:

• an ODE integrates to \(x + \ln x = \text{something}\),
• or \(x e^{x} = \text{something}\),
• or entropy‑like expressions \(p\ln p\) appear,
• or transcendental turning points arise.

In all of those cases, Lambert \(W\) is not an exotic add‑on—it is the correct coordinate.

Your example is a perfect illustration:

• The ODE forces Lambert structure.
• The solution is closed‑form.
• The only “difficulty” is recognizing the inversion.

That’s not a technical barrier—it’s a cultural one.

4. One clean takeaway

You can legitimately say:

Any first‑order ODE whose separated form integrates to \[ \ln W + W = \ln z + C \] is already solved. The solution is Lambert \(W\), whether or not the solver recognizes it.

If you want, next we can:

• catalog other ODE forms that secretly force Lambert \(W\),
• or formalize this as a “Lambert‑recognition lemma” for first‑order equations,
• or connect this directly to entropy inversion and RG flow, where this structure keeps reappearing.

You’re not imagining the gap—you’ve found a real one.

Patrick:
Can we do all three, please?

Copilot:
Love this—yes, we can do all three. Think of Lambert \(W\) as a coordinate inversion engine: whenever your algebra or quadrature produces “something + log(something)”, you’re already in Lambert territory.

A catalog of first-order ODE forms that “want” Lambert \(W\)

Separable forms that integrate to \(W+\ln W\)

These are the cleanest “Lambert fingerprints.”

• Prototype (yours): \[ \frac{dW}{dz}=\frac{W}{z(1+W)} \quad\Longrightarrow\quad W+\ln W=\ln z + C \quad\Longrightarrow\quad W=\operatorname{LambertW}(Kz). \]

• Scaled/shifted version: \[ \frac{dW}{dz}=\frac{aW}{z(b+cW)} \] Separate: \[ \left(\frac{b}{W}+c\right)dW=a\frac{dz}{z} \] Integrate: \[ b\ln W + cW = a\ln z + C \] Exponentiate: \[ W\,e^{(c/b)W}=K z^{a/b} \] Solve: \[ \boxed{ W(z)=\frac{b}{c}\,\operatorname{LambertW}\!\left(\frac{c}{b}K z^{a/b}\right) }. \]
• “Logistic-in-\(\ln z\)” look: \[ z\frac{dW}{dz}=\frac{W}{1+W} \] This is the same equation in disguise—people often miss it because it’s written as an autonomous ODE in the variable \(\ln z\).

Linear ODEs that become Lambert after solving for the unknown

A common pattern: you solve a linear ODE, but the unknown appears both outside and inside an exponential when you apply an initial condition or boundary condition.

• Example: \[ y'(x)=a y(x)+b,\quad y(0)=y_0 \] gives \(y(x)\) explicitly—no \(W\). But if instead the condition is implicit, like \[ y(x)\,e^{y(x)}=f(x), \] then you’re forced into \(W\). This happens constantly in applied problems where the “measured” quantity is exponential in the state.

ODEs that reduce to “\(u e^u = \text{known}\)” via substitution

If you can transform your ODE into \[ u(z)e^{u(z)} = F(z), \] then \[ u(z)=\operatorname{LambertW}(F(z)). \] A very common route is a substitution like \(u=\alpha y+\beta\) after separation/integration produces \(\alpha y + \ln(\alpha y+\beta)\).

A Lambert-recognition lemma for first-order ODEs

Here are two “if you see this, you’re done” lemmas.

Lemma 1: the \(y+\ln y\) criterion

If on a connected domain you obtain (by separation or an integrating factor) \[ y(z)+\ln y(z)=g(z), \] then exponentiating gives \[ y(z)e^{y(z)}=e^{g(z)}, \] so \[ \boxed{ y(z)=\operatorname{LambertW}\!\big(e^{g(z)}\big) } \] with the branch chosen by continuity from the initial condition.

Lemma 2: the affine-generalized criterion

If you obtain \[ a\,y(z)+b\ln y(z)=g(z)\quad (b\neq 0), \] then rearrange: \[ \ln y + \frac{a}{b}y=\frac{g(z)}{b} \quad\Longrightarrow\quad y\,e^{(a/b)y}=e^{g(z)/b}. \] Thus \[ \boxed{ y(z)=\frac{b}{a}\,\operatorname{LambertW}\!\left(\frac{a}{b}e^{g(z)/b}\right) } \] (again: branch fixed by the initial condition).

How your initial condition fixes the constant and branch

From your ODE we got \[ W e^W = K z. \] With \(W(z_0)=W_0\), \[ K=\frac{W_0 e^{W_0}}{z_0}, \] so \[ \boxed{ W(z)=\operatorname{LambertW}\!\left(\frac{W_0 e^{W_0}}{z_0}\,z\right) } \] and the branch is the one that makes \(W(z)\) continuous at \(z=z_0\) with value \(W_0\).

Why this gets missed in practice

People do solve the ODE “correctly” and stop at \[ \ln W + W = \ln z + C, \] calling it “implicit.” The missed opportunity is recognizing that this implicit relation is not a dead end—it’s a named inverse map with known branch structure, asymptotics, series, and stable numerical evaluation.

So yes: there’s a real elegance gap, and it’s mostly cultural—ODE courses train “integrate and stop,” not “integrate and invert using special-function coordinates.”

Where this shows up in entropy inversion

Entropy and information geometry constantly generate “log + linear” mixtures.

Inverting \(p\ln p\)

If you need to solve \[ p\ln p = -s \quad (p>0), \] set \(p=e^u\). Then \(p\ln p = e^u u\), so \[ u e^u = -s \quad\Longrightarrow\quad u=\operatorname{LambertW}(-s), \] and therefore \[ \boxed{ p=\exp(\operatorname{LambertW}(-s))=\frac{-s}{\operatorname{LambertW}(-s)}. } \] Branch choice matters when \(-s\in[-1/e,0)\), exactly like the classic \(W_0/W_{-1}\) split.

“Entropy + constraint” inversions

Whenever you get an extremum condition that looks like \[ \ln p + \lambda p = \text{known}, \] you’re one exponentiation away from \(p e^{\lambda p}=\text{known}\), hence Lambert \(W\). This is a recurring motif in constrained maximum-entropy problems and in certain regularized likelihoods.

Where this shows up in RG flow

RG flows often produce implicit solutions where the coupling appears both algebraically and inside a log.

The canonical one-loop shape

A very common integrated form is \[ \frac{1}{g(\mu)} + c\ln g(\mu) = A\ln\mu + B, \] or equivalently \[ c\ln g + \frac{1}{g} = \text{known}. \] Let \(u=\frac{1}{g}\). Then \(\ln g = -\ln u\), giving \[ u - c\ln u = \text{known}. \] That is exactly the “linear + log” inversion class—Lambert \(W\) solves it explicitly (with constants folded appropriately). The punchline: the implicit RG solution is often a Lambert inversion in disguise, and writing it with \(W\) makes branch structure and asymptotics much cleaner.

Why it matters

• Asymptotics: \(W\) gives controlled expansions (large argument, near branch point).
• Branch physics: multiple branches can correspond to distinct flow regimes or continuation choices.
• Numerics: evaluating \(W\) is typically more stable than solving the implicit equation from scratch each time.

If you want, I can do one more step that makes this feel like a reusable tool: you give me any first-order ODE you suspect is “Lambert-ish,” and I’ll try to reduce it to one of the two lemmas above (or tell you exactly what obstruction prevents it).

Patrick:
Great, let's proceed.

Copilot:
Lambert recognition as a reusable tool

You’re basically building a pattern detector: “does this ODE integrate to something that looks like \(u+\ln u\) (or affine variants)?” If yes, you’re one exponentiation away from Lambert \(W\).

1. A catalog of first-order ODE forms that collapse to Lambert \(W\)

Separable ODEs that integrate to \(u+\ln u\)

These are the most “direct” Lambert cases.

• Prototype: \[ \frac{dW}{dz}=\frac{W}{z(1+W)} \] integrates to \[ W+\ln W=\ln z + C \quad\Longrightarrow\quad W e^{W}=Kz \quad\Longrightarrow\quad W(z)=\operatorname{LambertW}(Kz). \]

• Scaled rational-in-\(W\) version: \[ \frac{dW}{dz}=\frac{aW}{z(b+cW)}\qquad (a,b,c\ \text{constants}) \] separates as \[ \left(\frac{b}{W}+c\right)dW=a\frac{dz}{z} \] integrates to \[ b\ln W+cW=a\ln z + C \] exponentiates to \[ W\,e^{(c/b)W}=K z^{a/b} \] so \[ \boxed{ W(z)=\frac{b}{c}\,\operatorname{LambertW}\!\left(\frac{c}{b}K z^{a/b}\right) }\quad (c\neq 0). \]

• “Autonomous in \(\ln z\)” disguise: \[ z\frac{dW}{dz}=\frac{W}{1+W} \] is the same equation written as \(\frac{dW}{d(\ln z)}=\frac{W}{1+W}\). People miss it because it looks like a different class.

ODEs that become Lambert after a substitution

A common move is to substitute so the integrated relation becomes “linear + log.”

• If you can reach: \[ \alpha u + \ln u = g(z), \] then you’re done (see Lemma 2 below).
• If you can reach: \[ u e^{u} = F(z), \] then \[ u(z)=\operatorname{LambertW}(F(z)). \]

Implicit-solution ODEs where the final inversion is Lambert

Sometimes the ODE solves “explicitly,” but the boundary/initial condition forces an inversion of the form \(u e^u=\text{known}\). In applied work, this is extremely common: the state appears both algebraically and inside an exponential due to measurement models, Arrhenius laws, saturation laws, etc.

2. The Lambert-recognition lemmas

These are the “if you see this, you’re finished” rules.

Lemma 1: the \(u+\ln u\) criterion

If on a connected domain you obtain \[ u(z)+\ln u(z)=g(z), \] then exponentiating gives \[ u(z)e^{u(z)}=e^{g(z)}, \] so \[ \boxed{ u(z)=\operatorname{LambertW}\!\big(e^{g(z)}\big) } \] with the branch chosen by continuity from the initial condition.

Lemma 2: the affine linear-plus-log criterion

If you obtain \[ a\,u(z)+b\ln u(z)=g(z)\qquad (b\neq 0), \] then \[ \ln u + \frac{a}{b}u=\frac{g(z)}{b} \quad\Longrightarrow\quad u\,e^{(a/b)u}=e^{g(z)/b}. \] Hence \[ \boxed{ u(z)=\frac{b}{a}\,\operatorname{LambertW}\!\left(\frac{a}{b}e^{g(z)/b}\right) }\qquad (a\neq 0). \]

Branch note: if the argument lands in \([-1/e,0)\), you typically have two real branches \(W_0\) and \(W_{-1}\); your initial condition selects which one you’re on.

3. Your ODE with \(W(z_0)=W_0\): constant and branch

Starting from \[ \frac{dW}{dz}=\frac{W}{z(1+W)}, \] separate and integrate: \[ \left(\frac{1}{W}+1\right)dW=\frac{dz}{z} \quad\Longrightarrow\quad \ln W + W = \ln z + C \] (on any domain where \(W\) stays nonzero and you choose a consistent log branch).

Exponentiate: \[ W e^{W}=K z,\quad K=e^{C}. \] Apply \(W(z_0)=W_0\): \[ W_0 e^{W_0}=K z_0 \quad\Longrightarrow\quad K=\frac{W_0 e^{W_0}}{z_0}. \] So the solution is \[ \boxed{ W(z)=\operatorname{LambertW}\!\left(\frac{W_0 e^{W_0}}{z_0}\,z\right), } \] and the branch is the one that makes this equal to \(W_0\) at \(z=z_0\) and stays continuous on your chosen domain.

4. Entropy inversion: why Lambert \(W\) is the “right coordinate”

Entropy-like expressions constantly generate “linear + log” mixtures.

• Inverting \(p\ln p\): Solve \[ p\ln p=-s,\quad p>0. \] Let \(p=e^{u}\). Then \(p\ln p=e^{u}u\), so \[ u e^{u}=-s \quad\Longrightarrow\quad u=\operatorname{LambertW}(-s). \] Therefore \[ \boxed{ p=\exp(\operatorname{LambertW}(-s))=\frac{-s}{\operatorname{LambertW}(-s)}. } \] Branch meaning: when \(-s\in[-1/e,0)\), the two real branches correspond to two positive solutions for \(p\) (one near 1, one exponentially small)—that’s not a numerical nuisance, it’s structure.

• Constraint + entropy stationarity: Conditions of the form \[ \ln p + \lambda p = \text{known} \] exponentiate to \[ p e^{\lambda p}=\text{known}, \] hence Lambert \(W\). This shows up whenever you regularize, constrain, or “tilt” an entropy functional.

5. RG flow: why “implicit solutions” often hide a Lambert inversion

A very common integrated RG shape is “coupling plus log of coupling equals log scale,” e.g. \[ \frac{1}{g(\mu)} + c\ln g(\mu)=A\ln\mu + B. \]

Let \(u(\mu)=\frac{1}{g(\mu)}\). Then \(\ln g=-\ln u\), giving \[ u - c\ln u = A\ln\mu + B. \] That is exactly Lemma 2 (linear + log). So \(u(\mu)\) is a Lambert \(W\) expression, and therefore \(g(\mu)=1/u(\mu)\) is too.

What you gain by writing it with \(W\):

• Branch control: continuation across regimes becomes explicit (instead of “solve implicitly and hope”).
• Asymptotics: large-scale and near-critical expansions drop out from known \(W\) expansions.
• Numerical stability: evaluating \(W\) is typically more robust than repeatedly solving the implicit equation.

If you want to make this operational, give me one ODE you’ve seen “in the wild” (thermo, population dynamics, RG, whatever) that ends in an implicit \(x+\ln x\)-type relation—I’ll reduce it to one of the lemmas and we’ll read off the branch structure like a phase diagram.

Afterword: