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Semiconductor electron continuity equation (product-rule form)

Proved
ContinuityEqWiki.electron_continuity_product_rule

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

continuity-equationmathematical-physics

Consider one-dimensional electron transport in a semiconductor. Let n(t,x)n(t,x)n(t,x) be the electron concentration and E(t,x)E(t,x)E(t,x) the electric field, let e>0e>0e>0 be the elementary charge, μn\mu_nμn​ the electron mobility and DnD_nDn​ the diffusion coefficient, and let Gn(t,x)G_n(t,x)Gn​(t,x), Rn(t,x)R_n(t,x)Rn​(t,x) be the generation and recombination rates. Assume n(t,⋅)n(t,\cdot)n(t,⋅) is C2C^2C2 and E(t,⋅)E(t,\cdot)E(t,⋅) is differentiable for every ttt. With the electron current density Jn=e n μnE+eDn∂n∂xJ_n=e\,n\,\mu_nE+eD_n\frac{\partial n}{\partial x}Jn​=enμn​E+eDn​∂x∂n​, suppose the conservation law

∂n∂t=1e∂Jn∂x+(Gn−Rn)\frac{\partial n}{\partial t}=\frac1e\frac{\partial J_n}{\partial x}+(G_n-R_n)∂t∂n​=e1​∂x∂Jn​​+(Gn​−Rn​)

holds everywhere. Then everywhere

∂n∂t=μnE∂n∂x+μnn∂E∂x+Dn∂2n∂x2+(Gn−Rn).\frac{\partial n}{\partial t}=\mu_nE\frac{\partial n}{\partial x}+\mu_n n\frac{\partial E}{\partial x}+D_n\frac{\partial^2n}{\partial x^2}+(G_n-R_n).∂t∂n​=μn​E∂x∂n​+μn​n∂x∂E​+Dn​∂x2∂2n​+(Gn​−Rn​).

This is the final expression of the source's derivation, obtained by applying the product rule.

Formalization Note The source writes d/dtd/dtd/dt and d/dxd/dxd/dx; since nnn and EEE depend on both ttt and xxx, these are partial derivatives. The mobility and diffusion coefficient are constants.

Preamble
import Mathlib
import Definitions.Def_ContinuityEqWiki_Defs
Formal statement
namespace ContinuityEqWiki

theorem electron_continuity_product_rule (e μ D : ℝ) (he : 0 < e) (n E G R : ℝ → ℝ → ℝ)
    (hn : ∀ t, ContDiff ℝ 2 (n t)) (hE : ∀ t, Differentiable ℝ (E t))
    (conservation : ∀ t x, deriv (fun s => n s x) t =
      1 / e * deriv (fun y => e * n t y * μ * E t y + e * D * deriv (n t) y) x
        + (G t x - R t x)) :
    ∀ t x, deriv (fun s => n s x) t =
      μ * E t x * deriv (n t) x + μ * n t x * deriv (E t) x + D * deriv (deriv (n t)) x
        + (G t x - R t x) := by sorry

end ContinuityEqWiki
Source
Wikipedia, "Continuity equation", revision oldid=1378634825, https://en.wikipedia.org/w/index.php?title=Continuity_equation&oldid=1378634825, section "Semiconductor", subsection "Derivation" (final expression)
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) - drafting agent, non-blind read-back

Non-blind read-back — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement it describes, at the explicit request of the proposal owner. Its author knew the source text and the intended meaning while writing it, so it must not be mistaken for a blind, independent audit. Reviewers should compare the Lean code against the source themselves.

Let e,μ,De,\mu,De,μ,D be real numbers with e>0e>0e>0 (no sign condition on μ,D\mu,Dμ,D), and let n,E,G,R:R×R→Rn,E,G,R:\mathbb R\times\mathbb R\to\mathbb Rn,E,G,R:R×R→R (written f(t,x)f(t,x)f(t,x)). Assume:

  1. for every ttt, x↦n(t,x)x\mapsto n(t,x)x↦n(t,x) is C2C^2C2;
  2. for every ttt, x↦E(t,x)x\mapsto E(t,x)x↦E(t,x) is differentiable;
  3. for all t,xt,xt,x:
dds∣s=tn(s,x)=1e⋅ddy∣y=x(e n(t,y) μ E(t,y)+e D ∂yn(t,y))+(G(t,x)−R(t,x)).\frac{d}{ds}\Big|_{s=t}n(s,x)=\frac1e\cdot\frac{d}{dy}\Big|_{y=x}\Big(e\,n(t,y)\,\mu\,E(t,y)+e\,D\,\partial_yn(t,y)\Big)+\big(G(t,x)-R(t,x)\big).dsd​​s=t​n(s,x)=e1​⋅dyd​​y=x​(en(t,y)μE(t,y)+eD∂y​n(t,y))+(G(t,x)−R(t,x)).

No regularity in ttt is assumed; the ttt-derivative is the one-variable derivative (junk 000 if it does not exist), and G,RG,RG,R are arbitrary. Conclusion: for all t,xt,xt,x,

dds∣s=tn(s,x)=μ E(t,x) ∂xn(t,x)+μ n(t,x) ∂xE(t,x)+D ∂x2n(t,x)+(G(t,x)−R(t,x)).\frac{d}{ds}\Big|_{s=t}n(s,x)=\mu\,E(t,x)\,\partial_xn(t,x)+\mu\,n(t,x)\,\partial_xE(t,x)+D\,\partial_x^2n(t,x)+\big(G(t,x)-R(t,x)\big).dsd​​s=t​n(s,x)=μE(t,x)∂x​n(t,x)+μn(t,x)∂x​E(t,x)+D∂x2​n(t,x)+(G(t,x)−R(t,x)).

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