Editing Openai/6935580c-34e4-8001-9136-7bbdf1499790 (section)

=== Assistant: Below is a clean, fully rigorous account of how one passes from classical to mild solutions and back, while explicitly … ===
Below is a clean, fully rigorous account of how one passes from classical to mild solutions and back, while explicitly respecting the fact that the Newton–Leibniz formula holds only almost everywhere for Banach-valued functions.

I will be precise about what is true pointwise, what is true a.e., and why this is sufficient.

==== 1. Functional setting (what is assumed and why) ====

Fix a Banach space XXX (e.g. L2L^2L2, Hs−1H^{s-1}Hs−1, HsH^sHs).
Let A=νΔA=\nu\DeltaA=νΔ (with domain D(A)⊂XD(A)\subset XD(A)⊂X) and let etAe^{tA}etA be the heat semigroup.

Assume:
* u∈C([0,T];X)u \in C([0,T];X)u∈C([0,T];X),
* F∈L1(0,T;X)F \in L^1(0,T;X)F∈L1(0,T;X).

These are minimal assumptions under which the Duhamel integral

∫0te(t−s)AF(s) ds\int_0^t e^{(t-s)A}F(s)\,ds∫0te(t−s)AF(s)ds
is well-defined as a Bochner integral.

==== 2. Mild solution (definition) ====

A function u∈C([0,T];X)u\in C([0,T];X)u∈C([0,T];X) is a mild solution of

∂tu=Au+F(t)\partial_t u = Au + F(t)∂tu=Au+F(t)
with initial data u(0)=u0∈Xu(0)=u_0\in Xu(0)=u0∈X if

u(t)=etAu0+∫0te(t−s)AF(s) ds∀t∈[0,T].\boxed{
u(t)
=
e^{tA}u_0
+
\int_0^t e^{(t-s)A}F(s)\,ds
\qquad\forall t\in[0,T].
}u(t)=etAu0+∫0te(t−s)AF(s)ds∀t∈[0,T].
This definition involves no derivatives in time, only integrals.

==== 3. Key analytic fact: Newton–Leibniz in Banach spaces ====

Let g∈L1(0,T;X)g \in L^1(0,T;X)g∈L1(0,T;X). Define

G(t):=∫0tg(s) ds(Bochner integral).G(t) := \int_0^t g(s)\,ds \quad\text{(Bochner integral)}.G(t):=∫0tg(s)ds(Bochner integral).
Then:
# G∈C([0,T];X)G \in C([0,T];X)G∈C([0,T];X),
# GGG is absolutely continuous,
# The derivative exists almost everywhere and G′(t)=g(t)for a.e. t.G'(t) = g(t)\quad\text{for a.e. }t.G′(t)=g(t)for a.e. t.

This is the Banach-valued Newton–Leibniz theorem.

Crucially:
* the derivative need not exist everywhere,
* but equality a.e. is sufficient for PDE theory.

==== 4. Mild ⇒ differential equation (a.e.) ====

Assume uuu is a mild solution:

u(t)=etAu0+∫0te(t−s)AF(s) ds.u(t)=e^{tA}u_0+\int_0^t e^{(t-s)A}F(s)\,ds.u(t)=etAu0+∫0te(t−s)AF(s)ds.
===== Step 1 — Differentiate the linear term =====

For all t>0t>0t>0,

ddtetAu0=AetAu0\frac{d}{dt}e^{tA}u_0 = A e^{tA}u_0dtdetAu0=AetAu0
(classical semigroup theory).

===== Step 2 — Differentiate the integral term =====

Define

G(t):=∫0te(t−s)AF(s) ds.G(t):=\int_0^t e^{(t-s)A}F(s)\,ds.G(t):=∫0te(t−s)AF(s)ds.
Write it as

G(t)=∫0th(t,s) dswithh(t,s)=e(t−s)AF(s).G(t)=\int_0^t h(t,s)\,ds
\quad\text{with}\quad
h(t,s)=e^{(t-s)A}F(s).G(t)=∫0th(t,s)dswithh(t,s)=e(t−s)AF(s).
Then:
* h(⋅,s)h(\cdot,s)h(⋅,s) is differentiable for each s<ts<ts<t,
* ∂th(t,s)=Ae(t−s)AF(s)\partial_t h(t,s)=A e^{(t-s)A}F(s)∂th(t,s)=Ae(t−s)AF(s),
* F∈L1F\in L^1F∈L1 ensures Bochner integrability.

By standard differentiation-under-the-integral results for Bochner integrals:

ddtG(t)=F(t)+∫0tAe(t−s)AF(s) dsfor a.e. t.\frac{d}{dt}G(t)
=
F(t)+\int_0^t A e^{(t-s)A}F(s)\,ds
\quad\text{for a.e. }t.dtdG(t)=F(t)+∫0tAe(t−s)AF(s)dsfor a.e. t.
===== Step 3 — Combine =====

For almost every t∈(0,T)t\in(0,T)t∈(0,T),

∂tu(t)=Au(t)+F(t).\partial_t u(t)
=
Au(t)+F(t).∂tu(t)=Au(t)+F(t).
Thus:

: Every mild solution satisfies the PDE almost everywhere in time.

This is the strongest statement possible without extra regularity.

==== 5. Classical ⇒ mild (exact identity) ====

Now assume uuu is a classical solution, meaning:
* u∈C([0,T];D(A))∩C1((0,T);X)u\in C([0,T];D(A))\cap C^1((0,T);X)u∈C([0,T];D(A))∩C1((0,T);X),
* ∂tu=Au+F(t)\partial_t u = Au+F(t)∂tu=Au+F(t) holds pointwise.

Multiply the equation by e−(t−s)Ae^{-(t-s)A}e−(t−s)A and integrate in time:

ddt(e−tAu(t))=e−tAF(t).\frac{d}{dt}\big(e^{-tA}u(t)\big)=e^{-tA}F(t).dtd(e−tAu(t))=e−tAF(t).
Integrate from 000 to ttt:

e−tAu(t)−u0=∫0te−sAF(s) ds.e^{-tA}u(t)-u_0=\int_0^t e^{-sA}F(s)\,ds.e−tAu(t)−u0=∫0te−sAF(s)ds.
Multiply by etAe^{tA}etA:

u(t)=etAu0+∫0te(t−s)AF(s) ds.u(t)=e^{tA}u_0+\int_0^t e^{(t-s)A}F(s)\,ds.u(t)=etAu0+∫0te(t−s)AF(s)ds.
This identity holds for all ttt.

So:

: Every classical solution is automatically a mild solution.

==== 6. Why “a.e. equality” is enough in Navier–Stokes ====

In Navier–Stokes:

F(t)=−P∇⋅(u⊗u)(t),F(t)=-P\nabla\cdot(u\otimes u)(t),F(t)=−P∇⋅(u⊗u)(t),
and we only ever test the equation against smooth test functions or integrate in time.

All weak / mild / strong formulations rely on:
* u∈Lt∞Xu\in L^\infty_t Xu∈Lt∞X,
* ∂tu∈Lt1X\partial_t u\in L^1_t X∂tu∈Lt1X,
* equations holding almost everywhere.

Pointwise-in-time validity is not required and cannot be expected at low regularity.

==== 7. Summary (logical flow) ====
# Mild formulation uses only integrals → well-defined under F∈L1F\in L^1F∈L1.
# Banach-valued Newton–Leibniz gives time derivative a.e..
# Hence mild solutions satisfy the PDE a.e. in time.
# Classical solutions satisfy the PDE pointwise and automatically satisfy the mild formula.
# “Almost everywhere” is the natural and correct level of rigor for Navier–Stokes.

==== 8. Key takeaway (the misconception to avoid) ====

: You do not need Newton–Leibniz to hold everywhere.
: PDE theory only requires it to hold almost everywhere, and Banach-valued integration provides exactly that.

Nothing essential is lost.

If you want, I can next:
* write the full argument explicitly for Navier–Stokes with all operators,
* show exactly where additional regularity upgrades “a.e.” to pointwise,
* or show why assuming pointwise differentiability from the start is illegitimate.

Just say which.