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

===== Define v(t):=G−tu(t)v(t):=G_{-t}u(t)v(t):=G−tu(t), i.e. =====

v(t):=e−νtΔu(t).v(t) := e^{-\nu t\Delta}u(t).v(t):=e−νtΔu(t).
Here G−tG_{-t}G−t is the inverse semigroup operator (well defined as a bounded operator on the range of GtG_tGt); to be precise, work with the strongly continuous semigroup GtG_tGt and use the identity below in integrated form rather than applying a formal inverse when not bounded — a cleaner route is the integral identity that follows.

Differentiate vvv in the weak (distributional) sense. Using the product rule in Bochner/distribution sense and the generator property of the semigroup, we obtain

ddtv(t)=ddt(e−νtΔu(t))=−νΔe−νtΔu(t)+e−νtΔ∂tu(t).\frac{d}{dt}v(t) = \frac{d}{dt}\big(e^{-\nu t\Delta}u(t)\big)
= -\nu\Delta e^{-\nu t\Delta}u(t) + e^{-\nu t\Delta}\partial_t u(t).dtdv(t)=dtd(e−νtΔu(t))=−νΔe−νtΔu(t)+e−νtΔ∂tu(t).
Substitute ∂tu(t)=νΔu(t)−F(t)\partial_t u(t)=\nu\Delta u(t)-F(t)∂tu(t)=νΔu(t)−F(t) (distributionally):

ddtv(t)=−νΔe−νtΔu(t)+e−νtΔ(νΔu(t)−F(t)).\frac{d}{dt}v(t) = -\nu\Delta e^{-\nu t\Delta}u(t) + e^{-\nu t\Delta}\big(\nu\Delta u(t)-F(t)\big).dtdv(t)=−νΔe−νtΔu(t)+e−νtΔ(νΔu(t)−F(t)).
The νΔ\nu\DeltaνΔ terms cancel (this is the algebraic magic of the conjugation), leaving

ddtv(t)=−e−νtΔF(t),\frac{d}{dt} v(t) = - e^{-\nu t\Delta} F(t),dtdv(t)=−e−νtΔF(t),
where the equality holds in the Bochner / distributional sense (the right-hand side is in L1(0,T;X)L^1(0,T;X)L1(0,T;X) because F∈L1(0,T;X)F\in L^1(0,T;X)F∈L1(0,T;X) and the semigroup is bounded on XXX).