# The derivative of a $C_0$-semigroup with respect to a perturbation parameter

Let $$H$$ be a Hilbert space, and $$A : H \to H$$ be the (semi-bounded) generator of the $$1$$-parameter $$C_0$$-semigroup $$[0, \infty) \ni t \mapsto \mathrm e ^{-t A}$$. Let $$B : H \to H$$ be a bounded operator, and consider the "perturbation" $$[0,1] \ni \varepsilon \mapsto A + \varepsilon B$$. I would like to use the formula

$$\mathrm e ^{-t (A + B)} - \mathrm e ^{-t A} = \int _0 ^1 \frac {\mathrm d} {\mathrm d \varepsilon} \ \mathrm e ^{-t (A + \varepsilon B)} \ \mathrm d \varepsilon$$

which I believe is true, but I do not know where to find. (In order to be true, it might be necessary to consider the formula in a strong sense, i.e. applied on some arbitrary $$v \in H$$.)

Davies' "One-parameter semigroups" doesn't have it. Please also notice that I am not interested in a proof, but only in a citable reference.

Furthermore, is it true that $$\frac {\mathrm d} {\mathrm d \varepsilon} \ \mathrm e ^{-t (A + \varepsilon B)} = -t \ \mathrm e ^{-t (A + \varepsilon B)} B$$ ?

• A standard reference is Semigroups of Linear Operators and Applications to Partial Differential Equations by Pazy. Jun 19, 2020 at 9:54
• The last statement is not true, unless $A$ and $B$ commute. Jun 19, 2020 at 11:39
• @MichaelRenardy: If it is not true, do you know the correct statement? Is it like the Duhamel formula: $\frac {\mathrm d} {\mathrm d \varepsilon} \ \mathrm e ^{-t (A + \varepsilon B)} = \int_0^1 \mathrm e ^{-s t (A + \varepsilon B)} (-t B) \mathrm e ^{-(1-s) t (A + \varepsilon B)} \ \mathrm d s$? Jun 19, 2020 at 12:25
• @AlexM. have a look at the Kato reference I gave. You can extract the answer from there. Jun 21, 2020 at 7:38

$$\mathrm e ^{-t(A+B)} v - \mathrm e ^{-tA} v = - \int _0 ^t \mathrm e ^{-(t-s) (A+B)} B \ \mathrm e ^{-s A} v \ \mathrm d s$$
for all $$v \in H$$.