4 L. D. Faddeev and O. A. Yakubovski˘ı
We find a differential equation that the function ft(q, p) satisfies.
To this end, we differentiate the identity fs+t(µ) = ft(Gsµ) with
respect to the variable s and set s = 0:
∂fs+t(µ)
∂s
s=0
=
∂ft(µ)
∂t
,
∂ft(Gsµ)
∂s
s=0
= ∇ft(µ) · v(µ) =
∂ft
∂q
∂H
∂p

∂ft
∂p
∂H
∂q
.
Thus, the function ft(q, p) satisfies the differential equation
(13)
∂ft
∂t
=
∂H
∂p
∂ft
∂q

∂H
∂q
∂ft
∂p
and the initial condition
(14) ft(q, p)|t=0 = f(q, p).
The equation (13) with the initial condition (14) has a unique solu-
tion, which can be obtained by the formula (12); that is, to construct
the solutions of (13) it suffices to know the solutions of the Hamil-
tonian equations.
We can rewrite (13) in the form
(15)
dft
dt
= {H, ft},
where {H, ft} is the Poisson bracket of the functions H and ft. For
arbitrary observables f and g the Poisson bracket is defined by
{f, g} =
∂f
∂p
∂g
∂q

∂f
∂q
∂g
∂p
,
and in the case of a system with n degrees of freedom
{f, g} =
n
i=1
∂f
∂pi
∂g
∂qi

∂f
∂qi
∂g
∂pi
.
We list the basic properties of Poisson brackets:
1) {f, g + λh} = {f, g} + λ{f, h} (linearity);
2) {f, g} = −{g, f} (skew symmetry);
3) {f,{g, h}} + {g,{h, f}} + {h,{f, g}} = 0 (Jacobi identity);
4) {f, gh} = g{f, h} + {f, g}h.
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