98 CHAPTER 2. FUNDAMENTAL PRINCIPLES OF PHYSICS
Theorem 2.46(Energy Conservation).Let(u,v)be the solutions of (2.6.33), andHdoes not
explicitly contain time t. Then the energy H(u,v)is conserved, i.e. H(u,v)satisfies (2.6.27)
with q=u and p=v, ifΩ=R^3. In addition, ifΩ 6 =R^3 , then we have
(2.6.34)
dH
dt=
∫∂Ω[
∂H
∂ ξij∂uj
∂t+
∂H
∂ ζij∂vj
∂t]
nids,where n= (n 1 ,n 2 ,n 3 )is the unit outward normal at∂Ω.
Proof.For (2.6.32), we have
(2.6.35)
dH
dt=
∫Ω[
∂H
∂uk∂uk
∂t+
∂H
∂ ξij∂i(
∂uj
∂t)
+
∂H
∂vk∂vk
∂t+
∂H
∂ ζij∂i(
∂vj
∂t)]
dx.By the Gauss formula,
∫Ω∂H
∂ ξij∂i(
∂uj
∂t)
dx=∫∂Ω∂H
∂ ξij∂uj
∂tnids−∫Ω∂i(
∂H
∂ ξij)
∂uj
∂tdx,∂H
∂ ζij∂i(
∂vj
∂t)
dx=∫∂Ω∂H
∂ ζij∂vj
∂tnids−∫Ω∂i(
∂H
∂ ζij)
∂vj
∂tdx.Hence (2.6.35) is rewritten as
d
dtH=
∫Ω[(
∂H
∂uk
−∂i(
∂H
∂ ξik))
∂uk
∂t+
(
∂H
∂vk
−∂i(
∂H
∂ ζik))
∂vk
∂t]
(2.6.36) dx
+
∫∂Ω[
∂H
∂ ξij∂uj
∂t+
∂H
∂ ζij∂vj
∂t]
nids.Since(u,v)is a solution of (2.6.33), then (2.6.36) becomes
dH
dt=
∫∂Ω[
∂H
∂ ξij∂uj
∂t+
∂H
∂ ζij∂vj
∂t]
nids,and (2.6.34) follows.
IfΩ=R^3 (i.e.∂Ω=/0) oru=v=0 on∂Ω, then
d
dt
H(u,v) = 0 for(u,v)satisfy( 2. 6. 33 ).
The proof is complete.
Remark 2.47.The integral functions in (2.6.34):
(2.6.37)
P= (P 1 ,P 2 ,P 3 ),
Pk=−[
∂H
∂ ξk j∂uj
∂t+
∂H
∂ ζk j∂vj
∂t]
,
are the energy fluxes. Hence (2.6.34) can be expressed as
d
dtH(u,v) =−∫∂ΩP·nds,