Mathematical Principles of Theoretical Physics

(Rick Simeone) #1

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

=




[


∂H


∂uk

∂uk
∂t

+


∂H


∂ ξij

∂i

(


∂uj
∂t

)


+


∂H


∂vk

∂vk
∂t

+


∂H


∂ ζij

∂i

(


∂vj
∂t

)]


dx.

By the Gauss formula,




∂H


∂ ξij

∂i

(


∂uj
∂t

)


dx=


∂Ω

∂H


∂ ξij

∂uj
∂t

nids−



∂i

(


∂H


∂ ξij

)


∂uj
∂t

dx,

∂H
∂ ζij

∂i

(


∂vj
∂t

)


dx=


∂Ω

∂H


∂ ζij

∂vj
∂t

nids−



∂i

(


∂H


∂ ζij

)


∂vj
∂t

dx.

Hence (2.6.35) is rewritten as


d
dt

H=




[(


∂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
dt

H(u,v) =−


∂Ω

P·nds,
Free download pdf