Statement of the balance of energy [ edit | edit source ] The balance of energy can be expressed as:
ρ
e
˙
−
σ
:
(
∇
v
)
+
∇
∙
q
−
ρ
s
=
0
{\displaystyle \rho ~{\dot {e}}-{\boldsymbol {\sigma }}:({\boldsymbol {\nabla }}\mathbf {v} )+{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0}
where
ρ
(
x
,
t
)
{\displaystyle \rho (\mathbf {x} ,t)}
e
(
x
,
t
)
{\displaystyle e(\mathbf {x} ,t)}
σ
(
x
,
t
)
{\displaystyle {\boldsymbol {\sigma }}(\mathbf {x} ,t)}
v
(
x
,
t
)
{\displaystyle \mathbf {v} (\mathbf {x} ,t)}
q
{\displaystyle \mathbf {q} }
s
{\displaystyle s}
Recall the general balance equation
d
d
t
[
∫
Ω
f
(
x
,
t
)
dV
]
=
∫
∂
Ω
f
(
x
,
t
)
[
u
n
(
x
,
t
)
−
v
(
x
,
t
)
⋅
n
(
x
,
t
)
]
dA
+
∫
∂
Ω
g
(
x
,
t
)
dA
+
∫
Ω
h
(
x
,
t
)
dV
.
{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }f(\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\partial {\Omega }}f(\mathbf {x} ,t)[u_{n}(\mathbf {x} ,t)-\mathbf {v} (\mathbf {x} ,t)\cdot \mathbf {n} (\mathbf {x} ,t)]~{\text{dA}}+\int _{\partial {\Omega }}g(\mathbf {x} ,t)~{\text{dA}}+\int _{\Omega }h(\mathbf {x} ,t)~{\text{dV}}~.}
In this case, the physical quantity to be conserved the total energy density
which is the sum of the internal energy density and the kinetic energy
density, i.e.,
f
=
ρ
e
+
1
/
2
ρ
|
v
⋅
v
|
{\displaystyle f=\rho ~e+1/2~\rho ~|\mathbf {v} \cdot \mathbf {v} |}
g
=
v
⋅
t
−
q
⋅
n
{\displaystyle g=\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} }
n
{\displaystyle \mathbf {n} }
h
=
v
⋅
(
ρ
b
)
+
ρ
s
{\displaystyle h=\mathbf {v} \cdot (\rho \mathbf {b} )+\rho ~s}
Hence we have
d
d
t
[
∫
Ω
ρ
(
e
+
1
2
v
⋅
v
)
dV
]
=
∫
∂
Ω
ρ
(
e
+
1
2
v
⋅
v
)
(
u
n
−
v
⋅
n
)
dA
+
∫
∂
Ω
(
v
⋅
t
−
q
⋅
n
)
dA
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\text{dV}}\right]=\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(u_{n}-\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}(\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
Let
Ω
{\displaystyle \Omega }
∫
Ω
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
dV
=
−
∫
∂
Ω
ρ
(
e
+
1
2
v
⋅
v
)
(
v
⋅
n
)
dA
+
∫
∂
Ω
(
v
⋅
t
−
q
⋅
n
)
dA
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}(\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
Using the relation
t
=
σ
⋅
n
{\displaystyle \mathbf {t} ={\boldsymbol {\sigma }}\cdot \mathbf {n} }
v
⋅
(
σ
⋅
n
)
=
(
σ
T
⋅
v
)
⋅
n
{\displaystyle \mathbf {v} \cdot ({\boldsymbol {\sigma }}\cdot \mathbf {n} )=({\boldsymbol {\sigma }}^{T}\cdot \mathbf {v} )\cdot \mathbf {n} }
∫
Ω
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
dV
=
−
∫
∂
Ω
ρ
(
e
+
1
2
v
⋅
v
)
(
v
⋅
n
)
dA
+
∫
∂
Ω
(
σ
⋅
v
−
q
)
⋅
n
dA
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}({\boldsymbol {\sigma }}\cdot \mathbf {v} -\mathbf {q} )\cdot \mathbf {n} ~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
We now apply the divergence theorem to the surface integrals to get
∫
Ω
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
dV
=
−
∫
Ω
∇
∙
[
ρ
(
e
+
1
2
v
⋅
v
)
v
]
dV
+
∫
Ω
∇
∙
(
σ
⋅
v
)
dV
−
∫
Ω
∇
∙
q
dV
+
∫
Ω
ρ
(
v
⋅
b
+
s
)
dV
.
{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\Omega }{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]~{\text{dV}}+\int _{\Omega }{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )~{\text{dV}}-\int _{\Omega }{\boldsymbol {\nabla }}\bullet \mathbf {q} ~{\text{dV}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}
Since
Ω
{\displaystyle \Omega }
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
=
−
∇
∙
[
ρ
(
e
+
1
2
v
⋅
v
)
v
]
+
∇
∙
(
σ
⋅
v
)
−
∇
∙
q
+
ρ
(
v
⋅
b
+
s
)
.
{\displaystyle {\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]=-{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]+{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )-{\boldsymbol {\nabla }}\bullet \mathbf {q} +\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~.}
Expanding out the left hand side, we have
∂
∂
t
[
ρ
(
e
+
1
2
v
⋅
v
)
]
=
∂
ρ
∂
t
(
e
+
1
2
v
⋅
v
)
+
ρ
(
∂
e
∂
t
+
1
2
∂
∂
t
(
v
⋅
v
)
)
=
∂
ρ
∂
t
(
e
+
1
2
v
⋅
v
)
+
ρ
∂
e
∂
t
+
ρ
∂
v
∂
t
⋅
v
.
{\displaystyle {\begin{aligned}{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]&={\frac {\partial \rho }{\partial t}}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\rho ~\left({\frac {\partial e}{\partial t}}+{\frac {1}{2}}~{\frac {\partial }{\partial t}}(\mathbf {v} \cdot \mathbf {v} )\right)\\&={\frac {\partial \rho }{\partial t}}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\rho ~{\frac {\partial e}{\partial t}}+\rho ~{\frac {\partial \mathbf {v} }{\partial t}}\cdot \mathbf {v} ~.\end{aligned}}}
For the first term on the right hand side, we use the identity
∇
∙
(
φ
v
)
=
φ
∇
∙
v
+
∇
φ
⋅
v
{\displaystyle {\boldsymbol {\nabla }}\bullet (\varphi ~\mathbf {v} )=\varphi ~{\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\varphi \cdot \mathbf {v} }
∇
∙
[
ρ
(
e
+
1
2
v
⋅
v
)
v
]
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
∇
[
ρ
(
e
+
1
2
v
⋅
v
)
]
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
(
e
+
1
2
v
⋅
v
)
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
e
⋅
v
+
1
2
ρ
∇
(
v
⋅
v
)
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
e
⋅
v
+
ρ
(
∇
v
T
⋅
v
)
⋅
v
=
ρ
(
e
+
1
2
v
⋅
v
)
∇
∙
v
+
(
e
+
1
2
v
⋅
v
)
∇
ρ
⋅
v
+
ρ
∇
e
⋅
v
+
ρ
(
∇
v
⋅
v
)
⋅
v
.
{\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+{\frac {1}{2}}~\rho ~{\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {v} )\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+\rho ~({\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {v} )\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+\rho ~({\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )\cdot \mathbf {v} ~.\end{aligned}}}
For the second term on the right we use the identity
∇
∙
(
S
T
⋅
v
)
=
S
:
∇
v
+
(
∇
∙
S
)
⋅
v
{\displaystyle {\boldsymbol {\nabla }}\bullet ({\boldsymbol {S}}^{T}\cdot \mathbf {v} )={\boldsymbol {S}}:{\boldsymbol {\nabla }}\mathbf {v} +({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}})\cdot \mathbf {v} }
∇
∙
(
σ
⋅
v
)
=
σ
:
∇
v
+
(
∇
∙
σ
)
⋅
v
.
{\displaystyle {\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )={\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +({\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }})\cdot \mathbf {v} ~.}
After collecting terms and rearranging, we get
(
∂
ρ
∂
t
+
ρ
∇
∙
v
+
∇
ρ
⋅
v
)
(
e
+
1
2
v
⋅
v
)
+
(
ρ
∂
v
∂
t
+
ρ
∇
v
⋅
v
−
∇
∙
σ
−
ρ
b
)
⋅
v
+
ρ
(
∂
e
∂
t
+
∇
e
⋅
v
)
+
−
σ
:
∇
v
+
∇
∙
q
−
ρ
s
=
0
.
{\displaystyle \left({\frac {\partial \rho }{\partial t}}+\rho ~{\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\rho \cdot \mathbf {v} \right)\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\left(\rho ~{\frac {\partial \mathbf {v} }{\partial t}}+\rho ~{\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} -{\boldsymbol {\nabla }}\bullet {\boldsymbol {\sigma }}-\rho ~\mathbf {b} \right)\cdot \mathbf {v} +\rho ~\left({\frac {\partial e}{\partial t}}+{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }\right)+-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0~.}
Applying the balance of mass to the first term and the balance of linear
momentum to the second term, and using the material time derivative of
the internal energy
e
˙
=
∂
e
∂
t
+
∇
e
⋅
v
{\displaystyle {\dot {e}}={\frac {\partial e}{\partial t}}+{\boldsymbol {\nabla }}e\cdot \mathbf {v} }
we get the final form of the balance of energy:
ρ
e
˙
−
σ
:
∇
v
+
∇
∙
q
−
ρ
s
=
0
.
{\displaystyle {\rho ~{\dot {e}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} +{\boldsymbol {\nabla }}\bullet \mathbf {q} -\rho ~s=0~.}}
![{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }f(\mathbf {x} ,t)~{\text{dV}}\right]=\int _{\partial {\Omega }}f(\mathbf {x} ,t)[u_{n}(\mathbf {x} ,t)-\mathbf {v} (\mathbf {x} ,t)\cdot \mathbf {n} (\mathbf {x} ,t)]~{\text{dA}}+\int _{\partial {\Omega }}g(\mathbf {x} ,t)~{\text{dA}}+\int _{\Omega }h(\mathbf {x} ,t)~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fcb120ab19e9001d85370ccb115c509cec72cf33)
![{\displaystyle {\cfrac {d}{dt}}\left[\int _{\Omega }\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\text{dV}}\right]=\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(u_{n}-\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}(\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47d1dcadabc2275df387b8c7cff8aedeae28ac73)
![{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}(\mathbf {v} \cdot \mathbf {t} -\mathbf {q} \cdot \mathbf {n} )~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba847b01545ab9e45d01b76fa692c2c3a55412fc)
![{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\partial {\Omega }}\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)(\mathbf {v} \cdot \mathbf {n} )~{\text{dA}}+\int _{\partial {\Omega }}({\boldsymbol {\sigma }}\cdot \mathbf {v} -\mathbf {q} )\cdot \mathbf {n} ~{\text{dA}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66ac123c32c10ef05d3feaabf166083fefb6797b)
![{\displaystyle \int _{\Omega }{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]~{\text{dV}}=-\int _{\Omega }{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]~{\text{dV}}+\int _{\Omega }{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )~{\text{dV}}-\int _{\Omega }{\boldsymbol {\nabla }}\bullet \mathbf {q} ~{\text{dV}}+\int _{\Omega }\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~{\text{dV}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/233958693d5b0c3bbf83c18f8157b098efd7308b)
![{\displaystyle {\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]=-{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]+{\boldsymbol {\nabla }}\bullet ({\boldsymbol {\sigma }}\cdot \mathbf {v} )-{\boldsymbol {\nabla }}\bullet \mathbf {q} +\rho ~(\mathbf {v} \cdot \mathbf {b} +s)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bc8199e3c8a61be4e93fe0d53e1be2d944b3d0f)
![{\displaystyle {\begin{aligned}{\frac {\partial }{\partial t}}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]&={\frac {\partial \rho }{\partial t}}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\rho ~\left({\frac {\partial e}{\partial t}}+{\frac {1}{2}}~{\frac {\partial }{\partial t}}(\mathbf {v} \cdot \mathbf {v} )\right)\\&={\frac {\partial \rho }{\partial t}}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)+\rho ~{\frac {\partial e}{\partial t}}+\rho ~{\frac {\partial \mathbf {v} }{\partial t}}\cdot \mathbf {v} ~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48a151f052ce4eea10255c868dad85a8f571b587)
![{\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet \left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\mathbf {v} \right]&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +{\boldsymbol {\nabla }}\left[\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\right]\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+{\frac {1}{2}}~\rho ~{\boldsymbol {\nabla }}(\mathbf {v} \cdot \mathbf {v} )\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+\rho ~({\boldsymbol {\nabla }}\mathbf {v} ^{T}\cdot \mathbf {v} )\cdot \mathbf {v} \\&=\rho ~\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right)~{\boldsymbol {\nabla }}\bullet \mathbf {v} +\left(e+{\frac {1}{2}}~\mathbf {v} \cdot \mathbf {v} \right){\boldsymbol {\nabla }}\rho \cdot \mathbf {v} +\rho ~{\boldsymbol {\nabla }}e\cdot {\mathbf {v} }+\rho ~({\boldsymbol {\nabla }}\mathbf {v} \cdot \mathbf {v} )\cdot \mathbf {v} ~.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ed8aa7cabb7f64b6991ca999f9dd47b5ad6765a)
