Introduction to Mathematical Fluid Dynamics-II
Balance of Momentum
Meng Xu
Department of Mathematics
University of Wyoming
Bergische Universität Wuppertal
Math Fluid Dynamics-II
Fluid path
Consider the path followed by a fluid particle flows inside a
domain W .
x(t) = (x(t), y(t), z(t))
Then the velocity field becomes
u(x(t), y(t), z(t), t) = (ẋ(t), ẏ(t), ż(t))
or
u(x(t), t) =
Bergische Universität Wuppertal
dx
(t)
dt
Math Fluid Dynamics-II
Fluid path
Consider the path followed by a fluid particle flows inside a
domain W .
x(t) = (x(t), y(t), z(t))
Then the velocity field becomes
u(x(t), y(t), z(t), t) = (ẋ(t), ẏ(t), ż(t))
or
u(x(t), t) =
Bergische Universität Wuppertal
dx
(t)
dt
Math Fluid Dynamics-II
Fluid path
Consider the path followed by a fluid particle flows inside a
domain W .
x(t) = (x(t), y(t), z(t))
Then the velocity field becomes
u(x(t), y(t), z(t), t) = (ẋ(t), ẏ(t), ż(t))
or
u(x(t), t) =
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dx
(t)
dt
Math Fluid Dynamics-II
Acceleration of fluid particle
Another physical quantity in fluid mechanics is the acceleration
of the fluid particle
a(t) =
Denote ux =
∂u
∂x ,..
d
d2
x(t) = u(x(t), y(t), z(t))
2
dt
dt
∂u
∂u
∂u
∂u
=
ẋ +
ẏ +
ż +
∂x
∂y
∂z
∂t
ut =
∂u
∂t
and
u(x, y, z, t) = (u(x, y, z, t), v (x, y, z, t), w(x, y, z, t))
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Math Fluid Dynamics-II
Acceleration of fluid particle
Another physical quantity in fluid mechanics is the acceleration
of the fluid particle
a(t) =
Denote ux =
∂u
∂x ,..
d
d2
x(t) = u(x(t), y(t), z(t))
2
dt
dt
∂u
∂u
∂u
∂u
=
ẋ +
ẏ +
ż +
∂x
∂y
∂z
∂t
ut =
∂u
∂t
and
u(x, y, z, t) = (u(x, y, z, t), v (x, y, z, t), w(x, y, z, t))
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Math Fluid Dynamics-II
Material derivative
From the above notation, we can rewrite
a(t) = uux + v uy + wuz + ut
= ∂t u + u · ∇u
We will frequently use the operator
D
= ∂t + u · ∇
Dt
Operator (1) is called the material derivative.
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Math Fluid Dynamics-II
(1)
Material derivative
From the above notation, we can rewrite
a(t) = uux + v uy + wuz + ut
= ∂t u + u · ∇u
We will frequently use the operator
D
= ∂t + u · ∇
Dt
Operator (1) is called the material derivative.
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Math Fluid Dynamics-II
(1)
Ideal fluid
Ideal Fluid
For any motion of the fluid in a region W , there is a function
p(x, t) called the pressure, such that ∂W is a surface in the fluid
with a chosen unit normal n, the force of stress exerted across
the surface ∂W per unit area at x ∈ ∂W at time t is p(x, t)n.
Remark: The absence of tangential forces implies that there is
no rotation for fluid in W .
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Math Fluid Dynamics-II
Ideal fluid
Ideal Fluid
For any motion of the fluid in a region W , there is a function
p(x, t) called the pressure, such that ∂W is a surface in the fluid
with a chosen unit normal n, the force of stress exerted across
the surface ∂W per unit area at x ∈ ∂W at time t is p(x, t)n.
Remark: The absence of tangential forces implies that there is
no rotation for fluid in W .
Bergische Universität Wuppertal
Math Fluid Dynamics-II
Ideal fluid
Ideal Fluid
For any motion of the fluid in a region W , there is a function
p(x, t) called the pressure, such that ∂W is a surface in the fluid
with a chosen unit normal n, the force of stress exerted across
the surface ∂W per unit area at x ∈ ∂W at time t is p(x, t)n.
Remark: The absence of tangential forces implies that there is
no rotation for fluid in W .
Bergische Universität Wuppertal
Math Fluid Dynamics-II
Ideal fluid
Ideal Fluid
For any motion of the fluid in a region W , there is a function
p(x, t) called the pressure, such that ∂W is a surface in the fluid
with a chosen unit normal n, the force of stress exerted across
the surface ∂W per unit area at x ∈ ∂W at time t is p(x, t)n.
Remark: The absence of tangential forces implies that there is
no rotation for fluid in W .
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Math Fluid Dynamics-II
Force on the boundary
For ideal fluid, the total force on the fluid inside W by means of
stress on its boundary is
Z
S∂W = {force on W } = −
pndA
∂W
For any fixed vector e, divergence theorem gives us
Z
e · S∂W = −
pe · ndA
∂W
Z
=−
div(pe)dV
ZW
=−
(gradp) · edV
W
Hence
Z
S∂W = −
gradpdV
W
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Math Fluid Dynamics-II
Force on the boundary
For ideal fluid, the total force on the fluid inside W by means of
stress on its boundary is
Z
S∂W = {force on W } = −
pndA
∂W
For any fixed vector e, divergence theorem gives us
Z
e · S∂W = −
pe · ndA
∂W
Z
=−
div(pe)dV
ZW
=−
(gradp) · edV
W
Hence
Z
S∂W = −
gradpdV
W
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Math Fluid Dynamics-II
Force on the boundary
For ideal fluid, the total force on the fluid inside W by means of
stress on its boundary is
Z
S∂W = {force on W } = −
pndA
∂W
For any fixed vector e, divergence theorem gives us
Z
e · S∂W = −
pe · ndA
∂W
Z
=−
div(pe)dV
ZW
=−
(gradp) · edV
W
Hence
Z
S∂W = −
gradpdV
W
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Math Fluid Dynamics-II
Balance of momentum
Denote b(x, t) as the given body force per unit mass, then the
total body force is
Z
B=
ρbdV
W
In all, force per unit volume is equal to
−gradp + ρb
Balance of Momentum(Differential Form)
By the principle of momentum balance (Newton’s second law),
ρ
Du
= −gradp + ρb
Dt
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Math Fluid Dynamics-II
Balance of momentum
Denote b(x, t) as the given body force per unit mass, then the
total body force is
Z
B=
ρbdV
W
In all, force per unit volume is equal to
−gradp + ρb
Balance of Momentum(Differential Form)
By the principle of momentum balance (Newton’s second law),
ρ
Du
= −gradp + ρb
Dt
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Math Fluid Dynamics-II
Balance of momentum
Denote b(x, t) as the given body force per unit mass, then the
total body force is
Z
B=
ρbdV
W
In all, force per unit volume is equal to
−gradp + ρb
Balance of Momentum(Differential Form)
By the principle of momentum balance (Newton’s second law),
ρ
Du
= −gradp + ρb
Dt
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Math Fluid Dynamics-II
Integral form
An integral form of the balance of momentum can be derived
for general fluid:
Balance of Momentum(Integral Form)
By the principle of momentum balance,
Z
Z
d
ρudV = S∂Wt +
ρbdV
dt Wt
Wt
Here Wt is a region at time t and S∂Wt represents the total force
exerted on the surface ∂Wt .
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Math Fluid Dynamics-II
Flow map
Write ϕ(x, t) as the trajectory followed by the particle at point x
and time t. Assume the flow is smooth enough. Then we can
define a mapping
ϕt : x 7→ ϕ(x, t)
Given a region W ⊂ D, ϕt (W ) = Wt is the volume W at time t.
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Math Fluid Dynamics-II
Flow map
Write ϕ(x, t) as the trajectory followed by the particle at point x
and time t. Assume the flow is smooth enough. Then we can
define a mapping
ϕt : x 7→ ϕ(x, t)
Given a region W ⊂ D, ϕt (W ) = Wt is the volume W at time t.
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Math Fluid Dynamics-II
First lemma
The first lemma before we continue is the following
Lemma 1
Define J(x, t) as the Jacobian determinant of the map ϕt , we
have
∂
J(x, t) = J(x, t) [divu(ϕ(x, t), t)]
∂t
We give a sketch of proof for this lemma.
Write the components of ϕ as ξ(x, t),η(x, t) and ζ(x, t). Then its
Jacobian determinant can be written as


J(x, t) =
∂ξ
∂x
 ∂ξ
 ∂y
∂ξ
∂z
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∂η
∂x
∂η
∂y
∂η
∂z
∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z
Math Fluid Dynamics-II
First lemma
The first lemma before we continue is the following
Lemma 1
Define J(x, t) as the Jacobian determinant of the map ϕt , we
have
∂
J(x, t) = J(x, t) [divu(ϕ(x, t), t)]
∂t
We give a sketch of proof for this lemma.
Write the components of ϕ as ξ(x, t),η(x, t) and ζ(x, t). Then its
Jacobian determinant can be written as


J(x, t) =
∂ξ
∂x
 ∂ξ
 ∂y
∂ξ
∂z
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∂η
∂x
∂η
∂y
∂η
∂z
∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z
Math Fluid Dynamics-II
First lemma
The first lemma before we continue is the following
Lemma 1
Define J(x, t) as the Jacobian determinant of the map ϕt , we
have
∂
J(x, t) = J(x, t) [divu(ϕ(x, t), t)]
∂t
We give a sketch of proof for this lemma.
Write the components of ϕ as ξ(x, t),η(x, t) and ζ(x, t). Then its
Jacobian determinant can be written as


J(x, t) =
∂ξ
∂x
 ∂ξ
 ∂y
∂ξ
∂z
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∂η
∂x
∂η
∂y
∂η
∂z
∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z
Math Fluid Dynamics-II
First lemma
The first lemma before we continue is the following
Lemma 1
Define J(x, t) as the Jacobian determinant of the map ϕt , we
have
∂
J(x, t) = J(x, t) [divu(ϕ(x, t), t)]
∂t
We give a sketch of proof for this lemma.
Write the components of ϕ as ξ(x, t),η(x, t) and ζ(x, t). Then its
Jacobian determinant can be written as


J(x, t) =
∂ξ
∂x
 ∂ξ
 ∂y
∂ξ
∂z
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∂η
∂x
∂η
∂y
∂η
∂z
∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z
Math Fluid Dynamics-II
Proof of lemma 1
For fixed x,

∂
J=
∂t
=
∂ξ
∂  ∂x
∂ξ

∂t ∂y
∂ξ
∂z

∂ ∂ξ
∂t ∂x
 ∂ ∂ξ
 ∂t ∂y
∂ ∂ξ
∂t ∂z
∂η
∂x
∂η
∂y
∂η
∂z
∂η
∂x
∂η
∂y
∂η
∂z
+

∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z

∂ζ
∂x
∂ζ 
∂y  +
∂ζ
 ∂z
∂η
∂ξ
∂x
 ∂x
∂ξ
∂η
 ∂y ∂y
∂ξ
∂η
∂z
∂z
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
∂ξ
∂
∂x
∂t
 ∂ξ ∂
 ∂y ∂t
∂ξ
∂
∂z
∂t

∂ ∂ζ
∂t ∂x
∂ ∂ζ 
∂t ∂y 
∂ ∂ζ
∂t ∂z
∂η
∂x
∂η
∂y
∂η
∂z
Math Fluid Dynamics-II

∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z
By definition of the velocity field
∂
ϕ(x, t) = u(ϕ(x, t), t)
∂t
Thus
∂ ∂ξ
∂ ∂ξ
∂
=
=
u(ϕ(x, t), t)
∂t ∂x
∂x ∂t
∂x
∂ ∂ξ
∂
∂ ∂ξ
=
=
u(ϕ(x, t), t)
∂t ∂y
∂y ∂t
∂y
.................................
∂ ∂ζ
∂ ∂ζ
∂
=
=
w(ϕ(x, t), t)
∂t ∂z
∂z ∂t
∂z
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Math Fluid Dynamics-II
By definition of the velocity field
∂
ϕ(x, t) = u(ϕ(x, t), t)
∂t
Thus
∂ ∂ξ
∂ ∂ξ
∂
=
=
u(ϕ(x, t), t)
∂t ∂x
∂x ∂t
∂x
∂ ∂ξ
∂
∂ ∂ξ
=
=
u(ϕ(x, t), t)
∂t ∂y
∂y ∂t
∂y
.................................
∂ ∂ζ
∂ ∂ζ
∂
=
=
w(ϕ(x, t), t)
∂t ∂z
∂z ∂t
∂z
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Math Fluid Dynamics-II
Moreover
∂
∂u ∂ξ
∂u ∂η ∂u ∂ζ
u(ϕ(x, t), t) =
+
+
,
∂x
∂ξ ∂x ∂η ∂x ∂ζ ∂x
.................................
∂w ∂ξ ∂w ∂η ∂w ∂ζ
∂
w(ϕ(x, t), t) =
+
+
,
∂z
∂ξ ∂z
∂η ∂z
∂ζ ∂z
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Math Fluid Dynamics-II
Now plug these expressions into
 ∂u ∂ξ
∂
J=
∂t
=
=
∂ξ ∂x
 ∂u ∂ξ
 ∂ξ ∂y
∂u ∂ξ
∂ξ ∂z
 ∂u ∂ξ
∂ξ ∂x
 ∂u ∂ξ
 ∂ξ ∂y
∂u ∂ξ
∂ξ ∂z
+
+
+
∂u
∂η
∂u
∂η
∂u
∂η
∂η
∂x
∂η
∂y
∂η
∂z
∂
∂t J,
∂η
∂η
∂u ∂ζ
∂x + ∂ζ ∂x
∂x
∂η
∂η
∂u ∂ζ
∂y + ∂ζ ∂y
∂y
∂η
∂η
∂u ∂ζ
+
∂z
∂ζ ∂z
∂z
  ∂ξ
∂ζ
∂x
∂x
 ∂ξ
∂ζ 
+
 ∂y
∂y 
∂ζ
∂ξ
∂z
∂z
we get
∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z
∂v ∂η
∂η ∂x
∂v ∂η
∂η ∂y
∂v ∂η
∂η ∂z

+ ......
∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z
 ∂ξ

+
∂u
∂v
∂w
J+
J+
J = [divu(ϕ(x, t), t)] J
∂ξ
∂η
∂ζ
The proof is complete.
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Math Fluid Dynamics-II
∂x
 ∂ξ
 ∂y
∂ξ
∂z
∂η
∂x
∂η
∂y
∂η
∂z
∂w
∂ζ
∂w
∂ζ
∂w
∂ζ
∂ζ
∂x
∂ζ 
∂y 
∂ζ
∂z

Second lemma
Lemma 2
Given a scalar or vector function f (x, t), we have
Z
Z
d
∂f
+ div(f u) dV
f (x, t)dV = Wt
dt Wt
∂t
(2)
A similar result can be proved and is called the transport
theorem.
Transport Theorem
d
dt
Z
Z
ρudV =
Wt
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ρ
Wt
Du
dV
Dt
Math Fluid Dynamics-II
(3)
Second lemma
Lemma 2
Given a scalar or vector function f (x, t), we have
Z
Z
d
∂f
+ div(f u) dV
f (x, t)dV = Wt
dt Wt
∂t
(2)
A similar result can be proved and is called the transport
theorem.
Transport Theorem
d
dt
Z
Z
ρudV =
Wt
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ρ
Wt
Du
dV
Dt
Math Fluid Dynamics-II
(3)
Second lemma
Lemma 2
Given a scalar or vector function f (x, t), we have
Z
Z
d
∂f
+ div(f u) dV
f (x, t)dV = Wt
dt Wt
∂t
(2)
A similar result can be proved and is called the transport
theorem.
Transport Theorem
d
dt
Z
Z
ρudV =
Wt
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ρ
Wt
Du
dV
Dt
Math Fluid Dynamics-II
(3)
Proof of Lemma 2
Let us prove (2) first.
By change of variables formula and the first lemma
Z
d
LHS =
f (ϕ(x, t), t)J(x, t)dV
dt
Z W
df
∂J
=
(ϕ(x, t), t)J + f (ϕ(x, t), t)
dV
∂t
W dt
Z Df
=
(ϕ(x, t), t) + divuf JdV
W Dt
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Proof of Lemma 2
Let us prove (2) first.
By change of variables formula and the first lemma
Z
d
LHS =
f (ϕ(x, t), t)J(x, t)dV
dt
Z W
df
∂J
=
(ϕ(x, t), t)J + f (ϕ(x, t), t)
dV
∂t
W dt
Z Df
=
(ϕ(x, t), t) + divuf JdV
W Dt
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Z
=
Wt
Z
=
Wt
Z
=
Wt
Df
+ divuf dV
Dt
∂f
+ uf + divuf dV
∂t
∂f
+ div(f u) dV
∂t
Thus (2) is proved.
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To prove (3), we first observe that
d
D
(ρu)(ϕ(x, t), t) =
(ρu)(ϕ(x, t), t)
dt
Dt
This is because the time derivative takes into account the fact
that the fluid is moving and that the positions of fluid particles
change with time. So, if f (x, y, z, t) is any function of position
and time, then by the chain rule
d
f (x(t), y(t), z(t), t)
dt
= ∂t f + u · ∇f
Df
=
(x(t), y(t), z(t), t)
Dt
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Math Fluid Dynamics-II
To prove (3), we first observe that
d
D
(ρu)(ϕ(x, t), t) =
(ρu)(ϕ(x, t), t)
dt
Dt
This is because the time derivative takes into account the fact
that the fluid is moving and that the positions of fluid particles
change with time. So, if f (x, y, z, t) is any function of position
and time, then by the chain rule
d
f (x(t), y(t), z(t), t)
dt
= ∂t f + u · ∇f
Df
=
(x(t), y(t), z(t), t)
Dt
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Math Fluid Dynamics-II
To prove (3), we first observe that
d
D
(ρu)(ϕ(x, t), t) =
(ρu)(ϕ(x, t), t)
dt
Dt
This is because the time derivative takes into account the fact
that the fluid is moving and that the positions of fluid particles
change with time. So, if f (x, y, z, t) is any function of position
and time, then by the chain rule
d
f (x(t), y(t), z(t), t)
dt
= ∂t f + u · ∇f
Df
=
(x(t), y(t), z(t), t)
Dt
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Math Fluid Dynamics-II
Using Lemma 1, we have
Z
Z
Z
d
d
d
[(ρu)J]dV
ρudV =
(ρu)JdV =
dt Wt
dt W
dt
W
Z
∂
D
(ρu)(ϕ(x, t), t)J + (ρu)(ϕ(x, t), t) J(x, t)dV
=
Dt
∂t
ZW D
=
(ρu) + (ρdivu)u JdV
Dt
W
By the conservation of mass
Dρ
∂ρ
+ ρdivu =
+ div(ρu) = 0
Dt
∂t
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Using Lemma 1, we have
Z
Z
Z
d
d
d
[(ρu)J]dV
ρudV =
(ρu)JdV =
dt Wt
dt W
dt
W
Z
∂
D
(ρu)(ϕ(x, t), t)J + (ρu)(ϕ(x, t), t) J(x, t)dV
=
Dt
∂t
ZW D
=
(ρu) + (ρdivu)u JdV
Dt
W
By the conservation of mass
Dρ
∂ρ
+ ρdivu =
+ div(ρu) = 0
Dt
∂t
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Math Fluid Dynamics-II
Thus
d
dt
Z
Z
ρudV =
Wt
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ρ
Wt
Du
dV
Dt
Math Fluid Dynamics-II
Incompressible fluid
Definition
We call a flow incompressible if for any fluid subregion W ,
Z
volume(Wt ) =
dV = constant in t
Wt
From the first lemma, we know
Z
Z
d
d
0=
dV =
JdV
dt Wt
dt W
Z
Z
=
(divu)JdV =
(divu)dV
W
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Wt
Math Fluid Dynamics-II
Incompressible fluid
Definition
We call a flow incompressible if for any fluid subregion W ,
Z
volume(Wt ) =
dV = constant in t
Wt
From the first lemma, we know
Z
Z
d
d
0=
dV =
JdV
dt Wt
dt W
Z
Z
=
(divu)JdV =
(divu)dV
W
Bergische Universität Wuppertal
Wt
Math Fluid Dynamics-II
Incompressible fluid
Definition
We call a flow incompressible if for any fluid subregion W ,
Z
volume(Wt ) =
dV = constant in t
Wt
From the first lemma, we know
Z
Z
d
d
0=
dV =
JdV
dt Wt
dt W
Z
Z
=
(divu)JdV =
(divu)dV
W
Bergische Universität Wuppertal
Wt
Math Fluid Dynamics-II
The following statements are equivalent:
the fluid is incompressible.
divu = 0
J≡1
Previous slide shows that the first and second statements are
equivalent. To show J ≡ 1 for incompressible fluid, recall the
first lemma and divergence free condition,
Z
Z
Z
dV = C =
JdV = J
dV
Wt
W
W
Since the volume of Wt remains the same, we get
J≡1
Bergische Universität Wuppertal
Math Fluid Dynamics-II
The following statements are equivalent:
the fluid is incompressible.
divu = 0
J≡1
Previous slide shows that the first and second statements are
equivalent. To show J ≡ 1 for incompressible fluid, recall the
first lemma and divergence free condition,
Z
Z
Z
dV = C =
JdV = J
dV
Wt
W
W
Since the volume of Wt remains the same, we get
J≡1
Bergische Universität Wuppertal
Math Fluid Dynamics-II
The following statements are equivalent:
the fluid is incompressible.
divu = 0
J≡1
Previous slide shows that the first and second statements are
equivalent. To show J ≡ 1 for incompressible fluid, recall the
first lemma and divergence free condition,
Z
Z
Z
dV = C =
JdV = J
dV
Wt
W
W
Since the volume of Wt remains the same, we get
J≡1
Bergische Universität Wuppertal
Math Fluid Dynamics-II
The following statements are equivalent:
the fluid is incompressible.
divu = 0
J≡1
Previous slide shows that the first and second statements are
equivalent. To show J ≡ 1 for incompressible fluid, recall the
first lemma and divergence free condition,
Z
Z
Z
dV = C =
JdV = J
dV
Wt
W
W
Since the volume of Wt remains the same, we get
J≡1
Bergische Universität Wuppertal
Math Fluid Dynamics-II
The following statements are equivalent:
the fluid is incompressible.
divu = 0
J≡1
Previous slide shows that the first and second statements are
equivalent. To show J ≡ 1 for incompressible fluid, recall the
first lemma and divergence free condition,
Z
Z
Z
dV = C =
JdV = J
dV
Wt
W
W
Since the volume of Wt remains the same, we get
J≡1
Bergische Universität Wuppertal
Math Fluid Dynamics-II
The following statements are equivalent:
the fluid is incompressible.
divu = 0
J≡1
Previous slide shows that the first and second statements are
equivalent. To show J ≡ 1 for incompressible fluid, recall the
first lemma and divergence free condition,
Z
Z
Z
dV = C =
JdV = J
dV
Wt
W
W
Since the volume of Wt remains the same, we get
J≡1
Bergische Universität Wuppertal
Math Fluid Dynamics-II
The following statements are equivalent:
the fluid is incompressible.
divu = 0
J≡1
Previous slide shows that the first and second statements are
equivalent. To show J ≡ 1 for incompressible fluid, recall the
first lemma and divergence free condition,
Z
Z
Z
dV = C =
JdV = J
dV
Wt
W
W
Since the volume of Wt remains the same, we get
J≡1
Bergische Universität Wuppertal
Math Fluid Dynamics-II
Continuity equation for incompressible fluid
Recall the continuity equation
Dρ
+ ρdivu = 0
Dt
For incompressible fluid, it reduces to
Dρ
=0
Dt
Bergische Universität Wuppertal
Math Fluid Dynamics-II
Continuity equation for incompressible fluid
Recall the continuity equation
Dρ
+ ρdivu = 0
Dt
For incompressible fluid, it reduces to
Dρ
=0
Dt
Bergische Universität Wuppertal
Math Fluid Dynamics-II