NPTEL – Mechanical Engineering – Continuum Mechanics
Module-3: Kinematics
Lecture-22: The Reynolds’ Transport Theorem
Here, we discuss the transformation between material and spatial integrals. We also
present the famous Reynolds’ transport theorem which plays a crucial role in the derivation of differential form of governing equation from the integral form of balance laws.
Transformation of integrals from spatial to material and vice versa is equivalent to
the change of variable in the case of one-variable calculus. Therefore, we first present an
example to understand the change of variable in the integration.
Example 1. Let {e1 , e2 } be orthonormal basis to the two dimensional vector space and
x = (x1 , x2 ) be a position vector. Let g(x) = (−x1 , x2 ) be a given vector field. Let C be a
quarter circle with unit radius in the first quadrant of plane as shown in Fig. (1). Then
show that the line integral
Z
g(x) · dx = −1,
C
on given curve of quarter circular arch using direct method and also change of variables.
Consider the starting and ending points of the curve as (0, 1) and (1, 0) (see Fig. (1)).
x2
(0,1)
x1
(0,0)
(1,0)
Figure 1: A quarter circle in 2-D space
Solution: We first present the direct solution. Since the given curve x21 + x22 = 1, taking
the differentiation, we get
x2 dx2 = −x1 dx1 .
Given vector valued function g(x) = (−x1 , x2 ) with (0, 1) and (1, 0) as starting and ending
points, respectively. Therefore, the integration along the curve
Z
g(x) · dx =
C
=
Z
(−x1 dx1 + x2 dx2 )
C
Z 1
0
=
h
−2x1 dx1
−x21
(Since x2 dx2 = −x1 dx1 )
i1
0
= −1
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We now use change of variables to evaluate the integration. Let x1 = cos ξ and x2 = sin ξ.
Then x1 and x2 satisfy the curve x21 + x22 = 1 automatically. The parameter ξ = π/2 and
ξ = 0 represent the starting and ending points of the curve as (x1 , x2 ) = (0, 1) at ξ = π/2
and (x1 , x2 ) = (1, 0) at ξ = 0. Since both x1 and x2 are functions of ξ, we can have
dx1 = − sin ξ and dx2 = cos ξ. Using change of variables, we can evaluate integral along
the curve
Z
C
Z 0
dx
dξ
dξ
Z 0
dx1
dx2
= π (−x1 (ξ)
+ x2 (ξ)
) dξ
dξ
dξ
2
g(x) · dx =
=
=
g(x(ξ)) ·
π
2
Z 0
π
2
((cos ξ)(sin ξ) + (sin ξ)(cos ξ)) dξ
Z 0
π
2
sin(2ξ) dξ
"
cos(2ξ)
= −
2
#0
π
2
= −1
Clearly, the evaluation of integration by direct method and change of variables are
equivalent. Hence, we observe the following points from this simple example:
• The integrand should be transformed using changed variables.
• The differential element in the integration should be transformed.
• The integral limits or domain of integration should be transformed.
These three points need to be accounted in order to transform integration from the reference domain to spatial domain or vice versa.
In the transformation of the integrals, let Ω0 be a reference configuration and Ω be
a deformed configuration at an instant of time t. Let us also consider x = χ(X, t)
be a transformation from Ω0 to Ω and F (X, t) be the deformation gradient. Then the
transformation of line, surface and volume integrals are stated below.
Transformation of line integrals:
The line integrals are always evaluated along a curve for a given vector field. Let g(x, t)
be a vector field defined over Ω and Ct be a curve in the deformed configuration. Then
the line integral is given by
Z
IL =
g(x, t) · dx.
Ct
In order to transform the spatial line integral IL , we need to transform spatial vector field
g(x, t), the differential line element dx and the spatial curve Ct into material description.
Let g(x, t) = g̃(X, t) for every x = χ(X, t), i.e.,
g(x, t) = g(χ(X, t), t) = g̃(X, t).
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Since F (X, t) is the deformation gradient, the transformation of differential line element
can be written as
dx = F (X, t) dX.
Let material curve C in Ω0 be a transformation of spatial curve Ct . Then the transformation of line integral can be written as
Z
Z
g(x, t) · dx =
Ct
ZC
=
g̃(X, t) · F dX
F T g̃(X, t) · dX.
(1)
C
Transformation of surface integrals:
Let Γ0 be a boundary of reference domain Ω0 and Γ be a boundary of deformed domain
Ω. Let g(x, t) be a vector field defined over Γ. Then the surface integral is defined by
IS =
Z
g(x, t) · n ∂Γ,
(2)
Γ
where n is outward unit normal to the surface Γ. Similar to previous case, we can
transform g(x, t) = g̃(X, t) for x = χ(X, t). From our earlier discussion (see Problem-3
in Lecture-19), the transformation to area element is given by
n ∂Γ = (cof F )n0 ∂Γ0 = JF −T n0 ∂Γ0 ,
where n0 is normal to the surface Γ0 and J is the determinant of F .
Substituting the expression for g and n ∂Γ in Eq. (2), we can get the following transformation for surface integrals of vector field.
Z
g(x, t) · n ∂Γ =
Z
Γ
Γ0
g̃(X, t) · (cof F )n0 ∂Γ0 =
Z
Γ0
g̃(X, t) · JF −T n0 ∂Γ0 .
(3)
If G(x, t) is a tensor field defined over Ω and G(x, t) = G̃(X, t) for x = χ(X, t) then we
can have the following relation for surface integral of tensor field.
Z
G(x, t)n ∂Γ =
Z
Γ
G̃(X, t)(cof F )n0 ∂Γ0 .
Γ0
(4)
Transformation of volume integrals:
Let φ(x, t), g(x, t) and G(x, t) be scalar, vector and tensor fields. Let J be the determinant of deformation gradient F . Let φ(x, t) = φ̃(X, t), g(x, t) = g̃(X, t) and
G(x, t) = G̃(X, t) for x = χ(X, t). Then we have
Z
φ(x, t) ∂Ω =
Z
Ω
Z
Ω0
g(x, t) ∂Ω =
Z
Ω
Z
Ω
G(x, t) ∂Ω =
Ω0
Z
Ω0
φ̃(X, t)J ∂Ω0 ,
(5)
g̃(X, t)J ∂Ω0 ,
(6)
G̃(X, t)J ∂Ω0 .
(7)
The transformation of volume element ∂Ω = J∂Ω0 can refer to a section on “volume
element” in Lecture-19.
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We now want to prove an important result called Reynolds’ transport theorem which
plays an important role in balance laws. However, before going to show the general
result, we want to discuss the Leibniz rule of integration which is one dimensional version
of Reynolds’ transport theorem.
Leibniz rule of integration:
Let f (ξ, t) be a function. Let α(t) and β(t) be direct function of time t. Then we have
the following result known as Leibniz rule of integration.
Z β(t)
∂f
dβ
dα
d Z β(t)
f (ξ, t) dξ =
dξ + f (β(t), t)
− f (α(t), t)
dt α(t)
dt
dt
α(t) ∂t
(8)
Proof. Using the definition of differentiation, we can write
!
Z β(t)
d Z β(t)
1 Z β(t+)
f (ξ, t) dξ = lim
f (ξ, t + ) dξ −
f (ξ, t) dξ .
→0 dt α(t)
α(t+)
α(t)
Substituting Taylor’s expansion of α(t + ) and β(t + ), we get
!
dβ
Z β(t)
d Z β(t)
1 Z β(t)+ dt +o()
f (ξ, t) dξ
f (ξ, t) dξ = lim
f (ξ, t + ) dξ −
→0 dt α(t)
α(t)
α(t)+ dα
+o()
dt
Z β(t)
1 Z α(t)
= lim
f (ξ, t + ) dξ
f (ξ, t + ) dξ +
→0 α(t)+ dα
α(t)
+o()
dt
+
Z β(t)+ dβ +o()
dt
f (ξ, t + ) dξ −
Z β(t)
!
f (ξ, t) dξ
α(t)
β(t)
dα
Z α(t)+ +o()
1
dt
−
f (ξ, t + ) dξ
→0 α(t)
= lim
+
Z β(t)+ dβ +o()
dt
β(t)
f (ξ, t + ) dξ +
Z β(t)
!
f (ξ, t + ) − f (ξ, t) dξ
α(t)
!
dβ
Z α(t)+ dα +o()
1 Z β(t)+ dt +o()
dt
= lim
f (ξ, t + ) dξ −
f (ξ, t + ) dξ
→0 β(t)
α(t)
!
Z β(t)
f (ξ, t + ) − f (ξ, t)
lim
+
dξ
→0
α(t)
!
!!
1
dβ
dα
= lim
f (β(t), t + )
+ o() − f (α(t), t + )
+ o()
→0 dt
dt
Z β(t)
∂f
+
dξ
α(t) ∂t
Z β(t)
∂f
dβ
dα
=
dξ + f (β(t), t)
− f (α(t), t)
dt
dt
α(t) ∂t
This completes the proof.
We now present an example to understand the theorem.
Example 2. Show that
d Z cos t
1
(1 + tξ)dξ = cos(2t) − t sin(2t) − sin t − cos t.
dt sin t
2
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Solution: Let us evaluate integral using direct integration.
"
#cos t
d Z cos t
d
ξ2
(1 + tξ)dξ =
ξ+t
dt sin t
dt
2 sin t
d
t
2
2
=
(cos t − sin t) + (cos t − sin t)
dt
2
1
t
= (− sin t − cos t) + (cos2 t − sin2 t) + (−2 cos t sin t − 2 sin t cos t)
2
2
1
cos(2t) − t sin(2t) − sin t − cos t.
=
2
We now evaluate integral using Leibniz rule and show that it is indeed same as that of
direct evaluation. Let α(t) = sin t, β(t) = cos t and f (ξ, t) = 1 + tξ. Then
∂f
= ξ,
∂t
dα
= cos t,
dt
and
dβ
= − sin t.
dt
Substituting in Leibniz rule, we get
Z cos t
d Z cos t
(1 + tξ)dξ =
ξ dξ + (1 + t cos t)(− sin t) − (1 + t sin t)(cos t)
dt sin t
sin t
" #cos t
ξ2
− sin t − t cos t sin t − cos t − t sin t cos t
=
2 sin t
1
=
(cos2 t − sin2 t) − 2t sin t cos t − − sin t − cos t
2
1
=
cos(2t) − t sin(2t) − sin t − cos t.
2
dα
dβ
The solution obtained in both ways is same. In case of Leibniz rule
and
represent
dt
dt
velocity of boundaries. Similarly, we now see that the effect of boundary velocity appears
in the following Reynolds’ transport theorem.
Reynolds’ transport theorem:
Let Ω0 be a reference configuration with Γ0 as its boundary and Ω be a deformed configuration with Γ as its boundary. Let φ(x, t) be a scalar field defined over Ω. Then the
R
material time derivative of integral Ω φ(x, t) ∂Ω is given by
Z
Z
d Z
∂φ
φ ∂Ω =
∂Ω + φ v · n ∂Γ,
dt Ω
Ω ∂t
Γ
(9)
where n is normal to Γ.
Proof. We note that the deformed configuration Ω changes its shape with time. This is
similar to the earlier integral (i.e., the integral on the left hand side of Eq. (8)) where
the integration limits are function of time. Analogous to earlier integration, the time
derivative cannot be taken inside the spatial integration as Ω changes with time. On
the other hand, the undeformed domain that is independent of time. Therefore, we
follow three steps to evaluate integral: (i) transform the spatial integration to material
integral, (ii) take the total time derivative inside the integration as reference domain (i.e.,
undeformed domain) is independent of time (iii) transform the material integration again
back to spatial integration.
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Let x = χ(X, t) be a map from Ω0 to Ω. Let J be a determinant of deformation
gradient F . Let φ̃(X, t) = φ(x, t) for x = χ(X, t), i.e., φ(x, t) = φ(χ(X, t), t) = φ̃(X, t).
Then
d Z
d Z
φ̃(X, t)J ∂Ω0
φ(x, t) ∂Ω =
dt Ω
dt Ω0
Z
∂(φ̃J)
=
∂Ω0
∂t
Ω0
!
Z
∂ φ̃
∂J
=
J + φ̃
∂Ω0
∂t
∂t
Ω0
=
=
=
=
=
!
∂ φ̃ φ̃ ∂J
+
J ∂Ω0
∂t
J ∂t
Ω0
!
Z
Dφ φ DJ
+
∂Ω
Dt
J Dt
Ω
!
Z
∂φ
+ (∇x φ) · v + φ(∇x · v) ∂Ω
∂t
Ω
Z
Z
∂φ
∂Ω + ∇x · (φv) ∂Ω
Ω ∂t
Ω
Z
Z
∂φ
∂Ω + φ v · n ∂Γ.
Ω ∂t
Γ
Z
Clearly, the second term on the right hand side of equation represents contribution from
velocity of the boundary.
Let w(x, t) be a vector field. Then, following similar steps as that of previous result,
we can prove the following Reynolds’ transport theorem for the vector field.
Z
Z
∂w
d Z
w ∂Ω =
∂Ω + w(v · n) ∂Γ.
dt Ω
Ω ∂t
Γ
(10)
Reference
1. C. S. Jog, Foundations and Applications of Mechanics: Continuum Mechanics,
Volume-I, 2007, Narosa Publishing House Pvt. Ltd., New Delhi.
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