Comparing this equation with that which determines the value of Q
Q
=
A
.
(
v
∂
w
¯
∂
x
+
u
′
∂
w
¯
∂
y
)
{\displaystyle Q=A.\left(v{\frac {\partial {\bar {w}}}{\partial x}}+u^{\prime }{\frac {\partial {\bar {w}}}{\partial y}}\right)}
or substituting for
w
¯
{\displaystyle {\bar {w}}}
Q
=
−
2
∂
f
(
y
,
t
)
∂
y
.
A
.
(
v
∂
∂
x
+
u
′
∂
∂
y
)
.
∇
−
2
∂
v
∂
x
{\displaystyle Q=-2{\frac {\partial f(y,t)}{\partial y}}.A.\left(v{\frac {\partial }{\partial x}}+u^{\prime }{\frac {\partial }{\partial y}}\right).\nabla ^{-2}{\frac {\partial v}{\partial x}}}
The isotropy with respect to x and z gives the equation
2
A
.
(
v
0
∂
∂
x
+
u
0
′
∂
∂
y
)
.
∇
−
2
∂
v
0
∂
x
=
A
.
{
v
0
(
∂
2
∂
x
2
+
∂
2
∂
z
2
)
{\displaystyle 2A.\left(v_{0}{\frac {\partial }{\partial x}}+u_{0}^{\prime }{\frac {\partial }{\partial y}}\right).\nabla ^{-2}{\frac {\partial v_{0}}{\partial x}}=A.\left\{v_{0}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)\right.\qquad \qquad }
+
(
u
0
′
∂
∂
x
+
w
0
∂
∂
z
)
∂
∂
y
}
∇
−
2
v
0
{\displaystyle \left.+\left(u_{0}^{\prime }{\frac {\partial }{\partial x}}+w_{0}{\frac {\partial }{\partial z}}\right){\frac {\partial }{\partial y}}\right\}\nabla ^{-2}v_{0}}
But by integration by parts we obtain the equation
(
u
0
′
∂
∂
x
+
w
0
∂
∂
z
)
∂
∂
y
.
∇
−
2
v
0
=
−
A
(
∂
u
0
′
∂
x
+
∂
w
0
∂
z
)
∂
∂
y
∇
−
2
v
0
{\displaystyle \left(u_{0}^{\prime }{\frac {\partial }{\partial x}}+w_{0}{\frac {\partial }{\partial z}}\right){\frac {\partial }{\partial y}}.\nabla ^{-2}v_{0}=-A\left({\frac {\partial u_{0}^{\prime }}{\partial x}}+{\frac {\partial w_{0}}{\partial z}}\right){\frac {\partial }{\partial y}}\nabla ^{-2}v_{0}}
and by the condition of incompressibility the second member may be written
A
.
(
∂
v
0
/
∂
y
)
.
(
∂
/
∂
y
)
.
∇
−
2
v
0
,
{\displaystyle A.(\partial v_{0}/\partial y).(\partial /\partial y).\nabla ^{-2}v_{0},}
−
A
.
v
0
.
(
∂
/
2
∂
y
2
)
.
∇
−
2
v
0
{\displaystyle -A.v_{0}.(\partial /^{2}\partial y^{2}).\nabla ^{-2}v_{0}}
[errata 1]
So we have
Q
0
=
∂
f
(
y
,
t
)
∂
y
A
.
{
v
0
(
∂
2
∂
x
2
+
∂
2
∂
z
2
−
∂
2
∂
y
2
)
}
.
∇
−
2
v
0
{\displaystyle Q_{0}={\frac {\partial f(y,t)}{\partial y}}A.\left\{v_{0}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}-{\frac {\partial ^{2}}{\partial y^{2}}}\right)\right\}.\nabla ^{-2}v_{0}}
On account of the isotropy, we may write 1 / 3
(
∂
2
∂
x
2
−
∂
2
∂
y
2
+
∂
2
∂
z
2
)
∇
−
2
{\displaystyle \left({\frac {\partial ^{2}}{\partial x^{2}}}-{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)\nabla ^{-2}}
so
Q
0
=
1
9
R
2
∂
f
(
y
,
t
)
∂
y
{\displaystyle Q_{0}={\frac {1}{9}}R^{2}{\frac {\partial f(y,t)}{\partial y}}}
∂
∂
t
(
t
=
0
)
{
A
.
(
e
′
v
)
}
=
−
2
9
R
2
{
∂
f
(
y
,
t
)
∂
y
}
t
=
0
{\displaystyle {\frac {\partial }{\partial t}}_{(}t=0)\{A.(e^{\prime }v)\}=-{\tfrac {2}{9}}R^{2}\left\{{\frac {\partial f(y,t)}{\partial y}}\right\}_{t=0}}
The deviation from isotropy shown by this equation is very small, because of the smallness of ∂f (y,t )/∂y . The equation is therefore not restricted to the initial values of the two members,
↑ Correction:
−
A
.
v
0
.
(
∂
/
2
∂
y
2
)
.
∇
−
2
v
0
{\displaystyle -A.v_{0}.(\partial /^{2}\partial y^{2}).\nabla ^{-2}v_{0}}
−
A
.
v
0
.
(
∂
2
/
∂
y
2
)
.
∇
−
2
v
0
{\displaystyle -A.v_{0}.(\partial ^{2}/\partial y^{2}).\nabla ^{-2}v_{0}}
