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## Q. 10.14.4

Consider a steady motion of an incompressible viscous fluid under a conservative body force. If

$H_{0}=\frac{1}{2}v^2+\frac{p}{\rho}+\chi$                    (10.14.33)

prove the following

(i) $H_{0}$ is constant along the field lines of the vector

$\textbf{f}=(\textbf{v}\times\textbf{w})\times curl\ \textbf{w}$                (10.14.34)

(ii)                            $\textbf{v}.\triangledown H_{0}=v(\triangledown^2 H_{0}-\textbf{w}^2)$                      (10.14.35)

## Verified Solution

For the motion considered, the relation (10.14.17) holds. Using (10.4.33), this relation can be rewritten as

$\textbf{v}\times \textbf{w}=\triangledown\left(\frac{p}{\rho}+\frac{1}{2}v^2+\chi\right)+v\ curl\ \textbf{w}$               (10.14.17)

$\triangledown H_{0}=(\textbf{v}\times\textbf{w})-v\ curl\ \textbf{w}$                  (10.14.36)

From (10.14.34) and (10.14.36) we readily see that $\textbf{f}.\triangledown H_{0}=0.\ Thus,\ \triangledown H_{0}$ is orthogonal to f and hence to the field line of f. But $\triangledown H_{0}$ is always orthogonal to the surfaces of constant $H_{0}$. Hence f must be tangential to a surface of constant $H_{0}$. That is, $H_{0}$ is constant along the field lines of f.
From (10.14.36), we get

$\triangledown^2H_{0}=div(\textbf{v}\times\textbf{w})=\textbf{w}^2-\textbf{v}.\ curl\ \textbf{w}$                      (10.14.37)

and

$\textbf{v}.\triangledown H_{0}=-v\textbf{v}.\ curl\ \textbf{w}$                      (10.14.38)

These relations together yield the relation (10.14.35).