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## Q. 10.14.6

For an incompressible viscous fluid moving under a conservative body force, prove the following:

$\begin{matrix}(i)&\triangledown^2\left(\frac{p}{\rho}+\chi+\frac{1}{2}v^2\right)=div(\textbf{v}\times\textbf{w})&(10.14.41)\\(ii)&\triangledown^2\left(\frac{p}{\rho}+\chi\right)=\frac{1}{2}\textbf{w}^2-\textbf{D.D}&(10.14.42)\end{matrix}$

Further, if the motion is irrotational, deduce that

$\begin{matrix}(iii)&\triangledown^2v^2=2\textbf{D.D}\geq 0&(10.14.43)\end{matrix}$

## Verified Solution

(i) For an incompressible viscous fluid moving under a conservative body force, the Navier-Stokes equation is given by (10.14.9).

$v\triangledown^2\textbf{v}-\triangledown\left(\frac{p}{\rho}+\chi+\frac{1}{2}v^2\right)=\frac{\partial\textbf{v}}{\partial t}+\textbf{w}\times\textbf{v}$                (10.14.9)
If we take divergence of this equation and note that div v = 0, we get the relation (10.14.41).
(ii) Substituting for $div(\textbf{v}\times\textbf{w})$ from the identity (6.3.21) (with div v = 0) in the righthand side of the relation (10.14.41), we get the relation (10.14.42).

$div(\textbf{v}\times\textbf{w})=\frac{1}{2}(\triangledown^2|\textbf{v}^2|+|\textbf{w}^2|)\\-\left\{|\textbf{D}^2|+(\textbf{v}.\triangledown)(div\ \textbf{v})\right\}$            (6.3.21)
(iii) For an irrotational motion, w = 0. In this case, subtracting (10.14.42) from (10.14.41) we get $\triangledown^2v^2=2\textbf{D.D}.Since\ \textbf{D.D}=d_{ij}d_{ij}\geq 0$, we obtain the inequality $\triangledown v^2\geq 0$. Thus, (10.14.43) is proven.