Question

For the function $\varphi =a{x}^{2}y-y^3$ to represent the velocity potential of an ideal fluid, ${▽}^2\varphi$ should be equal to zero. In that case, the value of ${}^{\prime }{a}^{\prime }$ has to be

• A. $-1$
• B. $1$
• C. $-3$
• D. $3$

Answer 1

The correct option is D $3$

$\varphi =a{x}^{2}y-y^3$
$\frac{\partial \varphi }{\partial x}=2axy,\frac{{\partial }^2\varphi }{{\partial }^{2}x}=2ay$
$\frac{\partial \varphi }{\partial y}=a{x}^2-3y^2,\frac{{\partial }^2\varphi }{{\partial }^{2}y}=-6y$
${▽}^2\varphi =\frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}=0$ (given)
$\Rightarrow 2ay-6y=0$
$a=3$