A stream function in a 2-D flow is y = 2xy. Show that the flow is irrotational. Also determine the corresponding velocity potential Φ .
Writing the Laplace equation (4.18) of stream function in 2-D, we have
\left[\frac{\pmb{\delta} ^{2}\Psi }{\pmb{\delta} x^{2} }+\frac{\pmb{\delta} ^{2}\Psi }{\pmb{\delta} y^{2} } \right]=0 (i)
Substituting the value of Ψ in Eq. (i), we get
\left[\frac{\pmb{\delta} ^{2}(2xy) }{\pmb{\delta} x^{2} }+\frac{\pmb{\delta} ^{2}(2xy) }{\pmb{\delta} y^{2} } \right]=0
Hence, the flow is irrotational.
We know that
\frac{\pmb{\delta} \pmb{\phi} }{\pmb{\delta} x}=-u =\frac{\pmb{\delta} \pmb{\Psi} }{\pmb{\delta} y}=\frac{\pmb{\delta} (2xy)}{\pmb{\delta} y}=2x (ii)
and \frac{\pmb{\delta} \pmb{\phi} }{\pmb{\delta} y}=-v =-\frac{\pmb{\delta} \pmb{\Psi} }{\pmb{\delta} x}=-\frac{\pmb{\delta} (2xy)}{\pmb{\delta} x}=-2y (iii)
Integrating Eq. (ii), we get
\pmb{\phi} =x^{2}+f(y) (iv)
Differentiating Eq. (iv) with respect to y, we get
\frac{\pmb{\delta} \pmb{\phi} }{\pmb{\delta} y}=\acute{f}(y) (v)
Equating dΦ/dy from Eqs. (iii) and (v), we get
\acute{f}(y)=- 2y (vi)
Integrating Eq. (vi), we get
{f}(y)=- y^{2}+C
where C is a constant of integration.
Substituting f(y) in Eq. (iv), we get
\pmb{\phi}=x^{2}-y^{2}+C