## Q. 4.3

A stream function in a 2-D flow is y = 2xy. Show that the flow is irrotational. Also determine the corresponding velocity potential Φ .

## Step-by-Step

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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

$\pmb{\phi}=x^{2}-y^{2}+C$

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