Given data:
Velocity potential        \phi=2 x y

For irrotational flow, the velocity potential (Φ) is defined as

u=\frac{\partial \phi}{\partial x}

v=\frac{\partial \phi}{\partial y}

Thus, the velocity components become

u=\frac{\partial \phi}{\partial x}=\frac{\partial}{\partial x}(2 x y)=2 y

v=\frac{\partial \phi}{\partial y}=\frac{\partial}{\partial y}(2 x y)=2 x

The velocity is then        \vec{V}=u \hat{i}+v \hat{j}=2 y \hat{i}+2 x \hat{j}

The velocity at (1, 2) is          \left.\vec{V}\right|_{(1.2)}=(2 \times 2) \hat{i}+(2 \times 1) \hat{j}=4 \hat{i}+2 \hat{j}

The velocity at (2, 2) is      \left.\vec{V}\right|_{(2,2)}=(2 \times 2) \hat{i}+(2 \times 2) \hat{j}=4 \hat{i}+4 \hat{j}

Hence,                    \frac{\partial u}{\partial x}=0

\frac{\partial v}{\partial y}=0

\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0

The above velocity field satisfies the continuity equation for incompressible flow. Hence, stream function exists.

From the definition of stream function ψ, we get

u=\frac{\partial \psi}{\partial y}

or                          \psi=\int u d y=\int 2 y d y

or                      \psi=2 \frac{y^2}{2}+f(x)=y^2+f(x)           (5.75)

and                                          v=\frac{\partial \psi}{\partial x}

\psi=-\int v d x=-\int 2 x d x

or                      \psi=-2 \frac{x^2}{2}+g(y)=-x^2+g(y)           (5.76)

Comparing Eqs. (5.75) and (5.76), we have

\psi=y^2-x^2

Hence, the stream function for the flow is

\psi=y^2-x^2

The stream function at (1, 2)        \left.\psi\right|_{(1,2)}=2^2-1^2=3 The stream function at (2, 2) is  \left.\psi\right|_{(2,2)}=2^2-2^2=0

The discharge per unit width passing between streamlines passing through these points is given by Eq. (5.64) as

q=\psi_2-\psi_1=3-0=3 \text{ units } ]