Given data:

(a) Velocity profile is given as u=\frac{1}{2 u} \frac{d p}{d x}\left(y^2-H^2\right)

Differentiating with respect to y, we have

\frac{d u}{d y}=\frac{1}{2 \mu} \frac{d p}{d x} 2 y=\frac{1}{\mu} \frac{d p}{d x} y

The wall shear stress is found to be

\tau_w=\left.\mu \frac{d u}{d y}\right|_{y=H}=\left.\mu \frac{1}{2 \mu} \frac{d p}{d x} 2 y\right|_{y=H}=\frac{d p}{d x} H

(b) The average velocity is found to be

u_{a v}=\frac{\int_{-H}^H u d y}{2 H}

or          u_{a v}=\frac{2}{2 H} \int_0^H u d y

=\frac{1}{H} \int_0^H \frac{1}{2 \mu} \frac{d p}{d x}\left(y^2-H^2\right) d y

=\frac{1}{2 \mu H} \frac{d p}{d x} \int_0^H\left(y^2-H^2\right) d y

=\frac{1}{2 \mu H} \frac{d p}{d x}\left[\frac{y^3}{3}-H^2 y\right]_0^H

=\frac{1}{2 \mu H} \frac{d p}{d x}\left[\frac{H^3}{3}-H^3\right]

(c) From the definition of stream function ψ, we get

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

or                                      \psi=\int u d y

or                                      \psi=\int \frac{1}{2 \mu} \frac{d p}{d x}\left(y^2-H^2\right) d y

or                                      \psi=\frac{1}{2 \mu} \frac{d p}{d x}\left(\frac{y^3}{3}-H^2 y\right)+f(x)                 (5.82)

Again,                          v=\frac{\partial \psi}{\partial x}

\psi=-\int v d x

or                                    \psi=g(y)               (5.83)

Comparing Eqs. (5.82) and (5.83), we have

\psi=\frac{1}{2 \mu} \frac{d p}{d x}\left(\frac{y^3}{3}-H^2 y\right)

Hence, the stream function for the flow is

\psi=\frac{1}{2 \mu} \frac{d p}{d x}\left(\frac{y^3}{3}-H^2 y\right)

(d) The vorticity is given by

\Omega_{x y}=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0-\frac{1}{\mu} \frac{d p}{d x} y=-\frac{1}{\mu} \frac{d p}{d x} y

(e) Since the vorticity is not equal to zero, the flow is rotational. Therefore, the velocity potential does not exist.