Stream function and velocity potential problem

Velocity components

Using $u = \dfrac{\partial\psi}{\partial y}$ and $v = -\dfrac{\partial\psi}{\partial x}$:

$$\begin{aligned} u &= 3x^2 - 3y^2 \\ v &= -6xy \end{aligned}$$

(a) Continuity

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 6x - 6x = 0$$

Continuity is satisfied (as it always is when a stream function exists).

(b) Irrotationality

$$\zeta_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = -6y - (-6y) = 0$$

The vorticity is zero, so the flow is irrotational.

(c) Velocity potential

Since $u = \partial\phi/\partial x$:

$$\phi = \int (3x^2 - 3y^2)\,dx = x^3 - 3xy^2 + f(y)$$

Then $\partial\phi/\partial y = -6xy + f'(y)$. This must equal $v = -6xy$, so $f'(y) = 0$. Taking the constant as zero,

$$\phi = x^3 - 3xy^2$$

(d) At the point (1, 2)

$$\begin{aligned} u &= 3(1) - 3(4) = -9\ \text{m/s} \\ v &= -6(1)(2) = -12\ \text{m/s} \\ V &= \sqrt{9^2 + 12^2} = 15\ \text{m/s} \end{aligned}$$

Discharge between (1, 1) and (2, 1)

$$\begin{aligned} \psi_1 &= 3(1)(1) - 1 = 2 \\ \psi_2 &= 3(4)(1) - 1 = 11 \\ q &= \psi_2 - \psi_1 = 9\ \text{m}^2/\text{s per unit depth} \end{aligned}$$

Answer: $u = 3x^2 - 3y^2$, $v = -6xy$; irrotational; $\phi = x^3 - 3xy^2$; at (1, 2) $V = 15\ \text{m/s}$; $q = 9\ \text{m}^2/\text{s}$