Following is the velocity potential function for the two dimensional irrotational flow in the cylinder coordinates: - $ϕ=\dfrac{m \cosθ}{r}$ - Determine the conjugate function (Stream function) -

Consider the velocity potential function for the two dimensional irrotational flow,

$$ϕ=\dfrac{m \cosθ}{r}$$

We know that by the definition of velocity potential function in cylindrical coordinates,

$$\dfrac{∂ϕ}{∂r}=u_r$$

$∴u_r=\dfrac{∂ϕ}{∂r} \dfrac{m \cosθ}{r} \\ ∴u_r=-\dfrac{m \cosθ}{r^2}$

And

$\dfrac 1r \dfrac{∂ϕ}{∂θ}=u_θ \\ ∴u_θ=\dfrac 1r \dfrac{∂}{∂θ}\bigg(\dfrac{m \cosθ}{r}\bigg) \\ ∴u_θ=-\dfrac{m \cosθ}{r^2}$

We also know that by the definition of stream function in cylindrical coordinates,

$u_r=\dfrac 1r \dfrac{∂ψ}{∂θ} \\ ∴\dfrac{∂ψ}{∂θ}=r\bigg(-\dfrac{m \cosθ}{r^2}\bigg) \\ ∴\dfrac{∂ψ}{∂θ}=\bigg(-\dfrac{m \cosθ}{r}\bigg)$

Integrating with respect to θ treating r as a constant,

$∴ψ=\int\bigg(-\dfrac{m \cosθ}{r}\bigg) dθ \\ ∴ψ=-\dfrac{m \sinθ}{r}+fn(r)$

And

$u_θ=-\dfrac{∂ψ}{∂r} \\ ∴\dfrac{∂ψ}{∂r}=-\bigg(-\dfrac{m \sinθ}{r^2} \bigg) \\ ∴\dfrac{∂ψ}{∂r}=\bigg(\dfrac{m \sinθ}{r^2} \bigg)$

Integrating with respect to r treating θ as a constant,

$∴ψ=∫\bigg(\dfrac{m \sinθ}{r^2} \bigg) dr \\ ∴ψ=-\dfrac{m \sinθ}{r}+fn(θ)$

Comparing the two equations of ψ for common and uncommon terms, we get,

$ψ=-\dfrac{m \sinθ}{r}$

…This is the required Conjugate function (Stream function) for the given Velocity Potential function.