PotentialFlow

Potential flow in two dimensions is simple to analyze using conformal mapping, by the use of transformations of the complex plane. However, use of complex numbers is not required, as for example in the classical analysis of fluid flow past a cylinder. It is not possible to solve a potential flow using complex numbers in three dimensions.[10]

The basic idea is to use a holomorphic (also called analytic) or meromorphic function f, which maps the physical domain (x, y) to the transformed domain (φ, ψ). While x, y, φ and ψ are all real valued, it is convenient to define the complex quantities

$$ z = x + i y , and w = φ + i ψ . {\displaystyle {\begin{aligned}z&=x+iy,,{\text{ and }}&w&=\varphi +i\psi ,.\end{aligned}}}$$

Now, if we write the mapping f as[10]

$$ f ( x + i y ) = φ + i ψ , or f ( z ) = w . {\displaystyle {\begin{aligned}f(x+iy)&=\varphi +i\psi ,,{\text{ or }}&f(z)&=w,.\end{aligned}}}$$

Then, because f is a holomorphic or meromorphic function, it has to satisfy the Cauchy–Riemann equations[10]

$$ ∂ φ ∂ x = ∂ ψ ∂ y , ∂ φ ∂ y = − ∂ ψ ∂ x . {\displaystyle {\begin{aligned}{\frac {\partial \varphi }{\partial x}}&={\frac {\partial \psi }{\partial y}},,&{\frac {\partial \varphi }{\partial y}}&=-{\frac {\partial \psi }{\partial x}},.\end{aligned}}}$$

The velocity components (u, v), in the (x, y) directions respectively, can be obtained directly from f by differentiating with respect to z. That is[10]

$$ d f d z = u − i v {\displaystyle {\frac {df}{dz}}=u-iv}$$

So the velocity field v = (u, v) is specified by[10]

$$ u = ∂ φ ∂ x = ∂ ψ ∂ y , v = ∂ φ ∂ y = − ∂ ψ ∂ x . {\displaystyle {\begin{aligned}u&={\frac {\partial \varphi }{\partial x}}={\frac {\partial \psi }{\partial y}},&v&={\frac {\partial \varphi }{\partial y}}=-{\frac {\partial \psi }{\partial x}},.\end{aligned}}} $$

Both φ and ψ then satisfy Laplace's equation:[10]

$$ Δ φ = ∂ 2 φ ∂ x 2 + ∂ 2 φ ∂ y 2 = 0 , and Δ ψ = ∂ 2 ψ ∂ x 2 + ∂ 2 ψ ∂ y 2 = 0 $$

$$ {\displaystyle {\begin{aligned}\Delta \varphi &={\frac {\partial ^{2}\varphi }{\partial x^{2}}}+{\frac {\partial ^{2}\varphi }{\partial y^{2}}}=0,,{\text{ and }}&\Delta \psi &={\frac {\partial ^{2}\psi }{\partial x^{2}}}+{\frac {\partial ^{2}\psi }{\partial y^{2}}}=0,.\end{aligned}}}$$

So φ can be identified as the velocity potential and ψ is called the stream function.[10] Lines of constant ψ are known as streamlines and lines of constant φ are known as equipotential lines (see equipotential surface).

Streamlines and equipotential lines are orthogonal to each other, since[10]

$$∇ φ ⋅ ∇ ψ = ∂ φ ∂ x ∂ ψ ∂ x + ∂ φ ∂ y ∂ ψ ∂ y = ∂ ψ ∂ y ∂ ψ ∂ x − ∂ ψ ∂ x ∂ ψ ∂ y = 0 $$ $$ {\displaystyle \nabla \varphi \cdot \nabla \psi ={\frac {\partial \varphi }{\partial x}}{\frac {\partial \psi }{\partial x}}+{\frac {\partial \varphi }{\partial y}}{\frac {\partial \psi }{\partial y}}={\frac {\partial \psi }{\partial y}}{\frac {\partial \psi }{\partial x}}-{\frac {\partial \psi }{\partial x}}{\frac {\partial \psi }{\partial y}}=0}$$

Thus the flow occurs along the lines of constant ψ and at right angles to the lines of constant φ.[10]

Δψ = 0 is also satisfied, this relation being equivalent to ∇ × v = 0. So the flow is irrotational. The automatic condition ∂2Ψ/∂x ∂y = ∂2Ψ/∂y ∂x then gives the incompressibility constraint ∇ · v = 0.