demonstration_NS.md

Demonstration of the Dimensionless $\omega-\psi$ Formulation

Navier-Stokes Equations

We consider an incompressible fluid of density $\rho$, dynamic viscosity $\mu$, and kinematic viscosity $\nu$. Let $\vec{v} = v_x \vec{e}_x + v_y \vec{e}_y$ be the 2D velocity field, $p$ the pressure, and $\vec{f}_V$ the body force. The conservation of mass and momentum is written as:

• Mass conservation

$$ \vec{\nabla} \cdot (\vec{v}) = 0 $$

• Momentum conservation

$$ \rho \left[ \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \vec{\nabla} \vec{v} \right] = \vec{\nabla} p + \mu \Delta\vec{v} + \rho \vec{f}_V $$

$\omega-\psi$ Formulation

Here, we solve the Navier-Stokes equations using the $\omega-\psi$ formulation (stream function and vorticity formulation), which in 2D reduces the system from 3 equations to solve for $u(x,y,t)$, $v(x,y,t)$, and $p(x,y,t)$ to a system of 2 equations for the vorticity $\omega(x,y,t)$ and the stream function $\psi(x,y,t)$. These two quantities are defined in 2D as:

• Vorticity

$$ \vec{\omega} = \vec{\nabla} \wedge \vec{v} = \left[ \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right] \vec{e_z} = \omega \vec{e_z} $$

• Stream function

$$ \vec{v} = \vec{\nabla} \wedge \psi \vec{e_z} \begin{aligned} = \begin{cases} u = \frac{\partial \psi}{\partial y} \\ v = - \frac{\partial \psi}{\partial x} \end{cases} \end{aligned} $$

The $\omega-\psi$ formulation of the Navier-Stokes equations is written as:

$$ \begin{aligned} \begin{cases} \Delta \psi = - \omega \\ \frac{\partial \omega}{\partial t} + \frac{\partial \psi}{\partial y} \frac{\partial \omega}{\partial x} -\frac{\partial \psi}{\partial x} \frac{\partial \omega}{\partial y} = \nu \Delta \omega \end{cases} \end{aligned} $$

Demonstration

To obtain the $\omega-\psi$ formulation, we proceed in several steps:

1. First, we note that by definition of the stream function, the mass conservation is automatically satisfied in 2D:

$$ \vec{\nabla} \cdot \vec{v} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = \frac{\partial^2 \psi}{\partial x\partial y} - \frac{\partial^2 \psi}{\partial y\partial x} = 0 $$

2. Next, we take the curl of the momentum equation:

$$ \vec{\nabla} \wedge \left(\rho \left[ \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \vec{\nabla} \vec{v} \right] = -\vec{\nabla} p + \mu \Delta\vec{v} + \rho \vec{f}_V\right) $$

3. The pressure term disappears because $\vec{\nabla} \wedge \vec{\nabla} p = 0$

4. For the left-hand side, using the definition of vorticity:

$$ \rho \frac{\partial \omega}{\partial t} + \rho \vec{\nabla} \wedge (\vec{v} \cdot \vec{\nabla} \vec{v}) $$

5. The nonlinear term is developed using vector identities to expand $(\vec{v} \cdot \vec{\nabla} \vec{v})$:

$$ \vec{\nabla} \wedge (\vec{v} \cdot \vec{\nabla} \vec{v}) = \vec{\nabla} \wedge \left[ \vec{\omega} \wedge \vec{v} + \vec{\nabla} \left( \frac{\lVert \vec{x} \rVert^2}{2} \right) \right] = \vec{\nabla} \wedge ( \vec{\omega} \wedge \vec{v} ) + \vec{\nabla} \wedge \vec{\nabla} \left( \frac{\lVert \vec{x} \rVert^2}{2} \right) $$

       Again, using vector identities, this term becomes (the last term is zero by the identity of the curl):

$$ \vec{\nabla} \wedge ( \vec{\omega} \wedge \vec{v} ) = \vec{v} \cdot \vec{\nabla} \vec{\omega} - \vec{\omega} \cdot \vec{\nabla} \vec{v} + \vec{\omega}(\vec{\nabla} \cdot \vec{v}) - \vec{v}(\vec{\nabla} \cdot \vec{\omega}) $$

       The last two terms are zero by the identity of the curl and because the fluid is incompressible, so the nonlinear term becomes:

$$ \vec{\nabla} \wedge (\vec{v} \cdot \vec{\nabla} \vec{v}) = \vec{v} \cdot \vec{\nabla} \vec{\omega} - \vec{\omega} \cdot \vec{\nabla} \vec{v} $$

6. For the right-hand side (assuming the body force is irrotational):

$$\mu \vec{\nabla} \wedge (\Delta\vec{v}) = \mu \Delta\omega$$

7. Dividing by $\rho$ and setting $\nu = \mu/\rho$, we obtain the vorticity transport equation:

$$\frac{\partial \vec{\omega}}{\partial t} + \vec{v} \cdot \vec{\nabla} \vec{\omega} = \vec{\omega} \cdot \vec{\nabla} \vec{v} + \nu \Delta \vec{\omega}$$

8. The term $\vec{\omega} \cdot \vec{\nabla} \vec{v}$ becomes zero in 2D and the only nonzero equation is the one projected onto $\vec{e_z}$, so the equation becomes, with the definition of $\psi$:

$$ \frac{\partial \omega}{\partial t} + \frac{\partial \psi}{\partial y} \frac{\partial \omega}{\partial x} -\frac{\partial \psi}{\partial x} \frac{\partial \omega}{\partial y} = \nu \Delta \omega $$

9. The relation between $\psi$ and $\omega$ comes directly from the definition of vorticity:

$$\omega = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = -\left(\frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2}\right) = -\Delta \psi$$

Hence the final system:

$$ \begin{aligned} \begin{cases} \Delta \psi = - \omega \ \frac{\partial \omega}{\partial t} + \frac{\partial \psi}{\partial y} \frac{\partial \omega}{\partial x} -\frac{\partial \psi}{\partial x} \frac{\partial \omega}{\partial y} = \nu \Delta \omega \end{cases} \end{aligned} $$

Non-dimensionalization

To non-dimensionalize this system, we introduce the following characteristic quantities:

• $L$: characteristic length
• $U$: characteristic velocity
• $T = L/U$: characteristic time

The dimensionless variables are defined by:

$$ \begin{aligned} \tilde{x} &= \frac{1}{L} , x \\ \tilde{y} &= \frac{1}{L} , y \\ \tilde{t} &= \frac{U}{L} , t \\ \tilde{\psi} &= \frac{1}{UL} , \psi \\ \tilde{\omega} &= \frac{L}{U} , \omega \end{aligned} $$

The dimensionless system then becomes:

$$ \begin{aligned} \begin{cases} \tilde{\Delta} \tilde{\psi} = - \tilde{\omega} \\ \frac{\partial \tilde{\omega}}{\partial \tilde{t}} + \frac{\partial \tilde{\psi}}{\partial \tilde{y}} \frac{\partial \tilde{\omega}}{\partial \tilde{x}} -\frac{\partial \tilde{\psi}}{\partial \tilde{x}} \frac{\partial \tilde{\omega}}{\partial \tilde{y}} = Re^{-1} \tilde{\Delta} \tilde{\omega} \end{cases} \end{aligned} $$

where $Re = \frac{UL}{\nu}$ is the Reynolds number and $\tilde{\Delta}$ is the dimensionless Laplacian operator.

Note: Hereafter, we will omit the tildes $\tilde{(\cdot)}$ to simplify the notation.