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Navier–Stokes Equation

The Navier–Stokes equation applies Newton's second law to a moving fluid: pressure, viscosity, and external forces determine its acceleration. The u⃗(x,t)\vec{u}(x,t) is a vector field giving the fluid's velocity at position xx and time tt.
INCOMPRESSIBLE NAVIER–STOKES EQUATIONS
Let Ω⊂Rd\Omega\subset\mathbb{R}^d, with d=2d=2 or 33, be an open fluid region, and let t>0t>0. For constant mass density ρ>0\rho>0 and dynamic viscosity μ>0\mu>0, the velocity u⃗(x,t)\vec{u}(x,t) and pressure p(x,t)p(x,t) satisfy
{ρ[∂u⃗∂t+(u⃗⋅∇)u⃗]=−∇p+μ∇2u⃗+ρf⃗,∇⋅u⃗=0.\begin{equation*}\begin{cases}\displaystyle \rho\left[\frac{\partial\vec{u}}{\partial t}+(\vec{u}\cdot\nabla)\vec{u}\right] = -\nabla p+\mu\nabla^2\vec{u}+\rho\vec{f},\\\nabla\cdot\vec{u} = 0.\end{cases}\end{equation*}

Derivation

Newton's second law

Follow a small element of fluid with volume V>0V>0 and constant density ρ\rho. Its mass is m=ρVm=\rho V, so Newton's second law gives
∑F⃗=ma⃗=ρVa⃗⟹ρa⃗=∑F⃗V.\begin{equation*}\sum\vec{F}=m\vec{a}=\rho V\vec{a} \quad\Longrightarrow\quad \rho\vec{a}=\frac{\sum\vec{F}}{V}.\end{equation*}
Here a⃗\vec{a} is the element's centre-of-mass acceleration, and ∑F⃗\sum\vec{F} is the total force acting on it. Dividing by volume expresses the force per unit volume. As the element shrinks to a point, this becomes a local balance: density times fluid acceleration equals the force per unit volume due to pressure, viscosity, and external body forces. This is the starting point for the Navier–Stokes equation.

Acceleration

Let x(t)x(t) be the path of a fluid particle. Its velocity is u⃗(x(t),t)=dx/dt\vec{u}(x(t),t)=\mathrm{d}x/\mathrm{d}t and the acceleration is
a⃗(t)=dv⃗dt=ddtu⃗(x(t),t).\begin{equation*}\vec{a}(t)=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\vec{u}(x(t),t).\end{equation*}
Writing u⃗=(u1,…,ud)\vec{u}=(u_1,\ldots,u_d), the chain rule gives
a⃗(t)=∂u⃗∂t(x(t),t)+∑j=1d∂u⃗∂xj(x(t),t)dxjdt=∂u⃗∂t(x(t),t)+∑j=1duj(x(t),t)∂u⃗∂xj(x(t),t)⏟(u⃗⋅∇)u⃗.\begin{equation*}\vec{a}(t) = \frac{\partial\vec{u}}{\partial t}(x(t),t)+\sum_{j=1}^d\frac{\partial\vec{u}}{\partial x_j}(x(t),t)\frac{\mathrm{d}x_j}{\mathrm{d}t}= \frac{\partial\vec{u}}{\partial t}(x(t),t)+\underbrace{\sum_{j=1}^d u_j(x(t),t)\frac{\partial\vec{u}}{\partial x_j}(x(t),t)}_{(\vec{u}\cdot\nabla)\vec{u}}.\end{equation*}
where the gradient operator is
∇=(∂∂x1,∂∂x2,…,∂∂xd)T.\begin{equation*}\nabla=\left(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\ldots,\frac{\partial}{\partial x_d}\right)^{\mathsf{T}}.\end{equation*}

Pressure force

Pressure acts normally to a surface. For uniform pressure pp on a face of area AA, the force magnitude satisfies
p=FA⟹F=pA.\begin{equation*}p=\frac{F}{A}\quad\Longrightarrow\quad F=pA.\end{equation*}
At a fixed time, consider the pressure profile p(x)p(x) shown in the animation. The fluid element lies between xx and x+Δxx+\Delta x, with equal face areas AA and volume V=AΔxV=A\Delta x. The left face is pushed to the right by p(x)Ap(x)A, while the right face is pushed to the left by p(x+Δx)Ap(x+\Delta x)A. Thus the net pressure force in the xx-direction is
Fpressure,x=p(x)A−p(x+Δx)A=−A[p(x+Δx)−p(x)].\begin{equation*}F_{\mathrm{pressure},x}=p(x)A-p(x+\Delta x)A=-A[p(x+\Delta x)-p(x)].\end{equation*}
Divide by the volume, substitute V=AΔxV=A\Delta x, and cancel the common area factor:
Fpressure,xV=−A[p(x+Δx)−p(x)]V=−A[p(x+Δx)−p(x)]AΔx=−p(x+Δx)−p(x)Δx.\begin{align*}\frac{F_{\mathrm{pressure},x}}{V} &= -\frac{A[p(x+\Delta x)-p(x)]}{V}\\&= -\frac{A[p(x+\Delta x)-p(x)]}{A\Delta x}\\&= -\frac{p(x+\Delta x)-p(x)}{\Delta x}.\end{align*}
As the element shrinks to a point, the difference quotient becomes a partial derivative:
lim⁡Δx→0Fpressure,xV=−∂p∂x.\begin{equation*}\lim_{\Delta x\to 0}\frac{F_{\mathrm{pressure},x}}{V}=-\frac{\partial p}{\partial x}.\end{equation*}
Repeating this argument in each coordinate direction gives the pressure force per unit volume −∇p-\nabla p. The minus sign means that the net pressure force points from higher pressure toward lower pressure.

Viscous acceleration

Consider a Newtonian fluid with constant dynamic viscosity μ\mu and parallel shear flow u⃗=u(y,t)e⃗x\vec{u}=u(y,t)\vec{e}_x. Take a small element whose horizontal faces have area AA and lie at y1y_1 and y2y_2. Its thickness is Δy=y2−y1\Delta y=y_2-y_1, so its volume is V=AΔyV=A\Delta y and its mass is ρAΔy\rho A\Delta y.
Viscosity causes neighbouring layers to exert tangential forces on the element. For the positive velocity gradients shown, the top force points right and the bottom force points left, with magnitudes
Ftop=Aμ∂u∂y∣y2,Fbottom=Aμ∂u∂y∣y1.\begin{align*}F_{\mathrm{top}} &= A\mu\left.\frac{\partial u}{\partial y}\right|_{y_2},\\F_{\mathrm{bottom}} &= A\mu\left.\frac{\partial u}{\partial y}\right|_{y_1}.\end{align*}
Their difference gives the net viscous force in the xx-direction:
Fnet,x=Ftop−Fbottom=Aμ[∂u∂y∣y2−∂u∂y∣y1].\begin{equation*}F_{\mathrm{net},x}=F_{\mathrm{top}}-F_{\mathrm{bottom}}=A\mu\left[\left.\frac{\partial u}{\partial y}\right|_{y_2}-\left.\frac{\partial u}{\partial y}\right|_{y_1}\right].\end{equation*}
Dividing by the element's mass and taking the limit as its thickness tends to zero gives the local viscous acceleration:
avisc,x=lim⁡Δy→0Fnet,xρAΔy=μρ∂2u∂y2.\begin{equation*}a_{\mathrm{visc},x}=\lim_{\Delta y\to 0}\frac{F_{\mathrm{net},x}}{\rho A\Delta y}=\frac{\mu}{\rho}\frac{\partial^2 u}{\partial y^2}.\end{equation*}
A uniform velocity gradient produces shear but no net viscous acceleration; the acceleration comes from changes in that gradient. For a general incompressible Newtonian flow with constant μ\mu, the viscous force per unit volume is μ∇2u⃗\mu\nabla^2\vec{u}, giving
a⃗visc=μρ∇2u⃗.\begin{equation*}\vec{a}_{\mathrm{visc}}=\frac{\mu}{\rho}\nabla^2\vec{u}.\end{equation*}

References