14: The Governing Equations

We derive here the governing equations for the velocity $\mathbf{u}=\mathbf{u}(\mathbf{x}, t)$ of a flowing fluid.

Multi-variable calculus

Fluid flows typically take place in three-dimensional space, and the governing equations will contain derivatives in all three directions. The mathematics learned in a multi-variable calculus course will therefore be useful. Here, I summarize some of this mathematics.

Vector algebra

Examples of vectors will be the position vector $\mathbf{x}$ and the velocity vector $\mathbf{u}$. We will use the Cartesian coordinate system to write vectors in terms of their components as

$$\mathbf{x}=(x, y, z), \quad \mathbf{u}=(u, v, w) \nonumber$$

or sometimes as

$$\mathbf{x}=\left(x_{1}, x_{2}, x_{3}\right), \quad \mathbf{u}=\left(u_{1}, u_{2}, u_{3}\right) \nonumber$$

Another notation makes use of the cartesian unit vectors, $\hat{\mathbf{x}}, \hat{\mathbf{y}}$, and $\hat{\mathbf{z}}$ :

$$\mathbf{x}=x \hat{\mathbf{x}}+y \hat{\mathbf{y}}+z \hat{\mathbf{z}}, \quad \mathbf{u}=u \hat{\mathbf{x}}+v \hat{\mathbf{y}}+w \hat{\mathbf{z}} . \nonumber$$

The velocity $\mathbf{u}$ is called a vector field because it is a vector that is a function of the position vector $x$.

The dot product between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by

$$\begin{aligned} \mathbf{u} \cdot \mathbf{v} &=u_{1} v_{1}+u_{2} v_{2}+u_{3} v_{3} \\ &=u_{i} v_{i} \end{aligned} \nonumber$$

where in the last expression we use the Einstein summation convention: when an index occurs twice in a single term, it is summed over. From hereon, the Einstein summation convention will be assumed unless explicitly stated otherwise.

The cross product between two vectors is given by a determinant:

$$\begin{aligned} \mathbf{u} \times \mathbf{v} &=\left|\begin{array}{ccc} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{array}\right| \\ &=\left(u_{2} v_{3}-u_{3} v_{2}, u_{3} v_{1}-u_{1} v_{3}, u_{1} v_{2}-u_{2} v_{1}\right) \end{aligned} \nonumber$$

The cross-product of two vectors is a vector, and the components of the cross product can be written more succinctly using the Levi-Civita tensor, defined as

$$\epsilon_{i j k}=\left\{\begin{aligned} 1 & \text { if }(i, j, k) \text { is an even permutation of }(1,2,3), \\ -1 & \text { if }(i, j, k) \text { is an odd permutation of }(1,2,3), \\ 0 & \text { if any index is repeated. } \end{aligned}\right. \nonumber$$

Using the Levi-Civita tensor, the $i$-th component of the cross product can be written as

$$(\mathbf{u} \times \mathbf{v})_{i}=\epsilon_{i j k} u_{j} v_{k} \nonumber$$

Another useful tensor is the Kronecker delta, defined as

$$\delta_{i j}= \begin{cases}0 & \text { if } i \neq j \\ 1 & \text { if } i=j\end{cases} \nonumber$$

Note that $v_{i} \delta_{i j}=v_{j}$ and that $\delta_{i i}=3$. A useful identity between the Levi-Civita tensor and the Kronecker delta is given by

$$\epsilon_{i j k} \epsilon_{i m n}=\delta_{j m} \delta_{k n}-\delta_{j n} \delta_{k m} . \nonumber$$

Gauss’s theorem (or the divergence theorem) and Stokes’ theorem are usually introduced in a course on multi-variable calculus. We will state these theorems here.

First, Gauss’s theorem. Let $V$ be a three-dimensional volume bounded by a smooth surface $S$, and let $\mathbf{F}$ be a vector field in $V$. Then Gauss’s theorem states that

$$\int_{S} \mathbf{F} \cdot \hat{\mathbf{n}} d S=\int_{V} \nabla \cdot \mathbf{F} d V \nonumber$$

where $\hat{\mathbf{n}}$ is the outward facing unit normal vector to the bounding surface $S$.

Second, Stokes’ theorem. Let $S$ be a smooth surface bounded by a simple closed curve $C$ with positive orientation. Then Stokes’ theorem states that

$$\oint_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{S}(\boldsymbol{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d S \nonumber$$

Continuity equation

We consider a control volume $V$ of fluid bounded by a smooth surface $S$. The continuity equation expresses the conservation of mass. The time-derivative of the total mass of the fluid contained in the volume $V$ is equal to the (negative) of the total mass of fluid that flows out of the boundary of $V$; that is;

$$\frac{d}{d t} \int_{V} \rho(\mathbf{x}, t) d V=-\int_{S} \rho(\mathbf{x}, t) \mathbf{u}(\mathbf{x}, t) \cdot \hat{\mathbf{n}} d S \nonumber$$

The integral on the right-hand-side represents the flux of mass through the boundary $S$ and has units of mass per unit time. We now apply the divergence theorem to the integral on the right-hand-side:

$$\int_{S} \rho(\mathbf{x}, t) \mathbf{u}(\mathbf{x}, t) \cdot \hat{\mathbf{n}} d S=\int_{V} \nabla \cdot(\rho \mathbf{u}) d V \nonumber$$

Combining the left- and right-hand-sides, we have

$$\int_{V}\left(\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{u})\right) d V=0 . \nonumber$$

Because the control volume is arbitrary, the integrand must vanish identically, and we thus obtain the continuity equation

$$\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{u})=0 \nonumber$$

We will only be considering here incompressible fluids, where we may assume that $\rho(\mathbf{x}, t)$ is a constant, independent of both space and time. The continuity equation (14.12) then becomes an equation for conservation of fluid volume, and is given by

$$\boldsymbol{\nabla} \cdot \mathbf{u}=0 \nonumber$$

This equation is called the incompressibility condition.

Momentum equation

Material derivative

The Navier-Stokes equation is derived from applying Newton’s law $F=m a$ to a fluid flow. We first consider the acceleration of a fluid element. The velocity of the fluid at a fixed position $x$ is given by $\mathbf{u}(\mathbf{x}, t)$, but the fluid element is not at a fixed position but follows the fluid in motion. Now a general application of the chain rule yields

$$\frac{d}{d t} \mathbf{u}(\mathbf{x}, t)=\frac{\partial \mathbf{u}}{\partial t}+\frac{\partial \mathbf{u}}{\partial x_{j}} \frac{\partial x_{j}}{\partial t} \nonumber$$

If the position $\mathbf{x}$ is fixed, then $\partial x_{j} / \partial t=0$. But if $\mathbf{x}=\mathbf{x}(\mathbf{t})$ represents the position of the fluid element, then $\partial x_{j} / \partial t=u_{j} .$ The latter assumption is called the material derivative and is written as

$$\begin{aligned} \frac{D \mathbf{u}}{D t} &=\frac{\partial \mathbf{u}}{\partial t}+u_{j} \frac{\partial \mathbf{u}}{\partial x_{j}} \\ &=\frac{\partial \mathbf{u}}{\partial t}+(\mathbf{u} \cdot \nabla) \mathbf{u} \end{aligned} \nonumber$$

and represents the acceleration of a fluid element as it flows with the fluid. Instead of the mass of the fluid element, we consider the mass per unit volume, and the right-hand-side of $F=m a$ becomes

$$\rho\left(\frac{\partial \mathbf{u}}{\partial t}+(\mathbf{u} \cdot \nabla) \mathbf{u}\right) \nonumber$$

We now need find the forces per unit volume acting on the flowing fluid element. We consider both pressure forces and viscous forces.

Pressure forces

We consider the normal pressure forces acting on two opposing faces of a control volume of fluid. With $A$ the area of the rectangular face at fixed $y$, and $d y$ the depth, the volume of the box is $A d y$, and the net pressure force per unit volume acting on the control volume in the $y$-direction is given by

$$\begin{aligned} f_{p} &=\frac{p A-(p+d p) A}{A d y} \\ &=-\frac{d p}{d y} \end{aligned} \nonumber$$

Similar considerations for the $x$ and $z$ directions yield the pressure force vector per unit volume to be

$$\begin{aligned} \mathbf{f}_{p} &=-\left(\frac{\partial p}{\partial x}, \frac{\partial p}{\partial y}, \frac{\partial p}{\partial z}\right) \\ &=-\nabla p \end{aligned} \nonumber$$

Viscous forces

The viscosity of a fluid measures its internal resistance to flow. Consider a fluid confined between two very large plates of surface area $A$, separated by a small distance $d y$. Suppose the bottom plate is stationary and the top plate move with velocity $d u$ in the $x$ - direction. The applied force per unit area required to keep the top surface in motion is empirically given by

$$\frac{F}{A}=\mu \frac{d u}{d y}, \nonumber$$

where $\mu$ is called the dynamic viscosity. Of course there is also an opposite force required to keep the bottom surface stationary. The difference between these two forces is the niscous force on the fluid element. Taking the difference, the resulting net force per unit area will be proportional to the second derivative of the velocity. Now the viscous forces act in all directions and on all the faces of the control volume. Without going into further technical details, we present the general form (for a so-called Newtonian fluid) of the viscous force vector per unit volume:

$$\begin{aligned} \mathbf{f}_{v} &=\mu\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}, \frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}+\frac{\partial^{2} v}{\partial z^{2}}, \frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}}\right) \\ &=\mu \nabla^{2} \mathbf{u} . \end{aligned} \nonumber$$

Navier-Stokes equation

Putting together all the terms, the Navier-Stokes equation is written as

$$\rho\left(\frac{\partial \mathbf{u}}{\partial t}+(\mathbf{u} \cdot \nabla) \mathbf{u}\right)=-\nabla p+\mu \nabla^{2} \mathbf{u} \nonumber$$

Now, instead of the dynamic viscosity $\mu$, one usually defines the kinematic viscosity $v=\mu / \rho$. The governing equations of fluid mechanics for a so-called incompressible Newtonian fluid, then, are given by both the continuity equation and the Navier-Stokes equation; that is,

$$\begin{align} \boldsymbol{\nabla} \cdot \mathbf{u} &=0 \\ \frac{\partial \mathbf{u}}{\partial t}+(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} &=-\frac{1}{\rho} \nabla p+v \nabla^{2} \mathbf{u} \end{align} \nonumber$$

Boundary conditions

Boundary conditions must be prescribed when flows contact solid surfaces. We will assume rigid, impermeable surfaces. If $\hat{\mathbf{n}}$ is the normal unit vector to the surface, and if there is no motion of the surface in the direction of its normal vector, then the condition of impermeability yields

$$\mathbf{u} \cdot \hat{\mathbf{n}}=0 \nonumber$$

We will also assume the no-slip condition: a viscous fluid should have zero velocity relative to a solid surface. In other words, a stationary or moving solid surface drags along the fluid touching it with the same velocity. The no-slip condition can be expressed mathematically as

$$\mathbf{u} \times \hat{\mathbf{n}}=\mathbf{V} \times \hat{\mathbf{n}} \nonumber$$

where $\mathbf{u}$ is the velocity of the fluid, $\mathbf{V}$ is the velocity of the surface, and $\hat{\mathbf{n}}$ is the normal vector to the surface.

Boundary conditions may also be prescribed far from any wall or obstacle. The free-stream boundary condition states that $\mathbf{u}=\mathbf{U}$ at infinity, where $\mathbf{U}$ is called the free-stream velocity.