Derivation of Euler's Equation in Fluid Dynamics
Theorem
$$ \mathbf{u} = \mathbf{u} \left( t ; \mathbf{x} \right) = \left( u_{1} \left( t ; \mathbf{x} \right) , u_{2} \left( t ; \mathbf{x} \right) , u_{3} \left( t ; \mathbf{x} \right) \right) $$ In particular, in three-dimensional space suppose the velocity field at time $t$ and spatial coordinate $\mathbf{x} = \left( x_{1} , x_{2} , x_{3} \right)$ is represented by the velocity vector as above. Similarly, $p : \mathbb{R}^{3} \to \mathbb{R}$ represents the pressure $p = p \left( \mathbf{x} \right)$ exerted at each coordinate. If $\mathbf{u}$ is the velocity of an inviscid and incompressible fluid, it satisfies the following governing equation. $$ {\frac{ \partial \mathbf{u} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) \mathbf{u} = - {\frac{ 1 }{ \rho }} \nabla p + \mathbf{g} $$ Here $\nabla \cdot$ is the divergence, $\rho$ is the density, and $\mathbf{g}$ is the gravitational acceleration.
Explanation
The first term on the right-hand side is often written as $\nabla w = \nabla p / \rho$, i.e., thermodynamic work.
Euler’s equation of motion can be regarded as an application of Newton’s second law of motion to fluid mechanics.
Derivation 1
The right-hand side of the material derivative contains two types of terms as follows. $$ {\frac{ D \mathbf{u} }{ D t }} = {\color{red} {\frac{ \partial \mathbf{u} }{ \partial t }}} + {\color{blue} \left( \mathbf{u} \cdot \nabla \right) \mathbf{u}} $$ Here the first red $\color{red} \partial \mathbf{u} / \partial t$ may be called the local acceleration or more simply the inertial term. The second blue $\color{blue} \left( \mathbf{u} \cdot \nabla \right) \mathbf{u}$ is the convective acceleration or more simply the convective term.
In fluid mechanics the acceleration of a fluid particle is expressed by the material derivative, and since $\mathbf{u}$ is already a velocity vector, taking the material derivative of $\mathbf{u}$ yields the sum of some acceleration $\mathbf{a}$ and the gravitational acceleration $\mathbf{g}$ as follows. $$ {\frac{ D \mathbf{u} }{ D t }} = \mathbf{a} + \mathbf{g} $$
Now let us compute $\mathbf{a}$. Without loss of generality (../2720), consider a differential cuboid whose width, length, and height changes are $dx_{1} , dx_{2} , dx_{3}$ respectively and whose mass is $m$; suppose a force $\mathbf{F} = \left( F_{1} , F_{2} , F_{3} \right)$ acts inward, normal to the faces.
Pressure–volume stress: For an object with surface area $A$ subject to a force $F$, define $p = F / A$ as the pressure. The change in pressure $\Delta p$ is called the volume stress. The ratio of the object’s original volume $V_{0}$ to its change in volume $\Delta V$, $\Delta V / V_{0}$, is called the volume strain. The ratio of volume stress to volume strain $B$ is called the bulk modulus. $$ B := \frac{\text{volume stress}}{\text{volume strain}} = - \frac{\Delta F / A}{\Delta V / V_{0}} = - \frac{\Delta P}{\Delta V / V_{0}} $$ Here the minus sign in the definition of $B$ is needed so that $B$ is positive, reflecting that an increase in pressure leads to a decrease in volume.
Since a force can be expressed as the product of pressure and area, $F = p A$, when a pressure $p$ is applied, $F = F_{1}$ can be expressed as the product $F_{1} = p d x_{2} d x_{3}$ of pressure $p$ and area $A = d x_{2} d_{3}$. Because the change in volume must be negative when pressure is applied, and $dV = - dx_{1} dx_{2} dx_{3}$, taking the partial derivative of both sides with respect to $x_{1}$ yields: $$ \begin{align*} & {\frac{ \partial F_{1} }{ \partial x_{1} }} = {\frac{ \partial p }{ \partial x_{1} }} d x_{2} d x_{3} \\ \implies & d F_{1} = {\frac{ \partial p }{ \partial x_{1} }} d x_{2} d x_{3} d x_{1} \\ \implies& d F_{1} = - {\frac{ \partial p }{ \partial x_{1} }} d V \end{align*} $$ This also holds for $d F_{2}$ and $d F_{3}$, so grouping terms appropriately gives the vector form: $$ \begin{align*} \begin{bmatrix} d F_{1} \\ d F_{2} \\ d F_{3} \end{bmatrix} =& - \begin{bmatrix} {\frac{ \partial p }{ \partial x_{1} }} \\ {\frac{ \partial p }{ \partial x_{2} }} \\ {\frac{ \partial p }{ \partial x_{3} }} \end{bmatrix} d V \\ \implies d \mathbf{F} =& - \nabla p d V \end{align*} $$
Newton’s law of motion: $$ \mathbf{F}=m\mathbf{a} $$
Density is defined as the ratio of mass to volume as $\rho = m / V$, so dividing both sides of Newton’s second law by the mass $m$ ($d \mathbf{F} / dm = \mathbf{a}$) yields: $$ \begin{align*} \mathbf{a} =& {\frac{ d \mathbf{F} }{ d m }} \\ =& - {\frac{ \nabla p d V }{ \rho d V }} \\ =& - {\frac{ 1 }{ \rho }} \nabla p \end{align*} $$ Finally, substituting $\mathbf{a}$ computed above into $D \mathbf{u} / Dt = \mathbf{a} + \mathbf{g}$ expressed using the material derivative, and rewriting the material derivative as the sum of the time derivative and the divergence term, we obtain the Euler equations as follows: $$ {\frac{ \partial \mathbf{u} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) \mathbf{u} = - {\frac{ 1 }{ \rho }} \nabla p + \mathbf{g} $$
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