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## Math 505, Mathematical Fluid Mechanics: Notes 2

I go on with some basic concepts and classical results in fluid dynamics [numbering is in accordance with the previous notes]. Throughout this section, I consider compressible barotropic ideal fluids with the pressure law ${p = p(\rho)}$ or incompressible ideal fluids with constant density ${\rho = \rho_0}$ (and hence, the pressure is an unknown function in the incompressible case).

To unify the two cases in the presentation, we write

$\displaystyle \frac{\nabla p}{\rho} = \nabla P$

with ${P = p/\rho_0}$ (incompressible) or ${P = \int_1^\rho s^{-1} p'(s)\; ds}$ (compressible). The momentum equation then reads

$\displaystyle (\partial_t + u \cdot \nabla )u + \nabla P = 0.$

See Section 3.4 in the previous notes.

4.1. Vorticity

Recall that the particle trajectory is defined by ${\dot X(t) = u(X(t),t)}$. To compare the dynamics of two particles that are infinitesimally close to one another, one looks at ${Y(t) = X(t) + h(t)}$, for sufficiently small ${h\ll1}$. One has

$\displaystyle \dot h = \dot Y - \dot X = u(X+h, t) - u(X,t) \approx \nabla_x u (X,t) h,$

in which by convention, ${\nabla u = [\nabla u_1, \nabla u_2, \nabla u_3]^{t}}$. We write

$\displaystyle \nabla u = \frac12 (\nabla u + \nabla u^t) + \frac12 (\nabla u - \nabla u^t)$

the sum of symmetric and anti-symmetric matrices. For constant symmetric matrices ${A}$, the map ${h_0 \mapsto e^{At}h_0}$ represents stretching or compressing, depending on the sign of eigenvalues of ${A}$. For the anti-symmetric part, we write

$\displaystyle \frac12 (\nabla u - \nabla u^t) h = \frac12 \omega \times h$

with ${\omega}$ denoting the vorticity of the fluid motion, defined by

$\displaystyle \omega = \nabla \times u.$

For constant vorticity ${\omega}$, the solution map ${h_0 \mapsto h(t)}$ of the ODE ${\dot h = \frac12 \omega \times h}$, with ${h(0)=h_0}$, preserves its length and is in fact a rigid rotation around the ${\omega}$-axis with an angle ${\frac12 |\omega| t}$. That is, vorticity is responsible for rotation in the fluid motion.

Lemma 4 Assume sufficient integrability of ${u}$ and ${u_{\vert_{\partial\Omega}} =0}$. There holds

$\displaystyle \int_\Omega |\nabla u|^2\; dx = \int_\Omega (|\omega|^2 + |\mathrm{div} u|^2)\; dx.$

Proof: One computes

\displaystyle \begin{aligned} \int_\Omega |\omega|^2 \; dx &= \int_\Omega (\nabla \times u)\cdot (\nabla \times u)\; dx \\ &= \int_\Omega u \cdot \nabla \times (\nabla \times u) + \int_{\partial \Omega} (n \times u)\cdot (\nabla \times u)\; dx. \end{aligned}

By using the identity ${ \nabla \times (\nabla \times u) = - \Delta u + \nabla (\mathrm{div} u)}$ and the zero boundary condition on ${u}$, the lemma is proved. $\Box$

4.2. Evolution of vorticity

With ${\omega = \nabla \times u}$, we can compute

\displaystyle \begin{aligned} u \cdot \nabla u &= \sum_j u_j \partial_{x_j} u = \sum_j u_j [\partial_{x_j} u - \nabla u_j] + \sum_j u_j \nabla u_j \\ &= \omega \times u + \nabla \Big(\frac{|u|^2}{2}\Big). \end{aligned}

That is, the momentum equation now reads

$\displaystyle \partial_t u + \omega \times u + \nabla \Big(\frac{|u|^2}{2} + P\Big) =0.\ \ \ \ \ (12)$

Theorem 5 (Bernoulli theorem) In the steady flow of a incompressible fluid with constant density or compressible barotropic fluid, the Bernoulli function ${P + \frac{|u|^2}{2}}$ is constant on the streamlines of the flow.

Proof: It follows directly from (12) that ${u \cdot \nabla \Big(\frac{|u|^2}{2} + P\Big) =0}$, which is the statement of the theorem. $\Box$

Taking the curl of the momentum equation (12) and using the fact that ${\nabla \times \nabla \phi =0}$, ${\nabla \cdot (\nabla \times u)=0}$, and

$\displaystyle \nabla \times (a\times b) =b \cdot \nabla a -a\cdot \nabla b + a \nabla \cdot b -b\nabla \cdot b,$

one gets the evolution of vorticity in the fluid:

$\displaystyle \partial_t \omega + u \cdot \nabla \omega - w\cdot \nabla u + \omega \nabla \cdot u =0.\ \ \ \ \ (13)$

For incompressible fluids, the vorticity equation reads

$\displaystyle ( \partial_t + u \cdot \nabla ) \omega = w\cdot \nabla u.\ \ \ \ \ (14)$

For compressible barotropic fluids, we use the continuity equation ${\nabla \cdot u = \rho^{-1} D_t \rho}$ into the vorticity equation, yielding

$\displaystyle (\partial_t + u \cdot \nabla ) \frac{\omega}{\rho} = \frac{w}{\rho} \cdot \nabla u.\ \ \ \ \ (15)$

Remark 1 (Vorticity in 2D) In 2D, vorticity ${\omega}$ is a scalar function, and the right-hand side term ${\omega \cdot \nabla u =0}$. This proves that the vorticity or ${\omega /\rho}$ is transported along the flow.

Remark 2 (Vortex stretching) In 3D, the term ${\omega \cdot \nabla u}$ is sometimes referred to as the vortex stretching term. The presence of this term makes the dynamics of vorticity in 3D greatly different and much delicate than that of vorticity in 2D. In fact, there is a well-known Beale-Kato-Majda’s criterium which asserts that the sup norm of vorticity controls the breakdown of smoothness of the solutions. In the other words, smooth solutions can be continued past a time ${T}$ as long as ${\int_0^T \|\omega(t)\|_{L^\infty}\; dt }$ remains finite. In 2D, ${\|\omega(t)\|_{L^\infty} = \| \omega(0)\|_{L^\infty}}$, the global smooth solutions exist. It remains an outstanding open problem for 3D to prove or disprove the breakdown of the smoothness of solutions in finite time (which is a million dollar problem, by the way).

Theorem 6 (Helmholtz law for vorticity) Let ${\phi^t}$ be the Lagrangian map: ${\phi^t (x) = X(t;x)}$.

• For incompressible fluids with constant density, vorticity moves with the fluid in the sense

$\displaystyle \omega (\phi^t(x),t) = \nabla_x \phi^t (x) \omega_0(x).\ \ \ \ \ (16)$

• For compressible barotropic fluids, ${\omega/\rho}$ moves with the fluid.

Proof: The theorem follows from a direct computation. Indeed, for the incompressible flows, we compute

$\displaystyle \frac{d}{dt} \omega (\phi^t(x),t) = (\partial_t + u \cdot \nabla )\omega (\phi^t(x),t)= \omega \cdot \nabla u (\phi^t(x),t)=( \nabla u) \omega (\phi^t(x),t)$

and

$\displaystyle \frac{d}{dt} (\nabla \phi^t (x)\omega_0(x)) = \nabla u (\phi^t(x),t) \nabla \phi^t \omega_0(x) = (\nabla u) \omega (\phi^t(x),t) .$

This yields the identity (16), as it holds initially at time ${t=0}$. Similar computation holds for the compressible case, upon using the continuity equation. $\Box$

Remark 3 For ideal incompressible fluids with constant density, there is no vorticity in the fluids, if none initially (unless there are non-potential forces taken into account in the flows). This is unlike in the viscous fluids where vorticity can be instantaneously created due to the presence of a boundary.

Theorem 6bis Vortex lines are material, that is, they move with the flow: image of a vortex line under the Lagrangian map ${x \mapsto \phi^t(x)}$ is also a vortex line.

Proof: By definition, vortex lines are curves in space, for each time, that are tangent to the vorticity at every point on the curve. Precisely, let ${\gamma}$ be a vortex line, with parametrization ${\gamma(s)}$ for ${0\le s\le 1}$. By definition,

$\displaystyle \gamma'(s) = c_0(s) \omega_0(\gamma(s))$

for some scalar function ${c(s)}$. Let ${\gamma_t(s) = \phi^t (\gamma(s))}$, the image of ${\gamma}$ under the map ${\phi^t}$. The tangent vector at each point on ${\gamma_t}$ reads

$\displaystyle \frac{d}{ds} \gamma_t (s) = \nabla_x \phi^t (\gamma(s)) \gamma'(s) = c_0(s) \nabla_x \phi^t (\gamma(s)) \omega_0(\gamma(s)) = c_0(s) \omega (\gamma_t(s),t),$

in which the last identity was due to the Helmholtz theorem. This proves that ${\gamma_t}$ remains a vortex line. $\Box$

4.3. Circulation

Let ${\gamma}$ be a simple, smooth, oriented closed contour which is a deformation of a circle. The fluid circulation of the velocity field ${u}$ on the contour ${\gamma}$ is defined by the line integral

$\displaystyle \Gamma_\gamma: = \oint_\gamma u \cdot d{\bf s} = \iint_S \nabla \times u \cdot d {\bf S},\ \ \ \ \ (17)$

in which the last identity is due to the Stokes theorem. That is, the circulation yields a computation of vorticity flux through the surface ${S}$, whose boundary is the contour ${\gamma}$.

Theorem 7 (Kelvin’s circulation theorem) For compressible barotropic fluids or incompressible fluids with constant density, the circulation of ${\gamma^t = \phi^t (\gamma)}$ is invariant under the flow:

$\displaystyle \frac{d}{dt} \Gamma_{\gamma^t} =0.$

Proof: We parametrize the contour ${\gamma}$ by ${\gamma(s)}$ for ${0\le s\le 1}$, with ${\gamma(0) = \gamma(1)}$. Then, ${\gamma^t (s)= \phi^t (\gamma(s))}$, and therefore we can compute

\displaystyle \begin{aligned} \frac{d}{dt} \Gamma_{\gamma^t} &= \frac{d}{dt} \int_0^1 u (\phi^t (\gamma(s)), t) \cdot \frac{d}{ds} \phi^t (\gamma(s)) \; ds \\ &= \sum_j \int_0^1\Big[ D_t u_j (\phi^t (\gamma(s)), t) \frac{d}{ds} \phi^t_j (\gamma(s)) + u_j (\phi^t (\gamma(s)), t) \frac{d}{ds} u_j (\phi^t (\gamma(s)), t)\Big] \; ds \\ &= \int_0^1\Big[ \nabla P (\phi^t (\gamma(s)), t) \frac{d}{ds} \phi^t (\gamma(s)) + \frac12 \frac{d}{ds} |u (\phi^t (\gamma(s)), t)|^2\Big] \; ds \\ &= \int_0^1\Big[ \frac{d}{ds} P (\phi^t (\gamma(s)), t) + \frac12 \frac{d}{ds} |u (\phi^t (\gamma(s)), t)|^2\Big] \; ds \\& =0,\end{aligned}

since ${\gamma(s)}$ is a closed curve. $\Box$

4.4. Potential flows

We consider the ideal incompressible fluids with constant density. Potential flows are those with zero vorticity: ${\omega = \nabla \times u =0}$. Equivalently, we can write

$\displaystyle u = \nabla_x \phi.$

The incompressibility yields that ${\phi}$ is a harmonic function; namely,

$\displaystyle \Delta \phi = 0,\qquad \mbox{with} \qquad \frac{\partial \phi}{\partial n} _{\vert_{\partial \Omega}} = h(x,t).$

That is, the time dependence arises only through the boundary condition on ${u\cdot n = h(x,t)}$. For a given ${h}$, ${\phi}$ is uniquely determined, up to a constant, from the integral formula

$\displaystyle \phi(x,t) = - \int_{\partial \Omega} G(x,y) h(y,t) \; d\sigma(y),\ \ \ \ \ (18)$

in which ${G(x,y)}$ denotes the Green kernel of the Laplacian with the Neumann boundary condition. It is useful to recall that the Green kernel in the three dimensional case satisfies

$\displaystyle |\partial_x^k G(x,y)|\le C_0 |x-y|^{2-d-k}, \qquad \forall ~ x\not = y.$

In 2D, the same bound holds, except for the case ${k=0}$ where a factor of ${\log |x-y|}$ enters in the estimate. As for the pressure ${P}$, the momentum equation (12) yields

$\displaystyle \nabla \Big[ \partial_t \phi + \frac{|\nabla \phi|^2}{2} + P\Big] =0.$

That is, up to a constant, the pressure is computed through the Bernoulli equation for unsteady flows (cf. Theorem 5 for steady flows):

$\displaystyle \partial_t \phi + \frac{|\nabla \phi|^2}{2} + P =0.$

Theorem 8 (d’Alembert paradox) For any bounded smooth solid obstacle ${\Omega\subset \mathbb{R}^3}$ that moves with a constant velocity ${U}$ within incompressible, irrotational fluids that are at rest infinitely far away, the total force exerted on the obstacle is zero.

Proof: To fix the fluid domain (to be stationary), we work with the moving frame ${x\mapsto x + Ut}$. Hence, the obstacle is at rest and the fluid is moving at the far field with constant velocity ${U}$. By assumption (and the fact that ${\mathbb{R}^3\setminus \Omega}$ is simply connected), the fluid is of the form ${u = \nabla \phi}$, with ${\phi}$ solving

$\displaystyle \Delta \phi =0 \qquad \mbox{on} \quad \mathbb{R}^3\setminus \Omega, \qquad \mbox{with}\quad \frac{\partial \phi}{\partial n}_{\vert_{\partial \Omega}} = 0$

and

$\displaystyle \lim_{x\rightarrow \infty}\nabla \phi = U.$

The boundary condition on ${\partial \Omega}$ is to assure that the fluid does not enter the obstacle. Up to a constant, we write

$\displaystyle \phi = U \cdot x + \tilde \phi,$

with ${\tilde \phi}$ being the harmonic function with nonzero boundary condition ${\partial_n \tilde \phi = - U\cdot n}$. By a view of (18), we have

$\displaystyle |\partial_x^k \tilde \phi (x) |\le C_0 |x|^{-1-k}, \qquad \mbox{as}\quad x\rightarrow \infty,$

for ${k\ge 1}$. In fact, we have the expansion near infinity for ${\tilde \phi}$, and hence for ${u}$:

$\displaystyle u = U + a_0 \frac{x}{|x|^3} + \mathcal{O}(|x|^{-3})\ \ \ \ \ (19)$

for some constant ${a_0}$. In addition, the Bernoulli relation yields ${ p = \rho_0 \frac{|u|^2}{2}.}$ The total force acting on the obstacle is computed by the surface integral

$\displaystyle F = - \int_{\partial \Omega} pn \; d\sigma(x) .$

Recall that ${u\cdot n =0}$ on the boundary ${\partial \Omega}$. For sufficiently large ${R}$, we compute

\displaystyle \begin{aligned} F &= - \int_{\partial \Omega} pn \; d\sigma(x) = - \int_{\partial \Omega} \Big[ \rho_0(u\cdot n) u + pn\Big] \; d\sigma(x) \\ &= \int_{B_R \setminus \Omega} \Big[ \rho_0 u \cdot \nabla u + \nabla p\Big] \; dx - \int_{\partial B_R} \Big[ \rho_0u \cdot n u + pn\Big]\; d\sigma(x) \\ &= - \int_{\partial B_R} \Big[ \rho_0 u \cdot n u + pn\Big]\; d\sigma(x), \end{aligned}

in which both incompressibility and irrationality were used. Now, when ${u=U}$ and ${p = \rho_0 \frac{|U|^2}{2}}$, the above integral is zero, by the divergence theorem. Using (19) and the fact that ${n = \frac{x}{|x|}}$, we have

\displaystyle \begin{aligned} F &= - \rho_0 \int_{\partial B_R} \Big[ a_0U \cdot n \frac{x}{|x|^3} + u \cdot n U - a_0 U \cdot \frac{x}{|x|^3} n\Big]\; d\sigma(x) + \int_{\partial B_R} \mathcal{O}(\frac{1}{R^3}) d\sigma(x) \\&= - \rho_0 U \int_{\partial B_R} u \cdot n \; d\sigma(x) + \mathcal{O}(\frac{1}{R}) = \mathcal{O}(\frac{1}{R}). \end{aligned}

Again, the incompressibility was used. The theorem follows, by taking ${R \rightarrow \infty}$.

$\Box$

Remark 4 The d’Alembert’s paradox asserts that any solid body emerged in stationary potential flows feels neither lift nor drag acting on it (in the layman words, birds can’t fly!). The paradox is precisely due to the neglect of viscous effects. It was Prandtl in the beginning of the 20th century who showed the significance of viscosity near the boundary, despite it being sufficiently small. However, the mathematical justification of the Prandtl boundary layer theory remains an open problem (I shall discuss this subject later on in the course).

4.5. 2D steady potential flows

We study 2D steady flows which are both incompressible and irrotational. Denote the velocity vector field by ${{\bf u}=(u,v)}$ and the 2D coordinates by ${(x,y)}$. By the incompressibility ${\nabla \cdot {\bf u}=0}$ and irrationality ${\nabla \cdot {\bf u}^\perp =0}$, we write

$\displaystyle (u,v) = \nabla \phi = \nabla^\perp \psi\ \ \ \ \ (20)$

for some potential function ${\phi}$ and some stream function ${\psi}$ (which represent the trajectories of particles). Here, ${\nabla^\perp = (\partial_y, -\partial_x)^t}$. Both ${\phi}$ and ${\psi}$ are harmonic functions. To make use of the power of complex analysis, we introduce complex potential

$\displaystyle w (z)= \phi(x,y) + i \psi(x,y)$

for ${z =x+iy}$. The equations (20) are the Cauchy-Riemann equations for ${w}$. The complex velocity of the fluid is defined by

$\displaystyle w'(z) = u(x,y) - i v(x,y).$

Conversely, given any complex potential function ${w(z)}$, the pair ${(u,v)}$, constructed by ${u = \Re w'}$ and ${v = -\Im w'}$, defines a steady potential flow.

We are interested in computing the forces exerted by the fluid on a solid body. Let ${\Omega}$ be a bounded open domain in ${\mathbb{R}^2}$ and let ${C = \partial \Omega}$. Assume that ${C}$ is not self-intersecting. The force on the body ${\Omega}$ is equal to the pressure force acting on the boundary ${C}$, and is computed by the line integral

$\displaystyle F = - \int_C pn \; dz = i \int_C p (dx + i dy).$

Here, we identify a 2D vector with a complex number in the usual way ${F = F_1 + i F_2}$. Blasius, who was in fact a student of Prandtl, then gave a force calculation in term of complex potential via a complex contour integral as follows.

Theorem 9 (Blasius theorem) Consider the incompressible, steady, and irrotational fluids in the exterior of a solid bounded body ${\Omega}$ in ${\mathbb{R}^2}$. Then, the force exerted by the fluid on the body is computed by

$\displaystyle F = -\frac{i \rho_0}{2} \overline{ \int_C w'(z)^2 \; dz}.$

Proof: By the Bernoulli relation for potential flows, we have ${p = - \rho_0\frac{u^2 + v^2}{2}}$ and so

$\displaystyle F = - \int_C pn dz = - \frac{i\rho_0}{2} \int_C (u^2 + v^2) (dx + i dy).$

On the other hand, since ${{\bf u} \cdot n =0}$ or ${udy = vdx}$ on the boundary ${C}$, one computes

$\displaystyle w'(z)^2 \; dz = (u - iv)^2 (dx + i dy) = (u^2 + v^2) (dx + i dy),$

which yields the theorem. $\Box$

Next, we obtain the following.

Theorem 10 (Kutta-Joukowski theorem) Consider the incompressible, steady, and irrotational fluids in the exterior of a solid bounded body ${\Omega}$ in ${\mathbb{R}^2}$. Assume that the fluid moves with a constant velocity ${ (U,V)}$ at infinity. Then, the force exerted by the fluid on ${\Omega}$ is computed by

$\displaystyle F = -\rho_0 \Gamma_C (V - iU)$

where ${\Gamma_C}$ is the circulation around ${\Omega}$; see (17).

Proof: Let ${w(z)}$ be the complex potential associated with the fluid. Since ${w}$ is analytic and the complex velocity ${w'(z)\rightarrow U- iV}$, there holds the Laurent’s asymptotic expansion near infinity

$\displaystyle w'(z) = a_0 + \frac{a_1}{z} + \frac{a_2}{z^2} + \cdots$

in which ${a_0 = U - iV}$. To compute ${a_1}$, one uses the Cauchy’s theorem, yielding

\displaystyle \begin{aligned} a_1 &= \frac{1}{2\pi i} \int_C w'(z) \; dz = \frac{1}{2\pi i} \int_C (u- iv) (dx + i dy) \\& = \frac{1}{2\pi i} \int_C u dx + v dy = \frac{1}{2\pi i} \int_C {\bf u} \cdot d{\bf s} \\& = \frac{1}{2\pi i} \Gamma_C, \end{aligned}

in which the boundary condition ${udy = v dx}$ was used. Similarly, using the Laurent series, we compute

$\displaystyle \frac{1}{2\pi i}\int_C w'(z)^2 \; dz = 2 a_0 a_1 = \frac{1}{\pi i} (U- iV ) \Gamma_C .$

The theorem is proved, upon using the Blasius theorem. $\Box$

Remark 5 The force exerted by the fluid on the solid body ${\Omega}$ is normal to the direction of the flow ${(U,V)}$ and is proportional to the circulation around ${\Omega}$. In particular, the body experiences no drag, that is, no force opposing the flow. This is of course again a paradox, in contradiction to observation and experiment. In addition, in the absence of the circulation, there is no net force at all acting on the body (neither drag nor lift). This is the 2D version of the d’Alembert paradox, which occurs due to the neglect of viscous effects in the fluid.