A steady smooth Euler flow with support in the vicinity of a helix

TL;DR

This paper constructs a smooth Euler flow localized near a helix, extending previous work on circular support to more complex geometries, demonstrating the flexibility of Euler flows around curved structures.

Contribution

It introduces a new class of smooth Euler flows supported near a helix, generalizing earlier solutions supported on circles.

Findings

Flow is supported in a neighborhood of a helix.

Generalizes previous circular support solutions.

Demonstrates the existence of smooth Euler flows around complex curves.

Abstract

In this article we construct a smooth Euler flow supported in a neighborhood of a helix. It may be considered a generalization of a similar solution found by the author for a circle.

Equations

$$\left\{\begin{array}[]{rl}u_{\rho}\frac{\partial}{\partial\rho}u_{\rho}+\frac{1}{\rho}u_{\varphi}\frac{\partial}{\partial\varphi}u_{\rho}+u_{z}\frac{\partial}{\partial z}u_{\rho}-\frac{1}{\rho}u_{\varphi}^{2}=-\frac{\partial}{\partial\rho}p,\,\,\,\,\,\,\,\,\,\,(2a)\\ u_{\rho}\frac{\partial}{\partial\rho}u_{\varphi}+\frac{1}{\rho}u_{\varphi}\frac{\partial}{\partial\varphi}u_{\varphi}+u_{z}\frac{\partial}{\partial z}u_{\varphi}+\frac{1}{\rho}u_{\rho}u_{\varphi}=-\frac{1}{\rho}\frac{\partial}{\partial\varphi}p,\,\,\,\,\,\,\,\,\,\,(2b)\\ u_{\rho}\frac{\partial}{\partial\rho}u_{z}+\frac{1}{\rho}u_{\varphi}\frac{\partial}{\partial\varphi}u_{z}+u_{z}\frac{\partial}{\partial z}u_{z}=-\frac{\partial}{\partial z}p.\,\,\,\,\,\,\,\,\,\,(2c)\end{array}\right.$$

$$u\cdot \nabla \left(\frac{1}{2}|u{|}^2+p\right)=0$$

$$x=\rho ,y=z-k\phi .$$

$$\rho (u_{\rho }\frac{\partial }{\partial \rho }{u}_{\phi }+\frac{1}{\rho }{u}_{\phi }\frac{\partial }{\partial \phi }{u}_{\phi }+u_z\frac{\partial }{\partial z}{u}_{\phi }+\frac{1}{\rho }{u}_{\rho }{u}_{\phi })+k(u_{\rho }\frac{\partial }{\partial \rho }{u}_z+\frac{1}{\rho }{u}_{\phi }\frac{\partial }{\partial \phi }{u}_z+u_z\frac{\partial }{\partial z}{u}_z)=-\xi (p)=0,$$

$$u\cdot \nabla h=0.$$

$$\frac{\partial }{\partial x}t=F,{\left(\frac{\partial }{\partial y}t\right)}^2=G$$

$$\frac{\partial }{\partial x}\left(\frac{G}{x^2}\right)=-\frac{2F}{1+k^2}+\frac{2}{x}{\left(\frac{xh+kF}{x^2+k^2}\right)}^2,$$

$$S=h^2(1+k^2)-3t(c+k^2).$$

$$h(0)=0,c(0)=1.$$

$$h=s+\frac{8k}{3(1+k^2)}{s}^2+\frac{82k^2-189}{36(1+k^2{)}^2}{s}^3+\frac{k(154k^2-369)}{27(1+k^2{)}^3}{s}^4+O(s^5),$$

$$4(1+k^2)(X^2-2s^2)+4(3+k^2)X^3-16k{X}^{2}s-4(2k^2+3)X{s}^2+\frac{64}{3}k{s}^3\le O(X^4+s^4).$$

$$t\ge t_{min}(X)\ge 0$$

$$\tilde{u}=\omega u,d\tilde{p}={\omega }^{2}dp.$$

$$\alpha =4t,H(\alpha )=16h^2(t).$$

$$\frac{\partial }{\partial x}t=F=\frac{1}{2}x(x^2-c);$$

Full text

1. Introduction

In this article we construct a smooth Euler flow in ${\mathbb{R}}^{3}$ supported in a neighborhood of a helix. While it may be considered a generalization of the solution of the Euler equation found by the author in [2], it is not of very much interest by itself. (The whole point of [2] was to find a smooth Euler flow with compact support. Of course, a helix is completely useless for this purpose.) Maybe this solution could be a stepping-stone to more interesting generalizations.

It is convenient to interpret this new flow as a modification of the old one, and for this reason the notation we use here is similar to [2]. We consider a helix ${\mathcal{C}}\subset{\mathbb{R}}^{3}$ described in the standard cylindrical coordinates $(\rho,\varphi,z)$ by the equations $\rho=1,\,z=k\varphi$ where $k>0$ is the slope. This curve has a group of isometries with the generator $\xi=\partial_{\varphi}+k\partial_{z}$, taking advantage of this we construct a flow which is invariant under this isometries. This flow retains simple topology described by the Arnold’s theorem [1, Ch. II, Theorem 1.2] except invariant tori become invariant cylinders.

2. The Euler equation in cylindrical coordinates

2.1. Preliminaries

A vector field in cylindrical coordinates is usually written using the local basis $e_{\rho}=\partial_{\rho},\,e_{z}=\partial_{z},\,e_{\varphi}=\frac{1}{\rho}\partial_{\varphi}$. In this coordinates the incompressibility condition $\mathop{div}\nolimits u=0$ for $u=u_{\rho}e_{\rho}+u_{z}e_{z}+u_{\varphi}e_{\varphi}$ becomes

[TABLE]

and the Euler equation itself $(u\cdot\nabla)u=-\nabla p$ is [3, §15]

[TABLE]

Following [2], we assume that

[TABLE]

(Note that with this additional equation the system (1-3) becomes overdetermined.) In this case the Bernoulli law

[TABLE]

implies $u\cdot\nabla p=0$. In general, we will call an Euler flow satisfying the latter condition localizable (for reasons explained in §4.2).

2.2. The flow

We are looking for a solution of the form

[TABLE]

where $t$ and $h$ are functions of the variables

[TABLE]

This field is obviously invariant under the isometries, and it is not difficult to check that $\mathop{div}\nolimits u=0$. Also, it follows directly from (4) that $u\cdot\nabla t=0$, hence to satisfy one of the necessary conditions $u\cdot\nabla p=0$ it is sufficient to assume111In fact, this assumption is more or less unavoidable. that $p=p(t)$. There is some leeway in chosing this function (due to the modification discussed in Sec. 4.2), a convenient choice is

[TABLE]

2.3. The equations

Combining (2b) and (2c) we have

[TABLE]

or simply

[TABLE]

This equation would follow if we assume222This is also not very much of an assumption because it is not difficult to see that $dt\wedge dh=0$. that $h=h(t)$ is also a function of $t$.

[TABLE]

What is left is the remaining pair of the Euler equations together with (3), which takes the form

[TABLE]

Changing the variables in (2a), (2c) from $(\rho,z)$ to $(x,y)$ we then have

[TABLE]

[TABLE]

3. The solution

3.1. The change of variables

Following the same tactic as in [2] we assume that

[TABLE]

where $F,G$ are functions of $(x,t)$. This assumption obviously implies that this two functions satisfy the following partial differential equation

[TABLE]

What we want is to rewrite the equations (6-7) in terms of $F$ and $G$. To begin with, (6) simply turns into an algebraic relation

[TABLE]

Taking into account that $t_{x}=F,\,t_{xx}=F_{x}+FF_{t},\,\,t_{xy}=t_{y}F_{t},$ we may write (7b) as

[TABLE]

This linear differential equation has a solution

[TABLE]

where $c=c(t)$. Finally, using $t_{y}^{2}=G,\,\,t_{yy}=\frac{1}{2}G_{t}$ and (8), the last equation (7a) may be rewritten as

[TABLE]

which is actually a consequence of (9) and (10).

3.2. The ODE

Under the assumptions we have made all the original equations (1-3) are satisfied. However, there is also the new one (8) which is not done yet. This equation contains two unknown functions of $t$, namely $h$ and $c$. After substituting (9) and (11) into (8) and obvious algebraic transformations, it is possible to get rid of the variable $x$ and reduce this PDE to two (rather cumbersome) ordinary differential equations,

[TABLE]

where

[TABLE]

We are interested in a solution of this system with initial condition

[TABLE]

Note that the denominator at this point becomes zero, so this is a singular Cauchy problem. It has no analytic solutions, but one can show333Apparently, it is not possible to reduce this system to a Briot-Bouquet equation the way it is done in [2]. We have to prove this fact using the series directly, which is straightforward but somewhat bothersome. that it has a solution in the form of a Puiseux series, analytic as a function of $s=\sqrt{t}$

[TABLE]

[TABLE]

4. Completing the construction

4.1. The variable $y$

Now we have to change the variables from $(x,t)$ back to $(x,y)$. This part is slightly more complicated then in [2] because in this case we have two completely different solutions instead of just one. It is easy to see why this happens if we take a closer look at the geometry. The streamlines of the original flow $u$ in [2] have the form of slightly deformed helices winding around the circle ${\mathcal{C}}$. We still have the same picture when ${\mathcal{C}}$ itself becomes a helix, except in this case it does matter if the helicity of ‘‘small’’ helices is the same as the ‘‘big’’ one or the opposite. The first choice corresponds to $s>0$, and the second one to $s<0$.

The function $G$ given by (9) is obviously real analytic as a function of $x$ and $s=\pm\sqrt{t}$ at the point $(x,s)=(1,0)$. However, as it is supposed to be equal to the square of $\frac{\partial t}{\partial y}$, the region $G<0$ is forbidden. A direct computation shows that the condition $G\geq 0$ is equivalent to

[TABLE]

which means (using $X=x-1$)

[TABLE]

This domain consists of two parts corresponding to $s>0$ and $s<0$. (With the common point $X=s=0$; note that at $s=0$ the left hand side of (14) factors as $(x^{2}-1)^{2}(x^{2}+k^{2})$.) In variables $(X,t)$ it may be described somewhat more explicitly as

[TABLE]

where $t_{min}(X)=\frac{1}{2}X^{2}+O(X^{4})$, but the function $t_{min}$ depends on the choice between $s=\sqrt{t}$ and $s=-\sqrt{t}$.

The rest of the construction is similar to [2]. We introduce the function $y$ by

[TABLE]

the form on the right hand side of (15) is closed because of (8) and exact because the domain may be chosen simply connected. We may assume that $y=0$ for $t=t_{min}$, which allows us to extend $t$ to a function $t(x,y)=t(x,-y)$ analytic near $(x,y)=(1,0)$ except for the point itself. (Not all of this is immediately obvious, but the argument is the same as in [2, Lemma 3].)

4.2. The modification

As explained in [2, §3], a localizable (i.e. satisfying $u\cdot\nabla p=0$) Euler flow $(u,p)$ can be modified to obtain another Euler flow,

[TABLE]

Choosing a smooth function $\omega$ such that $\omega(t)=0$ for $t\not\in[\varepsilon,2\varepsilon]$, we can obtain a smooth Euler flow $\widetilde{u}$ with support near the helix. (This is why we call such a flow ‘‘localizable’’: it allows modifications which can reduce its support.)

4.3. A comparison with the circle

It is possible to consider the Euler flow $u$ constructed in [2] as a degenerate case corresponding to $k=0$ (when our helix turns into a circle). In terms of [2] for our flow we have $R=1$ and

[TABLE]

If $k=0$ then

[TABLE]

comparing this with

[TABLE]

in [2] we must conclude that $c(t)=\psi(\alpha)$. Now (12) become

[TABLE]

excluding $h^{2}$ from this system we have a second order equation

[TABLE]

the same as in [2, Lemma 1].

It should be noted that for $k=0$ the function $t=t(x,y)=\frac{1}{4}\alpha$ is analytic at the point $(1,0)$ while for $k>0$ it is not, although this difference cannot be seen from just the main term of the asymptotic,

[TABLE]

This fact is related to the choice between two solutions mentioned above, which actually correspond to different analytic branches of this function. (The branching curve of $t$ in the complex $(x,y)$ plane has only one real point, so outside of it this function is real analytic.)

5. Some observations

5.1. Beltrami flows

The author would like to point out that an Euler flow satisfying the condition444Or any relation of the form $|u|^{2}=f(p)$ for that matter. $|u|^{2}=3p$ may be interpreted as a special case of a Beltrami flow. Indeed, a modification

[TABLE]

with $\omega=p^{-\frac{5}{6}}$ satisfies $\frac{1}{2}|\widetilde{u}|^{2}+\widetilde{p}=0,$ so by the known identity [3, §2]

[TABLE]

we have $\widetilde{u}\times\mathop{curl}\nolimits\widetilde{u}=0$ i.e. $\mathop{curl}\nolimits\widetilde{u}=\lambda\widetilde{u}$ for some function $\lambda$. This Beltrami flow is localizable because $|u|^{2}=3p$ implies $u\cdot\nabla p=0$. Conversely, given a localizable Beltrami flow $\widetilde{u}$ one can modify it to obtain a solution with $|u|^{2}=3p$. From the theoretical perspective Beltrami flows are more convenient to consider, so we will take this point of view in this section.

5.2. The Grad-Shafranov equation

Let

[TABLE]

be the Killing vector field from above. Under close examination, the field

[TABLE]

is another Beltrami flow,

[TABLE]

The Beltrami modification $\widetilde{u}$ of the field $u$ we have constructed in §4 may be written in the form

[TABLE]

where $\psi=t^{\frac{1}{6}},\,\chi=\frac{1}{6}t^{-\frac{5}{6}}h(t)$.

Now we may change the perspective and ask the following question: if $\widetilde{u}$ is some field given by (18) with $\xi(\psi)=\xi(\chi)=0$, what additional conditions this two functions must satisfy to make it a Beltrami flow? Note that we have $\mathop{div}\nolimits\widetilde{u}=0$ automatically, so this is a question about the vorticity. Using the formula555Where $[\cdot,\cdot]$ is the Lie bracket. [1, Ch II, §1]

[TABLE]

we have

[TABLE]

Note that $\xi$ is a generator of isometry and $\xi(\psi)=0$. It follows that $[\xi,\nabla\psi]=0$ hence $[a,\nabla\psi]=2(\nabla\log|\xi|,\nabla\psi)a$ and

[TABLE]

Assuming that

[TABLE]

and taking into account that both gradients are orthogonal to $a$, we must conclude that

[TABLE]

which essentially implies $\chi=h(\psi),\,\lambda=-h^{\prime}(\psi)$ (for some function $h$). Then the condition can be written in the form

[TABLE]

In the case $k=0$ we have $|\xi|=\rho$ and the equation becomes

[TABLE]

which is known as (a special case of) the Grad-Shafranov equation.

One can see that our construction was actually built on a Killing field. If we drop the assumption that $\xi$ is a generator of isometry then, apparently, we have no means to control the Lie bracket and the whole construction falls apart. It looks like the axial or helical symmetry of a flow was not merely a simplification to make the calculation easy but is necessary to make the ends meet. If there are any localizable Euler flows which are not symmetric the author does not really know.

5.3. Special Beltrami flows on Riemannian manifolds?

Unfortunately, the Euclidean space has no one-parameter isometry groups besides what we have already considered. However, we may take a more broad view and ask about possible generalizations of the above construction to Riemannian manifolds. The Euler equation on a Riemannian manifold is the same as in the Euclidean space, except $\nabla u$ must now be interpreted as the covariant derivative of $u$ with respect to the Levi-Civita connection [1]. A Belirami flow on an oriented manifold of dimension three is defined as usual, it is a vector field $u$ satisfying

[TABLE]

The fact that such a flow obeys the Euler equation666With $p=-\frac{1}{2}|u|^{2}+\text{const}$. follows from the formula (16). We assume forth that our Riemannian manifold has a Killing vector field $\xi$. In this case (§5.4)

[TABLE]

and the field $a=|\xi|^{-2}\xi$ is again a Beltrami flow. Indeed, $\mathop{div}\nolimits a=0$ because $\xi(|\xi|^{-2})=0$ and $\mathop{div}\nolimits\xi=0$, and

[TABLE]

because the vectors $\xi$ and $\nabla|\xi|^{2}$ are orthogonal. Thus, we have

[TABLE]

for some function $\mu$.

We may again try to construct a Belirami flow of the form

[TABLE]

assuming that $\xi(\psi)=\xi(\chi)=0$. In this case $\mathop{div}\nolimits u=0$ and

[TABLE]

Repeating the computation from §5.3, we have $\chi=h(\psi)$ and

[TABLE]

which may be considered a generalization of the Grad-Shafranov equation (21).

The way to make this Beltrami flow localizable is to assume that $|u|^{2}$ depends on $\psi$. We obviously have $|u|^{2}=|\xi|^{-2}(|\nabla\psi|^{2}+h^{2}(\psi)),$ so this condition means

[TABLE]

for some function $f$. All the variables in (23,24) are invariant under isometries generated by $\xi$, so this is, in fact, a PDE in the (two-dimensional) space of orbits rather then in the original manifold. The problem is that it is overdetermined and does not seem easy to handle. To overcome this obstacle in our special case we have de facto introduced a somewhat contrived vector field $U(\psi)$ depending on $\psi$ as a parameter, and then showed that both (23) and (24) follow from the same equation

[TABLE]

However, it may be difficult (if possible at all) to pull off the same trick in the general case.

5.4. Proofs of some formulas

We will prove here vector calculus formulas (16, 19, 22) used in the last section. All of this proofs are very simple, but the formulas are important and the author has yet to see them derived properly in the literature. So, he decided to write them down for the sake of a reader’s convenience.

We are dealing with an oriented Riemannian manifold of dimension three. Let $(x_{1},x_{2},x_{3})$ be local coordinates and $\partial_{i}=\frac{\partial}{\partial x^{i}}$ be the corresponding vector fields; naturally, we assume that the frame $(\partial_{1},\partial_{2},\partial_{3})$ agrees with the orientation. For a vector field $u=u^{i}\partial_{i}$ we have the following basic formulas

[TABLE]

where $\epsilon$ is the Levi-Civita tensor777Not the Levi-Civita symbol. In this notation, $\epsilon_{123}=\sqrt{|g|}$ and $\epsilon^{123}=1/\sqrt{|g|}$..

Unfortunately, from the presentation in the book [1, Ch. II, §1] it is unclear if the formula (16) was only meant for the Euclidean space or is valid regardless of the metric. However, it is not difficult to see that the latter is the case (there are no higher order derivatives of the metric, so curvature terms do not appear). Indeed, we have

[TABLE]

[TABLE]

The formula for the curl of a cross product is invariably omitted from vector calculus textbooks because of the Lie bracket (which, apparently, is considered inappropriate for undergraduates). If $C=A\times B$, then

[TABLE]

because $\nabla\epsilon=0$. Thus

[TABLE]

[TABLE]

A proof of the last one is not any more complicated. If $\xi$ is a Killing vector, then $\nabla_{j}\xi_{i}+\nabla_{i}\xi_{j}=0$ by definition, hence

[TABLE]

Comparing this equality with (16), we have

[TABLE]