A harmonic maps approach to fluid flows

We obtain a complete solution to the problem of classifying all two-dimensional ideal fluid flows with harmonic Lagrangian labelling maps; thus, we explicitly provide all solutions, with the specified structural property, to the incompressible two-dimensional Euler equations (in Lagrangian variables).

capture important aspects of the physical reality are of great importance [3]. Several publications in the applied mathematics, engineering and physics research literature (see [6,7,[10][11][12]15,18] and references therein) exploited a remarkable feature shared by some celebrated explicit solutions to the two-dimensional incompressible Euler equations, in Lagrangian variables (such as Kirchhoff's elliptical vortex [16] found in 1876, Gerstner's flow [9] found in 1809 and re-discovered in 1863 by Rankine [19], and the Ptolemaic vortices found in 1984 by Abrashkin and Yakubovich [1]), namely that in all of them the labelling map is harmonic at all times.
Recently, in [2], the authors proposed a complex analysis approach aimed at classifying all such flows. While new explicit solutions were obtained, the exhaustion of all possibilities was reduced to an explicit nonlinear ordinary differential system in C 4 . Solving this system in full generality proved elusive so far.
We propose a different approach that provides a complete solution to the original problem of finding all flows with harmonic labelling maps. Our approach is based on ideas from the theory of harmonic mappings, more precisely, on the fact that it is possible to characterize the relationship between planar harmonic maps having the same Jacobian-a property of the labelling maps that is a consequence of the equation of mass conservation, expressed in Lagrangian variables. Apart from achieving the full picture, our considerations provide an illustration of the deep links between the fields of complex analysis and fluid mechanics.

The governing equations
The Eulerian description of the two-dimensional motion of an ideal homogeneous fluid is obtained by imposing the law of mass conservation and Euler's equation of motion where u(t, x, y), v(t, x, y) is the velocity field in the time and space variables (t, x, y) and the scalar function P(t, x, y) represents the pressure. Since the reference density in hydrodynamics is 1 g/cm 3 , we normalize the constant fluid density to 1.
The most complete flow representation is provided in (material) Lagrangian coordinates, in which one describes the motion of all fluid particles. For a given velocity field u(t, x, y), v(t, x, y) , the motion of the individual particles x(t), y(t) is obtained by integrating the system of ordinary differential equations whereas the knowledge of the particle path t → x(t), y(t) provides by differentiation with respect to t the velocity field at time t and at the location x(t), y(t) .
Starting with a simply connected domain 0 , representing the labelling domain, each label (a, b) ∈ 0 identifies by means of the injective map the evolution in time of a specific particle, the fluid domain at time t, (t), being the image of 0 under the map (3). To write the governing equations in Lagrangian coordinates, we use the following relations: where J is the Jacobian of the transformation given by The local injectivity of the transformation between the Eulerian and Lagrangian coordinates is expressed by (5). In view of (4) and so that the Euler equation (2), in Lagrangian variables, becomes Since the Jacobian of the above system does not vanish, due to (5), we can solve for the gradient of P in the label space, obtaining P a = − x a x tt − y a y tt , in 0 . The domain 0 being simply connected, the above system is equivalent to the compatibility condition P ab = P ba , that is, x a x btt + y a y btt = x b x att + y b y att , or, equivalently, These considerations show that, in Lagrangian coordinates, the governing equations are equivalent to (7) and (8), under the constraint that, at any instant t, the map (3) is a global diffeomorphism from the label domain 0 to the fluid domain (t).

Harmonic labellings
As already mentioned in Introduction, the common structural property of the known explicit solutions to the governing equations (Kirchhoff's vortex [16], Gerstner's wave [9], the Ptolemaic solutions [1]) is that the labelling map (3) is harmonic at every fixed time t. Our aim is to explicitly find all solutions having this property. Most ideas in this section are inspired by [2] and are included for the sake of completeness. Since from now on our methods will rely exclusively on complex analysis, it is convenient to adapt the notation accordingly. Therefore we look for solutions to (7) and (8) having the form with z → F(t, z) and z → G(t, z) analytic in the simply connected domain 0 ⊂ C, at every time t. Recall that ∂ F ∂z = 0 characterizes analyticity and that The Jacobian of the harmonic map (9) is J = |F | 2 − |G | 2 . Therefore the equation of mass conservation (7) can be re-written as F F − G G t = 0, that is, Relation (10) together with (9) yield A lengthy but straightforward calculation using the above relations shows that (8) is In view of (11) and (12), the governing equations reduce to the single equation for some real-valued function ν. Remark 1 Using the notation F 0 := F(0, ·), G 0 := G(0, ·), the equation of mass conservation expressed as J t = 0 means that the Jacobian of the labelling map is constant in time, and hence it is given by J = |F 0 | 2 − |G 0 | 2 . Since 0 is simply connected and J does not vanish in 0 , we deduce that we either have J > 0 (i.e. F 0 + G 0 is sense preserving) or J < 0 (i.e. F 0 + G 0 is sense reversing) throughout 0 . From now on we shall assume without loss of generality that F 0 + G 0 is sense preserving. Indeed, if F 0 + G 0 is sense reversing, we can replace the label domain by We start by proving the following lemma that might have some independent interest and will be used later on.

Lemma 1 Let ϕ and ψ be two analytic functions in a simply connected domain
in for some real numbers r and s different from zero if and only if there exist two complex constants c 1 and c 2 with |c 1 | 2 = r |c 2 | 2 + s such that ϕ ≡ c 1 and ψ ≡ c 2 .
Proof Note that for (14) to hold, it is necessary that r |ψ(z)| 2 + s ≥ 0 for all z ∈ . If r |ψ| 2 + s ≡ 0, then a direct application of the open mapping theorem for analytic functions gives us the desired result. Assume now that r |ψ| 2 + s is not identically zero. Then, there exists a disk D = D(z 0 , R) centered at some z 0 ∈ and radius R > 0 such that D ⊂ and r |ψ(z)| 2 + s > 0 for all z ∈ D. The forthcoming analysis will be done in this disk.
We take logarithms in (14) to get Now, the function of the left-hand side in (15) is harmonic so that the one on the right-hand side must be harmonic as well. The Laplacian of this latter function equals so that ψ is identically zero in D and hence there exists a non-zero constant c 2 ∈ C such that ψ ≡ c 2 . Bearing in mind (14), we get that ϕ equals a constant c 1 too, with |c 1 | 2 = r |c 2 | 2 + s. A direct application of the identity principle for analytic functions completes the proof.

The case of linear dependence
We first treat the easier case when the harmonic mapping F + G is such that F and G are linearly dependent.
then there exist constants α, β ∈ C such that F 2 = α F 1 and G 2 = β F 1 , with Proof Let us first notice that F 1 has no zeros in since the Jacobian J 1 of F 1 + G 1 is strictly positive. Dividing by |F 1 | 2 in (16), we get We can now apply Lemma 1 to deduce that there exist α, β ∈ C such that F 2 = α F 1 and Taking the modulus on both sides, we get F 1 (z) − F 1 (w) = 0, since |α| = |β|. The univalence of F 1 forces z = w, thus proving the claim.

Linearly independent case
We now investigate the generic setting.
If |k| = 1, then |F 2 | = |F 1 | and the above relation implies |G 1 | = |G 2 |. In view of the open mapping theorem we must have F 2 = e is 1 F 1 and G 2 = e is 2 G 1 for some real constants s 1 and s 2 , and then (17) holds with β = 0, α = e is 1 , and ξ = s 1 + s 2 .
If |k| = 1, then by Lemma 1 we have that ω 1 is constant, which contradicts our assumption. Thus, we can assume that ϕ is not constant, so that its derivative is not identically zero. Therefore, there exists a disk D = D(z 0 , R), centered at some z 0 ∈ and with radius R > 0, contained in such that ϕ (z) = 0 for all z ∈ D. The forthcoming analysis is done in this disk.
As a consequence, we see that unless e iθ p 1 + |m| 2 − mn = 0, the values of ϕ lie on a line. This is not possible for non-constant ϕ. Therefore, we have and also |n| 2 = 1 + |p| 2 .

Remark 2
Assume that F 1 + G 1 and F 2 + G 2 are two sense preserving harmonic mappings in the simply connected domain , related by (17). In the case when F 1 +G 1 is univalent and ξ = 0, then F 2 + G 2 is obtained by composing with a sense preserving affine transformation. Since this affine transformation preserves univalence, F 2 + G 2 is univalent as well. This is not true if ξ = 0. For example, the so-called harmonic Koebe function introduced in [5] (see also [8,Sect. 5 3 and g(z) = is univalent in the unit disk D but there exists |μ| = 1 such that f + μg is not univalent (see [13,Theorem 7.1]).

The solutions
In what follows we will set and use the notation F 0 + G 0 for the function F(0, ·) + G(0, ·), which is supposed to be univalent in the simply connected domain 0 . Also, we write f 0 = f (0, ·) and g 0 = g(0, ·).

Solutions in the linearly dependent case
We start by finding the solutions f ≡ 0 and g ≡ 0 such that the governing Eq. (13) holds under the additional assumption that the initial harmonic (sense preserving) labelling map F 0 + G 0 is such that F 0 and G 0 are linearly dependent, or, in other words, F 0 + G 0 has constant dilatation. Indeed, since the harmonic map F 0 + G 0 is sense preserving, the linear dependence translates into the fact that there exists a constant λ ∈ C with |λ| < 1 such that G 0 = λF 0 .

Theorem 3
Let 0 ⊂ C be a simply connected domain. If the initial harmonic (univalent, sense preserving) labelling map F 0 + G 0 satisfies G 0 = λF 0 for some λ ∈ C, then the particle motion (9) of a fluid flow, defined by means of (24), is given by where β : [0, ∞) → C is a C 1 function with β(0) = λ, and ν 0 ∈ R is an arbitrary constant.

Example 1 Kirchhoff's solution
where A and k are non-zero real constants and λ ∈ (0, 1). The condition on the univalence of F 0 requires that 0 does not contain points z and w with I m{z} = I m{w} and Re{z} = Re{w} + 2mπ k for some integer m.

Theorem 4
Let 0 ⊂ C be a simply connected domain. Assume that the initial harmonic (univalent, sense preserving) labelling map F 0 + G 0 is such that F 0 and G 0 are linearly independent. The particle motion (9) of a fluid flow, defined by means of (24), is either described by where β : [0, ∞) → C is a C 1 function and ν 0 ∈ R, or by where ν 0 is as above and ξ 0 ∈ R = {0}. Moreover, for the solutions (28), univalence at any time is ensured once it holds at time t = 0. Univalence holds for the solutions (29) if and only if F 0 + λG 0 is univalent for all λ with |λ| = 1.
Proof By Theorem 2, we know that there exist C 1 functions α, β : [0, ∞) → C with |α(t)| 2 − |β(t)| 2 = 1 for all t ≥ 0 and ξ : [0, ∞) → R such that Note that since f (0, ·) = f 0 , g(0, ·) = g 0 , and the dilatation of F 0 + G 0 is nonconstant, we have the initial conditions A straightforward calculation shows that Now, by (13), we know that f t f − g g t = iν(z, z). Since the function | f 0 | 2 − |g 0 | 2 only depends on z and z and is always different from zero, we obtain that where ν = ν/((| f 0 | 2 −|g 0 |) 2 ). Let us re-write the previous equation using that Note that since we are assuming that ω is not constant, we have that there exists an open disk D ⊂ 0 such that ω (z) = 0 for all z ∈ 0 . From now on, we will assume that z ∈ D. Within these terms (33) becomes Taking derivatives with respect to z in (34), we obtain that the function only depends on z and on z. Hence, so does which has derivative with respect to z equal to Since ω is supposed to be non-zero, we have that where γ is a function of z and z. This fact finally implies (again taking derivatives with respect to z in (35) and dividing out by ω ) the equation for some constant c 1 and, in particular, by using this information in (35) we get that (|α| 2 +|β| 2 )ξ t is constant as well. Moreover, by (32), we see that αα t −ββ t is constant too (indeed, using that |α| 2 − |β| 2 = 1, it is easy to check that α t α − β t β is a purely imaginary complex number). In other words, we have seen that the following system must be satisfied by the functions α, β, and ξ (here, c 1 , c 2 , and c 3 are certain constants with c 3 ∈ R): Now, note that the second and fourth equations in (36) can be re-written as which gives On the other hand, using the first equation in (36), a direct consequence of (37) is that This shows that ξ 2 t (hence ξ t ) must be a constant function. We distinguish between two types of solutions. Case 1 ξ t ≡ 0. Then ξ must be constant and hence, by (31), we see that ξ ≡ 0. Moreover, in this case (36) becomes α t α − β t β = ic 3 , |α| 2 − |β| 2 = 1.