Burst of Point Vortices and Non-uniqueness of 2D Euler Equations

Abstract

We give a rigorous construction of solutions to the Euler point vortices system in which three vortices burst out of a single one in a configuration of many vortices; equivalently we show that there exist configurations of arbitrarily many vortices in which three of them collapse in finite time. As an intermediate step, we show that well-known self-similar bursts and collapses of three isolated vortices in the plane persist under a sufficiently regular external perturbation. We also discuss how our results produce examples of non-unique weak solutions to 2-dimensional Euler’s equations—in the sense introduced by Schochet—in which energy is dissipated.

Appendix A. Proofs of Technical Lemmas

1.1 Proof of Lemma 2.3

First of all, we discuss the existence of suitable parameters $a>0$, $b\in {\mathbb {R}}$, $a_1,a_2,a_3\in {\mathbb {C}}$ as in the statement of the lemma. The three equations of (2.3) (for $j=1,2,3$) are not linearly independent, and setting for brevity $\chi =\frac{6\pi i}{\xi }(a-ib)$, they provide two independent relations between $a_1,a_2,a_3$:

$$\begin{aligned} \chi {{\bar{a}}}_2&=\left( \frac{2}{a_2-a_3}-\frac{1}{a_2-a_1}\right) ,\\ \chi {{\bar{a}}}_3&=\left( \frac{2}{a_3-a_2}-\frac{1}{a_3-a_1}\right) . \end{aligned}$$

We can reduce a variable by imposing that the centre of vorticity is $0=\xi _1z_1+\xi _2z_2+\xi _3z_3$, since this implies, for our choice of intensities, that $a_1=2(a_2+a_3)$. Simple passages then lead to an equation involving only $a_2,a_3$:

$$\begin{aligned} |a_2|^2+|a_3|^2+4{{\,\mathrm{Re}\,}}({{\bar{a}}}_2 a_3)=0. \end{aligned}$$

Solutions $a_2,a_3$ to this last equation are of course not unique. From now on, since our aim is only existence of a set of parameters $a,b,a_1,a_2,a_3$ as in the statement, we can impose further conditions to reduce computations to particular, amenable cases.

Let us thus impose that $a_3=\lambda \in {\mathbb {R}}$ is real (in fact this is without loss of generality, by rotation invariance of the system), so that if $a_2=A+iB$ we want to solve

$$\begin{aligned} \lambda ^2+A^2+B^2+4\lambda A=0, \end{aligned}$$

and from relations above given such $\lambda ,A,B$ one can retrieve $a_1$ and

$$\begin{aligned} \chi =\frac{6\pi i}{\xi }(a-ib)=-\frac{3}{\lambda }\cdot \frac{A+i B+\lambda }{(A+i B-\lambda )(2A+2iB+\lambda )}. \end{aligned}$$

This last relation is important: since $a>0$, if $\lambda ,A,B$ can be chosen so that the imaginary part of $\chi$ can take any value in ${\mathbb {R}}{\smallsetminus }\left\{ 0\right\}$, so will $\xi$, as we are requiring. Elementary computations reveal that the following two sets of parameters

$$\begin{aligned} \xi >0,\,a= & {} \frac{\sqrt{3}}{84\pi }\xi ,\,b=\frac{5}{84\pi }\xi ,\, a_1=-2+i2\sqrt{3},\,a_2=-2+i\sqrt{3},\,a_3=1;\\ \xi <0,\,a= & {} -\frac{\sqrt{3}}{84\pi }\xi ,\,b=\frac{5}{84\pi }\xi ,\, a_1=2+i2\sqrt{3},\,a_2=2+i\sqrt{3},\,a_3=-1; \end{aligned}$$

satisfy the statement of the Lemma, allowing a choice of $a,b,a_1,a_2,a_3$ for any $\xi \in {\mathbb {R}}{\smallsetminus }\left\{ 0\right\}$.

Concerning the Hölder regularity of such a solution, for $t>s>0$,

$$\begin{aligned} |Z(t)-Z(s)| \leqq&|Z(t) - \sqrt{2as}\,e^{i \frac{b}{2a} \log t}| + |\sqrt{2as}\,e^{i \frac{b}{2a} \log t}-Z(s)| \\ =&\sqrt{2at} - \sqrt{2as} + \sqrt{2as}\, |e^{i \frac{b}{2a} \log t}-e^{i \frac{b}{2a} \log s}| \\ \leqq&\sqrt{2a(t-s)} + \sqrt{2as}\, \frac{|b|}{2a}\left( \log t - \log s \right) \\ =&\sqrt{2a(t-s)} + \sqrt{2as}\, \frac{|b|}{2a} \int _s^t \frac{\hbox {d}r}{r} \leqq \sqrt{2a(t-s)} + \sqrt{2a}\, \frac{|b|}{2a} \int _s^t \frac{\hbox {d}r}{r^{1/2}} \\ =&\sqrt{2a(t-s)} + \sqrt{2a}\frac{|b|}{2a} \frac{\sqrt{t}-\sqrt{s}}{2} \leqq \sqrt{2 a} \left( 1+\frac{|b|}{4a}\right) \sqrt{t-s}, \end{aligned}$$

which concludes the proof.

1.2 Proof of Lemma 2.7: Expansions and Leading Orders

Let the notation established in Section 2 prevail. In order to express the rational functions of $z_1,z_2,z_3$ in terms of the new coordinates, we will make use of elementary Taylor expansions.

Let us consider

$$\begin{aligned} Q_j(x)=\frac{1}{1-x-{a_j}/{a_1}}, \quad x \in {\mathbb {C}}, \quad j=2,3 \end{aligned}$$

for small values of |x|. Since $a_j \ne a_1$, $Q_j$ is holomorphic for $|x|<\rho _j= |1-a_j/a_1|/2$, and for the latter values $Q_j$ has a convergent power series expansion $Q_j(x)=\sum _{n=0}^{\infty } c_{j,n} x^n$. Remainders of the series,

$$\begin{aligned} R_{j,m+1}(x) =Q_j(x) - \sum _{n=0}^{m} c_{j,n} x^n =\sum _{n=m+1}^{\infty } c_{j,n} x^n, \end{aligned}$$

(A.1)

are holomorphic functions satisfying, for $|x|<\rho _j / 2$,

$$\begin{aligned} |R_{j,m+1}(x)| \leqq 2 \rho _j^{-m-2} |x|^{m+1},\quad |R'_{j,m+1}(x)| \leqq C_m \rho _j^{-m-2}|x|^{m}, \end{aligned}$$

(A.2)

$C_m>0$ depending only on m. Similar estimates hold for the functions

$$\begin{aligned} Q_{j,k}(x) = \frac{1}{x+\frac{a_j}{a_1}- \frac{a_k}{a_1}}, \quad |x|<|a_j-a_k|/{4|a_1|}; \end{aligned}$$

we denote by $R_{j,k,m}$ remainders of $Q_{j,k}$.

The dynamics of r and $\theta$ are the simpler ones. For the former we have

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}(r^2) =&\frac{1}{|a_1^2|} \frac{\hbox {d}}{\hbox {d}t}(z_1 {\bar{z}}_1) =\frac{2}{|a_1^2|} \text {Re} \left( \frac{z_1}{2\pi i} \sum _{k \ne 1} \frac{\xi _k}{z_1-z_k} + z_1 f(t,z_1)\right) \\ =&\frac{2}{|a_1^2|} \text {Re} \left( \frac{1}{2\pi i} \sum _{k \ne 1} \frac{\xi _k}{1-x_k-\frac{a_k}{a_1}} \right) + \frac{2}{|a_1^2|} \text {Re} \left( z_1 f(t,z_1) \right) . \end{aligned}$$

Recalling (A.1) and making use of (2.3), we rewrite the expression above as

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}(r^2) =&\frac{2}{|a_1^2|} \text {Re} \left( \frac{1}{2\pi i} \sum _{k \ne 1} \frac{a_1\xi _k}{a_1-a_k} + \frac{1}{2\pi i} \sum _{k \ne 1} R_{k,1}(x_k) \right) + \frac{2}{|a_1^2|} \text {Re} \left( z_1 f(t,z_1) \right) \\ =&2a + \frac{2}{|a_1^2|} \text {Re} \left( \frac{1}{2\pi i} \sum _{k \ne 1} R_{k,1}(x_k) \right) + \frac{2}{|a_1^2|} \text {Re} \left( z_1 f(t,z_1) \right) \\ =&2a + \omega _{r}(x_2,x_3) + \frac{2}{|a_1^2|} \text {Re} \left( z_1 f(t,z_1) \right) , \end{aligned}$$

where, by (A.2), there exist $C,\rho '>$ depending only on $\xi ,a_1,a_2,a_3$ such that

$$\begin{aligned} \forall |x_2|, |x_3| < \rho ',\quad |\omega _{r}(x_2,x_3)| \leqq C \left( |x_2| + |x_3|\right) ,\quad |\nabla \omega _{r}(x_2,x_3)| \leqq C. \end{aligned}$$

By an analogous computation,

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\theta&=-i \frac{\hbox {d}}{\hbox {d}t} \log \left( \frac{z_1}{a_1 r}\right) =-i\frac{r}{z_1} \frac{\hbox {d}}{\hbox {d}t}\left( \frac{z_1}{r}\right) \\&={{\,\mathrm{Im}\,}}\left( \frac{\dot{z}_1}{z_1}-\frac{\dot{r}}{r}\right) ={{\,\mathrm{Im}\,}}\left( \frac{\dot{z}_1}{z_1}\right) \\&=-{{\,\mathrm{Im}\,}}\left( \frac{1}{2\pi i {{\bar{z}}}_1} \sum _{k \ne 1} \frac{\xi _k}{z_1-z_k} \right) + {{\,\mathrm{Im}\,}}\left( \frac{\overline{f(t,z_1)}}{z_1} \right) \\&=\frac{b}{r^2} -{{\,\mathrm{Im}\,}}\left( \frac{|a_1|^2}{2\pi i r^2} \sum _{k \ne 1} R_{k,1}(x_k)\right) +{{\,\mathrm{Im}\,}}\left( \frac{\overline{f(t,z_1)}}{z_1} \right) \\&=\frac{b}{r^2} + \frac{\omega _{\theta }(x_2,x_3)}{r^2} +{{\,\mathrm{Im}\,}}\left( \frac{\overline{f(t,z_1)}}{z_1} \right) , \end{aligned}$$

where we have used the fact that ${{\,\mathrm{Re}\,}}\left( \frac{\dot{z}_1}{z_1}\right) =\frac{\dot{r}}{r}$ (which is easily verified by the definitions). Here, again by (A.2), $\omega _{\theta }$ satisifies

$$\begin{aligned} \forall |x_2|, |x_3| < \rho ', \quad |\omega _{\theta }(x_2,x_3)| \leqq C \left( |x_2| + |x_3|\right) ,\quad |\nabla \omega _{\theta }(x_2,x_3)| \leqq C, \end{aligned}$$

possibly reducing constants $C,\rho '>0$ depending again only on $\xi ,a_1,a_2,a_3$.

The evolution of $x_j$, $j=2,3,$ requires more care. It holds that

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t} x_j&= \frac{1}{z_1^2}\left( z_1 \frac{\hbox {d}}{\hbox {d}t} z_j - z_j \frac{\hbox {d}}{\hbox {d}t} z_1\right) \\&=\frac{1}{z_1} \overline{\left( \frac{1}{2\pi i} \left( \frac{\xi _k}{z_j-z_k}+ \frac{\xi _1}{z_j-z_1} \right) + f(t,z_j)\right) }\\&\quad -\frac{z_j}{z_1^2}\overline{\left( \frac{1}{2\pi i} \sum _{\ell \ne 1} \frac{\xi _\ell }{z_1-z_\ell } + f(t,z_1)\right) }. \end{aligned}$$

Let us first study the part of vector field due to vortices interactions; we have

$$\begin{aligned} \frac{1}{2\pi i}\left( \frac{\xi _k}{z_j-z_k}+ \frac{\xi _1}{z_j-z_1}\right) =&\frac{1}{2\pi i z_1} \left( \frac{\xi _k}{x_j-x_k + \frac{a_j}{a_1}- \frac{a_k}{a_1}}+ \frac{\xi _1}{x_j+ \frac{a_j}{a_1} -1} \right) \\ =&\frac{1}{2\pi i z_1} \left( Q_{j,k}(x_j-x_k) - \frac{\xi _1}{\xi _j} Q_j(x_j)\right) . \end{aligned}$$

Therefore,

$$\begin{aligned}&\frac{1}{2\pi i}\left( \frac{\xi _k}{z_j-z_k}+ \frac{\xi _1}{z_j-z_1}\right) \\&\quad = \frac{1}{2\pi i z_1} \left( \frac{a_1 \xi _k}{a_j- a_k} -\frac{a_1^2 \xi _k}{(a_j- a_k)^2}(x_j-x_k) +R_{j,k,2}(x_j-x_k)\right. \\&\qquad \left. +\frac{a_1 \xi _1}{a_j- a_1}- \frac{a_1^2 \xi _1}{(a_j- a_1)^2}x_j -\frac{\xi _1}{\xi _j}R_{j,2}(x_k)\right) \\&\quad = \frac{1}{2\pi i z_1} \left( 2\pi i \overline{a_j}a_1(a-i b)- \frac{a_1^2 \xi _k}{(a_j- a_k)^2}(x_j-x_k)- \frac{a_1^2 \xi _1}{(a_j- a_1)^2}x_j \right. \\&\qquad \left. +R_{j,k,2}(x_j-x_k) -\frac{\xi _1}{\xi _j}R_{j,2}(x_k)\right) , \end{aligned}$$

hence

$$\begin{aligned}&\frac{1}{z_1}\overline{\frac{1}{2\pi i} \left( \frac{\xi _k}{z_j-z_k}+ \frac{\xi _1}{z_j-z_1} \right) }\\&\quad = \frac{a_j}{a_1}\frac{|a_1|^2}{|z_1|^2} (a+ib) + \frac{1}{2\pi i |z_1|^2} \overline{\left( \frac{a_1^2 \xi _k}{(a_j- a_k)^2}(x_j-x_k)+ \frac{a_1^2 \xi _1}{(a_j- a_1)^2}x_j \right) } \\&\qquad + \frac{1}{2\pi i |z_1|^2} \overline{\left( R_{j,k,2}(x_j-x_k) - \frac{\xi _1}{\xi _j}R_{j,2}(x_k) \right) }. \end{aligned}$$

We move now to the contribution given by ${\dot{z}}_1$:

$$\begin{aligned} \frac{1}{2\pi i} \sum _{\ell \ne 1} \frac{\xi _\ell }{z_1-z_\ell } =&\frac{1}{2\pi i z_1} \left( \frac{\xi _k}{1-x_k- \frac{a_k}{a_1}}+ \frac{\xi _j}{1-x_j- \frac{a_j}{a_1}} \right) \\ =&\frac{1}{2\pi i z_1} \left( \frac{a_1 \xi _j}{a_1- a_j} + \frac{a_1^2 \xi _j}{(a_1- a_j)^2}x_j + \frac{a_1 \xi _k}{a_1- a_k} + \frac{a_1^2 \xi _k}{(a_1- a_k)^2}x_k\right. \\&\quad \left. + R_{j,2}(x_j) + R_{k,2}(x_k) \right) \\ =&\frac{1}{2\pi i z_1} \left( 2 \pi i |a_1|^2 (a-ib) + \frac{a_1^2 \xi _j}{(a_1- a_j)^2}x_j + \frac{a_1^2 \xi _k}{(a_1- a_k)^2}x_k\right. \\&\quad \left. + R_{j,2}(x_j) + R_{k,2}(x_k) \right) . \end{aligned}$$

Therefore

$$\begin{aligned}&-\frac{z_j}{z_1^2} \overline{ \left( \frac{1}{2\pi i} \sum _{\ell \ne 1} \frac{\xi _\ell }{z_1-z_\ell } \right) } \\&\quad = -\frac{z_j}{z_1}\frac{|a_1|^2}{|z_1|^2} (a+ib) \\&\qquad +\frac{z_j}{z_1} \frac{1}{2\pi i |z_1|^2} \overline{\left( \frac{a_1^2 \xi _j}{(a_1- a_j)^2}x_j + \frac{a_1^2 \xi _k}{(a_1- a_k)^2}x_k + R_{j,2}(x_j) + R_{k,2}(x_k) \right) }. \end{aligned}$$

All in all, we get

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t} x_j =&- x_j \frac{a+ib}{r^2} \\&+ \frac{1}{2\pi i |z_1|^2} \overline{\left( \frac{a_1^2 \xi _k}{(a_j- a_k)^2}(x_j-x_k)+ \frac{a_1^2 \xi _1}{(a_j- a_1)^2}x_j \right) } \\&+ \left( x_j + \frac{a_j}{a_1} \right) \frac{1}{2\pi i |z_1|^2} \overline{\left( \frac{a_1^2 \xi _j}{(a_1- a_j)^2}x_j + \frac{a_1^2 \xi _k}{(a_1- a_k)^2}x_k \right) } \\&+ \frac{1}{2\pi i |z_1|^2} \overline{\left( R_{j,k,2}(x_j-x_k) - \frac{\xi _1}{\xi _j}R_{j,2}(x_k)\right) } \\&+ \left( x_j + \frac{a_j}{a_1} \right) \frac{1}{2\pi i |z_1|^2} \overline{\left( R_{j,2}(x_j) + R_{k,2}(x_k) \right) } \\&+\frac{1}{z_1} \overline{f(t,z_j)} -\frac{z_j}{z_1^2}\overline{f(t,z_1)}. \end{aligned}$$

We arrive at the expression in Lemma 2.7, that is

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}x_j = \frac{L_{j}(x_2,x_3,\overline{x_2},\overline{x_3})}{r^2} + \frac{\omega _j(x_2,x_3,x_2-x_3)}{r^2} +\frac{1}{z_1} \overline{f(t,z_j)} -\frac{z_j}{z_1^2}\overline{f(t,z_1)}. \end{aligned}$$

by collecting linear terms into $L_2,L_3$ and the remainders R into holomorphic functions $\omega _j$ satisifying, for all $|x_2|, |x_3|, |x_2-x_3| < \rho '$,

$$\begin{aligned}&|\omega _{j}(x_2,x_3,x_2-x_3)| \leqq C \left( |x_2|^2 + |x_3|^2\right) ,\\&|\nabla \omega _{j}(x_2,x_3,x_2-x_3)|\leqq C \left( |x_2| + |x_3|\right) , \end{aligned}$$

where we redefined for the last time constants $C,\rho '>0$ depending only on $\xi ,a_1,a_2,a_3$. Notice that the proof has used, up to this point, only the fact that parameters $a,b,a_1,a_2,a_3$ satisfy (2.3) (rather than the particular choice in the end of proof of Lemma 2.3).

1.3 Proof of Lemma 2.7: Eigenvalues

We are left to prove the statement on eigenvalues of the matrix

$$\begin{aligned} L = \left( L_2,L_3,{{\bar{L}}}_2,{{\bar{L}}}_3\right) =\left( \begin{array}{cccc} -a-i b &{} 0 &{} L_{13} &{} L_{14} \\ 0 &{} -a-i b &{} L_{23} &{} L_{24} \\ \overline{L_{13}} &{} \overline{L_{14}} &{} -a + i b &{} 0 \\ \overline{L_{23}} &{} \overline{L_{24}} &{} 0 &{} -a + i b \end{array}\right) , \end{aligned}$$

where

$$\begin{aligned} L_{13} =&\frac{1}{2\pi i |a_1|^2} \overline{ \left( \frac{a_1^2 \xi _3}{(a_2-a_3)^2}+ \frac{a_1^2 \xi _1}{(a_2-a_1)^2}+ \frac{\overline{a_2}}{\overline{a_1}}\frac{a_1^2 \xi _2}{(a_1-a_2)^2} \right) }, \\ L_{14} =&\frac{1}{2\pi i |a_1|^2} \overline{ \left( -\frac{a_1^2 \xi _3}{(a_2-a_3)^2}+ \frac{\overline{a_2}}{\overline{a_1}}\frac{a_1^2 \xi _3}{(a_1-a_3)^2} \right) }, \\ L_{23} =&\frac{1}{2\pi i |a_1|^2} \overline{ \left( -\frac{a_1^2 \xi _2}{(a_3-a_2)^2}+ \frac{\overline{a_3}}{\overline{a_1}}\frac{a_1^2 \xi _2}{(a_1-a_2)^2} \right) }, \\ L_{24} =&\frac{1}{2\pi i |a_1|^2} \overline{ \left( \frac{a_1^2 \xi _2}{(a_3-a_2)^2}+ \frac{a_1^2 \xi _1}{(a_3-a_1)^2}+ \frac{\overline{a_3}}{\overline{a_1}}\frac{a_1^2 \xi _3}{(a_1-a_3)^2} \right) }. \end{aligned}$$

The eigenvalues of L coincide with the roots of its characteristic polynomial, which is given by

$$\begin{aligned} p(\lambda )&= y^2 - y c_1 + c_2,\\ y&=(-a-i b -\lambda )(-a+i b -\lambda ) = (a+\lambda )^2 + b^2,\\ c_1&=L_{23}\overline{L_{14}} + L_{24}\overline{L_{24}} + L_{13}\overline{L_{13}} + L_{14}\overline{L_{23}}, \\ c_2&=L_{13}\overline{L_{13}}L_{24}\overline{L_{24}}+ L_{23}\overline{L_{23}}L_{14}\overline{L_{14}} - L_{14}\overline{L_{13}}L_{23}\overline{L_{24}} - L_{13}\overline{L_{14}}L_{24}\overline{L_{23}}. \end{aligned}$$

Notice that $c_1,c_2 \in {\mathbb {R}}$. We now impose that the eigenvalues have the form $-a + i \mu$, $\mu \in {\mathbb {R}}$. This is true if and only if $y=b^2-\mu ^2$ solves

$$\begin{aligned} y^2 - y c_1 + c_2 = 0, \end{aligned}$$

which, in terms of $\mu$, becomes

$$\begin{aligned} \mu ^4 + \mu ^2 (c_1 - 2b^2) + b^4 - b^2 c_1 + c_2 = 0. \end{aligned}$$

From this, we deduce that if it holds that

$$\begin{aligned} 2b^2 - c_1 \pm \sqrt{(2b^2 - c_1)^2 - 4(b^4 - b^2 c_1 + c_2)} > 0, \end{aligned}$$

(A.3)

there are 4 distinct solutions $\mu$, producing 4 distinct roots of p, and thus all and only the eigenvalues. Only at this point of the proof we restrict ourselves to the particular choice of parameters made at the end of the proof of Lemma 2.3. Indeed, with such choice, (A.3) is satisfied independently from the sign of $\xi$:

$$\begin{aligned} 2b^2 - c_1 \sim \xi ^2 \times 3.4463 \times 10^{-4}, \quad b^4 - b^2 c_1 + c_2 \sim \xi ^4 \times 2.7035 \times 10^{-9}, \end{aligned}$$

and this concludes the proof.