Non-conservative Solutions of the Euler-$\alpha$ Equations

Abstract

The Euler-$\alpha$ equations model the averaged motion of an ideal incompressible fluid when filtering over spatial scales smaller than $\alpha$. We show that there exists $\beta >1$ such that weak solutions to the two and three dimensional Euler-$\alpha$ equations in the class $C^0_t H^\beta _x$ are not unique and may not conserve the Hamiltonian of the system, thus demonstrating flexibility in this regularity class. The construction utilizes a Nash-style intermittent convex integration scheme. We also formulate an appropriate version of the Onsager conjecture for Euler-$\alpha$, postulating that the threshold between rigidity and flexibility is the regularity class .

Appendices

Proof of Lemma 1.5

In this section we provide a proof of Lemma 1.5.

Proof

We follow the proof from [20]. Mollifying (1.8) in space with a standard radial, compactly supported mollifier $\varphi _{\varepsilon }$, assuming differentiability in time, and integrating in space against $u_i^{\varepsilon }=u *\varphi _\varepsilon$ gives

$$\begin{aligned}&\frac{1}{2} \frac{d}{dt} \left( \Vert u_i^{\varepsilon } \Vert _{L^2}^2 + \alpha ^2 \Vert \nabla u_i^{\varepsilon }\Vert _{L^2}^2 \right) \nonumber \\&= \int _{\mathbb {T}^n } (u_j u_i)^{\varepsilon } \partial _j u_i^{\varepsilon } + \alpha ^2 (u_j \partial _k u_i )^{\varepsilon } \partial _j \partial _k u_i^{\varepsilon } + \alpha ^2 (\partial _k u_j \partial _k u_i)^{\varepsilon } \partial _ju_i^{\varepsilon } - \alpha ^2 (\partial _k u_j \partial _i u_j)^{\varepsilon } \partial _k u_i^{\varepsilon } \, dx \, \end{aligned}$$

(A.1)

where we have used Remark 1.3.

We wish to show that the right-hand side vanishes when $\varepsilon$ is sent to 0. We recall some standard estimates for mean-zero $f \in B_{3, \infty }^{s}$ where ${1<s<2}$; see for example [2, 20].

$$\begin{aligned}&\Vert \nabla f(\cdot ) - \nabla f(\cdot - y)\Vert _{L^3} \lesssim |y|^{s -1} \Vert \nabla f \Vert _{B_{3 , \infty }^{s - 1}} \end{aligned}$$

(A.2a)

$$\begin{aligned}&\Vert f - f^{\varepsilon } \Vert _{L^3} \lesssim \varepsilon \Vert f \Vert _{B_{3, \infty }^{s}} \end{aligned}$$

(A.2b)

$$\begin{aligned}&\Vert \nabla f - \nabla f^{\varepsilon } \Vert _{L^3} \lesssim \varepsilon ^{s -1} \Vert \nabla f \Vert _{B_{3, \infty }^{s-1}} \end{aligned}$$

(A.2c)

$$\begin{aligned}&\Vert \nabla ^2 f^{\varepsilon } \Vert _{L^3} \lesssim \varepsilon ^{s-2} \Vert \nabla f \Vert _{B_{3, \infty }^{s - 1}} \end{aligned}$$

(A.2d)

$$\begin{aligned}&{\left\| \nabla f \right\| _{L^3} \lesssim \left\| \nabla f \right\| _{B^{s-1}_{3,\infty }}} \end{aligned}$$

(A.2e)

$$\begin{aligned}&\left\| \nabla f \right\| _{B^{s-1}_{3,\infty }} \lesssim \left\| f \right\| _{B^{s}_{3,\infty }} \end{aligned}$$

(A.2f)

$$\begin{aligned}&{\left\| f^\varepsilon \right\| _{B^{s}_{3,\infty }}\lesssim \left\| f \right\| _{B^{s}_{3,\infty }}} \, . \end{aligned}$$

(A.2g)

Finally, recall the double commutator identity from [20]:

$$\begin{aligned} (fg)^{\varepsilon } = f^{\varepsilon }g^{\varepsilon } - (f - f^{\varepsilon })(g -g^{\varepsilon }) + r_{\varepsilon }(f,g) \, , \end{aligned}$$

(A.3)

where $f^{\varepsilon }(x) = f * \varphi _{\varepsilon }(x)$ and

$$\begin{aligned} r_{\varepsilon }(f,g) = \int _{\mathbb {T}^n} \varphi _{\varepsilon }(y) (f(x - y) - f(x))(g(x - y) - g(x)) \, dy \, . \end{aligned}$$

(A.4)

Then the right-hand side of (A.1) can be written as

$$\begin{aligned}&\int _{\mathbb {T}^n } (u_j u_i)^{\varepsilon } \partial _j u_i^{\varepsilon } + \alpha ^2 (u_j \partial _k u_i )^{\varepsilon } \partial _j \partial _k u_i^{\varepsilon } + \alpha ^2 (\partial _k u_j \partial _k u_i)^{\varepsilon } \partial _ju_i^{\varepsilon } - \alpha ^2 (\partial _k u_j \partial _i u_j)^{\varepsilon } \partial _k u_i^{\varepsilon } dx \nonumber \\&\quad = \int _{\mathbb {T}^n} u_j^{\varepsilon }u_i^{\varepsilon } \partial _j u_i^{\varepsilon } + \alpha ^2\left[ u_j^{\varepsilon }\partial _ku_i^{\varepsilon } \partial _j \partial _k u_i^{\varepsilon } + \partial _ku_j^{\varepsilon } \partial _k u_i^{\varepsilon } \partial _j u_i^{\varepsilon } - \partial _ku_j^{\varepsilon } \partial _i u_j^{\varepsilon } \partial _k u_i^{\varepsilon } \right] dx \nonumber \\&\qquad - \int _{\mathbb {T}^n} (u_j - u_j^{\varepsilon })(u_i - u_i^{\varepsilon }) \partial _j u_i^{\varepsilon } dx \nonumber \\&\qquad - \alpha ^2\int _{\mathbb {T}^n} \left[ ( u_j - u_j^{\varepsilon })(\partial _k u_i - \partial _k u_i^{ \varepsilon })\partial _j \partial _ku_i^{\varepsilon } + (\partial _k u_j - \partial _k u_j^{\varepsilon }) (\partial _k u_i - \partial _k u_i^{\varepsilon }) \partial _j u_i^{\varepsilon } \right. \nonumber \\&\qquad \left. - (\partial _k u_j - \partial _k u_j^{\varepsilon } )(\partial _i u_j - \partial _i u_j^{\varepsilon }) \partial _k u_i^{\varepsilon } \right] dx \nonumber \\&\qquad + \int _{\mathbb {T}^n} r_{\varepsilon }(u_j, u_i) \partial _j u_i^{\varepsilon } + \alpha ^2\left[ r_{\varepsilon } (u_j, \partial _ku_i ) \partial _j \partial _k u_i^{\varepsilon } + r_{\varepsilon } (\partial _ku_j ,\partial _k u_i) \partial _j u_i^{\varepsilon } - r_{\varepsilon }(\partial _ku_j , \partial _i u_j) \partial _k u_i^{\varepsilon } \right] dx \, . \end{aligned}$$

(A.5)

We have

$$\begin{aligned} \int _{\mathbb {T}^n} u_j^{\varepsilon } u_i^{\varepsilon } \partial _j u_i^{\varepsilon } dx = \int _{\mathbb {T}^n} u_j^{\varepsilon } \partial _k u_i^{\varepsilon } \partial _j \partial _k u_i^{\varepsilon } = 0 \end{aligned}$$

using integration by parts and that $\textrm{div}u = 0$. Furthermore, the equality

$$\begin{aligned} \partial _k u_j^{\varepsilon } \partial _ku_i^{\varepsilon } \partial _j u_i^{\varepsilon } = \partial _k u_j^{\varepsilon } \partial _iu_j^{\varepsilon } \partial _k u_i^{\varepsilon } \end{aligned}$$

follows from switching the roles of i and j. Therefore the first four terms after the equal sign on the right-hand side of (A.5) vanish, and we can bound the remaining terms by

$$\begin{aligned}&\lesssim \Vert u - u^{\varepsilon }\Vert _{L^3}^2 \Vert \nabla u^{\varepsilon } \Vert _{L^3} + \Vert u - u^{\varepsilon }\Vert _{L^3} \Vert \nabla u - \nabla u^{\varepsilon } \Vert _{L^3} \Vert \nabla ^2 u^{\varepsilon } \Vert _{L^3} + \Vert \nabla u - \nabla u^{\varepsilon } \Vert _{L^3}^2 \Vert \nabla u^{\varepsilon } \Vert _{L^3} \\&\quad + \Vert r_{\varepsilon } (u, u) \Vert _{L^{\frac{3}{2}}} \Vert \nabla u^{\varepsilon } \Vert _{L^3} + \Vert r_{\varepsilon } (u, \nabla u)\Vert _{L^{\frac{3}{2}}} \Vert \nabla ^2 u^{\varepsilon }\Vert _{L^3} + \Vert r_{\varepsilon }( \nabla u, \nabla u) \Vert _{L^{\frac{3}{2}}} \Vert \nabla u^{\varepsilon } \Vert _{L^3 } \\&\lesssim \varepsilon ^{{2}} \Vert u \Vert _{B_{3, \infty }^{s} }^3 + \varepsilon ^{ {2(s-1)}} \Vert u \Vert _{B_{3 ,\infty }^{s}}^3. \end{aligned}$$

Thus, taking $s > 1$ shows that the right-hand side of (A.1) converges to zero when $\varepsilon \rightarrow 0$. Concluding the proof in the case of continuity in time may be done analogously as for the classical Euler equations, and we omit further details.

Proof of Lemma 4.1

In this section we provide a proof of Lemma 4.1.

Proof

We first construct $\mathcal {K}_0$ and $c_i^0$ by hand while allowing $\left\| {\mathring{R}}\right\| _{0}\le 1$, afterwards constructing $\mathcal {K}_n$ for $1\le n \le N$ and choosing $\varepsilon$. Consider a symmetric traceless matrix ${\mathring{R}}$, which without loss of generality may be written as

$$\begin{aligned} {\mathring{R}}= \begin{pmatrix} a &{} c &{} d\\ c &{} b &{} e\\ d &{} e &{} -a-b \end{pmatrix} \, . \end{aligned}$$

(B.1)

Notice that the set of such matrices is 5-dimensional, and combined with the condition on the sum of $(c_i^0)^2$ in (4.1) will require a set of at least six vectors. The extra three vectors will ensure that the coefficients are all positive. Ignoring for now the upper subscript 0 on the vectors $k_i^0\in \mathcal {K}_0$, we set

$$\begin{aligned} k_1 = e_1 \, , \qquad k_2 = e_2 \, , \qquad k_3 = e_3 \, . \end{aligned}$$

For ease of notation, let us define

$$\begin{aligned} f(k_i) = 3k_i\otimes k_i - \text {Id} \, . \end{aligned}$$

Then it is clear that $f(k_1)$, $f(k_2)$, and $f(k_3)$ are symmetric traceless matrices which only contain entries on the diagonal; specifically, we have

$$\begin{aligned} f(k_1) = \begin{pmatrix} 2 &{} 0 &{} 0\\ 0 &{} -1 &{} 0\\ 0 &{} 0 &{} -1 \end{pmatrix}\, , \qquad f(k_2) =\begin{pmatrix} -1 &{} 0 &{} 0\\ 0 &{} 2 &{} 0\\ 0 &{} 0 &{} -1 \end{pmatrix}\, , \qquad f(k_3) = \begin{pmatrix} -1 &{} 0 &{} 0\\ 0 &{} -1 &{} 0\\ 0 &{} 0 &{} 2 \end{pmatrix}\, . \qquad \end{aligned}$$

(B.2)

We shall use $f(k_1)$ and $f(k_2)$ to engender the entries of the matrix on the diagonal in (B.1), while we shall use a “balanced” sum of the form $c(f(k_1)+f(k_2)+f(k_3))$ to ensure the second condition in (4.1). We further set

$$\begin{aligned} k_4&= \frac{3e_1+4e_2}{5}\, , \qquad k_5=\frac{3e_1+4e_3}{5}\, ,\qquad k_6=\frac{3e_2+4e_3}{5} \\ k_7&= \frac{3e_1-4e_2}{5}\, , \qquad k_8=\frac{3e_1-4e_3}{5}\, ,\qquad k_9=\frac{3e_2-4e_3}{5}\, . \end{aligned}$$

Then

$$\begin{aligned} f(k_4)= & {} \frac{1}{25}\begin{pmatrix} 2 &{} 36 &{} 0\\ 36 &{} 23 &{} 0\\ 0 &{} 0 &{} -25 \end{pmatrix}\, , \qquad f(k_5) = \frac{1}{25}\begin{pmatrix} 2 &{} 0 &{} 36\\ 0 &{} -25 &{} 0\\ 36 &{} 0 &{} 23 \end{pmatrix}\, , \qquad f(k_6) = \frac{1}{25}\begin{pmatrix} -25 &{} 0 &{} 0\\ 0 &{} 2 &{} 36\\ 0 &{} 36 &{} 23 \end{pmatrix}\nonumber \\ f(k_7)= & {} \frac{1}{25}\begin{pmatrix} 2 &{} -36 &{} 0\\ -36 &{} 23 &{} 0\\ 0 &{} 0 &{} -25 \end{pmatrix}\, , \quad f(k_8) = \frac{1}{25}\begin{pmatrix} 2 &{} 0 &{} -36\\ 0 &{} -25 &{} 0\\ -36 &{} 0 &{} 23 \end{pmatrix}\, , \quad f(k_9) = \frac{1}{25}\begin{pmatrix} -25 &{} 0 &{} 0\\ 0 &{} 2 &{} -36\\ 0 &{} -36 &{} 23 \end{pmatrix}. \nonumber \\ \end{aligned}$$

(B.3)

We shall use $f(k_4)$, $f(k_5)$, $f(k_6)$, $f(k_7)$, $f(k_8)$, and $f(k_9)$ to engender the entries of (B.1) off the diagonal.

It is simple to check that the set $\{f(k_1),f(k_2),f(k_4),f(k_5),f(k_6)\}$ is a linearly independent set in the space of symmetric traceless matrices. Therefore, there exist smooth functions $\{\tilde{c}_1, \tilde{c}_2, \tilde{c}_4, \tilde{c}_5, \tilde{c}_6\}$ given by the solutions of a linear system of equations such that for all symmetric traceless matrices ${\mathring{R}}$ satisfying $\left\| {\mathring{R}}\right\| _0\le 1$,

$$\begin{aligned} \tilde{c}_1 f(k_1) + \tilde{c}_2 f(k_2) + \tilde{c}_4 f(k_4) + \tilde{c}_5 f(k_5) + \tilde{c}_6 f(k_6) = {\mathring{R}}\, . \end{aligned}$$

(B.4)

Unfortunately the functions $\tilde{c}_i$ are not strictly positive. Let us define the auxiliary parameter

$$\begin{aligned} c_0 = \max _{\begin{array}{c} \left\| {\mathring{R}}\right\| _0\le 1, i=4,5,6 \end{array}} |\tilde{c}_i({\mathring{R}})| \, . \end{aligned}$$

Then from (B.4) and (B.3), we have that

$$\begin{aligned} \left( 1 -\delta _{ij} \right)&\bigg ( \tilde{c}_1 f(k_1) + \tilde{c}_2 f(k_2) + \tilde{c}_4 f(k_4) + \tilde{c}_5 f(k_5) + \tilde{c}_6 f(k_6) \\&\qquad + 2c_0 \left( f(k_4) + f(k_5) + f(k_6) + f(k_7) + f(k_8) + f(k_9) \right) \bigg )^{ij} = \left( 1 -\delta _{ij} \right) {\mathring{R}}^{ij}\, ; \end{aligned}$$

that is, off the diagonal, the matrix on the left hand side of the equation is equal to ${\mathring{R}}$. The advantage now is that the coefficients on $f(k_i)$ for $4\le i \le 9$ are all positive. In order to ensure equality on the diagonal, we may replace the coefficients $\tilde{c}_1$ and $\tilde{c}_2$ on $f(k_1)$ and $f(k_2)$ with new coefficients $\dot{c}_1$ and $\dot{c}_2$ such that

$$\begin{aligned}&\bigg (\dot{c}_1 f(k_1) + \dot{c}_2 f(k_2) + \tilde{c}_4 f(k_4) + \tilde{c}_5 f(k_5) + \tilde{c}_6 f(k_6) \nonumber \\&\qquad \qquad + 2c_0 \left( f(k_4) + f(k_5) + f(k_6) + f(k_7) + f(k_8) + f(k_9) \right) \bigg )^{ij} = {\mathring{R}}^{ij}\, \end{aligned}$$

(B.5)

for all $1\le i,j\le 3$. Since the coefficients $\dot{c}_1$ and $\dot{c}_2$ are not necessarily positive, we set

$$\begin{aligned} \dot{c}_0 = \max _{\left\| {\mathring{R}} \right\| _0\le 1,i=1,2} \left| \dot{c}_i ({\mathring{R}}) \right| \, . \end{aligned}$$

Then from (B.2) and (B.5), we have that

$$\begin{aligned}&\bigg (\dot{c}_1 f(k_1) + \dot{c}_2 f(k_2) + 2\dot{c}_0 \left( f(k_1)+f(k_2)+f(k_3) \right) \\&\qquad + \tilde{c}_4 f(k_4) + \tilde{c}_5 f(k_5) + \tilde{c}_6 f(k_6) + 2c_0 \left( f(k_4) + f(k_5) + f(k_6) + f(k_7) + f(k_8) + f(k_9) \right) \bigg )^{ij} = {\mathring{R}}^{ij}\, \end{aligned}$$

Aggregating coefficients for each of the nine tensors on the left-hand side, we see that each is strictly positive. We still have to ensure the second condition in (4.1); however, this is easily achieved by replacing the constant $\dot{c}_0$ with a non-negative, bounded function $c_0({\mathring{R}}):B_1(0)\rightarrow [0,\infty )$ which depends on ${\mathring{R}}$ and imposes that the sum of the coefficients is equal for all ${\mathring{R}}\in B_1(0)$. Thus we have achieved everything in (4.1) for $n=0$. Note that the value of $\mathcal {C}_\textrm{sum}$ could in principle be computed explicitly and depends on the choice of the function $c_0({\mathring{R}})$.

To construct $\mathcal {K}_n$ for $n=1$, choose a rational rotation matrix $O_1$ with $0 < \left\| O_1 - \text {Id} \right\| _0 \ll 1$, and set $\mathcal {K}_1= O_1 \mathcal {K}_0$. Plugging the rotated vectors $O_1 k_i^0$ into (4.1), we find that the effect of $O_1$ is that

$$\begin{aligned} \sum _{i=1}^9 \left( c_i^0\right) ^2\left( {\mathring{R}}\right) \left( 3 k_i^0 \otimes k_i^0 - \text {Id} \right) = {\mathring{R}}\quad \rightarrow \quad \sum _{i=1}^9 \left( c_i^0\right) ^2\left( {\mathring{R}}\right) O_1\left( 3 k_i^0 \otimes k_i^0 - \text {Id} \right) O_1^T = O_1 {\mathring{R}}O_1^T \, . \end{aligned}$$

Using the equalities $\text {tr}(AB)=\text {tr}(BA)$ and $O^T=O^{-1}$, we have that $O_1 {\mathring{R}}O_1^T$ is still traceless and symmetric. Define

$$\begin{aligned} c^1_i\left( {\mathring{R}}\right) = c^0_i\left( O_1^T {\mathring{R}}O_1 \right) \, \end{aligned}$$

for $\left\| {\mathring{R}}\right\|$ sufficiently small so that $c^0_i\left( O_1 {\mathring{R}}O_1^T\right)$ is well-defined and strictly positive for each i. Then this specifies a value $\varepsilon _1$ such that for all $\left\| {\mathring{R}}\right\| _0 \le \varepsilon _1$,

$$\begin{aligned} \sum _{i=1}^9 \left( c_i^1\right) ^2\left( {\mathring{R}}\right) \left( 3 k_i^1 \otimes k_i^1 - \text {Id} \right)&= \sum _{i=1}^9 \left( c^0_i\right) ^2 \left( O_1^T {\mathring{R}}O_1\right) O_1 \left( 3 k_i^0 \otimes k_i^0 - \text {Id} \right) O_1^T \\&= {\mathring{R}}\, , \end{aligned}$$

showing that the first equality in (4.1) is satisfied for $n=1$. The second equality in (4.1) comes immediately from the construction of the $c_i^0$’s and $c^1_i$’s. In addition, $\mathcal {K}_0$ and $\mathcal {K}_1$ will be disjoint if $0 < \left\| O_1 - \text {Id} \right\| _0\ll 1$. Iterating this procedure and taking the minimum value of $\varepsilon _n$ concludes the proof.

$\square$