Weak Solutions of Ideal MHD Which Do Not Conserve Magnetic Helicity

Abstract

We construct weak solutions to the ideal magneto-hydrodynamic (MHD) equations which have finite total energy, and whose magnetic helicity is not a constant function of time. In view of Taylor’s conjecture, this proves that there exist finite energy weak solutions to ideal MHD which cannot be attained in the infinite conductivity and zero viscosity limit. Our proof is based on a Nash-type convex integration scheme with intermittent building blocks adapted to the geometry of the MHD system.

Appendices

Appendix

1.1 Proof of Geometric Lemmas

In this section, we will provide proofs of Lemmas 4.1 and 4.2, following the classical arguments of [8, 28].

Proof of Lemma 4.1

Let $\Lambda _B = \{ e_1, e_2, e_3, \frac{3}{5}e_1 + \frac{4}{5}e_2, -\frac{4}{5}e_2 - \frac{3}{5}e_3 \}$ and to these vectors, consider the orthonormal bases given by

We define

$$\begin{aligned} A_1&:= e_2 \otimes e_3 - e_3 \otimes e_2, \quad A_2 := e_3 \otimes e_1 - e_1 \otimes e_3, \quad A_3 := e_1 \otimes e_2 - e_2 \otimes e_1,\\ A_4&:= \left( \frac{4}{5}e_1 - \frac{3}{5}e_2 \right) \otimes e_3 - e_3 \otimes \left( \frac{4}{5}e_1 - \frac{3}{5}e_2 \right) , \quad \\ A_5&:= \left( \frac{3}{5}e_2 - \frac{4}{5}e_3 \right) \otimes e_1 - e_1 \otimes \left( \frac{3}{5}e_2 - \frac{4}{5}e_3 \right) \,. \end{aligned}$$

Using these matrices we can write

$$\begin{aligned} \frac{7}{4}A_1 + \frac{11}{3}A_2 + A_3 + \frac{35}{12}A_4 + \frac{5}{3}A_5 = 0 \,. \end{aligned}$$

(A.1)

Since $A_1, A_2, A_3$ form a basis for the 3 $\times$ 3 skew-symmetric matrices, we can express any skew-symmetric matrix A as a unique linear combination $A = c_1A_1 + c_2A_2 + c_3A_3$. Combining this with (A.1) gives

$$\begin{aligned} \left( \frac{7}{4} + c_1 \right) A_1 + \left( \frac{11}{3} + c_2 \right) A_2 + \left( 1 + c_3 \right) A_3 + \frac{35}{12}A_4 + \frac{5}{3}A_5 = A \,. \end{aligned}$$

Therefore we can define

$$\begin{aligned} \gamma _{1,B}&= \sqrt{\frac{7}{4} + c_1}, \quad \gamma _{2,B} = \sqrt{\frac{11}{3} + c_2}, \quad \gamma _{3,B} = \sqrt{ 1 + c_3}, \quad \gamma _{4,B} = \sqrt{\frac{35}{12}}, \\ \gamma _{5,B}&= \sqrt{\frac{5}{3}} \, . \end{aligned}$$

For $\varepsilon _B < \sqrt{2}$, the $\gamma _i$ will be smooth. Therefore it suffices to take $\varepsilon _B = 1$. $\square$

Proof of Lemma 4.2

Proceeding as before let $\Lambda _u = \{ \frac{5}{13}e_1 \pm \frac{12}{13}e_2, \frac{12}{13}e_1 \pm \frac{5}{13}e_3, \frac{5}{13}e_2 \pm \frac{12}{13}e_3 \}$ and to these vectors, consider the orthonormal bases given by

Note that $\Lambda _u \cap \Lambda _B = \emptyset$. Next, note that $\sum _{k\in \Lambda _u} \frac{1}{2} k_1\otimes k_1 = \mathrm {Id}$, and thus by the implicit function theorem, there exists $\varepsilon _u$ such that for $S \in B_{\varepsilon _u}(\mathrm {Id})$, S can be expressed as a linear combination of the $S_i$ with positive coefficients. See [8, 28] for further details. $\square$

Proof of Magnetic Helicity Conservation

In this appendix we give the proof of Theorem 1.3. For $u, B \in L^3(0,T;L^3(\mathbb {T}^3))$ we have magnetic helicity conservation for (1.1), as in [42]. A simple modification of this argument shows that Leray-Hopf solutions of (1.2) satisfy a magnetic helicity balance (by interpolation we have that $u, B \in L_{x,t}^{\frac{10}{3}}({\mathbb {T}}^3)$):

$$\begin{aligned} \int _{{\mathbb {T}}^3 } A \cdot B(t)dx + 2 \mu \int _0^t \int _{{\mathbb {T}}^3} \mathrm {curl\,}B \cdot B(s) dx ds = \int _{{\mathbb {T}}^3} A \cdot B (0) dx. \end{aligned}$$

(B.1)

Assume that $(u_j, B_j)$ is a weak ideal sequence and that $\mu _j \rightarrow 0$. Using the uniform bounds coming from the total energy inequality (1.3) we have that

$$\begin{aligned} \mu _j \int _0^t \int _{{\mathbb {T}}^3} |\mathrm {curl\,}B_j \cdot B_j| dx ds \le t\mu _j^{\frac{1}{2}} \Vert \mu _j^{\frac{1}{2}} (\mathrm {curl\,}B_j) \Vert _{L_t^{\infty }L_x^2} \Vert B_j\Vert _{L_t^{\infty }L_x^2 } \rightarrow 0 \quad \text { as } j {\rightarrow } \infty \, . \end{aligned}$$

Therefore, passing to the limit in (B.1) (for $A_j$ and $B_j$), we obtain

$$\begin{aligned} \liminf _{j \rightarrow \infty } \int _{{\mathbb {T}}^3 } A_j \cdot B_j (t)dx = \liminf _{j \rightarrow \infty } \int _{{\mathbb {T}}^3 } A_j \cdot B_j (0)dx = \int _{{\mathbb {T}}^3} A \cdot B(0)dx \end{aligned}$$

where the last equality comes from the fact that since $B_j(0) \rightharpoonup B(0)$ in $L^2$, $A_j(0) \rightarrow A(0)$ in $L^2$ and the product of a weakly convergent sequence and a strongly convergent sequence converges. By Aubin-Lions Lemma with the triple $L^2 \subset H^{-\frac{1}{2}} \subset H^{-3}$ applied to $B_j$, we conclude that $B_j(t)$ has a strongly convergent subsequence in $C([0,T]; H^{-\frac{1}{2}})$ (also denoted $B_j(t)$). This implies $A_j(t)$ is strongly convergent in $C([0,T]; {\dot{H}}^{\frac{1}{2}})$. Along this subsequence

$$\begin{aligned} \int _{{\mathbb {T}}^3} A_j \cdot B_j(t) dx= & {} \int _{{\mathbb {T}}^3} |\nabla |^{\frac{1}{2}}A_j \cdot |\nabla |^{-\frac{1}{2}} B_j(t) dx \rightarrow \int _{{\mathbb {T}}^3} |\nabla |^{\frac{1}{2}}A\cdot |\nabla |^{-\frac{1}{2}} B(t) dx \\= & {} \int _{{\mathbb {T}}^3} A \cdot B(t) dx \end{aligned}$$

where we are using that limit of the strongly convergent subsequence must coincide with the weak ideal limit by uniqueness of weak-* limits. Furthermore, we can extend this to the entire sequence to conclude

$$\begin{aligned} \int _{{\mathbb {T}}^3} A(t) \cdot B(t) dx = \int _{{\mathbb {T}}^3} A(0) \cdot B(0) dx \end{aligned}$$

as desired.