Edit: This question has been substantially modified on January 12th, 2015.

I have been studying Michael Struwe's paper "On Partial Results for the Navier-Stokes Equations", Comm. Pure Appl. Math 41 (1988), no. 4, 437-458, and I have a question regarding the rigorous justification of the formal calculations in the proof of the Serrin condition in Section 3 of the paper. I have also studied the original proof by Serrin.

Consider the weak solutions of the Navier-Stokes system. Let's assume $n=3$ for simplicity. A weak solution $u \in L^{2, \infty}(Q; \mathbb R^3)$, with $|\nabla u| \in L^2(Q)$, is defined by requiring that $\nabla \cdot u=0$ and $$ \int_Q -u \cdot \partial_t \varphi + \nabla u \cdot \nabla \varphi - [(u \cdot \nabla)\varphi \cdot u] \, dx \, dt =0 $$ for every $\varphi \in C_0^\infty(Q)$ with $\nabla \cdot \varphi=0$.

By choosing a test function $\varphi = \nabla \times \psi$ with a compactly supported smooth $\psi=(\psi_1, \psi_2, \psi_3)$ it is possible to show that the vorticity $\omega = \text{curl}\ u$ satisfies the vorticity equation $$\tag 2 -\sum_{i=1}^3 \int_Q \omega_i\partial_t\psi_i+\omega_i\Delta \psi_i+ \sum_{j=1}^3 (u_i\omega_j-\omega_iu_j)\frac{\partial \psi_i}{\partial x_j} \, dx \, dt= 0. $$

Then, in the proof by Struwe, one formally chooses a test function $\psi=\omega|\omega|^{2s-2}\zeta^2$, where $\zeta$ is a standard cut-off function. In order to rigorously justify this formal choice of the test function, one considers an approximated vorticity equation (2) with $u_i$ replaced by $u_i^k$, where $u_i^k$ is any smooth approximation of $u_i$ in $L^2$-norm. Then one obtains an approximative solution $\omega_i^k$, which has enough regularity for the formal calculations to be rigorously justified.

The final step of the proof is to show that $\omega_i^k \to \omega_i$ weakly in $L^2$. By some analysis, it is possible to show that $\omega_i^k$, indeed, converge to a function $\tilde \omega_i$, and also that for each $k$, the function $\omega_i^k$ is a unique solution of the approximative equation.

Now my question is that how do we know that $\omega_i = \tilde\omega_i$?

I believe that it is possible to choose the approximative sequence $u_i^k$ to be smooth solutions of the Navier-Stokes equation. In this case, it is clear that $\omega_i^k = \nabla \times u_i^k$ and, as discussed below in the comments section, then we may use the continuity of differential operators in the distributional topology to conclude that $\omega_i^k = \nabla \times u_i^k \to \nabla \times u = \omega_i$ as $k \to \infty$. However, it seems like a non-trivial result that $u_i^k$ can be chosen to be smooth solutions of the Navier-Stokes equation, and Struwe does not state or prove such result.

I am mostly interested in the methodology rather than the result itself and, thus, I would like to understand if it is possible to avoid using the fact that $u_i^k$ solves the equation. In particular, if $u_i^k$ is any smooth approximation of $u_i$, I do not see why $\omega_i^k$ needs to satisfy $\omega_i^k = \nabla \times u_i^k$. Perhaps this is something which follows from the fact that $\omega_i^k$ is a unique solution of the approximative equation? My first guess would be to try showing by Biot-Savart law that $\nabla \times u_i^k$ is always a solution and then to rely on uniqueness, but I would prefer not to reinvent the wheel if someone knows how to do this.

  • 2
    $\begingroup$ One has $u_h \to \partial_{x_i} u$ in the sense of distributions for any distribution $u$, so I don't think the linear diffusive term is problematic as long as one works at a distributional level. To derive the vorticity equation in weak form, I would start with the weak form of the Navier-Stokes equation with $\varphi$ set equal to $\nabla \times \psi$ for some test function $\psi$ and then move the curl over to the $u$ terms; the fact that $u$ is in spacetime $W^{1,2}$ seems to be sufficient regularity to make this work by the usual limiting argument, I think. $\endgroup$
    – Terry Tao
    Commented Jun 3, 2014 at 23:49
  • $\begingroup$ @TerryTao: Thank you for this clarification. I was trying to obtain the result by differentiating the equation, while the trick seems to be choosing the test functions as you pointed out. I think this more or less settles my original question. I also have some additional related questions on the topic and thus the question is now modified accordingly. $\endgroup$ Commented Jun 4, 2014 at 22:00
  • 1
    $\begingroup$ I don't have access to Struwe's paper, but I seriously doubt he is claiming (or using) any uniqueness for weak solutions here, as such claims are generally either open or false. Perhaps he is just using the fact that differential operators are continuous in the distributional topology, so that if $u^k$ converges in the sense of distributions to $u$, then $\omega^k = \nabla \times u^k$ converges in the sense of distributions to $\omega = \nabla \times u$. Note that the distributional topology is Hausdorff, so limits are unique. $\endgroup$
    – Terry Tao
    Commented Jun 4, 2014 at 22:41
  • $\begingroup$ @TerryTao: Indeed, Struwe is not claiming uniqueness, and apparently he is not using it either; I was just assuming since I thought this is the standard way to prove such results for the limits. But of course, this is different from many other cases since $\omega^k$ is not just any sequence, but $\omega^k = \nabla \times u^k$ as you pointed out. So this seems to settle the question. Thank you, again. $\endgroup$ Commented Jun 4, 2014 at 22:54
  • $\begingroup$ @TerryTao: But now that I think about it, this would naturally require that we know that indeed $\omega^k=\nabla \times u^k$. But in the argument the vorticity equation (2) is replaced by the same equation with $u$ replaced by $u^k$. Then $\omega^k$ is the solution of the equation with initial and boundary data given by $\omega^k=\omega$ on the parabolic boundary. The unique existence of the solution probably follows from the standard PDE theory for smooth $u^k$, but is it clear that $\omega^k$ is indeed given by $\omega^k=\nabla \times u^k$? $\endgroup$ Commented Jun 4, 2014 at 23:06

1 Answer 1


If you read one more line down from equation (3.1), Struwe helpfully points you to

(see Serrin [15], (11), p.190)

which happens to be this paper and indeed, around page 190 you find the claim proved in detail. Please note that the notion of weak solution in this particular part of the manuscript may be slightly different from the sense you included in the statement of you questions; see also the first few pages of Serrin's article for what he meant by weak solutions.

  • $\begingroup$ Yes, Struwe refers to Serrin's original paper where the same question is commented by saying that eq. (1) "being clear if $u$ and $\omega$ are of class $C^2$, and therefore true in present case by standard limiting arguments (observe that by Hölder's inequality $u\omega \in L^{1,1})$". As I said, I have probably just missed something, but in any case, I have had some difficulties to prove this starting from the definition of the weak solution. The definition and assumptions in my question are directly from Struwe's paper. $\endgroup$ Commented May 20, 2014 at 8:55
  • $\begingroup$ Typically also the equivalence between the different concepts of solutions is not a trivial issue. So as I am new to Navier-Stokes equations I do not know the literature too well and that's why I am looking for the references. $\endgroup$ Commented May 20, 2014 at 8:58

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