Uniqueness for Weak Solutions to Navier Stokes We proved the following existence result for Navier-Stokes:

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1 Uniqueness for Weak Solutions to Navier Stokes We prove the following existence result for Navier-Stokes: heorem (Existence of a weak solution, n 4) For every f L 0, : V each u 0 H there exists at least one u L 0, : V that is a weak solution of the N-S system In aition, this weak solution satisfies: (a) u L 0, : H u L 4/n 0, : V (b) u t is weakly continuous in H; Uniqueness in the Case n When n, the weak solution has aitional regularity is, in fact, unique heorem (Uniqueness of the weak solution, n ) In the case n, the weak solution of the Navier Stokes equations satisfies: i) u L 0, : V u L 0, : V hence u ~ ũ t C 0, : H ii) u t is unique Proof- We begin by repeating an estimate for the nonlinear term in the N-S system For arbitrary u,v,w H 0 1 U n, we have (by the extene Holer ineqality followe by the C-S inequality) b u,v,w i,j 1 U u i i v j w j x i,j 1 u i 4 i v j w j 4 i,j 1 i v j 1/ i 1 u i 4 1/ j 1 w j 4 1/ In aition, the Sobolev inequality implies where u q C u p u r 1 u C 0 U 0 1, 1/q 1/p 1/n 1 1/r If we choose q 4 p r so that n/4, then with n, we have u 4 C u 1/ u 1/, combine with the previous estimate, 1

2 i 1 u i 4 C i 1 u i u i C u H u V his leas to the following estimate on b u,v,w b u,v,w C u H u V 1/ v V w H w V b u,u,v b u,v,u C u H u V v V u,v V 1/ his implies that B u t V C u H u V Since the weak solution, u t, belongs to L 0, : H L 0, : V, we have 0 B u t V t C 0 u t H t C u u L 0,:V hence B u V C u u L 0,:V B u t L 0, : V hen u t also belongs to L 0, : V then it follows from heorem 3, p87 in Evans that u L 0, : V u L 0, : V implies u t ~ ũ t C 0, : H Now, for u L 0, : V u L 0, : V we have u t,u t V V 1 u t t H hen, if u 1 t, u t are two weak solutions for N-S, let u t u 1 t u t hen u t Au t B u t B u 1 t, u 0 0, u t t H u t V b u t,u t,u t b u 1 t,u 1 t,u t But b u t,u t,u t b u 1 t,u 1 t,u t b u 1 t,u 1 t,u t b u t,u t,u t b u 1 t,u 1 t,u t b u t,u 1 t,u t b u t,u 1 t,u t b u t,u t,u t b u 1 t,u 1 t,u t b u t,u 1 t,u t b u t,u 1 t,u t b u t,u 1 t,u t b u t,u t,u t b u t,u t,u t 0 hen u t t H u t V b u t,u 1 t,u t

3 But as we showe previously, ( using abc C a C b c ) b u t,u 1 t,u t C u t H u 1 t V C C u t V C u t H u 1 t V therefore, u t t H C u t H u 1 t V Since u 1 t V L 1 0,, we get t u t H exp C 0 t u 1 s V s 0 ie, u t H u 0 H 0, or u 1 t u t Summarizing, when n the NS system has a unique solution u t C 0, : H L 0, : V, with u t L 0, : V In aition, u t L 4 U Since the solution is unique in the case n is not unique in the case n 3, it must be that the assumptions uner which a physical flow can be treate as a -imensional flow are also the conitions that preclue turbulence the consequent loss of uniqueness hus a flow which can be treate as -imensional uner some conitions (eg, shallow flow with low flow velocity ) may uner other conitions behave in ways that cannot be viewe as -imensional, even though the flow may still be a shallow flow Uniqueness in the Case n 3 Recall the corollary to the Sobolev inequality implies u 4 C u n/4 u 1 n/4 so that, in the case n 3, we have u 4 C u 3/4 u 1/4 his leas to the result that the weak solution of N-S satisfies, u t L 8/3 0, : L 4 U n, u t L 4/3 0, : V Note that this is strictly weaker than what was true of the solution in the case n 3

4 heorem Suppose n 3 the weak solution of the N-S system satisfies: u t L 0, : H L 0, : V u t L 8 0, : L 4 U n, hen, uner these conitions, u t is unique u t C 0, : H Proof- Supppose u t is a weak solution for N-S with the aitional regularity inicate above hen b u t,u t,v C u t 4 v V u t,v V, implies B u t V C u t 4 0 B u t p V t C 0 u t p 4 t C 0 u t np/ V t It follows that B u t L p 0, : V if u t L 0, : V np/ ; ie, B u t L 4/n 0, : V L 4/3 0, : V On the other h, if u t L 8 0, : L 4 U n, then 0 u t 4 p t for p 4 B u t L 0, : V (at least) hen it follows from the N-S equation that u t L 0, : V this, together with u t L 0, : V imply u t C 0, : H Now observe that b u t,u t,v C 0 u t 4 v 4 C 1 u t 1/4 3/4 v 4 C 1 u t H 1/4 7/4 v 4 Now suppose u 1 t, u t are two weak solutions for N-S both of which have the aitional regularity of the hypotheses Let u t u 1 t u t, note that as in the n proof 4

5 u t t H u t V b u t,u t,u t Also so b u t,u t,u t C 1 u t H 1/4 7/4 u t 4 C u t H u t 4 8 u t t H C u t 8 4 u t H Since u t 4 8 L 1 0,, we can finish the proof as in the n case In the step, u t H 1/4 7/4 u t 4 C u t H u t 4 8 we use the following version of Young s inequality, ab ap p bq q a p C b q, 1 p 1 q 1, a,b 0 5