I am studying Majda-Bertozzi book about incompressible flows. I have applied Gronwall lemma several times, but I do not know how to do in the following case: We have $$ |\nabla v(\cdot,t)|_{L^{\infty}}\le C\left(1+\int_0^t|\nabla v(\cdot,s)|_{L^{\infty}}\right)\left(1+|w(\cdot,t)|_{L^{\infty}}\right) $$ and we must use Gronwall lemma to get: $$ |\nabla v(\cdot,t)|_{L^{\infty}}\le|\nabla v_0|_0 \exp\left(\int_0^t|w(\cdot,s)|_{L^{\infty}}ds\right). $$ Thanks a lot!

  • $\begingroup$ Where is this, presumably the energy methods chapter? You must need some constant in the final line right $\endgroup$ – Calvin Khor Oct 16 '18 at 7:46
  • $\begingroup$ Probably yes. The goal (the proof of the theorem 3.6) is achieved if there is a constant in the final line right. But I do not know how to obtain the rhs of the conclusion. $\endgroup$ – Alex Oct 16 '18 at 7:57
  • $\begingroup$ If I understand correctly, this is from the paper of Beale-Kato-Majda but the inequality in the paper there is different. projecteuclid.org/euclid.cmp/1103941230 maybe you can figure something out $\endgroup$ – Calvin Khor Oct 16 '18 at 8:56

I can't quite obtain that inequality, but I can get something that is enough to finish the proof of the theorem. Majda-Bertozzi quotes lemma 3.1 as Gronwall,

$$q(t) \le c(t) + \int_0^t u(s) q(s) ds\implies q(t) \le c(0) \exp\left(\int_0^t u(s) ds\right) + \int_0^t c'(s)\left(\exp\int_s^t u(\tau)d\tau \right)ds$$ To put it in this form, you can set $$ q(t) = \frac{|\nabla v(t)|_{L^\infty}}{1+|\omega(t)|_{L^\infty}}, u(t) = C(1+|\omega(t)|_{L^\infty}), c(t) = C$$ Then we have $$ \frac{|\nabla v(t)|_{L^\infty}}{1+|\omega(t)|_{L^\infty}} \le C\exp\left(C\int _0^t 1+ |\omega(s)|_{L^\infty}ds \right) $$ i.e. $$ |\nabla v(t)|_{L^\infty} \le C (1+|\omega(t)|_{L^\infty})\exp\left(C\int _0^t 1+ |\omega(s)|_{L^\infty}ds \right) = \frac{d}{dt} \exp \left(C\int _0^t 1+ |\omega(s)|_{L^\infty}ds \right)$$ And hence $$\int_0^t |\nabla v(s)|_{L^\infty} ds \le C\exp \left(C\int _0^t 1+ |\omega(s)|_{L^\infty}ds \right) $$ which you can plug into the $H^m$ energy estimate (3.79), $$ \|u(T)\|_m \le \|u_0\|_m \exp\left(c_m\int_0^T|\nabla v(t)|_{L^\infty} dt \right)$$ to get the required a priori estimate.

  • $\begingroup$ (the BKM paper also gets a double exponential bound) $\endgroup$ – Calvin Khor Oct 17 '18 at 8:42

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