2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.