# Applying Gronwall lemma in Majda-Bertozzi book.

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!

• Where is this, presumably the energy methods chapter? You must need some constant in the final line right – Calvin Khor Oct 16 '18 at 7:46
• 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. – Alex Oct 16 '18 at 7:57
• 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 – Calvin Khor Oct 16 '18 at 8:56

$$q(t) \le c(t) + \int_0^t u(s) q(s) ds\implies q(t) \le c(0) \exp\left(\int_0^t u(s) ds\right) + \int_0^t c'(s)\left(\exp\int_s^t u(\tau)d\tau \right)ds$$ 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.