A small calculating problem in the article by M. Gage and R. S. Hamilton In the article " The heat equation shrinking convex plane curves " by M. Gage and R. S. Hamilton, I didn't finish the calculation in 4.3.4:$$\frac{\partial }{{\partial t}}\int\limits_0^{2\pi } {\log k(\theta ,t)d\theta  = \int\limits_0^L {\left[ {{k^2} - {{\left( {\frac{{\partial k}}{{\partial \theta }}} \right)}^2}} \right]d\theta } }$$
I have already known that it's supposed to use the evolution equation in 4.1.4:$$
\frac{{\partial k}}{{\partial t}} = {k^2}\frac{{{\partial ^2}k}}{{\partial {\theta ^2}}} + {k^3}$$
And the article remind us to use integration by parts, but I just fail to finish the calculation and fall into disorder. Could you please provide the detailed steps, or provide another method to simplify? Thanks a lot.
 A: \begin{align}
\frac{\partial }{\partial t} \int_0^{2\pi} \log k(\theta ,t)d\theta  &= \int_0^{2\pi} \frac{\partial}{\partial t} \log k\, d\theta \\
&= \int_0^{2\pi} \frac{1}{k} \frac{\partial k}{\partial t} d\theta \\
&= \int_0^{2\pi} \left( k \frac{\partial ^2k}{\partial \theta^2} + k^2\right) d\theta\\
&= \int_0^{2\pi}\left( - \left( \frac{\partial k}{\partial \theta } \right)^2 
 + k^2 \right)d\theta 
\end{align}
The integration by part is used at the last step: to be explicit, by product rule $gf' = (gf)' - f'g$,
$$\int_0^{2\pi} k \frac{\partial^2 k}{\partial \theta^2} d\theta= \int_0^{2\pi} \left( \frac{\partial}{\partial \theta } \left( k \frac{\partial k}{\partial\theta}\right) - \frac{\partial k}{\partial\theta}\frac{\partial k}{\partial\theta}\right) d\theta$$
then the fundamental theorem of calculus (and the fact that the curve is closed) implies
$$\int_0^{2\pi} \frac{\partial}{\partial \theta } \left( k \frac{\partial k}{\partial\theta}\right) d\theta = 0.$$
Remark: For any periodic function $f(\theta) = f(\theta + 2\pi)$, we have
$$ \int_0^{2\pi} \frac{\partial }{\partial\theta} f(\theta) d\theta = f(2\pi) - f(0) = f(0) - f(0) = 0.$$
