12: A change of variables in (8)
$$(8)\ u(x,t)=\sum_{n=-\infty}^{\infty}a_{n}e^{-4\pi^{2}n^{2}t}e^{2\pi inx}=(f*H_{t})(x)$$ $$H_{t}(x)=\sum_{n=-\infty}^{\infty}e^{-4\pi^{2}n^{2}t}e^{2\pi inx}$$ leads to the solution $$u(\theta,r)=\sum a_{n}e^{-n^{2}r}e^{in\theta}=(f*h_{r})(\theta)$$ of the equation $$\frac{\partial u}{\partial r}=\frac{\partial^{2}u}{\partial \theta^{2}}\ \ with\ 0\le\theta\le2\pi\ and\ r>0,$$ with boundary condition $u(\theta,0)=f(\theta) \sim \sum a_{n}e^{in\theta}.$ Here $h_{r}(\theta)=\sum_{n=-\infty}^{\infty}e^{-n^{2}r}e^{in\theta}.$ This version of the heat kerna=el on $[0,2\pi]$ is the analogue of the Poisson kernel, which can be written as $P_{r}(\theta)=\sum_{n=-\infty}^{\infty}e^{-|n|r}e^{in\theta}$ with $r=e^{-r}$ (and so $0<r<1$ corresponds to $r>0$). I don't understand the part where the heat kernel is the analogue of the Poisson kernel with $r=e^{-r}$.