Problem about convolution and diffusion equation (Fourier)

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I am trying to understand how to solve a problem (It's in a Fourier Chapter) .

Here the statement :

If $u(\theta,t)$ is the temperature at a time $t$ at the point $(R,\theta)$ in polar coordinates on a conduction circle. And that the temperature is following the diffusion equation : $\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial \theta^2}$

Express $u$ as a convolution by $\theta$ :

$u(\theta,t)=g(\theta,t) \circledast u(\theta,0)$

where $g$ need to be precised and $\circledast$ is the circular convolution :

$(x \circledast y)(\theta):=\frac{1}{2\pi}\int_{0}^{2\pi}x(u)y(\theta-u)du$.


Thank by advance for your help !

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Hint

This is the standard Heat equation and you don't need to use convolution as it throws you off the simple and convenient way that is most often taken to solve such equations like this or the Wave equation $u_{tt}=u_{xx}$. Try $$u=f(t)g(\theta)$$therefore $$u_t=u_{\theta\theta}\implies {f'(t)\over f(t)}={g''(\theta)\over g(\theta)}$$