I'm trying to solve the following heat equation
$$ \begin{cases} u_t = \frac{1}{10}u_{\theta,\theta} \\ u_\theta(0,t) = u_\theta(\pi, t) = 0 \\ u(\theta,0) = 3 - 4\cos 2\theta \end{cases} $$
I've succeeded to see that the solution must be $$ u(\theta, t) = 3 - 4e^{-\frac{4}{10}t}\cos 2 \theta $$ using Fourier series, but I'm not using the condition $u_\theta(0,t) = u_\theta(\pi,t) = 0$. Here is my argument:
We are looking for a function $u:\mathbb{T}\times [0, +\infty) \longrightarrow \mathbb{R}$, where $\mathbb{T} = [-\pi, \pi]$, which satisfies the expressions above. Notice that $u$ is going to be a continuous function, continuously differentiable with respect to $t$ and twice continuously differentiable with respect to $\theta$. In particular, for each $t \in [0, +\infty)$, the functions $u(\cdotp, t)$ and $u_t(\cdotp,t)$ are continuous in $[0,\pi]$, so $u(\cdotp,t), u_t(\cdotp,t) \in L^1(\mathbb{T})$, and it makes sense to consider its Fourier coefficients. Thus, the equation $$ u_t = \frac{1}{10}u_{\theta,\theta} $$ can be written, in terms of Fourier coefficients with respect to the $\theta$ variable, as $$ \widehat{u_t}(n) = \frac{1}{10} \widehat{u_{\theta,\theta}}(n) = - \frac{n^2}{10}\widehat{u}(n) $$ Also, because $u$ is continuous and so is $u_t$, derivation under the integral sign holds and we have $$ \widehat{u_t}(n) = \frac{1}{2\pi} \int_{-\pi}^\pi \frac{\partial u(\theta,t)}{\partial t} e^{-in\theta} \ d\theta = \frac{d}{d t} \left(\frac{1}{2\pi} \int_{-\pi}^\pi u(\theta,t) e^{-in\theta} \ d\theta \right) = \frac{d}{dt} \widehat{u}(n) $$ so we get $$ \frac{d}{dt}\widehat{u}(n) = - \frac{n^2}{10}\widehat{u}(n) $$ and the solution to this equation is $\hat{u}(n) = K_n e^{-\frac{n^2}{10}t}$.
Now, consider the Fourier series $$ S[u](\theta,t) = \sum_{n \in \mathbb{Z}} K_n e^{-\frac{n^2}{10}t}e^{in\theta} $$ For each $t \in [0, + \infty)$, this series converges uniformly to $u(\cdotp, t)$, because $u(\cdotp,t)$ is differentiable. In particular, we have that $$ u(\theta, 0) = \sum_{k \in \mathbb{Z}} K_n e^{in\theta} = 3 - 4\cos 2\theta $$ meaning that we must have $K_0 = 3$, $K_2 = K_{-2} = -2$, and $K_n = 0$ for every $n \neq 0, 2,-2$. That is, $$ u(\theta,t) = S[u](\theta,t) = 3 - 4e^{-\frac{4}{10}t}\cos(2\theta) $$
Am I doing something wrong, or using the condition $u_\theta(0,t) = u_\theta(\pi,t) = 0$ implicitely somewhere?
Thank you very much!