I was reading the following pdf about the heat equation on an interval $[0,l]$ with Neumann conditions,
http://texas.math.ttu.edu/~gilliam/fall03/m4354_f03/heat_N_web/heat_ex_homo_neum.pdf
i.e. $$\partial_t u = u'' \quad u(0,x)=\varphi(x), \quad \partial_t u(t,0)=\partial_t u(t,l)=0.$$
The solution is given explicitly as, $$u(t,x) = a_0 + \sum_{n\geq 1} a_n e^{\lambda_n t} \cos \left(\frac{\pi}{l}nx\right)$$ with $$a_0 = \frac{1}{l} \quad a_n = \frac{2}{l}\int_0^l \varphi(x) \cos\left( \frac{\pi}{l}n x\right) dx.$$
My question is: What happens if you start from a Dirac distribution $\varphi = \delta_{l/2}$. I get 0 but I do not understand why. If one starts from $\delta_0$ or $\delta_l$ then you don't get the constant 0. Moreover, if $\varphi=\delta_{l/2}$ then $$u(t,x)=a_0=\frac{1}{l}$$ which does not satisfy $u(t,x)=\varphi(x)=\delta_{l/2}(x)$.
Where is the mistake? Thanks!
You should not get $0$; the heat equation with homogeneous Neumann boundary conditions preserves mass, and the Dirac delta has mass. In particular, at $l/2$ the odd frequency eigenfunctions are zero but the even frequency eigenfunctions are not zero: they are $\pm 1$. That is, you should have
$$a_0=\frac{1}{l} \\ a_n = \frac{2}{l} \int_0^l \delta_{l/2}(x) \cos \left ( \frac{n \pi x}{l} \right ) dx = \frac{2}{l} \cos \left ( \frac{n \pi}{2} \right ).$$
So $a_{2n}=\frac{2 (-1)^n}{l}$ for $n \geq 1$.