I'm having some issues to solve this laplace's equation $u(x,t)$:
$\frac{∂u}{∂t}=\alpha^2 \frac{∂^2u}{∂x^2} $, $t>0$ and $0 < x < \pi$
with the following boundary conditions:
$\frac{∂u}{∂x}(0,t)= 0$, $u(\pi, t) = 0 $, $u(x,0) = g(x)$
I know at the end I gonna use the Fourier series, but I don't understand how to deal with the first boundary condition to get that. Any help?
This is usually known as the heat equation, Laplace's equation involves a second derivative in $t$.
Anyway, assuming you know something about using a Fourier series to solve this, you start with the usual separation of variables $u(t,x) = T(t)X(x)$ to arrive at $$ \frac{T'}{\alpha^2 T} = \frac{X''}{X} = -\lambda $$ so that $X$ and $T$ satisfy ODEs $$ T' = -\lambda \alpha^2 T, \\ X'' = -\lambda X. $$ The solutions of the second equation are of the form $$ X = c_1 x + c_2 \quad : \quad \lambda = 0 \\ X = c_1 e^{\sqrt{-\lambda} x} + c_2e^{-\sqrt{-\lambda}x} \quad : \quad \lambda < 0 \\ X = c_1 \cos(\sqrt{\lambda}x) + c_2\sin(\sqrt{\lambda}x) \quad : \quad \lambda > 0 \\ $$
Now use the boundary conditions: $$ \partial_x u(0,t) = X'(0) T(t) = 0 \Rightarrow X'(0) = 0, \\ u(\pi,t) = X(\pi) T(t) = 0 \Rightarrow X(\pi) = 0. $$
Apply these to the various cases above to narrow down your choices for $\lambda$. In particular you should be able to eliminate $\lambda \leq 0$, and also find that it constrains your choices of $\lambda > 0$ as well: $$ X'(0) = -c_1 \sin(0) + c_2 \cos(0) = c_2 = 0 \\ X(\pi) = c_1 \cos(\sqrt{\lambda}\pi) = 0 $$ so that $\sqrt{\lambda}\pi$ is a zero of cosine, e.g. $$ \sqrt{\lambda} = \frac{2n + 1}{2}. $$
You can use Fourier's transformation, let $$\widehat{u}(\xi,t)=\int_{\mathbb{R}}{u(x,t)e^{-i\xi x}dx},$$you have that $\widehat{\frac{\partial u}{\partial x}}=i\xi\widehat{u}$ and $\widehat{\frac{\partial u}{\partial t}}=\frac{\partial\widehat{u}}{\partial t}$. Hence $$\frac{\partial\widehat{u}}{\partial t}+\alpha^2\xi^2\widehat{u}=0$$ It is a differential equation that you can easily solve : $$\widehat{u}(\xi,t)=\widehat{u}(\xi,0)e^{-\alpha^2\xi^2t}=\widehat{g}(\xi)e^{-\alpha^2\xi^2t}$$ where $$\widehat{g}(\xi)=\int_{\mathbb{R}}{g(x)e^{-i\xi x}dx}$$ You can finally use Fourier's inversion formula : $$u(x,t)=\frac{1}{2\pi}\int_{\mathbb{R}}{\widehat{u}(\xi,t)e^{i\xi x}d\xi}=\frac{1}{2\pi}\int_{\mathbb{R}}{\widehat{g}(\xi)e^{-\alpha^2\xi^2 t+i\xi t}d\xi}.$$ You can't easily compute this integral, but you can calculate it by derivating it as a function of the variable $t$ verifying a differential equation. In the end you find $$u(x,t)=\frac{1}{\sqrt{4\pi\alpha^2 t}}\int_{\mathbb{R}}{f(y)e^{-\frac{(x-y)^2}{4\alpha^2 t}}dy} $$ Notice that this proof only works if you suppose that $\widehat{u}$ and $\widehat{g}$ are well defined and that $\widehat{u}(\cdot,t)\in\mathbb{L}^1(\mathbb{R})$ for all $t\in\mathbb{R}^+$ in order to use Fourier's inversion formula.
