I have the following equation in $[0,L]$
\begin{equation} \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} = 0 \end{equation} \begin{equation} f(0,t) = 1 \end{equation}
And I know it should be equivalent to the PDE (in the distributional sense):
\begin{equation} \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} = \delta(x) \\ f(0,t) = 0 \end{equation}
Both with the initial condition $f(x,0) = f_0(x)$
However I do not get how this is true rigorously.
Help!
On
This is a badly phrased problem (by whoever made it) as a delta function on the boundary of your domain is ambiguous. With $\delta(x) = \frac{dH(x)}{dx}$ with $H(x)$ being the Heaviside step-function the ambiguity comes from the fact that the $H(x) = 0$ for $x<0$ and $H(x) = 1$ for $x>0$, but is not uniquely defined at $x=0$. See e.g. this. To get the desired result here you must require $H(0) \equiv 0$. Maybe your textbook deals with it this way, but is a technicality I thought I'd mention as it's easy to get confused by this.
Given the following PDE on $(t,x)\in\mathbb{R}_+\times[0,L]$ $$f_t + f_x = \delta(x) \text{ with } f(0,t) = 0 \text{ and } f(x,0) = s(x)$$ then you can take $f(x,t) = H(x) + g(x,t) - 1$ then $g(x,t)$ satisfy $$g_t + g_x = 0 \text{ with } g(0,t) = 1 - H(0) \text{ and } g(x,0) = s(x) + (1- H(x))$$
Now since $H(0) = 0$ and $1 - H(x) = 0$ for all $x\geq 0$ the initial conditions for these two PDEs agree and the boundary conditions differ by $1$ and you have your equivalence.
Let $f : [0,L] \to \mathbb{R}$ be a solution to the PDE and let $\bar{f} : \mathbb{R} \to \mathbb{R}$ be defined by $$ \bar{f}(x) = \begin{cases} f(x) & \text{if } x \in [0, L] \\ 0 & \text{if } x < 0 \\ f(L) & \text{if } x > L \\ \end{cases} $$
Then, if $f(0) \neq 0$, then $\bar{f}$ has a discontinuity at $x=0,$ but is continuous elsewhere, and its distributional partial derivative w.r.t. $x$ is given by $$ \partial_x \bar{f} = \partial_x f \, \chi_{[0,L]} + f(0) \, \delta, $$ where $\chi_{[0,L]}(x) = 1$ if $x \in [0,L]$ and $=0$ elsewhere.