Characterization of weak solutions to a(n) (simple) ODE Fix $f \in L^1_{loc}(a,b)$ and define the operator $L_f: C_c^\infty(a,b) \to \mathbb{R}$ given by
$$L_f(\phi) = -\int_a^b f\phi'.$$ It is well-known that $f$ is a weak solution to the differential equation $u' = 0$. Moreover, in this case $f$ is constant a.e. I am sort of new to this idea of weak derivatives, but how could we formulate a similar operator so that $f$ is a weak solution to say for example the differential equation $u'' = 0$? In this case how would one find all solutions?
 A: If you had a strong solution then you would have
$$\int_a^b u'' f dx = 0 \Rightarrow -\int_a^b u' f' dx = 0 \Rightarrow \int_a^b u f'' dx = 0$$
where the boundary terms drop because you assumed $f \in C^\infty_c$. So now for a weak solution you just take this equation as the definition of a solution.
That said, in an ODE like this, which is linear with leading coefficient $1$ and the other coefficients/forcing are smooth, the notion of weak solution doesn't buy you anything analytically. That's because you can prove existence/uniqueness of a strong solution directly, and all weak solutions are equal a.e. to a strong solution.
By the way, there's an important detail in the definitions here. Specifically, $f \in C^\infty_c((a,b))$ means that the support of $f$ is contained in some $[c,d]$ with $a<c<d<b$. If $f$ just vanished at the boundary, then the boundary terms in the first integration by parts would vanish, but the boundary terms in the second integral would not necessarily vanish (cf. something like $(x-a)(x-b)$ to see why).
