I want to show that the Neumann boundary value problem $$ \begin{cases} - u''(x) = H(x), & x \in (-1, 1)\\ u'(-1) = u'(1) = 0, \end{cases} $$ where $$H(x) = \begin{cases} 1, & x > 0, \\ 0, & x < 0 \end{cases}$$ is the Heaviside function, is not uniquely solvable in the weak sense, by finding infinitely many solutions.
For the test space I choose $$H^1((-1, 1); \mathbb R) := \{ u \in L^2((-1, 1); \mathbb R): u' \in L^2((-1, 1); \mathbb R) \},$$ where $u'$ is the weak derivative of $u$. We have $$ - \int_{-1}^{1} u''(x) v(x) dx = \int_{-1}^{1} u'(x) v'(x) dx $$ and $$ \int_{0}^{1} v(x) dx = \int_{-1}^{1} H(x) v(x) dx $$ Thus the weak formulation is: find $u \in H^1((-1, 1); \mathbb R)$ such that $$ \int_{-1}^{1} u'(x) v'(x) dx = \int_{0}^{1} v(x) dx $$ for all $v \in H^1((-1, 1); \mathbb R)$.
My idea is to write the right side as $$ \int_{0}^{1} v(x) dx = v(1) - \int_{-1}^{1} \max(0, x) v'(x) dx. $$ If $v(1)$ we zero (which its not) the solutions would be $$ u_c(x) = \begin{cases} c, & x < 0, \\ c - \frac{x^2}{2}, & x \ge 0 \end{cases} $$ for any constant $c \in \mathbb R$, as the weak derivative of $u_c$ is $$ u_c'(x) = - \max(0, x). $$ I know that I can represent $v(1) = \int_{-1}^{1} \delta(x - 1) v(x) dx$, where $\delta$ is the Dirac distribution, but this hasn't helped. Can I maybe use the Fundamental Theorem of the Calculus of Variations (that is, $\int_{a}^{b} u(x) v'(x) dx = 0$ for all $v \in \mathbb{C}_{0}^{\infty}([a, b]; \mathbb R)$ implies the existence of a $c \in \mathbb R$ such that $u(x) = c$ almost everywhere)?
Generally, when you have a linear (inhomogeneous) equation and you want to know how many solutions there are, one thing to try is suppose $u_{1}$ and $u_{2}$ are solutions and then ask, "What equation does $u_{1} - u_{2}$ solve?" This is true even at the level of linear algebra (not just differential equations), as you probably know.
If we do that here, if $u_{1}$ and $u_{2}$ solve your equation, then $w = u_{1} - u_{2}$ is a solution of the equation \begin{equation*} -w'' = 0 \quad \text{in} \, \, (-1,1), \quad w'(-1) = w'(1) = 0. \end{equation*} Notice that any such function must be constant due to the boundary conditions. (Look at the equation that $f = w'$ solves; or play clever tricks with the weak formulation.) That is, $w \equiv C$ for some $C \in \mathbb{R}$. Hence $u_{1} \equiv u_{2} + C$.
I did not prove that there were two solutions; I just showed that any two solutions differ by a constant. At the same time, rerunning the previous argument in reverse, we see that if $u$ is a solution, then so is $u + C$, no matter which constant $C$ I take. That is, as soon as we have one solution, we have infinitely many. (Again, this is consistent with linear algebra: a linear system has no solutions, a unique solution, or infinitely many.)
(The question now is: do you know that there is at least one solution? The answer is yes at any rate.)