The problem is as follows,
$$G_{xx}(x,k) = - \delta(x-k) \tag{1} $$ $$ G_x(0,k) = G_x(1,k) = 0 \tag{2} $$
Is this possible? From what I understand,
(1) implies $G(x,k)$ is continuous and piecewise linear, but the conditions in (2) then imply that $G_x = 0$.
Indeed, no such solution can exist: if it did, we would have by the Fundamental Theorem of Calculus $$ 0 = G_x(1,k) - G_x(0,k) = \int_0^1 G_{xx}(x',k) \, dx' = \int_0^1 -\delta(x'-k) dx' = -1 ,$$ which is obviously wrong.