= Solution
{c}
Put $e=u-u_h$ and $V=\{v\in H^1(0,1):v(1)=0\}$. The primal solution and its <finite element> approximation lie in the same affine space with value one at the right endpoint, so $e\in V$. Use
$$
a(w,v)=\int_0^1(pw'v'+qwv).
$$
Define the dual solution by
$$
\boxed{-(pz')'+qz=e,\qquad z'(0)=0,\quad z(1)=0.}
$$
Equivalently $z\in V$ satisfies $a(v,z)=(e,v)$ for every $v\in V$. Positivity of $p,q$ gives a unique solution, and one-dimensional regularity gives $z\in H^2$.
Then $\|e\|_{L^2}^2=a(e,z)$. The nodal interpolant $I_hz$ belongs to the homogeneous <finite element> test space, so <Galerkin orthogonality> gives $a(e,I_hz)=0$. Set $w=z-I_hz$. Using the primal weak equation,
$$
\|e\|_{L^2}^2=(f,w)-a(u_h,w).
$$
Integrate the <derivative> term separately on each element. Since $w$ vanishes at every <mesh> node, all endpoint flux and flux-jump terms vanish. Therefore
$$
\boxed{\|e\|_{L^2}^2
=\sum_i\int_{x_{i-1}}^{x_i}R_h(z-I_hz)\,dx.}
$$
The absence of jump terms here follows specifically from one-dimensional nodal <interpolation>; it is not an assumption that the <finite element> flux itself is continuous.
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