 Solution

Let $V_h$ be the continuous piecewise linear space including both endpoint hat functions. Robin conditions are natural conditions and do not restrict the endpoint values of this space. The finite element problem is $u_h\in V_h$, $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. coercivity restricted to $V_h$ gives existence and uniqueness.

Expanding $u_h={\sum }_{j=0}^N{U}_j{\varphi }_j$ gives $\mathsf{A}U=F$, whereThe stiffness contribution has diagonal $2/h$ at interior nodes, $1/h$ at endpoints, and adjacent entries $-1/h$. For completeness, the weighted mass contribution has interior diagonal $2h{x}_i^2/3+h^3/15$, left endpoint diagonal $h{a}^2/3+a{h}^2/6+h^3/30$, right endpoint diagonal $h{b}^2/3-b{h}^2/6+h^3/30$, and adjacent entry $h{x}_i^2/6+h^2{x}_i/6+h^3/20$. All more distant entries vanish. The Robin contribution adds one to each endpoint diagonal.

Symmetry follows from symmetry of the bilinear form. If $V\ne 0$, nodal independence makes $v_h={\sum }_i{V}_i{\varphi }_i$ nonzero, and $V^T\mathsf{A}V=a(v_h,v_h)\ge \gamma \Vert v_h{\Vert }_{H^1}^2>0$. Thus the matrix is symmetric positive definite.