I'm solving the differential equation $$ \left( \sigma^{2}(x) u ''(x) \right)'' = f(x), \;\;\; 0 \leqslant x \leqslant 1 $$ with initial conditions $u(0) = u(1) = 0$, $u''(0) = u''(1) = 0$. Here $\sigma(x) \geqslant \sigma_{0} > 0$ is parameter. In operator form we can rewrite the differential equation as $Au = f$, where operator $A$ is positive definite.
Following FEM scheme, I reduce my problem to an optimisation problem $$ J(u) = (Au,u) - 2(f,u) \to \min_{u} $$ I introduce finite elements $h_{k}(x)$ as $$ v_{k}(x) = \left\{ \begin{array}{rl} 1 - \left( \frac{x-x_{k}}{h} \right)^2, & x \in [x_{k-1},x_{k+1}] \\ 0, & \text{otherwise} \end{array} \right. $$ for any $k = 1,\ldots,n-1$, where $x_{k} = hk$, $h = \frac{1}{n}$. Finite elements $v_{0}(x)$ and $v_{n}(x)$ are introduced similarly.
I try to find numericaly the vector $\alpha$ such that $u(x) = \sum_{k=0}^{n} \alpha_{k} v_{k}(x)$ solves the optimisation problem. We have $$ J(u) = \sum\limits_{i=0}^{n} \sum\limits_{j=0}^{n} \alpha_{i} \alpha_{j} (Av_{i},v_{j}) - \sum\limits_{i=0}^{n} 2\alpha_{i} (v_{i},f) = \alpha^{T} V \alpha - 2\alpha^{T} b \to \min\limits_{\alpha}, $$ where $b_{i} = (f,v_{i})$ and $V_{i,j} = (Av_{i},v_{j})$. After differentiation with respect to $\alpha$ I receive $$ V\alpha = b, $$ but here the stiffness matrix $V$ is singular. So what I have to do? Maybe I have to choose other finite elements?