We multiply the strong formulation by functions from the testing space and integrate
\( -\int_0^l \left( \left(a(x)u'(x)\right)'+b(x)u'(x)+c(x)u(x)\right)v(x)dx=\int_0^l f(x)v(x)dx \quad \forall v \in V=\{u\in H^1(\Omega): u(0)=0 \} \)
We integrate the first term through parts
\( \int_0^l a(x)u'(x)v'(x)dx-a(0)u'(0)v(0)-a(l)u'(l)v(l)+\int_0^l b(x)u'(x)v(x)dx+\int_0^l c(x)u(x)v(x)dx= \\ = \int_0^l f(x)v(x)dx \quad \forall v \in L^2(0,l) \)
We use the fact that \( v \ in V \) or \( v (0) = 0 \), and substitute the boundary condition \( a (l) u '(l) + \ beta_l u ( l) = \ gamma_l \).
We mark \( B(u,v)=\int_0^l \left( a(x) u'(x) v'(x) +b(x)u'(x)v(x)+c(x)u(x)v(x)\right)dx + \beta u(l)v(l) \) and \( L(v)=\int_0^l f(x) v(x)dx + \gamma v(l) \).
We substitute \( w=u-\tilde{u} \) to get \( w(0)=0 \) and \( B(w,v)=L(v)-B(\tilde{u},v) \). For simplicity of notation instead of a symbol \( w \) we use once again the symbol \( u \).