Yes, that is one of the whole points of using weak formulation: the solution is defined in a weaker way. Consider $-\Delta u = f$. A classical solution needs $u$ to be twice differentiable. But a weak formulation of this problem may be $\int_\Omega \nabla u \nabla v = \int_\Omega fv$ (holding for all $v \in H^1_0$) and this only needs one space derivative. You may ask: why is such a notion of solution good enough? For some answers see this thread.