Does operator have to be linear for weak formulation?

Question

Wikipedia says, that we find a weak formulation of the equation $$ Au = f $$ by defining a bilinear form $a(u,v)$. And in the examples, the operator $A$ is always linear, however it doesn't specify this in the requirements of the definition. Is it possible, to find a bilinear form, also when $A$ is not linear?

Answer 1

When $A$ is not linear, there is a form, but it's not a bilinear form. For example, consider the $p$-Laplacian.

The Wikipedia page you link to only considers linear equations, and their weak formulations lead to quadratic forms. If you start with something non-linear, you might have a weak solution, and you can expect it to be something not-quite-quadratic.

Answer 2

There are lots of different sorts of weak formulations of PDE. For a linear problem, we indeed recast $Au=f$ as $a(u,v)=b(f,v)+c(u,v)$, where $a$ and $b$ are bilinear forms and $c$ is a boundary term. Nonlinear PDE have weak formulations too, but there is no reason to expect them to involve this bilinear structure. Read Evans.