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Let $A: H^s(\Omega) \to H^1(\Omega)$ be a bounded linear map. Let $u \in H^s(\Omega)$. Let $f:\mathbb{R} \to \mathbb{R}$ be nonlinear and Lipschitz such that $f(u) \in H^s(\Omega)$.

Is it possible to write the weak gradient $$\nabla(A(f(u))$$ in terms of the derivative of $f$, the (possibly Frechet) derivative of $A$ and (possibly?) $\nabla (Au)$? Basically I am looking for a chain rule for such kinds of expressions but preferably with a gradient term which is linear in $u$.

Motivation: the motivation is that this gradient term comes up in a PDE, and I wish to linearise the PDE and look for a fixed point. So to take another example, instead of considering $\nabla g(u)$ I would write it as $g'(u)\nabla u$, then linearise: $g'(v)\nabla u$ then look for a fixed point of the map $v \mapsto u$ where $u$ is the solution of an equation with the coefficient $g'(v)$.

This is related to Nemytskii maps but it is not quite the same.

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  • $\begingroup$ I don't quite understand how you are planning to interchange $A$ and $f$? This is not just the chain rule anymore. $\endgroup$ Commented Feb 21, 2015 at 10:17
  • $\begingroup$ @AlexDegtyarev I was hoping to rewrite it so as to have a gradient term which is linear in $u$, but I'd be happy for whatever the right expression should be. $\endgroup$
    – MainW
    Commented Feb 21, 2015 at 10:29
  • $\begingroup$ @AlexDegtyarev For example, let $g = A \circ f$. Then, although it is wrong, one could think of $\nabla g(u)$ as $g'(u)\nabla u$ (this obviously does not make sense since $ u \notin H^1$ but this is the kind of thing I wanted to think about. $\endgroup$
    – MainW
    Commented Feb 21, 2015 at 10:31
  • $\begingroup$ There is no simple formula. You need to use whatever you know about $A$. You can see this by defining it to be "multiply by a fixed non-constant function" or more generally a linear PDO with variable coefficients. $\endgroup$
    – Deane Yang
    Commented Feb 21, 2015 at 14:27
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    $\begingroup$ $w = Af(u)$ satisfies a PDE boundary value problem in terms of $f$, its derivatives, $u$, and its derivatives. So does its gradient. You have to work with that to infer whatever you need, including linearization. $\endgroup$
    – Deane Yang
    Commented Feb 22, 2015 at 18:32

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