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Suppose $\Omega\subset\mathbb{R}^n$ is a regular open set, $f\in L^2(\Omega)$ and consider the following elliptic problem. $$-\Delta u + u=f'(u) , \;\;u_{|\partial \Omega}=0,$$ where $f'$ is the derivative of a function $f:\mathbb{R}\to\mathbb{R}$ and $f'(u)\in L^2(\Omega)$ for all $u\in H^1_0(\Omega)$ (choose any assumptions you like on $f'$ to obtain this). Suppose also that that $f''$ exists.

It is well known that $u$ minimizes the energy functional $J:H^1_0(\Omega)\to \mathbb{R}$ given by $$J(u)=\frac{1}{2}\int_\Omega|\nabla u|^2 +\frac{1}{2}\int_\Omega u^2-\int_\Omega f(u).$$ Now this also means that $u$ is a root of the Frechet (equiv to Gâteaux in this case) derivative $J':H^1_0(\Omega)\to H^{-1}(\Omega)$. In particular, this means that $$J'(u)v=\int_\Omega\nabla u\cdot\nabla v+\int_\Omega u v-\int_\Omega f'(u)v=0, \;\;\;\text{ for all }v\in H^1_0(\Omega).$$ (Notice that this is nothing but the variational formulation of the pde)

Suppose now that we wish to apply the Newton method in Banach spaces to minimize the function $J'$. I will assume that $J'':H^1_0(\Omega)\to L(H^1_0(\Omega),H^{-1}(\Omega)) $ exists and that $J''(u)$ be identified with the continuous bilinear form $J''(u)(\cdot,\cdot)\in \mathcal{B}(H^1_0(\Omega))$ as
$$J''(u)(v,w)=\int_\Omega\nabla v\cdot\nabla w+\int_\Omega vw-\int_\Omega f''(u)vw, \;\;\;\text{ for all }v,w\in H^1_0(\Omega).$$ Starting with a guess $u_0\in H^1_0(\Omega)$, the usual Newton method iterative procedure to minimize $J'$ is then to solve to following "abstract" equation $$J''(u_n)(u_{n+1}-u_n)=-J'(u_n). \;\;\;\text{(Equality is in $H^{-1}(\Omega)$)}.\tag{1}\label{1}$$

  1. I would like to check whether or not \eqref{1} is equivalent to $$J''(u_n)(u_{n+1},v)-J''(u_n)(u_n,v)=-J'(u_n)v, \;\;\;\text{ for all }v\in H^1_0(\Omega).$$
  2. If the above is true, to apply Newton method, is is necessary that that $$J''(u)^{-1}:L(H^{-1}(\Omega),H^1_0(\Omega))\simeq \mathcal{B}(H^1_0(\Omega))\to H^1_0(\Omega)$$ exists is bounded in the operator norm. How does one even start thinking about such a problem?

Important Fix: Question 2 is wrong. What we need to show is that $J''(u)$ is invertible for every $u$, ie, $J''(u)^{-1}:H^{-1}(\Omega)\to H^1_0(\Omega)$ exists.

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  • $\begingroup$ Q1 seems trivial to me: (1) is set in dual space, so it is equivlaent to the formulation below Q1. Why/Where do you have doubts? $\endgroup$
    – daw
    Oct 5 at 12:04
  • $\begingroup$ Now that I think about it is trivial so need for an answer there. $\endgroup$
    – UserA
    Oct 5 at 12:10
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Let me write out the equation $J''(u)w = g$ for $g\in H^{-1}$ and $w\in H^1_0(\Omega)$. This is equivalent to $$ \int_\Omega \nabla w \cdot \nabla v - f''(u)wv = g(v) \quad \forall v\in H^1_0(\Omega). $$ This is the weak formulation of $$ -\Delta w - f''(u)w = g$$ plus boundary conditions. To show existence of solutions, you need some assumptions on $f''$ to be able to invoke Lax-Milgram theorem or other existence theorems. The easiest one would be to require $f''(u) \le0$ for all $u$.

In order to apply the convergence theory of Newton's method, you will need that $J''(u)^{-1}$ is a bounded linear operator, and that the inverses $J''(u)^{-1}$ are bounded on a neightborhood of the reference solution.

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  • $\begingroup$ By the way, in the paper I linked by Bartle, the condition is that $\|J''(u)^{-1}\|<\infty$. Is is guaranteed? It seems to be false on $H^{-1}(\Omega)$ so maybe we have restrict our search in a ball instead? $\endgroup$
    – UserA
    Oct 5 at 12:15
  • $\begingroup$ More precisely, maybe we should've taken $J'$ as $J':B\to H^{-1}(\Omega)$ where $B$ is a ball containing $u\in H^1_0(\Omega)$ ($u$ being the minimizer). This way maybe we can have $\|J''(u)^{-1}\|<\infty$? $\endgroup$
    – UserA
    Oct 5 at 12:21
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    $\begingroup$ $J''(u)$ is a linear continuous map from $H^1_0$ to $H^{-1}$. It is invertible as soon as the linear pde in my answer is uniquely solvable. This is for instance the case if $f$ is concave. $\endgroup$
    – daw
    Oct 5 at 12:26
  • $\begingroup$ Okay thanks for the help! $\endgroup$
    – UserA
    Oct 5 at 12:33
  • $\begingroup$ But then again how do you get a bound on $\|J''(u)^{-1}\|$? It seems the convergence of the Newton method requires this bound. $\endgroup$
    – UserA
    Oct 5 at 12:35

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