# Semi-linear elliptic problem, energy functionals, Fréchet derivatives and the Newton method in Banach spaces

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.

• 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?
– daw
Oct 5, 2021 at 12:04
• Now that I think about it is trivial so need for an answer there. Oct 5, 2021 at 12:10

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.
• 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? Oct 5, 2021 at 12:15
• 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 $\|J''(u)^{-1}\|<\infty$? Oct 5, 2021 at 12:21
• $J''(u)$ 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.
• But then again how do you get a bound on $\|J''(u)^{-1}\|$? It seems the convergence of the Newton method requires this bound. Oct 5, 2021 at 12:35