Ngô Quốc Anh

March 1, 2011

The implicit function theorem: A PDE example

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 23:29

This entry devotes an existence result for the following semilinear elliptic equation

-\Delta u + u = u^p+f(x)

in the whole space \mathbb R^n where 0<u \in H^1(\mathbb R^n).

Our aim is to apply the implicit function theorem. It is known in the literature that

Theorem (implicit function theorem). Let X, Y, Z be Banach spaces. Let the mapping f:X\times Y\to Z be continuously Fréchet differentiable.


(x_0,y_0)\in X\times Y, \quad F(x_0,y_0) = 0,


y\mapsto DF(x_0,y_0)(0,y)

is a Banach space isomorphism from Y onto Z, then there exist neighborhoods U of x_0 and V of y_0 and a Frechet differentiable function g:U\to V such that

F(x,g(x)) = 0

and F(x,y) = 0 if and only if y = g(x), for all (x,y)\in U\times V.

Let us now consider

X=L^2(\mathbb R^n), \quad Y=H_+^2(\mathbb R^n), \quad Z=L^2(\mathbb R^n).

Let us define

F(f,u)=-\Delta u + u - u^p-f(x), \quad f \in X, \quad u \in Y, \quad x \in \mathbb R^n.

It is not hard to see that Fréchet derivative of F at (f,u) with respect to u in the direction v is given by

{D_u}F(f,u)v = - \Delta v + v - p{u^{p - 1}}v.

Since -\Delta +I defines an isomorphism from Y to Z, it is clear to see that our PDE is solvable for f small enough in the X-norm.

February 11, 2011

The implicit function theorem: An ODE example

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 23:35

We want to continue the series of notes involving some applications of  the implicit function theorem. As in the previous note, here we consider the solvability of the following ODE

x''+\mu x+f(x)=0, \quad J= [0,1]

with the following boundary conditions


We assume that f \in C^1(\mathbb R) and f(0)=0. For the sake of convenience, let us recall the standard implicit function theorem

Theorem (implicit function theorem). Let X, Y, Z be Banachspaces, U\subset X and V\subset Y neighbourhoods of x_0 and y_0 respectively, F: U\times V\to Z continuous and continuously differentiable with respect to y. Suppose also that

F(x_o, y_o) = 0,\quad F_y^{-1}(x_o,y_o) \in L(Z, Y).

Then  there exist balls \overline B_r(x_o) \subset U, \overline B_r (y_o) \subset V and exactly one map T: B_r(x_o) \to B_r (y_o) such that

Tx_o = y_o and F(x, Tx) = 0 on B_r(x_o).

This map T is continuous.


July 27, 2010

The implicit function theorem: How to prove a continuously dependence on parameters for solutions of ODEs

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 0:25

It is clear that the implicit function theorem plays an important role in analysis. From now on, I am going to demonstrate this significant matter from the theory of differential equations, both ODE and PDE, point of view.

Let us start with the following ODE

-u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}=0

on some domain \Omega \subset \mathbb R^n with \alpha \not\equiv 0 and \beta \not\equiv 0. We assume the existence result on W_+^{2,p} is proved for some p>1. We prove the following

Theorem. The solution u \in W_+^{2,p} depends continuously on (\alpha, \beta) \in L^\infty \times L^\infty.

Proof. Consider the map

\mathcal N : W_+^{2,p} \times (L^\infty \times L^\infty) \to L^p


(u,\alpha,\beta) \mapsto -u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}.

This map is evidently continuous (since W_+^{2,p} is an algebra). One readily shows that its Fréchet derivative at (u, \alpha, \beta) with respect to u in the direction h is

\mathcal N'[u,\alpha ,\beta ]h = - h'' + \left[ {(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}} \right]h.

The continuity of the map

(u,\alpha,\beta) \mapsto \mathcal N'[u,\alpha ,\beta ]

follows from the fact that W_+^{2,p} is an algebra continuously embedded in C^0(\Omega).

Since \alpha \not\equiv 0 and \beta \not\equiv 0, the potential

V={(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}}

is not identically zero. Thus it is well-known that the map

-\Delta +V : W^{2,p} \to L^p

is an isomorphism.

The implicit function theorem then implies that if u_0 is a solution for data (\alpha_0, \beta_0), there is a continuous map defined near (\alpha_0, \beta_0) taking (\alpha, \beta) to the corresponding solution of the ODE. This establishes the conclusion.

For the more details, we prefer the reader to this preprint.

Blog at