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This is an exercise in "Variation et optimisation des formes", chapter 3, Ex. 3.8. The preliminaries are: $$D=(0,1)^2,\ f \in L^2(D),\ x_{ij}=(i/n, j/n),\ 0<i,j<n,$$ $$\Omega_n = D\setminus \bigcup_{0<i,j<n} \overline{B}(x_{ij},r_n), u_n=u_{\Omega_n}^f.$$

We pick $r_n=e^{-dn^2}$.

The notation $u_\Omega^f$ is the unique solution of the problem $-\Delta u=f$, which means $$ \int_\Omega fv=\int_\Omega \nabla u \nabla v,\ \forall v \in H_0^1(\Omega).$$

1) Prove that there exists $u^* \in H_0^1(D)$ and a subsequence of $u_n$ which converges weakly in $H_0^1(D)$ to $u^*$ and strongly in $L^2(D)$ to the same function.

This is not very hard, and there are some theorems in the book which help.

2) This is the part where it becomes messy. Consider the squares centred at $x_{ij}$ with sides $1/n$. Consider further the circles inscribed in these squares and denote with $C_{ij}^n$ the annulus formed by the inscribed circle in the square (centred in $x_{ij}$ with radius $1/2n$) and the circle of center $x_{ij}$ and radius $r_n$.

Define the following functions $z_n \in H^1(D)$ with the properties

  • $\Delta z_n=0$ on $C_{ij}^n$;
  • $z_n=0$ on $B(x_{ij},r_n)$;
  • $z_n=1$ elsewhere.

Determine the expression of $z_n$ for each small square (cell)[I managed to do this assuming that the solutions are radially constant around $x_{ij}$]. Here comes the part I can't get:

Prove that for each small square $-\Delta z_n=\mu_n-\nu_n$ where $\mu_n,\nu_n$ are the positive measures with support respectively the circles $C(x_{ij},1/2n)$ and $C(x_{ij},r_n)$.

The problem has more parts, but to get there I need to understand this one. Thank you.

[edit:] To clarify the aspects in Florian's comment: I don't understand why $-\Delta z_n$ has the given formula, and don't know how the measures $\mu_n,\nu_n$ are defined. In my intuition, $-\Delta z_n=0$ since this is the condition on $C_{ij}^n$ and outside this annulus $z_n$ is constant.

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Are you asking what it means that a differential operator applied to a function is a measure, or do you know that and you just can't do it? Or in other words, is your question about terminology or about calculations? –  Florian Jul 14 '11 at 12:17
    
@Florian: My question is mostly about terminology. –  Beni Bogosel Jul 14 '11 at 14:46

1 Answer 1

up vote 5 down vote accepted

First of all you have to know what distributions and distributional derivatives are. It's best to look at the Wikipedia article.

The important points are: Distributions are continuous functionals on a space of test functions (usually $C_c ^\infty(\Omega)$ for some domain $\Omega$, that is, smooth functions with compact support inside $\Omega$). A function $u$ (which has to be at least in $L_{\rm loc}^1$) is identified with the distribution $\varphi \mapsto \int_\Omega u \varphi$. More generally, a Radon measure $\mu$ is identified with the distribution $\varphi\mapsto \int \varphi d \mu$.

Any distribution can be differentiated: if $T$ is a distribution and $s\in \{1,...,n\}$ and $\varphi\in C_c^\infty(\Omega)$ then $\partial_s T(\varphi):=-T(\partial_s \varphi)$. Thus $\Delta T(\varphi)=T(\Delta \varphi)$. This makes it possible to differentiate any function; the derivatives might be functions or measures (that is, distributions which can be identified with a function or a measure) but in general they are not.

When you are asked to show that $\Delta z = \mu - \nu$ with measures $\mu,\nu$, this has to be interpreted in the distributional sense explained above. Unwrapping all definitions, you have to show that there exist positive Radon measures $\mu$ and $\nu$ such that for all $\varphi\in C_c ^\infty(\Omega)$, $$\int_\Omega z \Delta \varphi = \int \varphi d \mu - \int \varphi d \nu$$ and since you have an explicit formula for $z$, you can work this out (Hint: Gauss theorem or Green's identities might help).

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Thank you for explaining this to me. :) –  Beni Bogosel Jul 14 '11 at 16:10

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