On the existence of Parker’s ideal bodies

Abstract

After a short introduction on the historical context, the paper deals with the existence of the solution of Parker’s ideal body problem, namely the body of minimum constant density generating a given external potential. A crucial element of the proof is the use of a recently introduced topological space of closed sets, closed and compact with the distance defined as the Lebesgue measure of the symmetric difference of a couple of sets. Such a space is indeed smaller than that of all closed sets of a given B, but larger than that of star-shaped Lipschitz domains, where previous studies of the inverse gravimetric problem (with constant density) have been conducted. However, with the present knowledge, it is only in this class that a uniqueness theorem holds.

Appendix A

Here we present, for the sake of completeness, a sketch of the proof of uniqueness by mixing the two approaches of Novikov (1938) and Barzaghi and Sansò (1986).

The following Theorem is based on two preliminary results:

• (a) given a bounded, star-shaped Lipschitz domain D, the Dirichlet problem

  $$\begin{aligned} \left\{ \begin{array}{ll} \Delta v = 0 &{} \text{ in } D \\ v\mid _S =f &{} \text{ on } S=\partial D \end{array}\right. \end{aligned}$$

  (A1)

  with $f \in L^p (S), 1 < p \le +\infty$, has one and only one solution (see Dhalberg 1977; McLean 2000), where the boundary relation of (A1) has to be understood in the sense that

  $$\begin{aligned} \lim _{t\rightarrow 1} \int _S \mid v (t{\pmb {x}}) - f({\pmb {x}})\mid ^p dS_x=0 \ ; \end{aligned}$$

  (A2)

  let us recall that if $D\in {\mathcal {F}}_{s,a}$, then it is also Lipschitz, namely $R_\sigma$ is a Lipschitz function, as proved in Proposition 4,

• (b) that any density function $\rho ({\pmb {x}}),{\pmb {x}}\in D$, that is in $L^2(D)$ and generates a zero potential outside $S=\partial B$, has to be $L^2$-orthogonal to all functions v harmonic in B that are also belonging to $L^2(D)$,

  $$\begin{aligned} \left\langle \rho , \frac{1}{\ell _{{\pmb {x}},{\pmb {y}}}}\right\rangle _{L^2(D)} = 0 \Leftrightarrow <v,\rho >_{L^2 (D)} =0 \quad \forall v , \ (\Delta v=0, v\in L^2(D)) \ , \end{aligned}$$

  (A3)

  (see Ballani and Stromeyer 1983; Freeden and Nashed 2018b and the large literature reported in Michel and Fokas (2008)); let us recall that the space of harmonic functions $\in L^2(S)$, namely the Hardy space $H^2(D)$, is contained into the space of harmonic functions square integrable on D, namely the Bergman space $L^2(D)$,

  $$\begin{aligned} \Delta v=0 \text{ in } D \ , \ \int _D v^2 ({\pmb {x}}) dD \le c \int _S v^2 ({\pmb {x}}) dS \ . \end{aligned}$$

  (A4)

Theorem 6

Let two domains $D_1, D_2 \in {\mathcal {F}}_{s,a}, D_1,D_2 \subset B$ be support of a mass distribution with a constant density $\rho$ which generates the same Newtonian potential on $S=\partial B$, and out of it; then we have

$$\begin{aligned} D_1=D_2 \ . \end{aligned}$$

(A5)

Proof

Since $u({\pmb {x}})$ is homogeneous in the constant $\rho$, we can suppose $\rho =1$. We must have

$$\begin{aligned} \forall {\pmb {x}}\in S \ , \quad \int _{D_1} \frac{1}{\ell _{{\pmb {x}}{\pmb {y}}}} dD = \int _{D_2}\frac{1}{\ell _{{\pmb {x}}{\pmb {y}}}} dD \ , \end{aligned}$$

(A6)

so that by the uniqueness of the solution of the outer Dirichlet problem, we have that (A6) holds $\forall {\pmb {x}}\in \Omega = D^c$ too (Fig. 4).

Fig. 4

The body B, the two “equivalent” domains $D_1,D_2$ and the decomposition of $D_1\div D_2$ into $\delta D_+ \cup \delta D_-$

By the unique continuation property of harmonic functions (see e.g. Sansò and Sideris 2013) we have that

$$\begin{aligned} v({\pmb {x}}) = \int _{D_1} \frac{1}{\ell _{{\pmb {x}}{\pmb {y}}}} dD - \int _{D_2} \frac{1}{\ell _{{\pmb {x}}{\pmb {y}}}} dD \end{aligned}$$

(A7)

is zero not only in $\Omega$ but also in $B\backslash (D_1 \cup D_2)$, which is connected to $\Omega$. On the other hand $v({\pmb {x}})$, as defined by (A7), is continuous everywhere in $R^3$ and so it is zero even on the boundary of $D_1\cup D_2$. We will call

$$\begin{aligned} S_+ =\partial (D_1\cup D_2) \ , \ S_- = \partial (D_1\cap D_2) \ ; \end{aligned}$$

(A8)

Also from Fig. 3 it is clear that, calling

$$\begin{aligned} R_{+\sigma } = \max (R_{1,\sigma }, R_{2,\sigma }) , \ R_{-\sigma }= \min (R_{1,\sigma }, R_{2,\sigma }) \ , \end{aligned}$$

(A9)

we have that

$$\begin{aligned} \{ r=R_{+\sigma }\} \equiv \partial (D_1\cup D_2) , \ \{r=R_{-\sigma }\} \equiv \partial (D_1\cap D_2) \ . \end{aligned}$$

(A10)

In case $R_{1\sigma } = R_{2\sigma }$ (i.e. at the crossing of $S_1$ and $S_2$) we conventionally attribute the point to one of them, e.g. to $S_1$.

We denote

$$\begin{aligned} \left\{ \begin{array}{l} \delta D_+ \equiv D_1\backslash D_2 \equiv \{ (r,\sigma ) \ ; \ R_{2,\sigma } \le r \le R_{1,\sigma } \} \\ \delta D_- \equiv D_2\backslash D_1 \equiv \{ (r,\sigma ) \ ; \ R_{1,\sigma } \le r \le R_{2,\sigma } \} \end{array}\right. \end{aligned}$$

(A11)

and notice that the symmetric difference of $D_1$ and $D_2$ is given by

$$\begin{aligned} D_1\div D_2 \equiv \delta D_+ \cup \delta D_- \ . \end{aligned}$$

Moreover the potential (A7) can be written, in terms of the variable density

$$\begin{aligned} \rho ({\pmb {x}}) = \{ \chi _{\delta D_+} ({\pmb {x}}) - \chi _{\delta D_-} ({\pmb {x}}) \} \ , \end{aligned}$$

(A12)

as

$$\begin{aligned} v({\pmb {x}}) = \int _B \rho ({\pmb {y}}) \frac{1}{\ell _{{\pmb {x}}{\pmb {y}}}} dB \ . \end{aligned}$$

(A13)

Note that $\rho ({\pmb {x}})$ so defined attains the values $\pm 1$ when ${\pmb {x}}$ is $\delta D_\pm$ and it is zero in $D_1\cap D_2$.

One useful Remark is that, since $S_{1,2}$ enjoy a radial cone condition, the same is true for $S_\pm$. Moreover, it can very well be that

$$\begin{aligned} R_{1,\sigma } \equiv R_{2,\sigma } \end{aligned}$$

on a set $\Gamma$ of the unit sphere of non-zero surface measure. In any way the following proof holds irrespectively of the validity of the above condition.

Let us split the unit sphere $\Sigma$ into

$$\begin{aligned} \Sigma _+ \equiv \{ \sigma \ ; \ R_{1,\sigma } = R_{+\sigma }\} , \ \Sigma _- \equiv \{ \sigma \ ; \ R_{2,\sigma } = R_{+\sigma }\} \ . \end{aligned}$$

(A14)

Further consider functions $f_n(\sigma )$ sufficiently smooth, that the Dirichlet problem with boundary data $f_n$ gives solutions $w_n \in H^{1,2} (B)$ (i.e. $\nabla w_n \in L^2(B))$ and on the same time $\mid f_n (\sigma )\mid \le 1$ and

$$\begin{aligned} n\rightarrow \infty \ , \quad f_n(\sigma )\rightarrow \chi _{\Sigma _+} (\sigma ) - \chi _{\Sigma _-} (\sigma ) \ , \end{aligned}$$

(A15)

in $L^2(\Sigma )$.

This is possible because $S_+$ satisfies a cone condition so that the traces of $w_n$ on $S_+$ are in $H^{1/2} (S)$ and such a space is dense in $L^2(\Sigma )$ (see McLean 2000).

Note that the above implies

$$\begin{aligned} w_n\mid _{S_+} = w_n (R_{+\sigma },\sigma )\equiv f_n(\sigma ) \ . \end{aligned}$$

(A16)

Since $w_n \in H^{1,2}$ we have that ${\pmb {x}}\cdot \nabla w_n=r \frac{\partial w_n}{\partial r}$ are harmonic too and belong to $L^2(B)$.

Therefore $\rho ({\pmb {x}})$ given by (A12) should be such that

$$\begin{aligned} 0\equiv & {} \int _B \left( r\frac{\partial }{\partial r} w_n \right) \rho (x) dD = \int _{D_1\div D_2} \left( r\frac{\partial }{\partial r} w_n\right) \rho ({\pmb {x}}) dD \nonumber \\= & {} \int d\sigma \int ^{R_+\sigma }_{R_-\sigma } \left( r\frac{\partial }{\partial r} w_n\right) r^2 \rho ({\pmb {x}}) dr \ . \end{aligned}$$

(A17)

We integrate by parts in r and note that between $R_{-\sigma }$ and $R_{+\sigma }$, $\rho$ is constant in r so that its derivative is zero.

Therefore we get

$$\begin{aligned} 0\equiv & {} \int d\sigma (\rho _+ w_{n+} R^3_{+\sigma } - \rho _- w_{n-} R^3_{-\sigma } ) \nonumber \\&-3\int d\sigma \int _{R_-\sigma }^{R_+\sigma } w_n r^2 \rho (x) dx \ . \end{aligned}$$

(A18)

On the other hand

$$\begin{aligned} \int d\sigma \int _{R_-\sigma }^{R_+ \sigma } w_n \rho ({\pmb {x}}) r^2 dr = \int _{_{D_1\div D_2}} w_n \rho ({\pmb {x}}) dD \equiv \int _B w_n \rho ({\pmb {x}}) dD \equiv 0 \ , \end{aligned}$$

(A19)

because $\rho$ has to be $L^2(B)$ orthogonal to all square integrable harmonic functions in B.

So we are left with

$$\begin{aligned} 0 \equiv \int d\sigma (\rho _+ f_n (\sigma ) R^3_{+\sigma } - \rho _- w_{n-} R^3_{-\sigma } ) \ . \end{aligned}$$

(A20)

Next, by the maximum principle (see Sansò and Sideris 2013; Yosida 1980)

$$\begin{aligned} \mid w_{n-}\mid \le \sup \mid f_n (\sigma )\mid \le 1 \end{aligned}$$

and

$$\begin{aligned} \mid \rho _-\mid =1 \ , \end{aligned}$$

so that from (A20) we derive

$$\begin{aligned} 0 \ge \int d\sigma (\rho _+ f_n (\sigma ) R^3_{+\sigma } - R^3_{-\sigma }) \ . \end{aligned}$$

(A21)

On the other hand, from (A12) we see that

$$\begin{aligned} \left. \rho \right| _{S_+} = \rho _+=1 \quad \text{ on } \Sigma _+ \ ; \ \left. \rho \right| _{S_+} = \rho _+=-1 \quad \text{ on } \Sigma _- \end{aligned}$$

so that, from (A15),

$$\begin{aligned} f_n (\sigma ) \underset{L^2(\Sigma )}{\rightarrow } \rho _+ \ \ ; \ f_n (\sigma ) \rho _+ \underset{L^2(\Sigma )}{\rightarrow } 1 \ . \end{aligned}$$

(A22)

Therefore, passing to the limit for $n\rightarrow \infty$ in (A21) we obtain

$$\begin{aligned} 0 \ge \int d\sigma (R^3_{+\sigma } - R^3_{-\sigma }) \ . \end{aligned}$$

(A23)

Since $R_{+\sigma } \ge R_{-\sigma }$, this can only be if

$$\begin{aligned} R_{+\sigma } \equiv R_{-\sigma } \text{ a.e. } \end{aligned}$$

namely

$$\begin{aligned} R_{1,\sigma } \equiv R_{2,\sigma } \Rightarrow D_1=D_2 \ . \end{aligned}$$

$\square$