Geodetic Boundary Value Problem

Abstract

The paper starts with an introduction recalling the historical definitions of the geodetic boundary value problem to arrive at the foundation of the so-called Scalar Molodensky Boundary Value Problem, in its fully nonlinear form (SMP).

The problem is then linearized and slightly modified by introducing further data in terms of the first degrees of harmonic coefficients of the asymptotic expansion of the potential and a suitable set of unknowns, along with the idea of Hörmander’s analysis of the problem.

In Sect. 3 a simplified formulation of the Linear Scalar Molodensky Problem (LSMP) based on a spherical approximation is introduced, together with suitable topologies for data and unknowns.

Such simplified Molodensky Problem is then analyzed proving the existence, uniqueness, and stability of the solution. Finally the LSMP is analyzed too by a perturbation of the simplified version.

The geometric conditions that emerge from the above analysis show roughly that the basic theorem holds when the approximate telluroid $\tilde{S}$ used for the linearization has a maximum inclination of 60^∘ with respect to the vertical and the coefficients of the first 12 degrees are also provided as data.

Appendix

In this appendix we aim to prove the subsequent proposition.

Proposition 4.

Let $u \in \overline{H}_{1}$ , namely,

$$\displaystyle\begin{array}{rcl} \parallel u \parallel _{1} = \left \{\int _{\tilde{S}}\vert \nabla u\vert ^{2}R_{\sigma }^{2}d\sigma \right \}^{\frac{1} {2} } < +\infty & &{}\end{array}$$

(129)

and

$$\displaystyle\begin{array}{rcl} & & <\psi _{jk},u > _{0} =\int u(\overline{R},\sigma )Y _{jk}(\sigma )d\sigma = 0 \\ & & 0 \leq j \leq L\,\ -j \leq k \leq j\; {}\end{array}$$

(130)

then

$$\displaystyle\begin{array}{rcl} \parallel u \parallel _{0} = \left \{\int u^{2}(R_{\sigma },\sigma )R_{\sigma }d\sigma \right \}^{\frac{1} {2} } \leq C_{OL} \parallel u \parallel _{1}& &{}\end{array}$$

(131)

where

$$\displaystyle\begin{array}{rcl} C_{OL} = \left ( \frac{\delta R} {R_{+}} + \frac{1} {L + 2}\right )J_{+}.& &{}\end{array}$$

(132)

Proof.

Let us put

$$\displaystyle\begin{array}{rcl} u_{+}(\sigma ) = u(R_{+},\sigma )& &{}\end{array}$$

(133)

and observe that

$$\displaystyle\begin{array}{rcl} & & \parallel u \parallel _{0} \leq \parallel u - u_{+} \parallel _{0}+ \parallel u_{+} \parallel _{0} \equiv \\ & &\quad \equiv \left \{\int [u(R_{\sigma },\sigma ) - u(R_{+},\sigma )]^{2}R_{\sigma }d\sigma \right \}^{\frac{1} {2} } \\ & & \quad \quad + \left \{\int u(R_{+},\sigma )^{2}R_{\sigma }d\sigma \right \}^{\frac{1} {2} }. {}\end{array}$$

(134)

On the other hand, we can write

$$\displaystyle\begin{array}{rcl} \vert u - u_{+}\vert ^{2}& =& \left \vert \int _{ R_{\sigma }}^{R_{+} }u'dr\right \vert ^{2} \leq \int _{ R_{\sigma }}^{R_{+} }u^{{\prime}2}rdr\left ( \frac{1} {R_{\sigma }} - \frac{1} {R_{+}}\right ) {}\\ & & \leq \frac{\delta R} {R_{\sigma }R_{+}}\int _{R_{\sigma }}^{R_{+} }\vert \nabla u\vert ^{2}r^{2}dr\; {}\\ & & {}\\ \end{array}$$

therefore multiplying by R _σ and integrating over the unit sphere, we get

$$\displaystyle\begin{array}{rcl} \parallel u - u_{+} \parallel _{0}^{2} \leq \frac{\delta R} {R_{+}}\int _{\varOmega \setminus \varOmega _{+}}\vert \nabla u\vert ^{2}d\varOmega & &{}\end{array}$$

(135)

where

$$\displaystyle\begin{array}{rcl} \varOmega _{+} \equiv \{ R_{+} \leq r\}& &{}\end{array}$$

(136)

$$\displaystyle\begin{array}{rcl} \varOmega \setminus \varOmega _{+} \equiv \{ R_{\sigma } \leq r \leq R_{+}\}.& &{}\end{array}$$

(137)

Furthermore, we can write

$$\displaystyle\begin{array}{rcl} \parallel u_{+} \parallel _{0}^{2} \leq R_{ +}4\pi \sum _{n,m=L+1}^{+\infty }u_{ nm}^{+^{2} } = R_{+}\int u(R_{+},\sigma )^{2}d\sigma \,& &{}\end{array}$$

(138)

with

$$\displaystyle\begin{array}{rcl} u_{nm}^{+} = \frac{1} {4\pi }\int u(R_{+},\sigma )Y _{nm}(\sigma )d\sigma.& &{}\end{array}$$

(139)

Therefore we can claim too that

$$\displaystyle\begin{array}{rcl} \parallel u_{+} \parallel _{0}^{2} \leq \frac{1} {L + 2}\left (R_{+}4\pi \sum _{n,m=L+1}^{+\infty }(n + 1)u_{ nm}^{+^{2} }\right ).& &{}\end{array}$$

(140)

On the other hand it is easy to verify that

$$\displaystyle\begin{array}{rcl} \int _{\varOmega _{+}}\vert \nabla u\vert ^{2}d\varOmega = -\int _{ S_{+}}uu'R_{+}^{2}d\sigma = R_{ +}4\pi \sum _{n,m=L+1}^{+\infty }(n + 1)u_{ nm}^{+^{2} }& & {}\\ \end{array}$$

so that (140) can be put into the form

$$\displaystyle\begin{array}{rcl} \parallel u_{+} \parallel _{0}^{2} \leq \frac{1} {L + 2}\int _{\varOmega _{+}}\vert \nabla u\vert ^{2}d\varOmega.& &{}\end{array}$$

(141)

Collecting (134)–(136) yields

$$\displaystyle\begin{array}{rcl} \parallel u \parallel _{0}& \leq & \sqrt{ \frac{\delta R} {R_{+}}}\sqrt{ \int _{\delta \varOmega }\vert \nabla u\vert ^{2}d\varOmega } + \frac{1} {\sqrt{L + 2}}\sqrt{\int _{\varOmega _{+ } } \vert \nabla u\vert ^{2 } d\varOmega } {}\\ & \leq & \sqrt{ \frac{\delta R} {R_{+}} + \frac{1} {L + 2}}\sqrt{ \int _{\varOmega }\vert \nabla u\vert ^{2}d\varOmega }. {}\\ \end{array}$$

So, by applying the Gauss theorem,

$$\displaystyle\begin{array}{rcl} \parallel u \parallel _{0}^{2}& \leq & \left ( \frac{\delta R} {R_{+}} + \frac{1} {L + 2}\right )\left (-\int _{S}uu_{n}R_{\sigma }^{2}Jd\sigma \right ) \\ & \leq & C_{OL} \parallel u \parallel _{0} \parallel u_{n}R_{\sigma } \parallel _{0} \\ & \leq & C_{OL} \parallel u \parallel _{0} \parallel u \parallel _{1}. {}\end{array}$$

(142)

Dividing both members of (142) by $\parallel u \parallel _{0}$, we get (131). □