Research Institute of Geodesy, Topography and Cartography Zdiby, Prague-East, Czech Republic

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1 Leibniz Society of Science at Berlin, Scientific Colloquium Geodesy - Mathematic - Physics - Geophysics in honour of Erik W. Grafarend on the occasion of his 75th birthday Berlin, Germany, 3 February 05 Summation of Series and an Approximation of Legendre s Functions in Constructing Integral Kernels for the Exterior of an Ellipsoid: Application to Boundary Value Problems in Physical Geodesy Petr Holota Research Institute of Geodesy, Topography and Cartography Zdiby, Prague-East, Czech Republic

2 . Kernels in the Classical Solution In geodesy we are dealing with a number of interesting problems of mathematical physics. We follow classical as well as weak solution concept and often use an integral representation of the solution, Green s function method or also Galerkin s approximations of the solution. The classical example is the famous Stokes function G( x, y ). It is associated with the problem to find T such that T + T = x R f for x = R and T = g for x > R Apart from the first degree harmonics T ( x ) the function enables to get the solution of the problem in the following form T( x) = T ( x) G( x, y) f( y) dys G( x, y) g( y) dy 4π 4π y = R y > R For the exterior of the sphere of radius R Stokes s function can even be nicely expressed in a closed form. We have

3 where R R x y G( x, y) = + 3 x y x x y x y 3 R cosψ R cos ψ x y 5 3ln + + x y x y y R x = x and x x = R x and ψ is the angle between the position vectors x and y. Remark. G( x, y ) above is a more general function in comparison with the function usually applied in physical geodesy. It enables to solve the geodetic boundary value problem not only for Laplace s equation but also for Poissons partial differential equation. 3

4 Another example is Neumann s function N( x, y ). Their use has now reasonable motivations in physical geodesy. It is associated with the problem to find T such that T = f for x = R and T = g for x > R x We also have a closed form expression for this function R x y + x y cosψ N( x, y) = + ln x y x x y R y ( cos ψ ) Remark. For y = R, it is obvious that N = N( x, ψ ) and we easily deduce that x y + R x cosψ N = ln x y R x ( cos ψ ) In the case that x = y = R, we even have N + sin( ψ /) = N( ψ ) = ln R sin ( ψ / ) sin ( ψ / ) 4

5 . Reproducing Kernel and the Weak Solution In geodesy kernel functions have their position also within the concept of the weak solution of boundary value problems. Good example is the reproducing kernel. Nevertheless, its construction corresponds not only to the geometry of the solution domain, but also to the scalar product associated with the particular Hilbert s space of functions used. () H ( S ) of functions harmonic in the Consider e.g. the space R exterior of the sphere of radius R which is equipped with the following scalar product ( uv, ) Auv (, ) =, where A( uv, ) = 3 i= S i i One can show that there exists a function K( x, y ) such that R u x v x dx 5

6 3 K( x, y) v( x) x x i= S i i R and that in this case n= 0 dx = v( ) y holds for all n + n+ K( x, y ) = z Pn (cos ψ ), 4π R n+ () v H ( S ) R R z = x y where P n is the usual Legendre polynomial of degree n, and ψ is the angle between the position vectors x and y. Moreover, it is not extremely difficult to find that where z L+ z cosψ K( x, y ) = ln 4πR L cosψ L = zcosψ + z 6

7 The advantage of using the kernel function K( x, y ) can be seen from the construction of Galerkin s approximations to the solution of Neumann s problem. Recall, therefore, that in this case we are looking for u such that u = 0 in R S and u n = where is Laplace s operator and in the direction of the unit (outer) normal n of f on SR / n denotes the derivative S. In the numerical solution the function u is approximated by the linear combination n ( n) n = j j j= u c v where v j are members of a function basis of () R H ( S R ). 7

8 The coefficients c ( n) j can be obtained from the respective Galerkin system where Putting now n ( n) j ( j, k) = k j= S c A v v v f ds, k =,, n R 3 v j vk A( vj, vk) = dx x x v j i= S i i R ( x) = K( x, y ) we can immediately deduce that in our Galerkin system the elements A( v, v ) are given by j k A( v, v ) = K( y, y ) j k j k j 8

9 3. Use of Elementary Potentials The advantage of using the function K( x, y ) definitely attracts attention. Nevertheless, it is also tempting to use elementary potentials for constructing Galerkin s approximations. This means that the basis functions will be of the following form v ( x) = j x y In this case one can show that the diagonal elements are given by R R j Av ( j, vj) π + y = + ln R j j R j y y y Concerning the off diagonal elements, however, the computation is more laborious. j 9

10 In particular, we have to use a series representation of v j ( x ), i.e. This enables us to get n j j x = y n+ n j n= 0 x v ( ) P (cos ψ ) or 4π n + Av (, v) z P(cos ) n j k = n ψ jk R n= 0 n+ where j k z = y y R π 4π Av (, v) z P(cos ) z P(cos ) n n j k = n ψ jk + n ψ jk R n= 0 R n= 0 n+ Here, however, we are faced by the problem to find a closed expression for the off diagonal element A( v, v ) j k 0

11 and what we only can get is where π 4π A( vj, vk) = + S R zcosψ + z R S = jk ( cosψ ) jk + z z z z dz is an elliptic integral.

12 4. Transition to an Ellipsoid of Revolution We have seen that the mathematical apparatus related to the exterior of the sphere is fairly developed and transparent. However, the sphere is rather far from reality, i.e. from the real surface of the Earth. Therefore, our aim is discuss the construction of the apparatus for the exterior of an oblate ellipsoid of revolution., are the semi-axes of an oblate ellipsoid of revolution and will consider the ellipsoidal coordinates u, β, λ related to x, x, x 3 by the equations We will suppose that a and b, a b x = u + E cosβcosλ, x = u + E cosβsin λ, x3 = usinβ E = a b denotes the linear eccentricity, see e.g. Ω is then where Heiskanen and Moritz (967). Clearly, the boundary defined by u = b. ell

13 Going now back to K( x, y ), or more precisely to K ( x, y ), and referring to Holota (004, 0,04), we can deduce that with K ( x, y) = (n+ ) K P (sin β ) P (sin β ) + ell n0xy n x n y 4π b n= 0 n ( n m)! + K xy P (sin βx) P (sin βy )cos m ( λx λy ) ( n+ m)! K m= xy ieb Q ( z ) Q ( z ) dq ( z ) x y 0 = a Q( z0) Q( z0) Q( z0) dz where P and Q are associated Legendre s functions of the st and the nd kind, respectively; while z x iu E x =, z y iu E y =, z0 ell ib = and i = E 3

14 The problem, however, is the summation of the series representing the kernel Kell ( x, y ). In this connection, inspecting the right hand side of K xy ieb Q ( z ) Q ( z ) dq ( z ) x y 0 = a Q( z0) Q( z0) Q( z0) dz it is tempting to use the limit layer theory as discussed in Sona (995) or Sansò and Sona (00) and also analyzed in Holota (003a). 4

15 As it is known is a solution of 5. Legendre s Differential Equation Q ( z) u b v ( u) = = Q i Q i E Q( z0) E dv dv Em ( u + E ) + u n( n ) v 0 + = du du u + E which can be shown to be equivalent to Legendre s equation by using the pure imaginary variable z = iu / E. This equation has two solutions: Legendre s functions of the st kind Legendre s functions of the nd kind P and Q The functions Q are the suitable solutions in case that we deal with harmonic functions in an unbounded domain. 5

16 6. Approximation Limit Layer Theory As in Sona (995) we will work now with the variable s = u/ b. It is obvious that v ( u) = v ( sb) = v( s) and we can deduce that for v Legendre s equation transforms into dv dv em ( s + e ) + s n( n ) v 0 + = ds ds s + e Now, confining ourselves to s, s bound for s, such that es and e, where max, where max ( s + e ) ( + e ) e = E a s is an upper can be taken for admissible approximations, we can simplify the equation above. It transforms into d w dw e m ( + e ) s + s n( n ) w 0 + = ds ds + e 6

17 The solution () (995), we have ws obviously differs from () ws () A = where A is a constant s α vs. Following Sona and α has to be one of the two roots, α n and α n +, of the quadratic equation em ( + e ) α ( e ) α n( n+ ) + = 0 + e Clearly, α is the suitable solution in our case and we obtain ( n+ )( n+ ) + m ws () =, ε = e n + ε s n + where A has been fixed so that w () =. Note in particular that vs ( ) = ws ( ) = for s =. As regards the differenced of the two solutions, it has been estimated by means of Gronwall s inequality in Holota (003a). 7

18 7. Coefficients in the Expansion of Kell ( x, y) Within the limit layer theory, we can write approximately that Q( z ) x b Q( z0) u x n + ε and subsequently derive that 0, Q( z ) y b Q( z0) u dq( z0) E dw ie n y n + ε () = ( + ε ) Q ( z ) dz ib ds b which holds, but again with some accuracy only. Now using some algebra, the right hand side of the equation above can be substantially modified. We obtain dq( z0) Eb E ( n+ )( n ) m i ( n+ ) + ( 0) a ( + )( + ) Q z dz ab n n 8

19 Thus in K ( x, y) = (n+ ) K P (sin β ) P (sin β ) + ell n0xy n x n y 4π b n= 0 n ( n m)! + K xy P (sin βx) P (sin βy )cos m ( λx λy ) ( n+ m)! m= we will have with K n+ ε n+ b b b κ ε ln κ uu x y uu x y uu x y xy E ( n+ )( n+ ) + m ε = a n+ and κ E ( n+ )( n ) m = n+ ab ( n+ )(n+ ) 9

20 Now, denoting by ψ the angular distance of points ( βx, λ y) and ( β, λ ) on a sphere, when β and λ are interpreted as spherical y y latitude and longitude, respectively and using the well-known Legendre s addition theorem, we can deduce that () E () E (3) Kell ( x, y) K ( x, y) K ( x, y) + K ( x, y) 4π b 4πab 4πab E b (4) E b (5) ln (, ) ln (, ) K x y + K x y 4πab uu x y 4πab uu x y with K () n + n+ b ( x, y ) = Pn (cos ψ ) n= 0 n+ uxu y 0

21 K () n + n b ( x, y ) = Pn (cos ψ ) n= 0 n+ uxu y and K (3) n + b P n n= ( n + ) uu x y ( x, y ) = n + (cos ψ ) (4) b K ( x, y ) = ( n+ ) Pn (cos ψ ) n= 0 uu x y K (5) n + b Pn (cos ψ ) ( x, y ) = n= n+ uxu y λ λ For terms containing the derivative of n P with respect to λ Holota (003b) has been used.

22 8. An Alternative Concept for our Approach In our approach we so far used an exact solution of an approximate differential equation. Let us try now to change this philosophy and work with an approximate solution of an exact differential equation. In this case we recall that n Q n+ () z m n!( n+ m)! n+ m+ n m+ n+ 3 = ( ) ( z ) F,, ; (n + )! z where F is a hypergeometric function and that ( a) n( b) n Fz = F a, b, c; = + z n= () c n n! z with ( a) n = a( a+ ) ( a+ n ), n=,,3 [ similarly for ( b ) n and ( c ) n ]. = n

23 After elementary modifications we have ab Fz = Fz c z z0 aa ( + ) bb ( + ) + + cc ( + ) z z0 Thus, e.g., in a 0km layer close above the ellipsoid and 4 e = 0, z z0 6 e = z z0 0, etc. In the sequel, therefore, considering these estimates, we can simply put 3

24 Q ( ) ( ) z Q z x Q ( z ) Q ( z ) y n ρ Similarly recalling that and putting we arrive at Hence where ρ = dq ( n+ ) z n m+ = Q Q dz z z n+, m Qn+, m( z0) n+ m+ E Q ( z ) n+ 3 ia 0 a x + y + u E u E dq( z0) Eb E ( n + ) m i ( n+ ) + ( 0) a ( + )( + 3) Q z dz ab n n xy n+ with K ρ κ κ E ( n+ ) m = n+ ab ( n+ )(n+ 3) 4

25 Now we return to the reproducing kernel Kell ( x, y ). Denoting again by ψ the angular distance of points ( βx, λ y) and ( β, λ ) on a sphere, when β and λ are interpreted as spherical y y latitude and longitude, respectively, we can deduce that () E () E (3) Kell ( x, y) K ( x, y) K ( x, y) + K ( x, y ) 4π b 4π ab 4π ab with K K n + ( x, y ) = ρ Pn (cos ψ ) n + () n+ n= 0 () n + n+ K ( x, y ) = ρ Pn (cos ψ ) n= 0 n + 3 n + n+ Pn (cos ψ ) ( x, y ) = ρ ( n+ ) (n+ 3) λ (3) n= see Holota (0, 004). For the expression of the last term Holota (003b) has been used. 5

26 Thank you for your attention! 6