Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation

Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation

The straight lines three-dimensional vector space realize the shortest distance for various metrics. This property is reformulated in terms of the inverse problem of the calculus of variations and closely related to the ultrahyperbolic equation with four independent variables. The interrelation is u...

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Дата:2010
Автор: Chrastinova, V.
Формат: Стаття
Мова:English
Опубліковано: Кримський науковий центр НАН України і МОН України 2010
Назва видання:Таврический вестник информатики и математики
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Цитувати:Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation / V. Chrastinova // Таврический вестник информатики и математики. — 2010. — № 1. — С. 35-49. — Бібліогр.: 11 назв. — англ.

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spelling irk-123456789-181852013-02-13T03:08:00Z Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation Chrastinova, V. The straight lines three-dimensional vector space realize the shortest distance for various metrics. This property is reformulated in terms of the inverse problem of the calculus of variations and closely related to the ultrahyperbolic equation with four independent variables. The interrelation is useful in both directions. For instans, polynomial solutions of the ultrahyperbolic equation provide all polynomial metrics with extremals the straight lines and conversely, a slight generalization of the Hilbert metrics leads to rather nontrivial (multi-valued of focusing) solutions of the ultrahyperbolic equation. In general, the article clarifies some well-known achievements concerning the 4th Hilbert Problem. 2010 Article Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation / V. Chrastinova // Таврический вестник информатики и математики. — 2010. — № 1. — С. 35-49. — Бібліогр.: 11 назв. — англ. 1729-3901 http://dspace.nbuv.gov.ua/handle/123456789/18185 en Таврический вестник информатики и математики Кримський науковий центр НАН України і МОН України
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
language English
description The straight lines three-dimensional vector space realize the shortest distance for various metrics. This property is reformulated in terms of the inverse problem of the calculus of variations and closely related to the ultrahyperbolic equation with four independent variables. The interrelation is useful in both directions. For instans, polynomial solutions of the ultrahyperbolic equation provide all polynomial metrics with extremals the straight lines and conversely, a slight generalization of the Hilbert metrics leads to rather nontrivial (multi-valued of focusing) solutions of the ultrahyperbolic equation. In general, the article clarifies some well-known achievements concerning the 4th Hilbert Problem.
format Article
author Chrastinova, V.
spellingShingle Chrastinova, V.
Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation
Таврический вестник информатики и математики
author_facet Chrastinova, V.
author_sort Chrastinova, V.
title Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation
title_short Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation
title_full Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation
title_fullStr Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation
title_full_unstemmed Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation
title_sort straight lines in three-dimensional space and the ultrahyperbolic equation
publisher Кримський науковий центр НАН України і МОН України
publishDate 2010
url http://dspace.nbuv.gov.ua/handle/123456789/18185
citation_txt Straight Lines in Three-Dimensional Space and the Ultrahyperbolic Equation / V. Chrastinova // Таврический вестник информатики и математики. — 2010. — № 1. — С. 35-49. — Бібліогр.: 11 назв. — англ.
series Таврический вестник информатики и математики
work_keys_str_mv AT chrastinovav straightlinesinthreedimensionalspaceandtheultrahyperbolicequation
first_indexed 2025-07-02T19:17:37Z
last_indexed 2025-07-02T19:17:37Z
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fulltext STRAIGHT LINES IN THREE-DIMENSIONAL SPACE ANDTHE ULTRAHYPERBOLIC EQUATION Chrastinov�a V.Brno University of Te hnology, Veve�r�� 331/95, 602 00 Brno,Cze h Republi .Department of Mathemati s, Fa ulty of Civil Engineering,e-mail: hrastinova.v�f e.vutbr. zAbstra t. The straight lines in three-dimensional ve tor spa e realize the shortest distan e forvarious metri s. This property is reformulated in terms of the inverse problem of the al ulus of variationsand losely related to the ultrahyperboli equation with four independent variables. The interrelation isuseful in both dire tions. For instan e, polynomial solutions of the ultrahyperboli equation provide allpolynomial metri s with extremals the straight lines and onversely, a slight generalization of the Hilbertmetri s leads to rather nontrivial (multi-valued or fo using) solutions of the ultrahyperboli equation. Ingeneral, the arti le lari�es some well-known a hievements on erning the 4th Hilbert Problem.Introdu tionThe history of our topi goes ba k to the famous Hilbert Problems [1℄, namely to the4th Problem on erning the determination of all metri s in the open subsets of Pn thathave the straight lines as the shortest urves and the study of the relevant geometries. Inthis strong version, it is still far from a omplete solution [2℄. With additional smoothnessassumptions, a lose onne tion to the inverse problem of the al ulus of variations (I P)and the prominent role of the ultrahyperboli equation (U H ) was soon indi ated [3℄.Let us re all that I P onsists in determination of the variational integral if theextremals are given in advan e. In two dimensions, for the integral R f(x; y; v)dx (v = y0),the solution is rather easy [4℄. Espe ially in the parti ular ase of extremal straightlines the formula fvv = U(v; y � vx) with arbitrary U resolves the problem. Thisresult was adapted to three dimensions [3℄ with the following result. The variationalintegral R f(x; y; z; v; w)dx (v = y0; w = z0) has the straight lines for extremals if and onlyif fvv = U; fvw = V; fww = W are fun tions of the variables� = v; � = y � vx; = w; Æ = z � wx:One an he k the ompatibility onditionsU = V�; V =W�; UÆ = V�; UÆ =W�and they imply the U H equation �2(�)=��� = �2(�)=���Æ for all fun tions (�) = U; Vand W: Then the fun tion f an be re onstru ted from U; V;W by double quadrature.Subsequently other solutions of I P were dis ussed. In the ingenious arti le [5℄, thethree-dimensional sub ase was thoroughly analysed in full generality. However, in theparti ular ase of extremal straight lines the path from U H (formula (8.22)) to thekernel fun tion f (pages 82-84) is not quite easy. The re ent general solution [6℄ of I Prests on non-elementary tools, the variational bi omplex, and the straight lines are notseparately mentioned.In this arti le we follow the geometri al approa h [7℄ based on the systemati alappli ation of the Poin ar�e-Cartan (PC ) forms [8℄ with intentional use of quite 36 Chrastinov�a V.elementary methods of algorithmi al nature. Our result is as follows: for the variationalintegral R f(x; y; z; v; w)dx (v = y0; w = z0) with extremals the straight lines, everyfun tion �f = f( ; y; z; v; w) ( = onst) satis�es U H �2 �f=�y�w = �2 �f=�z�v and onversely, every solution �f = �f(y; z; v; w) of this U H permits to re onstru t the kernelfun tion f of the variational integral (whi h redu es to �f if x = is kept �xed for a given onstant ) by a quadrature.On this o asion, a few examples are presented. The polynomial ase related tothe Bessel fun tions, a far going generalizations of the Hilbert proje tive metri s [9℄on Riemannian surfa es with the multivalued and fo using solutions of U H ; and�nally the proof of analyti ity of the ellipti al Hilbert metri s employing very advan edresults [10, 11℄.We will establish a lose relationship between the familiar property of the straightlines y = Ax + B; z = Cx + D in the spa e R3 with oordinates x; y; z; i.e., that theyrepresent the shortest urves for ertain metri s, and the solutions �f = �f(y; z; v; w) of theultrahyperboli equation �2 �f=�y�w = �2 �f=�z�v.In order to employ the ommon tools of di�erential al ulus, we shall deal withmetri s � su h that the limitlim"!0 1"�((x; y; z); (x+ u"; y + v"; z + w")) = F (x; y; z; u; v; w) (1)is a smooth (in�nitely-di�erentiable) fun tion whenever juj+ jvj+ jwj 6= 0. In geometri alterms, F is the rate of hange of the distan e at the point (x; y; z) 2 R3 as one moves inthe dire tion (u; v; w). Equation (1) reads�((x; y; z); (x+ u"; y + v"; z + w")) = (F (x; y; z; u; v; w) + o("))"and it follows that the length of a smooth urve(x(t); y(t); z(t)) 2 R3 (a � t � b; jx0(t)j+ jy0(t)j+ jz0(t)j 6= 0) (2)is represented by a Riemannian integral as follows. The sum of the distan es between theneighbouring points of a partitionxi = x(ti); yi = y(ti); zi = z(ti); a < � � � < ti < ti+1 < � � � < bof the urve approximates the Riemannian integral sum and has a limit as the norm ofthe partition tends to zero:limX �((xi; yi; zi); (xi+1; yi+1; zi+1)) = bZa F (x(t); y(t); z(t); x0(t); y0(t); z0(t))dt:We speak of a generalized length L = bZa Fdt of the urve (2) and our aim is to deal withmetri s � su h that the straight lines realize the shortest urves onne ting two given(su� iently lose) points.The generalized length L is independent of the parametrization of the urve (2) whi hmay ause some te hni al di� ulties, however, on every su� iently short segment of the¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �1' 2010 Straight Lines in Three-Dimensional Spa e and the Ultrahyperboli Equation 37 urve one an hoose one of the oordinates x; y; z for a new parameter. We shall mostlyuse the parameter x assuming x0(t) 6= 0 in (2). Then the urves under onsideration aregiven by the equations y = y(x); z = z(x) and the generalized length is represented bythe integralL = bZa f(x; y(x); z(x); y0(x); z0(x))dx; f(x; y; z; v; w) = F (x; y; z; 1; v; w): (3)With this adaptation, the methods of the lassi al al ulus of variations an be omfortably applied.It is well-known that the urves of the minimal length onne ting two given pointssatisfy the Euler-Lagrange (E L ) systemfy(� � � ) = ddxfv(� � � ); fz(� � � ) = ddxfw(� � � ); where (� � � ) = (x; y; z; y0; z0):Re all that the solutions of E L system are alled extremals. We wish to determinefun tions f su h that the straight lines are just the extremals. Re all that E L systemrepresents only the ne essary onditions and the lo al minimum property of the straightlines is ensured if moreover the familiar Legendre ondition holds true, we shall howeverfo us our interest just on the E L system.With these preparations, our task an be explained in quite simple terms.The E L system readsfy = fvx + fvyy 0 + fvzz 0 + fvvy 00 + fvwz 00;fz = fwx + fwyy 0 + fwzz 0 + fwvy 00 + fwwz 00:It follows that all straight lines y = vx + B; z = wx + D with the variableparameters A = v, C = w are extremals (solutions of the E L system) if and only ifthe identities fy(�) = fvx(�) + fvy(�)v + fvz(�)w;fz(�) = fwx(�) + fwy(�)v + fwz(�)w;where (�) = (x; y; z; v; w) hold true. They provide a system of the se ond order partialdi�erential equations for the fun tion f = f(x; y; z; v; w) and we will also use thealternative trans riptionfy = Xfv; fz = Xfw �X = ��x + v ��y + w ��z� (4)of these equations in future. One an then observe thatfyw = fvxw + fvywv + fvzww + fvz;fzv = fwxv + fwyvv + fwy + fwzvw;when e the ultrahyperboli equation fyw = fzv follows. The oordinate x appears as a mereparameter, so it is of interest to onsider the equation�fyw = �fzv ( �f = �f(y; z; v; w) = f( ; y; z; v; w)) (5)¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �1' 2010 38 Chrastinov�a V.with an arbitrary onstant. We shall see that the equation (5) for the fun tion�f = �f(y; z; v; w) of four variables is in ertain sense equivalent to the system (4). Morepre isely: a solution �f of (5) together with the hoi e of a onstant permits us tore onstru t the original fun tion f = f(x; y; z; v; w) satisfying (4).In this arti le a fun tion f = f(x; y; z; v; w) is alled resolving if (4) is sa�s�ed, i.e., ifall straight lines are extremals. We will determine all resolving fun tions f by using thesolutions �f of (5). The onverse setting will also be quite interesting; ertain resolvingfun tions f will be obtained by dire t geometri al onstru tion whi h provides rathernontrivial solutions �f of the ultrahyperboli equation (5).PrerequisitiesOur reasonings will be arried out in an open subset of the spa e R5 with oordinatesdenoted x; y; z; v; w: We also use the alternative oordinatesx; � = v; � = y � vx; = w; Æ = z � wx:The oordinates v; w orrespond to the derivatives, therefore we onsider straight linesgiven by the equationsy = Ax +B; z = Cx +D; v = A; w = C (A;B;C;D are onstants) (6)and, in terms of the alternative oordinates the equations (6) read:� = A; � = B; = C; Æ = D (A;B;C;D are onstants): (7)For a given fun tion g = g(x; y; z; v; w) learlydg = Xgdx + gy(dy � vdx) + gz(dz � wdx) + gvdv + dgwdw == Xgdx+ gyd� + gzdÆ + (gv + xgy)d� + (gw + xgz)d : (8)The fun tions �; �; ; Æ are redu ed to �� = v; �� = y� v; � = w; � = z� w if they are onsidered on the hyperplane x = (a onstant). In general, a fun tion g = g(x; y; z; v; w)is redu ed to �g = g( ; y; z; v; w) = �g(y; z; v; w): In the alternative oordinates, a fun tionh = h(x; �; �; ; Æ) is redu ed to�h = h( ; ��; ��; � ; �Æ) = h( ; v; y � v; w; z � w)and this redu tion will be again denoted �h = �h(y; z; v; w) when regarded as a fun tion ofthe original oordinates.Conversely, every fun tion h = h(�; �; ; Æ) independent of x (better: expressible interms of �; �; ; Æ) an be restored from its restri tion �h(y; z; v; w) expressed in terms ofthe original oordinates sin eh(�; �; ; Æ) = �h(� + �; Æ + ; �; ):Indeed, the restri tion of the fun tion is �h( �� + ��; �Æ + � ; ��; � ) = �h(y; z; v; w); use theformulae for ��; ��; � ; �Æ given above.Analogous pro edure an be applied to di�erential forms. A di�erential form anbe redu ed to the form denoted � and onversely, every di�erential form expressibleonly in terms the fun tions �; �; ; Æ (without the use of x) an be restored from theredu tion � expressed in terms of y; z; v; w if the fun tions � + �; Æ + ; �; are¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �1' 2010 Straight Lines in Three-Dimensional Spa e and the Ultrahyperboli Equation 39substituted for y; z; v; w, respe tively. One an verify that the di�erential is preserved:d�h = dh; d � = d : We omit the simple dire t proof (alternatively: redu tions andrestorations are spe ial pull-ba ks).De�nition. For every fun tion f = f(x; y; z; v; w) we introdu e the Poin ar�e-Cartan(PC ) form �' = fdx + fv(dy � vdx) + fw(dz � wdx): (9)The exterior di�erential � = d �' has the obvious properties d� = 0 (� is a losed form)and � �= 0 (mod dx; dy; dz)): If f is a resolving fun tion, we speak of a resolving PCform. These resolving PC forms will be alternatively hara terized in the followinglemmas.Dire t Lemma. If �' is a resolving PC form, then � = d �' an be expressed interms of the fun tions �; �; ; Æ (without the use of the oordinate x and the di�erentialdx).Proof. Obviously dy � vdx = d� + xd�; dz � wdx = dÆ + xd ; dv = d�; dw = d ;therefore� = df ^ dx + dfv ^ (d� + xd�) + dfw ^ (dÆ + xd )� (fvd� + fwd ) ^ dx:Using (8) for g = f; fv; fw; if follows that� = ((fy�Xfv)(d�+xd�)+ (fz�Xfw)(dÆ+xd ))^dx+ f��d�^d�+ � � �+ f Æd ^dÆwith ertain oe� ients f��; � � � ; f Æ (they need not be expli itly stated). By virtue of (4),we obtain � = f��d� ^ d� + � � �+ f Æd ^ dÆ;where the oe� ients an be expressed in terms of the alternative oordinates x; �; �; ; Æ:However, they are in fa t independent of x as follows from the identity d� = 0: �Converse Lemma. Let be a losed 2-form satisfying �= 0 (mod dx; dy; dz):If is expressible only in terms of �; �; ; Æ then there (lo ally) exists a resolving PCform �' su h that d �' = :Proof. By virtue of the Poin ar�e lemma, = d for an appropriate form = Kdx+ Ldy +Rdz +Mdv +Ndw:The ongruen e = d �= 0 implies Mw = Nv and there exists g = g(x; y; z; v; w) su hthat gv =M; gw = N: Clearly d( � dg) = where the orre ted form � dg = Udx+ V dy +Wdz (U = K � gx; V = L� gy; W = R� gz)¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �1' 2010 40 Chrastinov�a V.has no terms in dv; dw: We will see that � dg is identi al to the sought resolving PCform �': Obviously � dg = fdx + V (dy � V dx) +W (dz � wdx)where f = U + vV + wW;when e = d( � dg) = df ^ dx + dV ^ (d� + xd�) + dW ^ (dÆ + xd )� (V d� +Wd ) ^ dx:Using (8), one an verify that = ((fy �XV )(d� + xd�) + (fz �XW )(dÆ + xd )++(fv � V )d� + (fw �W )d ) ^ dx+ � � �where all the produ ts d� ^ d�; � � � ; d ^ dÆ are negle ted. Sin e is expressible interms of �; �; ; Æ, we on lude that fv = V; fw = W; hen e is indeed a PC form.Moreover fy = XV = Xfv; fz = XW = Xfw; therefore �' is resolving. The proof is done.�Cru ial Lemma. Let be a 2-form expressible only in terms of �; �; ; Æ: Thenthe ongruen e �= 0 (mod dx; dy; dz) holds true if and only if the redu tion � to some(equivalently: to any) hyperplane x = is of the spe ial kind� = (Mdv + Ldw) ^ dy + (Ldv +Ndw) ^ dz: (10)Proof. Assuming (10), we obtain = (Md� + Ld ) ^ d(� + �) + (Ld� +Nd ) ^ (dÆ + )by restoration. Then the desired ongruen e an be dire tly veri�ed.In order to prove the onverse, we use the formula = f��d� ^ d� + � � � + f Æd ^ dÆ == f��dv ^ d(y � vx) + � � �+ f Ædw ^ d(z � wx) �=�= (f� � x (f�Æ + f� ) + x2f�Æ)du ^ dv (mod dx; dy; dz):Assuming the ongruen e �= 0, it follows that f� = f�Æ + f� = f�Æ = 0; hen e theredu tion is � = �f��d�� ^ d�� + � � � + �f Æd� ^ d�Æ == ( �f��dv + �f�Ædw) ^ dy + ( �f�Ædv + �f Ædw) ^ dzafter some al ulations. This is exa tly (10).(Alternative proof. Clearly X� = � � � = XÆ = 0; hen e LXd� = � � � = LXdÆ = 0and therefore LX = 0 where LX denotes the Lie derivative. Let us denote� = dx ^ dy ^ dz ^ : Then LX� = dx ^ (dv ^ dz + dy ^ dw) ^ ; L 2X� = 2dx ^ dv ^ dw ^ ; L 3X� = 0:¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �1' 2010 Straight Lines in Three-Dimensional Spa e and the Ultrahyperboli Equation 41The form� satis�es a third order linear di�erential equation, therefore� is determinedif the initial values �x= = dx ^ dy ^ dz ^ � ; LX�x= = dx ^ (dv ^ dz + dy ^ dw) ^ � ; L 2X�x= = dx ^ dv ^ dw ^ � ;at the point x = are given. In parti ular � = 0 identi ally if and only if the initial valuesare all vanishing, however, this is true just for the forms � of the spe ial kind (10). Finally,the identity � = 0 is a mere trans ription of the ongruen e �= 0 (mod dx; dy; dz) andtherefore the last ongruen e is equivalent to (10). The alternative proof is not elementary.However, it may be applied even to the general I P where the given extremals need notbe the straight lines [7℄. As for the other Lemmas, there need not be any hange.) �Final resultsLet us overview our a hievements and add some remarks.By virtue of the Converse Lemma (better: the proof), if is su h a 1-form thatthe di�erential = d is expressible in terms of �; �; ; Æ and moreover satis�es �= 0 (mod dx; dy; dz), then � dg �= 0 (mod dx; dy; dz) for an appropriate fun tion gand for every orre tion dg, the result �' = � dg is a resolving PC form. The Dire tLemma ensures that every resolving PC form an be obtained in this manner (triviality:put = �', dg = 0).In order to determine all resolving PC forms �', we have to sear h for allforms = Kdx + Ldy + Rdz + Mdv + Ndw with the above two properties of thedi�erential = d . The �rst property will be guaranteed if is hosen as a restorationof a form � = �Ldy + �Rdz + �Mdv + �Ndw; the se ond property by the use of the Cru ialLemma (namely � = d � must be of the form (10)). In fa t it is su� ient to deal with theredu tions of the spe ial kind� = �Ldy + �Rdz = V (y; z; v; w)dy +W (y; z; v; w)dz: (11)Indeed, let us suppose that we sear h for a resolving PC form �' (not yet expli itlyknown). Clearly ��' �= 0 (mod dy; dz) for the redu tion and so we may put � = ��' whi his just of the form (11) (with V;W as yet unknown). Let be the restoration of theform (11). Then = d is the restoration of� = d � = d ��' = �� (� = d �');hen e = � (restorations are uniquely determined) whi h means d = d �' and �' = �dgfor appropriate orre tion dg therefore we do not lose any possible resolving PC form �'.Altogether, in order to obtain an arbitrary resolvingPC form �'; it is su� ient to startwith the restorations of the spe ial di�erential form (11). So let us hoose a di�erentialform (11). Then the restoration , hen e the di�erential = d an be expressed in termsof �; �; ; Æ; the �rst requirement is satis�ed. The ongruen e �= 0 (moddx; dy; dz); thatis, the se ond requirement is ensured if the redu tion is of the form (10). However learly� = d � = (Wy � Vz)dy ^ dz + (Vvdv + Vwdw) ^ dy + (Wvdv +Wwdw) ^ dz¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �1' 2010 42 Chrastinov�a V.and we have the onditions Wy = Vz; Wv = Vw for the oe� ients in (11). Now we re allthe U H equation (5). If �f is a solution, the fun tions V = �fv;W = �fw learly satisfythe onditions and the onverse also holds true. We are done.Our a hievements an be summarized as follows.Main Theorem. We start with an arbitrary form� = �fv(y; z; v; w)dy + �fw(y; z; v; w)dz (12)where �f = �f(y; z; v; w) is a solution of (5). Then the restoration = �fv(� � � )d(� + � ) + �fw(� � � )d(Æ + ) == �fv(� � � )(dy � vdx) + �fw(� � � )(dz � wdx) + ( � x)( �fv(� � � )dv + �fw(� � � )dw) (13)where (� � � ) = (� + � ; Æ + ; �; ) = (y + ( � x)v; z + ( � x)w; v; w) (14)needs the orre tion �dg(x; y; z; v; w) su h thatgv = ( � x) �fv(� � � ); gw = ( � x) �fw(� � � ) (15)in order to obtain the resolving PC form�' = � dg = �Xgdx + ( �fv(� � � )� gy)(dy � vdx) + ( �fw(� � � )� gz)(dz � wdx) (16)when e f = �Xg is the desired resolving fun tion.Equations (15) for the unknown fun tion g an be resolved by the line integrals inthe two-dimensional subspa es x = x0; y = y0; z = z0 (with oordinates v; w) of the total�ve-dimensional spa e with the oordinates x; y; z; v; w: For instan eg(x0; y0; z0; v; w) = g0 + ( � x) 1Z0 ( �fv(� � � )(v � v0) + �fw(� � � )(w � w0))dt (17)where x0; y0; z0; v0 + t(v � v0); w0 + t(w� w0) is substituted for x; y; z; v; w (respe tively)into the arguments (14). The integral is taken along the straight line segment with theendpoints (v0; w0) and (v; w): Arbitrary smooth fun tions of the parameters x0; y0; z0 anbe in prin iple hosen for g0; v0; w0; however, one an also assume v0; w0 are onstantswithout any loss of generality.If we are not interested in PC forms but only in the orresponding resolvingfun tion f = �Xg; the formulaf(x0; y0; z0; v; w) = �( ��x0 + v ��y0 + w ��z0 )g0(x0; y0; z0)++(v � v0) 1Z0 �fv(� � � )dt+ (w � w0) 1Z0 �fw(� � � )dt��( � v0)((v � v0)2 1Z0 �fvy(� � � )(1� t)dt+ 2(v � v0)(w � w0) 1Z0 �fvz(� � � )(1� t)dt+¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �1' 2010 Straight Lines in Three-Dimensional Spa e and the Ultrahyperboli Equation 43+(w � w0)2 1Z0 �fwz(� � � )(1� t)dt (18)dire tly follows from (17). For onvenien e, we re all the arguments (� � � ) =(y0 + ( � x0)(v0 + t(v� v0)); z0 + ( � x0)(w0 + t(w�w0)); v0 + t(v� v0); w0 + t(w�w0))appearing in the integrands (17) and (18).A few appli ationsFirst Example.We shall deal with the resolving fun tions f whi h are polynomials.For this aim, it is su� ient to determine all solutions �f of the equation (5) that arehomogeneous polynomialsPn =XP aibj kdln yaizbjv kwdl (ai + bj + k + dl = n) (19)of given degree n. The equation (5) is satis�ed if and only if all the equalities�n�yai�zbj�v k�wdl Pn = �n�yai�1�zbj+1�v k+1�wdl�1Pnbetween n-th order derivatives are valid whenever ai; dl � 1: However�n�yai�zbj�v k�wdl Pn = ai! bj! k! dl!P aibj kdlnand so we have the onditionsai! bj! k! dl!P aibj kdln = (ai � 1)! (bj + 1)! ( k + 1)! (dl � 1)!P ai�1; bj+1; k+1; dl�1nfor the oe� ients. The produ ts yw and zv play a signi� ant role here whi h is not learlyexpressed. Let us therefore rearrange (19) as followsPn =X yAizBjvCkwDlPAiBjCkDl2m (Ai +Bj + Ck +Dl + 2m = n) (20)where the homogeneous polynomialsPAiBjCkDl2m =XPAiBjCkDl rs2m (yw)r(zv)s (r + s = m)involve all fa tors yw and zv, i.e, we suppose AiDl = BjCk = 0 in the sum (20). Afterthis arrangement, we have the onditions(Ai + r)!(Bj + s)!(Ck + s)!(Dl + r)!PAiBjCkDl rs2m == (Ai + r � 1)!(Bj + s+ 1)!(Ck + s+ 1)!(Dl + r � 1)!PAiBjCkDl;r�1;s+12m (21)whenever r � 1: They an be expli itly resolved:PAiBjCkDl2m = onstXr;s (yw)r(zv)s(Ai + r)!(Dl + r)!(Bj + s)!(Ck + s)! (r + s = m) (22)where onst = CAiBjCkDl : Substituting (22) into (20), we obtain all polynomialsolutions �f = Pn of the ultrahyperboli equation (5).¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �1' 2010 44 Chrastinov�a V.A ertain relationship to Bessel fun tions is worth mentioning. For the parti ular ase Ai = � � � = Dl = 0 this is simple sin e1Xr=0 (yw�)rr!2 1Xs=0 (zv�)ss!2 = 1Xr=0 P0:::02m �m (23)whi h reads I0(2(yw�)1=2)I0(2(zv�)1=2) = PP0:::02m �m: As the general ase is on erned,re all that Ai; Bj; Ck; Dl are natural numbers su h that either Ai; Dl or Bj; Ck arevanishing. Assume Ai = Ck = 0: Then the obvious identity1Xr=0 (yw�)rr!(Dl + r)! 1Xs=0 (zv�)s(Bj + s)!s! =XP0Bj0Dl2m �mreads IDl(2(yw�)1=2)IBj (2(zv�)1=2) = (2(yw�))Dl(2(zv�))Bj XP0Bj0Dl2m �mwhere I�(z) = (z=2)�P(z=2)2k=(k!�(�+k+1)) are the Bessel fun tions. We have obtainedthe generating fun tions for the ultrahyperboli polynomials.The generating fun tions are useful if one al ulates the resolving fun tions f: Wemention only the parti ular ase when Ai = Bj = Ck = Dl = 0 here. Then, applying (17)with the fun tions �f = PP0:::02m �m and hoosing g0 = v0 = w0 = = 0 for simpli ity, weobtain the result f(x; y; z; v; w;�) == ( ��x + v ��y + w ��z ) 1Xi;j;r;s(�x)i+j yr�izs�jvs+iwr+ji!(r � i)!j!(s� j)!r!s! r + sr + s+ i + j �r+s(sum over i; j; s; r � 0 and moreover r � i; s � j; r + s + i + j > 0). The oe� ientof �m on the right-hand side is the resolving fun tion whi h orresponds to the parti ularsolution �f = P0:::02m of the ultrahyperboli equation (5).By using the �general solution� �f = 1Z�1 �(t; y � vt; z � wt)dt of U H with the hoi e� = g(t)(y � tv)p(z � tw)q (p; q;= 0; 1; � � � )where g(t) is an arbitrary fun tion with a ompa t support, we obtain the abovepolynomial solutions as well. However, this is in fa t a misleading strategy: the proofs ofthe most interesting geometri al results to follow rest on quite other prin iples.Se ond Example. Together with the primary oordinates x; y; z; v; w we shall usethe alternative oordinates x; �; �; ; Æ: Then the straight lines (6) have the alternativeequations (7). Let � = �(�; �; ; Æ)be a given fun tion. In virtue of (7), � may be regardedas a fun tion on the family of straight lines. We will determine all resolving fun tionsof the form f = F (x; �; �; ; Æ) = h(x; �): One an verify that onditions (4) for ourresolving fun tion read 2F� = F�x � xF�x; 2FÆ = F x � xFÆx (24)¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �1' 2010 Straight Lines in Three-Dimensional Spa e and the Ultrahyperboli Equation 45in the alternative oordinates. Continuing in this way we obtain the system of di�erentialequations 2��h� = (�� � x��)h�x; 2�Æ = (� � x�Æ)h�xfor the fun tions h; � where the variables an be separated as���� = � �Æ = 2H + x �H = h�xh� � : (25)We have omitted the ex eptional (and quite simple) ases when some of thefun tions ��; �Æ; h�; H vanish. It follows that �(2=H + x)=�x = 0 when eH = � 2x� p(�) ; h� = q(�)(x� p(�))2 ; h = Z q(�)(x� p(�))2d� + r(x)where p(�); q(�); r(x) are arbitrary fun tions. With this fun tion h; onditions (25) redu eto the system �� = p(�)��; � = p(�)�Æ: (26)The solution is given by the impli it equation� =M(� + p(�)�; Æ + p(�) ) (27)and we have the result.If a fun tion � = �(�; �; ; Æ) satis�es the impli it equation (27) and p(�); q(�); r(x)are arbitrary, then h(x; �) = Z q(�)(x� p(�))2d� + r(x) (28)is a resolving fun tion. The fun tion r(x) is immaterial here and may be omitted.Without mu h loss of generality, we may assume p(�) = �: Then (27) simpli�es as� =M(� + ��; Æ + ��): (29)In terms of the primary oordinates, we have the equation~x =M(y � (x� ~x)v; z � (x� ~x)w) (~x = �(v; y � vx; w; z � wx)) (30)and a ni e geometri al interpretation is as follows. XXXXXXXXXXXXz (1; v; w)à~P = (~x; ~y; ~z)̀aP = (x; y; z)~x =M(~y; ~z) y � ~y = (x� ~x)vz � ~z = (x� ~x)w Let a smooth surfa e x =M(y; z) in the spa e R3 with the oordinates x; y; z be given,the domain of M being a ertain open subset of R2 : Given x; y; z; v; w; we onsider the¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �1' 2010 46 Chrastinov�a V.point ~P = (~x; ~y; ~z) of interse tion of the straight line y�~y = (x�~x)v; z�~z = (x�~x)w withthis surfa e; then the interse tion point ~P is determined just by the impli it equation (30).Smooth dependen e of the interse tion ~P on the variables x; y; z; v; w is ensured if theinequality �(~x�M)�~x = 1�My(~y; ~z)v �Mz(~y; ~z)w 6= 0 (31)holds true (i.e., if the line transversally interse ts the surfa e).Assuming p(�) = �, the formula (28) also greatly simpli�es and leads to a huge familyof resolving fun tions. We shall however deal only with the simplest possible ase q(�) = 1,hen e h(x; �) = Z d�(x� �)2 = 1x� �(�; �; ; Æ) = 1x� ~x(x; y; z; v; w) (32)from now on.Let us re all the U H equation (5). In general, if f = �f(y; z; v; w) is a solution then� �2�y�w � �2�z�v� k( �f) = k00( �f)( �fy �fw � �fz �fv)for any fun tion k. In parti ular, let us hoose the fun tion (32), hen e �f = 1=(x � ~x)where x is regarded as a mere parameter. Then �fy �fw = �fz �fv identi ally. (Dire tveri� ation: the formulae~xy = My� ; ~xz = Mz� ; ~xv = �(x� ~x)My� ; ~xw = �(x� ~x)Mz�with � = 1� vMy�wMz easily follow from (30).) So we have the result: every omposedfun tion k(1=(x � ~x)) is a solution of U H (if x = is kept �xed) and therefore every omposed fun tion of the form K(~x) and in parti ular ~x itself is a solution of U H .A ording to the geometri al meaning of the fun tion ~x = ~x(x; y; z; v; w) whi h is the x- oordinate of the interse tion point ~P , we have very lear insight into a huge the familyof (generalized) solutions of U H with the singularities at the ex eptional �fo usingpoints�where the inequality (31) is not satis�ed.By employing the latter result, one an obtain a ertain ounterpart of the formula (18)by applying (17) with the hoi e �f = K(~x( ; y; z; v; w)). Clearlyg = ( � x) 1Z0 K 0(~x( ; : : :))(~xv( ; : : :)(v � v0) + ~xw( ; : : :)(w � w0))dt:In the ase v0 = w0 = 0 we may substitute~xvv + ~xww = ( � ~x)�Myv �Mzw� = ( � ~x)�� 1�and therefore the �nal resultg = ( � x) 1Z0 K 0(~x( ; : : :))( � ~x( ; : : :))�� 1� dt¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �1' 2010 Straight Lines in Three-Dimensional Spa e and the Ultrahyperboli Equation 47with the arguments (� � � ) = (y0 + ( � x0)tv; z + ( � x0)tw; tv; tw) looks fairly good.Third Example. We shall deal with the omplex�valued fun tions f = f1 + if2of the ommon real variables (either of the original ones x; y; z; v; w or thealternative x; �; �; ; Æ). Su h a fun tion is alled resolving if (4) is satis�ed (with fsubstituted for f) and this is true if and only if the omponents f1; f2 are resolvingfun tions in the ommon sense. Nontrivial results an be nevertheless obtained if onedeals with the generalized length R fdx. We wish to obtain real values after appropriateadaptations.Re all that every (real) straight line is determined by the real onstants � = A; � = B; = C; Æ = D in terms of the alternative oordinates �; �; ; Æ and the equations of thestraight line are as before y = Ax +B; z = Cx+Din terms of the original oordinates. However we will hoose a omplex point ~P = (~x; ~y; ~z)on the real straight line (hen e ~y = A~x+B; ~z = C~x+D): It is learly determined by the(arbitrary) hoi e of a omplex-valued fun tion ~x = �(�; �; ; Æ) of real variables �; �; ; Æ:Theorem. Assuming Im � 6= 0; then the fun tion f = 1=(x� �) is resolving if and onlyif � is a holomorphi fun tion su h that (ex ept for some degenerate ases) a ertainimpli it equation � =M(�+ ��; Æ+ � ) is satis�ed, whereM =M(y; z) is a holomorphi fun tion.One an observe that we again deal with the point of interse tion ~P = (~x; ~y; ~z)where ~x = �; with the omplex surfa e ~x =M(~y; ~z):Proof is a mere slight adaptation of the above reasonings. Let usdenote f = F(x; �; �; ; Æ) in terms of the alternative oordinates. Then f is resolvingif (24) is satis�ed for the ( omplex-valued) fun tion F instead of the previous F: Inparti ular, assuming F = 1=(x � �); we obtain the requirement (26) with p(�) = � andthe new, omplex-valued fun tion � (instead of the previous real �). However, � = �1+ i�2and separation of the real and imaginary omponents provides the systems�1��1�� � �2��2�� = ��1�� ; �1��2�� + �2��1�� = ��2�� ; (33)�1��1�Æ � �2��2�Æ = ��1� ; �1��2�Æ + �2��1�Æ = ��2� : (34)Both (33) and (34) are ellipti al systems (see below) if �2 6= 0; therefore any lassi alsolution �1; �2 is in reality an analyti al fun tion of the variables �; �; ; Æ (whi h maybe therefore extended to omplex variables). Then equations (26) ensure the existen eof ertain nontrivial identity M(�; � + ��; Æ + � ) = 0 where M is appropriate analyti fun tion. The proof is done.Note to the ellipti ity. We mention the parti ular ase of the quasilinearsystem PAjik��i=�yk = Bj (i; j = 1; : : : ; m; k = 1; : : : ; n) with the holomorphi oe� ients Ajik; Bj: The system is alled ellipti al if det(PAjiktk) 6= 0 for all real¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �1' 2010 48 Chrastinov�a V.nonvanishing ve tors (t1; : : : ; tn): In our parti ular ase, i.e., for the system (33), we havedenoted y1 = �; y2 = � (n = m = 2) and thenXA11ktk =XA22ktk = �1t2 � t1;XA21ktk = �XA12ktk = �2t2when e det(PAjiktk) = (�1t2 � t1)2 + (�2t2)2 6= 0 if �2 6= 0 and (t1; t2) 6= (0; 0):Passing to the example proper, one an see that the distan es (lengths of the segmentsalong a �xed straight line where � = onst) are given by the omplex-valued and multi-valued logarithmZ fdx = Z dxx� � = Ln(x� �) = ln jx� �j+ iArg(x� �):It is desirable to introdu e the real-valued omponentsRe Z fdx = 12 Z � 1x� �1 � i�2 + 1x� �1 + i�2� dx = Z x� �1jx� �j2dx = ln jx� �j;ImZ fdx = 12i Z � 1x� �1 � i�2 � 1x� �1 + i�2� dx = Z �2jx� �j2dx = Arg(x� �):Both are rather interesting. For instan e, if one takes the imaginary spherex2 + y2 + z2 = �1 for the surfa e ~x =M(~y; ~z); the resolving fun tionsRe f = ((y � vx)v + (z � wx)w)=(1 + x2 + y2 + z2);Im f = (1 + v2 + w2 + ((y � vz)2 + (z � wx)2 � (vz + wy)2)=(1 + x2 + y2 + z2)appear. The latter one provides the distan e in the non-Eu lidean ellipti al geometry.One an observe that(in ontrast to the hyperboli ase) the distan es are de�ned ontotal lines and for the omponent Im f also at the points of in�nity.A knowledgements. This resear h has been ondu ted at the Department ofMathemati s as a part of the resear h proje t CEZ MSM 2611 00009.Referen es1. Hilbert, D. Mathematis he Probleme, Vortrag gehalten auf dem internationalen MathematikerCongress, Paris 1900.2. Mathemati al developments arising from Hilbert problems, Pro . Symp. in Pure and Appl. Math.AMS, Vol XXVII, 1976.3. Hamel, G. �Uber die Geometrien, in denen die Geraden die K�urzesten sind, Mathematis he AnnalenLVII, 1903.4. Darboux, G. Le� ons sur la th�eorie g�en�erale des surfa es III, Paris 1894.5. Douglas, J. Solution of the inverse problem of the al ulus of variations, Transa tions AMS 50, 1941.6. Anderson, I., Thompson, G. The inverse problem of the al ulus of variations for ordinary di�erentialequations, Memoires AMS 473, 1992.7. Chrastina, J. Solution of the inverse problem of the al ulus of variations, Mathemati a Bohemi a119, 1994.8. Cartan, E. Le� ons sur les invariant integrals, Hermann 1922.9. Busemann, H., Kelly, P.J. Proje tive geometry and proje tive metri s, A ademi Press 1953.¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �1' 2010 Straight Lines in Three-Dimensional Spa e and the Ultrahyperboli Equation 4910. Morrey, B. Multiple integrals in the al ulus of variations, Grundlehren Math. Wiss. 130, SpringerVerlag 1966(pages 170-186).11. Petrovskij, I. G. On the analyti ity of solutions of systems of partial di�erential equations (Russian),Matem. Sbornik 5(47), 1939. Ñòàòüÿ ïîñòóïèëà â ðåäàêöèþ 26.11.2009 ¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �1' 2010

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