ENGINIERING RESEARCH' INSTITUTE UNlVERS TT OFI XBTWGAHN ANN ARB0R TECHNICAL REPOTRT NO 1 NEW OPERATIONAL 1mATEMATICS THE OPERATIONAL CALCULUS OF LEGENIRE TRANSFORM -By.;f...RCIL L Pof eassy ofE^Mlatheatics - P jet 21537 ORDNANCE CRPSU, S ARMY C ONIT O:NOE TDA.Oi8-:.RD- 12 916 Auust, 1955

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- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN THE OPERATIONAL CALCULUS OF LEGENDRE TRANSFORMS* 1. Introduction. The sequence of numbers f(n) defined by the equation (1) f(n) = fsF(x)Pn(x)dx (n 0,1,2,-), where Pn(x) denotes the Legendre polynomial of degree n, is the Legendre transform of the function F(x). The integral transformation here will be represented by the symbol T [F(x)]. For functions F(x) satisfying well-known conditions on the interval -1 x < 1 the inverse of this transformation is represented by the expansion of F(x) in series of the Legendre polynomials, (2) F(x) = (n + 1/2) f(n)Pn(x) T1 [f(n) (-1 < x < ). n=O Let R[F] denote the differential form (3) R[F(x)] = [(1-x2) d F(x)] 'When the integral T [R[F]] is integrated successively by parts and -n (n + l) Pn(x) is substituted for R[Pn(x)] in accordance with Legendre's differential equation, the following result is easily obtained. THEOREM 1: Let F(x) denote a function that satisfies these conditiObn F'(x) is continuous and FE (x) is bounded and integrable over each interval interior to the interval -1 < x < 1; T IF(x)} exists and *The research recorded here was conducted in part under a contract between the Office of Ordnance Research oftt the he Army and: the -University of 'Michigan. I - 1

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN lim (1-x2) F(x) = lim (1-x2) F'(x) = 0. x -+ + 1:x + 1 Then T [R:[F(x) 3 exists and (4) T [R[F(x)]3 = -n(n + 1) f(n) (n = 0^2,- ). Formula (4) represents the basic operational property of the Legendre transformation T under which the differential operation R[F] defined by equation (3) is replaced by the algebraic operation -n(n + 1) f(n). The established forms of operational calculus for solving problems in differential equations are based on integral transformations of Fourier type, transformations whose kernels are exponential functions or linear combinations of such functions. These transformations consist of the various Laplace and Fourier integral transformations. Transforms of the other types, including Legendre transforms [1], have been recognized as bases for other forms of operational calculus [2,5,4], but to date the operational methods have not been developed beyond the stage of applying the basic operational property corresponding to formula (4) and the application of the inverse transformation. The type of boundary value problem that can be reduced by an integral transformation is of course governed by the kernel and the interval of integration [5]. A convolution property, one that gives directly the image of the operation of taking the product of the transforms of two functions, is now known for Legendre transforms [61. Additional operational properties will be noted here and a short table of Legendre transforms will be developed. The operational calculus will then be illustrated by applying it to some classical boundary value problems in partial differential equations. The application to the third boundary value problem for the potential inside a sphere leads to a simple expression for the solution of that problem 2

- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN in terms of the solution of a corresponding Dirichlet problem. This formula. together with the accompanying extension of the Poisson integral formula to the problem of the third kind, should be known; but a search of the literature and inquiry among colleagues has not yet revealed these formulas. 2. Operational Properties. If each of the functions R[F(x)] and F(x) satisfy the sufficient conditions stated for the validity of formula (4) then the transform of the iterated differential form R[R[F]] can be written at once as (5) T R2[F(x)]] = n2(n+1)2 Similarly for iterations of higher order. The substitution of (n + 1/2)2- 1/4 for n(n + 1) in formula: () leads. to this modification of that operational property: (6) (n + 1/2)2 f(n) = T [1/4 F(x) - R[F(x)]. It should be noted that under the substitution x = cos Q (0 = 0 < r) our transformation T [F(cos Q) } = f F )(co s ) Pn( ) sin 0 d" = f(n) JO transforms the differential form (7) R[F(cos Q)] = 1 d [sin 0 d F(cos,)] sin V do d8g into -n(n + 1) f(n), according to formula (4). Consider now the transform of the function R1[F], where Rl.a is the inverse of the differential operator R. Let Y(x) denote the function R'l[F(x)]; 3

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN - then Y(x) is a solution of the differential equation (8) R[Y(x)] = F(x). Suppose that F(x) is a function of bounded variation on each interval I<|x 1Ix<l, and that (9): F(x) dx = 0; 1 -that is, f(0) = 0, The first integral of equation (8), x (.-x2) Y'(x) = f F(t) dt is then a continuous function of x (Ixl ' l) with limit zero as x + + 1. The second integral (10) Y(x) = 1 - '. F(t)dtds +C C R"[F], where C is an arbitrary constant, is continuous when IxI <1 and it is easy to show that (1 - x2) Y(x) vanishes as x + 1; in fact IY(x)l is of the order of (1 - x2)-k for each positive constant k < 1 as x + + 1 and hence T [Y] exists, According to TheOrem 1 and equation (8) then T [R[Y] = -n(n + l) T [Y = f(n); thus (11) TT [R' [FI] = - (n) (n 12 nn n+-1)

- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN The value of the transform of R'1[F] at n=O depends on the value assigned to the constant C; if F(x) is an odd function the value is easily shown to be 2C. The operational property concerning R'1 can be stated as follows. TTBEORE2: Let F(x) denote a function of bounded variation in each subinterval of the interval -1 < x < 1 and let f(0) = 0. Then f(n) exists and for each constant C, (12) T'i {f(n1) n(n + 1) = R-[F(x)] = f 2 (t)dtds + C o-(n - 1,2 (n = 112, W*). The convolution property can be stated as follows [6]. tIEORIEM 3: Let F(x) and G(x) denote bounded integrable functions on the interval -1 = x 1. Then the product f(n)g(n) of their Legendre transforms is the transform of the function H(x); that is, (13) T-1 f(n)g(n)} = H(x) where H(x) is described by any one of these formulas: (14) H(cos @) _ 1 fF(cos Q) sin t' G(cos:x) dd' " Jo — Jo where cos x = cos C cos 6' + sin Q sin $' cos A; * F * ' (.15) S(cos) = 1 2 i Fsin' Fsi n(,) ]G[sin sin(- )G[sin dis; _) (1- x- y. z2 2 (16) H(x) 1 ff F(y) G(z) (1 - x2 y2 z2 + 2xyz)"l/2 dz X JEs - - 5

- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN where, for each fixed x (-1 < x <.1), Ex is the region interior to the ellipse y2 + z2 2y 1 - x2. Formula. (14) has a geometrical interpretation. Let (r,~') denote spherical polar coordinates. If P denotes any point (1,0, 0) on the boundary of a hemispherical surface of unit radius and P: any point (1,jlQ') on that surfacei then according to the cosine law for spherical triangles X is the arc PP' of a great circle. If S denotes the hemisphere r = 1, 0 < ' <, 0 < then formula (.14) can be written (17) E(co$ H F(cos ) = F(cos ') G(cos k) dS; that is, H(cos Q) is half the mean value of F(cos @') G(cos X) over the hemisphere. For Q = 0 and ~ = it formula (14) reduces to 1 1 (18) H(1) = I F(t) G(t) dt, H(-l) = f F(t) G(-t) dt, respectively. In view of formula (2) and the fact that Pn(l) = 1 the first of equations.(18) is seen to be the Parseval relation for the orthogonal set of functions Pn(x). THEREM 4: If T [F(x)] and T [G(x)] exist then (19) T [C1 F(x) + C2 G(x)) = C1 f(n) + C2 g(n), where C1 and C2 are constants; also (20) T [F(-x)} = (-l)n f(n). I I - - 6

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN These are obvious properties of the transformation. since T is clearly linear and since Pn(-x) = (-1) Pn(x). When G(x) = 1 then g(n) = 0 (n = 1,2, ") and g(0) = 2; according to equation (19) then, if C is a constant, T |F(x) + C] = f(n) whenn = 1,2, ', = f(0) + 2C when n = O. The images under T of the operations of differentiation and indefinite integration involve finite differences. Let F(x) be a sectionally continuous function on the interval -1 = x 1 and let G(x) denote the continuous function X G(x) = f F(t) dt. Then 1i f(n) = G'(x) Pn(x) dx = G(1) - 1 G(x) ~P(x) dx and it follows from a differential recurrence relation for Pn(x) that 1 (21) f(n - 1) - f(n + 1) = G(x) [PA+i (x) - PA(x)] dx 1 = (2n + 1) g(n) (n = 1,2-.*); also, f(O) = G(l)and since Pl(x) = x, f(l) = G() - g(O) so that f(O) - f(l) = g(o). The solution of the difference equation (21) for g(n) and f(n) in turn leads to these conclusions: 7

[ ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN TIEREM 5: If F(x) is sectionally continuous then (22) T F F(t) dt = f(n - ) - f(n + ) 2n + 1 where, for n = 0, f(n - 1) is to be replaced by f(0). If G(x) is continuous and G' (x) sectionally continuous and if G(-l) = 0 then (23) T [G'(x)} = G() - (2n - 1) g(n - 1) - (2n - 5) g(n- 3) -..- g(O) (n = 1,35,'), = G(l) - (2n - l:1g(n- 1) - (2n - 5) g(n- 3) -* --- g(l) (n = 2,4,.)), = G(1) (n. 0). Other recurrence formulas for Pn(x) lead to operational properties of the. transforms, but none seem simple or promising. 3. Transforms of Particular Functions, A short table of Legendre transform is presented here. Some methods of computing transforms in the table will now be indicated. It follows at once from the orthogonality properties of the Legendre polynomials that when F(x) = P (x) (m = OJ1.22, -) then f(n) = 0 (n m), f(m) = m+l-12 From the well-known representation of the function G(x) = xm in a finite series of Legendre polynomials of degree a and lower it follows that g(n) = 0 when n > a. The formula for g(n) when n m is nosimple (cf. [7]). If F(x) is any polynomial of degree m then f(n) - 0 when n > m. I 8

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN The function F(x) = log (1 - x) satisfies the condition RIF] = -1^ where R is the differential operator involved in Theorem 1, But F' (x) does rot satisfy the limit condition at x = 1 in that theorem. Integration by parts, however, gives the equation T [R[F]] = - (1 + x) Pn(x) -(1 - x2) log (1 - x) PA(x) -n(n + l)f:(nt I-1 - that is, T [-.1. = -2 -n(n + 1) f(n) = 0 (n = 1,2,.). zbqz Direct integration gives f(0) =^ -2; hence T [log ( - x)] 2 - (n > 0), = 2 log 2 - 2o). The transform of log (1 + x) follows from Theorem 4. From the uniformly convergent expansion of the generating function co.. L. n=0 it follows that n. (24). A1 (1 - 2tx + t2)1/2 P(x) dx< I) J-l n + 1/2 When t > 0 the integrand here is a continuous function of t and x except at x =t = 1 and it is dominated by the function (1 - x2) 1/2 Hence the inte/a 0 gral is uniformly convergent with respect to t up to t = + 1 and it represents a continuous function of t there. Thus the transformation is valid when t = 1; that is,. I 9

- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN T (1 - x)-1/2} n + 1/2" The transform of (1 + x)1/2 fllows from Theorem 4. Differentiation of both members of equation (24) with respect to the parameter t followed by multiplication by t leads to the equation _ ( t 2 ntn <. T [(1 - 2tx + t2)5/2 (2tx - 2t2)} = nt (Itl <1) 2 n +. Since 2tx - 2t2 = -(1 - 2tx + t2) + 1 - t2 it follows with the aid of formula (24) that (25) T [(1 - 2tx + t2)-3/21 2 tn (It < l) A continuation of the process leads to a slightly involved expression for T-1 ntn} when Itl <1. The transformation I T [log a - x + (1 - 2ax + a2)1/23 1 - x n+1 A 'a.' (n + 1/2) (n +"" 1) ( can be verified by integrating both members of equation (24) with respect to t, From the partial fractions expansion of the coefficient of an+l here and from formula (24) with t = a (la 1) it follows that 2T. a x+og (l 2ax+a2)l/2 (, 2ax + 2)a/2 (Ial a i). From equation (24) it follows that 1 1t'l (L1 - 2tx + t2)-1/2 t-l] Pn(x) dx = tn n Wn + 1/2 - 0 - 10 (n =.1,2, -w) (n = 0), -

[ - ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN I TABLE OF LEG 'EMI 2RANSFCRMS f (n (n s 0,02:,a F(x) -i < x < 1J I........ 1.:. if n = m; 0 if n i m lM.1/2 Pm.(x) (m = 0,1,2,'. ) 2, 1 if n > 0; n(n+l 1-log 2 if n=O I. l o g ( 1 - ) 2 3. n (-InI 3, n +'^I') if n > O; 1-log 2 if n=O 4. an ( lal< 1) -a2 (1-2ax+a2) 2, an if n> O n 2 if n=O (la< 1) (l-2ax+a2r1/2 -.log l-+ax+(12ax+a2) / - log "MW 22 I 6. an n+1/2 (lal l) (l.2ax+a2)'1 2 I I. I an+l 7. + an+l (lal.1) a (1-2ax+a2) / 21 2 /2 log a-x+-(32,axa2 )a/2 l-x 8. (n+l/2) (n+l) an 9-,. — a. if n n(n+1/2): (n+l/2) (n+c) (lal<.1) > O; 0 if n=O (c > o, lal<.).(o) ~ — if n > 0; log a-x+ (1-2ax+a2)/ l.-x -log l-ax+(l-2ax+a2) ) ( lal l) 0.. tcldt1 O (12 tx+t 2)1 O;W- x < 0;- 1 if x > O M 11. Pn-(O) - n 2n+l 1 if n=O 11

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN and the integration of this equation with respect to t leads to the transformation T-l 1 - a + (1 - 2 + a)/ = a (n- = L 2 |.,A 1) ~ ( ~ o...... Ti ~2,......~2 n(n + 1/2) = 0 (n =-0; laal< ). From the representation of the coefficient of an here in partial fractions the inverse transform of an/n follows with the aid of formula (24). Transform No. 10 in the table can be found by multiplying equation (24) by tC-l and integrating with respect to t. Transforms of step functions, illustrated by No. 11 in the:table, are easily written with the aid of — a wellknown integration formula for Pn(x). Combinations and special cases of the transforms in the table- as well as applications of the operational properties of the preceding sections, lead to a considerable extension of the list of transforms 4. Applications, Dirichlet Problem for Sphere. Let V(r, cos 8) denote the potential function interior to the unit sphere when the potential on the surface r = 1 is a prescribed function F(cos Q) of the spherical coordinate. only (0 < = ai). The interior is free from sources so that V satisfies Laplace.s equation V2V = 1 (r V), +- 1- (sin V.)9 = 0 r r r2 sin ' when r < 1. If x = cos 9 this equation can be written (26) r(r V)r + [(C - 2) Vx]x = 0 (r< 1)..r.. I The function V is also bounded in the region and satisfies tbhe' boundary ~oitifon ----------------:12 a

- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN - (27) V(l x) = F(x) (-1 x ). To solve this boundary value problem formally by means of the operational calculus developed in the foregoing sections the problem is written in terms of the transforms v(r,n) = T [V(rx), f(n) = T {F(x)3 (n = 0,1,2,' ) with respect to the variable x. Let the operatr T be applied to the members of equations (26) and (27) and let the order of differentiating V with respect to r and then integrating with respect to x be reversed. In view of the basic operational property (Theorem 1) it follows that v(r, n) satisfies the conditions d2 r d (r v) - n(n 1) = 0, (l n) f( and v(r, n) is bounded when 0 r < 1 The solution of the simple problem in v here is v(r, n) = f(n) rn and the inversion formula (2) can be used to represent V(r, x) by an infinite series in the functions r (x), the form of the solution that would be ob" tained by separating variables in the boundary value problem (26) (27). The function V(r, x) can be written in closed form, however, with the aid of the cnolution property (Theorem 3) and the inverse transform of rn (No. 4 of the table). According to formula (14) for the inverse transform the product of two transforms the inverse transform of v(r,: n) is 13

- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN - (28) V(rtco. @) r F(cOs O') sin ' (+r os )3/2 0 0 (r < 1), where cos x = cos @9 cos O' + sin Q sin ~' cos,. This is the well-knLow Poisson integral formxla for the potential inside a sphere in this special case in which the ", tial is. a function of r and 0 only. When F(x) is a linear combination of any of the functions listed among the first nine items of the table, then f(nrn can be written as a linear combination of the transforms listed in the table and the function V can be written in a simple form free from integrals. In the corresponding problem for the potential W(r, cos ') in the region exerior t the sphere r =. the solutio of the transfored probl is' w (r, n) = f(n) r r n r.. (r. 1). Hence (29) W (r, cos ~) 1 v(iL cos ).r r where V denotes the above potential function for the interior region. The Poisson integral formula for the potential W follows at once from formulas (29) and (28). When the potential is not independent of the spherical coordinate 0 the Laplacian involves a differential tor with rspect to cos 9 (x = cos).14,.

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN that is more inv led than the oprator R of equation (3) and the transfOrma-, tion T d es not eliminate derivatives with resect to x from the Laplacian. 5. INeuann r blebm for Sphere. Let U(r, cos 0) denote the potential in the region interior to the sphere r = I when the nrml deriyative is a prescribed function F(cos Q), so. that (30) V " — = 0 (r <), Ur (1, ) = F(x) where x = cos Q, and U is bounded wen 0 r < i, Then u(r, n), the transform of U(r, x), is bounded and formally satisfies the conditions -r:(r u)'" - n(n+ 1) u = 0, ut(i, n) = f(n) (n = O12' ) where the primes denote differentiation with resct t r. When n = 0 the ided slution of the differential eqtation here is u(r, O) = CQ where C is a constant. Hence f(0) = 0 if th boundary condition is to be satisfied; thus f F(cos @) sin @ d = 0; O0 that is, the mean value of F(eos ) over the surface is zero, a well-knwn necessary condition for the soutiion f the Neuann problem. The transfred problem has the solutin (31) A(r, n) = n) ) r n ( 1,2,. ) = C (n 0) where C is an arbitrary constant,. Thus the tential U(r, x) is determined up to an additive constant. It can b e- ressed in a series:f the functions 15 ---------

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN - rtP(x) with the aid of the inrsin r (2); but it can be written in the -clsed f ormt - 1 tog [I-r cos X-+ (+r2 -2r cos X)1 /. d0'.d' + C, 2.,..I where: cos C Z cO s COs ' + sin Qsin Q cos ', with the aid of the conwoiution fo la (P$) and. tranStfrm No. 5 in the table. Let Q denote any- pint (1, ', o') on the hemisphere S, r = i, 0 = ^ o It, 0 > and let P denote any oint (r, 0, ) of the region boundd by the boundary circle of S. Then the abive variable X represents the angle between the radii through:P 'and":' Q-.d the length:of the line seg ment PQ is p = (1 + r2 - 2r cos X)/2. With the aid f formula (17) for the coniuti an alternate form of equation (32) n-o can b written as (33) U(rcos @) J f F(cos @)[2 _- lo(l+p - r cos x) dS + C. This is a known form [8]. of the PoissOn integral solution of the 'Ne.umannprobblem -for the sphere. From equation (51) it follows that r u'(r,: n) = f(n) rn = v(r, n) (n = 12,-**)I ".16

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN where v(r, n) is the transform of the Dirichlet prob (26), (27), with the same function F(x) there as here. Thus r0 r u(r n) frtrt n) - (n =1-2 = X C; (n = 0) and the inverse. transfrmsatin gives the expressioa (34) (r,, cos ) f,r( )dr + C ^0 5= fv(rt.; c @ @) d c for the s6lution of theNeumnan pr (30) in terms of the solution o'f the Dirichelt plroble (26), (27). This elationMhp bee n the sautionss can be verified directly and it is not eesary * as that the potential funo;tions and the function F are indpendnt of the coordinate.. 6.?rblem of Third. fer Thia - &w let U(r, cos,) represent the bonded. tential futi fr t reio finterior to the sphere r = 1 satisfying the ixed type of bundary c iti at the surface, Specificly | (35) V7% -0 (r < 1), Ur(lx) + (k + 1) U(lx) = F(x) where x = cs Q and. k 1 is a psitive constant The function U can be interrpreted as the steady-state temperatures at pints in a solid sphere which is subje~cted to linear heat transfer at its surface into surroundings whose temaeture- is pr rtil to F(cs )i The transfr U(r n) U(r, x) is the nded function that satioS | fies the conditions 7 -------------.. --- —--------

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN - r(ru)" - n(n + 1) u = Ot u'(1, n) + (k + l)'u(l, n) f(n); hence (36) u(r, n) = f(n) rn + (0 < r. 1, 1 It follows that r'k (rk+l u) = f(n) rn v(r, n), where v(r, n) is the transform of the solution V(r, x) of the Dirichlet lem (26), (27)ad4 therefore u(r, n) = ( v(rts, n) = (rt, n)tk dt 0 Q k > -1). prob I! The inverse transformation then gives the formula (37) U(r, cos Q) 1 = f V(rt, cos e) tk dt O0 for the solution of problem (5) of the third type in terms of the solution of the Dirichlet problem (26), (27) with the same function F(x). A generalization f formula (7) can be verified easily.. Let the above functions U. V and F be replaced here by U(ry,, @), V(r, 0, Q) and: F(), 9)* For the sake of simplicity assume that F is a eontinuous function Of-its two variables. THEOREM 6: The potential function U(r, 0 9) of the third boundary value problem for the sphere, (38) 2U = O (r <1), Ur(1l,e) + (k+l) U(1,,) = F(,e) I? L 18

ENGINEERING RESEARCH INSTITUTE ~ UNIVERSITY OF MICHIGAN where k > -1 and constant and the function F is continuous, is given in terms of the solution V(r, 0, Q) of the Dirichlet problem (59) VV = 0 (r < 1), V(l,, @) = F(0, 0) by the formula O k (40) U(r~ ~, Q) = (rt~ 9 G)tk dt. 0I The verification that the function U given by formula (40) is harimonic in the region r < 1 when V is harmonic there is straightforward. To show that the function U satisfies the boundary condition in problem (38) let rt = s and note that r Ur(r, 03, 0) = f Vs(s, 0 ) tk+l rdt = Vt(s,, a0) tk+l dt.0 o0 when r < 1. Hence (41) r Ur(r,. 80) + (k + 1) U(rs, 0) = J [tk+l Vt(S,,0) + (k+ 11 tk V(s,,^)] dt V(r,,G0) (r <l) It is known that the function V is continuous in the region r 1 and it'.ollws from formula (40) that U is continuous there. According to equation (41) then rUr is also continuous there. When r = 1 equation (41) reduces to the boundary condition in problem (38). A Poisson integral formula for the solution of the third boundry value problem (38) follows at once by substituting the known Poisson integral that represents the solution V of the Dirichlet problem into formula (40), -n (40 y9

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN Other forms of the Poisson integral formula for the special case (35) in which U is independent of 0 can be found from the frula (36) for u(r, n) with the aid of the convolution property. For the inverse transform of rn(n + k + 1)-1 can be found either from transforms No. 10 and No. 6 in the table or by inte-. grating the transformation No. 4 after multiplying by ak. The relation (40) between the solutions of the two types of proble for the sphere is so simple that it should be known. Neither that relation rW a Poisson integral formula for the problem Of the thrd kind for the sphere nor the corresponding results in logarithmic potential for the circle, which follow similarly from finite Fourier transformations have been found yet in the literature * If Y(r, cos 8) denotes the potential in the exterior region r > 1 for the third problem (42) V2Y 0 (r > Yr(lr x) - kY(1,x) x where k > 0 here and constant and x = cos Q, the transform of Y is easily found to be y(r, n) = f(n) rn U u n) (rtl) r n -+ k'+.l r r where u(r n) is the transform (36) of the solution of the third prOblem (35) for the interior of the sphere. Hence (43) Y(r, cos Q) 2U (1 cos @) (r >). r r Moreover the procedure used to arrive at formula (37) is easily ap plied to the above formula for y(r, n) to arrive at the formula * o n. i. 1 1 y ~ \ * i i i iiii i ii,, i 1 1 i i i i i i. i i i i i i in i i ip - - - meu

ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN - OO (4) YY(r, cos e) = W(rt, cos Q) t'k' dt where W(r, cos 0) is the potential (29) in the Dirichlet problem for the eXt*; rior region, V2W - 0 (r > 1)- W(I. cos 9) F(cos G). The verification of formulas (43) and (44) shows that tlw ptentials and the function F involved need not be independent of the coordinate -. 7. Distributed Sources in phere. As a final illustrative appli. cation of the transforms consider the proble,. v 2, X*, v 1 (45) V-V(r, x) = F(r, x) (r < 1) V(l, X) = 0 where x = cos 0. The transformed problem is L r,n), v(l,n). r2 v"(rn) + 2r vT(rn) - n(n+l) v(r,n) = r2 f( It,s solution can be written by elementary method in the form 1 n2ds l. (46) v(rn) = (rs f(sn) s2 ds r' f(sn) 2n-+" r n-J + 10s n - rens'n f(sIn) -s ds, Jr,2n+.l According to transform No- 6 in the table, s2' -ds. T G(a, x)}. n ~~~f': ~ ( aj is.1) where G(a, x) = -. (1 + a2.2ax)'/ 2 21

- ENGINEERING RESEARCH INSTITUTE * UNIVERSITY OF MICHIGAN If H(a, s, x) denOtes the conW'1ution (Teorem 3) of tshe two functions G(a, x) and F(s, x) as functions of x the f irversen transformti~n 'of the emers Of equation (46) gives the forula| ~~I:~r 1 S. V()ro rsx) H,, te 1fr8s os X. ds - H(~,Sdx) a 0Q r ' for the solution of proble (45) REFERENCES.1 Tranter, J. C.,. Integral Transfos in: Mathea tic Physicsy Wiley, NewYork, 1951. — 2. Sneddo, I. N., Fourier Transf'ors, McGrw. l New York, 1951 35 Burgat,. P. "tResolution de 'rob:-ni- s aux limites au moyen de transformations fonctionmelles".: ' eh:s; ' eAh..-, 146-.i.i953) 4. Scott, E. J., "Jacbi Transforms", w J. Math. (Oxford) 4, 6-40 (1953), V 5* ChurhilU;.? V'. '"Integral Transforms and Buidary Value Prbl.ems" A"ier. Math. Mnehy 59,.49155 (1952) and Do. p L T for~w Of Tr''cts. -o6 Churc-hill, R. V, and. I,ad Ci L.., 'Inverse Tranfors Of duct8s of Legendre Integral Transforms" (Abstract ), Bul., A:er. Math. Sbc. 8, 634 (1952). ~ '. 7. Churchii, R.. V.,. Fourier Series Mand." o.day Value Pobles McGravl'i., New York,11; i18Yw 8, Ke llgg, 0. D. Po-tential Thery, Sri.' r, Berlin l1929. 99; O.. DWI~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ - Brln 22

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