297 Internat. J. Math. & Math. Sci. VOL. 12 NO. 2 (1989) 297-304 ON THE NUMBER OF SOLUTIONS OF SOME INTEGRAL ECIUATIONS ARISING IN RADIATIVE TRANSFER IOANNIS K. ARGYROS Department of Mathematics New Mexico State University Las Cruces, NM 88003 (Received on March I0, 1988 and in revised form on August 4, 1988) We discuss the number of solutions of some nonlinear integral equations ABSTRACT. neutron transport and in the kinetic arising in the theories of radiative transfer, theory of gases. KEYWORDS AND PHRASES. Radiative transfer, integral equation. 65R20, 45LI0. A.M.S. 1980 CLASSIFICATION CODES: I. In INTRODUCTION. the theories [3], [4] an important role transport of radiative transfer [I] [2] and neutron is played by nonlinear integral equations of the form H(x) The known + function xH(x)/ is (t)H(t) { assumed (I.I) dt. to be nonegative, and measurable on bounded, [0,I], and a positive, continuous solution H of (I.I) is sought. Chandrasekhar’s treatment of (I.I) can be found in [2]. The first proof however of the existence of a solution of (I.I) was given by M. Crum, equation in the complex plane [5]. Crum also showed that if who considered the f (t)dt I/2 then (I.I) has at most two solutions which are bounded in [0,I] and in case 0 (t)dt =I/2 there is only one such solution. in order to prove existence of solutions of are not necessarily solutions of (I.I). (I.I)[I]. C. C. Fox [6] solved simpler equations But the solution of Fox’s equation Stuart [7] gave a nonconstructive existence proof for (I.I) using the Leray-Schauder degree theory but did not discuss the number or location of solutions. Darbo for a set B. Cahlon and M. Eskin [3] used a theorem of contraction map to prove a nonconstructive existence theorem for (1.1). Finally, C. Kelley [8] had solved some interesting generalizations of (I.I) using the solutions of finite rank approximations of solutions of (I.I). I.K. ARGYROS 298 Here we consider the generalized equation: xH(x)o + H(x) The known kernal function (a) 0 < k(x,t) < k(x,t) (1.2) k(x,t)(t)H(t)dt. is a measurable function on [0,I] x [0,I] satisfying [0,I], for all x, t and (b) k(x,t) + k(t,x) for all x, t I/2 (t)dt We show that whenever a minimal solution H can be found using a specific iteration. Flnally, assumptlon, same the under we provide a way of constructing new nonmlnlmal solutions H of (I.I) in terms of the minimal solution. 2. BASIC RESULTS. We denote by C[0,1] the Banach space of all real continuous functions on [0,I] with the maximum norm I" u(t) max Otl We now llst the following well-known theorem whose proof can be found in u [2, pp. 106-107]. If H is a solution of (1.2), then either THEOREM I. or A condition that (Io2) has a solution is tt necessa A function H e C[0,1] satisfies the equation If and only if H satisfies (1.2) and (2.1). Chandrasekr in [2], after proving tt a solution H of (I.I) satisfies either (2.1) or (2.2), clal tt, in fact, H st satisfy (2.1). is claim is not true because as we sh, there always exists a solution H satisfying (2.1), but In ny cases there exists a second solution H satisfying (2.2) and not (2.1). t C[O,I], tt is, if (P2(X) for all x [0,I] and be the natural partial ordering on C[0,1], then pl PI’ P2 P2 if Pl(X) follong: d- D tz- zS {p C[0,1llp(x) ,(t)dt Y d, x [0,11 }, the operator R:D R(p(x) and for d > + p(x) 0, define the operator F f D C[0,1] by k(x,t)(t)p(t)dt, p C[0,1] by D define the INTEGRAL EQUATIONS ARISING IN RADIATIVE TRANSFER /0 k(t’x)(t)(p(t))-Idt’ d + F(p(x)) It is routine to verify that R is isotone, that is if and F is antltone, that is if Finally, denote Pl then P2 F(P2) (respectively, by d) p D. Pl P2 299 then R(Pl) R(P2) F(Pl). the function with constant value (respectively, d). We can now prove the proposition: PROPOSITION. Asstne that the kernel function k(x,t) is as in the introduction and satisfies the condition and some b PROOF. A {p > Ix- ,I 0. I, 2, n is equlcontinuous. Let H be a solution satisfying (1.2) and (2.1) and H}. C[0,1]/I p C[O,I] by A Define Q /0 k(x,t)(t)p(t)dt, Q(p(x)) > Let a Rn(l), Then the sequence (t)H(t)dt 0 < then there exists a, 0 < and < + > @(t)A(t)dt 0. a < [0,11, p x A. I, such that [0,I], Then for x, y = se A) is equicontinuous. Let p A and x [0,1], then < (t)HCt)dt Therefore there exists c, 0 x c < 1. < I, such that Q(p(x)) c for all p A and [o,]. For any Ig(x)- g(y)l The < 1- d function exists 0 < 0 > 60 0, there exists -1 (I- c) g0 if IH[[ R(1) 61 60 is continuous > 0 such that for every g (Q(A), Ix- Yl and < hence 6 (since 0 uniformly Q(A) continuous, such that R((x)) -R((Y))I < 0 f x is -yl < , equlcontlnuous). therefore there I.K. ARGYROS 300 We shall show that the same 61 works for sk(l(x)) -sk (I(Y))I < Rn(1). Set p= ]R(p(x)) Then if if 0 and Rn+l(1) Ix- y[ < 61 if I, 2, for k n. Ix- y] < 61 p(y)Q(p(y))] [p(x)Q(p(x)) R(p(y))] 0 Ip(x)Q(p(x)) p(x)Q(p(y))] p(x)[Q(p(x)) Q(P(Y))! + ]p(x)Q(p(y)) + p(y)Q(p(y))) P(Y)I Q(p(y))]p(x) that is, < ,{ o which completes the induction and the proof of the proposition. TTEOREM. 2. Assume that the kernel function k(x,t) is as in the proposition. Then the following are true: (a) equation (1.2) has exactly one solution H satisfying (2.1) if and only if converges to (2.3) holds. 0, I, 2 Moreover, the increasing sequence Rn(1), n and H; 0, I, 2 (b) if inequality holds in (2.3), the sequence Fn(d), n -I H converges to and ]tt-l(x) IFn Fn(d(x))] F n+l (d(x)) (d(x))], fO If (1.2) has a solution H, then by Theorem I, (A). CASE I. PROOF. (t)dt Assume I/2 It can easily be [0,I]. x $(t)dt I/2 that verified (2.5) since F is antitone: F7(d) F5(d) F3(d) F2(d) F4(d) F6(d) d F(d). Working as in the proposition we can easily show that the bounded set {F(p)/d N is equlcontinuous. and F2n+1(d), n 0, I, 2, have functions v and w respectively. p F2n(d), Then the sequences n convergent F(d)} I, 2, subsequences converging to the From the monotonlcity of the above sequences and the continuity of F we obtain F2n(d) F 2n+l v, (d) w, dvw, F(v) w F(w) v. and The function v has largest ntunber q, 0 < q minimum value I, with qw v. greater than If q zero, I, then that so w v there exists w, that is v w. a INTEGRAL EQUATIONS ARISING IN RADIATIVE TRANSFER < If q I, define on the domain of F the operator F F(p) F l(p) 301 by d. Then d + v d + Fl(W) Fl(q-lv) (I -q)d + q(d + d + qFl(v) (I -q)d + qw Fl(V)) ew + qw-- (e +q)w, for some e > O. But this contradicts the maxlmallty of q. F(v) Therefore, w, v -I H _=v (1.2), satisfying (2.1), and the sequence is a solution of converges to H -I Fn(d), 0, 1, 2, n Inequality (2.5) follows from the fact that F2k(d) -I H CASE 2. Assume that F2k+l(d), ’0 I, 2, 3 for k (t)dt =I/2 ..... let {c }, I, 2 n be a strictly increasing ’sequence of positive numbers converging to 1, and consider the functions c n , n I, 2, 3, Si nce Cn(t)dt =1/2 Cn < 1/2’ fO has a solution Hn for n it follows from Case + H(x) H(x)fo k(x,t)Cn(t)H(t)dt I, 2, 3 n 1, 2, 3, Set M= {p Define F: M > C[0,1)/r p(x) C[0,1] by F(p(x)) fo + Cnf01 k(t,x)(t)dt clf Therefore, there exists r [0,I] Then for each x Cn n that the equation k(t,x) and Cn(t)Hn(t)dt 0 k(t,x)(t)dt 0 such that (H (x)) n I, x hn(X) [0,I]}. -1 r for each x Then H -I k(t,x)(t)(p(t)) -Idt, P each I, 2 M, n n [0,I] and M. It is easy to verify that the set F(M) is bounded and equlcontlnuous. Also, for each n, H Since n F(HI) n [I-2fO Cn(t)dt]I/2+ fO k(t,x) (x) F(M) for each n, some subsequence F(Hn ), j I, 2 .... converges to -I F(Hnj HO F(H 1) converges in C[0,1] to some point I, 2, sequence -1 c n (t)Hn(t)dt I, 2, ), so that H and H -I n. of H 0 that is, F(Hnl), Then the 302 I.K. ARGYROS F(H HO Then H satisfies (1.2), (2.1) and (2.2). 0 Therefore there exists a positive (B). (2.3) Assume R(1), H and Since holds, H satisfying suppose H satisfies (1.2) (2.1) and Rn(1), n R2(1) R3(1) I, 2, n is a convergent subsequence, say R continuity of R that R(h) k h H, and, since the sequence -I (2.1). Rn(1), f k(t,x)0(t)H(t)dt n 0, I, It follows from the h. (t)dt]I/2+ 2f [I- and H. Now h must satisfy either (2.1) H, h must satisfy (2.1). Therefore, for x [0,I], h that is, h (1.2) is uniformly bounded and equicontlnuous there the entire sequence converges to h. is nondecreasing, since 0 and it follows from the fact that R is isotone that I R(1) Since the sequences 2, function satisfies (2.3). whenever or (2.2), and H-l(x), -I H Together with the inequality h H, this implies h H. We have proved that H is the only function satisfying both (1.2) and (2.1), and that the increasing sequence Rn(1), n 0, I, 2, converges to H which completes the proof of the theorem. COROLLARY. [0,I] such that f (t)dt I and 2 are Suppose that l(t) I/2, 2(t) I, 2. i corresponding to $i’ nonnegatlve, bounded, measurable functions on almost everywhere in Let H i [0,I] and such that be the unique solution of equations (1.2) and (2.1) I, 2. i The n, H PROOF. C[0,1] C[0,1], Ri(p(x)) + p(x) Define R i If Pl and H 2 I, 2, by f k(x,t)i(t)p(t)dt are nonnegatlve functions in P2 RI(P 1) R2(P2). i-- I-nce R(1) R(1). Since follows that HI H2. RI(I the R2(1), C[0,1] R21(1) R(1), increasing sequence C[0,1]. p with Pl P2’ then and in general, R(1), converges to H i =I/2 i I, 2, it Note that if it follows from the previous results that the (t)dt function H satisfying (1.2) and (2.1) is the unique solution of (1.2), since, in this case (2.1) and (2.2) reduce the to same equation. However, if 0 (t)dt <I/2 THEOREM 3. As sume: (a) (t)dt equation (1.2) may have two distinct solutions. <I/2and H is the unique solution of (1.2) and (2.1); INTEGRAL EQUATIONS ARISING IN RADIATIVE TRANSFER (b) . 303 the following estimate is true (t) > H(t)dt l-t (2.6) and (c) k(x,t)[ there exists functions 2(x)(1 (1 + -kt) WI(X) + kx I’ + l(t))] > C[0,1] such that W2’ 3 O3(x) --0, for all x, t (0,I], O, for all x i(0) (2.7) [0,1] (2.8) 0 and +l(X))[ @2(x)(H(x)- (1 (HI(x) I)(I +O3(x)H(x)] 1) kx) for all x [0,I] 6 (2.9) where k is the unique number in (0,I) for which 1, and the function H is given by l(x) 1+ H (x) Then H 1-kx is a solution of (2.11) [0,1]. H(x), x (1.2) and (2.2) and Hi(x) > PROOF. (2.10) H(t)dt l-kt H(x), x HI(0) (0,1], H(0). By the monotone convergence theorem $(t) 1-k t lira k+l since (I kt) -1 f H(t)dt increases monotonleally with k, 0 (t 1-t < k < H t )d t I, If (2.6) holds, since I-0.? and since the function f H(t)dt-- (0,I) deflned by f(k) is strictly increasing, )dt] I/2 [I /01 (t) l-kt H(t)dt there exists a unique k Let H be defined as in (2.11). each x e [0,I] (0,i), for which (2.10) holds. Applying a trick used in [5], [9], we find that for ( t )d t 304 I.K. ARGYROS 02(x) 2 (x) [I k(x,t)(t)H(t)dt + ----x-’] + H’x 3 O3(x) (x) (by (2.7) and (2.9)) that is, H satisfies (1.2). Since H must satisfy either (2.1) (2.2) and since Hl(X) > H(x), x [0,I] (by (2.8)), H satisfies (2.2) and the or proof of the theorem is completed. REMARK. By choosing the kernel function k(x,t) to be x k(x, t) x+t x, t [0,I] we observe that the conditions (a) and (b) in the introduction are satisfied and that the equaion (1.2) reduces to equation (I.I). 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