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The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term

The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the... Article The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term Mikhail Sgibnev Sobolev Institute of Mathematics, 630090 Novosibirsk, Russia; sgibnev@math.nsc.ru Abstract: We consider the inhomogeneous Wiener–Hopf equation whose kernel is a nonarithmetic probability distribution with positive mean. The inhomogeneous term behaves like a submultiplica- tive function. We establish asymptotic properties of the solution to which the successive approxima- tions converge. These properties depend on the asymptotics of the submultiplicative function. Keywords: Wiener–Hopf equation; inhomogeneous equation; nonarithmetic probability distribution; positive mean; submultiplicative function; asymptotic behavior MSC: 45E10; 60K05 1. Introduction The classical Wiener–Hopf equation has the form z(x) = k(x y)z(y) dy + g(x), x  0, or, equivalently, z(x) = z(x y)k(y) dy + g(x), x  0. Citation: Sgibnev, M. The Wiener–Hopf Equation with We shall consider the inhomogeneous generalized Wiener–Hopf equation Probability Kernel and Submultiplicative Asymptotics of the z(x) = z(x y) F(dy) + g(x), x  0, (1) Inhomogeneous Term. AppliedMath ¥ 2022, 2, 501–511. https://doi.org/ where z is the function sought, F is a given probability distribution on R, and the inho- 10.3390/appliedmath2030029 mogeneous term g is a known complex function. A probability distribution G on R is Received: 17 August 2022 called nonarithmetic if it is not concentrated on the set of points of the form 0, l, 2l, Accepted: 14 September 2022 . . . (see Section V.2, Definition 3 of [1]). Let R be the set of all nonnegative numbers and Published: 19 September 2022 R := RnR be the set of all negative numbers. For c 2 C, we assume that c/¥ is equal to zero. The relation a(x)  cb(x) as x ! ¥ means that a(x)/b(x) ! c as x ! ¥; if c = 0, Publisher’s Note: MDPI stays neutral then a(x) = o(b(x)). with regard to jurisdictional claims in published maps and institutional affil- iations. Definition 1. A positive function j(x), x 2 R, is called submultiplicative if it is finite, Borel measurable, and satisfies the conditions: j(0) = 1, j(x + y)  j(x) j(y), x, y 2 R. The following properties are valid for submultiplicative functions defined on the Copyright: © 2022 by the author. whole line (Theorem 7.6.2) of [2]: Licensee MDPI, Basel, Switzerland. This article is an open access article log j(x) log j(x) ¥ < r := lim = sup distributed under the terms and x!¥ x x x<0 conditions of the Creative Commons log j(x) log j(x) Attribution (CC BY) license (https://  inf = lim =: r < ¥. (2) x>0 x x!¥ x creativecommons.org/licenses/by/ 4.0/). AppliedMath 2022, 2, 501–511. https://doi.org/10.3390/appliedmath2030029 https://www.mdpi.com/journal/appliedmath AppliedMath 2022, 2 502 Here are some examples of submultiplicative function on R : (i) j(x) = (x + 1) , r > 0; (ii) j(x) = exp(cx ), where c > 0 and 0 < b < 1; and (iii) j(x) = exp(gx), where g 2 R. In (i) and (ii), r = 0, while in (iii), r = g. The product of a finite number of + + submultiplicative function is again a submultiplicative function. In the present paper, we investigate the asymptotic behavior of the solution to Equation (1), where F is a nonarithmetic probability distribution with finite positive mean m := x F(dx) and the function g(x) is asymptotically equivalent (up to a constant fac- tor) to a nondecreasing submultiplicative function j(x) tending to infinity as x ! ¥: g(x)  cj(x) as x ! ¥. In the main theorems (Theorems 2 and 3), j(x), x 2 R , is a nondecreasing submultiplicative function for which there exists lim j(x + y)/j(x) for x!¥ each y 2 R. If such a limit exists, then it is equal to exp(r y). Earlier [3], the asymptotic behavior of z was studied in detail under the following assumptions: (i) m 2 (0, +¥] and (ii) g belong to either g 2 L (0, ¥) or g 2 L (0, ¥). 1 ¥ Roughly speaking, if g 2 L (0, ¥), then z(x) tends to a specific finite limit as x ! ¥. Moreover, under appropriate conditions, a submultiplicative rate of convergence was given in the form o(1/j(x)). If g 2 L (0, ¥), then z(x) = O(x) or even z(x) = f (¥)x/m as x ! ¥, provided f (¥) := lim f (x) exists. x!¥ The existence of the solution to Equation (1) and its explicit form (5) were established in [4] for g 2 L (0, ¥) and arbitrary probability distributions F, regardless of whether F is of oscillating or drifting type. If m = 0 and if some other hypotheses are fulfilled, then z(x) tends to a specific finite limit as x ! ¥ (Theorem 4 of [4]). The stability of an integro-differential equation with a convolution type kernel was studied in [5,6]. 2. Preliminaries Consider the collection S(j) of all complex-valued measures {, such that k{k := j(x)j{j(dx) < ¥; here, j{j stands for the total variation of {. The collection S(j) is a Banach algebra with norm kk by the usual operations of addition and scalar multiplication of measures; the product of two elements n and { of S(j) is defined as their convolution n{ (Section 4.16) of [2]. The unit element of S(j) is the measure d of unit mass concentrated at zero. Define the Laplace transform of a measure { as {b(s) := exp(sx){(dx). It follows from (2) that the Laplace transform of any { 2 S(j) converges absolutely with respect to j{j for all s in the strip P(r , r ) := fs 2 C : r  <s  r g. Let n and { be two complex-valued + + measures on the s-algebra B of Borel sets in R. Their convolution is the measure ZZ Z n{( A) := n(dx){(dy) = n( A x){(dx), A 2 B, fx+y2 Ag provided the integrals make sense; here, A x := fy 2 R : x + y 2 Ag. Denote by F the n-th convolution power of F: 0 1 (n+1) n F := d , F := F, F := F  F, n  1. Let U be the renewal measure generated by F: U := F . n=0 Let X , k  1, be independent random variables with the same distribution F not concentrated at zero. These variables generate the random walk S = 0, S = X + . . . + X , 0 n 1 n n  1. Put T := min n  1 : S  0 . The random variable H := S is called the + n + first weak ascending ladder height. Similarly, T := min n  1 : S < 0 and H := S AppliedMath 2022, 2 503 is the first strong descending ladder height. We have the factorization identity (the symbol E stands for “expectation”). sX T sH T sH + + 1 xE(e ) = 1 E x e 1 E x e , jxj  1, <s = 0. (3) This can easily be deduced from an analogous identity in Section XVIII.3 of [1] for another collection of ladder variables. Denote by F the distributions of the random variables H , respectively. It follows from the identity (3) that d F = (d F ) (d F ). (4) 0 0 0 + ¥ k Let U := F be the renewal measures generated by the distributions F , k=0 respectively. Denote by 1 the indicator of the subset R in R: 1 (x) = 1 for x 2 R and R + R + + + 1 (x) = 0 for x 2 R . Extend the function g onto the whole line: g(x) := 0, x < 0. This convention will be valid throughout. Let n be a measure defined on B, and a(x), x 2 R, a function. Define the convolution n a(x) as the function a(x y) n(dy), x 2 R. The following theorem has been proven in [4]. Theorem 1. Let F be a probability distribution and g 2 L (R ). Then, the function 1 + z(x) = U  (U  g)1 (x), x 2 R , (5) + R + is the solution to Equation (1), which coincides with the solution obtained by successive approxima- tions. If m is finite and positive, then m := x F (dx) is also finite and positive (Sec- + + tion XII.2, Theorem 2 of [1]). We have m = m (1 F (R )), U (R [f0g) = . (6) 1 F (R ) In fact, pass in (4) to Laplace transforms and divide both sides by s. We get b b 1 F(s) 1 F (s) = 1 F (s) , s 6= 0, <s = 0. s s Let s tend to zero. Then, the fractions on both sides will tend to m and m , respectively. The second equality in (6) is a consequence of the fact that the distribution F is defective, i.e., F (R ) < 1. Lemma 1. Let F be a nonarithmetic probability distribution, such that m = x F(dx) 2 (0, ¥) and let j(x), x 2 R, be a submultiplicative function with r  0  r . Assume that j(x)F((¥, x]) dx < ¥. Suppose additionally that F(r ) < 1 if r < 0. Then U 2 S(j). Proof. By Theorem 4 in [7] with n = 1 and Remark 5 therein, we have j(x) F (dx) < ¥, ¥ AppliedMath 2022, 2 504 i.e., F 2 S(j). Let us prove that the element n := d F is invertible in S(j). Let n = n + n + n be the decomposition of n into absolutely continuous, discrete, and c d s singular components. By Theorem 1 of [8], the element n 2 S(j) has an inverse if n(s) 6= 0 for all s 2 P(r , r ), and if 1 2 d d inf nb (s) > max jn j(r ), jn j(r ) . (7) d s s + s2P(r ,r ) c d s Let F = F + F + F be the decomposition of F 2 S(j) into absolutely continuous, d d s s discrete, and singular components. Then, n = d F and n = F . We have c d c d inf nb (s)  1 sup F (s) = 1 F (r ). s2P(r ,r ) s2P(r ,r ) d d c On the other hand, max jn j(r ), jn j(r ) = F (r ). Hence, in order to prove (7), it s s + suffices to show that c d s c c 1 F (r ) F (r )  1 F (r ) > 0. If r = 0, this follows from the fact that the distribution F is defective. Let r < 0. b b By assumption, F(r ) < 1 and, obviously, F (r ) < 1. Relation (4) implies b c c 1 F(s) = 1 F (s) 1 F (s) , s 2 P(r , r ), (8) + + whence 1 F (r ) > 0 and (7) follows. Finally, c c c jnb(s)j  1jF (s)j  1 F (jsj)  1 F (r ) > 0, s 2 P(r , r ). Therefore, by Theorem 1 in [8], the measure d F is invertible in the Banach algebra S(j) and U = (d F ) 2 S(j). The proof of the lemma is complete. Lemma 2. Let a(x), x 2 R , be a monotone nondecreasing positive function. Suppose that lim a(x + y)/a(x) = 1 for each y 2 R. Then, x!¥ a(x) = o a(y) dy as x ! ¥. Proof. Let M > 0 be arbitrary. We have Z Z Z x x x a(y) a(y) a(x M) a(x M) dy  dy  dy = M . a(x) a(x) a(x) a(x) 0 x M x M It follows that lim inf a(y) dy/a(x) = ¥. The proof of the lemma is com- x!¥ plete. Lemma 3. Let G be a nonarithmetic probability distribution on R , such that m := x G(dx) 2 (0, ¥) and let U be the corresponding renewal measure: U := G . Suppose that a(x) and b(x), G G n=0 x 2 R , are nonnegative functions such that a(x)  b(x) as x ! ¥.Then, I(x) := U  a(x)  U  b(x) =: J(x) as x ! ¥. G G Proof. Given # > 0, choose A > 0, such that (1 #)b(x)  a(x)  (1 + #)b(x), x  A. AppliedMath 2022, 2 505 Let Z Z x A x I(x) = + a(x y) U (dy) =: I (x) + I (x). 1 2 0 x A Similarly, let J(x) = J (x) + J (x). Obviously, I (x) I (x) 1 1 1 #  lim inf  lim sup  1 + #. x!¥ J (x) J (x) 1 x!¥ 1 Since # is arbitrary, lim I (x)/ J (x) = 1, i.e., I (x)  J (x) as x ! ¥. Moreover, x!¥ 1 1 1 1 I (x)  a(x A)U ([0, x A]) ! ¥ as x ! ¥ by the elementary renewal theorem for the 1 G measure U : U ([0, x])  x/m as x ! ¥ (see Section 1.2 of [9]). According to Blackwell’s G G G theorem (Section XI.1, Theorem 1 of [1]), I (x)  a( A)U ((x A, x]) ! a( A) A/m as x ! ¥. 2 G G Hence, I(x)  I (x) as x ! ¥. A similar relation also holds for J(x), which completes the proof of the lemma. Lemma 4. Let j(x), x 2 R , be a submultiplicative function, such that there exists n(y) := lim j(x + y)/j(x) for each y 2 R. Then n(y) = exp(r y), y 2 R. x!¥ + Proof. By the Corollary of Theorem 4.17.3 in Section 4.17 of [2], n(y) = exp(ay) for some j(n + 1) a 2 R. Given # > 0, there exists n = n (#), such that log  a + # for n  n . 0 0 0 j(n) m(a+#) Hence, j(n + m)  j(n )e and 0 0 log j(n + m) log j(n ) m(a + #) 0 0 r = lim  lim + lim = a + #. m!¥ m m!¥ m m!¥ m Similarly, r  a #. Since # > 0 is arbitrary, a = r . The proof of the lemma is + + complete. 3. Main Results Theorem 2. Let F be a nonarithmetic probability distribution, such that m = x F(dx) 2 (0, ¥) and let j(x), x 2 R , be a nondecreasing continuous submultiplicative function tending to infinity as x ! ¥, such that r = 0 and there exists lim j(x + y)/j(x) for each y 2 R. Suppose + x!¥ that the inhomogeneous term g(x), x 2 R , is bounded on finite intervals and satisfies the relation g(x)  cj(x) as x ! ¥, where c 2 C. Assume that j(jxj)F((¥, x]) dx < ¥. Then, the function z(x), x 2 R , defined by (5) is a solution to Equation (1) and satisfies the asymptotic relation z(x)  j(y) dy as x ! ¥. Proof. Put M(x) = j(y) dy. By Lemma 4, lim j(x + y)/j(x) = 1 for each y 2 R. x!¥ Extend the function j(x) onto the whole line R by setting j(x) = j(jxj) for x 2 R . The extended function retains the submultiplicative property and r = 0. To prove the AppliedMath 2022, 2 506 first statement of the theorem, it suffices to assume g  0. Choose C > 0, such that g(x)  Cj(x), x 2 R . The function z(x) defined by (5) is finite, since 0+ U  g(x)  CU  j(x) = C j(x y) U (dy)  Cj(x)kU k , z(x)  CkU k j(x y) U (dy)  CkU k j(x)U ([0, x]) < ¥ j + j + for all x 2 R . Let n be a natural number. Denote by 1 the indicator of [0, n]. Consider [0,n] Equation (1) with the inhomogeneous term g (x) = g(x)1 (x). Let z be the solution to n n [0,n] the equation z (x) = z (x y) F(dy) + g (x), x 2 R , (9) n n n + defined by formula (5): z (x) = U  (U  g )1 (x), x 2 R . (10) n + n + The integral in (9) can be written as z (x y)1 (y) F(dy)  z (x)  z(x) < ¥. n n [0,x] The last two inequalities are consequences of (5). Obviously, z (x) " as n ". By Section 27, Theorem B of [10], the integral tends to z(x y) F(dy) as n " ¥. Letting n " ¥ in (9) and (10), we get that z is a solution to (1). Let us prove the assertion of the theorem for the solution z to (1) for g = j. Let us show that U  j(x) ! U (R [f0g) as x ! ¥. (11) j(x) We have U  j(x) j(x y) = U (dy). (12) j(x) j(x) By Lemma 4, the integrand tends to 1 as x ! ¥ and it is majorized by the U - integrable function j(y), since j(x y) j(y) = j(y) j(x) and U 2 S(j) by Lemma 1. Applying Lebesgue’s bounded convergence theorem (Sec- tion 26, Theorem D of [10]), we can pass to the limit under the integral sign in (12), which proves (11). Apply Lemma 3 with the following choice of G, a(x) and b(x): G := F , a(x) := 1 (x)U  j(x), b(x) := U (R [f0g)1 (x)j(x). R R + + We get Z Z x x z (x) = U  j(x y) U (dy)  U (R [f0g) j(x y) U (dy) as x ! ¥. j + + 0 0 Recalling (6), we see that in order to prove the theorem for z , it suffices to establish Z Z x x 1 1 U  (1 j)(x) = j(x y) U (dy)  j(y) dy = M(x) as x ! ¥. (13) + + m m 0 + 0 + AppliedMath 2022, 2 507 Integrating by parts, we get Z Z x x j(x y) U (dy) = j(x y)U ([0, y]) U ([0, y]) d j(x y) + + + y y=0 0 0 = U ([0, x]) j(x) U ([0, y]) d j(x y). (14) + + y The following three estimates hold: j(x), x, U ([0, x]) = o( M(x)) as x ! ¥. (15) The first estimate follows from Lemma 2 with a(x) = j(x). The second one follows from the assumption j(y) ! ¥ as y ! ¥. The third estimate follows from the second one and the elementary renewal theorem for the measure U : U ([0, x])  x/m as x ! ¥. + + + Show that Z Z x x U ([0, y]) d j(x y)  y d j(x y) as x ! ¥, (16) + y y 0 + 0 1 1 y d j(x y)  M(x) as x ! ¥. (17) m m + 0 + We prove first (17). This follows from the second estimate in (15) and the equality Z Z x x y d j(x y) = yj(x y) + j(x y) dy = x + M(x). y=0 0 0 Let # > 0 be arbitrary. Use the elementary renewal theorem and choose y = y (#), 0 0 such that (1 #)U ([0, y])   (1 + #)U ([0, y]), y  y . + + 0 Write the left-hand side of (16) in the form Z Z y x + U ([0, y]) d j(x y) =: K (x) + K (x), + y 1 2 0 y and let M (x) + M (x) be a similar decomposition for the right-hand side. Obviously, 1 2 (1 #) M (x)  K (x)  (1 + #) M (x). (18) 2 2 2 Let us prove that, as x ! ¥, both sides in (16) are asymptotically equivalent to K (x) and M (x), respectively. We have Z Z x x x 1 y 1 M (x) = y d j(x y) = j(x y) + j(x y) dy 2 y m m m y y=y y + + 0 + 0 0 xy x y 1 = + j(x y ) + j(y) dy. m m m + + + Let us show that xy M (x) := j(y) dy  M(x) as x ! ¥. Using the first estimate in (15), we get Z Z x x j(y) dy  j(x) j(y x) dy xy xy 0 0 = j(x) j(y) dy  j(x)j(y )y = o( M(x)) as x ! ¥. 0 0 0 AppliedMath 2022, 2 508 Finally, Z Z x x M (x) 1 = j(y) dy j(y) dy M(X) M(x) 0 xy = 1 j(y) dy = 1 o(1) ! 1 as x ! ¥, M(x) xy which establishes the desired equivalence M (x)  M(x) as x ! ¥. Taking into account the estimates in (15), we see that M (x)  M(x)/m as x ! ¥. Moreover, 2 + y j(x y ) 1 0 0 M (x) = + j(u) du. m m + + xy The integral is estimated by y j(x)/m . Thus, M (x) = o( M(x)) as x ! ¥ (see (15)). 0 1 Relation (17) is proven. Now, divide all parts of (18) by M (x) and let x tend to infinity. We obtain K (x) K (x) 2 2 1 #  lim inf  lim sup  1 + #. x!¥ M (x) M (x) 2 2 x!¥ Hence, K (x)  M (x)  M(x) as x ! ¥. Relation (16) is proven, since, as x ! ¥, 2 2 K (x)  U ([0, y ]) d j(x y) 1 + 0 y = U ([0, y ])[j(x) j(x y )]  U ([0, y ])j(x) = o( M(x)). + + 0 0 0 The equivalence (13) now follows from (14)–(17), which proves the theorem in the particular case g = j. Let g satisfy the hypotheses of the theorem. If, for some C > 0, jg(x)j  Cj(x), x 2 R , then lim supjz(x)j j(y) dy  . x!¥ It follows that if c = 0, then z(x) = o(z (x)) as x ! ¥. To see this, choose a small # > 0 and a natural number n, such that jg(x)j  #j(x), x  n. Write g = 1 g + (g 1 g) =: g + g . [0,n] [0,n] 1 Let z and z be the solutions to (1) corresponding to g and g , respectively. Then, 2 2 1 1 z = z + z and jz (x)j  #z (x), x 2 R . By Theorem 6.2 in [3], z (x) = o(x) as x ! ¥. 1 2 2 j + 1 Since j(x)  1, x 2 R , it follows that z (x) = o j(y) dy as x ! ¥. Therefore, lim supjz(x)j j(y) dy  . x!¥ Since # > 0 is arbitrary, the assertion of the theorem is true for c = 0. Let c 6= 0. Write g in the form g = cj + g . Then, g (x) = o(j(x)) as x ! ¥, and we have z = cz + z , 1 1 j 1 where z is the solution to Equation (1) with the inhomogeneous term g . The proof of the 1 1 theorem is complete. Theorem 3. Let F be a nonarithmetic probability distribution, such that m = x F(dx) 2 (0, ¥), and let j(x), x 2 R , be a nondecreasing submultiplicative function, such that r > 0, and + + there exists lim j(x + y)/j(x) for each y 2 R. Suppose that the inhomogeneous term g(x), x!¥ AppliedMath 2022, 2 509 x 2 R , is bounded on finite intervals and satisfies the relation g(x)  cj(x) as x ! ¥, where c 2 C. Assume that j(jxj)F((¥, x]) dx < ¥ and F(r ) < 1. Then, the function z(x), x 2 R , defined by (5) is a solution to Equation (1) + + and satisfies the asymptotic relation z(x)  j(x) as x ! ¥. 1 F(r ) Proof. As in the proof of the preceding theorem, we verify that z(x) is a solution to (1). First, let us prove the assertion of the theorem for the solution z to (1) corresponding to g = j, i.e., let us prove that, as x ! ¥, z (x) U  j(x y) 1 b b = U (dy) ! U (r )U (r ) = . (19) + + + + j(x) j(x) 0 1 F(r ) Write the integrand in the form U  j(x y) j(x y) I(x, y) := 1 (y) , y 2 R . [0,x] j(x y) j(x) Notice that U  j(x) j(x y) = U (dy) ! U (r ) as x ! ¥. (20) j(x) j(x) r y In fact, j(x y)/j(x) ! e as x ! ¥ by Lemma 4 and, according to Lemma 1, this ratio is majorized by the U -integrable function j(y), y 2 R : Z Z 0 0 U  j(x) j(x y) = U (dy)  j(jyj) U (dy) = kU k < ¥. j(x) j(x) ¥ ¥ Relation (20) now follows from Lebesgue’s bounded convergence theorem. Our further actions are as follows. We will pick out a majorant for the function I(x, y), y 2 R , in the by form Me with b 2 (r , 0). Then, by Lebesgue’s theorem, we pass to the limit under the integral sign in the left-side integral in (19) as x ! ¥, and thus prove relation (19). Put f (x) = log j(x) r x. By hypothesis, we have f (x y) f (x) = log j(x y) log j(x) + r x ! 0 as x ! ¥ (21) for each y 2 R. According to Lemma 1.1 in [11], relation (21) is fulfilled uniformly in y 2 [0, 1]. Hence, j(x y) exp(r y) ! 1 as x ! ¥ j(x) uniformly in y 2 [0, 1]. Choose a small # > 0 such that b := log(1 + #) r < 0. Let N = N(#) > 0 be an integer such that j(x y) exp(r y) 1 + #, x  N, y 2 [0, 1]. j(x) AppliedMath 2022, 2 510 Denote by [x] the integral part of a real number x; i.e., [x] is the maximal integer not exceeding x: x = [x] + J, J 2 [0, 1). For y 2 [l, l + 1], l = 0, . . . , [x] N 1, we have j(x y) j(x l (y l)) j(x l) = , j(x) j(x l) j(x) j(x l (y l)) (1 + #) exp(r (y l)), j(x l) j(x l) j(x l) j(x l + 1) j(x 1) = . . .  (1 + #) exp(lr ). j(x) j(x l + 1) j(x l + 2) j(x) Ultimately, j(x y) l+1 l+1 (1 + #) exp(r (y l)) exp(lr ) = (1 + #) exp(r y) + + + j(x) (1 + #) exp(by), y 2 [l, l + 1], l = 0, . . . , [x] N 1. Now, let y 2 ([x] N 1, x]. We have j(x y) j(N + 2) j(N + 2) j(N + 2) j(N + 2) exp(by). j(x) j(x) exp(r x) exp(r y) + + Thus, the U -integrable majorant sought for the function I(x, y), y 2 R , which does + + not depend on x, is of the form kU k maxf(1 + #), j(N + 2)g exp(by), y 2 R . j + Now, in order to prove relation (19), it suffices, by Lebesgue’s theorem, to pass to the limit under the integral sign in (19). The last equality in (19) is a consequence of (8) for <s = r : 1 1 1 b b b U(s) = = = U (s)U (s), b b b 1 F(s) 1 F (s) 1 F (s) which is admissible, since b b b b jF(s)j  F(r ) < 1, jF (s)j  F (r ) < 1, <s = r . +   + + In the general case, it suffices to repeat the concluding reasoning of the previous proof using the estimate jz(x)j C lim sup j(x) 1 F(r ) x!¥ for jg(x)j  Cj(x), x 2 R , and, considering the case c = 0, take into account the relation r x z (x) = o(x) as x ! ¥ and all the more z (x) = o(j(x)) as x ! ¥, since x  e  j(x), 1 1 x 2 R . 4. Conclusions We have established the asymptotic behavior of the solution z of the generalized Wiener–Hopf Equation (1), where the inhomogeneous term g behaves like an unbounded submultiplicative function, up to a constant factor, i.e., g(x)  cj(x) as x ! ¥. Depending on whether r = 0 or r > 0, there are two different types of asymptotics for z (Theorems 2 + + and 3): either z(x)  c j(y) dy or z(x)  c j(x) as x ! ¥, where c and c are specific 1 2 1 2 constants. Here are two simple examples (c = 1): (i) If j(x) = (x + 1) , r > 0, then r+1 z(x)  as x ! ¥; m(r + 1) AppliedMath 2022, 2 511 (ii) If j(x) = exp(gx), g > 0, then gx z(x)  as x ! ¥. 1 F(r ) Funding: This research received no external funding. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: The work was carried out within the framework of the State Task to the Sobolev Institute of Mathematics (Project FWNF-2022-0004). Conflicts of Interest: The author declares no conflict of interest. References 1. Feller, W. An Introduction to Probability Theory and Its Applications; Wiley: New York, NY, USA, 1966; Volume 2. 2. Hille, E.; Phillips, R.S. Functional Analysis and Semi-Groups; American Mathematical Society Colloquium Publications; American Mathematical Society: Providence, RI, USA, 1957; Volume 31. 3. Sgibnev, M.S. Wiener–Hopf equation whose kernel is a probability distribution. Differ. Equ. 2017, 53, 1174–1196. [CrossRef] 4. Sgibnev, M.S. The Wiener–Hopf equation with probability kernel of oscillating type. Sib. Èlektron. Mat. Izv. 2020, 17, 1288–1298. [CrossRef] 5. Inoan, D.; Marian, D. Semi-Hyers–Ulam–Rassias Stability of a Volterra Integro-Differential Equation of Order I with a Convolution Type Kernel via Laplace Transform. Symmetry 2021, 13, 2181. [CrossRef] 6. Inoan, D.; Marian, D. Semi-Hyers–Ulam–Rassias Stability via Laplace Transform, for an Integro-Differential Equation of the Second Order. Mathematics 2022, 10, 1893. [CrossRef] 7. Sgibnev, M.S. Semimultiplicative moments of factors in Wiener—Hopf matrix factorization. Sb. Math. 2008, 199, 277–290. [CrossRef] 8. Sgibnev, M.S. On invertibilty conditions for elements of Banach algebras of measures. Math. Notes 2013, 93, 763–765. [CrossRef] 9. Smith, W.L. Renewal theory and its ramifications. J. R. Stat. Soc. Ser. B 1958, 20, 243–302. [CrossRef] 10. Halmos, P.R. Measure Theory; Springer: New York, NY, USA, 1974. 11. Seneta, E. Regularly Varying Functions; Springer: Berlin, Germany, 1976. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png AppliedMath Multidisciplinary Digital Publishing Institute

The Wiener&ndash;Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term

Sgibnev, Mikhail
AppliedMath , Volume 2 (3) – Sep 19, 2022
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Article The Wiener–Hopf Equation with Probability Kernel and Submultiplicative Asymptotics of the Inhomogeneous Term Mikhail Sgibnev Sobolev Institute of Mathematics, 630090 Novosibirsk, Russia; sgibnev@math.nsc.ru Abstract: We consider the inhomogeneous Wiener–Hopf equation whose kernel is a nonarithmetic probability distribution with positive mean. The inhomogeneous term behaves like a submultiplica- tive function. We establish asymptotic properties of the solution to which the successive approxima- tions converge. These properties depend on the asymptotics of the submultiplicative function. Keywords: Wiener–Hopf equation; inhomogeneous equation; nonarithmetic probability distribution; positive mean; submultiplicative function; asymptotic behavior MSC: 45E10; 60K05 1. Introduction The classical Wiener–Hopf equation has the form z(x) = k(x y)z(y) dy + g(x), x  0, or, equivalently, z(x) = z(x y)k(y) dy + g(x), x  0. Citation: Sgibnev, M. The Wiener–Hopf Equation with We shall consider the inhomogeneous generalized Wiener–Hopf equation Probability Kernel and Submultiplicative Asymptotics of the z(x) = z(x y) F(dy) + g(x), x  0, (1) Inhomogeneous Term. AppliedMath ¥ 2022, 2, 501–511. https://doi.org/ where z is the function sought, F is a given probability distribution on R, and the inho- 10.3390/appliedmath2030029 mogeneous term g is a known complex function. A probability distribution G on R is Received: 17 August 2022 called nonarithmetic if it is not concentrated on the set of points of the form 0, l, 2l, Accepted: 14 September 2022 . . . (see Section V.2, Definition 3 of [1]). Let R be the set of all nonnegative numbers and Published: 19 September 2022 R := RnR be the set of all negative numbers. For c 2 C, we assume that c/¥ is equal to zero. The relation a(x)  cb(x) as x ! ¥ means that a(x)/b(x) ! c as x ! ¥; if c = 0, Publisher’s Note: MDPI stays neutral then a(x) = o(b(x)). with regard to jurisdictional claims in published maps and institutional affil- iations. Definition 1. A positive function j(x), x 2 R, is called submultiplicative if it is finite, Borel measurable, and satisfies the conditions: j(0) = 1, j(x + y)  j(x) j(y), x, y 2 R. The following properties are valid for submultiplicative functions defined on the Copyright: © 2022 by the author. whole line (Theorem 7.6.2) of [2]: Licensee MDPI, Basel, Switzerland. This article is an open access article log j(x) log j(x) ¥ < r := lim = sup distributed under the terms and x!¥ x x x<0 conditions of the Creative Commons log j(x) log j(x) Attribution (CC BY) license (https://  inf = lim =: r < ¥. (2) x>0 x x!¥ x creativecommons.org/licenses/by/ 4.0/). AppliedMath 2022, 2, 501–511. https://doi.org/10.3390/appliedmath2030029 https://www.mdpi.com/journal/appliedmath AppliedMath 2022, 2 502 Here are some examples of submultiplicative function on R : (i) j(x) = (x + 1) , r > 0; (ii) j(x) = exp(cx ), where c > 0 and 0 < b < 1; and (iii) j(x) = exp(gx), where g 2 R. In (i) and (ii), r = 0, while in (iii), r = g. The product of a finite number of + + submultiplicative function is again a submultiplicative function. In the present paper, we investigate the asymptotic behavior of the solution to Equation (1), where F is a nonarithmetic probability distribution with finite positive mean m := x F(dx) and the function g(x) is asymptotically equivalent (up to a constant fac- tor) to a nondecreasing submultiplicative function j(x) tending to infinity as x ! ¥: g(x)  cj(x) as x ! ¥. In the main theorems (Theorems 2 and 3), j(x), x 2 R , is a nondecreasing submultiplicative function for which there exists lim j(x + y)/j(x) for x!¥ each y 2 R. If such a limit exists, then it is equal to exp(r y). Earlier [3], the asymptotic behavior of z was studied in detail under the following assumptions: (i) m 2 (0, +¥] and (ii) g belong to either g 2 L (0, ¥) or g 2 L (0, ¥). 1 ¥ Roughly speaking, if g 2 L (0, ¥), then z(x) tends to a specific finite limit as x ! ¥. Moreover, under appropriate conditions, a submultiplicative rate of convergence was given in the form o(1/j(x)). If g 2 L (0, ¥), then z(x) = O(x) or even z(x) = f (¥)x/m as x ! ¥, provided f (¥) := lim f (x) exists. x!¥ The existence of the solution to Equation (1) and its explicit form (5) were established in [4] for g 2 L (0, ¥) and arbitrary probability distributions F, regardless of whether F is of oscillating or drifting type. If m = 0 and if some other hypotheses are fulfilled, then z(x) tends to a specific finite limit as x ! ¥ (Theorem 4 of [4]). The stability of an integro-differential equation with a convolution type kernel was studied in [5,6]. 2. Preliminaries Consider the collection S(j) of all complex-valued measures {, such that k{k := j(x)j{j(dx) < ¥; here, j{j stands for the total variation of {. The collection S(j) is a Banach algebra with norm kk by the usual operations of addition and scalar multiplication of measures; the product of two elements n and { of S(j) is defined as their convolution n{ (Section 4.16) of [2]. The unit element of S(j) is the measure d of unit mass concentrated at zero. Define the Laplace transform of a measure { as {b(s) := exp(sx){(dx). It follows from (2) that the Laplace transform of any { 2 S(j) converges absolutely with respect to j{j for all s in the strip P(r , r ) := fs 2 C : r  <s  r g. Let n and { be two complex-valued + + measures on the s-algebra B of Borel sets in R. Their convolution is the measure ZZ Z n{( A) := n(dx){(dy) = n( A x){(dx), A 2 B, fx+y2 Ag provided the integrals make sense; here, A x := fy 2 R : x + y 2 Ag. Denote by F the n-th convolution power of F: 0 1 (n+1) n F := d , F := F, F := F  F, n  1. Let U be the renewal measure generated by F: U := F . n=0 Let X , k  1, be independent random variables with the same distribution F not concentrated at zero. These variables generate the random walk S = 0, S = X + . . . + X , 0 n 1 n n  1. Put T := min n  1 : S  0 . The random variable H := S is called the + n + first weak ascending ladder height. Similarly, T := min n  1 : S < 0 and H := S AppliedMath 2022, 2 503 is the first strong descending ladder height. We have the factorization identity (the symbol E stands for “expectation”). sX T sH T sH + + 1 xE(e ) = 1 E x e 1 E x e , jxj  1, <s = 0. (3) This can easily be deduced from an analogous identity in Section XVIII.3 of [1] for another collection of ladder variables. Denote by F the distributions of the random variables H , respectively. It follows from the identity (3) that d F = (d F ) (d F ). (4) 0 0 0 + ¥ k Let U := F be the renewal measures generated by the distributions F , k=0 respectively. Denote by 1 the indicator of the subset R in R: 1 (x) = 1 for x 2 R and R + R + + + 1 (x) = 0 for x 2 R . Extend the function g onto the whole line: g(x) := 0, x < 0. This convention will be valid throughout. Let n be a measure defined on B, and a(x), x 2 R, a function. Define the convolution n a(x) as the function a(x y) n(dy), x 2 R. The following theorem has been proven in [4]. Theorem 1. Let F be a probability distribution and g 2 L (R ). Then, the function 1 + z(x) = U  (U  g)1 (x), x 2 R , (5) + R + is the solution to Equation (1), which coincides with the solution obtained by successive approxima- tions. If m is finite and positive, then m := x F (dx) is also finite and positive (Sec- + + tion XII.2, Theorem 2 of [1]). We have m = m (1 F (R )), U (R [f0g) = . (6) 1 F (R ) In fact, pass in (4) to Laplace transforms and divide both sides by s. We get b b 1 F(s) 1 F (s) = 1 F (s) , s 6= 0, <s = 0. s s Let s tend to zero. Then, the fractions on both sides will tend to m and m , respectively. The second equality in (6) is a consequence of the fact that the distribution F is defective, i.e., F (R ) < 1. Lemma 1. Let F be a nonarithmetic probability distribution, such that m = x F(dx) 2 (0, ¥) and let j(x), x 2 R, be a submultiplicative function with r  0  r . Assume that j(x)F((¥, x]) dx < ¥. Suppose additionally that F(r ) < 1 if r < 0. Then U 2 S(j). Proof. By Theorem 4 in [7] with n = 1 and Remark 5 therein, we have j(x) F (dx) < ¥, ¥ AppliedMath 2022, 2 504 i.e., F 2 S(j). Let us prove that the element n := d F is invertible in S(j). Let n = n + n + n be the decomposition of n into absolutely continuous, discrete, and c d s singular components. By Theorem 1 of [8], the element n 2 S(j) has an inverse if n(s) 6= 0 for all s 2 P(r , r ), and if 1 2 d d inf nb (s) > max jn j(r ), jn j(r ) . (7) d s s + s2P(r ,r ) c d s Let F = F + F + F be the decomposition of F 2 S(j) into absolutely continuous, d d s s discrete, and singular components. Then, n = d F and n = F . We have c d c d inf nb (s)  1 sup F (s) = 1 F (r ). s2P(r ,r ) s2P(r ,r ) d d c On the other hand, max jn j(r ), jn j(r ) = F (r ). Hence, in order to prove (7), it s s + suffices to show that c d s c c 1 F (r ) F (r )  1 F (r ) > 0. If r = 0, this follows from the fact that the distribution F is defective. Let r < 0. b b By assumption, F(r ) < 1 and, obviously, F (r ) < 1. Relation (4) implies b c c 1 F(s) = 1 F (s) 1 F (s) , s 2 P(r , r ), (8) + + whence 1 F (r ) > 0 and (7) follows. Finally, c c c jnb(s)j  1jF (s)j  1 F (jsj)  1 F (r ) > 0, s 2 P(r , r ). Therefore, by Theorem 1 in [8], the measure d F is invertible in the Banach algebra S(j) and U = (d F ) 2 S(j). The proof of the lemma is complete. Lemma 2. Let a(x), x 2 R , be a monotone nondecreasing positive function. Suppose that lim a(x + y)/a(x) = 1 for each y 2 R. Then, x!¥ a(x) = o a(y) dy as x ! ¥. Proof. Let M > 0 be arbitrary. We have Z Z Z x x x a(y) a(y) a(x M) a(x M) dy  dy  dy = M . a(x) a(x) a(x) a(x) 0 x M x M It follows that lim inf a(y) dy/a(x) = ¥. The proof of the lemma is com- x!¥ plete. Lemma 3. Let G be a nonarithmetic probability distribution on R , such that m := x G(dx) 2 (0, ¥) and let U be the corresponding renewal measure: U := G . Suppose that a(x) and b(x), G G n=0 x 2 R , are nonnegative functions such that a(x)  b(x) as x ! ¥.Then, I(x) := U  a(x)  U  b(x) =: J(x) as x ! ¥. G G Proof. Given # > 0, choose A > 0, such that (1 #)b(x)  a(x)  (1 + #)b(x), x  A. AppliedMath 2022, 2 505 Let Z Z x A x I(x) = + a(x y) U (dy) =: I (x) + I (x). 1 2 0 x A Similarly, let J(x) = J (x) + J (x). Obviously, I (x) I (x) 1 1 1 #  lim inf  lim sup  1 + #. x!¥ J (x) J (x) 1 x!¥ 1 Since # is arbitrary, lim I (x)/ J (x) = 1, i.e., I (x)  J (x) as x ! ¥. Moreover, x!¥ 1 1 1 1 I (x)  a(x A)U ([0, x A]) ! ¥ as x ! ¥ by the elementary renewal theorem for the 1 G measure U : U ([0, x])  x/m as x ! ¥ (see Section 1.2 of [9]). According to Blackwell’s G G G theorem (Section XI.1, Theorem 1 of [1]), I (x)  a( A)U ((x A, x]) ! a( A) A/m as x ! ¥. 2 G G Hence, I(x)  I (x) as x ! ¥. A similar relation also holds for J(x), which completes the proof of the lemma. Lemma 4. Let j(x), x 2 R , be a submultiplicative function, such that there exists n(y) := lim j(x + y)/j(x) for each y 2 R. Then n(y) = exp(r y), y 2 R. x!¥ + Proof. By the Corollary of Theorem 4.17.3 in Section 4.17 of [2], n(y) = exp(ay) for some j(n + 1) a 2 R. Given # > 0, there exists n = n (#), such that log  a + # for n  n . 0 0 0 j(n) m(a+#) Hence, j(n + m)  j(n )e and 0 0 log j(n + m) log j(n ) m(a + #) 0 0 r = lim  lim + lim = a + #. m!¥ m m!¥ m m!¥ m Similarly, r  a #. Since # > 0 is arbitrary, a = r . The proof of the lemma is + + complete. 3. Main Results Theorem 2. Let F be a nonarithmetic probability distribution, such that m = x F(dx) 2 (0, ¥) and let j(x), x 2 R , be a nondecreasing continuous submultiplicative function tending to infinity as x ! ¥, such that r = 0 and there exists lim j(x + y)/j(x) for each y 2 R. Suppose + x!¥ that the inhomogeneous term g(x), x 2 R , is bounded on finite intervals and satisfies the relation g(x)  cj(x) as x ! ¥, where c 2 C. Assume that j(jxj)F((¥, x]) dx < ¥. Then, the function z(x), x 2 R , defined by (5) is a solution to Equation (1) and satisfies the asymptotic relation z(x)  j(y) dy as x ! ¥. Proof. Put M(x) = j(y) dy. By Lemma 4, lim j(x + y)/j(x) = 1 for each y 2 R. x!¥ Extend the function j(x) onto the whole line R by setting j(x) = j(jxj) for x 2 R . The extended function retains the submultiplicative property and r = 0. To prove the AppliedMath 2022, 2 506 first statement of the theorem, it suffices to assume g  0. Choose C > 0, such that g(x)  Cj(x), x 2 R . The function z(x) defined by (5) is finite, since 0+ U  g(x)  CU  j(x) = C j(x y) U (dy)  Cj(x)kU k , z(x)  CkU k j(x y) U (dy)  CkU k j(x)U ([0, x]) < ¥ j + j + for all x 2 R . Let n be a natural number. Denote by 1 the indicator of [0, n]. Consider [0,n] Equation (1) with the inhomogeneous term g (x) = g(x)1 (x). Let z be the solution to n n [0,n] the equation z (x) = z (x y) F(dy) + g (x), x 2 R , (9) n n n + defined by formula (5): z (x) = U  (U  g )1 (x), x 2 R . (10) n + n + The integral in (9) can be written as z (x y)1 (y) F(dy)  z (x)  z(x) < ¥. n n [0,x] The last two inequalities are consequences of (5). Obviously, z (x) " as n ". By Section 27, Theorem B of [10], the integral tends to z(x y) F(dy) as n " ¥. Letting n " ¥ in (9) and (10), we get that z is a solution to (1). Let us prove the assertion of the theorem for the solution z to (1) for g = j. Let us show that U  j(x) ! U (R [f0g) as x ! ¥. (11) j(x) We have U  j(x) j(x y) = U (dy). (12) j(x) j(x) By Lemma 4, the integrand tends to 1 as x ! ¥ and it is majorized by the U - integrable function j(y), since j(x y) j(y) = j(y) j(x) and U 2 S(j) by Lemma 1. Applying Lebesgue’s bounded convergence theorem (Sec- tion 26, Theorem D of [10]), we can pass to the limit under the integral sign in (12), which proves (11). Apply Lemma 3 with the following choice of G, a(x) and b(x): G := F , a(x) := 1 (x)U  j(x), b(x) := U (R [f0g)1 (x)j(x). R R + + We get Z Z x x z (x) = U  j(x y) U (dy)  U (R [f0g) j(x y) U (dy) as x ! ¥. j + + 0 0 Recalling (6), we see that in order to prove the theorem for z , it suffices to establish Z Z x x 1 1 U  (1 j)(x) = j(x y) U (dy)  j(y) dy = M(x) as x ! ¥. (13) + + m m 0 + 0 + AppliedMath 2022, 2 507 Integrating by parts, we get Z Z x x j(x y) U (dy) = j(x y)U ([0, y]) U ([0, y]) d j(x y) + + + y y=0 0 0 = U ([0, x]) j(x) U ([0, y]) d j(x y). (14) + + y The following three estimates hold: j(x), x, U ([0, x]) = o( M(x)) as x ! ¥. (15) The first estimate follows from Lemma 2 with a(x) = j(x). The second one follows from the assumption j(y) ! ¥ as y ! ¥. The third estimate follows from the second one and the elementary renewal theorem for the measure U : U ([0, x])  x/m as x ! ¥. + + + Show that Z Z x x U ([0, y]) d j(x y)  y d j(x y) as x ! ¥, (16) + y y 0 + 0 1 1 y d j(x y)  M(x) as x ! ¥. (17) m m + 0 + We prove first (17). This follows from the second estimate in (15) and the equality Z Z x x y d j(x y) = yj(x y) + j(x y) dy = x + M(x). y=0 0 0 Let # > 0 be arbitrary. Use the elementary renewal theorem and choose y = y (#), 0 0 such that (1 #)U ([0, y])   (1 + #)U ([0, y]), y  y . + + 0 Write the left-hand side of (16) in the form Z Z y x + U ([0, y]) d j(x y) =: K (x) + K (x), + y 1 2 0 y and let M (x) + M (x) be a similar decomposition for the right-hand side. Obviously, 1 2 (1 #) M (x)  K (x)  (1 + #) M (x). (18) 2 2 2 Let us prove that, as x ! ¥, both sides in (16) are asymptotically equivalent to K (x) and M (x), respectively. We have Z Z x x x 1 y 1 M (x) = y d j(x y) = j(x y) + j(x y) dy 2 y m m m y y=y y + + 0 + 0 0 xy x y 1 = + j(x y ) + j(y) dy. m m m + + + Let us show that xy M (x) := j(y) dy  M(x) as x ! ¥. Using the first estimate in (15), we get Z Z x x j(y) dy  j(x) j(y x) dy xy xy 0 0 = j(x) j(y) dy  j(x)j(y )y = o( M(x)) as x ! ¥. 0 0 0 AppliedMath 2022, 2 508 Finally, Z Z x x M (x) 1 = j(y) dy j(y) dy M(X) M(x) 0 xy = 1 j(y) dy = 1 o(1) ! 1 as x ! ¥, M(x) xy which establishes the desired equivalence M (x)  M(x) as x ! ¥. Taking into account the estimates in (15), we see that M (x)  M(x)/m as x ! ¥. Moreover, 2 + y j(x y ) 1 0 0 M (x) = + j(u) du. m m + + xy The integral is estimated by y j(x)/m . Thus, M (x) = o( M(x)) as x ! ¥ (see (15)). 0 1 Relation (17) is proven. Now, divide all parts of (18) by M (x) and let x tend to infinity. We obtain K (x) K (x) 2 2 1 #  lim inf  lim sup  1 + #. x!¥ M (x) M (x) 2 2 x!¥ Hence, K (x)  M (x)  M(x) as x ! ¥. Relation (16) is proven, since, as x ! ¥, 2 2 K (x)  U ([0, y ]) d j(x y) 1 + 0 y = U ([0, y ])[j(x) j(x y )]  U ([0, y ])j(x) = o( M(x)). + + 0 0 0 The equivalence (13) now follows from (14)–(17), which proves the theorem in the particular case g = j. Let g satisfy the hypotheses of the theorem. If, for some C > 0, jg(x)j  Cj(x), x 2 R , then lim supjz(x)j j(y) dy  . x!¥ It follows that if c = 0, then z(x) = o(z (x)) as x ! ¥. To see this, choose a small # > 0 and a natural number n, such that jg(x)j  #j(x), x  n. Write g = 1 g + (g 1 g) =: g + g . [0,n] [0,n] 1 Let z and z be the solutions to (1) corresponding to g and g , respectively. Then, 2 2 1 1 z = z + z and jz (x)j  #z (x), x 2 R . By Theorem 6.2 in [3], z (x) = o(x) as x ! ¥. 1 2 2 j + 1 Since j(x)  1, x 2 R , it follows that z (x) = o j(y) dy as x ! ¥. Therefore, lim supjz(x)j j(y) dy  . x!¥ Since # > 0 is arbitrary, the assertion of the theorem is true for c = 0. Let c 6= 0. Write g in the form g = cj + g . Then, g (x) = o(j(x)) as x ! ¥, and we have z = cz + z , 1 1 j 1 where z is the solution to Equation (1) with the inhomogeneous term g . The proof of the 1 1 theorem is complete. Theorem 3. Let F be a nonarithmetic probability distribution, such that m = x F(dx) 2 (0, ¥), and let j(x), x 2 R , be a nondecreasing submultiplicative function, such that r > 0, and + + there exists lim j(x + y)/j(x) for each y 2 R. Suppose that the inhomogeneous term g(x), x!¥ AppliedMath 2022, 2 509 x 2 R , is bounded on finite intervals and satisfies the relation g(x)  cj(x) as x ! ¥, where c 2 C. Assume that j(jxj)F((¥, x]) dx < ¥ and F(r ) < 1. Then, the function z(x), x 2 R , defined by (5) is a solution to Equation (1) + + and satisfies the asymptotic relation z(x)  j(x) as x ! ¥. 1 F(r ) Proof. As in the proof of the preceding theorem, we verify that z(x) is a solution to (1). First, let us prove the assertion of the theorem for the solution z to (1) corresponding to g = j, i.e., let us prove that, as x ! ¥, z (x) U  j(x y) 1 b b = U (dy) ! U (r )U (r ) = . (19) + + + + j(x) j(x) 0 1 F(r ) Write the integrand in the form U  j(x y) j(x y) I(x, y) := 1 (y) , y 2 R . [0,x] j(x y) j(x) Notice that U  j(x) j(x y) = U (dy) ! U (r ) as x ! ¥. (20) j(x) j(x) r y In fact, j(x y)/j(x) ! e as x ! ¥ by Lemma 4 and, according to Lemma 1, this ratio is majorized by the U -integrable function j(y), y 2 R : Z Z 0 0 U  j(x) j(x y) = U (dy)  j(jyj) U (dy) = kU k < ¥. j(x) j(x) ¥ ¥ Relation (20) now follows from Lebesgue’s bounded convergence theorem. Our further actions are as follows. We will pick out a majorant for the function I(x, y), y 2 R , in the by form Me with b 2 (r , 0). Then, by Lebesgue’s theorem, we pass to the limit under the integral sign in the left-side integral in (19) as x ! ¥, and thus prove relation (19). Put f (x) = log j(x) r x. By hypothesis, we have f (x y) f (x) = log j(x y) log j(x) + r x ! 0 as x ! ¥ (21) for each y 2 R. According to Lemma 1.1 in [11], relation (21) is fulfilled uniformly in y 2 [0, 1]. Hence, j(x y) exp(r y) ! 1 as x ! ¥ j(x) uniformly in y 2 [0, 1]. Choose a small # > 0 such that b := log(1 + #) r < 0. Let N = N(#) > 0 be an integer such that j(x y) exp(r y) 1 + #, x  N, y 2 [0, 1]. j(x) AppliedMath 2022, 2 510 Denote by [x] the integral part of a real number x; i.e., [x] is the maximal integer not exceeding x: x = [x] + J, J 2 [0, 1). For y 2 [l, l + 1], l = 0, . . . , [x] N 1, we have j(x y) j(x l (y l)) j(x l) = , j(x) j(x l) j(x) j(x l (y l)) (1 + #) exp(r (y l)), j(x l) j(x l) j(x l) j(x l + 1) j(x 1) = . . .  (1 + #) exp(lr ). j(x) j(x l + 1) j(x l + 2) j(x) Ultimately, j(x y) l+1 l+1 (1 + #) exp(r (y l)) exp(lr ) = (1 + #) exp(r y) + + + j(x) (1 + #) exp(by), y 2 [l, l + 1], l = 0, . . . , [x] N 1. Now, let y 2 ([x] N 1, x]. We have j(x y) j(N + 2) j(N + 2) j(N + 2) j(N + 2) exp(by). j(x) j(x) exp(r x) exp(r y) + + Thus, the U -integrable majorant sought for the function I(x, y), y 2 R , which does + + not depend on x, is of the form kU k maxf(1 + #), j(N + 2)g exp(by), y 2 R . j + Now, in order to prove relation (19), it suffices, by Lebesgue’s theorem, to pass to the limit under the integral sign in (19). The last equality in (19) is a consequence of (8) for <s = r : 1 1 1 b b b U(s) = = = U (s)U (s), b b b 1 F(s) 1 F (s) 1 F (s) which is admissible, since b b b b jF(s)j  F(r ) < 1, jF (s)j  F (r ) < 1, <s = r . +   + + In the general case, it suffices to repeat the concluding reasoning of the previous proof using the estimate jz(x)j C lim sup j(x) 1 F(r ) x!¥ for jg(x)j  Cj(x), x 2 R , and, considering the case c = 0, take into account the relation r x z (x) = o(x) as x ! ¥ and all the more z (x) = o(j(x)) as x ! ¥, since x  e  j(x), 1 1 x 2 R . 4. Conclusions We have established the asymptotic behavior of the solution z of the generalized Wiener–Hopf Equation (1), where the inhomogeneous term g behaves like an unbounded submultiplicative function, up to a constant factor, i.e., g(x)  cj(x) as x ! ¥. Depending on whether r = 0 or r > 0, there are two different types of asymptotics for z (Theorems 2 + + and 3): either z(x)  c j(y) dy or z(x)  c j(x) as x ! ¥, where c and c are specific 1 2 1 2 constants. Here are two simple examples (c = 1): (i) If j(x) = (x + 1) , r > 0, then r+1 z(x)  as x ! ¥; m(r + 1) AppliedMath 2022, 2 511 (ii) If j(x) = exp(gx), g > 0, then gx z(x)  as x ! ¥. 1 F(r ) Funding: This research received no external funding. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: The work was carried out within the framework of the State Task to the Sobolev Institute of Mathematics (Project FWNF-2022-0004). Conflicts of Interest: The author declares no conflict of interest. References 1. Feller, W. An Introduction to Probability Theory and Its Applications; Wiley: New York, NY, USA, 1966; Volume 2. 2. Hille, E.; Phillips, R.S. Functional Analysis and Semi-Groups; American Mathematical Society Colloquium Publications; American Mathematical Society: Providence, RI, USA, 1957; Volume 31. 3. Sgibnev, M.S. Wiener–Hopf equation whose kernel is a probability distribution. Differ. Equ. 2017, 53, 1174–1196. [CrossRef] 4. Sgibnev, M.S. The Wiener–Hopf equation with probability kernel of oscillating type. Sib. Èlektron. Mat. Izv. 2020, 17, 1288–1298. [CrossRef] 5. Inoan, D.; Marian, D. Semi-Hyers–Ulam–Rassias Stability of a Volterra Integro-Differential Equation of Order I with a Convolution Type Kernel via Laplace Transform. Symmetry 2021, 13, 2181. [CrossRef] 6. Inoan, D.; Marian, D. Semi-Hyers–Ulam–Rassias Stability via Laplace Transform, for an Integro-Differential Equation of the Second Order. Mathematics 2022, 10, 1893. [CrossRef] 7. Sgibnev, M.S. Semimultiplicative moments of factors in Wiener—Hopf matrix factorization. Sb. Math. 2008, 199, 277–290. [CrossRef] 8. Sgibnev, M.S. On invertibilty conditions for elements of Banach algebras of measures. Math. Notes 2013, 93, 763–765. [CrossRef] 9. Smith, W.L. Renewal theory and its ramifications. J. R. Stat. Soc. Ser. B 1958, 20, 243–302. [CrossRef] 10. Halmos, P.R. Measure Theory; Springer: New York, NY, USA, 1974. 11. Seneta, E. Regularly Varying Functions; Springer: Berlin, Germany, 1976.

Journal

AppliedMath – Multidisciplinary Digital Publishing Institute

Published: Sep 19, 2022

Keywords: Wiener–Hopf equation; inhomogeneous equation; nonarithmetic probability distribution; positive mean; submultiplicative function; asymptotic behavior

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