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МАТЕМАТИКА И МЕХАНИКА
DOI: https://doi.Org/10.15688/mpcm.jvolsu.2019.1.1
UDC 517.442 LBC 511
THE REGULARITY OF THE LAPLACE TRANSFORM
Andrey Valerianovich Pavlov
Candidate of Physical and Mathematical Sciences, Associate Professor, Department of Higher Mathematics-1, MIREA — Russian Technological University a_pavlov@mirea.ru
Prosp. Vernadskogo, 78, 119454 Moscow, Russian Federation
Abstract. The paper proves the regularity of the double Laplace transform in the neighborhood of zero. The class of the transform of Laplace from the transform of Fourier is considered from the functions without a regularity in null.
Key words: Transform of Fourier, transform of Laplace, regularity of the double transform of Laplace, regularity of the transform of Laplace from the transform Fourier.
In the memory of B.V. Gnedenko and A.D. Solovjev 1. Introduction
We consider the regularity of the double transform of Laplace (the theorem 1, the remarks 1, 2). The theorem proved in this part have a general-mathematical character and are easily checked up. With help of the theorem 1 and the remark 2 it is simply to prove a 2 some theorems related with the transform of Fourier and Laplace [3-6] (for instance, about S the inverse operator of the transform of Laplace, using only positive values of the transform >r of Laplace on the [0, [5]). The theorems are not by the theme of the the article and require the separate study in connection with the theorem 1. Some results in the direction is were formulated in the works [4-7].
£ The fact about double decomposition on the elementary fractions is considered conclu-@ sion. In opinion of author the fact underlines interest to the remark 2.
By definition,
L±Z(t)(x) = e±xtZ(t)dt,x G [0, ж),
(we will use L±Z(t)(^)(x) = L±Z(t)(x) too),
oo
F±u(t)(^)(p) = j e±pitu(t)dt,p e (-w, w),L+ = L,
—o
o o
C°u(t)(^)(x) = J cosxtu(t)dt, S°u(t)(^)(x) = j sinxtu(t)dt,x e (-w, w), 0 0
o
F±u(t)(^)(p) = j e±pitu(t)dt,p e (-w, w). 0
2. The regularity of Laplace transform in |z| < a > 0
In the section we use the Y1 condition.
The Y1 condition takes place for the u(p) function, if the u(p) function is regular for all p without only k points z\,... ,Zk, Zj e (-w, iw, iw),k = 0,1,..., u(0) = 0,
and
max[|«(p)|, ldu(p)/dpl, ld2u(p)/p2|]|p2+5| ^ 0, |p| ^ w,
6 > 0, 6 = const.
We use the Ch1 condition too. Ch1 condition.
The u(p) function is regular in K++ = {p : Imp > 0^ P|{p : Rep > 0} or in K+— = {p : Imp < 0}f]{p : Rep > 0}. Theorem 1. The functions
oo oo
LF0+ u(x)t)(z) = j e—ztdtj eitxu(x)dx = iLLu(x)^)(iz), LLu(x)(^)(z), 00
are regular in the area {z : |z| < e} for some e > 0, if for the u(p) function the Y1 condition takes place, and Reu(t) = u(t),t e [0,<x).
Proof. We can use the proposition 1 [5; 6]. Proposition 1. The equalities
LF0+ u(x)(^)(v) = iF— Lu(x)(^)(v),v e [0, w),
LC0u(x)(^)(v) = S0Lu(x)t)(v),LS0u(x)(^)(v) = C0Lu(x)(^)(v),v e [0, w)
take place, if for the function u(p) the Y1 condition takes place. The similar equality
LF°u(x)(^)(v) = -iF^Lu(x)(^)(v),v e [0, w) takes place too.
Proof. We get the first formula after the change of order of integration in both parts of the first equality. If u(0) = 0, it is obviously with help of the expressions
IF0u(x)^)(t)l < l(du(x)/dxlx=0)/t2l + l(l/t2)F0(d2u(x)/dx2)(-)(i)| < Ci/t2,t ^ m, cl = const, cl < m (see [8]).
With help of the proposition 1 we obtain, that the F0Lu(x)(^)(p) = l_(p) function is defined for all Imp < 0, and
lim F0Lu(x)()(p) = F0Lu(x)(^)(iy),y E (-m, m),
p—yi y,lm p<0
(it is obviously, if u(0) = 0; as in the proposition 1 we use the formula of integration on parts [8]).
Similar facts take place for a similar function l+ (p) = F0Lu(x)(^)(p); the function is definite from other side of plane Imp > 0.
We suppose u(-p) = -u(p).
We can write
F0Lu(x)()(p) + F0Lu(x)^)(p) = 2C°Lu(x)t)(p) = F(p), p = y, y E [0, m), if u(-p) = -u(p), or
l- (p) + l+(p) = F (p),
where F(jp) are regular in {p : | Im rpl < ^^ UIp : I Re Pi < A] for some A > 0, if function u(p) is regular as in the Y1 condition (the fact is well-known [2;5;6]).
To prove the fact for p = y E (-m, 0] we can define a new functions 1(p),l(p):
li0+ (p) = l- (p), Im p < 0, l = l+(P), Im p > 0,
where l- + (p) is an analytical continuation of the l_(pi), Im p < 0 function from the lower part of plane to the overhead part of plane {p : Imp > 0]; 1(p) is an analytical continuation of the l+ (p), Im p > 0 function from from the overhead part of plane to the lower part of plane {p : Imp < 0] [2].
The equality 1(p)+l+ (p) = F(p),Imp > 0 repeats the main equality l_(p)+1+ (p) = = F(p), but in the {p : Im p > 0] area; the equality l_(p) + ll°~(p) = F(p) repeats the main equality l_(p) +1+(p) = F(p), but in the {p : Imp < 0] area, where the F(p) function is regular in {p : | Imp | < A]\J{p : | Rep | < A] for the u(-p) = -u(p) function [2]. We obtain, that
t°+(p) + l+ (p) = l-(p) + l'r(p),p = y E [0, m).
But the same equality takes place and for the p = y E (-m, 0] (we use, that both functions l_+(p) + l+(p) = F(p),l~(p) + l(p) = F(p) are equal to the regular F(p) function in different parts of the plane for the u(-p) = -u(p) function [2]).
We get
/i0+ (p) +1+ (p) = I-(p) +1(p),p = У e Ж).
it i l
The functions l(p),l+ (p),l~(p),l(p) are the transforms of Laplace in area of definition, and the functions are regular in area of definition [2] (from the proposition 1) with values on the boundary. The l(p),l(p)) functions are regular in the area of regularity of the sums from the the lemma 1.
Lemma 1. 1. The function LF0u(x)(^)(p) = iF-Lu(x)(^)(p) = il-(p) = ilt-+ (p) is regular for all [p : Imp > 0} (we consider the branch of the function, passing through [p : p = = iy,y = Imp > 0} and [p : p = x,x = Rep > 0}, where the F0Lu(x)(^)(p)) values not defined), if for the u(p) function or for the Vi(p) functions the Y1,Ch1 conditions take place i = 1,2, u(p) = V1(p) + V2(p) (see the part 2 of the remark 2 too).
2. The function LF0u(x)(^)(p) = —iF0Lu(x)(^)(p) = —il+(p) = (p) is regular for all [p : Imp < 0} (we consider the branch of the function, passing through [p : p = = iy,y = Imp < 0} and [p : p = x,x = Rep > 0}, where the F0Lu(x)(^)(j>)) values not defined), if for the u(p) function or for the Vi(p) functions the Y1, Ch1 conditions take place i = 1,2, u(p) = V1(p) + V2(p) (see the part 2 of the remark 2 too).
Proof. If the u(p) function is regular in K++ = [p :Imp > 0} P|[p : Re p > 0}, after integration along the line L+ = L^L2[jL3 of the (1/ip + z)u(z) function anticlockwise , L1 = [0, R], L2 = [p : p = Reiq, 0 < q < n/2}, L3 = [iR, 0], we obtain, that
I(p) = I-(p) = F0Lu(x)^)(p) = J (1/ip + x)u(x)dx =
o
oo
= j(1/ip + ix1)u(ix1)dix1 = (1/i) L Lu(ix1)(^)(p),p G (0, o
as for [p : Rep G (—<x, 0)} so as for all [p : Imp > 0}. We use the Y1, Ch1 conditions (the u(p) function is regular in K++ = [p : Imp > 0} P|[p : Rep > 0}; for Imp =0 we use the u(0) = 0 condition for proof of continuity on the (—<x, <x) axis with help of the proposition 1 [8]).
We obtain, that the such sum is regular for all [p :Imp > 0, Re ip < 0}, and the function
oo
J(1/ip + ix1)u(ix1 )dix1 = (1/i)LLu(x)(^)(p),Rep G 0) o
is regular in [p :Im p > 0, Re ip < 0} with the values on the boundary line [p : Im p = = 0, Re p < 0} (with help of the formula of integration on parts as in proposition 1 [2; 8]).
For the function F0Lu(x)(^)(p) = l+(p) = I (p), Imp < 0 we use I (p) = l- + (p), as the branches of the functions l+(p) = l~(p) [2] (with help of the formula F0Lu(x)(^)(p) = = F0Lu(x)(^)(p) on the [0, line) by the theorem of Riemann about the analytical continuation across the (—<x, <x) line [2], and the function I+(p) = 1(p) = (p)(p) is defined and regular in [p : Imp < 0} (or for the u(p) = V1(p) + V2(p) function in connection with the part 2 of the remark 2 too) .
If the function u(p) is regular in K+- = [p : Im p < 0} P|[p : Re p > 0}, after integration along the line L+ = L1\}L2^L3* of the (1/ip + z)u(z) function anticlockwise, L1 = [0, R], L2* = [p :p = Reiq, —n/2 < q < 0}, L3* = [—iR, 0], we obtain, that
oo
F0Lu(x)(^)(p) = l+(p) = — j (1/ip — x)u(x) dx =
o
= - J (1/ip - ixl)u(ixl)dixl = o
+^
= J (1/ip + ix2)u(—ix2)dix2 = LLu(—ixl)(^)(p),p E (0, +m). o
The further proof of lemma repeats proof of the first part (we use, that Re u(t) = = u(t),t E [0,m), but u(p) = Vl(p) + V2(p)).
With help of lemma 1 we get
\l- + (p) + l+ (p^ < C = const, Im p > 0; l_(p) + l(p) < C = const, Im p < 0,
C < m. Both sums ll°+ (p) + l+(p), l-(p) + ll°~(p) are regular in area of definition and continuous on the boundary [2; 8]. We proved
+ (p) + l+ (p) = C, C = const, C < m
for all p (see [2]) and 2C0Lu(x)(^)(p) = F(p) = ll°+ (p) + l+ (p) = Cl = const, Cl < m [2; 6] or 1(p) = —l+ (p) for all p including p E (—m, m).
We obtain, that the F(p) function is regular [2] with the values 2C0Lu(x)(^)(p) = = F(p),p E (—m, m).
We can use, that the function LF0u(x)(^)(p) = iF0Lu(x)(^)(p) is regular in
{p : | Re p| < e] : | Im p| < e]
for some e > 0. It is well-known fact [2; 5; 6] in Y1 condition for the function u(p). We proved, that
F0Lu(x)^)(p) = F(p) - F0Lu(x)^)(p)
is the analytical continuation from from one side of plane on other [2]. For the u(-p) = u(p) we can use
F0Lu(x)^)(p) - F0Lu(x)t)(p) = 2iS0Lu(x)(^)(p) = F(p),p = y,y E [0, m),
further by analogy with the first part (with help of the lemma 1 and the u(0) = 0 condition). The theorem 1 is proved.
From the theorem 1 we obtain the remark 1.
Remark 1. The theorem 1 takes place for the functions u(p) = v(p) + v(-p), u(p) = v(p) - v(-p), if all the essential points [2] of the v(p) function are placed or in K+ + (or all the essential points are in K_ +).
From the remark 1 we get the first part of the remark 2.
The second part is easily proved by the methods of the work [1;6].
Remark 2.
1) The theorem 1 takes place for the function u(p) = (vl(p)-vl(-p)) + (v2(p)-v2(-p)), if the essential points [2] of the function Vi(p) are placed or in K++ or in K_+, i = 1, 2, and for the u(p) functions the conditions of the theorem 1 take place.
2) The function u(-p) = —u(p) can be presented in the form u(p) = (vl(p) — vl(—p)) + + {v2{p) — v2{—p)), if for the function u(p) the Y1 condition takes place.
We will get a similar result, if to apply the zl = 1/z inversion and the w = e%^zl, cp = = n/4, function; we use, that the same result we obtain for the functions in the reverse order with cp = —n/2 (by analogy for —
3. Conclusion
We will mark the fact about double decomposition on the elementary fractions:
r 1 1,1 1 Pi—T ] = —T +
p — 1 p +1 p — 1 p +1' p p 11
(p -1)2 p2 -1 (p -1)2 p2 -1'
p = 1,-1. The fact in opinion of author underlines interest to the theorem 1 and to the consequences of the theorem 1.
Probably, the equality IC0S0u(t)(-)(x)l = IS0C0u(t)(-)(x)l ensues from theorem 1 and remark 2, x E (0, ro).
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РЕГУЛЯРНОСТЬ ПРЕОБРАЗОВАНИЯ ЛАПЛАСА
Андрей Валерианович Павлов
Кандидат физико-математических наук, доцент кафедры высшей математики-1,
МИРЭА — Российский технологический университет
a_pavlov@mirea.ru
просп. Вернадского, 78, 119454 г. Москва, Российская Федерация
Аннотация. Доказана регулярность двойного преобразования Лапласа в окрестности нуля. Рассматривается класс функций с нарушением регулярности в нуле, преобразование Лапласа от преобразования Фурье от которых регулярно в окрестности нуля.
Ключевые слова: преобразование Фурье, преобразование Лапласа, регулярность двойного преобразования Лапласа, регулярность преобразования Лапласа от преобразования Фурье.