Citation: Najeeb Alam Khan, Samreen Ahmad. Framework for treating non-Linear multi-term fractional differential equations with reasonable spectrum of two-point boundary conditions[J]. AIMS Mathematics, 2019, 4(4): 1181-1202. doi: 10.3934/math.2019.4.1181
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It is well known that the convexity [1,2,3,4,5,6,8,9,11,15,16,40,55,63,64], monotonicity [7,12,13,14,41,42,43,44,45,46,47,48,49,50,51,52,53] and complete monotonicity [58,59,61,62] have widely applications in many branches of pure and applied mathematics [19,24,28,32,35,38,65]. In particular, many important inequalities [20,25,30,33,37,39,69] can be discovered by use of the convexity, monotonicity and complete monotonicity. The concept of complete monotonicity can be traced back to 1920s [18]. Recently, the complete monotonicity has attracted the attention of many researchers [23,34,56,67] due to it has become an important tool to study geometric function theory [26,31,36], its definition can be simply stated as follows.
Definition 1.1. Let I⊆R be an interval. Then a real-valued function f:I↦R is said to be completely monotonic on I if f has derivatives of all orders on I and satisfies
(−1)nf(n)(x)≥0 | (1.1) |
for all x∈I and n=0,1,2,⋯.
If I=(0,∞), then a necessary and sufficient condition for the complete monotonicity can be found in the literature [54]: the real-valued function f:(0,∞)↦R is completely monotonic on (0,∞) if and only if
f(x)=∫∞0e−xtdα(t) | (1.2) |
is a Laplace transform, where α(t) is non-decreasing and such that the integral of (1.2) converges for 0<x<∞.
In 1997, Alzer [10] studied a class of completely monotonic functions involving the classical Euler gamma function [21,22,60,66,68] and obtained the following result.
Theorem 1.1. Let n≥0 be an integer, κ(x) and fn(x) be defined on (0,∞) by
κ(x)=lnΓ(x)−(x−12)lnx+x−12ln(2π) | (1.3) |
and
fn(x)={κ(x)−n∑k=1B2k2k(2k−1)x2k−1,n≥1,κ(x),n=0, | (1.4) |
where Bn denotes the Bernoulli number. Then both the functions x↦f2n(x) and x↦−f2n+1(x) are strictly completely monotonic on (0,∞).
In 2009, Koumandos and Pedersen [27] first introduced the concept of completely monotonic functions of order r. In 2012, Guo and Qi [17] proposed the concept of completely monotonic degree of nonnegative functions on (0,∞). Since the completely monotonic degrees of many functions are integers, in this paper we introduce the concept of the completely monotonic integer degree as follows.
Definition 1.2. Let f(x) be a completely monotonic function on (0,∞) and denote f(∞)=limx→∞f(x). If there is a most non-negative integer k (≤∞) such that the function xk[f(x)−f(∞)] is completely monotonic on (0,∞), then k is called the completely monotonic integer degree of f(x) and denoted as degxcmi[f(x)]=k.
Recently, Qi and Liu [29] gave a number of conjectures about the completely monotonic degrees of these fairly broad classes of functions. Based on thirty six figures of the completely monotonic degrees, the following conjectures for the functions (−1)mR(m)n(x)=(−1)m[(−1)nfn(x)](m)=(−1)m+nf(m)n(x) are shown in [29]:
(ⅰ) If m=0, then
degxcmi[Rn(x)]={0,if n=01,if n=12(n−1),if n≥2; | (1.5) |
(ⅱ) If m=1, then
degxcmi[−R′n(x)]={1,if n=02,if n=12n−1,if n≥2; | (1.6) |
(ⅲ) If m≥1, then
degxcmi[(−1)mR(m)n(x)]={m−1,if n=0m,if n=1m+2(n−1),if n≥2. | (1.7) |
In this paper, we get the complete monotonicity of lower-order derivative and lower-scalar functions (−1)mR(m)n(x) and their completely monotonic integer degrees using the Definition 1.2 and a common sense in Laplace transform that the original function has the one-to-one correspondence with the image function, and demonstrated the correctness of the existing conjectures by using a elementary simple method. The negative conclusion to the second clause of (1.7) is given. Finally, we propose some operational conjectures which involve the completely monotonic integer degrees for the functions (−1)mR(m)n(x) for m=0,1,2,⋯.
In order to prove our main results, we need several lemmas and a corollary which we present in this section.
Lemma 2.1. If the function xnf(x) (n≥1) is completely monotonic on (0,∞), so is the function xn−1f(x).
Proof. Since the function 1/x is completely monotonic on (0,∞), we have xn−1f(x)=(1/x)[xnf(x)] is completely monotonic on (0,∞) too.
Corollary 2.1. Let α(t)≥0 be given in (1.2). Then the functions xi−1f(x) for i=n,n−1,⋯,2,1 are completely monotonic on (0,∞) if the function xnf(x) (n∈N) is completely monotonic on (0,∞).
The above Corollary 2.1 is a theoretical cornerstone to find the completely monotonic integer degree of a function f(x). According to this theory and Definition 1.2, we only need to find a nonnegative integer k such that xkf(x) is completely monotonic on (0,∞) and xk+1f(x) is not, then degxcmi[f(x)]=k.
The following lemma comes from Yang [57]:
Lemma 2.2. Let fn(x) be defined as (1.4). Then fn(x) can be written as
fn(x)=14∫∞0pn(t2)e−xtdt, | (2.1) |
where
pn(t)=cothtt−n∑k=022kB2k(2k)!t2k−2. | (2.2) |
Lemma 2.3. Let m,r≥0, n≥1, fn(x) and pn(t) be defined as (2.1) and (2.2). Then
xr(−1)mR(m)n(x)=xr(−1)m+nf(m)n(x)=14∫∞0[(−1)ntmpn(t2)](r)e−xtdt. | (2.3) |
Proof. It follows from (2.1) that
x(−1)mR(m)n(x)=x(−1)m+nf(m)n(x)=x(−1)m+n14∫∞0(−t)mpn(t2)e−xtdt=x(−1)n14∫∞0tmpn(t2)e−xtdt=(−1)n−114∫∞0tmpn(t2)de−xt=(−1)n−114{[tmpn(t2)e−xt]∞t=0−∫∞0[tmpn(t2)]′e−xtdt}=(−1)n14∫∞0[tmpn(t2)]′e−xtdt. |
Repeat above process. Then we come to the conclusion that
xr(−1)mR(m)n(x)=(−1)n14∫∞0[tmpn(t2)](r)e−xtdt, |
which completes the proof of Lemma 2.3.
In recent paper [70] the reslut degxcmi[R1(x)]=degxcmi[−f1(x)]=1 was proved. In this section, we mainly discuss degxcmi[R2(x)] and degxcmi[R3(x)]. Then discuss whether the most general conclusion exists about degxcmi[Rn(x)].
Theorem 3.1. The function x3R2(x) is not completely monotonic on (0,∞), and
degxcmi[R2(x)]=degxcmi[f2(x)]=2. |
Proof. Note that the function x2R2(x) is completely monotonic on (0,∞) due to
x2R2(x)=14∫∞0p(2)2(t2)e−xtdt,p2(t)=cothtt+145t2−1t2−13,p′2(t)=245t−1tsinh2t+2t3−1t2coshtsinht,p′′2(t)=45A(t)+180t2B(t)+t4C(t)90t4sinh3t>0, |
where
A(t)=tcosh3t−3sinh3t+9sinht−tcosht=∞∑n=52(n−4)(32n−1)(2n+1)!t2n+1>0,B(t)=tcosht+sinht>0,C(t)=sinh3t−3sinht>0. |
So degxcmi[R2(x)]≥2.
On the other hand, we can prove that the function x3f2(x)=x3R2(x) is not completely monotonic on (0,∞). By (2.3) we have
x3f2(x)=14∫∞0p(3)2(t2)e−xtdt, |
then by (1.2), we can complete the staged argument since we can verify
p(3)2(t2)>0⟺p(3)2(t)>0 |
is not true for all t>0 due to
p′′′2(t)=2tsinh2t−2t3sinh2t+4t3+24t5−6tcosh2tsinh4t−4t3cosh2tsinh2t−2t2cosh3tsinh3t+2t2coshtsinht−4t2coshtsinh3t−6t4coshtsinht |
with p′′′2(10)=−0.00036⋯.
Theorem 3.2. The function x4R3(x) is completely monotonic on (0,∞), and
degxcmi[R3(x)]=degxcmi[−f3(x)]=4. |
Proof. By (2.3) we obtain that
x4R3(x)=∫∞0[−p(4)3(t2)]e−xtdt. |
From (2.2) we clearly see that
p3(t)=cothtt−1t2+145t2−2945t4−13, |
p(4)3(t)=:−1630H(t)t6sinh5t, |
or
−p(4)3(t)=1630H(t)t6sinh5t, |
where
H(t)=(2t6+4725)sinh5t−945tcosh5t−(1260t5+3780t3−2835t)cosh3t−(10t6+2520t4+3780t2+23625)sinh3t−(13860t5−3780t3+1890t)cosht+(20t6−7560t4+11340t2+47250)sinht:=∞∑n=5hn(2n+3)!t2n+3 |
with
hn=2125[64n6+96n5−80n4−120n3+16n2−2953101n+32484375]52n−1027[64n6+96n5+5968n4+105720n3+393136n2+400515n+1760535]32n+20[64n6+96n5−11168n4−9696n3+9592n2+13191n+7749]>0 |
for all n≥5. So x4R3(x) is completely monotonic on (0,∞), which implies degxcmi[R3(x)]≥4.
Then we shall prove x5R3(x)=−x5f3(x) is not completely monotonic on (0,∞). Since
x5R3(x)=∫∞0[−p(5)3(t2)]e−xtdt, |
and
−p(5)3(t)=14K(t)t7sinh6t, |
where
K(t)=540cosh4t−1350cosh2t−90cosh6t−240t2cosh2t+60t2cosh4t+80t4cosh2t+40t4cosh4t+208t6cosh2t+8t6cosh4t−120t3sinh2t+60t3sinh4t+200t5sinh2t+20t5sinh4t+75tsinh2t−60tsinh4t+15tsinh6t+180t2−120t4+264t6+900. |
We find K(5)≈−2.6315×1013<0, which means −p(5)3(5)<0. So the function x5R3(x)=−x5f3(x) is not completely monotonic on (0,∞).
In a word, degxcmi[R3(x)]=degxcmi[−f3(x)]=4.
Remark 3.1. So far, we have the results about the completely monotonic integer degrees of such functions, that is, degxcmi[R1(x)]=1 and degxcmi[Rn(x)]=2(n−1) for n=2,3, and find that the existing conclusions support the conjecture (1.5).
In this section, we shall calculate the completely monotonic degrees of the functions (−1)mR(m)n(x), where m=1 and 1≤n≤3.
Theorem 4.1 The function −x2R′1(x)=x2f′1(x) is completely monotonic on (0,∞), and
degxcmi[(−1)1R′1(x)]=2. |
Proof. By the integral representation (2.3) we obtain
x2f′1(x)=14∫∞0[−tp1(t2)]′′e−xtdt. |
So we complete the proof of result that x2f′1(x) is completely monotonic on (0,∞) when proving
[−tp1(t2)]′′>0⟺[tp1(t2)]′′<0⟺[tp1(t)]′′<0. |
In fact,
tp1(t)=t(cothtt−1t2−13)=coshtsinht−13t−1t,[tp1(t)]′=1t2−cosh2tsinh2t+23,[tp1(t)]′′=2sinh3t[cosht−(sinhtt)3]<0. |
Then we have degxcmi[(−1)1R′1(x)]≥2.
Here −x3R′1(x)=x3f′1(x) is not completely monotonic on (0,∞). By (2.2) and (2.3) we have
x3f′1(x)=14∫∞0[−tp1(t2)]′′′e−xtdt, |
and
[−tp1(t)]′′′=23t4cosh2t−t4sinh2t−3sinh4tt4sinh4t |
with [−tp1(t)]′′′|t=2≈−3.6237×10−2<0.
Theorem 4.2. The function −x3R′2(x) is completely monotonic on (0,∞), and
degxcmi[−R′2(x)]=degxcmi[−f′2(x)]=3. |
Proof. First, we can prove that the function −x3R′2(x) is completely monotonic on (0,∞). Using the integral representation (2.3) we obtain
−x3f′2(x)=14∫∞0[tp2(t2)](3)e−xtdt, |
and complete the proof of the staged argument when proving
[tp2(t2)](3)>0⟺[tp2(t)](3)>0. |
In fact,
p2(t)=cothtt+145t2−1t2−13,L(t):=tp2(t)=coshtsinht−13t−1t+145t3, |
L′′′(t)=−180cosh2t+45cosh4t−124t4cosh2t+t4cosh4t−237t4+13560t4sinh4t=160t4sinh4t[∞∑n=322n+2bn(2n+4)!t2n+4]>0, |
where
bn=22n(4n4+20n3+35n2+25n+2886)−4(775n+1085n2+620n3+124n4+366)>0 |
for all n≥3.
On the other hand, by (2.3) we obtain
x4(−1)1R′2(x)=−x4f′2(x)=14∫∞0[tp2(t2)](4)e−xtdt, |
and
L(4)(t)=[tp2(t)](4)=16coshtsinht−40cosh3tsinh3t+24cosh5tsinh5t−24t5 |
is not positive on (0,∞) due to L(4)(10)≈−2.3993×10−4<0, we have that −x4R′2(x) is not completely monotonic on (0,∞).
Theorem 4.3. The function −x5R′3(x) is completely monotonic on (0,∞), and
degxcmi[−R′3(x)]=degxcmi[f′3(x)]=5. |
Proof. We shall prove that −x5R′3(x)=x5f′3(x) is completely monotonic on (0,∞) and −x6R′3(x)=x6f3(x) is not. By (2.2) and (2.3) we obtain
xrf′3(x)=14∫∞0[−tp3(t2)](r)e−xtdt, r≥0. |
and
p3(t)=cothtt−1t2+145t2−2945t4−13,M(t):=tp3(t)=coshtsinht−13t−1t+145t3−2945t5,M(5)(t)=−1252p(t)t6sinh6t, |
we have
[−M(t)](5)=1252p(t)t6sinh6t,[−M(t)](6)=14q(t)t7sinh7t, |
where
p(t)=−14175cosh2t+5670cosh4t−945cosh6t+13134t6cosh2t+492t6cosh4t+2t6cosh6t+16612t6+9450,q(t)=945sinh3t−315sinh5t+45sinh7t−1575sinht−456t7cosh3t−8t7cosh5t−2416t7cosht. |
Since
p(t)=∞∑n=42⋅62n+492⋅42n+13134⋅22n(2n)!t2n+6−∞∑n=4945⋅62n+6−5670⋅42n+6+14175⋅22n+6(2n+6)!t2n+6>0,q(0.1)≈−2.9625×10−5<0, |
we obtain the expected conclusions.
Remark 4.1. The experimental results show that the conjecture (1.6) may be true.
Theorem 5.1. The function x3R′′1(x) is completely monotonic on (0,∞), and
degxcmi[R′′1(x)]=degxcmi[−f′′1(x)]=3. | (5.1) |
Proof. By (2.2) and (2.3) we obtain
x3R′′1(x)=−x3f′′1(x)=14∫∞0[−t2p1(t2)]′′′e−xtdt, |
and
t2p1(t)=tcoshtsinht−13t2−1, |
[−t2p1(t)]′′′=23sinh4t[∞∑n=23(n−1)22n+1(2n+1)!t2n+1]>0. |
So x3R′′1(x) is completely monotonic on (0,∞).
But x4R′′1(x) is not completely monotonic on (0,∞) due to
x4R′′1(x)=14∫∞0[−t2p1(t2)](4)e−xtdt, |
and
[−t2p1(t)](4)=1sinh5t(4sinh3t+12sinht−22tcosht−2tcosh3t) |
with [−t2p1(t)](4)|t=10≈−5.2766×10−7<0.
So
degxcmi[R′′1(x)]=degxcmi[−f′′1(x)]=3. |
Remark 5.1. Here, we actually give a negative answer to the second paragraph of conjecture (1.7).
Theorem 5.2. The function x4R′′2(x) is completely monotonic on (0,∞), and
degxcmi[R′′2(x)]=degxcmi[f′′2(x)]=4. |
Proof. By (2.2) and (2.3) we
x4f′′2(x)=14∫∞0[t2p2(t2)](4)e−xtdt, |
and
t2p2(t)=145t4−13t2+tcoshtsinht−1,[t2p2(t)](4)=130(−125sinh3t+sinh5t−350sinht+660tcosht+60tcosh3t)sinh5t=130sinh5t[∞∑n=35(52n+(24n−63)32n+264n+62)(2n+1)!t2n+1]>0. |
Since
x5f′′2(x)=14∫∞0[t2p2(t2)](5)e−xtdt, |
and
[t2p2(t)](5)=50sinh2t−66t+5sinh4t−52tcosh2t−2tcosh4tsinh6t |
with [t2p2(t)](5)|t=10≈−9.8935×10−7<0, we have that x5f′′2(x) is not completely monotonic on (0,∞). So
degxcmi[R′′2(x)]=4. |
Theorem 5.3. The function x6R′′3(x) is completely monotonic on (0,∞), and
degxcmi[R′′3(x)]=degxcmi[−f′′3(x)]=6. |
Proof. By the integral representation (2.3) we obtain
−x6f′′3(x)=14∫∞0[−t2p3(t2)](6)e−xtdt,−x7f′′3(x)=14∫∞0[−t2p3(t2)](7)e−xtdt. |
It follows from (2.2) that
p3(t)=cothtt−1t2+145t2−2945t4−13,N(t):=t2p3(t)=145t4−13t2−2945t6+tcoshtsinht−1,−N(6)(t)=142r(t)sinh7t,−N(7)(t)=(2416t−1715sinh2t−392sinh4t−7sinh6t+2382tcosh2t+240tcosh4t+2tcosh6t)sinh8t, |
where
r(t)=6321sinh3t+245sinh5t+sinh7t+10045sinht−25368tcosht−4788tcosh3t−84tcosh5t=∞∑n=4cn(2n+1)!t2n+1 |
with
cn=7⋅72n−(168n−1141)52n−(9576n−14175)32n−(50736n+15323). |
Since ci>0 for i=4,5,6,7, and
cn+1−49cn=(4032n−31584)52n+(383040n−653184)32n+2435328n+684768>0 |
for all n≥8. So cn>0 for all n≥4. Then r(t)>0 and −N(6)(t)>0 for all t>0. So x6R′′3(x) is completely monotonic on (0,∞).
In view of −N(7)(1.5)≈−0.57982<0, we get x7R′′3(x) is not completely monotonic on (0,∞). The proof of this theorem is complete.
Remark 5.2. The experimental results show that the conjecture (1.7) may be true for n,m≥2.
In this way, the first two paragraphs for conjectures (1.5) and (1.6) have been confirmed, leaving the following conjectures to be confirmed:
degxcmi[Rn(x)]=2(n−1), n≥4; | (6.1) |
degxcmi[−R′n(x)]=2n−1, n≥4; | (6.2) |
and for m≥1,
degxcmi[(−1)mR(m)n(x)]={m,if n=0m+1,if n=1m+2(n−1),if n≥2, | (6.3) |
where the first formula and second formula in (6.3) are two new conjectures which are different from the original ones.
By the relationship (2.3) we propose the following operational conjectures.
Conjecture 6.1. Let n≥4, and pn(t) be defined as (2.2). Then
[(−1)npn(t)](2n−2)>0 | (6.4) |
holds for all t∈(0,∞) and
[(−1)npn(t)](2n−1)>0 | (6.5) |
is not true for all t∈(0,∞).
Conjecture 6.2. Let n≥4, and pn(t) be defined as (2.2). Then
(−1)n[tpn(t)](2n−1)>0 | (6.6) |
holds for all t∈(0,∞) and
(−1)n[tpn(t)](2n)>0 | (6.7) |
is not true for all t∈(0,∞).
Conjecture 6.3. Let m≥1, and pn(t) be defined as (2.2). Then
[tmp0(t)](m)>0, | (6.8) |
[−tmp1(t)](m+1)>0 | (6.9) |
hold for all t∈(0,∞), and
[tmp0(t)](m+1)>0, | (6.10) |
[−tmp1(t)](m+2)>0 | (6.11) |
are not true for all t∈(0,∞).
Conjecture 6.4. Let m≥1, n≥2, and pn(t) be defined as (2.2). Then
(−1)n[tmpn(t)](m+2n−2)>0 | (6.12) |
holds for all t∈(0,∞) and
(−1)n[tmpn(t)](m+2n−1)>0 | (6.13) |
is not true for all t∈(0,∞).
The author would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.
The research is supported by the Natural Science Foundation of China (Grant No. 61772025).
The author declares no conflict of interest in this paper.
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