Citation: Chang-Jian Zhao, Wing-Sum Cheung. On Opial-Traple type inequalities for β-partial derivatives[J]. AIMS Mathematics, 2020, 5(6): 5716-5723. doi: 10.3934/math.2020366
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The fractional derivative of a function to order a is often now defined by means of the Fourier or Mellin integral transforms. Various types of fractional derivatives were introduced: Riemann-Liouville, Caputo, Hadamard, Erdelyi-Kober, Grunwald-Letnikov, Marchaud and Riesz are just a few to name [1,2,3,4]. For instance in more recent times a new local, limit-based definition of a conformable derivative has been introduced in [5,6,7], with several follow-up papers [8,9,10,11]. Recently a new local, limit-based definition of a so-called α-conformable derivative has been formulated in [4,12,13,14,15,16,17].
In the paper, we give a new concept of β-partial derivatives as the most natural extension of the familiar limit definition of the partial derivative. We show also that the β-partial derivatives obeys classical properties including: continuity, linearity, product rule, quotient rule, power rule, chain rule and vanishing derivatives for constant functions. As applications, we establish some new Opial-Traple type inequalities for the β-partial derivatives.
There exist a quite few definitions of fractional derivatives in the literatures, we will present one definition. Given a function f:[0,∞]→R. Then for all β∈(0,1] and x∈(0,∞), the β-derivative, defined by (see [18])
A0Dβx(f(x))=limε→0f(x+ε(x+1Γ(β))1−β)−f(x)ε, | (2.1) |
provided the limits exist, where Γ(⋅) is the usual Γ function. A function f is β-differentiable at a point x≥0, if the limits in (2.1) exist and are finite.
In this section, we give a new definition as the most natural extensions of the limit definitions of the partial derivative and the β-derivative. To this end, we start with the following definition which is a generalization of the classical partial derivative and β-derivative, respectively.
Definition 2.1 (β-partial derivatives) Let f(x,y) be a function, such that f(x,y):[a,∞)×[b,∞)→R.
(ⅰ) the β-partial x derivative of a function f(x,y) is defined as
AaPβx(f(x,y))=limε→0f(x+ε(x+1Γ(β))1−β,y)−f(x,y)ε(1−ax1−β), | (2.2) |
for all x≥a and β∈(0,1]. If the limit of the above exists, then f(x,y) is said to be β-partial x differentiable and call AaPβx(f(x,y)) as β-partial x derivatives of f(x,y).
(ⅱ) the beta-partial y derivative of a function f(x,y) is defined as
AbPβy(f(x,y))=limε→0f(x,y+ε(y+1Γ(β))1−β)−f(x,y)ε(1−by1−β), | (2.3) |
for all y≥b and β∈(0,1]. If the limit of the above exists, then f(x,y) is said to be β-partial y differentiable and call AbPβy(f(x,y)) as β-partial y derivatives of f(x,y). β-partial x, and β-partial y differentiable are collectively called β-partial differentiable.
Remark 2.2 Putting β=1 and a=b=0 in (2.2) and (2.3), the β-partial derivatives AaP1x(f(x,y)) and AbP1y(f(x,y)) just are the usual partial derivatives ∂f(x,y)∂x and ∂f(x,y)∂y, respectively.
Let f(x,y) become f(x) and with a proper transformation in (2.2), and let a=0, the β-partial x derivatives AaPβx(f(x,y)) reduces to the well-known β-derivatives A0Dβx(f(x)).
In this section, we give several results for the β-partial derivatives such as the continuity, linearity, product rule, quotient rule, power rule, chain rule and vanishing derivatives for constant functions.
Theorem 3.1 (Continuity) If f(x,y):[a,∞)×[b,∞)→R is β-partial differentiable and β∈(0,1], then f(x,y) is continuous at (x0,y0).
Proof Since f(s,t) is β-partial differentiable, so
limε→0f(x0+ε(x0+1Γ(β))1−β,y0)−f(x0,y0)ε(1−ax1−β0)=AaPβx(f(x,y))|(x0,y0), | (3.1) |
and
limε→0f(x0,y0+ε(y0+1Γ(β))1−β)−f(x0,y0)ε(1−by1−β0)=AbPβy(f(x,y))|(x0,y0). | (3.2) |
From (3.1), (3.2), and let h=ε(x0+1Γ(β))1−β and k=ε(y0+1Γ(β))1−β, we have
limε→0[f(x0+h,y0+k)−f(x0,y0)]=limh→0,k→0[f(x0+h,y0+k)−f(x0,y0)]=limε→0[f(x0+h,y0+k)−f(x0,y0+k)ε(1−ax1−β0)⋅ε(1−ax1−β0)]+limε→0[f(x0,y0+k)−f(x0,y0)ε(1−by1−β0)⋅ε(1−by1−β0)]=AaPβx(f(x,y))|(x0,y0)limε→0ε(1−ax1−β0)+AbPβy(f(x,y))|(x0,y0)limε→0ε(1−by1−β0)=0, |
which implies that f(x,y) is continuous at (x0,y0).
This completes the proof.
Theorem 3.2 Assuming that f(x,y) and g(x,y) are two β-partial x differentiable functions with β∈(0,1], then the following relations can be satisfied:
(ⅰ) AaPβx(af(x,y)±bg(x,y))=a⋅AaPβx(f(x,y))+b⋅AaPβx(g(x,y)), for all a and b real number.
(ⅱ) AaPβx(f(x,y))=(x+1Γ(β))1−β1−ax1−β⋅limε→0f(x+ε,y)−f(x,y)ε.
(ⅲ) AaPβx(f(x,y)g(x.y))=g(x,y)⋅AaPβx(f(x,y)+f(x,y)⋅AaPβx(g(x.y)).
(ⅳ) AaPβx(f(x,y)g(x.y))=f(s,t)⋅AaPβx(g(x.y))−g(x,y)⋅AaPβx(f(x,y))g(x,y)2, where g(x,y)≠0.
(v) AaPβx(λ)=0 for λ any given constant.
Proof Obviously, the (ⅰ) and (ⅴ) follow immediately from Definition 2.1.
Let
u=x+ε(x+1Γ(β))1−β. |
Noting that f(x,y) is continuous on [a,∞) at x≥a, we have
limε→0∂f(u,y)∂u=∂f(x,y)∂xandlimε→0∂u∂ε=(x+1Γ(β))1−β. |
Since f(x,y) is β-partial x differentiable at x≥a, and by using L'Hospital rule, we obtain
AaPβx(f(x,y))=limε→0f(x+ε(x+1Γ(β))1−β,y)−f(x,y)ε(1−ax1−β),=(1−ax1−β)−1limε→0∂f(u.y)∂u⋅∂u∂ε=(x+1Γ(β))1−β1−ax1−β⋅limε→0f(x+ε,y)−f(x,y)ε. |
This completes the proof of (ⅱ).
On the other hand, from (ⅱ), we have
AaPβx(f(x,y)⋅g(x,y))=(x+1Γ(β))1−β1−ax1−β⋅∂(f(x,y)⋅g(x,y))∂x=f(x,y)⋅(x+1Γ(β))1−β1−ax1−β⋅∂g(x,y)∂x+g(x,y)⋅(x+1Γ(β))1−β1−ax1−β⋅∂f(x,y)∂x=f(x,y)⋅AaPβx(g(x,y))+g(x,y)⋅AaPβx(f(x,y)). |
This completes the proof of (ⅲ).
The proof of the (ⅳ) is similar to (ⅲ). Here, we omit this details.
This completes the proof.
Theorem 3.3 Assuming that f(x,y) and g(x,y) are two β-partial y differentiable functions with β∈(0,1], then the following relations can be satisfied:
(ⅰ) AbPβy(af(x,y)±bg(x,y))=a⋅AbPβy(f(x,y))+b⋅AbPβy(g(x,y)), for all a and b real number.
(ⅱ) AbPβy(f(x,y))=(y+1Γ(β))1−β1−ay1−β⋅limε→0f(x,y+ε)−f(x,y)ε
(ⅲ) AbPβy(f(x,y)g(x.y))=g(x,y)⋅AbPβy(f(x,y)+f(x,y)⋅AbPβy(g(x.y)).
(ⅳ) AbPβy(f(x,y)g(x.y))=f(s,t)⋅AbPβy(g(x.y))−g(x,y)⋅AbPβy(f(x,y))g(x,y)2, where g(x,y)≠0.
(v) AbPβy(λ)=0, for λ any given constant.
Proof This follows immediately from the proof of Theorem 3.2 with a proper transformation.
Theorem 3.4 Let f(x,y):[a,∞)×[b,∞)→R be a function such that f(x,y) is β-partial x differentiable. If g(x,y) is a function defined in the range of f(x,y) and also β-partial x differentiable, then
AaPβx(f∘g(x,y))=f′g(g(x,y))⋅AaPβx(g(x,y)), | (3.3) |
where f′g(g(x,y)) denotes the derivative of function f to g(x,y).
Proof Let
v=g(x+ε(x+1Γ(β))1−β,y), |
and
u=ε(x+1Γ(β))1−β. |
Hence
limε→0f′(v)=f′g(g(x,y)),limε→0∂v∂u=∂g(x,y)∂x,andlimε→0∂u∂ε=(x+1Γ(β))1−β. | (3.4) |
Since f(x,y) and g(x,y) are two β-partial x differentiable, so f∘g is β-partial x differentiable, from (3.4) and by using L'Hospital rule, we obtain
AaPβx(f∘g(x,y))=limε→0f∘g(x+ε(x+1Γ(β))1−β,y)−f∘g(x,y)ε(1−ax1−β)=limε→0f(g(x+ε(x+1Γ(β))1−β,y))−f(g(x,y))ε(1−ax1−β)=(1−ax1−β)−1limε→0f′(v)⋅∂v∂u⋅∂u∂ε=f′g(g(x,y))⋅((x+1Γ(β))1−β1−ax1−β⋅∂g(x,y)∂x)=f′g(g(x,y))⋅AaPβx(g(x,y)). |
This completes the proof.
This chain rule theorem is important, but it is also understood. In order for the reader to better understand this theorem, we give another proof below.
Second proof Let
δ=g(x+ε(x+1Γ(β))1−β,y)−g(x,y). |
Obviously, if ε→0, then δ→0. From the hypotheses, we obtain
AaPβx(f∘g(x,y))=limε→0f(g(x+ε(x+1Γ(β))1−β,y))−f(g(x,y))ε(1−ax1−β)=limε→0f(g(x+ε(x+1Γ(β))1−β,y))−f(g(x,y))g(x+ε(x+1Γ(β))1−β,y)−g(x,y)×g(x+ε(x+1Γ(β))1−β,y)−g(x,y)ε(1−ax1−β)=limδ→0f(g(x,y)+δ)−f(g(x,y))δ⋅limε→0g(x+ε(x+1Γ(β))1−β,y)−g(x,y)ε(1−ax1−β)=f′g(g(x,y))⋅AaPβx(g(x,y)). |
This completes the proof.
Let f(x,y) and g(x,y) change f(x) and g(x) with a proper transformation in Theorem 3.4, it becomes the following result, which was established in [18].
Corollary 3.5 Let f(x):[0,∞)→R be a function such that f(x) is β-differentiable. If g(x,y) is a function defined in the range of f(x,y) and also differentiable, then
A0Dβx(f∘g(x))=(x+1Γ(β))1−βf′(x)g′(f(x)), | (3.5) |
where, A0Dβx(f(x)) denotes the β-derivatives of f(x).
Theorem 3.6 Let f(x,y):[a,∞)×[b,∞)→R be a function such that f(x,y) is β-partial y differentiable. If g(x,y) is a function defined in the range of f(x,y) and also β-partial y differentiable, then
AbPβy(f∘g(x,y))=f′g(g(x,y))⋅AbPβy(g(x,y)), | (3.6) |
where f′g(g(x,y)) denotes the derivative of function f to g(x,y).
Proof This follows immediately from the proof of Theorem 3.4 with a proper transformation.
In the section, we establish Opial-Traple type inequalities for the β-partial derivatives.
Definition 4.1 (β-conformable integrals) Let β∈(0,1], 0≤a<b and 0≤c<d. A function f(x,y):[a,b]×[c,d]→R is said β-conformable integrable on [a,b]×[c,d], if the integral
∫baf(x,y)dβx:=∫ba(1−ax1−β)(x+1Γ(β))β−1f(x,y)dx | (4.1) |
exists and is finite.
Theorem 4.2 (Opial-Traple inequality for β-partial derivatives) Let α∈(0,1], a≤s≤b, and c≤t≤d, and p(s,t) be nonnegative and continuous function on [a,b]×[c,d]. Let u(s,t) and p(s,t) be β-partial differentiable on [a,b]×[c,d] with u(a,t)=u(b,t)=0, then
∫bap(s,t)|u(s,t)|2dβs≤M(a,b,β)⋅∫bap(s,t)dβs⋅∫ba|AaPβs(u(s,t))|2dβs, | (4.2) |
where
M(a,b,β)=14∫ba(1−as1−β)(s+1Γ(β))β−1ds. |
Proof Let
y(s,t)=∫sa|AaPβσ(u(σ,t))|dβσ, |
and
z(s,t)=∫bs|AaPβσ(u(σ,t))|dβσ. |
From (4.1) and Theorem 3.3, we obtain
AaPβs(y(s,t))=|AaPβs(u(s,t))|=−AaPβs(z(s,t)), | (4.3) |
and for all (s,t)∈[a,b]×[c,d],
u(s,t)≤y(s,t),u(s,t)≤z(s,t). | (4.4) |
Hence
u(s,t)≤y(s,t)+z(s,t)2=12∫ba|AaPβσ(u(σ,t))|dβσ. | (4.5) |
From (4.5) and in view of Cauchy-Schwarz inequality, we obtain
∫bap(s,t)|u(s,t)|2dβs |
≤14∫bap(s,t)(∫ba|AaPβσ(u(σ,t))|dβσ)2dβs≤14(∫bap(s,t)dβs)(∫badβσ)(∫ba|AaPβσ(u(σ,t))|2dβσ)=14∫ba(1−as1−β)(s+1Γ(β))β−1ds⋅(∫bap(s,t)dβs)(∫ba|AaPβσ(u(σ,t))|2dβσ)=M(a,b,β)⋅(∫bap(s,t)dβs)(∫ba|AaPβs(u(s,t))|2dβs). |
This completes the proof.
Theorem 4.3 Let α∈(0,1], 0≤s≤b, and p(s) be nonnegative and continuous function on [0,b]. Let u(s) and p(s) be β-differentiable on [0,b] with u(0)=u(b)=0, then
∫b0p(s)|u(s)|2dβs≤N(b,β)⋅(∫b0p(s)dβs)(∫b0|A0Dβs(u(s))|2dβs), | (4.6) |
where A0Dβs is as in (2.1), and
N(b,β)=1β[(b+1Γ(β))β−(1Γ(β))β]. |
Proof Let u(s,t) and p(s,t) change to u(s) and p(s), respectively, and with a proper transformation, and let a=0, (4.6) follows immediately from (4.2).
Theorem 4.4 Let a≤s≤b, and c≤t≤d, and p(s,t) be nonnegative and continuous function on [a,b]×[c,d]. Let u(s,t) and p(s,t) be partial differentiable on [a,b]×[c,d] with u(a,t)=u(b,t)=0, then
∫bap(s,t)|u(s,t)|2ds≤b−a4(∫bap(s,t)ds)(∫ba|∂u(s,t)∂s|2ds). | (4.7) |
Proof This follows immediately from Theorem 4.2 with β=1.
Let p(s,t) and u(s,t) reduce to p(t) and u(t), respectively, and with suitable modifications, and let a=0 and b=h, (4.7) becomes the following result.
Corollary 4.5 Let p(t) be a nonnegative and continuous function on [0,h]. Let u(t) be an absolutely continuous function on [0,h] with u(0)=u(h)=0, then
∫h0p(t)|u(t)|2dsdt≤h4(∫h0p(t)dt)(∫h0|u′(t)|2dt). |
This is just an inequality which was established in [14]. Here, we call it Opial-Traple's inequality.
Research is supported by National Natural Science Foundation of China (11371334, 10971205). Research is partially supported by the Research Grants Council of the Hong Kong SAR, China (Project No.: HKU7017/05P).
The authors declare that they have no competing interests.
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