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Multivariable Conformable Fractional Calculus Nazlı Yazıcı Gözu¨toka, Ug˘ur Gözu¨toka,∗ aDepartment of Mathematics 7 Karadeniz Technical University 1 61080, Trabzon, TURKEY 0 2 n a J Abstract 3 Conformable fractional derivative is introduced by the authors Khalil et al. In ] A this study we develop their concept and introduce multivariable conformable C derivative for a vector valued function with several variables. . h Keywords: Conformable fractional derivative, multivariable conformable t a m calculus, conformable partial derivatives, conformable jacobian matrix [ 2010 MSC: 26B12,26A33 1 v 6 1. Introduction 1 6 0 For many years, many definitions of fractional derivative have been intro- 0 . duced by various researchers. One of them is the Riemann-Liouville fractional 1 0 derivative and the second one is the so-called Caputo derivative. But these 7 1 are not all purpose definitions. Recently, a new fractional derivative has been : v introducedin [1]andone cansee thatthe new derivative suggestedin these pa- i X pers satisfies all the properties of the standart one. However, definitions given r a in the literature are only for the real valued functions. In this paper, we give conformablefractionalderivativedefinitionfor the vectorvaluedfunctions with several real variables. Our paper has the following order: In Section 2, some basic definitions and theorems appeared in the literature are given. In Sec- tion 3, α derivative of a vector valued function, conformable Jacobian matrix − ∗Correspondingauthor Email addresses: [email protected] (NazlıYazıcıGözu¨tok), [email protected] (U˘gurGözu¨tok) Preprint submitted toElsevier January 4, 2017 are defined; relationbetween α derivative and usual derivative of a vector val- − ued function is revealed; chain rule for multivariable conformable derivative is given. In Section 4, conformable partial derivatives of a real valued function with n variables is defined and relation between conformable Jacobian matrix − and conformable partial derivatives is given. 2. Basic definitions and theorems In this section we will give some definitions and properties introduced in [1, 2] Definition 2.1. Given afunction f :[0, ) R. The conformable derivative ∞ −→ of the function f of order α is defined by f(x+hx1−α) f(x) T (f)(x)= lim − (1) α h→0 h for all x>0, α (0,1). ∈ Theorem 2.2. If a function f : [0, ) R is α differentiable at t > 0, 0 ∞ −→ − α (0,1], then f is continuous at t . 0 ∈ Theorem 2.3. Let α (0,1] and f,g be α differentiable at a point t > 0. ∈ − Then (1) T (af +bg)=aT (f)+bT (g), for all a,b R. α α α ∈ (2) T (tp)=ptp−α for all p R. α ∈ (3) T (λ)=0, for all constant functions f(t)=λ. α (4) T (fg)=fT (g)+gT (f). α α α f gT (f) fT (g) α α (5) T ( )= − . α g g2 d (6) If, in addition, f is differentiable, then T (f)(t)=t1−α f(t). α dt 2 Theorem 2.4. Assume f,g : (0, ) R be two α differentiable functions ∞ −→ − where α (0,1]. Then g f is α differentiable and for all t with t = 0 and ∈ ◦ − 6 f(t)=0 we have 6 T (g f)(t)=T (g)(f(t))T (f)(t)f(t)α−1. (2) α α α ◦ 3. α−Derivative of a Vector Valued Function Definition 3.1. Let f be a vector valued function with n real variables such that f(x ,...,x ) = (f (x ,...,x ),...,f (x ,...,x )). Then we say that f is 1 n 1 1 n m 1 n α differentiable at a = (a ,...,a ) Rn where each a > 0, if there is a linear 1 n i − ∈ transformation L:Rn Rm such that −→ f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) L(h) lim k 1 1 1 n n n − 1 n − k =0 (3) h→0 h k k where h = (h ,...,h ) and α (0,1]. The linear transformation is denoted by 1 n ∈ Dαf(a) and called the conformable derivative of f of order α at a. Remark 3.2. For m=n=1, Definition 3.1 equivalent to Definition 2.1. Theorem 3.3. Let f be a vector valued function with n variables. If f is α differentiable at a = (a ,...,a ) Rn, each a > 0, then there is a unique 1 n i − ∈ linear transformation L:Rn Rm such that −→ f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) L(h) lim k 1 1 1 n n n − 1 n − k =0. h→0 h k k Proof. Suppose M :Rn Rm satisfies −→ f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) M(h) lim k 1 1 1 n n n − 1 n − k =0. h→0 h k k Then, L(h) M(h) lim k − k h→0 h k k L(h) (f(a +h a1−α,...,a +h a1−α) f(a)) lim k − 1 1 1 n n n − k ≤h→0 h k k (f(a +h a1−α,...,a +h a1−α) f(a)) M(h) + lim k 1 1 1 n n n − − k =0 h→0 h k k 3 If x Rn, then tx 0 as t 0. Hence for x=0 we have ∈ → → 6 L(tx) M(tx) L(x) M(x) 0= lim k − k = k − k. h→0 tx x k k k k Therefore L(x)=M(x). Example 3.4. Let us consider the function f defined by f(x,y) = sinx and the point (a,b) R2 such that a,b > 0, then Dαf(a,b) = L satisfies L(x,y) = ∈ xa1−αcosa. To prove this, note that f(a+h a1−α,b+h b1−α) f(a,b) L(h ,h ) 1 2 1 2 lim | − − | (h1,h2)→(0,0) (h1,h2) k k sin(a+h a1−α) sina h a1−αcosa 1 1 = lim | − − | (h1,h2)→(0,0) ph21+h22 sin(a+h a1−α) sina h a1−αcosa 1 1 lim | − − | ≤h1→0 h1 | | sin(a+h a1−α) sina = lim 1 − a1−αcosa h1→0| h1 − | = a1−αcosa a1−αcosa =0 | − | Definition 3.5. Consider the matrix of the linear transformation Dαf(a) : Rn Rm with respect to the usual bases of Rn and Rm. This m n matrix −→ × is called the conformable Jacobian matrix of f at a, and denoted by fα(a). Example 3.6. If f(x,y)=sinx, then fα(a,b)= a1−αcosa 0 h i Theorem 3.7. Let f be α differentiable at a=(a ,...,a ) Rn, each a >0. 1 n i − ∈ If f is differentiable at a, then Dαf(a)=Df(a) L1−α ◦ a where Df(a) is the usual derivative of f and L1−α is the linear transformation a from Rn to Rn defined by L1−α(x ,...,x )=(a1−αx ,...,a1−αx ). a 1 n 1 1 n n Proof. It is suffices to show that f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) Df(a) L1−α(h) lim k 1 1 1 n n n − 1 n − ◦ a k =0. h→0 h k k 4 Let ǫ = (ǫ ,...,ǫ ) = (h a1−α,...,h a1−α), then ǫ 0 as h 0. On the other 1 n 1 1 n n → → hand, if we put M =max (a1−α)2 a >0,i=1,2,...,n >0, then { i | i } kǫk=qh21(a11−α)2+...+h2n(a1n−α)2 ≤qh21M +...+h2nM =√nMkhk. Hence we have 1 ǫ h . √nMk k≤k k Finally, f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) Df(a) L1−α(h) lim k 1 1 1 n n n − 1 n − ◦ a k h→0 h k k f(a +h a1−α,...,a +h a1−α) f(a) Df(a)(h a1−α,...,h a1−α) = lim k 1 1 1 n n n − − 1 1 n n k h→0 h k k f(a+ǫ) f(a) Df(a)(ǫ) lim k − − k ≤ǫ→0 1 ǫ √nMk k f(a+ǫ) f(a) Df(a)(ǫ) =√nM lim k − − k =√nM.0=0. ǫ→0 ǫ k k This completes the proof. Theorem 3.7 is the generalized case of the part (6) of Theorem 2.3. Also matrix form of the Theorem 3.7 is given by the following: a11−α 0 ... 0 fα(a)=f′(a). 0... a11−...α ...... 0... ,  0 0 ... a1−α n a11−α 0 ... 0 where f′(a) is the usual Jacobian of f and  0... a11−...α ...... 0...  is the matrix  0 0 ... a1−α corresponding to linear transformation L1−α. n a Theorem3.8. Ifavectorvaluedfunctionf withnvariables isα differentiable − at a=(a ,...,a ) Rn, each a >0, then f is continuous at a Rn. 1 n i ∈ ∈ 5 Proof. Since f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) k 1 1 1 n n n − 1 n k f(a +h a1−α,...,a +h a1−α) f(a) L(h)+L(h) = k 1 1 1 n n n − − k h h k k k k f(a +h a1−α,...,a +h a1−α) f(a) L(h) k 1 1 1 n n n − − k h + L(h) . ≤ h k k k k k k We have f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) k 1 1 1 n n n − 1 n k f(a +h a1−α,...,a +h a1−α) f(a) L(h) k 1 1 1 n n n − − k h + L(h) . ≤ h k k k k k k By taking limits of the two sides of the inequality as h 0, we have → lim f(a +h a1−α,...,a +h a1−α) f(a ,...,a ) h→0k 1 1 1 n n n − 1 n k f(a +h a1−α,...,a +h a1−α) f(a) L(h) lim k 1 1 1 n n n − − k lim h + lim L(h) ≤h→0 h h→0k k h→0k k k k =0. Let (ǫ ,...,ǫ )=(h a1−α,...,h a1−α), then ǫ 0 as h 0. Since 1 n 1 1 n n → → lim f(a+ǫ) f(a) 0 ǫ→0k − k≤ we have lim f(a+ǫ) f(a) =0. ǫ→0k − k Hence f is continuous at a Rn. ∈ Theorem 3.9. (Chain Rule) Let x Rn, y Rm. If f(x)=(f (x),...,f (x)) 1 m ∈ ∈ is α differentiable at a=(a ,...,a ) Rn, each a >0 such that α (0,1] and 1 n i − ∈ ∈ g(y) = (g (y),...,g (y)) is α differentiable at f(a) Rm, all f (a) > 0 such 1 p i − ∈ that α (0,1]. Then the composition g f is α differentiable at a and ∈ ◦ − Dα(g f)(a)=Dαg(f(a)) f(a)α−1 Dαf(a) (4) ◦ ◦ ◦ where f(a)α−1 is the linear transformation from Rm to Rm defined by f(a)α−1(x ,...,x )=(x f (a)α−1,...,x f (a)α−1). 1 m 1 1 m m 6 Proof. Let L=Dαf(a), M =Dαg(f(a)). If we define (i) ϕ(a +h a1−α,...,a +h a1−α) 1 1 1 n n n =f(a +h a1−α,...,a +h a1−α) f(a) L(h), 1 1 1 n n n − − (ii) ψ(f (a)+k f (a)1−α,...,f (a)+k f (a)1−α) 1 1 1 n n n =g(f (a)+k f (a)1−α,...,f (a)+k f (a)1−α) g(f(a)) M(k), 1 1 1 n n n − − (iii) ρ(a +h a1−α,...,a +h a1−α) 1 1 1 n n n =g f(a +h a1−α,...,a +h a1−α) g f(a) M f(a)α−1 L(h), ◦ 1 1 1 n n n − ◦ − ◦ ◦ then ϕ(a +h a1−α,...,a +h a1−α) (iv) limh→0 k 1 1 1 h n n n k =0, k k ψ(f (a)+k f (a)1−α,...,f (a)+k f (a)1−α) 1 1 1 n n n (v) limk→0 k k k =0, k k and we must show that ρ(a +h a1−α,...,a +h a1−α) limh→0k 1 1 1 h n n n k =0 k k Now, ρ(a +h a1−α,...,a +h a1−α) 1 1 1 n n n =g(f(a +h a1−α,...,a +h a1−α)) g(f(a)) M f(a)α−1 L(h) 1 1 1 n n n − − ◦ ◦ =g(f (a +h a1−α,...,a +h a1−α),...,f (a +h a1−α,...,a +h a1−α)) 1 1 1 1 n n n m 1 1 1 n n n g(f(a)) M f(a)α−1(f(a +h a1−α,...,a +h a1−α) f(a) − − ◦ 1 1 1 n n n − ϕ(a +h a1−α,...,a +h a1−α)) by(i) − 1 1 1 n n n =[g(f (a +h a1−α,...,a +h a1−α),...,f (a +h a1−α,...,a +h a1−α)) 1 1 1 1 n n n m 1 1 1 n n n g(f(a)) M(f(a)α−1(f(a +h a1−α,...,a +h a1−α) f(a)))] − − 1 1 1 n n n − +M f(a)α−1(ϕ(a +h a1−α,...,a +h a1−α)) ◦ 1 1 1 n n n =[g(f (a +h a1−α,...,a +h a1−α),...,f (a +h a1−α,...,a +h a1−α)) 1 1 1 1 n n n m 1 1 1 n n n g(f(a)) M(f(a)α−1(f (a +h a1−α,...,a +h a1−α) f (a),..., − − 1 1 1 1 n n n − 1 f (a +h a1−α,...,a +h a1−α) f (a)))] m 1 1 1 n n n − m +M f(a)α−1(ϕ(a +h a1−α,...,a +h a1−α)) ◦ 1 1 1 n n n 7 =[g(f (a +h a1−α,...,a +h a1−α),...,f (a +h a1−α,...,a +h a1−α)) 1 1 1 1 n n n m 1 1 1 n n n g(f(a)) M([f (a +h a1−α,...,a +h a1−α) f (a)]f (a)α−1,..., − − 1 1 1 1 n n n − 1 1 [f (a +h a1−α,...,a +h a1−α) f (a)]f (a)α−1)] m 1 1 1 n n n − m m +M f(a)α−1(ϕ(a +h a1−α,...,a +h a1−α)) ◦ 1 1 1 n n n If we put u = [f (a +h a1−α,...,a +h a1−α) f (a)]f (a)α−1, i = 1,...,n, i i 1 1 1 n n n − i i then we have f (a +h a1−α,...,a +h a1−α)=f (a)+u f (a)1−α and u 0 i 1 1 1 n n n i i i → as h 0. Therefore, → ρ(a +h a1−α,...,a +h a1−α) 1 1 1 n n n =[g(f (a +h a1−α,...,a +h a1−α),...,f (a +h a1−α,...,a +h a1−α)) 1 1 1 1 n n n m 1 1 1 n n n g(f(a)) M(u)]+M f(a)α−1(ϕ(a +h a1−α,...,a +h a1−α)) − − ◦ 1 1 1 n n n =ψ(f (a)+u f (a)1−α,...,f (a)+u f (a)1−α) 1 1 1 m m m +M f(a)α−1(ϕ(a +h a1−α,...,a +h a1−α)) by(ii). ◦ 1 1 1 n n n Thus we must show ψ(f (a)+u f (a)1−α,...,f (a)+u f (a)1−α) (vi) limu→0 k 1 1 1 u m m m k =0, k k M f(a)α−1(ϕ(a +h a1−α,...,a +h a1−α)) (vii) limh→0 k ◦ 1 1h1 n n n k =0. k k For (vi), it is obvious from (v). For (vii), the linear transformation satisfies M f(a)α−1(ϕ(a +h a1−α,...,a +h a1−α)) k ◦ 1 1 1 n n n k K ϕ(a +h a1−α,...,a +h a1−α) ≤ k 1 1 1 n n n k such that K >0. Hence, from (iv), (vii) holds. Corollary 3.10. For, m=n=p=1, Theorem 3.9 states that T (g f)(a)=T g(f(a))T f(a)f(a)α−1. α α α ◦ Above corollary says that Theorem 3.9 is the generalized case of Theorem 2.4. 8 Corollary 3.11. Let all conditions of Theorem 3.9 be satisfied. Then f1(a)α−1 0 ... 0 (g◦f)α(a)=gα(f(a)) 0... f2(a...)α−1 ...... 0... fα(a)  0 0 ... fm(a)α−1 f1(a)α−1 0 ... 0  0 f2(a)α−1 ... 0  where  ... ... ... ...  is the matrix corresponding to the linear  0 0 ... fm(a)α−1 transformation f(a)α−1. Theorem 3.12. Let f be a vector valued function with n variables such that f(x ,...,x ) = (f (x ,...,x ),...,f (x ,...,x )). Then f is α differentiable at 1 n 1 1 n m 1 n − a=(a ,...,a ) Rn, each a >0 if and only if each f is, and 1 n i i ∈ Dαf(a)=(Dαf (a),...,Dαf (a)). 1 m Proof. If eachf is α differentiable at a andL=(Dαf (a),...,Dαf (a)), then i 1 m − f(a +h a1−α,...,a +h a1−α) f(a) L(h) 1 1 1 n n n − − =(f (a +h a1−α,...,a +h a1−α) f (a) Dαf (a)(h),..., 1 1 1 1 n n n − 1 − 1 f (a +h a1−α,...,a +h a1−α) f (a) Dαf (a)(h)). m 1 1 1 n n n − m − m Therefore, f(a +h a1−α,...,a +h a1−α) f(a) L(h) lim k 1 1 1 n n n − − k h→0 h k k n f (a +h a1−α,...,a +h a1−α) f (a) Dαf (a)(h) lim k i 1 1 1 n n n − i − i k =0. ≤h→0X h i=1 k k If,ontheotherhand,f isα differentiableata,thenf =π f isα differentiable i i − − at a by Theorem 3.9. Theorem 3.13. Let α (0,1] and f,g be α differentiable at a point a = ∈ − (a ,...,a ) Rn, each a >0. Then 1 n i ∈ (i) Dα(λf +µg)(a)=λDαf(a)+µDαg(a) for all λ,µ R. ∈ (ii) Dα(fg)(a)=f(a)Dαg(a)+g(a)Dαf(a). 9 Proof. (i) follows from the definition, thus we omitted the proof of (i). For (ii), let a +h a1−α,...,a +h a1−α =A, then 1 1 1 n n n (fg)(A) (fg)(a) (f(a)Dαg(a)+g(a)Dαf(a))(h) lim k − − k h→0 h k k f(A)g(A) f(a)g(A) g(A)Dαf(a)(h) lim k − − k ≤h→0 h k k f(a)g(A) f(a)g(a) f(a)Dαg(a)(h) + lim k − − k h→0 h k k g(A)Dαf(a)(h) g(a)Dαf(a)(h) + lim k − k h→0 h k k f(A) f(a) Dαf(a)(h) = lim g(A) k − − k h→0k k h k k g(A) g(a) Dαg(a)(h) g(A) g(a) + lim f(a) k − − k + lim Dαf(a)(h) k − k h→0k k h h→0k k h k k k k g(A) g(a) lim K h k − k =0 ≤h→0 k k h k k This completes the proof. 4. Conformable Partial Derivatives We begin this section with a correction. In [3], authors define the conformable derivative of a function with m variables [Definition 2.9]. But the definition is not valid for α= 1 and x <0. We give the corrected definition of 2 i conformable partial derivative of a real valued function with n variables by the following. Definition 4.1. Let f be a real valued function with n variables and a = (a ,...,a ) Rn be a point whose ith component is positive. Then the limit 1 n ∈ f(a ,...,a +ǫa1−α,...,a ) f(a ,...,a ) lim 1 i i n − 1 n , (5) ǫ→0 ǫ ∂α ifit exists, is denotedby f(a), and called theithconformable partial deriva- ∂xα tive of f of order α (0,1] at a. ∈ Theorem 4.2. Let f be a vector valued function with n variables. If f is ∂α α differentiable at a=(a ,...,a ) Rn, each a >0, then f (a) exists for − 1 n ∈ i ∂xα i j 10