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Some Multiple Integral Inequalities Via the Divergence Theorem

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Some Multiple Integral Inequalities Via the Divergence Theorem Journal of Mathematical Inequalities Volume 14, Number 3 (2020), 661–671 doi:10.7153/jmi-2020-14-42 SOME MULTIPLE INTEGRAL INEQUALITIES VIA THE DIVERGENCE THEOREM SILVESTRU SEVER DRAGOMIR (Communicated by J. Pecari´ˇ c) Abstract. In this paper, by the use of the divergence theorem, we establish some inequalities for functions defined on closed and bounded subsets of the Euclidean space Rn, n 2. 1. Introduction Let ∂D be a simple, closed counterclockwise curve bounding a region D and f defined on an open set containing D and having continuous partial derivatives on D. In the recent paper [4], by the use of Green’s identity, we have shown among others that 1 f (x,y)dxdy− [(β − y) f (x,y)dx+(x − α) f (x,y)dy] D 2 ∂D 1 ∂ f (x,y) ∂ f (x,y) |α − x| + |β − y| dxdy =: M (α,β; f ) (1.1) 2 D ∂x ∂y for all α, β ∈ C and ⎧ ⎪ ∂ f |α − | + ∂ f |β − | ⎪ ∂ D x dxdy ∂ D y dxdy; ⎪ x D,∞ y D,∞ ⎪ ⎪ ⎨⎪ ∂ / ∂ / f ( |α − x|q dxdy)1 q + f ( |β − y|q dxdy)1 q (α,β ) ∂x , D ∂y , D M ; f ⎪ D p D p ⎪ where p, q > 1, 1 + 1 = 1; ⎪ p q ⎪ ⎪ ⎩⎪ |α − | ∂ f + |β − | ∂ f , sup(x,y)∈D x ∂ sup(x,y)∈D y ∂ x D,1 y B,1 (1.2) Mathematics subject classification (2010): 26D15. Keywords and phrases: Multiple integral inequalities, divergence theorem, Green identity, Gauss- Ostrogradsky identity. Ð c ,Zagreb 661 Paper JMI-14-42 662 S. S. DRAGOMIR · where D,p are the usual Lebesgue norms, we recall that ⎧ ⎨ ( | ( , )|p )1/p , D g x y dxdy p 1; = g D,p : ⎩ | ( , )|, = ∞. sup(x,y)∈D g x y p Applications for rectangles and disks were also provided in [4]. For some recent double integral inequalities see [1], [2]and[3]. We also considered similar inequalities for 3-dimensional bodies as follows, see [5]. Let B be a solid in the three dimensional space R3 bounded by an orientable closed surface ∂B.Iff : B → C is a continuously differentiable function defined on a open set containing B, then by making use of the Gauss-Ostrogradsky identity,we have obtained the following inequality 1 f (x,y,z)dxdydz− (x − α) f (x,y,z)dy∧ dz B 3 ∂B + (y − β) f (x,y,z)dz∧ dx+ (z − γ) f (x,y,z)dx∧ dy ∂B ∂B 1 ∂ f (x,y,z) ∂ f (x,y,z) ∂ f (x,y,z) |α − x| + |β − y| + |γ − z| dxdydz 3 B ∂x ∂y ∂z =:M (α,β,γ; f ) (1.3) for all α, β, γ complex numbers. Moreover, we have the bounds M (α,β,γ; f ) ⎧ ⎪ ∂ f |α − | + ∂ f |β − | ⎪ ∂ B x dxdydz ∂ B y dxdydz ⎪ x B,∞ y B,∞ ⎪ ⎪ + ∂ f |γ − | ⎪ ∂ B z dxdydz; ⎪ z B,∞ ⎪ ⎪ ⎪ ∂ / ∂ / ⎨ f ( |α − x|q dxdydz)1 q + f ( |β − y|q dxdydz)1 q 1 ∂x , B ∂y , B Bp B p (1.4) 3 ⎪ + ∂ f ( |γ − | )1/q , , > , 1 + 1 = ⎪ ∂z B z dxdydz p q 1 p q 1; ⎪ B,p ⎪ ⎪ ⎪ ∂ f ∂ f ⎪ sup( , , )∈ |α − x| + sup( , , )∈ |β − y| ⎪ x y z B ∂x , x y z B ∂y , ⎪ B1 B 1 ⎩ + |γ − | ∂ f . sup(x,y,z)∈B z ∂ z B,1 Applications for 3-dimensional balls were also given in [5]. For some triple inte- gral inequalities see [6]and[9]. Motivated by the above results, in this paper we establish several similar inequali- ties for multiple integrals for functions defined on bonded subsets of Rn (n 2) with smooth (or piecewise smooth) boundary ∂B. To achieve this goal we make use of the well known divergence theorem for multiple integrals as summarized below. SOME MULTIPLE INTEGRAL INEQUALITIES 663 2. Some preliminary facts Let B be a bounded open subset of Rn (n 2) with smooth (or piecewise smooth) n boundary ∂B.LetF =(F1,...,Fn) be a smooth vector field defined in R , or at least in B∪ ∂B.Letn be the unit outward-pointing normal of ∂B. Then the Divergence Theorem states, see for instance [8]: divFdV = F · ndA, (2.1) B ∂B where n ∂ = ∇ · = Fk , divF F ∑ ∂ k=1 xk dV is the element of volume in Rn and dA is the element of surface area on ∂B. If n =(n1,...,nn), x =(x1,...,xn) ∈ B and use the notation dx for dV we can write (2.1) more explicitly as n n ∂Fk (x) ∑ dx = ∑ Fk (x)nk (x)dA. (2.2) ∂ ∂ k=1 B xk k=1 B By taking the real and imaginary part, we can extend the above equality for com- plex valued functions Fk, k ∈{1,...,n} defined on B. If n = 2, the normal is obtained by rotating the tangent vector through 90◦ (in the correct direction so that it points out). The quantity tds can be written (dx1,dx2) along the surface, so that ndA := nds =(dx2,−dx1) Here t is the tangent vector along the boundary curve and ds is the element of arc-length. From (2.2)wegetforB ⊂ R2 that ∂F1 (x1,x2) ∂F2 (x1,x2) dx1dx2+ dx1dx2 = F1 (x1,x2)dx2− F2 (x1,x2)dx1, B ∂x1 B ∂x2 ∂B ∂B (2.3) which is Green’s theorem in plane. Ifn = 3andif∂B is described as a level-set of a function of 3 variables i.e. 3 ∂B = x1,x2,x3 ∈ R | G(x1,x2,x3)=0 , then a vector pointing in the direction of n is gradG. We shall use the case where G(x1,x2,x3)=x3 − g(x1,x2), (x1,x2) ∈ D, a domain in R2 for some differentiable function g on D and B corresponds to the inequality x3 < g(x1,x2), namely 3 B = (x1,x2,x3) ∈ R | x3 < g(x1,x2) . Then (−g ,−g ,1) / = x1 x2 , = + 2 + 2 1 2 n / dA 1 gx gx dx1dx2 1 + g2 + g2 1 2 1 2 x1 x2 664 S. S. DRAGOMIR and =(− ,− , ) . ndA gx1 gx2 1 dx1dx2 From (2.2)weget ∂F (x ,x ,x ) ∂F (x ,x ,x ) ∂F (x ,x ,x ) 1 1 2 3 + 2 1 2 3 + 3 1 2 3 dx dx dx ∂x ∂x ∂x 1 2 3 B 1 2 3 = − ( , , ( , )) ( , ) − ( , , ( , )) ( , ) F1 x1 x2 g x1 x2 gx1 x1 x2 dx1dx2 F1 x1 x2 g x1 x2 gx2 x1 x2 dx1dx2 D D + F3 (x1,x2,g(x1,x2))dx1dx2 (2.4) D which is the Gauss-Ostrogradsky theorem in space. 3. Identities of interest We have the following identity of interest: THEOREM 1. Let B be a bounded open subset of Rn (n 2) with smooth (or piecewise smooth) boundary ∂B. Let f be a continuously differentiable function de- n fined in R ,oratleastinB∪ ∂B and with complex values. If αk, βk ∈ C for ∈{ ,..., } ∑n α = , k 1 n with k=1 k 1 then n ∂ f (x) n f (x)dx= ∑ (βk − αkxk) dx+ ∑ (αkxk − βk) f (x)nk (x)dA. (3.1) ∂ ∂ B k=1 B xk k=1 B We also have 1 n ∂ f (x) 1 n f (x)dx = ∑ (γk − xk) dx + ∑ (xk − γk) f (x)nk (x)dA (3.2) ∂ ∂ B n k=1 B xk n k=1 B for all γk ∈ C,wherek∈{1,...,n}. Proof. Let x =(x1,...,xn) ∈ B. We consider Fk (x)=(αkxk − βk) f (x), k ∈{1,...,n} ∂ ( ) and take the partial derivatives Fk x to get ∂xk ∂Fk (x) ∂ f (x) = αk f (x)+(αkxk − βk) , k ∈{1,...,n}. ∂xk ∂xk If we sum this equality over k from 1 to n we get n ∂ ( ) n n ∂ ( ) n ∂ ( ) Fk x = α ( )+ (α − β ) f x = ( )+ (α − β ) f x ∑ ∂ ∑ k f x ∑ kxk k ∂ f x ∑ kxk k ∂ k=1 xk k=1 k=1 xk k=1 xk (3.3) SOME MULTIPLE INTEGRAL INEQUALITIES 665 for all x =(x1,...,xn) ∈ B. Now, if we take the integral in the equality (3.3) over (x1,...,xn) ∈ B we get n ∂ ( ) n ∂ ( ) Fk x = ( ) + (α − β ) f x . ∑ ∂ dx f x dx ∑ kxk k ∂ dx (3.4) B k=1 xk B k=1 B xk By the Divergence Theorem (2.2)wealsohave n n ∂Fk (x) ∑ dx = ∑ (αkxk − βk) f (x)nk (x)dA (3.5) ∂ ∂ B k=1 xk k=1 B and by making use of (3.4)and(3.5)weget n ∂ f (x) n f (x)dx + ∑ (αkxk − βk) dx = ∑ (αkxk − βk) f (x)nk (x)dA ∂ ∂ B k=1 B xk k=1 B which gives the desired representation (3.1). α = 1 β = 1 γ , ∈{ ,..., }. The identity (3.2) follows by (3.1)for k n and k n k k 1 n For the body B we consider the coordinates for the centre of gravity G(xB,1,...,xB,n) defined by 1 xB,k := xkdx, k ∈{1,...,n}, V (B) B where V (B) := xdx B is the volume of B. COROLLARY 1. With the assumptions of Theorem 1 we have n ∂ f (x) n f (x)dx = ∑ αk xB,k − xk dx + ∑ αk xk − xB,k f (x)nk (x)dA ∂ ∂ B k=1 B xk k=1 B (3.6) and, in particular, 1 n ∂ f (x) 1 n f (x)dx = ∑ xB,k − xk dx + ∑ xk − xB,k f (x)nk (x)dA. ∂ ∂ B n k=1 B xk n k=1 B (3.7) The proof follows by (3.1)ontakingβk = αkxB,k, k ∈{1,...,n}. For a function f as in Theorem 1 above, we define the points ∂ f (x) xk ∂ dx = B xk , ∈{ ,..., }, xB,∂ f ,k : ∂ ( ) k 1 n f x dx B ∂xk provided that all denominators are not zero.
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