We have obtained the best constant of the following Lp
Sobolev inequality
sup
0≤y≤1|
u(j)(y)|
≤C (∫ 01
 |
u(M)(x)|
p
dx)1/p
,
where u is a function satisfying u(M) ∈ Lp(0, 1), u(2i)(0) = 0 (0  ≤i ≤
[(M − 1)/2]) and u(2i+1)(1) = 0 (0 ≤ i ≤ [(M − 2)/2]), where u(i) is
the abbreviation of (d/dx)iu(x). In [9], the best constant of the above
inequality was obtained for the case of p = 2 and j = 0. This paper
extends the result of [9] under the conditions p > 1 and 0 ≤ j ≤ M −1.
The best constant is expressed by Bernoulli polynomials

Kametaka, Yoshinori

Watanabe, Kohtaro

Yamagishi, Hiroyuki

English

Okayama University Scientific Achievement Repository

Math. J. Okayama Univ. 56 (2014), 145–155THE BEST CONSTANT OF Lp SOBOLEV INEQUALITYCORRESPONDING TO DIRICHLET-NEUMANNBOUNDARY VALUE PROBLEMHiroyuki Yamagishi, Kohtaro Watanabe and Yoshinori KametakaAbstract. We have obtained the best constant of the following LpSobolev inequalitysup0≤y≤1˛˛˛u(j)(y)˛˛˛ ≤ C„Z 10˛˛˛u(M)(x)˛˛˛p dx«1/p,where u is a function satisfying u(M) ∈ Lp(0, 1), u(2i)(0) = 0 (0 ≤ i ≤[(M − 1)/2]) and u(2i+1)(1) = 0 (0 ≤ i ≤ [(M − 2)/2]), where u(i) isthe abbreviation of (d/dx)iu(x). In [9], the best constant of the aboveinequality was obtained for the case of p = 2 and j = 0. This paperextends the result of [9] under the conditions p > 1 and 0 ≤ j ≤M − 1.The best constant is expressed by Bernoulli polynomials.1. IntroductionForM = 1, 2, 3, · · · , let us consider the following 1-dim Sobolev inequality:sup0≤y≤1|u(y)| ≤ C(∫ 10∣∣∣u(M)(x)∣∣∣2 dx)1/2 ,where u is an element of Sobolev-Hilbert space H(X,M) = {u|u(M) ∈L2(0, 1), u satisfies A(X)}. Here the condition A(X) assumesA(P) : u(i)(1)− u(i)(0) = 0 (0 ≤ i ≤M − 1),∫ 10u(x)dx = 0,A(AP) : u(i)(1) + u(i)(0) = 0 (0 ≤ i ≤M − 1),A(C) : u(i)(0) = u(i)(1) = 0 (0 ≤ i ≤M − 1),A(D) : u(2i)(0) = u(2i)(1) = 0 (0 ≤ i ≤ [ (M − 1)/2 ]),A(N) : u(2i+1)(0) = u(2i+1)(1) = 0 (0 ≤ i ≤ [ (M − 2)/2 ]),∫ 10u(x)dx = 0,A(DN) : u(2i)(0) = 0 (0 ≤ i ≤ [ (M − 1)/2 ]),u(2i+1)(1) = 0 (0 ≤ i ≤ [ (M − 2)/2 ]),Mathematics Subject Classification. Primary 34B27; Secondary 46E35.Key words and phrases. Lp Sobolev inequality, Best constant, Green function, Repro-ducing kernel, Bernoulli polynomial, Ho¨lder inequality.145146 H. YAMAGISHI, K. WATANABE AND Y. KAMETAKABoundary condition of Sobolev space p = 2 1 < p <∞ (general case)P (Periodic) [10] [3]AP (Anti Periodic) [10] —C (Clamped) [8] M = 1, 2, 3 [7]D (Dirichlet) [10] M = 2m [4], M = 1, 3, 5 [5]N (Neumann) [10] [6]DN (Dirichlet-Neumann) [10] this paperTable 1. Various boundary conditions and best constants.etc., and u(i) denotes i-th derivative in a distributional sense. It should benoted that if M = 1 the boundary conditions for u in A(N) and for u onx = 1 in A(DN) are not required. In our previous work, we obtained thebest constant of the following Lp Sobolev inequality (1.1) in some boundaryconditions:sup0≤y≤1|u(y)| ≤ C(∫ 10∣∣∣u(M)(x)∣∣∣p dx)1/p ,(1.1)where u is an element ofW (X,M, p) = {u|u(M) ∈ Lp(0, 1), u satisfies A(X)}.From this table, we see that the difficulty in obtaining the best constantseems to increase in the case of p 6= 2. Here, we would like to stress thateach result in the case of p 6= 2 was obtained through a different method.The unified approach (maximizing the diagonal value of reproducing kernels;see [8, 10]) as in the case of p = 2 does not exist in the case of p 6= 2.This paper studies the best constant of the following j-th Lp Sobolevinequality:sup0≤y≤1∣∣∣u(j)(y)∣∣∣ ≤ C (∫ 10∣∣∣u(M)(x)∣∣∣p dx)1/p ,(1.2)where u ∈W (DN,M, p) for any fixed p > 1 and 0 ≤ j ≤M − 1. To see thatthe inequality (1.2) itself is valid, we express u as follows; see [1, Th.VIII.2]u(M−1)(y) =∫ y0u(M)(x)dx (M = 2n− 1),−∫ 1yu(M)(x)dx (M = 2n),where the boundary conditions u(M−1)(0) = 0 (M = 2n−1) and u(M−1)(1) =0 (M = 2n) are used. Applying Ho¨lder inequality, we obtain (1.2) forj = M − 1, and similar argument leads (1.2). Now, to state the result, letLp SOBOLEV INEQUALITY 147us introduce Bernoulli polynomials bj(x) defined byb0(x) = 1,b′j(x) = bj−1(x),∫ 10bj(x)dx = 0 (j = 1, 2, 3, · · · ).Hence, we haveb1(x) = x−12, b2(x) =x22−x2+112, b3(x) =x36−x24+x12, · · · .The main result is as follows:Theorem 1.1. Let M,m,n = 1, 2, 3, · · · be integers, j = 0, 1, · · · ,M − 1and l = 0, 1, 2, · · · . Then, the best constant C(M, j, q) of (1.2) is given byC(M, j, q) =(1.3) 22(M−j)−1(∫ 10∣∣∣∣(−1)M+1bM−j(1− x4)+ bM−j(1 + x4)∣∣∣∣qdx)1/q(j = 2l),22(M−j)−1(∫ 10∣∣∣∣bM−j (x4)+ (−1)M bM−j(2− x4)∣∣∣∣qdx)1/q(j = 2l + 1),where q satisfies 1/p + 1/q = 1. Moreover, the equality of (1.2) attained byu(x) =(1.4) (−1)l∫ 10∂zG(n;x, z)(∂zG(n − l; z, 1))q−1dz(M = 2n− 1, j = 2l),(−1)l∫ 10∂zG(n;x, z)(∂y∂zG(n− l; z, y)|y=0)q−1dz(M = 2n− 1, j = 2l + 1),(−1)l∫ 10G(n;x, z)(G(n − l; z, 1))q−1dz(M = 2n, j = 2l),(−1)l∫ 10G(n;x, z)(∂yG(n− l; z, y)|y=0)q−1dz(M = 2n, j = 2l + 1),whereG(m;x, y) =(−1)m+142m−1[b2m(|x− y|4)− b2m(x+ y4)+(1.5)148 H. YAMAGISHI, K. WATANABE AND Y. KAMETAKAb2m(12−x+ y4)− b2m(12−|x− y|4) ]is Green function of the following Drichlet-Neumann boundary value problemBVP (DN,m){(−1)mu(2m) = f(x) (0 < x < 1),u(2i)(0) = u(2i+1)(1) = 0 (0 ≤ i ≤ m− 1).The following are concrete forms of C(M, j, q) for small value of M .M\j 0 1 2 31 1 - - -2 (q + 1)−1/q 1 - -3 32pi12q(2−6q−1Γ(q+1)Γ(q+ 32 )) 1q(q + 1)−1/q 1 -4 12(2F1(−q, q+12 ;q+32; 13 )q+1)1q 32pi32q(− 2−6q−1 csc(piq)Γ(−q)Γ(q+ 32 )) 1q(q + 1)−1/q 1Table 2. C(M, j, q) forM = 1, 2, 3, 4, where 2F1 is Gaussianhypergeometric function; see [2, Section 5.5].2. LemmasIn this section, lemmas necessary for proving Theorem 1.1 are enumer-ated. We assume that M,m,n = 1, 2, 3, · · · , j = 0, 1, · · · ,M − 1 andl = 0, 1, 2, · · · . First, we prepare the lemma concerning the properties ofBernoulli polynomials.Lemma 2.1. u(x) = (−1)m+1b2m(x) satisfies the following properties:max0≤x≤1u(x) = u(0) = u(1) > 0, min0≤x≤1u(x) = u(1/2) < 0,u′(x) < 0 (0 < x < 1/2) > 0 (1/2 < x < 1),max0≤x≤1|u(x) | = u(0) = u(1).Proof. See; [2, Section 9.5]. Figure 1 shows the graphs of u for m = 1 and 2.In fact, we show the case of m = 1. For u(x) = b2(x) = x2/2− x/2 + 1/12,we have u(0) = u(1) = 1/12 and u(1/2) = −1/24. Since u′(x) = b1(x) =x − 1/2 < 0 (0 < x < 1/2), > 0 (1/2 < x < 1), we have max0≤x≤1|u(x)| =u(0) = u(1). Next, we introduce lemmas for G(m;x, y).Lp SOBOLEV INEQUALITY 1490.2 0.4 0.6 0.8 1.00.040.020.020.040.060.080.2 0.4 0.6 0.8 1.00.00100.00050.00050.0010(i) (ii)Figure 1. The graphs of (i) u(x) = b2(x) (m = 1) and (ii)u(x) = −b4(x) (m = 2).Lemma 2.2. For any bounded continuous function f(x) (0 < x < 1),BVP(DN,m) has a unique classical solution u(x) (0 < x < 1) expressedasu(x) =∫ 10G(m;x, y)f(y)dy,where G(m;x, y) (0 < x, y < 1) is given by (1.5).Proof. See; Yamagishi [9, Theorem 3.1]. Lemma 2.3. For any u ∈ W (DN,m, p) and for any fixed y (0 ≤ y ≤ 1),the following reproducing relation holds:u(y) =∫ 10u(m)(x)∂mx G(m;x, y)dx.(2.1)Proof. See; Yamagishi [9, Theorem 5.1] (although the proof of Theorem5.1 is written in the case of u ∈ H(DN,m), it still applies to our case ofu ∈W (DN,m, p) without modification). Lemma 2.4. The following relations hold in 0 < x, y < 1 and x 6= y.(1) ∂2xG(m;x, y) = ∂2yG(m;x, y) = −G(m− 1;x, y).(2) ∂jy∂Mx G(M ;x, y) =(−1)n−1+l ∂xG(n − l;x, y) (M = 2n− 1, j = 2l),(−1)n−1+l ∂y∂xG(n − l;x, y) (M = 2n− 1, j = 2l + 1),(−1)n+lG(n − l;x, y) (M = 2n, j = 2l),(−1)n+l ∂yG(n− l;x, y) (M = 2n, j = 2l + 1).Proof. Differentiating G(m;x, y) with respect to x twice, we have∂xG(m;x, y) =(2.2)150 H. YAMAGISHI, K. WATANABE AND Y. KAMETAKA(−1)m+142m−2[sgn(x− y) b2m−1(|x− y|4)− b2m−1(x+ y4)−b2m−1(12−x+ y4)+ sgn(x− y) b2m−1(12−|x− y|4) ],∂2xG(m;x, y) =− (−1)m42(m−1)−1[b2(m−1)(|x− y|4)− b2(m−1)(x+ y4)+b2(m−1)(12−x+ y4)− b2(m−1)(12−|x− y|4) ]=− G(m− 1;x, y).SinceG(m,x, y) = G(m; y, x), we have ∂2yG(m;x, y) = ∂2yG(m; y, x)= −G(m−1; y, x) = −G(m − 1;x, y). So, (1) is obtained. Using (1), we have the fol-lowing relation:∂jy∂Mx G(M ;x, y) =∂2ly ∂2(n−1)+1x G(2n − 1;x, y) =(−1)n−1+l ∂xG(2n − 1− (n− 1)− l;x, y),∂2l+1y ∂2(n−1)+1x G(2n − 1;x, y) =(−1)n−1+l ∂y∂xG(2n − 1− (n− 1)− l;x, y),∂2ly ∂2nx G(2n;x, y) = (−1)n+lG(2n − n− l;x, y),∂2l+1y ∂2nx G(2n;x, y) = (−1)n+l ∂yG(2n − n− l;x, y).This shows (2). Lemma 2.5. The following relations hold.G(m;x, y) > 0 (0 < x, y < 1).∂xG(m;x, y), ∂yG(m;x, y), ∂y∂xG(m;x, y) > 0(0 < x, y < 1, x 6= y).∂2xG(m;x, y), ∂2yG(m;x, y) < 0 (0 < x, y < 1, x 6= y).Proof. Considering Lemma 2.1 and0 ≤|x− y|4<x+ y4<12, 0 <12−x+ y4<12−|x− y|4≤12,we have(−1)m+1[b2m(|x− y|4)− b2m(x+ y4) ]> 0,Lp SOBOLEV INEQUALITY 151(−1)m+1[b2m(12−x+ y4)− b2m(12−|x− y|4) ]> 0.Hence we have G(m;x, y) > 0. Moreover, from Lemma 2.4 (1), we have∂2xG(m;x, y) = ∂2yG(m;x, y) = −G(m− 1;x, y) < 0(2.3)(0 < x, y < 1, x 6= y).For any fixed y (0 ≤ y ≤ 1), since ∂xG(m;x, y)∣∣x=1= 0 and (2.3), we have∂xG(m;x, y) > 0 (0 < x < 1, x 6= y). Similarly, for any fixed x (0 ≤ x ≤ 1),since ∂yG(m;x, y)∣∣y=1= 0 and (2.3), we have ∂yG(m;x, y) > 0 (0 < y <1, x 6= y). From (2.2) and Lemma 2.4 (2), we have∂y∂xG(m;x, y) =(2.4)(−1)m+142m−3[− b2m−2(|x− y|4)− b2m−2(x+ y4)+b2m−2(12−x+ y4)+ b2m−2(12−|x− y|4) ],∂2y∂xG(m;x, y) = − ∂xG(m− 1;x, y) < 0 (0 < x, y < 1, x 6= y).For any fixed x (0 ≤ x ≤ 1), from ∂y∂xG(m;x, y)∣∣y=1= 0 and (2.4), we have∂y∂xG(m;x, y) > 0 (0 < y < 1, x 6= y). Using these lemmas, we can prove the following lemma.Lemma 2.6. Let us defineg(M, j; y) =∫ 10∣∣∂jy∂Mx G(M ;x, y)∣∣q dx (0 ≤ y ≤ 1).Then, it holds thatmax0≤y≤1g(M, j; y) ={g(M, j; 1) (j = 2l),g(M, j; 0) (j = 2l + 1).Proof. By Lemma 2.4 (2) and 2.5, we obtain:g(2n − 1, 2l; y) =∫ 10(∂xG(n − l;x, y))qdx > 0,g′(2n− 1, 2l; y) =q∫ 10(∂xG(n− l;x, y))q−1dx ∂y∂xG(n − l;x, y) > 0,g(2n − 1, 2l + 1; y) =∫ 10(∂y∂xG(n − l;x, y))qdx > 0,g′(2n− 1, 2l + 1; y) =152 H. YAMAGISHI, K. WATANABE AND Y. KAMETAKA− q∫ 10(∂y∂xG(n− l;x, y))q−1dx ∂xG(n − l − 1;x, y) < 0,g(2n, 2l; y) =∫ 10(G(n − l;x, y))qdx > 0,g′(2n, 2l; y) = q∫ 10(G(n− l;x, y))q−1dx ∂yG(n− l;x, y) > 0,g(2n, 2l + 1; y) =∫ 10(∂yG(n− l;x, y))qdx > 0,g′(2n, 2l + 1; y) =− q∫ 10(∂yG(n− l;x, y))q−1dxG(n − l − 1;x, y) < 0,where g′ stands for the derivative with respect to y. Thus, if j is even,g(M, j, y) takes its maximum at y = 1, else at y = 0. Next lemma is necessary for the construction of the function, which at-tains the best constant. Note the difference from BVP(DN,m) (it is the oddorder differential equation and boundary conditions at x = 1 are slightlydifferent).Lemma 2.7. For any bounded continuous function f(x) (0 < x < 1), thefollowing boundary value problem:BVP (DN′,m)(−1)m−1u(2m−1) = f(x) (0 < x < 1),u(2i)(0) = 0 (0 ≤ i ≤ m− 1),u(2i+1)(1) = 0 (0 ≤ i ≤ m− 2),has a unique classical solution u(x) (0 < x < 1) expressed asu(x) =∫ 10∂yG(m;x, y)f(y)dy.(2.5)Proof. Integrating the equation f(y) = (−1)m−1u(2m−1)(y) with respect toy (0 < y < x) and noting u(2m−2)(0) = 0, we have∫ x0f(y)dy = (−1)m−1∫ x0u(2m−1)(y)dy = (−1)m−1u(2(m−1))(x).Since u satisfies u(2i)(0) = u(2i+1)(1) = 0 for (0 ≤ i ≤ m− 2), u is a solutionof BVP(DN,m− 1). Thus, from Lemma 2.2, we have the solution formulau(x) =∫ 10G(m− 1;x, z)∫ z0f(y)dydz =Lp SOBOLEV INEQUALITY 153∫ 10∫ 1yG(m− 1;x, z)dzf(y)dy.Using Lemma 2.4 (1), we have −∂2zG(m;x, z) = G(m − 1;x, z). So, weobtainu(x) =∫ 10∫ 1y(− ∂2zG(m;x, z))dzf(y)dy =∫ 10∂yG(m;x, y)f(y)dy,where we use ∂zG(m;x, z)∣∣z=1= 0. The uniqueness easily follows from theexpression (2.5). 3. Proof of Theorem 1.1Proof of Theorem 1.1 Let u ∈ W (DN,M, p). Differentiating (2.1) inLemma 2.3 with respect to y (0 ≤ y ≤ 1), j (j = 0, 1, · · · ,M − 1) times, weobtainu(j)(y) =∫ 10u(M)(x) ∂jy∂Mx G(M ;x, y)dx.Applying Ho¨lder inequality, we have∣∣∣ u(j)(y) ∣∣∣ ≤(∫ 10∣∣ ∂jy∂Mx G(M ;x, y) ∣∣q dx)1/q (∫ 10∣∣∣ u(M)(x) ∣∣∣p dx)1/p .Taking the supremum of the above relation and using Lemma 2.6, we havesup0≤y≤1∣∣∣u(j)(y) ∣∣∣ ≤(3.1) (∫ 10∣∣∣ ∂jy∂Mx G(M ;x, y)∣∣y=1∣∣∣q dx)1/q (∫ 10∣∣∣ u(M)(x) ∣∣∣p dx)1/p(j = 2l),(∫ 10∣∣∣ ∂jy∂Mx G(M ;x, y)∣∣y=0∣∣∣q dx)1/q (∫ 10∣∣∣ u(M)(x) ∣∣∣p dx)1/p(j = 2l + 1).The equality holds if u satisfiesu(M)(x) =(3.2) sgn(∂jy∂Mx G(M ;x, y)∣∣y=1) ∣∣∣∂jy∂Mx G(M ;x, y)∣∣y=1∣∣∣q−1 (j = 2l),sgn(∂jy∂Mx G(M ;x, y)∣∣y=0) ∣∣∣∂jy∂Mx G(M ;x, y)∣∣y=0∣∣∣q−1 (j = 2l + 1).154 H. YAMAGISHI, K. WATANABE AND Y. KAMETAKATherefore, if the equality holds, the best constant isC(M, j, q) =(3.3) (∫ 10∣∣∣ ∂jy∂Mx G(M ;x, y)∣∣y=1∣∣∣q dx)1/q (j = 2l),(∫ 10∣∣∣ ∂jy∂Mx G(M ;x, y)∣∣y=0∣∣∣q dx)1/q (j = 2l + 1).The concrete expression of (3.3) is (1.3). Finally, we see that the equalityof (3.1) is attained by (1.4). From the equation (3.2) and Lemma 2.4 (2),we haveu(2n−1)(x) =(−1)n−1+l(∂xG(n − l;x, y)∣∣y=1)q−1(M = 2n− 1, j = 2l),(∂y∂xG(n − l;x, y)|y=0)q−1(M = 2n− 1, j = 2l + 1),u(2n)(x) =(−1)n+l(G(n − l;x, y)∣∣y=1)q−1(M = 2n, j = 2l),(∂yG(n− l;x, y)|y=0)q−1(M = 2n, j = 2l + 1).Since u satisfies boundary condition A(DN), we have the following boundaryvalue problems:(−1)n−1u(2n−1) =(−1)l(∂xG(n− l;x, y)∣∣y=1)q−1(M = 2n− 1, j = 2l),(∂y∂xG(n − l;x, y)|y=0)q−1(M = 2n− 1, j = 2l + 1),u(2i)(0) = 0 (0 ≤ i ≤ n− 1), u(2i+1)(1) = 0 (0 ≤ i ≤ n− 2),(3.4)(−1)nu(2n) =(−1)l(G(n− l;x, y)∣∣y=1)q−1(M = 2n, j = 2l),(∂yG(n − l;x, y)|y=0)q−1(M = 2n, j = 2l + 1),u(2i)(0) = u(2i+1)(1) = 0 (0 ≤ i ≤ n− 1).(3.5)Thus u in (3.4) is the solution of BVP(DN′,n) and u in (3.5) is the solutionof BVP(DN,n). So, by Lemma 2.7 and Lemma 2.2, we have (1.4). References[1] H. Brezis, Analyse Functionnelle, Masson Editeur (1983).Lp SOBOLEV INEQUALITY 155[2] R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics, Addison-Wesley(1989).[3] Y. Kametaka, Y. Oshime, K. Watanabe, H. Yamagishi, A. Nagai and K. Takemura,The best constant of Lp Sobolev inequality corresponding to the periodic boundaryvalue problem for (−1)M (d/dx)2M , Sci. Math. Jpn. e-2007 (2007), 269–281.[4] Y. Oshime, Y. Kametaka, H. Yamagishi, The best constant of Lp Sobolev inequalitycorresponding to the Dirichlet boundary value problem for (d/dx)4m, Sci. Math. Jpn.68 (2008), 313-321.[5] Y. Oshime, K. Watanabe, The best constant of Lp Sobolev inequality correspondingto the Dirichlet boundary value problem II, Tokyo J. Math. 34 (2011), 115–133.[6] Y. Oshime, H. Yamagishi, K. Watanabe, The best constant of Lp Sobolev inequalitycorresponding to Neumann boundary value problem for (−1)M (d/dx)2M , HiroshimaMath. J. 42 (2012), 293–299.[7] K. Watanabe, Y. Kametaka, A. Nagai, H. Yamagishi, K. Takemura, Symmetrizationof functions and the best Constant of 1-DIM Lp Sobolev inequality, J. of Ineq. andAppl. Vol. 2009 (2009), Article ID 874631.[8] K.Watanabe, Y.Kametaka, H.Yamagishi, A.Nagai and K.Takemura, The best con-stant of Sobolev inequality corresponding to clamped boundary value problem, Bound.Value Probl. Vol. 2011 (2011), Article ID 875057.[9] H. Yamagishi, The best constant of Sobolev inequality corresponding to Dirichlet-Neumann boundary value problem for (−1)M (d/dx)2M , Hiroshima Math. J. 39, 3(2009), 421-442.[10] H. Yamagishi, Y. Kametaka, A. Nagai, K. Watanabe and K. Takemura, Riemann zetafunction and the best constants of five series of Sobolev inequalities, RIMS KokyurokuBessatsu B13 (2009), 125–139.Hiroyuki YamagishiTokyo Metropolitan College of Industrial Technology1-10-40 Higashi-ooi, Shinagawa Tokyo 140-0011, Japane-mail address: yamagisi@s.metro-cit.ac.jpKohtaro WatanabeDepartment of Computer Science, National Defense Academy,1-10-20 Yokosuka 239-8686, Japane-mail address: wata@nda.ac.jpYoshinori KametakaFaculty of Engineering Science, Osaka University,1-3 Machikaneyama-cho, Toyonaka 560-8531, Japane-mail address: kametaka@sigmath.es.osaka-u.ac.jp(Received April 25, 2011 )(Revised July 8, 2011 )

THE BEST CONSTANT OF L<sup>p</sup> SOBOLEV INEQUALITY CORRESPONDING TO DIRICHLET-NEUMANN BOUNDARY VALUE PROBLEM

Okayama University Digital Information Repository

http://ousar.lib.okayama-u.ac.jp/files/public/5/52074/20160528113240582435/miou_056_145_155.pdf

THE BEST CONSTANT OF L<sup>p</sup> SOBOLEV INEQUALITY CORRESPONDING TO DIRICHLET-NEUMANN BOUNDARY VALUE PROBLEM

Abstract

Similar works

Full text

Available Versions

Okayama University Scientific Achievement Repository

Okayama University Digital Information Repository