Green's function of the clamped boundary value problem for the differential operator on the interval is obtained. The best constant of corresponding Sobolev inequality is given by . In addition, it is shown that a reverse of the Sobolev best constant is the one which appears in the generalized Lyapunov inequality by Das and Vatsala (1975).

For , , let be a Sobolev (Hilbert) space associated with the inner product :

The fact that induces the equivalent norm to the standard norm of the Sobolev (Hilbert) space of th order follows from Poincaré inequality. Let us introduce the functional as follows:

To obtain the supremum of (i.e., the best constant of Sobolev inequality), we consider the following clamped boundary value problem:

Concerning the uniqueness and existence of the solution to , we have the following proposition. The result is expressed by the monomial :

Proposition 1.1.

For any bounded continuous function on an interval , has a unique classical solution expressed by

where Green's function is given by

is the determinant of matrix   , , and .

With the aid of Proposition 1.1, we obtain the following theorem. The proof of Proposition 1.1 is shown in Appendices A and B.

Theorem 1.2.

(i) The supremum (abbreviated as if there is no confusion) of the Sobolev functional is given by

Concretely,

(ii) is attained by , that is, .

Clearly, Theorem 1.2(i), (ii) is rewritten equivalently as follows.

Corollary 1.3.

Let , then the best constant of Sobolev inequality (corresponding to the embedding of into )

is . Moreover the best constant is attained by , where is an arbitrary complex number.

Next, we introduce a connection between the best constant of Sobolev- and Lyapunov-type inequalities. Let us consider the second-order differential equation

where . If the above equation has two points and in satisfying , then these points are said to be conjugate. It is wellknown that if there exists a pair of conjugate points in , then the classical Lyapunov inequality

holds, where . Various extensions and improvements for the above result have been attempted; see, for example, Ha [1], Yang [2], and references there in. Among these extensions, Levin [3] and Das and Vatsala [4] extended the result for higher order equation

For this case, we again call two distinct points and s₂conjugate if there exists a nontrivial solution of (1.12) satisfying

We point out that the constant which appears in the generalized Lyapunov inequality by Levin [3] and Das and Vatsala [4] is the reverse of the Sobolev best embedding constant.

Corollary 1.4.

If there exists a pair of conjugate points on with respect to (1.12), then

where is the best constant of the Sobolev inequality (1.9).

Without introducing auxiliary equation and the existence result of conjugate points as [2, 4], we can prove this corollary directly through the Sobolev inequality (the idea of the proof origins to Brown and Hinton [5, page 5]).

Proof of Corollary 1.4.

Consider

In the second inequality, the equality holds for the function which attains the Sobolev best constant, so especially it is not a constant function. Thus, for this function, the first inequality is strict, and hence we obtain

Since

we obtain the result.

Here, at the end of this section, we would like to mention some remarks about (1.12). The generalized Lyapunov inequality of the form (1.14) was firstly obtained by Levin [3] without proof; see Section 4 of Reid [6]. Later, Das and Vatsala [4] obtained the same inequality (1.14) by constructing Green's function for . The expression of the Green's function of Proposition 1.1 is different from that of [4]. The expression of [4, Theorem 2.1] is given by some finite series of and on the other hand, the expression of Proposition 1.1 is by the determinant. This complements the results of [7–9], where the concrete expressions of Green's functions for the equation but different boundary conditions are given, and all of them are expressed by determinants of certain matrices as Proposition 1.1.

First we enumerate the properties of Green's function of . has the following properties.

Lemma 2.1.

Consider the following:

(1)

(2)

(3)

(4)

Proof.

For and , , we have from (1.5)

For , noting the fact , we have (1). Next, for and , we have from (2.5)

Since , we have

Note that subtracting the th row from th row, the second equality holds. Equation is shown by the same way. Hence, we have (2). For , we have

where we used the fact , . So we have (3), and (4) follows from (3).

Using Lemma 2.1, we prove that the functional space associated with inner norm is a reproducing kernel Hilbert space.

Lemma 2.2.

For any , one has the reproducing property

Proof.

For functions and with arbitrarily fixed in , we have

Integrating this with respect to on intervals and , we have

Using (1), (2), and (4) in Lemma 2.1, we have (2.9).

In this section, we give a proof of Theorem 1.2 and Corollary 1.3.

Proof of Theorem 1.2 and Corollary 1.3.

Applying Schwarz inequality to (2.9), we have

Note that the last equality holds from (2.9); that is, substituting (2.9), . Let us assume that

holds (this will be proved in the next section). From definition of , we have

Substituting in to the above inequality, we have

Combining this and trivial inequality , we have

Hence, we have

which completes the proof of Theorem 1.2 and Corollary 1.3.

Thus, all we have to do is to prove (3.2).

In this section, we consider the diagonal value of Green's function, that is, . From Proposition 1.1, we have for

Thus, we can expect that takes the form . Precisely, we have the following proposition.

Proposition 4.1.

Consider

Hence,

where satisfy .

To prove this proposition, we prepare the following two lemmas.

Lemma 4.2.

Let , where

( satisfy ), then it holds that

Lemma 4.3.

Let , where , then it holds that (4.6) and .

Proof of Proposition 4.1.

From Lemmas 4.2 and 4.3, and satisfy BVP (in case of ). So we have

Inserting (4.9) into (4.8), we have Proposition 4.1.

Proof of Lemma 4.2.

Let

then differentiating    times we have

At first, for , we have

The first term vanishes because

The third term also vanishes because

Thus, we have

Hence, we have

by which we obtain (4.5). Next, for , we have

Since , we have . Thus, we have . For , we have

The first term vanishes because . Next, we show that the second term also vanishes. Let

Since , two rows, including the last row, coincide, and hence we have . Thus, we have . So we have obtained . By the same argument, we have . Hence, we have (4.6). Finally, we will show (4.7). For , noting , we have

where

Thus, we obtain . Hence we have

that is,

This completes the proof of Lemma 4.2.

Proof of Lemma 4.3.

Let

Differentiating times, we have

For , noting , , and , we have

Thus, we have (4.5). If , then we have

Since , we have . Hence, we have (4.6). If , then we have

This proves Lemma 4.3.

A. Deduction of (1.5)

In this section, (1.5) in Proposition 1.1 is deduced. Suppose that has a classical solution . Introducing the following notations:

is rewritten as

Let the fundamental solution be expressed as , where

then satisfy . satisfies the initial value problem , . is a unit matrix. Solving (A.2), we have

or equivalently, for , we have

Employing the boundary conditions (A.2), we have

In particular, if , then we have

On the other hand, using the boundary conditions (A.2) again, we have

Solving the above linear system of equations with respect to , , we have

Substituting (A.9) into (A.7), we have

Taking an average of the above two expressions and noting , we obtain (1.4), where Green's function is given by

Using properties , we have

where is Kronecker's delta defined by . Inserting these three relations into (A.11), we have

Applying the relation

where is any regular matrix and and are any matrices, we have (1.5).

B. Deduction of (1.6)

To prove (1.6), we show

Let . If (B.1) holds, substituting it to (1.5), replacing with , with , then we obtain (1.6). The case is shown in a similar way. Let be fixed, and let . Then satisfies

On the other hand, let

Differentiating times with respect to , we have

For , noticing , we have . For , we have

where we used . Similarly, for , we have . So satisfies

which is the same equation as (B.2). Hence, we have .

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