Exact solutions and bifurcation curves of nonlocal elliptic equations with convolutional Kirchhoff functions

Abstract

We study the one-dimensional nonlocal elliptic equation of Kirchhoff type with convolutional Kirchhoff functions. We establish the exact solutions $u_{\lambda}$ and bifurcation curves $\lambda (\alpha )$, where $\alpha := \Vert u_{\lambda}\Vert _{\infty}$.

1 Introduction

We consider the following one-dimensional nonlocal elliptic equation with convolutional Kirchhoff function:

$$ \textstyle\begin{cases} - ({\int _{0}^{1}} f(x)u(x)^{q} \,dx ) u''(x)= \lambda u(x)^{p}, \quad x \in I:= (0,1), \\ u(x) > 0, \quad x\in I, \\ u(0) = u(1) = 0, \end{cases} $$

(1.1)

where $f(x) = (1-x)^{n}$ ($n \in \mathbb{N}$) and $p, q > 1$ are given constants. Further, $\lambda > 0$ is a bifurcation parameter.

Equation (1.1) is motivated by the convolution nonlocal elliptic problem of Kirchhoff type in [9]:

$$ -J \bigl( \bigl(h *u^{q} \bigr) (1) \bigr)u''(x)= \lambda g \bigl(x, u(x) \bigr), \quad x \in I, $$

(1.2)

where J, h, and g are continuous functions and

$$ \bigl(h * u^{q} \bigr) (t) := \int _{0}^{t} h(t-s)u(s)^{q}\,ds. $$

If we put $J(y) = y$, $h(y) = y^{n}$ and $g(x, u) = u^{p}$, then we obtain (1.1).

The purpose of this paper is to obtain the exact solutions and bifurcation curves of problem (1.1) by concentrating on the typical convolutional equation (1.1). Our results are novel since there seem to be few results to treat such problem as (1.1) from a viewpoint of bifurcation analysis, and the results obtained here will be the good first step to understand well the structures of solutions and bifurcation curves in the field of nonlocal elliptic problems. Moreover, as far as the author knows, the bifurcation phenomena of nonlocal problem with the coefficient coming from the convolution have not been considered before.

It is known that there are so many results concerning nonlocal and related problems. We refer to [1–4, 6–11, 13–18, 23, 25] and the references therein. In addition to this, there is much interesting and significant motivation to study this kind of nonlocal problems as (1.2). We refer to [9] and the references therein to understand the background of this problem. On the other hand, although the analysis of bifurcation diagram is a very popular problem in nonlinear elliptic problems, there are few results concerning the bifurcation problems for nonlocal problems. We refer to [19–22, 24].

Before stating our results, we explain some notations. For $p> 1$, let

$$ \textstyle\begin{cases} {-}W''(x)= W(x)^{p}, \quad x \in I, \\ W(x) > 0, \quad x\in I, \\ W(0) = W(1) = 0. \end{cases} $$

(1.3)

We know from [5] that there exists a unique solution $W_{p}(x)$ of (1.3). For $d, k\ge 0$, we put

$$\begin{aligned} &L_{k,d}:= \int _{0}^{1} \frac{s^{d}}{\sqrt{1-s^{k+1}}}\,ds, \end{aligned}$$

(1.4)

$$\begin{aligned} &M_{k,d}:= \int _{0}^{1} (1-x)^{k}W_{p}(x)^{d} \,dx, \end{aligned}$$

(1.5)

$$\begin{aligned} &R_{k, d}:= \int _{0}^{1} x^{k} W_{p}(x)^{d}\,dx, \end{aligned}$$

(1.6)

$$\begin{aligned} &S_{k, d}:= \int _{0}^{1/2} x^{k} W_{p}(x)^{d}\,dx. \end{aligned}$$

(1.7)

Let $\Vert \cdot \Vert _{\beta}$ ($1 \le \beta \le \infty$) be the usual $L^{\beta}$-norm. We know from [22] that, for $p > 1$,

$$\begin{aligned} \xi _{p} := \Vert W_{p} \Vert _{\infty } =& \bigl(2(p+1) \bigr)^{1/(p-1)}L_{p,0}^{2/(p-1)}. \end{aligned}$$

(1.8)

Now, we state our results.

Theorem 1.1

Let $f(x) = (1-x)^{n}$ $(n \in \mathbb{N})$. Further, $\lambda > 0$ is a given constant.

(i) Assume that $p \neq q+1$. Then the solution $u_{\lambda}$ is given by

$$\begin{aligned} u_{\lambda }= \biggl(\frac{\lambda}{M_{n,q}} \biggr)^{1/(q-p+1)}W_{p}(x). \end{aligned}$$

(1.9)

(ii) Assume that $p = q+1$.

(a) Suppose that $\lambda = M_{n,q}$. Then all the solutions $u_{\lambda}$ are represented as $u_{\lambda }= tW_{p}$, where $t > 0$ is an arbitrary constant.

(b) Assume that there exists a solution $U_{\lambda}$ of (1.1). Then $U_{\lambda }= r_{\lambda }W_{p}$, where $r_{\lambda}:= (Q_{n,q}/\lambda )^{1/(p-1)}$ and

$$\begin{aligned} Q_{n,q}:= \int _{0}^{1} (1-x)^{n}u_{\lambda}(x)^{q} \,dx. \end{aligned}$$

(1.10)

Moreover, $\lambda = M_{n,q}$ holds. Therefore, by (i) above, all the solutions $u_{\lambda}$ of (1.1) are obtained as $u_{\lambda }= tW_{p}$, where $t > 0$ is an arbitrary constant.

(c) Assume that $\lambda \neq M_{n,q}$. Then (1.1) has no solutions.

By Theorem 1.1, we see that the essential point to obtain the solution $u_{\lambda}$ is to find $M_{n,q}$. In the following Theorems 1.2 and 1.3, by using Theorem 1.1 (i), we obtain the exact solution $u_{\lambda}$ for given $\lambda > 0$ and show that λ is parameterized by $\alpha := \Vert u_{\lambda}\Vert _{\infty}$, namely, $\lambda =\lambda (\alpha )$, and we establish the exact formula of $\lambda (\alpha )$.

Theorem 1.2

Let $f(x) = 1-x$ in (1.1). Assume that $p \neq q+1$. Then, for any given $\lambda > 0$,

$$\begin{aligned} u_{\lambda}(x) &= \lambda ^{1/(q-p+1)} \\ &\quad \times \bigl\{ 2^{(q-p+1)/(p-1)}(p+1)^{q/(p-1)}L_{p,0}^{(2q-p+1)/(p-1)} L_{p,q} \bigr\} ^{-1/(q-p+1)} W_{p}(x), \end{aligned}$$

(1.11)

$$\begin{aligned} \lambda (\alpha ) =& (p+1)L_{p,0}L_{p,q}\alpha ^{q-p+1}. \end{aligned}$$

(1.12)

Theorem 1.3

Let $f(x) = (1-x)^{2}$. Assume that $q = m(p+1)$ or $q = m(p+1)+p$, where $m \in \mathbb{N}$. Then

(i)

$$\begin{aligned} &u_{\lambda}(x) = \biggl(\frac{\lambda}{M_{2,q}} \biggr)^{1/(q-p+1)}W_{p}(x), \end{aligned}$$

(1.13)

$$\begin{aligned} &\lambda (\alpha ) = M_{2,q}\xi _{p}^{-(q-p+1)} \alpha ^{q-p+1}, \end{aligned}$$

(1.14)

where

$$\begin{aligned} &M_{2,q} := C_{0,m} + C_{0,1,m}S_{1,0} + C_{0,2,m}S_{2, 0}, \quad \bigl(q = m(p+1) \bigr), \end{aligned}$$

(1.15)

$$\begin{aligned} &M_{2,q} := C_{1,m} + C_{1,1,m}S_{1,p} + C_{1,2,m}S_{2, p}, \quad \bigl(q = m(p+1)+p \bigr), \end{aligned}$$

(1.16)

$C_{0,m}$, $C_{0,1,m}$, $C_{1,2,m}$, $C_{1,m}$, $C_{1,1,m}$, $C_{1,2,m}$ are constants that depend on $\xi _{p}$ and are obtained inductively. Here,

$$\begin{aligned} &S_{1,0} =\frac{1}{8}, \end{aligned}$$

(1.17)

$$\begin{aligned} &S_{2,0} = \frac{1}{24}, \end{aligned}$$

(1.18)

$$\begin{aligned} &S_{1,p} =\xi _{p}, \end{aligned}$$

(1.19)

$$\begin{aligned} &S_{2,p} = \xi _{p} - \sqrt{2(p+1)}\xi _{p}^{(3-p)/2}L_{p,1}. \end{aligned}$$

(1.20)

For the special case $m = 1$, the following (ii) and (iii) hold:

(ii) Let $q = p + 1$. Then

$$\begin{aligned} M_{2,p+1} =& R_{2,p+1} = S_{0,q} - 2S_{1,q} + 2S_{2,q} \\ =& \sqrt{\frac{p+1}{2}}\xi _{p}^{(p+3)/2}L_{p,p+1} - \frac{1}{3(p+1)}\xi _{p}^{p+1} - \frac{p+1}{p+3}\sqrt{2(p+1)}\xi _{p}^{(5-p)/2}L_{p,1}. \end{aligned}$$

(1.21)

(iii) Let $q = 2p+1 (= p+1 + p)$. Then

$$\begin{aligned} M_{2, 2p+1} =& R_{2,2p+1} = S_{0,q} - 2S_{1,q} + 2S_{2,q} \\ =& \sqrt{\frac{p+1}{2}}\xi _{p}^{3(p+1)/2}L_{p, 2p+1} - \frac{2}{3} \sqrt{2(p+1)}\xi _{p}^{(p+5)/2} \biggl(\frac{1}{p+2}L_{p,p+2} + 2L_{p,1} \biggr). \end{aligned}$$

(1.22)

Now, we consider the case $p = q$.

Theorem 1.4

Let $p = q > 1$.

(i) Assume that $n \ge 2$. Then $u_{\lambda}$ is obtained inductively by using $S_{k,1}$ ($1 \le k \le n-2$).

(ii) Especially, let $n = 3$. Then

$$\begin{aligned} M_{3,p} = \sqrt{\frac{2}{p+1}}\xi ^{(p+1)/2} - 3 \sqrt{2(p+1)} \xi _{p}^{(3-p)/2}L_{p,1} \end{aligned}$$

(1.23)

and (1.12), (1.13) hold by replacing $M_{2,q}$ and q with $M_{3,p}$ and p, respectively.

Remarks. (i) The case $n = 1$ in Theorem 1.4 is contained in Theorem 1.1.

(ii) The novelty of Theorem 1.4 is to give a scheme to obtain $u_{\lambda}$ inductively. Unfortunately, we are not able to calculate the concrete value of $S_{k,1}$ for $k \ge 1$. We only find that $M_{n,p}$ is expressed by $S_{k,1}$ ($1 \le k \le n-2$). Exceptionally, we have (cf. (4.3))

$$\begin{aligned} S_{0,1} = \sqrt{\frac{p+1}{2}}\xi _{p}^{(3-p)/2}L_{p,1}. \end{aligned}$$

(1.24)

The remainder of this paper is organized as follows. In Sect. 2, we first prove Theorem 1.1. Next, we explain the existence of $u_{\lambda}$ for $\lambda > 0$ and the fundamental properties of $W_{p}$. In Sects. 3, 4, and 5, the proofs of Theorems 1.2, 1.3, and 1.4 will be given respectively. The main tools of the proofs are time map argument and complicated direct calculation.

2 Proof of Theorem 1.1 and preliminaries

In what follows, we write $\xi = \xi _{p}$ for simplicity. In this section, we consider the case $f(x) = (1-x)^{n}$, where $n \in \mathbb{N}$. For given $\lambda > 0$, we look for the solution $u_{\lambda}$ of the form $u_{\lambda }= tW_{p}$ ($t > 0$).

Proof of Theorem 1.1

(i) Let $p \neq q+1$. We look for the solution of (1.1) of the form $u_{\lambda }= t_{\lambda }W_{p}$, where $t_{\lambda }> 0$ is a suitable constant determined in (2.2). By (1.1), we have

$$\begin{aligned} -t_{\lambda}^{q+1} \biggl( \int _{0}^{1} (1-x)^{n}W_{p}(x)^{q} \,dx \biggr)W_{p}''(x) = \lambda t_{\lambda}^{p} W_{p}(x)^{p}. \end{aligned}$$

(2.1)

Then there exists unique $t_{\lambda }> 0$ satisfying

$$\begin{aligned} t_{\lambda} = \biggl(\frac{\lambda}{M_{n,q}} \biggr)^{1/(q-p+1)}, \end{aligned}$$

(2.2)

and we find that $u_{\lambda }= t_{\lambda }W_{p}$ satisfies (1.1). We next show that, if there exists a solution $u_{\lambda}$ of (1.1), then $u_{\lambda }= t_{\lambda }W_{p}$, where $t_{\lambda}$ is a constant given in (2.2). Indeed, we put ${r_{\lambda}:= (\frac{Q_{n,q}}{\lambda} )^{1/(p-1)}}$ and $w_{\lambda}:= r_{\lambda}^{-1} u_{\lambda}$. Then we see from (1.1) and (1.10) that $w_{\lambda}$ satisfies (1.3). Namely, $w_{\lambda }= W_{p}$. Then, by the definition of $r_{\lambda}$ and (1.10), we obtain

$$\begin{aligned} \lambda r_{\lambda}^{p-1} =& Q_{n,q} = \int _{0}^{1} (1-x)^{n} u_{ \lambda}(x)^{q}\,dx \\ =& \int _{0}^{1} (1-x)^{n} r_{\lambda}^{q} W_{p}(x)^{q}\,dx = r_{ \lambda}^{q} M_{n,q}. \end{aligned}$$

(2.3)

By this and (2.2), we see that ${r_{\lambda }= (\frac{\lambda}{M_{n,q}} )^{1/(q-p+1)} = t_{\lambda}}$ and $u_{\lambda }= t_{\lambda }W_{p}$. Thus the proof is complete. □

(ii) Now assume that $p = q+1$.

(a) Assume that $\lambda = M_{n,q}$. For $t > 0$, we put $U_{\lambda }:= tW_{p}$ and substitute it into (1.1). Then we obtain

$$\begin{aligned} -t^{q+1} \biggl( \int _{0}^{1} (1-x)^{n}W_{p}(x)^{q} \,dx \biggr)W_{p}''(x) = \lambda t^{p} W_{p}(x)^{p}. \end{aligned}$$

(2.4)

Then we see that (2.4) drives us to (1.3). Therefore, $U_{\lambda }= tW_{p}$ is a solution of (1.1). Thus the proof of (ii) (a) is complete.  □

(b) The proof of (ii) (b) is the same as that of Theorem 1.1 (i). So, we omit the proof.  □

(c) By (a) and (b) above, a solution $u_{\lambda}$ of (1.1) exists if and only if the equality $\lambda = M_{n,q}$ holds. Therefore, if $\lambda \neq M_{n,q}$, then (1.1) has no solutions. Thus the proof of Theorem 1.1 is complete. □

To calculate $M_{n,q}$, we need some fundamental properties of $W_{p}$. Since (1.3) is autonomous, we know from [5] that

$$\begin{aligned} &W_{p}(x) = W_{p}(1-x), \quad 0 \le x \le \frac{1}{2}, \end{aligned}$$

(2.5)

$$\begin{aligned} &\xi = \Vert W_{p} \Vert _{\infty }= \max _{0\le x \le 1}W_{p}(x) = W_{p} \biggl( \frac{1}{2} \biggr), \end{aligned}$$

(2.6)

$$\begin{aligned} &W_{p}'(x) > 0, \quad 0 \le x < \frac{1}{2}. \end{aligned}$$

(2.7)

By (1.3), for $0 \le x \le 1$, we have

$$\begin{aligned} \bigl\{ W_{p}''(x) + W_{p}(x)^{p} \bigr\} W_{p}'(x) = 0. \end{aligned}$$

(2.8)

By this and (2.6), we have

$$\begin{aligned} \frac{1}{2}W_{p}'(x)^{2} + \frac{1}{p+1}W_{p}(x)^{p+1} = \text{constant} = \frac{1}{p+1} W_{p} \biggl(\frac{1}{2} \biggr)^{p+1} = \frac{1}{p+1}\xi ^{p+1}. \end{aligned}$$

(2.9)

By this and (2.8), for $0 \le x \le 1/2$, we have, using $\theta = \xi _{p} s$,

$$\begin{aligned} W_{p}'(x) = \sqrt{\frac{2}{p+1} \bigl(\xi ^{p+1} - W_{p}(x)^{p+1} \bigr)}. \end{aligned}$$

(2.10)

By (2.5)–(2.7), (2.10) and putting $W_{p}(x) = \xi s$, we have

$$\begin{aligned} \Vert W_{p} \Vert _{q}^{q} =& 2 \int _{0}^{1/2} W_{p}(x)^{q} \,dx \\ =& 2 \int _{0}^{1/2} W_{p}(x)^{q} \frac{W_{p}'(x)}{ \sqrt{\frac{2}{p+1}(\xi ^{p+1} - W_{p}(x)^{p+1})}}\,dx \\ =& 2\sqrt{\frac{p+1}{2}}\xi ^{(2q-p+1)/2} \int _{0}^{1} \frac{s^{q}}{\sqrt{1-s^{p+1}}}\,ds \\ =& 2\sqrt{\frac{p+1}{2}}\xi ^{(2q-p+1)/2}L_{p,q}. \end{aligned}$$

(2.11)

3 Proof of Theorem 1.2

Now we put $n = 1$ and consider the case $f(x) = 1-x$. By (1.6), (2.5), (2.11) and putting $s = 1-t$, we have

$$\begin{aligned} R_{1,q} =& \int _{0}^{1} sW_{p}(s)^{q} \,ds \\ =& \int _{0}^{1/2} sW_{p}(s)^{q} \,ds + \int _{1/2}^{1} sW_{p}(s)^{q} \,ds \\ =& \int _{0}^{1/2} sW_{p}(s)^{q} \,ds + \int _{0}^{1/2} (1-t)W_{p}(t)^{q} \,dt \\ =& \int _{0}^{1/2} W_{p}(s)^{q} \,ds \\ =& \sqrt{\frac{p+1}{2}}\xi ^{(2q-p+1)/2}L_{p,q}. \end{aligned}$$

(3.1)

By this, (1.5), and (2.11), we obtain

$$\begin{aligned} M_{1,q} =& \Vert W_{p} \Vert _{q}^{q} - R_{1,q} = \sqrt{\frac{p+1}{2}} \xi ^{(2q-p+1)/2}L_{p,q}. \end{aligned}$$

(3.2)

By this and (2.2), we have

$$\begin{aligned} u_{\lambda}(x) =& \biggl(\frac{\lambda}{M_{1,q}} \biggr)^{1/(q-p+1)}W_{p}(x) \\ =& \lambda ^{1/(q-p+1)} \biggl\{ \sqrt{\frac{p+1}{2}}\xi ^{(2q-p+1)/2}L_{p,q} \biggr\} ^{-1/(q-p+1)} W_{p}(x). \end{aligned}$$

(3.3)

This along with (1.8) implies (1.11). Now, we put $x = 1/2$ in (3.3). Then we have

$$\begin{aligned} \alpha =& \lambda ^{1/(q-p+1)} \biggl\{ \sqrt{\frac{p+1}{2}}L_{p,q} \biggr\} ^{-1/(q-p+1)}\xi ^{-(p-1)/(2(q-p+1))}. \end{aligned}$$

(3.4)

This along with (1.8) implies (1.12). Thus the proof of Theorem 1.2 is complete.  □

4 Proof of Theorem 1.3

In this section, let $n = 2$, namely, $f(x) = (1-x)^{2}$. As in Sect. 3, we look for the solution of (1.1) of the form $u_{\lambda }= t_{\lambda }W_{p}$, where $t_{\lambda }> 0$ is a constant defined by (2.4). By (1.5), (1.6), (2.11), and (3.1), we have

$$\begin{aligned} M_{2,q} =& \int _{0}^{1} W_{p}(x)^{q} \,dx -2 \int _{0}^{1} xW_{p}(x)^{q} \,dx + \int _{0}^{1} x^{2}W_{p}(x)^{q} \,dx \\ =& \int _{0}^{1} x^{2} W_{p}(x)^{q}\,dx = R_{2,q}. \end{aligned}$$

(4.1)

Lemma 4.1

Assume that $q = m(p+1)$ or $q = m(p+1)+ p$, where $m \in \mathbb{N}$. Then $R_{2,q}$ is explicitly determined inductively.

Proof

By (1.7), (2.5), (4.1) and putting $x = 1-t$, we have

$$\begin{aligned} R_{2,q} =& \int _{0}^{1/2} x^{2}W_{p}(x)^{q} \,dx + \int _{1/2}^{1} x^{2}W_{p}(x)^{q} \,dx \\ =& \int _{0}^{1/2} x^{2}W_{p}(x)^{q} \,dx + \int _{0}^{1/2} \bigl(1 - 2t + t^{2} \bigr)W_{p}(t)^{q}\,dt \\ =& \int _{0}^{1/2} W_{p}(x)^{q} \,dx - 2 \int _{0}^{1/2} xW_{p}(x)^{q} \,dx + 2 \int _{0}^{1/2} x^{2}W_{(}x)^{q} \,dx \\ =& S_{0,q} - 2S_{1,q} + 2S_{2,q}. \end{aligned}$$

(4.2)

By (2.6), we have $W_{p}'(1/2) = 0$. By this, (1.1), (1.3), (2.10), (2.11) and integration by parts, we have

$$\begin{aligned} &S_{0,q} = \int _{0}^{1/2} W_{p}(x)^{q} \,dx = \sqrt{\frac{p+1}{2}} \xi ^{(2q-p+1)/2}L_{p,q}, \end{aligned}$$

(4.3)

$$\begin{aligned} &S_{1,q} = \int _{0}^{1/2} xW_{p}(x)^{q-p}W_{p}(x)^{p} \,dx \\ &\hphantom{S_{1,q}} = - \int _{0}^{1/2} xW_{p}(x)^{q-p}W_{p}''(x) \,dx \\ &\hphantom{S_{1,q}}= - \bigl[xW_{p}(x)^{q-p}W_{p}'(x) \bigr]_{0}^{1/2} + \int _{0}^{1/2} \bigl(xW_{p}(x)^{q-p} \bigr)'W_{p}'(x)\,dx \\ &\hphantom{S_{1,q}}= \int _{0}^{1/2} \bigl\{ W_{p}(x)^{q-p} + x(q-p)W_{p}(x)^{q-p-1}W_{p}'(x) \bigr\} W_{p}'(x)\,dx \\ &\hphantom{S_{1,q}}= \frac{1}{q-p+1} \bigl[W_{p}(x)^{q-p+1} \bigr]_{0}^{1/2} + (q-p) \int _{0}^{1/2}xW_{p}(x)^{q-p-1} \bigl(W_{p}'(x) \bigr)^{2}\,dx \\ &\hphantom{S_{1,q}}= \frac{1}{q-p+1}\xi ^{q-p+1} + (q-p) \int _{0}^{1/2} xW_{p}(x)^{q-p-1} \frac{2}{p+1} \bigl(\xi ^{p+1} - W_{p}(x)^{p+1} \bigr)\,dx \\ &\hphantom{S_{1,q}}= \frac{1}{q-p+1}\xi ^{q-p+1} + \frac{2(q-p)}{p+1}\xi ^{p+1} \int _{0}^{1/2}xW_{p}(x)^{q-p-1} \,dx - \frac{2(q-p)}{p+1}S_{1,q}. \end{aligned}$$

(4.4)

By this, we obtain

$$\begin{aligned} S_{1,q} =& \frac{p+1}{2q-p+1} \biggl\{ \frac{1}{q-p+1}\xi ^{q-p+1} + \frac{2(q-p)}{p+1} \xi ^{p+1}S_{1, q-p-1} \biggr\} . \end{aligned}$$

(4.5)

Similar to the argument to derive (4.5), we obtain

$$\begin{aligned} S_{2,q} =& \int _{0}^{1/2} x^{2}W_{p}(x)^{q-p}W_{p}(x)^{p} \,dx \\ =& - \int _{0}^{1/2} x^{2}W_{p}(x)^{q-p}W_{p}''(x) \,dx \\ =& - \bigl[x^{2}W_{p}(x)^{q-p}W_{p}'(x) \bigr]_{0}^{1/2} + \int _{0}^{1/2} \bigl(x^{2}W_{p}(x)^{q-p} \bigr)'W_{p}'(x)\,dx \\ =& \int _{0}^{1/2} \bigl\{ 2xW_{p}(x)^{q-p} + x^{2}(q-p)W_{p}(x)^{q-p-1}W_{p}'(x) \bigr\} W_{p}'(x)\,dx \\ =& 2 \int _{0}^{1/2} xW_{p}(x)^{q-p} W_{p}'(x)\,dx + (q-p) \int _{0}^{1/2}x^{2}W_{p}(x)^{q-p-1} \bigl(W_{p}'(x) \bigr)^{2}\,dx \\ =& 2 \int _{0}^{1/2} xW_{p}(x)^{q-p} W_{p}'(x)\,dx \\ &{} + (q-p) \int _{0}^{1/2}x^{2}W_{p}(x)^{q-p-1} \frac{2}{p+1} \bigl(\xi ^{p+1} - W_{p}(x)^{p+1} \bigr)\,dx \\ =& 2 \int _{0}^{1/2} xW_{p}(x)^{q-p} W_{p}'(x)\,dx + \frac{2(q-p)}{p+1} \xi ^{p+1} \int _{0}^{1/2}x^{2}W_{p}(x)^{q-p-1} \,dx - \frac{2(q-p)}{p+1}S_{2,q}. \end{aligned}$$

(4.6)

By this, we have

$$\begin{aligned} \frac{2q-p+1}{p+1}S_{2,q} =& 2 \int _{0}^{1/2} xW_{p}(x)^{q-p} W_{p}'(x)\,dx \\ &{}+ \frac{2(q-p)}{p+1}\xi ^{p+1} \int _{0}^{1/2}x^{2}W_{p}(x)^{q-p-1} \,dx \\ :=& 2U_{1} + \frac{2(q-p)}{p+1}\xi ^{p+1}S_{2, q-p-1}. \end{aligned}$$

(4.7)

By (4.3) and integration by parts, we have

$$\begin{aligned} U_{1} =& \int _{0}^{1/2} x \biggl(\frac{1}{q-p+1}W_{p}(x)^{q-p+1} \biggr)'\,dx \\ =& \biggl[ x\frac{1}{q-p+1}W_{p}(x)^{q-p+1} \biggr]_{0}^{1/2} - \int _{0}^{1/2}\frac{1}{q-p+1} W_{p}(x)^{q-p+1}\,dx \\ =& \frac{1}{2(q-p+1)}\xi ^{q-p+1}-\frac{1}{q-p+1}S_{0,q-p+1} \\ =& \frac{1}{q-p+1} \biggl(\frac{1}{2}\xi ^{q-p+1}- \sqrt{\frac{p+1}{2}} \xi ^{(2q-3p+3)/2}L_{p,q-p+1} \biggr). \end{aligned}$$

(4.8)

By this and (4.7), we have

$$\begin{aligned} S_{2,q} =& \frac{p+1}{2q-p+1} \biggl\{ \frac{1}{q-p+1} \bigl(\xi ^{q-p+1}- \sqrt{2(p+1)}\xi ^{(2q-3p+3)/2}L_{p,q-p+1} \bigr) \\ &{} + \frac{2(q-p)}{p+1}\xi ^{p+1}S_{2,q-p-1} \biggr\} . \end{aligned}$$

(4.9)

We repeat the calculation (4.2)–(4.9). Then we obtain $R_{2,q}$ for $q = m(p+1)$ and $q = m(p+1) + p$ inductively. Indeed, assume that $q = m(p+1)$ (resp. $q = m(p+1)+p$). Then, by repeating m times the argument above, we have

$$\begin{aligned} &R_{2,q} := C_{0,m} + C_{0,1,m}S_{1,0} + C_{0,2,m}S_{2, 0}, \quad \bigl(q = m(p+1) \bigr), \end{aligned}$$

(4.10)

$$\begin{aligned} &R_{2,q} := C_{1,m} + C_{1,1,m}S_{1,p} + C_{1,2,m}S_{2, p}, \quad \bigl(q = m(p+1)+p \bigr), \end{aligned}$$

(4.11)

where $C_{0,m}$, $C_{0,1,m}$, $C_{1,2,m}$, $C_{1,m}$, $C_{1,1,m}$, $C_{1,2,m}$ are explicit constants containing ξ, which are obtained inductively. According to the case where $q = m(p+1)$, or $q = m(p+1)+p$, (4.10) and (4.11) are determined by

$$\begin{aligned} &S_{1,0} = \int _{0}^{1/2} x\,dx = \frac{1}{8}, \end{aligned}$$

(4.12)

$$\begin{aligned} &S_{2,0} = \int _{0}^{1/2} x^{2}\,dx = \frac{1}{24}, \end{aligned}$$

(4.13)

$$\begin{aligned} &S_{0,p} = \int _{0}^{1/2} W_{p}(x)^{p} \,dx = - \int _{0}^{1/2} W_{p}''(x) \,dx \end{aligned}$$

(4.14)

$$\begin{aligned} &\hphantom{S_{0,p}}= - \bigl[W_{p}'(x) \bigr]_{0}^{1/2} = W_{p}'(0) = \sqrt{ \frac{2}{p+1}}\xi ^{(p+1)/2}, \\ &S_{1,p} = \int _{0}^{1/2} xW_{p}(x)^{p} \,dx = \int _{0}^{1/2} x \bigl(-W_{p}''(x) \bigr)\,dx \end{aligned}$$

(4.15)

$$\begin{aligned} &\hphantom{S_{0,p}}= \bigl[x \bigl(-W_{p}'(x) \bigr) \bigr]_{0}^{1/2} + \int _{0}^{1/2} W_{p}'(x) \,dx \\ &\hphantom{S_{0,p}}= \bigl[W_{p}(x) \bigr]_{0}^{1/2} \\ &\hphantom{S_{0,p}}= \xi , \\ &S_{2,p} = \int _{0}^{1/2} x^{2}W_{p}(x)^{p} \,dx = \int _{0}^{1/2} x^{2} \bigl(-W_{p}''(x) \bigr)\,dx \\ &\hphantom{S_{0,p}}= \bigl[-x^{2} W_{p}'(x) \bigr]_{0}^{1/2} + \int _{0}^{1/2}2xW_{p}'(x) \,dx \\ &\hphantom{S_{0,p}}= \bigl[2xW_{p}(x) \bigr]_{0}^{1/2} - 2 \int _{0}^{1/2} W_{p}(x)\,dx \\ &\hphantom{S_{0,p}}= \xi - \sqrt{2(p+1)}\xi ^{(3-p)/2}L_{p,1}. \end{aligned}$$

(4.16)

Since $M_{2,q} = R_{2,q}$ by (4.1), we see from (4.12)–(4.16) that $M_{2,q}$ is explicitly determined inductively. Thus the proof is complete. □

Proof of Theorem 1.3.

(i) The proof of Theorem 1.3 follows from Lemma 4.1 immediately.

(ii) Let $q = p + 1$. Then by (4.3), (4.5), (4.9), (4.12), (4.13) and direct calculation, we have

$$\begin{aligned} &S_{0,p+1} =\sqrt{\frac{p+1}{2}}\xi ^{(p+3)/2}L_{p,p+1}, \end{aligned}$$

(4.17)

$$\begin{aligned} &S_{1,p+1} = \frac{p+1}{p+3} \biggl\{ \frac{1}{2}\xi ^{2} + \frac{1}{4(p+1)}\xi ^{p+1} \biggr\} , \end{aligned}$$

(4.18)

$$\begin{aligned} &S_{2,p+1} = \frac{p+1}{p+3} \biggl\{ \frac{1}{2} \bigl(\xi ^{2} - \sqrt{2(p+1)}\xi ^{(5-p)/2}L_{p,2} \bigr) + \frac{1}{12(p+1)}\xi ^{p+1} \biggr\} . \end{aligned}$$

(4.19)

By (4.17)–(4.19), we have

$$\begin{aligned} M_{2,p+1} =& R_{2,p+1} = S_{0,q} - 2S_{1,q} + 2S_{2,q} \\ =& \sqrt{\frac{p+1}{2}}\xi ^{(p+3)/2}L_{p,p+1} - \frac{1}{3(p+3)} \xi ^{p+1} - \frac{p+1}{p+3}\sqrt{2(p+1)} \xi ^{(5-p)/2}L_{p,2}. \end{aligned}$$

(4.20)

(iii) Let $q = 2p+1 (= p+1 + p)$. Then, by a similar calculation as above, we have

$$\begin{aligned} &S_{0,2p+1} =\sqrt{\frac{p+1}{2}}\xi ^{3(p+1)/2}L_{p, 2p+1}, \end{aligned}$$

(4.21)

$$\begin{aligned} &S_{1,2p+1} = \frac{2p+5}{3(p+2)}\xi ^{p+2}, \end{aligned}$$

(4.22)

$$\begin{aligned} &S_{2,2p+1} = \frac{2p+5}{3(p+2)}\xi ^{p+2} - \frac{1}{3}\sqrt{2(p+1)} \xi ^{(p+5)/2} \biggl( \frac{1}{p+2}L_{p,p+2} + 2L_{p,1} \biggr). \end{aligned}$$

(4.23)

By (4.22)–(4.24), we have

$$\begin{aligned} M_{2, 2p+1} =& R_{2,2p+1} = S_{0,q} - 2S_{1,q} + 2S_{2,q} \\ =& \sqrt{\frac{p+1}{2}}\xi ^{3(p+1)/2}L_{p, 2p+1} - \frac{2}{3}\sqrt{2(p+1)} \xi ^{(p+5)/2} \biggl( \frac{1}{p+2}L_{p,p+2} + 2L_{p,1} \biggr). \end{aligned}$$

(4.24)

Thus the proof of Theorem 1.3 is complete. □

5 Proof of Theorem 1.4

In this section we consider the case $p = q > 1$ and $f(x) = (1-x)^{n}$ for $n \in \mathbb{N}$ with $n \ge 2$. We show that we are able to obtain $M_{n,p}$ inductively by using the constants $S_{m,1}$ ($m \ge 0$).

Proof of Theorem 1.4 (i)

Case 1. Assume that $n = 2k + 1$, where $k \in \mathbb{N}$. We put $t = 1-s$. By (2.6), we have

$$\begin{aligned} M_{2k+1,p} =& \int _{0}^{1} (1 - s)^{2k+1} W_{p}(s)^{p}\,ds \\ =& \int _{0}^{1/2} (1 - s)^{2k+1} W_{p}(s)^{p}\,ds + \int _{1/2}^{1}(1-s)^{2k+1}W_{p}(s)^{p} \,ds \\ =& \sum_{r = 0}^{2k} (-1)^{r}{}_{2k+1}C_{r} \int _{0}^{1/2}s^{r} W_{p}(s)^{p}\,ds - \int _{0}^{1/2} s^{2k+1}W_{p}(s)^{p} \,ds + \int _{0}^{1/2} t^{2k+1} W_{p}(t)^{p}\,dt \\ =& \sum_{r = 0}^{2k} (-1)^{r}{}_{2k+1}C_{r} \int _{0}^{1/2}s^{r} W_{p}(s)^{p}\,ds. \end{aligned}$$

(5.1)

Therefore, $M_{2k+1,p}$ is obtained by $S_{r, p}$ ($0 \le r \le 2k$).

Case 2. Assume that $n = 2k$, where $k \in \mathbb{N}$. Then

$$\begin{aligned} M_{2k,p} =& \int _{0}^{1} (1 - s)^{2k} W_{p}(s)^{p}\,ds \\ =& \int _{0}^{1/2} (1 - s)^{2k} W_{p}(s)^{p}\,ds + \int _{1/2}^{1}(1-s)^{2k+1}W_{p}(s)^{p} \,ds \\ =& \sum_{r = 0}^{2k-1} (-1)^{r}{}_{2k}C_{r} \int _{0}^{1/2}s^{r} W_{p}(s)^{p}\,ds + \int _{0}^{1/2} s^{2k}W_{p}(s)^{p} \,ds + \int _{1/2}^{1} t^{2k} W_{p}(t)^{p}\,dt \\ =& \sum_{r = 0}^{2k-1} (-1)^{r}{}_{2k}C_{r} \int _{0}^{1/2}s^{r} W_{p}(s)^{p}\,ds + 2 \int _{0}^{1/2} s^{2k}W_{p}(s)^{p} \,ds. \end{aligned}$$

(5.2)

By (5.1) and (5.2), we find that $M_{2k,p}$ is obtained by $S_{r, p}$ ($0 \le r \le 2k$). By (2.7), for $r \ge 2$, we have

$$\begin{aligned} S_{r,p} =& \int _{0}^{1/2} x^{r} W_{p}(x)^{p}\,dx \\ =& \int _{0}^{1/2} x^{r} \bigl(-W_{p}(x)'' \bigr)\,dx = \bigl[x^{r} \bigl(-W_{p}'(x) \bigr) \bigr]_{0}^{1/2} + \int _{0}^{1/2} rx^{r-1}W_{p}'(x) \,dx \\ =& r \bigl[x^{r-1}W_{p}(x) \bigr]_{0}^{1/2} - r(r-1) \int _{0}^{1/2}x^{r-2}W_{p}(x) \,dx \\ =& r \biggl(\frac{1}{2} \biggr)^{r-1}\xi - r(r-1)S_{r-2, 1}. \end{aligned}$$

(5.3)

By this, we see that $S_{r,p}$ ($r \ge 2$) is represented by $S_{r-2, 1}$. By (5.2) and (5.3), we have

$$\begin{aligned} &M_{2k+1,p} = \sum_{r = 0}^{2k} (-1)^{r}{}_{2k+1}C_{r} \biggl\{ r \biggl( \frac{1}{2} \biggr)^{r-1}\xi - r(r-1)S_{r-2,1} \biggr\} , \end{aligned}$$

(5.4)

$$\begin{aligned} &M_{2k,p} = \sum_{r = 0}^{2k-1} (-1)^{r}{}_{2k}C_{r} \biggl\{ r \biggl( \frac{1}{2} \biggr)^{r-1}\xi - r(r-1)S_{r-2,1} \biggr\} \\ & \hphantom{M_{2k,p}}\quad + 2 \biggl\{ 2k \biggl(\frac{1}{2} \biggr)^{2k-1}\xi - 2k(2k-1)S_{2k-2,1} \biggr\} . \end{aligned}$$

(5.5)

By (5.4) and (5.5), we obtain our conclusion. Thus the proof is complete. □

Proof of Theorem 1.4 (ii)

By (4.14), (4.15), and (4.16) and putting $t = 1-x$, we have

$$\begin{aligned} M_{3,p} =& \int _{0}^{1} (1-x)^{3}W_{p}(x)^{p} \,dx \\ =& \int _{0}^{1/2} (1-x)^{3}W_{p}(x)^{p} \,dx + \int _{1/2}^{1} (1-x)^{3}W_{p}(x)^{p} \,dx \\ =& \int _{0}^{1/2} \bigl(1-3x+3x^{3}-x^{3} \bigr)W_{p}(x)^{p}\,dx + \int _{0}^{1/2} t^{3}W_{p}(t) \,dt \\ =& \int _{0}^{1/2} \bigl(1-3x+3x^{3} \bigr)W_{p}(x)^{p}\,dx = S_{0,p} - 3S_{1,p} + 3S_{2,p} \\ =& \sqrt{\frac{2}{p+1}}\xi ^{(p+1)/2} - 3\sqrt{2(p+1)}\xi ^{(3-p)/2}L_{p,1}. \end{aligned}$$

(5.6)

Thus the proof is complete. □