Asymptotic behavior of solution curves of nonlocal one-dimensional elliptic equations

Abstract

We study the one-dimensional nonlocal elliptic equation

$$\begin{aligned}& -A\bigl( \bigl\Vert u' \bigr\Vert _{p}^{p} \bigr) u''(x) = \lambda B\bigl( \bigl\Vert u' \bigr\Vert _{q}^{q} \bigr)u(x)^{r} , \quad x \in I:= (0,1), u(x) > 0, x \in I, \\& u(0) = u(1) = 0, \end{aligned}$$

where $A = A(y)$ and $B = B(y)$ are continuous functions, satisfying $A(y) > 0$, $B(y) > 0$ for $y > 0$, $p \ge 1$, $q \ge 1$, and $r > 1$ are given constants, and $\lambda > 0$ is a bifurcation parameter. We establish the global behavior of solution curves and precise asymptotic formulas for $u_{\lambda}(x)$ as $\lambda \to \infty$.

1 Introduction

We consider the following one-dimensional nonlocal elliptic equation

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

(1.1)

where $A = A(y)$ and $B = B(y)$ are continuous functions with $A(y) > 0$, $B(y) > 0$ for $y > 0$, while $p \ge 1$, $q \ge 1$, $r > 1$ are given constants, $\Vert \cdot \Vert _{m}$ ($m \ge 1$) denotes the usual $L^{m}$-norm of the real-valued functions on I, and $\lambda > 0$ is a bifurcation parameter. In this paper, we consider the following typical three cases.

• (i) $A(y) = y$, $B(y) = y$,

• (ii) $A(y) = e^{y}$, $B(y) = 1$, $p = 2$,

• (iii) $A(y) = e^{y}$, $B(y) = y$.

The case (i) is a modification of a nonlocal problem of Kirchhoff type, which is motivated by the following problem (1.2) in [13]

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

(1.2)

Besides, the cases (ii) and (iii) are motivated by the mean field equation and nonlocal Liouville-type equations.

Nonlocal problems have been of interest to many researchers from mathematical point of view, since many problems are derived from the phenomena of relevant physical, biological, and engineering problems. Therefore, nonlocal problems have been widely studied by many authors. We refer the reader to Goodrich [7–9], Lacey [11, 12], Stańczy [14], [2–4, 10], and the references therein. One of the main interest in this area is the existence of positive solutions. On the other hand, there seems to be a few studies on bifurcation problems. We refer the reader to [17] and the references therein. Roughly speaking, in [17], the case $A(y) = y^{j} + b$ and $B(y) = y^{k}$, where $j \ge 0$, $k > 0$ are constants, has been considered, and the existence of a branch of positive solutions bifurcating from infinity at $\lambda = 0$ has been discussed. Recently, the case, where $A(y) = y + b$, $B(y) \equiv 1$ and $p = 2$, has been studied in [15], where $b > 0$ is a constant, and the precise global behavior of solution curve has been obtained. For a standard bifurcation problems, we refer to [5].

The purpose of this paper is to consider more general nonlocal terms motivated by equations having background in physics and obtain the precise asymptotic behavior of bifurcation curves $\lambda = \lambda (\alpha )$ and $u_{\lambda}$ as $\lambda \to \infty$. Here, $\alpha := \alpha _{\lambda }=\Vert u_{\lambda}\Vert _{\infty}$ for given $\lambda > 0$. The main tool here is time map method, also known as quadrature technique (cf. [12]). One can see the simple example of time map method in the Appendix.

Now, we state our main Theorems 1.1, 1.2, and 1.3. To do this, we prepare the following notation. For $r > 1$, let

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

(1.3)

We know from [6] that there exists a unique solution $W_{r}(x)$ of (1.3). For $m \ge 1$, we put

$$ L_{m}:= \int _{0}^{1} \frac{1}{\sqrt{1-s^{m+1}}}\,ds ,\qquad M_{r,m}:= \int _{0}^{1} \bigl(1-s^{r+1} \bigr)^{(m-1)/2}\,ds . $$

(1.4)

We note that $L_{m}$ is finite since $\sqrt{1 - s^{2}} \le \sqrt{1 - s^{m+1}}$ for $0 \le s \le 1$. Then, we have

$$\begin{aligned}& \bigl\Vert W_{r}' \bigr\Vert _{m}^{m} = 2^{mr/(r-1)}(r+1)^{m/(r-1)}L_{r}^{(mr+m-r+1)/(r-1)}M_{r,m}, \end{aligned}$$

(1.5)

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

(1.6)

Equation (1.6) has been obtained in [16]. For completeness, the proof of (1.5) and (1.6) will be given in the Appendix.

We begin with the first main Theorem 1.1, which will be proved in Sect. 2.

Theorem 1.1

Let $A(y) = y$, $B(y) = y$ in (1.1). Assume that $p-q-r+1 \neq 0$. Then, there exists a unique solution $u_{\lambda}(x)$ of (1.1) for any $\lambda > 0$, and it is represented as

$$ u_{\lambda}(x) = \lambda ^{1/(p-q-r+1)} \frac{ \Vert W_{r}' \Vert _{q}^{q/(p-q-r+1)}}{ \Vert W_{r}' \Vert _{p}^{p/(p-q-r+1)}}W_{r}(x). $$

(1.7)

Further, λ is represented as the function of $\alpha := \Vert u_{\lambda}\Vert _{\infty}$ as

$$ \lambda = \lambda (\alpha ) = \frac{ \Vert W_{r}' \Vert _{p}^{p}}{ \Vert W_{r}' \Vert _{q}^{q} \Vert W_{r} \Vert _{\infty}^{p-q-r+1}}\alpha ^{p-q-r+1}. $$

(1.8)

Next, we consider the case $A(y) = e^{y}$, $B(y) = 1$ and $p = 2$ and state Theorem 1.2. The proof will be given in Sect. 3. For $r > 1$, we put

$$ R_{r} := \frac{r-1}{2} \bigl\{ 1-\log (r-1) +\log 2 + 2\log \bigl\Vert W_{r}' \bigr\Vert _{2} \bigr\} . $$

(1.9)

Theorem 1.2

Let $A(y) = e^{y}$, $B(y) = 1$ and $p = 2$. For $r > 1$, put $\lambda _{r}:= e^{R_{r}}$.

• (i) If $0 < \lambda < \lambda _{r}$, then there are no solutions of (1.1).

• (ii) If $\lambda = \lambda _{r}$, then (1.1) has a unique solution $u_{1,\lambda}$.

• (iii) If $\lambda > \lambda _{r}$, then there are exactly two solutions $u_{1,\lambda}$, $u_{2,\lambda}$ with $u_{1,\lambda}(x) < u_{2,\lambda}(x)$ for $x \in I$.

• (iv) Let $\lambda > \lambda _{r}$ be fixed. Then, there are two numbers $\alpha _{1,\lambda}:= \Vert u_{1,\lambda}\Vert _{\infty}$ and $\alpha _{2,\lambda}:= \Vert u_{2,\lambda}\Vert _{\infty}$ satisfying:

  $$ \lambda = 2(r+1)L_{r}^{2} A\bigl(4M_{r,2}L_{r} \alpha _{j,\lambda}^{2}\bigr) \alpha _{j,\lambda}^{1-r}\quad (j = 1,2). $$

  (1.10)

  If $\lambda = \lambda _{r}$, then $\alpha _{0}:=\alpha _{1,\lambda} = \alpha _{2,\lambda}$ in (1.10).

• (v) As $\lambda \to \infty$,

  $$\begin{aligned}& u_{1,\lambda}(x) = \lambda ^{-1/(r-1)} \biggl\{ 1 + \frac{1}{r-1} \bigl\Vert W_{r}' \bigr\Vert _{2}^{2} \lambda ^{-2/(r-1)}\bigl(1 + o(1)\bigr) \biggr\} \bigl\Vert W_{r}' \bigr\Vert _{2}^{-1}W_{r}(x), \end{aligned}$$

  (1.11)
  $$\begin{aligned}& u_{2,\lambda}(x) = \biggl\{ \log \lambda + \frac{r-1}{2}\log (\log \lambda ) \bigl(1 + o(1)\bigr) \biggr\} ^{1/2} \bigl\Vert W_{r}' \bigr\Vert _{2}^{-1}W_{r}(x). \end{aligned}$$

  (1.12)

We finally state Theorem 1.3, which will be proved in Sect. 4.

Theorem 1.3

Let $A(y) = e^{y}$, $B(y) = y$. Let

$$ C_{0} = \frac{q+r-1}{p} \biggl(1 - \log \frac{q+r-1}{p} + \log \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} - \frac{p}{q+r-1}\log \bigl\Vert W_{r}' \bigr\Vert _{q}^{q} \biggr) $$

(1.13)

and $\lambda _{0}:= e^{C_{0}}$.

• (i) If $0 < \lambda < \lambda _{0}$, then there are no solutions of (1.1).

• (ii) If $\lambda = \lambda _{0}$, then (1.1) has a unique solution $u_{1,\lambda}$.

• (iii) If $\lambda > \lambda _{0}$, then there are exactly two solutions $u_{1,\lambda}$, $u_{2,\lambda}$ with $u_{1,\lambda}(x) < u_{2,\lambda}(x)$ for $x \in I$.

• (iv) As $\lambda \to \infty$

  $$\begin{aligned}& u_{1,\lambda}(x) = \lambda ^{-1/(q+r-1))} \bigl\Vert W_{r}' \bigr\Vert _{q}^{-q/(q+r-1)} \\ \end{aligned}$$

  (1.14)
  $$\begin{aligned}& \hphantom{u_{1,\lambda}(x) ={}}{} \times \biggl\{ 1 + \frac{1}{q+r-1} \bigl\Vert W_{\lambda}' \bigr\Vert _{p}^{p} \bigl\Vert W_{\lambda}' \bigr\Vert _{q}^{pq/(q+r-1)} \lambda ^{-p/(q+r-1)}\bigl(1 + o(1)\bigr) \biggr\} W_{r}(x), \\& \alpha _{1,\lambda} = \lambda ^{-1/(q+r-1))} \bigl\Vert W_{r}' \bigr\Vert _{q}^{-q/(q+r-1)} \end{aligned}$$

  (1.15)
  $$\begin{aligned}& \hphantom{\alpha _{1,\lambda} ={}}{} \times \biggl\{ 1 + \frac{1}{q+r-1} \bigl\Vert W_{\lambda}' \bigr\Vert _{p}^{p} \bigl\Vert W_{\lambda}' \bigr\Vert _{q}^{pq/(q+r-1)} \lambda ^{-p/(q+r-1)}\bigl(1 + o(1)\bigr) \biggr\} \Vert W_{\lambda} \Vert _{\infty}, \end{aligned}$$

  (1.16)
  $$\begin{aligned}& u_{2,\lambda}(x) = \bigl\Vert W_{r}' \bigr\Vert _{q}^{-1}(\log \lambda )^{1/p} \biggl\{ 1 + \frac{p^{2}}{(q+r-1)^{3}} \frac{\log (\log \lambda )}{\log \lambda}\bigl(1 + o(1)\bigr) \biggr\} W_{r}(x), \\& \alpha _{2,\lambda} = \bigl\Vert W_{r}' \bigr\Vert _{q}^{-1}(\log \lambda )^{1/p} \biggl\{ 1 + \frac{p^{2}}{(q+r-1)^{3}} \frac{\log (\log \lambda )}{\log \lambda}\bigl(1 + o(1)\bigr) \biggr\} \Vert W_{ \lambda} \Vert _{\infty}. \end{aligned}$$

  (1.17)

The rest of this paper is organized as follows. We prove Theorems 1.1, 1.2, and 1.3 in Sects. 2, 3, and 4, respectively. In the proofs, the time map method and the argument from [1] play important roles. Finally, we prove (1.5) and (1.6) in the Appendix.

2 Proof of Theorem 1.1

Let $\lambda > 0$ be fixed. We write $w_{r}(x)$ as a unique solution of

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

(2.1)

We look for the solution $u_{\lambda}(x)$ of the form

$$ u_{\lambda}(x) = t_{\lambda }w_{r}(x), $$

(2.2)

where $t_{\lambda }> 0$ is a constant. By (1.1) and (2.1), we have

$$ -t_{\lambda}^{p} \bigl\Vert w_{r}' \bigr\Vert _{p}^{p} t_{\lambda }w_{r}''(x) = \lambda t_{\lambda}^{q} \bigl\Vert w_{r}' \bigr\Vert _{q}^{q} t_{\lambda}^{r} w_{r}(x)^{r}. $$

(2.3)

Since $w_{r}(x) = \lambda ^{-1/(r-1)}W_{r}(x)$, we find from (2.3) that if

$$ t_{\lambda}^{p-q-r+1} = \lambda ^{(p-q)/(r-1)} \frac{ \Vert W_{r}' \Vert _{q}^{q}}{ \Vert W_{r}' \Vert _{p}^{p}}, $$

(2.4)

then (2.2) satisfies (1.1). By this, we have

$$ t_{\lambda }= \lambda ^{(p-q)/((p-q-r+1)(r-1))} \frac{ \Vert W_{r}' \Vert _{q}^{q/(p-q-r+1)}}{ \Vert W_{r}' \Vert _{p}^{p/(p-q-r+1)}}. $$

(2.5)

By this and (2.2), we have

$$\begin{aligned} u_{\lambda}(x) =& \lambda ^{(p-q)/((p-q-r+1)(r-1))} \frac{ \Vert W_{r}' \Vert _{q}^{q/(p-q-r+1)}}{ \Vert W_{r}' \Vert _{p}^{p/(p-q-r+1)}}\lambda ^{-1/(r-1)}W_{r}(x) \\ =& \lambda ^{1/(p-q-r+1)} \frac{ \Vert W_{r}' \Vert _{q}^{q/(p-q-r+1)}}{ \Vert W_{r}' \Vert _{p}^{p/(p-q-r+1)}}W_{r}(x). \end{aligned}$$

(2.6)

This implies (1.7). Since $\alpha = \Vert u_{\lambda}\Vert _{\infty }= u_{\lambda}(1/2)$, we put $x = 1/2$ in (2.6). Then, we obtain

$$ \alpha = \lambda ^{(1/(p-q-r+1))} \frac{ \Vert W_{r}' \Vert _{q}^{q/(p-q-r+1)}}{ \Vert W_{r}' \Vert _{p}^{p/(p-q-r+1)}} \Vert W_{r} \Vert _{\infty}. $$

(2.7)

By this, we obtain (1.8). Thus, the proof of Theorem 1.1 is complete.

3 Proof of Theorem 1.2

We first show the existence of $u_{\lambda}$. We follow the argument in [1]. Let $t > 0$ and $A(t) = e^{t}$. We consider the following equation for $t > 0$.

$$ A(t) = \bigl\Vert w_{r}' \bigr\Vert _{2}^{1-r}t^{(r-1)/2}. $$

(3.1)

Assume that $t_{\lambda }> 0$ satisfies (3.1). We put $\gamma := t_{\lambda}^{1/2}\Vert w_{r}'\Vert _{2}^{-1}$ and $u_{\lambda}:= \gamma w_{r}$. Then, we have

$$ A\bigl( \bigl\Vert \gamma w_{r}' \bigr\Vert _{2}^{2}\bigr) = A(t_{\lambda}) = \gamma ^{r-1}. $$

By this, we have

$$\begin{aligned} -A\bigl( \bigl\Vert u_{\lambda}' \bigr\Vert _{2}^{2}\bigr)u_{\lambda}''(x) =& - A\bigl( \bigl\Vert \gamma w_{\lambda}' \bigr\Vert _{2}^{2}\bigr) \gamma w_{\lambda}''(x) \\ =& \gamma ^{r}\lambda w_{\lambda}(x)^{r} = \lambda u_{\lambda}(x)^{r}. \end{aligned}$$

(3.2)

Since $w_{r} = \lambda ^{1/(1-r)}W_{r}$, we have

$$ u_{\lambda}(x) = t_{\lambda}^{1/2} \bigl\Vert W_{r}' \bigr\Vert _{2}^{-1} W_{r}(x). $$

(3.3)

On the other hand, suppose that $u_{\lambda}$ satisfies (3.2). Then by putting $t_{\lambda }= \Vert u_{\lambda}'\Vert _{2}^{2}$, we see that $t_{\lambda}$ satisfies (3.1). Therefore, the number of the positive solutions t of the equation

$$ e^{t} = \lambda \bigl\Vert W_{r}' \bigr\Vert _{2}^{1-r} t^{(r-1)/2} $$

(3.4)

coincide with the number of the solutions of (3.2). So, we solve the equation (3.1). To do this, for $r > 1$, we set

$$ R_{r}:= \frac{r-1}{2} \bigl\{ 1-\log (r-1) +\log 2 + 2\log \bigl\Vert W_{r}' \bigr\Vert _{2} \bigr\} . $$

(3.5)

Lemma 3.1

For $r > 1$, let $\lambda _{r}:= e^{R_{r}}$.

• (i) If $0 < \lambda < \lambda _{r}$, then (3.4) has no solution.

• (ii) If $\lambda = \lambda _{r}$, then (3.4) has a unique solution $t_{0}$.

• (iii) If $\lambda > \lambda _{r}$, then (3.4) has exactly two solutions $t_{\lambda ,1}$, $t_{\lambda ,2}$ with $0 < t_{\lambda ,1} < t_{0} < t_{\lambda ,2}$.

Proof

We put $K_{\lambda ,r}:= \lambda \Vert W_{r}'\Vert _{2}^{1-r}$. By (3.4), we have the equation

$$ t = g(t):= \frac{r-1}{2}\log t + \log K_{\lambda ,r}. $$

(3.6)

Since $g'(t) = \frac{r-1}{2t}$, we see that $g'(t_{0}) = 1$, where $t_{0} = \frac{r-1}{2}$. Then the tangent line of $y = g(t)$ at $(t_{0}, g(t_{0}))$ is $g(t) - g(t_{0}) = t - t_{0}$. We see from this that if $g(t_{0}) = t_{0}$, namely,

$$ \frac{r-1}{2}\log \frac{r-1}{2} + \log K_{\lambda ,r} = \frac{r-1}{2}, $$

(3.7)

then the tangent line of $g(t)$ at $t = t_{0}$ is exactly the line $y = t$. Equation (3.7) implies that

$$ \log \lambda = R_{r}, $$

(3.8)

namely, $\lambda = e^{R_{r}}$. This implies (ii). Since logt is a concave function w.r.t. $t > 0$, and $\log K_{\lambda , r}$ is increasing function of $\lambda > 0$, if $0 < \lambda < e^{R_{r}}$ (resp. $\lambda > e^{R_{r}}$), then we obtain (i) and (iii), respectively. Thus, the proof is complete. □

Proof of Theorem 1.2

By Lemma 3.1, we obtain Theorem 1.2(i), (ii), and (iii). Now, we show (iv). Assume that $u_{\lambda}(x)$ is a solution of (1.1) for some $\lambda > 0$. We write $A = e^{\Vert u_{\lambda}'\Vert _{2}^{2}}$. By (1.1), we have

$$ \bigl\{ Au_{\lambda}''(x) + \lambda u_{\lambda}(x)^{r}\bigr\} u_{\lambda}'(x) = 0. $$

(3.9)

Recall that $\alpha := \Vert u_{\lambda}\Vert _{\infty}$. Then (3.9) implies that

$$ \frac{1}{2}Au_{\lambda}'(x)^{2} + \frac{1}{r+1}\lambda u_{\lambda}(x)^{r+1} = \frac{1}{r+1}\lambda \alpha ^{r+1}. $$

(3.10)

We know that $u_{\lambda}(x)$ is a positive solution of $-u''_{\lambda}(x) = (\lambda /A)u_{\lambda}(x)^{r}$ with the condition $u_{\lambda}(0) = u_{\lambda}(1) = 0$. Therefore, by the result of Gidas, Ni, and Nirenberg [6], we know that $u_{\lambda}(x) = u_{\lambda}(1-x)$ ($0 \le x \le 1/2$). By this, (3.10) implies that for $0 \le x \le 1/2$,

$$ u_{\lambda}'(x) = \sqrt{\frac{2\lambda}{(r+1)A}}\sqrt{\alpha ^{r+1} - u_{ \lambda}(x)^{r+1}}. $$

(3.11)

By this, we have

$$\begin{aligned} \bigl\Vert u_{\lambda}' \bigr\Vert _{2}^{2} =& 2 \int _{0}^{1/2} \sqrt{ \frac{2\lambda}{(r+1)A}} \sqrt{\alpha ^{r+1} - u_{\lambda}(x)^{r+1}}u_{ \lambda}'(x)\,dx \\ =& 2 \int _{0}^{\alpha} \sqrt{\frac{2\lambda}{(r+1)A}} \sqrt{\alpha ^{r+1}- \theta ^{r+1}}\,d\theta \\ =& 2\sqrt{\frac{2\lambda}{(r+1)A}}\alpha ^{(r+3)/2} \int _{0}^{1} \sqrt{1-s^{r+1}}\,ds \\ =& 2\sqrt{\frac{2\lambda}{(r+1)A}}M_{r,2}\alpha ^{(r+3)/2}. \end{aligned}$$

(3.12)

By this, we have

$$ A \bigl\Vert u_{\lambda}' \bigr\Vert _{2}^{4} = \frac{8\lambda}{r+1}M_{r,2}^{2} \alpha ^{r+3}. $$

(3.13)

By (3.11), we have

$$\begin{aligned} \frac{1}{2} =& \int _{0}^{1/2} \frac{u_{\lambda}'(x)}{ \sqrt{\frac{2\lambda}{(r+1)A}}\sqrt{\alpha ^{r+1} - u_{\lambda}(x)^{r+1}}}\,dx \\ =& \sqrt{\frac{(r+1)A}{2\lambda}} \int _{0}^{\alpha } \frac{1}{\sqrt{\alpha ^{r+1} - \theta ^{r+1}}}\,d\theta \\ =& \sqrt{\frac{(r+1)A}{2\lambda}}\alpha ^{(1-r)/2} \int _{0}^{1} \frac{1}{\sqrt{1-s^{r+1}}}\,ds \\ =& \sqrt{\frac{(r+1)A}{2\lambda}}L_{r}\alpha ^{(1-r)/2}. \end{aligned}$$

(3.14)

By this, we have

$$ A = \frac{\lambda}{2(r+1)L_{r}^{2}}\alpha ^{r-1}. $$

(3.15)

By (3.13) and (3.15), we have

$$ \bigl\Vert u_{\lambda}' \bigr\Vert _{2}^{2} = 4M_{r,2}{L_{r}}\alpha ^{2}. $$

(3.16)

By this and (3.15), we have

$$ A = A\bigl( \bigl\Vert u_{\lambda}' \bigr\Vert _{2}^{2}\bigr) = A\bigl(4M_{r,2}{L_{r}} \alpha ^{2}\bigr) = \frac{\lambda}{2(r+1)L_{r}^{2}}\alpha ^{r-1}. $$

(3.17)

This implies that

$$ \lambda = 2(r+1)L_{r}^{2}A\bigl(4M_{r,2}{L_{r}} \alpha ^{2}\bigr)\alpha ^{1-r}. $$

(3.18)

Thus, the proof of (iv) is complete. □

Now, we prove Theorem 1.2(v).

Lemma 3.2

As $\lambda \to \infty$,

$$\begin{aligned}& u_{1,\lambda}(x) = \lambda ^{-1/(r-1)} \biggl\{ 1 + \frac{1}{r-1} \bigl\Vert W_{r}' \bigr\Vert _{2}^{2} \lambda ^{-2/(r-1)}\bigl(1 + o(1)\bigr) \biggr\} \bigl\Vert W_{r}' \bigr\Vert _{2}^{-1}W_{r}(x), \end{aligned}$$

(3.19)

$$\begin{aligned}& u_{2,\lambda}(x) = \biggl\{ \log \lambda + \frac{r-1}{2}\log (\log \lambda ) \bigl(1 + o(1)\bigr) \biggr\} ^{1/2} \bigl\Vert W_{r}' \bigr\Vert _{2}^{-1}W_{r}(x). \end{aligned}$$

(3.20)

Proof

We first prove (3.19). Since $t_{0} = (r-1)/2$, by (3.6), we see that $t_{\lambda ,1} \to 0$ and $t_{\lambda ,2} \to \infty$ as $\lambda \to \infty$. By (3.4), we have

$$ t_{\lambda ,1} = \frac{r-1}{2}\log t_{\lambda ,1} + \log \lambda - (r-1) \log \bigl\Vert W_{r}' \bigr\Vert _{2}. $$

(3.21)

This implies that

$$ t_{\lambda ,1} = \bigl\Vert W_{r}' \bigr\Vert _{2}^{2}\lambda ^{-2/(r-1)}(1 + \delta ), $$

(3.22)

where $\delta \to 0$ as $\lambda \to \infty$. By this and (3.6), we have

$$\begin{aligned} \bigl\Vert W_{r}' \bigr\Vert _{2} \lambda ^{-2/(r-1)}(1 + \delta ) =& \frac{r-1}{2} \biggl\{ - \frac{2}{r-1}\log \lambda + 2\log \bigl\Vert W_{r}' \bigr\Vert _{2}^{2} + \log (1 + \delta ) \biggr\} \\ &{} + \log \lambda - (r-1) \bigl\Vert W_{r}' \bigr\Vert _{2}. \end{aligned}$$

(3.23)

By this and the Taylor expansion, we have

$$ \bigl\Vert W_{r}' \bigr\Vert _{2}^{2} \lambda ^{-2/(r-1)} = \frac{r-1}{2}\bigl(1 + o(1)\bigr) \log (1 + \delta ) = \frac{r-1}{2}\bigl(1 + o(1)\bigr)\delta . $$

(3.24)

By this, we have

$$ \delta = \frac{2}{r-1} \bigl\Vert W_{r}' \bigr\Vert _{2}^{2}\bigl(1 + o(1)\bigr)\lambda ^{-2/(r-1)}. $$

(3.25)

This implies that as $\lambda \to \infty$

$$\begin{aligned} t_{\lambda ,1}^{1/2} =& \lambda ^{-1/(r-1)} \biggl\{ 1 + \frac{2}{r-1} \bigl\Vert W_{r}' \bigr\Vert _{2}^{2}\bigl(1 + o(1)\bigr)\lambda ^{-2/(r-1)} \biggr\} ^{1/2} \\ =& \lambda ^{-1/(r-1)} \biggl\{ 1 + \frac{1}{r-1} \bigl\Vert W_{r}' \bigr\Vert _{2}^{2} \bigl(1 + o(1)\bigr) \lambda ^{-2/(r-1)} \biggr\} . \end{aligned}$$

(3.26)

By this and (3.3), we obtain (3.19). Now, we prove (3.20). Since $t_{\lambda ,2} \to \infty$ as $\lambda \to \infty$, we have

$$ t_{\lambda ,2} = \frac{r-1}{2}\log t_{\lambda ,2} + \log \lambda - (r-1) \log \bigl\Vert W_{r}' \bigr\Vert _{2}. $$

(3.27)

This implies that

$$ t_{\lambda ,2} = (1 + \epsilon )\log \lambda , $$

(3.28)

where $\epsilon \to 0$ as $\lambda \to \infty$. By this and (3.21), we obtain

$$\begin{aligned} t_{\lambda ,2} =& \log \lambda + \epsilon \log \lambda \\ =&\frac{r-1}{2} \bigl\{ \log (1 + \epsilon ) + \log (\log \lambda ) \bigr\} + \log \lambda - (r-1)\log \bigl\Vert W_{r}' \bigr\Vert _{2}. \end{aligned}$$

(3.29)

By this, we obtain

$$ \epsilon = \frac{r-1}{2}\bigl(1 + o(1)\bigr) \frac{\log (\log \lambda )}{\log \lambda}. $$

(3.30)

By this, (3.3) and (3.28), we obtain (3.20). Thus, the proof is complete. □

4 Proof of Theorem 1.3

Let $\lambda > 0$ be fixed. In what follows, C denotes various constants independent of λ. Following the idea of (3.1)–(3.3), we look for the solution of the form

$$ u_{\lambda}(x) = t w_{r}(x) = t\lambda ^{-1/(r-1)}W_{r}(x). $$

(4.1)

If (4.1) is the solution of (1.1) with $A(\Vert u_{\lambda}'\Vert _{p}^{p}) = e^{\Vert u_{\lambda}'\Vert _{p}^{p}}$ and $B(\Vert u_{\lambda}'\Vert _{q}^{q}) = \Vert u_{\lambda}'\Vert _{q}^{q}$, then we have

$$ -\exp \bigl(t^{p}\lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} \bigr) W_{r}''(x) = t^{q+r-1} \lambda ^{-q/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{q}^{q} W_{r}(x)^{r}. $$

(4.2)

This implies that

$$ \exp \bigl(t^{p}\lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} \bigr) = t^{q+r-1}\lambda ^{-q/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{q}^{q}. $$

(4.3)

We put $s:= t^{q+r-1}$. By taking log of the both side of (4.3), we have

$$ C_{1}s^{p/(q+r-1)} = \log s + C_{2}, $$

(4.4)

where

$$ C_{1}:= \lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p}, \qquad C_{2}:= - \frac{q}{r-1}\log \lambda + \log \bigl\Vert W_{r}' \bigr\Vert _{q}^{q}. $$

(4.5)

We put

$$ g(s):= C_{1}s^{p/(q+r-1)} - \log s - C_{2}. $$

(4.6)

We look for $s > 0$ satisfying $g(s) = 0$. To do this, we consider the graph of $g(s)$. We know that

$$ g'(s) = \frac{p}{q+r-1}C_{1}s^{(p-q-r+1)/(q+r-1)} - \frac{1}{s}. $$

(4.7)

By this, we find that $g'(s_{0}) = 0$, where

$$ s_{0}:= \biggl(\frac{q+r-1}{pC_{1}} \biggr)^{(q+r-1)/p} = \biggl( \frac{q+r-1}{p \Vert W_{r}' \Vert _{p}^{p}} \biggr)^{(q+r-1)/p}\lambda ^{(q+r-1)/(r-1)}. $$

(4.8)

By an elementary calculation, we see that if $0 < s < s_{0}$ (resp. $s > s_{0}$), then $g(s)$ is strictly decreasing (resp. strictly increasing) and $g(s_{0})$ is the minimum value of $g(s)$. By (4.5), (4.6), (4.8), and direct calculation, we have

$$\begin{aligned} g(s_{0}) =& -\log \lambda + \frac{q+r-1}{p} \biggl(1 - \log \frac{q+r-1}{p} + \log \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} - \frac{p}{q+r-1} \log \bigl\Vert W_{r}' \bigr\Vert _{q}^{q} \biggr) \\ =& -\log \lambda + C_{0}. \end{aligned}$$

(4.9)

We put $\lambda _{1}:= e^{C_{0}}$. Then, $g(s_{0}) > 0$ if $0 < \lambda < \lambda _{1}$, $g(s_{0}) = 0$ if $\lambda = \lambda _{1}$ and $g(s_{0}) < 0$ if $\lambda > \lambda _{1}$. Then, we see that if $0 < \lambda < \lambda _{1}$, then (4.6) (namely, (4.3) and (4.4)) has no solution, and if $\lambda = \lambda _{1}$, then (4.6) (namely, (4.3) and (4.4)) has a unique solution $s_{0}$, and if $\lambda > \lambda _{1}$, then (4.6) (namely, (4.3) and (4.4)) has exactly two solutions $s_{1}$, $s_{2}$ with $s_{1} < s_{0} < s_{2}$.

We see from the argument above that Theorem 1.3(i), (ii) hold. Moreover, let $t_{\lambda ,j}:= s_{j}^{1/(q+r-1)}$ ($j = 1,2$). By this and (4.1), we obtain Theorem 1.3(iii).

Now, we consider the case (iv). Since it is difficult to obtain $t_{\lambda ,j}:= s_{j}^{1/(q+r-1)}$ ($j = 1,2$) exactly, we first establish the asymptotic formula for $t_{\lambda ,j}$ for $\lambda \gg 1$.

Lemma 4.1

Assume that $\lambda \gg 1$. Then,

$$ t_{\lambda ,2} = \bigl\Vert W_{r}' \bigr\Vert _{q}^{-1}\lambda ^{1/(r-1)}( \log \lambda )^{1/p} \biggl\{ 1 + \frac{p^{2}}{(q+r-1)^{3}} \frac{\log (\log \lambda )}{\log \lambda}\bigl(1 + o(1)\bigr) \biggr\} . $$

(4.10)

Proof

We put $s_{\lambda ,2}:= t_{\lambda ,2}^{q+r-1}$. By (4.3), we have

$$ \exp \bigl(s_{\lambda ,2}^{p/(q+r-1)}\lambda ^{-p/(r-1)} \Vert W_{r}' \Vert _{p}^{p} \bigr)= s_{\lambda ,2}\lambda ^{-q/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{q}^{q}. $$

(4.11)

By this, we have

$$ \frac{q}{r-1}\log \lambda + s_{\lambda ,2}^{p/(q+r-1)}\lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} = \log s_{\lambda ,2} + q\log \bigl\Vert W_{r}' \bigr\Vert _{q}. $$

(4.12)

Then, three cases should be considered.

Case 1. Assume that there exists a subsequence of $\{\lambda \}$, which is denoted by $\{\lambda \}$ again, such that as $\lambda \to \infty$,

$$ \log \lambda \gg s_{\lambda ,2}^{p/(q+r-1)}{\lambda ^{-p/(r-1)}} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p}. $$

(4.13)

Then, by this and (4.12), we have

$$ \frac{q}{r-1}\bigl(1 + o(1)\bigr)\log \lambda = \log s_{\lambda ,2}. $$

(4.14)

This implies that

$$ s_{\lambda ,2} = \lambda ^{q/(r-1)}\bigl(1 + o(1)\bigr). $$

(4.15)

By this and (4.8), we have $s_{0} > s_{\lambda ,2}$. This is a contradiction.

Case 2. Assume that there exists a subsequence of $\{\lambda \}$, which is denoted by $\{\lambda \}$ again, such that as $\lambda \to \infty$,

$$ \log \lambda \ll s_{\lambda ,2}^{p/(q+r-1)}{\lambda ^{-p/(r-1)}} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p}. $$

(4.16)

Then, by (4.12), we have

$$ s_{\lambda ,2}^{p/(q+r-1)}{\lambda ^{-p/(r-1)}} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} = \bigl(1 + o(1)\bigr)\log s_{\lambda ,2}. $$

(4.17)

This implies that

$$\begin{aligned} s_{\lambda ,2} =& \bigl( \bigl\Vert W_{r}' \bigr\Vert _{p}^{-p}\bigl(1 + o(1)\bigr) \lambda ^{p/(r-1)}\log s_{\lambda ,2} \bigr)^{(q+r-1)/p} \\ =& \bigl\Vert W_{r}' \bigr\Vert _{p}^{-(q+r-1)}\bigl(1 + o(1)\bigr)\lambda ^{(q+r-1)/(r-1)}( \log s_{\lambda ,2})^{(q+r-1)/p} \\ =& \bigl\Vert W_{r}' \bigr\Vert _{p}^{-(q+r-1)}\bigl(1 + o(1)\bigr)\lambda ^{(q+r-1)/(r-1)} \biggl(\frac{q+r-1}{r-1}\log \lambda \biggr)^{(q+r-1)/p} \\ =& \biggl(\frac{q+r-1}{r-1} \biggr)^{(q+r-1)/p} \bigl\Vert W_{r}' \bigr\Vert _{p}^{-(q+r-1)} \lambda ^{(q+r-1)/(r-1)}(\log \lambda )^{(q+r-1)/p}\bigl(1 + o(1)\bigr). \end{aligned}$$

(4.18)

By this, (4.12), and (4.18), we have

$$ \frac{q}{r-1}\log \lambda + \frac{q+r-1}{r-1}\bigl(1 + o(1)\bigr) \log \lambda = \frac{q+r-1}{r-1}\bigl(1 + o(1)\bigr) \log \lambda . $$

(4.19)

This is a contradiction.

Case 3. Therefore, there exists a subsequence of $\{\lambda \}$, which is denoted by $\{\lambda \}$ again, such that as $\lambda \to \infty$,

$$ C^{-1} < \frac{\log \lambda}{ s_{\lambda ,2}^{p/(q+r-1)}{\lambda ^{-p/(r-1)}} \Vert W_{r}' \Vert _{p}^{p}} \le C. $$

(4.20)

By this and taking a subsequence of $\{\lambda \}$ again if necessary, we see that there exists a constant $C_{4} > 0$ such that as $\lambda \to \infty$,

$$ s_{\lambda ,2} = C_{4}\lambda ^{(q+r-1)/(r-1)}(\log \lambda )^{(q+r-1)/p}(1 + \delta _{1}), $$

(4.21)

where $\delta _{1} \to 0$ as $\lambda \to \infty$. By this and (4.12), we have

$$\begin{aligned} &\frac{q}{r-1}\log \lambda + s_{\lambda}^{p/(q+r-1)}\lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} \\ &\quad = \log C_{4} + \frac{q+r-1}{r-1}\log \lambda + \frac{q+r-1}{p}\log ( \log \lambda ) + \log (1 + \delta _{0}) + q\log \bigl\Vert W_{r}' \bigr\Vert _{q}. \end{aligned}$$

(4.22)

This implies that

$$ s_{\lambda ,2}^{p/(q+r-1)}\lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} = (1 + \delta _{1})\log \lambda , $$

(4.23)

where $\delta _{1} \to 0$ as $\lambda \to \infty$. This implies that

$$ s_{\lambda ,2} = \bigl\Vert W_{r}' \bigr\Vert _{p}^{-(q+r-1)}\lambda ^{(q+r-1)/(r-1)}( \log \lambda )^{(q+r-1)/p} (1 + \delta _{0}). $$

(4.24)

Namely, $C_{4} = \Vert W_{r}'\Vert _{p}^{-(q+r-1)}$. By (4.22) and (4.23), we have

$$ \delta _{1}\log \lambda = \log C_{4} + \frac{q+r-1}{p}\log (\log \lambda ) + \log (1 + \delta _{0}) + q\log \bigl\Vert W_{r}' \bigr\Vert _{q}. $$

(4.25)

By this, we have

$$ \delta _{1} = \frac{q+r-1}{p} \frac{\log (\log \lambda ) }{\log \lambda}\bigl(1 + o(1)\bigr). $$

(4.26)

By this, (4.23), and the Taylor expansion, we have

$$\begin{aligned} s_{\lambda ,2} =& \bigl\Vert W_{r}' \bigr\Vert _{p}^{-(q+r-1)}\lambda ^{(q+r-1)/(r-1)}( \log \lambda )^{(q+r-1)/p} (1 + \delta _{1})^{(q+r-1)/p} \\ =& \bigl\Vert W_{r}' \bigr\Vert _{p}^{-(q+r-1)}\lambda ^{(q+r-1)/(r-1)}(\log \lambda )^{(q+r-1)/p} \\ &{}\times \biggl\{ 1 + \biggl(\frac{p}{q+r-1} \biggr)^{2} \frac{\log (\log \lambda ) }{\log \lambda}\bigl(1 + o(1)\bigr) \biggr\} . \end{aligned}$$

(4.27)

Indeed, we see from (4.27) that $s_{0} < s_{\lambda ,2}$. Therefore, by (4.27), we obtain (4.10). Thus, the proof is complete. □

Lemma 4.2

Assume that $\lambda \gg 1$. Then

$$\begin{aligned} t_{1,\lambda} =& \lambda ^{q/((r-1)(q+r-1))} \bigl\Vert W_{r}' \bigr\Vert _{q}^{-q/(q+r-1)} \\ &{}\times \biggl\{ 1 + \frac{1}{q+r-1} \bigl\Vert W_{\lambda}' \bigr\Vert _{p}^{p} \bigl\Vert W_{\lambda}' \bigr\Vert _{q}^{pq/(q+r-1)} \lambda ^{-p/(q+r-1)}\bigl(1 + o(1)\bigr) \biggr\} . \end{aligned}$$

(4.28)

Proof

Since $s_{\lambda ,1} < s_{0}$, we find from Lemma 4.1 that as $\lambda \to \infty$,

$$ s_{\lambda ,1}^{p/(q+r-1)}\lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} \ll \log \lambda . $$

(4.29)

By this and (4.12), we have

$$ \bigl(1 + o(1)\bigr)\frac{q}{r-1}\log \lambda = \log s_{\lambda}. $$

(4.30)

This implies that

$$ s_{\lambda ,1} = \lambda ^{q/(r-1)(1+o(1))}. $$

(4.31)

By this and (4.8), we see that $s = s_{\lambda ,1}$ is determined by (4.31). By (4.31), we have

$$ s_{\lambda ,1}^{p/(q+r-1)}{\lambda ^{-p/(r-1)}} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} \le Cs_{1}^{-p(r-1)(1 + o(1))/(q(q+r-1))} \to 0 $$

(4.32)

as $\lambda \to \infty$. Now, we calculate $s_{1}$. By (4.30) and (4.12), we have

$$ \frac{q}{r-1}\log \lambda = \log s_{\lambda ,1} + \bigl(1 + o(1) \bigr)q\log \bigl\Vert W_{r}' \bigr\Vert _{q}. $$

(4.33)

By this, for $\lambda \gg 1$, we have

$$ \lambda = \bigl\Vert W_{r}' \bigr\Vert _{q}^{r-1}s_{1}^{(r-1)/q}(1 + \eta ), $$

(4.34)

where $\eta \to 0$ as $\lambda \to \infty$. By this, (4.12), and the Taylor expansion, we have

$$ \frac{q}{r-1}\bigl(1 + o(1)\bigr)\eta + s_{\lambda ,1}^{p/(q+r-1)} \lambda ^{-p/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{p}^{p} = 0. $$

(4.35)

By this and (4.34), we have

$$ \eta = -\frac{r-1}{q} \bigl\Vert W_{\lambda}' \bigr\Vert _{p}^{p} \bigl\Vert W_{\lambda}' \bigr\Vert _{q}^{-pq/(q+r-1)} \lambda ^{-p/(q+r-1)}\bigl(1 + o(1)\bigr). $$

(4.36)

By this, (4.20), and the Taylor expansion, we have

$$\begin{aligned} s_{\lambda ,1} =& \lambda ^{q/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{q}^{-q} \biggl(1 - \frac{q}{r-1}\eta \biggr) \\ =& \lambda ^{q/(r-1)} \bigl\Vert W_{r}' \bigr\Vert _{q}^{-q} \bigl\{ 1 + \bigl\Vert W_{ \lambda}' \bigr\Vert _{p}^{p} \bigl\Vert W_{\lambda}' \bigr\Vert _{q}^{-pq/(q+r-1)} \lambda ^{-p/(q+r-1)}\bigl(1 + o(1)\bigr) \bigr\} . \end{aligned}$$

(4.37)

By this, we have

$$\begin{aligned} t_{\lambda ,1} =& \lambda ^{q/((r-1)(q+r-1))} \bigl\Vert W_{r}' \bigr\Vert _{q}^{-q/(q+r-1)} \\ &{}\times \biggl\{ 1 + \frac{1}{q+r-1} \bigl\Vert W_{\lambda}' \bigr\Vert _{p}^{p} \bigl\Vert W_{\lambda}' \bigr\Vert _{q}^{-pq/(q+r-1)} \lambda ^{-p/(q+r-1)}\bigl(1 + o(1)\bigr) \biggr\} . \end{aligned}$$

(4.38)

Thus, we obtain (4.28). □

Proof of Theorem 1.3

By Lemma 4.2, for $\lambda \gg 1$, we obtain

$$\begin{aligned} u_{1,\lambda}(x) =& \lambda ^{-1/(q+r-1)} \bigl\Vert W_{r}' \bigr\Vert _{q}^{-q/(q+r-1)} \\ &{}\times \biggl\{ 1 + \frac{1}{q+r-1} \bigl\Vert W_{\lambda}' \bigr\Vert _{p}^{p} \bigl\Vert W_{\lambda}' \bigr\Vert _{q}^{pq/(q+r-1)} \lambda ^{-p/(q+r-1)}\bigl(1 + o(1)\bigr) \biggr\} W_{r}(x). \end{aligned}$$

(4.39)

This implies (1.14). Further, by Lemma 4.1, we obtain

$$ u_{2,\lambda}(x) = \bigl\Vert W_{r}' \bigr\Vert _{q}^{-1}(\log \lambda )^{1/p} \biggl\{ 1 + \frac{p^{2}}{(q+r-1)^{3}} \frac{\log (\log \lambda )}{\log \lambda}\bigl(1 + o(1)\bigr) \biggr\} W_{r}(x). $$

(4.40)

This implies (1.16). To obtain (1.15) and (1.17), we just put $x = 1/2$ in (4.39) and (4.40). Thus, the proof is complete. □

Appendix

Let $r > 1$. We first show (1.6), which was proved in [14] for completeness. We apply the time map argument to (1.3), cf. [12]. Since (1.3) is autonomous, we have

$$\begin{aligned}& W_{r}(x) = W_{r}(1-x),\quad x \in [0, 1/2], \end{aligned}$$

(A.1)

$$\begin{aligned}& W_{r}'(x) > 0, \quad x \in [0,1/2), \end{aligned}$$

(A.2)

$$\begin{aligned}& \xi := \Vert W_{r} \Vert _{\infty }= \max _{0\le x \le 1}W_{r}(x) = W_{r}(1/2). \end{aligned}$$

(A.3)

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

$$ \bigl\{ W_{r}''(x) + W_{r}(x)^{r}\bigr\} W_{r}'(x) = 0. $$

(A.4)

By this and (A.3), we have

$$ \frac{1}{2}W_{r}'(x)^{2} + \frac{1}{r+1}W_{r}(x)^{r+1} = \text{constant} = \frac{1}{r+1}W_{p}(1/2)^{r+1} = \frac{1}{r+1}\xi ^{r+1}. $$

(A.5)

By this and (A.2), for $0 \le x \le 1/2$, using $\theta = \xi s$, we have

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

(A.6)

By (A.1) and (A.6), we have

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

(A.7)

By this, we have

$$ \xi = \bigl(2(r+1)\bigr)^{1/(r-1)}L_{r}^{2/(r-1)}. $$

(A.8)

This implies (1.6). We next show (1.5). By (A.1), (A.3), and (A.6), we have

$$\begin{aligned} \bigl\Vert W_{r}' \bigr\Vert _{m}^{m} =& 2 \int _{0}^{1/2} W_{r}'(x)^{m-1}W_{r}'(x)\,dx \\ =& 2 \biggl(\frac{2}{r+1} \biggr)^{(m-1)/2} \int _{0}^{1/2} \bigl( \xi ^{r+1} - W_{r}(x)^{r+1} \bigr)^{(m-1)/2} W_{r}'(x)\,dx \\ =& 2^{(m+1)/2}(r+1)^{-(m-1)/2} \int _{0}^{\xi }\bigl(\xi ^{r+1} - \theta ^{r+1}\bigr)^{(m-1)/2}\,d\theta \\ =& 2^{(m+1)/2}(r+1)^{-(m-1)/2}\xi ^{(m-1)(r+1)/2 + 1} \int _{0}^{1} \bigl(1-s^{r+1} \bigr)^{(m-1)/2}\,ds \\ =& 2^{mr/(r-1)}(r+1)^{m/(r-1)}L_{r}^{(mr+m-r+1)/(r-1)}M_{r,m}. \end{aligned}$$

(A.9)

This implies (1.5).