Maximizing the total population with logistic growth in a patchy environment

Abstract

This paper is concerned with a nonlinear optimization problem that naturally arises in population biology. We consider the population of a single species with logistic growth residing in a patchy environment and study the effects of dispersal and spatial heterogeneity of patches on the total population at equilibrium. Our objective is to maximize the total population by redistributing the resources among the patches under the constraint that the total amount of resources is limited. It is shown that the global maximizer can be characterized for any number of patches when the diffusion rate is either sufficiently small or large. To show this, we compute the first variation of the total population with respect to resources in the two patches case. In the case of three or more patches, we compute the asymptotic expansion of all patches by using the Taylor expansion with respect to the diffusion rate. To characterize the shape of the global maximizer, we use a recurrence relation to determine all coefficients of all patches.

Appendix

1.1 Preliminaries for small $\delta$

In this Appendix, we choose $N \ge 2$ and $\varvec{}{m} \in \mathcal {M}$ arbitrarily.

Lemma 7

Any solution of (1.3) satisfies either ${{\,\mathrm{\underline{deg}}\,}}u_i = 0$ or $1/2 \le {{\,\mathrm{\underline{deg}}\,}}u_i < 1$ for each $i \in \varOmega$. Moreover, $m_i > 0$ if ${{\,\mathrm{\underline{deg}}\,}}u_i = 0$, and $m_i = 0$ if $1/2 \le {{\,\mathrm{\underline{deg}}\,}}u_i < 1$.

Proof

By using (3.1)–(3.3), $u_i \rightarrow m_i$ as $\delta \rightarrow 0$. This implies ${{\,\mathrm{\underline{deg}}\,}}u_i \ge 0$. First, we choose $i = 1$ or $i = N$. Assume $m_i > 0$. Then by (3.1), we have

$$\begin{aligned} \begin{aligned} u_1&= \frac{m_1 - \delta }{2} + \left( \frac{1}{4} (m_1^2 - 2m_1 \delta + \delta ^2) + \delta u_2 \right) ^{1/2}\\&= \frac{m_1 - \delta }{2} + \frac{m_1}{2}\left( 1 + \frac{2}{m_1^2} \left( u_2 - \frac{m_1}{2} \right) \delta +o(\delta ) \right) \\&= m_1 - \left( \frac{u_2}{m_1} -1 \right) \delta +o(\delta ). \end{aligned} \end{aligned}$$

Similarly, by (3.3), we have

$$\begin{aligned} \begin{aligned} u_N&= m_N - \left( \frac{u_{N-1}}{m_N} -1 \right) \delta +o(\delta ). \end{aligned} \end{aligned}$$

This indicates that if $m_i > 0$, then ${{\,\mathrm{\underline{deg}}\,}}u_i = 0$.

Next, assume $m_i = 0$. By using (3.1) and (3.3), we have

$$\begin{aligned} \begin{aligned} u_1 = - \frac{\delta }{2} + \left( \frac{1}{4} \delta ^2 + \delta u_2 \right) ^{1/2}, \quad u_N = - \frac{\delta }{2} + \left( \frac{1}{4} \delta ^2 + \delta u_{N-1} \right) ^{1/2}. \end{aligned} \end{aligned}$$

This gives

$$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_1= & {} \min \left\{ 1, \frac{1 + {{\,\mathrm{\underline{deg}}\,}}u_2}{2}\right\} , \end{aligned}$$

(A.1)

$$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_N= & {} \min \left\{ 1, \frac{1 + {{\,\mathrm{\underline{deg}}\,}}u_{N-1}}{2}\right\} . \end{aligned}$$

(A.2)

This argument and Proposition 1 indicate that if $m_i = 0$, then $1/2 \le {{\,\mathrm{\underline{deg}}\,}}u_i \le 1$. It follows that ${{\,\mathrm{\underline{deg}}\,}}u_i = 0$ is equivalent to $m_i > 0$. From the above argument, we can assert that $1/2 \le {{\,\mathrm{\underline{deg}}\,}}u_i \le 1$ is equivalent to $m_i = 0$.

Second, we choose $i \in \mathbb {Z} \cap (1,N)$. Assume $m_i > 0$. By (3.2), we have

$$\begin{aligned} u_i&= \frac{m_i - 2 \delta }{2} + \left( \frac{1}{4} m_i^2 - m_i \delta +\delta ^2 + \delta (u_{i-1} + u_{i+1}) \right) ^{1/2} \\&= \frac{m_i - 2 \delta }{2} + \frac{m_i}{2} \left( 1+ \frac{2}{m_i^2} ( u_{i-1} + u_{i+1} - m_i ) \delta +o(\delta ) \right) \\&= m_i + \left( \frac{u_{i-1} + u_{i+1}}{m_i} -2 \right) \delta + o(\delta ). \end{aligned}$$

This implies that if $m_i > 0$, then ${{\,\mathrm{\underline{deg}}\,}}u_i = 0$. Next, assume $m_i = 0$. We use (3.2) again to have

$$\begin{aligned} u_i = - \delta + \left( \delta ^2 + \delta (u_{i-1} + u_{i+1}) \right) ^{1/2}. \end{aligned}$$

This gives

$$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_i = \min \left\{ 1, \frac{1 + \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{i-1}, {{\,\mathrm{\underline{deg}}\,}}u_{i+1} \} }{2}\right\} . \end{aligned}$$

(A.3)

Similar argument to the first case shows that ${{\,\mathrm{\underline{deg}}\,}}u_i = 0$ is equivalent to $m_i > 0$. Furthermore, $1/2 \le {{\,\mathrm{\underline{deg}}\,}}u_i \le 1$ is equivalent to $m_i = 0$.

Therefore, it is sufficient to prove that $0 \le {{\,\mathrm{\underline{deg}}\,}}u_i < 1$ for all $i \in \varOmega$. We assume that there exists $i \in \varOmega$ such that ${{\,\mathrm{\underline{deg}}\,}}u_i = 1$. From (A.1)–(A.3), we have

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\,\mathrm{\underline{deg}}\,}}u_2 = 1 &{} \text{ if } i=1, \\ {{\,\mathrm{\underline{deg}}\,}}u_{i-1} = {{\,\mathrm{\underline{deg}}\,}}u_{i+1} = 1 &{} \text{ if } 1< i < N, \\ {{\,\mathrm{\underline{deg}}\,}}u_{N-1} = 1 &{} \text{ if } i=N. \end{array}\right. } \end{aligned}$$

Repeated application of (A.1)–(A.3) enables us to obtain $m_i = 0$ for all $i \in \varOmega$, which contradicts our assumption. $\square$

Recall that (6.1)–(6.3) are obtained by Lemma 7. We next show that ${{\,\mathrm{\underline{deg}}\,}}u_i$ takes a finite number of values. By Lemma 8, we only have to seek the largest coefficients of an appropriate order of $\delta$ to maximize the total population.

Lemma 8

Any solution of (1.3) satisfies

$$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_i \in \left\{ 1- 2^{-p} \mid p \in \mathbb {Z} \cap {[0, N-1]} \right\} \end{aligned}$$

for all $i \in \varOmega$.

Proof

We assume that there exists $i \in \varOmega$ such that

$$\begin{aligned} 1-2^{-p}< {{\,\mathrm{\underline{deg}}\,}}u_i < 1 - 2^{-(p+1)}, \end{aligned}$$

(A.4)

where $p \in \mathbb {Z} \cap [0, \min \{ i-1, N-i \}]$. By using (A.1)–(A.3), we calculate

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\,\mathrm{\underline{deg}}\,}}u_1 = (1 + {{\,\mathrm{\underline{deg}}\,}}u_2)/2 &{} \text{ if } i = 1, \\ {{\,\mathrm{\underline{deg}}\,}}u_i = (1 + \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{i-1}, {{\,\mathrm{\underline{deg}}\,}}u_{i+1} \})/2 &{} \text{ if } i \in \mathbb {Z} \cap (1,N), \\ {{\,\mathrm{\underline{deg}}\,}}u_N = ( 1 + {{\,\mathrm{\underline{deg}}\,}}u_{N-1})/2 &{} \text{ if } i = N. \end{array}\right. } \end{aligned}$$

(A.5)

Repeating substitution of (A.4) to (A.5) p times, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} 0< {{\,\mathrm{\underline{deg}}\,}}u_{p+1}< 1/2 &{} \text{ if } i = 1, \\ 0< \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{i-p}, {{\,\mathrm{\underline{deg}}\,}}u_{i+p} \}< 1/2 &{} \text{ if } i \in \mathbb {Z} \cap (1,N), \\ 0< {{\,\mathrm{\underline{deg}}\,}}u_{N-p} < 1/2 &{} \text{ if } i = N. \end{array}\right. } \end{aligned}$$

This contradicts Lemma 7. Finally, we assume that

$$\begin{aligned} 1-2^{-(N-1)} < {{\,\mathrm{\underline{deg}}\,}}u_i. \end{aligned}$$

Applying (A.1)–(A.3) repeatedly $\max \{ N-i , i-1 \}$ times to have $\min _{i \in \varOmega } \{ {{\,\mathrm{\underline{deg}}\,}}u_i\}$. Since $\max \{ N-i , i-1 \} \le N-1$, we have $\min _{i \in \varOmega } \{ {{\,\mathrm{\underline{deg}}\,}}u_i\} > 0$. This contradicts our assumption. $\square$

We next evaluate the difference of ${{\,\mathrm{\underline{deg}}\,}}u_i$ between two adjacent patches.

Lemma 9

Let $i \in \varOmega$ and $p \in \mathbb {Z} \cap [0, N-1]$. Suppose that a solution of (1.3) satisfies

$$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_i = \frac{2^p - 1}{2^p}. \end{aligned}$$

Then we have

$$\begin{aligned} \frac{2^{p-1} - 1}{2^{p-1}} \le \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{i-1}, {{\,\mathrm{\underline{deg}}\,}}u_{i+1} \}, \end{aligned}$$

(A.6)

where the equality holds if and only if $m_i = 0$, and

$$\begin{aligned} \max \{ {{\,\mathrm{\underline{deg}}\,}}u_{i-1}, {{\,\mathrm{\underline{deg}}\,}}u_{i+1} \} \le \frac{2^{p+1} -1}{2^{p+1}}, \end{aligned}$$

(A.7)

where the equality holds if and only if $\min \{ m_{i-1}, m_{i+1} \} = 0$.

Proof

It is clear that (A.6) follows immediately from (A.5). Suppose that

$$\begin{aligned} \max \{ {{\,\mathrm{\underline{deg}}\,}}u_{i-1}, {{\,\mathrm{\underline{deg}}\,}}u_{i+1} \} \ge \frac{2^{p+2} - 1}{2^{p+2}}. \end{aligned}$$

By (A.6), we have

$$\begin{aligned} \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{i-2}, {{\,\mathrm{\underline{deg}}\,}}u_{i} \} = \frac{2^{p+1} - 1}{2^{p+1}}, \text{ or } \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{i}, {{\,\mathrm{\underline{deg}}\,}}u_{i+2} \} = \frac{2^{p+1} - 1}{2^{p+1}}. \end{aligned}$$

This contradicts our assumption. This proves (A.7). $\square$

Repeated application of Lemma 9 enables us to have the following lemma.

Lemma 10

Let $i,k \in \varOmega$, and take $p \in \mathbb {Z} \cap [1, N-1]$ such that $k-p$, $k+p \in \varOmega$. If a positive solution of (1.3) satisfies ${{\,\mathrm{\underline{deg}}\,}}u_k = (2^p - 1)/2^p$, then

$$\begin{aligned} \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{\max \{1, k-p\}}, {{\,\mathrm{\underline{deg}}\,}}u_{\min \{k+p,N\}} \} = 0. \end{aligned}$$

Moreover, $m_i = 0$ for all $i \in ( k-p, k+p)$.

Lemma 10 allows us to examine the effect of the distance between favorable patches on ${{\,\mathrm{\underline{deg}}\,}}u_i$. The following proposition plays an important role in Sect. 6.

Lemma 11

Let k, i, q be positive integers such that $k, k+i, k+2q \in \varOmega$.

• (i) Suppose that $m \in \mathcal {M}$ satisfies $m_k > 0$, $m_{k+2q} > 0$, and $m_{k+i} = 0 \; (0<i< 2q)$. Then the following equalities hold:

  $$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_{k+i} = {{\,\mathrm{\underline{deg}}\,}}u_{k+2q-i } = \frac{2^i - 1}{2^i} \quad (0 \le i \le q). \end{aligned}$$

• (ii) Suppose that $\varvec{}{m} \in \mathcal {M}$ satisfies $m_k > 0$, $m_{k+2q} > 0$, and $m_{k+i} = 0 \; (0<i< 2q+1)$. Then the following equalities hold:

  $$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_{k+i} = {{\,\mathrm{\underline{deg}}\,}}u_{k+2q+1-i} = \frac{2^i - 1}{2^i} \quad (0 \le i \le q). \end{aligned}$$

Proof

(i) Assume that $i \in \mathbb {Z} \cap [1,q]$ and

$$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_{k+i} = \frac{2^{i-j} - 1}{2^{i-j}} < (2^i - 1)/2^i, \end{aligned}$$

where $j \in \mathbb {Z} \cap [1, i]$. Then we have

$$\begin{aligned} \min \{ {{\,\mathrm{\underline{deg}}\,}}u_{k+j}, {{\,\mathrm{\underline{deg}}\,}}u_{k+2i-j}\} = 0 \end{aligned}$$

by Lemma 10. Similarly, assuming that

$$\begin{aligned} {{\,\mathrm{\underline{deg}}\,}}u_{k+i} = \frac{2^{i+j} - 1}{2^{i+j}} > (2^i - 1)/2^i, \end{aligned}$$

where $j \in \mathbb {Z} \cap [1, N-1-i]$, we have $m_k = 0$ by Lemma 10. This contradicts our assumption $m_k>0$. We now apply this argument again, with $k+i$ replaced by $k + 2q - i$, to obtain the value of ${{\,\mathrm{\underline{deg}}\,}}u_{k+2q-i}$.

The case (ii) may be proved in much the same way as the case (i), so we omit the proof. $\square$

1.2 Expansion of the total population in N patches

To express the total population $U(\varvec{}{m}, \delta )$ in terms of $\delta$ for each $\varvec{}{m} \in \mathcal {M}$, we show the following result.

Proposition 7

Let k, i, q be positive integers such that $k, k+i, k+2q \in \varOmega$. Choose $\varvec{}{m} \in \mathcal {M}$ arbitrarily. Then the positive solution of (1.3) are expanded as follows:

(i):

If $\varvec{}{m} \in \mathcal {M}$ satisfies $m_k > 0$, $m_{k+2q} > 0$ and $m_{k+i} = 0, \; (1 \le i \le 2q-1)$, then

$$\begin{aligned} u_{k}&= {\left\{ \begin{array}{ll} m_k - \delta +o(\delta ), &{}\; (k = 1),\\ \displaystyle m_k + \left( \frac{m_{k-1}}{m_k} - 2 \right) \delta + o(\delta ), &{}\; (k>1), \end{array}\right. } \end{aligned}$$

(A.8)

$$\begin{aligned} u_{k+i}&= m_k^{1/2^i} \delta ^{(2^i - 1)/2^{-i}} - \delta + o(\delta ), \; (1 \le i \le q-1), \end{aligned}$$

(A.9)

$$\begin{aligned} u_{k+q}&= \sqrt{m_k^{1/2^{q-1}} + m_{k+2q}^{1/2^{q-1}}} \delta ^{(2^q - 1)/2^q} - \delta + o(\delta ), \end{aligned}$$

(A.10)

$$\begin{aligned} u_{k+2q-i}&= m_{k+2q}^{1/2^i} \delta ^{(2^i - 1)/2^i} - \delta + o(\delta ), \; (1 \le i \le q-1), \end{aligned}$$

(A.11)

$$\begin{aligned} u_{k+2q}&= {\left\{ \begin{array}{ll} m_{k+2q} - \delta +o(\delta ), &{}\; (k+2q = N), \\ \displaystyle m_{k+2q} + \left( \frac{m_{k+2q+1}}{m_{k+2q}} - 2 \right) \delta + o(\delta ), &{}\; (k+2q<N). \end{array}\right. } \end{aligned}$$

(A.12)

(ii):

If $\varvec{}{m} \in \mathcal {M}$ satisfies $m_k > 0$, $m_{k+2q+1} > 0$ and $m_{k+i} = 0, \; (1 \le i \le 2q)$, then

$$\begin{aligned} u_{k}= & {} {\left\{ \begin{array}{ll} m_k - \delta +o(\delta ), &{}\; (k = 1),\\ \displaystyle m_k + \left( \frac{m_{k-1}}{m_k} - 2 \right) \delta + o(\delta ), &{}\; (k>1), \end{array}\right. }\nonumber \\ u_{k+i}= & {} m_k^{1/2^i} \delta ^{(2^i - 1)/2^i} - \delta + o(\delta ), \; (1 \le i \le q-1),\nonumber \\ u_{k+q}= & {} m_k^{1/2^{q}} \delta ^{(2^q - 1)/2^q} + \left( \frac{1}{2} \left( \frac{m_{k+2q+1}}{m_k} \right) ^{1/2^{q}} -1 \right) \delta + o(\delta ), \end{aligned}$$

(A.13)

$$\begin{aligned} u_{k+q+1}= & {} m_{k+2q+1}^{1/2^{q}} \delta ^{(2^q - 1)/2^q} + \left( \frac{1}{2} \left( \frac{m_k}{m_{k+2q+1}} \right) ^{1/2^{q}} -1 \right) \delta + o(\delta ), \nonumber \\ u_{k+2q+1-i}= & {} m_{k+2q+1}^{1/2^i} \delta ^{(2^i - 1)/2^i} - \delta + o(\delta ), \; (1 \le i \le q-1),\nonumber \\ u_{k+2q+1}= & {} {\left\{ \begin{array}{ll} m_{k+2q+1} - \delta +o(\delta ), &{}\; (k+2q+1 = N), \\ \displaystyle m_{k+2q+1} + \left( \frac{m_{k+2q+2}}{m_{k+2q+1}} - 2 \right) \delta + o(\delta ), &{}\; (k+2q+1<N). \end{array}\right. } \end{aligned}$$

(A.14)

(iii):

If $\varvec{}{m} \in \mathcal {M}$ satisfies $m_k> 0, \; (k > 1)$, $m_i = 0, \; (1 \le i \le k-1)$, then

$$\begin{aligned} u_{1}= & {} m_k^{1/2^{k-1}} \delta ^{(2^{k-1} - 1)/2^{k-1}} - \frac{\delta }{2} + o(\delta ),\nonumber \\ u_{k-i}= & {} m_k^{1/2^{i}} \delta ^{(2^{i} - 1)/2^{i}} - \delta + o(\delta ),\nonumber \\ u_{k}= & {} {\left\{ \begin{array}{ll} m_k - \delta +o(\delta ), &{}\; (k = N),\\ \displaystyle m_k + \left( \frac{m_{k+1}}{m_k} - 2 \right) \delta + o(\delta ), &{}\; (k<N), \end{array}\right. } \end{aligned}$$

(A.15)

(iv):

If $\varvec{}{m} \in \mathcal {M}$ satisfies $m_k > 0, \; (k < N)$, $m_i = 0, \; (k +1 \le i \le N)$, then

$$\begin{aligned} \begin{aligned} u_{k}&= {\left\{ \begin{array}{ll} m_k - \delta +o(\delta ), &{}\; (k = 1),\\ \displaystyle m_k + \left( \frac{m_{k-1}}{m_k} - 2 \right) \delta + o(\delta ), &{}\; (k>1), \end{array}\right. }\\ u_{k+i}&= m_k^{1/2^{i}} \delta ^{(2^{i} - 1)/2^{i}} - \delta + o(\delta ) ,\\ u_{N}&= m_k^{1/2^{N-k}} \delta ^{(2^{N-k} - 1)/2^{N-k}} - \frac{\delta }{2} + o(\delta ). \end{aligned} \end{aligned}$$

Proof

(i) It is easy to verify (A.8) by noting ${{\,\mathrm{\underline{deg}}\,}}u_{k+1} = 1/2$. So we omit the proof. We now consider (A.9) by induction on i. By Lemma 11, there exist coefficients $C_{k+i}$ for all $i \in \{0, 1, \ldots , q\}$ such that

$$\begin{aligned} u_{k+i} = C_{k+i} \delta ^{(2^i - 1)/2^i} + o(\delta ^{(2^i - 1)/2^i}), \quad C_k = m_k. \end{aligned}$$

By (6.1) or (6.2), we have the base case as

$$\begin{aligned} \begin{aligned} u_{k+1}&= \left( m_k \delta + C_{k+2} \delta ^{7/4} + o(\delta ^{7/4} ) \right) ^{1/2}-\delta + o(\delta )\\&= (m_k \delta )^{1/2} \left( 1 + \frac{C_{k+2}}{2m_k} \delta ^{3/4} + o(\delta ^{3/4}) \right) - \delta + o(\delta )\\&= m_k^{1/2} \delta ^{1/2} - \delta + o(\delta ). \end{aligned} \end{aligned}$$

Therefore, we have the relation between $C_{k}$ and $C_{k+1}$ given by

$$\begin{aligned} C_{k+1} = C_k^{1/2}. \end{aligned}$$

Suppose that

$$\begin{aligned} u_{k+i-1} = C_{k+i-1} \delta ^{(2^{i-1} - 1)/2^{i-1}} - \delta +o(\delta ) \quad (3 \le i \le q-1). \end{aligned}$$

By (6.2), we have the induction step as

$$\begin{aligned} \begin{aligned} u_{k+i}&= \left( C_{k+i-1} \delta ^{(2^{i} - 1)/2^{i-1}} + C_{k+i+1} \delta ^{(2^{i+2} - 1)/2^{i+1}} + o(\delta ^{(2^{i+2} - 1)/2^{i+1}}) \right) ^{1/2} - \delta + o(\delta ) \\&= C^{1/2}_{k+i-1} \delta ^{(2^{i} - 1)/2^{i}} \left( 1 + \frac{C_{k+i+1}}{ 2 C_{k+i-1}} \delta ^{1/2^{i-1} - 1/2^{i+1} } + o(\delta ^{1/2^{i-1} - 1/2^{i+1} }) \right) - \delta + o(\delta ) \\&= C_{k+i-1}^{1/2} \delta ^{(2^{i} - 1)/2^{i}} - \delta + o(\delta ). \\ \end{aligned} \end{aligned}$$

Therefore, we have recurrence relation $C_{k+i} = C_{k+i-1}^{1/2}$ which gives (A.9). From this, we replace “k” by “$k+2q$” to have (A.11) and (A.12).

To prove (A.10), we use (A.9) and (A.11) to have

$$\begin{aligned} \begin{aligned} u_{k+q-1}&= m_k^{1/2^{q-1}} \delta ^{(2^{q-1} - 1)/2^{q-1}} - \delta + o(\delta ),\\ u_{k+q+1}&= m_{k+2q}^{1/2^{q-1}} \delta ^{(2^{q-1} - 1)/2^{q-1}} - \delta + o(\delta ). \end{aligned} \end{aligned}$$

By (6.2), we obtain

$$\begin{aligned} \begin{aligned}&u_{k+q} = \left( m_k^{1/2^{q-1}} \delta ^{(2^{q} - 1)/2^{q-1}} + m_{k+2q}^{1/2^{q-1}} \delta ^{(2^{q} - 1)/2^{q-1}} - 2 \delta ^2 + o(\delta ^2) \right) ^{1/2} - \delta + o(\delta ) \\&\quad = \left( m_k^{1/2^{q-1}} + m_{k+2q}^{1/2^{q-1}} \right) ^{1/2} \delta ^{(2^{q} - 1)/2^{q}} \left( 1 - \frac{ \delta ^{1/2^{q-1}}}{ m_k^{1/2^{q-1}} + m_{k+2q}^{1/2^{q-1}} } + o(\delta ^{1/2^{q-1}}) \right) \\&\quad - \delta + o(\delta ) \\&\quad = \left( m_k^{1/2^{q-1}} + m_{k+2q}^{1/2^{q-1}} \right) ^{1/2}\delta ^{(2^{q} - 1)/2^{q}} - \delta + o(\delta ). \end{aligned} \end{aligned}$$

(ii) We give the proof only for (A.13) and (A.14); the other cases can be proved by the same argument as above. We use Lemma 11 to have

$$\begin{aligned} \begin{aligned} u_{k+q}&= C_{k+q} \delta ^{(2^q - 1)/2^q} + o(\delta ^{(2^q - 1)/2^q}), \\ u_{k+q+1}&= C_{k+q+1} \delta ^{2^q - 1/2^q} + o(\delta ^{(2^q - 1)/2^q}). \end{aligned} \end{aligned}$$

We now compute $C_{k+q}$ and $C_{k+q+1}$ by using (6.2) as

$$\begin{aligned} \begin{aligned}&u_{k+q} = \left( m_k^{1/2^{q-1}} \delta ^{(2^{q} - 1)/2^{q-1}} + C_{k+q+1} \delta ^{(2^{q+1} - 1)/2^q} + o(\delta ^{(2^{q+1} - 1)/2^q}) \right) ^{1/2} - \delta + (\delta ) \\&\quad = m_k^{1/2^{q}} \delta ^{(2^{q} - 1)/2^{q}} \left( 1 + \frac{C_{k+q+1}}{2 m_k^{1/2^{q-1}}} \delta ^{1/2^q} + o(\delta ^{1/2^q}) \right) - \delta + (\delta ) \\&\quad = m_k^{1/2^{q}} \delta ^{(2^{q} - 1)/2^{q}} + \left( \frac{C_{k+q+1}}{2 m_k^{1/2^{q}}} - 1 \right) \delta + (\delta ). \\ \end{aligned} \end{aligned}$$

Similarly, we have

$$\begin{aligned} u_{k+q+1} = m_{k+2q+1}^{1/2^{q}} \delta ^{(2^{q} - 1)/2^{q}} + \left( \frac{C_{k+q}}{2 m_{k+2q+1}^{1/2^{q}}} - 1 \right) \delta + (\delta ). \end{aligned}$$

By the definition of $C_{k+q}$ and $C_{k+q+1}$, we have

$$\begin{aligned} C_{k+q} = m_{k}^{1/2^{q}}, \quad C_{k+q+1} = m_{k+2q+1}^{1/2^{q}}. \end{aligned}$$

Therefore, we obtain (A.13) and (A.14).

(iii) We use (6.1) to have

$$\begin{aligned} \begin{aligned}&u_1 = \left( \delta \left( m_k^{1/2^{k-2}} \delta ^{(2^{k-2} - 1)/2^{k-2}} - \delta + o(\delta ) \right) \right) ^{1/2} - \frac{\delta }{2} + o(\delta ) \\&\quad = m_k^{1/2^{k-1}} \delta ^{(2^{k-1}-1)/2^{k-1}} \left( 1 - \frac{1}{2 m_k^{1/2^{k-2}}} \delta ^{1/2^{k-2}} + o(\delta ^{1/2^{k-2}}) \right) - \frac{\delta }{2} + o(\delta ) \\&\quad = m_k^{1/2^{k-1}} \delta ^{(2^{k-1}-1)/2^{k-1}} - \frac{\delta }{2} + o(\delta ). \end{aligned} \end{aligned}$$

This proves (A.15) The other cases can be proved by the same argument.

(iv) This can be proved in much the same way as (iii). So we omit the proof. $\square$