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

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

(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

$$ -exp bigl(t^{p}lambda ^{-p/(r-1)} biglVert W_{r}’ bigrVert _{p}^{p} bigr) W_{r}”(x) = t^{q+r-1} lambda ^{-q/(r-1)} biglVert W_{r}’ bigrVert _{q}^{q} W_{r}(x)^{r}. $$

(4.2)

This implies that

$$ exp bigl(t^{p}lambda ^{-p/(r-1)} biglVert W_{r}’ bigrVert _{p}^{p} bigr) = t^{q+r-1}lambda ^{-q/(r-1)} biglVert W_{r}’ bigrVert _{q}^{q}. $$

(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

$$ C_{1}:= lambda ^{-p/(r-1)} biglVert W_{r}’ bigrVert _{p}^{p}, qquad C_{2}:= – frac{q}{r-1}log lambda + log biglVert W_{r}’ bigrVert _{q}^{q}. $$

(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

$$begin{aligned} g(s_{0}) =& -log lambda + frac{q+r-1}{p} biggl(1 – log frac{q+r-1}{p} + log biglVert W_{r}’ bigrVert _{p}^{p} – frac{p}{q+r-1} log biglVert W_{r}’ bigrVert _{q}^{q} biggr) \ =& -log lambda + C_{0}. end{aligned}$$

(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*,

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

(4.10)

### Proof

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

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

(4.11)

By this, we have

$$ frac{q}{r-1}log lambda + s_{lambda ,2}^{p/(q+r-1)}lambda ^{-p/(r-1)} biglVert W_{r}’ bigrVert _{p}^{p} = log s_{lambda ,2} + qlog biglVert W_{r}’ bigrVert _{q}. $$

(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 ),

$$ log lambda gg s_{lambda ,2}^{p/(q+r-1)}{lambda ^{-p/(r-1)}} biglVert W_{r}’ bigrVert _{p}^{p}. $$

(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 ),

$$ log lambda ll s_{lambda ,2}^{p/(q+r-1)}{lambda ^{-p/(r-1)}} biglVert W_{r}’ bigrVert _{p}^{p}. $$

(4.16)

Then, by (4.12), we have

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

(4.17)

This implies that

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

(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

$$begin{aligned} &frac{q}{r-1}log lambda + s_{lambda}^{p/(q+r-1)}lambda ^{-p/(r-1)} biglVert W_{r}’ bigrVert _{p}^{p} \ &quad = log C_{4} + frac{q+r-1}{r-1}log lambda + frac{q+r-1}{p}log ( log lambda ) + log (1 + delta _{0}) + qlog biglVert W_{r}’ bigrVert _{q}. end{aligned}$$

(4.22)

This implies that

$$ s_{lambda ,2}^{p/(q+r-1)}lambda ^{-p/(r-1)} biglVert W_{r}’ bigrVert _{p}^{p} = (1 + delta _{1})log lambda , $$

(4.23)

where (delta _{1} to 0) as (lambda to infty ). This implies that

$$ s_{lambda ,2} = biglVert W_{r}’ bigrVert _{p}^{-(q+r-1)}lambda ^{(q+r-1)/(r-1)}( log lambda )^{(q+r-1)/p} (1 + delta _{0}). $$

(4.24)

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

$$ delta _{1}log lambda = log C_{4} + frac{q+r-1}{p}log (log lambda ) + log (1 + delta _{0}) + qlog biglVert W_{r}’ bigrVert _{q}. $$

(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

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

(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*

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

(4.28)

### Proof

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

$$ s_{lambda ,1}^{p/(q+r-1)}lambda ^{-p/(r-1)} biglVert W_{r}’ bigrVert _{p}^{p} ll log lambda . $$

(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

$$ s_{lambda ,1}^{p/(q+r-1)}{lambda ^{-p/(r-1)}} biglVert W_{r}’ bigrVert _{p}^{p} le Cs_{1}^{-p(r-1)(1 + o(1))/(q(q+r-1))} to 0 $$

(4.32)

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

$$ frac{q}{r-1}log lambda = log s_{lambda ,1} + bigl(1 + o(1) bigr)qlog biglVert W_{r}’ bigrVert _{q}. $$

(4.33)

By this, for (lambda gg 1), we have

$$ lambda = biglVert W_{r}’ bigrVert _{q}^{r-1}s_{1}^{(r-1)/q}(1 + eta ), $$

(4.34)

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

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

(4.35)

By this and (4.34), we have

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

(4.36)

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

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

(4.37)

By this, we have

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

(4.38)

Thus, we obtain (4.28). □

### Proof of Theorem 1.3

By Lemma 4.2, for (lambda gg 1), we obtain

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

(4.39)

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

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

(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. □

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