On the asymptotic behavior of certain solutions of the Dirichlet problem for the equation $-\Delta _p u=\lambda |u|^{q-2}u$

Abstract

Let $p>1$. We study the behavior of certain positive and nodal solutions of the problem

$$\begin{aligned} \left\{ \,\, \begin{array}{lll} -\Delta _p u=\lambda |u|^{q-2}u \ \ &{}\mathrm{in} \ \ &{}{\varOmega } \\ u=0 &{}\mathrm{in} \ \ &{}\partial {\varOmega } \end{array}\right. \end{aligned}$$

on varying of the parameters $\lambda >0$ and $q>1$.

Appendix

Let us consider the well known $beta$-function $\beta (x,y)=\int _0^1t^{x-1}(1-t)^{y-1}$ defined for $x,y\in ]0,+\infty [$. In what follows, we will make use of the identity

$$\begin{aligned} \beta (1+x,1+y)=\frac{x}{1+x+y}\beta (x,y), \ \ \ \mathrm{for \ all} \ \ x,y\in ]0,+\infty [. \end{aligned}$$

(34)

We want to derive the explicit expression of the unique positive solution of problem $(P)$ in the one-dimensionale case. For simplicity, we assume $\lambda =1$.

For ${\varOmega }=]a,b[$, with $a,b\in \mathbb{R }$ and $a<b$, the positive (and symmetric) solution of problem $(P)$ satisfies

$$\begin{aligned} \left\{ \,\, \begin{array}{lll} -(|u'|^{p-2}u')'=u^{q-1} \ \ &{}\mathrm{in} \ \ &{}]a,b[ \\ u>0 &{}\mathrm{in} \ \ &{}]a,b[\\ u(a)=u(b)=0 &{} \ \ &{} \end{array}\right. . \end{aligned}$$

Set $t_0=\frac{a+b}{2}$ and multiply the equation $-(|u'|^{p-2}u')'=u^{q-1}$ by $u'$. Then, integrating on $[t,t_0]$ with $a<t<t_0$, we obtain

$$\begin{aligned} \frac{p-1}{p}(u'(t))^p=\frac{1}{q}(C^q-u(t)^q) \end{aligned}$$

(35)

where $C=\max _{[a,b]}u=u\left( \frac{a+b}{2}\right)$. Hence,

$$\begin{aligned} \int _0^{\frac{u(t)}{C}}\frac{dz}{(1-z^q)^{\frac{1}{p}}}=\frac{\sigma }{q} C^{\frac{q}{p}-1}(t-a), \ \ \ \mathrm{for \ all} \ \ t\in [a,t_0], \end{aligned}$$

(36)

where $\sigma =\left( \frac{p}{p-1}\right) ^{\frac{1}{p}}q^{1-\frac{1}{p}}$. If we put

$$\begin{aligned} G(s)=\int _0^s\frac{dz}{(1-z^q)^{\frac{1}{p}}}, \ \ \ \mathrm{for \ all} \ \ s\in [0,1], \end{aligned}$$

by (36) it follows

$$\begin{aligned} u(t)=CG^{-1}\left( \frac{\sigma }{q} C^{\frac{q}{p}-1}(t-a)\right) , \ \ \ \mathrm{for \ all} \ \ t\in [a,t_0], \end{aligned}$$

(37)

and, by symmetry, $u(t)=u(b-t)$, for all $t\in ]t_0,b]$.

Knowing the solution $u$ allows, thanks to Lemma 2, to determine the exact value of $c_q$ in the one-dimensional case.

Indeed, note that, for $t=t_0$ in (36), one has

$$\begin{aligned} \frac{1}{q}\beta \left( \frac{1}{q},1-\frac{1}{p}\right) =\int _0^{1}\frac{dz}{(1-z^q)^{\frac{1}{p}}}=G(1)=\frac{\sigma }{q} C^{\frac{q}{p}-1}\left( \frac{b-a}{2}\right) . \end{aligned}$$

(38)

Therefore, by (34), (36), (37) and (38), we obtain

$$\begin{aligned} \int _a^b|u(t)|^qdt&= 2\int _a^{t_0}|u(t)|^qdt=2C^q\int _a^{t_0}\left( G^{-1}\left( \frac{\sigma }{q} C^{\frac{q}{p}-1}(t-a)\right) \right) ^qdt\\&= \frac{2q}{\sigma }C^{q-\frac{q}{p}+1}\int _0^1t^q(1-t^q)^{-\frac{1}{p}}dt= \frac{2}{\sigma }C^{q-\frac{q}{p}+1}\beta \left( 1+\frac{1}{q},1-\frac{1}{p}\right) \\&= \frac{2}{\sigma }C^{q-\frac{q}{p}+1}\frac{\frac{1}{q}}{\frac{1}{1\!-\!\frac{1}{p}\!+\!\frac{1}{q}}} \beta \left( \frac{1}{q},1\!-\!\frac{1}{p}\right) \!=\!\frac{2}{\sigma }C^{q-\frac{q}{p}+1}\frac{p}{pq\!-\!q\!+\!p} \beta \left( \frac{1}{q},1\!-\!\frac{1}{p}\right) \\&= \frac{p}{pq-q+p}\left( 2\frac{\beta \left( \frac{1}{q},1-\frac{1}{p}\right) }{\sigma }\right) ^{\frac{qp}{q-p}} (b-a)^{1-\frac{qp}{q-p}}. \end{aligned}$$

Thus, by Lemma 2 and recalling the definition of $\sigma$,

$$\begin{aligned} c_q=\frac{p^{\frac{1}{p}}q^{1-\frac{1}{p}}(qp-q+p)^{\frac{1}{p}- \frac{1}{q}}}{2(p-1)^{\frac{1}{p}}\beta \left( \frac{1}{q},1-\frac{1}{p}\right) }(b-a)^{\frac{1}{q}- \frac{1}{p}+1}. \end{aligned}$$

Note also that, by (35) and (38) and the symmetry of $u$, we have

$$\begin{aligned} u\left( \frac{a+b}{2}\right) =u'(a)=-u'(b)=\left( \frac{p-1}{p}\right) ^{\frac{1}{q-p}}q^{\frac{1-q}{q-p}} \left( \frac{2}{b-a}\beta \left( \frac{1}{q},1-\frac{1}{p}\right) \right) ^{\frac{q}{q-p}}.\nonumber \\ \end{aligned}$$

(39)