A Note on Radial Solutions to the Critical Lane-Emden Equation with a Variable Coefficient

Abstract

In this note, we consider the following problem

$$\displaystyle \begin {cases} -\Delta u=(1+g(x))u^{\frac {N+2}{N-2}},\ u>0\text{ in }B,\\ u=0\text{ on }\partial B, \end {cases} $$

where N ≥ 3 and $B\subset \mathbb {R}^N$ is the unit ball centered at the origin and g(x) is a radial Hölder continuous function such that g(0) = 0. We prove the existence and nonexistence of radial solutions by the variational method with the concentration compactness analysis and the Pohozaev identity.

A Critical Case

In this appendix, we give a proof of Lemma 22 under the assumption in Corollary 17 for the readers’ convenience. Especially we will use only the condition (g2) which is weaker than (k2).

Lemma A1

Assume (g1), (g2) and (u _n) ⊂ H _r(B) is a (PS) _c sequence of

$$\displaystyle \begin{aligned} I(u)=\frac{1}{2}\|u\|{}^2-\frac{1}{2^*}\int_B(1+g)u_+^{2^*}dx. \end{aligned}$$

Then if $c<S^{\frac {N}{2}}/N$, (u _n) has a subsequence which strongly converges in H _r(B).

Proof

From the definition we have

$$\displaystyle \begin{aligned} c+o(1)&=I(u_n)-\frac{1}{2^*}\langle I'(u_n),u_n\rangle+o(1)\|u_n\|\\ &\ge \frac 1N \|u_n\|{}^2+o(1)\|u_n\|. \end{aligned}$$

This implies (u _n) is bounded in H _r(B). Then we can assume that there exists a nonnegative function u ∈ H _r(B) such that

up to a subsequence. By the concentration compactness lemma, we can suppose that there exist values μ ₀, ν ₀ ≥ 0 such that

in the measure sense, where δ ₀ denotes the Dirac delta measure concentrated at the origin with mass 1 as before. Furthermore, we have

$$\displaystyle \begin{aligned} S\nu_0^{\frac{2}{2^*}}\le\mu_0.{} \end{aligned} $$

(A.1)

We show ν ₀ = 0. To this end, we assume ν ₀ > 0 on the contrary. Then, for small ε > 0, we define a smooth test function ϕ as in the proof of Lemma 22. Since I′(u _n) → 0 in H ⁻¹(B) and (u _n) is bounded, we have

$$\displaystyle \begin{aligned} 0&=\lim_{n\to \infty}\langle I'(u_n),u_n\phi\rangle\\ &=\lim_{n\to \infty}\left(\int_B\nabla u_n \cdot \nabla (u_n\phi)dx-\int_B (1+g)(u_n)_+^{2^*}\phi dx\right)\\ &=\lim_{n\to \infty}\left(\int_B|\nabla u_n|{}^2\phi dx-\int_B (1+g)(u_n)_+^{2^*}\phi dx+\int_B u_n\nabla u_n \cdot \nabla \phi dx\right)\\ &= \int_{\overline{B}}\phi d\mu-\int_{\overline{B}}(1+g)\phi d\nu+o(1) \end{aligned}$$

where o(1) → 0 as ε → 0. Taking ε → 0 and noting g(0) = 0, we obtain

$$\displaystyle \begin{aligned} 0\ge\mu_0-\nu_0. \end{aligned}$$

Then using (A.1), we get

$$\displaystyle \begin{aligned} \nu_0\ge S^{\frac N2}.{} \end{aligned}$$

Finally, noting this estimate, we see

$$\displaystyle \begin{aligned} c&=\lim_{n\to \infty}\left(I(u_n)-\frac 12\langle I'(u_n),u_n\rangle\right)\\ &=\frac{1}{N}\lim_{n\to \infty}\int_{\overline{B}}(1+g)d\nu\\ &\ge \frac{S^{\frac{N}{2}}}{N} \end{aligned}$$

since g(0) = 0, which is a contradiction. It follows that

$$\displaystyle \begin{aligned} \lim_{n\to \infty}\int_B (1+g)(u_n)_+^{2^*}dx=\int_B (1+g)u_+^{2^*}dx. \end{aligned}$$

Then a standard argument shows that u _n → u in H _r(B). This completes the proof. □