Correction to: Some results on the p(u)-Laplacian problem

1 Correction to: Mathematische Annalen https://doi.org/10.1007/s00208-019-01803-w

In the Original Publication of the article, few errors have been identified in section 5 and acknowledgements section. The corrected section 5 and acknowledgements are given below:

2 Nonlocal problems

In this section we consider a real function p such that

$$\begin{aligned} p \text { is continuous, } \quad 1<\alpha < p \le \beta , \end{aligned}$$

(5.1)

for some constants $\alpha , \beta$. We denote by b a mapping from $W_0^{1,\alpha }(\Omega )$ into $\mathbb {R}$ such that

$$\begin{aligned} b \text { is continuous, } \quad b \text { is bounded, } \end{aligned}$$

(5.2)

i.e. b sends bounded sets of $W_0^{1,\alpha }(\Omega )$ into bounded sets of $\mathbb {R}$.

Definition 2

A function u is a weak solution to the problem (1.3) if

$$\begin{aligned} \left\{ \begin{array}{l} u\in W_0^{1,p(b(u))}(\Omega ), \\ \displaystyle \int _\Omega |\varvec{\nabla }u|^{p(b(u))-2}\varvec{\nabla }u\cdot \varvec{\nabla }v\,dx =\langle f,v\rangle \quad \forall \ v\in W_0^{1,p(b(u))}(\Omega ), \end{array}\right. \end{aligned}$$

(5.3)

where $\langle \cdot ,\cdot \rangle$ denotes the duality pairing between $(W_0^{1,p(b(u))}(\Omega ))'$ and $W_0^{1,p(b(u))}(\Omega )$.

One should notice that p(b(u)) is here a real number and not a function so that the Sobolev spaces involved are the classical ones. We refer to [5, 7–9] for more on nonlocal problems.

Then one has:

Theorem 5.1

Let $\Omega \subset \mathbb {R}^d$, $d\ge 2$, be a bounded domain and assume that (5.1) and (5.2) hold together with

$$\begin{aligned} f \in W^{-1,\alpha '}(\Omega ). \end{aligned}$$

Then there exists at least one weak solution to the problem (1.3) in the sense of Definition 2.

The proof of Theorem 5.1 is based on the following result.

Lemma 5.1

For $n\in \mathbb {N}$, let $u_n$ be the solution to the problem

$$\begin{aligned} \left\{ \begin{array}{l} u_n\in W_0^{1,p_n}(\Omega ), \\ \displaystyle \int _\Omega |\varvec{\nabla }u_n|^{p_n-2}\varvec{\nabla }u_n\cdot \varvec{\nabla }v\,dx =\langle f,v\rangle \quad \forall \quad v\in W_0^{1,p_n}(\Omega ), \end{array}\right. \end{aligned}$$

(5.4)

where $\langle \cdot ,\cdot \rangle$ denotes here the duality pairing between $(W_0^{1,p_n}(\Omega ))'$ and $W_0^{1,p_n}(\Omega )$. Suppose that

$$\begin{aligned}&p_n\rightarrow p,\quad \text{ as }\quad n\rightarrow \infty ,\quad \text{ where }\quad p\in (1,\infty ), \end{aligned}$$

(5.5)

$$\begin{aligned}&f\in W^{-1,q'}(\Omega )\quad \text{ for } \text{ some }\quad q<p. \end{aligned}$$

(5.6)

Then

$$\begin{aligned} u_n\rightarrow u\quad \text{ in } \quad W^{1,q}_0(\Omega ),\quad \text{ as }\quad n\rightarrow \infty , \end{aligned}$$

(5.7)

where u is the solution to the problem

$$\begin{aligned} \left\{ \begin{array}{l} u\in W^{1,p}_0(\Omega ), \\ \displaystyle \int _\Omega |\varvec{\nabla }u|^{p-2}\varvec{\nabla }u\cdot \varvec{\nabla }v\,dx = \langle f,v\rangle \quad \forall \ v\in W^{1,p}_0(\Omega ). \end{array}\right. \end{aligned}$$

(5.8)

Proof of Lemma 5.1

We shall split this proof into two steps.

1. Weak convergence: We first observe that, in view of $p_n\rightarrow p$, as $n\rightarrow \infty$, and $q<p$, we may assume that

$$\begin{aligned} p+1> p_n>q\quad \forall \quad n\in \mathbb {N}. \end{aligned}$$

(5.9)

Taking $v=u_n$ in the equation of (5.4) we get

$$\begin{aligned} \int _\Omega |\varvec{\nabla }u_n|^{p_n}dx\le \Vert f\Vert _{-1,q'}\Vert \varvec{\nabla }u_n\Vert _{q}. \end{aligned}$$

(5.10)

Recall that $\Vert f\Vert _{-1,q'}$ denotes the strong dual norm of f associated to the norm $\Vert \varvec{\nabla }\cdot \Vert _{q}$. On the other hand, by using Hölder’s inequality and (5.9), we have

$$\begin{aligned} \Vert \varvec{\nabla }u_n\Vert _{q}\le C\Vert \varvec{\nabla }u_n\Vert _{p_n}, \end{aligned}$$

(5.11)

for some positive constant $C=C(p,q,\Omega )$. Plugging (5.11) into (5.10) it comes

$$\begin{aligned} \Vert \varvec{\nabla }u_n\Vert _{p_n}\le C, \end{aligned}$$

(5.12)

for some other positive constant $C=C(p,q,\Omega , f)$. Combining (5.11) with (5.12), it follows that

$$\begin{aligned} \Vert \varvec{\nabla }u_n\Vert _{q}\le C, \end{aligned}$$

(5.13)

for some positive constant C independent of n. From (5.13) we deduce then that for some subsequence still labelled by n and for some $u \in W^{1,q}_0(\Omega )$

$$\begin{aligned} \varvec{\nabla }u_n \rightharpoonup \varvec{\nabla }u\quad \text{ in }\quad L^q(\Omega ),\quad \text{ as }\quad n\rightarrow \infty . \end{aligned}$$

(5.14)

Due to (5.5), (5.9), (5.12) and (5.14), we can also apply Lemma 3.1 so that

$$\begin{aligned} \liminf _{n\rightarrow \infty }\int _\Omega |\varvec{\nabla }u_n|^{p_n}dx\ge \int _\Omega |\varvec{\nabla }u|^{p}dx. \end{aligned}$$

As a consequence we have

$$\begin{aligned} u\in W^{1,p}_0(\Omega ). \end{aligned}$$

(5.15)

Clearly the equation in (5.4) is equivalent to

$$\begin{aligned} \int _\Omega |\varvec{\nabla }u_n|^{p_n-2}\varvec{\nabla }u_n\cdot \varvec{\nabla }(v-u_n)\,dx \ge \langle f,v-u_n\rangle \quad \forall \ v\in W_0^{1,p_n}(\Omega ). \end{aligned}$$

and by the Minty lemma to

$$\begin{aligned} \int _\Omega |\varvec{\nabla }v|^{p_n-2}\varvec{\nabla }v\cdot \varvec{\nabla }(v-u_n)\,dx \ge \langle f,v-u_n\rangle \quad \forall \ v\in W_0^{1,p_n}(\Omega ). \end{aligned}$$

(5.16)

Taking $v\in C_0^\infty (\Omega )$, one can use (5.5) and (5.14) to pass to the limit in (5.16), as $n\rightarrow \infty$, so that

$$\begin{aligned} \int _\Omega |\varvec{\nabla }v|^{p-2}\varvec{\nabla }v\cdot \varvec{\nabla }(v-u)\,dx \ge \langle f,v-u\rangle \quad \forall \ v\in C_0^\infty (\Omega ). \end{aligned}$$

(5.17)

Using the density of $C_0^\infty (\Omega )$ in $W_0^{1,p}(\Omega )$, we see that (5.17) also holds for all $v\in W_0^{1,p}(\Omega )$. In this case, taking $v=u\pm \delta z$, with $z\in W_0^{1,p}(\Omega )$ and $\delta >0$, and letting $\delta \rightarrow 0$ after simplifying the resulting inequality, one obtains

$$\begin{aligned} \int _\Omega |\varvec{\nabla }u|^{p-2}\varvec{\nabla }u\cdot \varvec{\nabla }z\,dx =\langle f,z\rangle \quad \forall \ z\in W_0^{1,p}(\Omega ). \end{aligned}$$

Thus u is the solution to the problem (5.8).

2. Strong convergence: We want to show that the convergence (5.14) is in fact strong. To prove this, we first note that, taking $v=u_n$ in the equation of (5.4) and using (5.14) to pass to the limit, we obtain

$$\begin{aligned} \int _\Omega |\varvec{\nabla }u_n|^{p_n}dx = \langle f,u_n\rangle \rightarrow \langle f,u\rangle = \int _\Omega |\varvec{\nabla }u|^{p}dx,\quad \text{ as }\quad n\rightarrow \infty . \end{aligned}$$

(5.18)

Consider the case of the $p_n$’s such that

$$\begin{aligned} p_n\ge p\quad \forall \quad n\in \mathbb {N}. \end{aligned}$$

One has by Hölder’s inequality

$$\begin{aligned} \int _\Omega |\varvec{\nabla }u_n|^{p}dx \le \left( \int _\Omega |\varvec{\nabla }u_n|^{p_n}dx\right) ^\frac{p}{p_n} |\Omega |^{1-\frac{p}{p_n}}, \end{aligned}$$

where $|\Omega |$ denotes the d-Lebesgue measure of $\Omega$. Thus by (5.18) for such a sequence

$$\begin{aligned} \limsup _{n\rightarrow \infty }\int _\Omega |\varvec{\nabla }u_n|^{p}dx\le \int _\Omega |\varvec{\nabla }u|^{p}dx\le \liminf _{n\rightarrow \infty }\int _\Omega |\varvec{\nabla }u_n|^{p}dx, \end{aligned}$$

which shows (since $\Vert \varvec{\nabla }u_n\Vert _{p} \rightarrow \Vert \varvec{\nabla }u\Vert _{p}$, as $n\rightarrow \infty$)

$$\begin{aligned} u_n \rightarrow u\quad \text{ strongly } \text{ in }\ W^{1,p}_0(\Omega ),\quad \text{ as }\quad n\rightarrow \infty . \end{aligned}$$

(5.19)

Since $W^{1,p}_0(\Omega )\subset W^{1,q}_0(\Omega )$, (5.19) implies (5.7).

Next, consider the $p_n$’s such that

$$\begin{aligned} q<p_n<p\quad \forall \ n\in \mathbb {N} \end{aligned}$$

(5.20)

and set

$$\begin{aligned} A_n:=\int _\Omega \left( |\varvec{\nabla }u_n|^{p_n-2}\varvec{\nabla }u_n -|\varvec{\nabla }u|^{p_n-2}\varvec{\nabla }u\right) \cdot \left( \varvec{\nabla }u_n - \varvec{\nabla }u\right) \,dx. \end{aligned}$$

(5.21)

Due to the monotonicity, $A_n\ge 0$ and, because of (5.4), one has

$$\begin{aligned} A_n=\langle f, u_n-u \rangle -\int _\Omega |\varvec{\nabla }u|^{p_n-2}\varvec{\nabla }u\cdot \varvec{\nabla }(u_n -u)\,dx. \end{aligned}$$

From (5.6) and (5.14), we have

$$\begin{aligned} \langle f, u_n-u \rangle \rightarrow 0,\quad \text{ as }\ n\rightarrow \infty . \end{aligned}$$

(5.22)

Moreover, from (5.15) one easily gets

$$\begin{aligned} \left| |\varvec{\nabla }u|^{p_n-2}\varvec{\nabla }u\right| \le \max \{1,|\varvec{\nabla }u|\}^{p-1}\in L^{p'}(\Omega ). \end{aligned}$$

(5.23)

Hence, (5.20), (5.22) and (5.23) ensure that

$$\begin{aligned} A_n\rightarrow 0,\quad \text{ as }\quad n\rightarrow \infty . \end{aligned}$$

(5.24)

Assume first that

$$\begin{aligned} p_n\ge 2. \end{aligned}$$

This allows us to use property (3.10) of Lemma 3.2 in (5.21) so that

$$\begin{aligned} A_n\ge \frac{1}{2^{p_n-1}}\int _\Omega |\varvec{\nabla }(u_n - u)|^{p_n}\,dx. \end{aligned}$$

(5.25)

Since, by (5.20), $p_n>q$, we have by Hölder’s inequality, (5.20), (5.24) and (5.25)

$$\begin{aligned} \int _\Omega |\varvec{\nabla }(u_n - u)|^{q}\,dx\le \Big (\int _\Omega |\varvec{\nabla }(u_n -u)|^{p_n}\,dx\Big )^\frac{q}{p_n}| \Omega |^{1-\frac{q}{p_n}}\rightarrow 0, \end{aligned}$$

when $n \rightarrow \infty$. This proves (5.7) in this case.

Consider now the case when

$$\begin{aligned} p_n<2. \end{aligned}$$

Here, we use Hölder’s inequality as follows

$$\begin{aligned}&\int _\Omega |\varvec{\nabla }(u_n - u)|^{p_n}\,dx \nonumber \\&\quad =\int _\Omega |\varvec{\nabla }(u_n - u)|^{p_n} \left( |\varvec{\nabla }u_n|+|\varvec{\nabla }u|\right) ^{\frac{(p_n-2)p_n}{2}} \left( |\varvec{\nabla }u_n|+|\varvec{\nabla }u|\right) ^{\frac{(2-p_n)p_n}{2}}\,dx\nonumber \\&\quad \le \left[ \int _\Omega |\varvec{\nabla }(u_n - u)|^{2}\left( |\varvec{\nabla }u_n| {+}|\varvec{\nabla }u|\right) ^{p_n-2}\,dx\right] ^{\frac{p_n}{2}} \left[ \int _\Omega \left( |\varvec{\nabla }u_n|{+}|\varvec{\nabla }u|\right) ^{p_n} \,dx\right] ^{1-\frac{p_n}{2}}.\nonumber \\ \end{aligned}$$

(5.26)

Using property (3.11) of Lemma 3.2 we have

$$\begin{aligned} A_n\ge C \int _\Omega |\varvec{\nabla }(u_n -u)|^{2}\left( |\varvec{\nabla }u_n|+|\varvec{\nabla }u|\right) ^{p_n-2}\,dx, \end{aligned}$$

(5.27)

for some positive constant $C=C(p_n)$. Now, by using (5.26), (5.27) together with (5.12) we deduce that

$$\begin{aligned} \int _\Omega |\varvec{\nabla }(u_n - u)|^{p_n}\,dx \rightarrow 0,\quad \text{ as }\ n\rightarrow \infty . \end{aligned}$$

Thus, as above, (5.7) holds true also in this case. $\square$

Let us now show how Lemma 5.1 can be applied to prove the existence of weak solutions to the nonlocal problem (1.3).

Proof of Theorem 5.1

Note that $f \in (W_0^{1,\alpha }(\Omega ))' \subset (W_0^{1,\delta }(\Omega ))'$ for any $\delta > \alpha$. Thus for each $\lambda \in \mathbb {R}$, there exists a unique solution $u=u_\lambda$ to the $p(\lambda )$-Laplacian problem

$$\begin{aligned} \left\{ \begin{array}{l} u\in W^{1,p(\lambda )}_0(\Omega ), \\ \displaystyle \int _\Omega |\varvec{\nabla }u|^{p(\lambda )-2}\varvec{\nabla }u\cdot \varvec{\nabla }v\,dx =\langle f,v\rangle \quad \forall \ v\in W^{1,p(\lambda )}_0(\Omega ). \end{array}\right. \end{aligned}$$

(5.28)

Taking $v=u=u_\lambda$ in (5.28) one derives

$$\begin{aligned} \int _\Omega |\varvec{\nabla }u_\lambda |^{p(\lambda )}\,dx\le \Vert f\Vert _{-1,\alpha '} \Vert \varvec{\nabla }u_\lambda \Vert _{\alpha }. \end{aligned}$$

(5.29)

By Hölder’s inequality one has

$$\begin{aligned} \Vert \varvec{\nabla }u_\lambda \Vert _{\alpha }\le \Vert \varvec{\nabla }u_\lambda \Vert _{p(\lambda )} |\Omega |^{\frac{1}{\alpha } -\frac{1}{p(\lambda )}}. \end{aligned}$$

(5.30)

Thus by (5.29) it comes

$$\begin{aligned} \Vert \varvec{\nabla }u_\lambda \Vert _{p(\lambda )}^{p(\lambda )-1}\le \Vert f\Vert _{-1,\alpha '}|\Omega |^{\frac{1}{\alpha }- \frac{1}{p(\lambda )}}. \end{aligned}$$

(5.31)

Gathering (5.30) and (5.31), and using (5.1) we obtain

$$\begin{aligned} \Vert \varvec{\nabla }u_\lambda \Vert _{\alpha }\le \Vert f\Vert _{-1,\alpha '}^\frac{1}{p(\lambda )-1}|\Omega |^{\left( \frac{1}{\alpha }- \frac{1}{p(\lambda )}\right) \frac{p(\lambda )}{p(\lambda )-1}} \le \max _{p\in [\alpha ,\beta ]} \Vert f\Vert _{-1,\alpha '}^\frac{1}{p-1}|\Omega |^{\left( \frac{1}{\alpha }- \frac{1}{p}\right) \frac{p}{p-1}} =C, \end{aligned}$$

(5.32)

for some positive constant $C=C(\alpha ,\beta ,\Omega ,f)$. Due to the boundedness of b, see (5.2), and to (5.32), there exists $L\in \mathbb {R}$ such that

$$\begin{aligned} b(u_\lambda )\in [-L,L]\quad \forall \ \lambda \in \mathbb {R}. \end{aligned}$$

Let us now consider the map

$$\begin{aligned} \lambda \mapsto b(u_\lambda ), \end{aligned}$$

(5.33)

from $[-L,L]$ into itself. This map is continuous. Indeed, if $\lambda _n\rightarrow \lambda$ as $n\rightarrow \infty$, due to (5.1), we have $p(\lambda _n)\rightarrow p(\lambda )$. Applying now Lemma 5.1 with $p_n=p(\lambda _n)$, it follows that

$$\begin{aligned} u_{\lambda _n}\rightarrow u_\lambda \quad \text{ in } \quad W^{1,\alpha }_0(\Omega ),\quad \text{ as }\quad n\rightarrow \infty . \end{aligned}$$

Now, b being continuous (see (5.2)), it follows that $b(u_{\lambda _n})\longrightarrow b(u_\lambda )$, as $n\rightarrow \infty$, and thus the map (5.33) is also continuous. It has then a fixed point $\lambda _0$ and $u_{\lambda _0}$ is then solution to (5.3). $\square$

The original article has been corrected.