Some local properties of subsolution and supersolutions for a doubly nonlinear nonlocal p-Laplace equation

Abstract

We establish a local boundedness estimate for weak subsolutions to a doubly nonlinear parabolic fractional p-Laplace equation. Our argument relies on energy estimates and a parabolic nonlocal version of De Giorgi’s method. Furthermore, by means of a new algebraic inequality, we show that positive weak supersolutions satisfy a reverse Hölder inequality. Finally, we also prove a logarithmic decay estimate for positive supersolutions.

Appendix

In this section, we prove Lemma 2.9. To this end, we establish the following auxiliary lemmas. Throughout this section, we assume $p>1$.

Lemma 6.1

Let $f,g\in C^1([a,b])$. Then

$$\begin{aligned} \frac{f(b)-f(a)}{b-a}+\Big |\frac{g(b)-g(a)}{b-a}\Big |^p\le \max _{t\in [a,b]}\big [f'(t)+|g'(t)|^p\big ]. \end{aligned}$$

Proof

Suppose the result does not hold, then by contradiction, we get

$$\begin{aligned} \frac{f(b)-f(a)}{b-a}+\Big |\frac{g(b)-g(a)}{b-a}\Big |^p>f'(t)+|g'(t)|^p, \end{aligned}$$

for all $t\in [a,b]$. Integrating over a to b, we obtain

$$\begin{aligned} \Big |\frac{g(b)-g(a)}{b-a}\Big |^p>\frac{1}{b-a}\int _{a}^{b}|g'(t)|^p\,dt, \end{aligned}$$

which contradicts Jensen’s inequality. $\square$

Lemma 6.2

Let $a,b>0$, $0<\epsilon <p-1$. Then we have

$$\begin{aligned} |b-a|^{p-2}(b-a)(a^{-\epsilon }-b^{-\epsilon })\ge \zeta (\epsilon )\Big |b^\frac{p-\epsilon -1}{p}-a^\frac{p-\epsilon -1}{p}\Big |^p, \end{aligned}$$

where $\zeta (\epsilon )=\frac{p^p\epsilon }{(p-\epsilon -1)^p}$. Moreover, if $0<p-\epsilon -1<1$, then we may choose $\zeta (\epsilon )=\frac{p^p\epsilon }{p-\epsilon -1}$.

Proof

Let $0<\epsilon <p-1$ and $\zeta (\epsilon )=\frac{p^p\epsilon }{(p-\epsilon -1)^p}$. Let $f(t)=\frac{t^{-\epsilon }}{\zeta (\epsilon )}$ and $g(t)=t^\frac{p-\epsilon -1}{p}$. By Lemma 6.1, we have

$$\begin{aligned} \frac{1}{\zeta (\epsilon )}\frac{b^{-\epsilon }-a^{-\epsilon }}{b-a}+\Big |\frac{b^\frac{p-\epsilon -1}{p}-a^\frac{p-\epsilon -1}{p}}{b-a}\Big |^p\le 0. \end{aligned}$$

If $b\ge a$, multiplying by $(b-a)^p$, we obtain

$$\begin{aligned} (b-a)^{p-1}(a^{-\epsilon }-b^{-\epsilon })\ge \zeta (\epsilon )\big |b^\frac{p-\epsilon -1}{p}-a^\frac{p-\epsilon -1}{p}\big |^p. \end{aligned}$$

(6.1)

If $b<a$, interchanging a and b, the Lemma follows. If $0<p-\epsilon -1<1$, then we have $0<(p-\epsilon -1)^p<p-\epsilon -1$, therefore $\zeta (\epsilon )\ge \frac{p^p\epsilon }{p-\epsilon -1}$ and (6.1) implies

$$\begin{aligned} (b-a)^{p-1}(a^{-\epsilon }-b^{-\epsilon })\ge \frac{p^p\epsilon }{p-\epsilon -1}\Big |b^\frac{p-\epsilon -1}{p}-a^\frac{p-\epsilon -1}{p}\Big |^p. \end{aligned}$$

Hence the claim follows with $\zeta (\epsilon )=\frac{p^p\epsilon }{p-\epsilon -1}$ when $0<p-\epsilon -1<1$. $\square$

1.1 Proof of Lemma 2.9

We denote the left-hand and right-hand sides of (2.3) by L.H.S and R.H.S, respectively. Let $\zeta _1(\epsilon )=\frac{\zeta (\epsilon )}{c(p)}$ and $\zeta _2(\epsilon )=\zeta (\epsilon )+1+\frac{1}{\epsilon ^{p-1}}$. Then $\zeta _1(\epsilon )-\zeta _2(\epsilon )<-1$ since $C(p)>1$ (to be finally chosen appropriately).

Case 1. If $\tau _1=\tau _2=0$, then (2.3) holds trivially.

Case 2. If $\tau _1>0$ and $\tau _2=0$. In this case, we note that if $b>a$, then

$$\begin{aligned} \text {L.H.S}=|b-a|^{p-2}(b-a)(\tau _1^{p}a^{-\epsilon }-\tau _2^{p}b^{-\epsilon })=(b-a)^{p-1}\tau _1^{p}a^{-\epsilon } \end{aligned}$$

and

$$\begin{aligned} \text {R.H.S}&=\zeta _1(\epsilon )\tau _1^{p}a^{p-\epsilon -1}-\zeta _2(\epsilon )\tau _1^{p}(b^{p-\epsilon -1}+a^{p-\epsilon -1})\\&=(\zeta _1(\epsilon )-\zeta _2(\epsilon ))\tau _1^{p}a^{p-\epsilon -1}-\zeta _2(\epsilon )\tau _1^{p}b^{p-\epsilon -1}. \end{aligned}$$

Now L.H.S is positive and since $\zeta _1(\epsilon )-\zeta _2(\epsilon )<0$ and $\zeta _2(\epsilon )>0$, the R.H.S is negative. Therefore, we have L.H.S $\ge$ R.H.S. On the other hand if $b\le a$, then

$$\begin{aligned} \text {L.H.S}&=-(a-b)^{p-1}\tau _1^{p}a^{-\epsilon }\ge -\tau _1^{p}a^{p-\epsilon -1}, \end{aligned}$$

and since $\zeta _1(\epsilon )-\zeta _2(\epsilon )<-1$ and $\zeta _2(\epsilon )>0$, we have

$$\begin{aligned} \text {R.H.S}=(\zeta _1(\epsilon )-\zeta _2(\epsilon ))\tau _1^{p}a^{p-\epsilon -1}-\zeta _2(\epsilon )\tau _1^{p}b^{p-\epsilon -1} <-\tau _1^{p}a^{p-\epsilon -1} \le \text {L.H.S}. \end{aligned}$$

Case 3. If $\tau _1=0$ and $\tau _2>0$. Then we have

$$\begin{aligned} \text {L.H.S}=-|b-a|^{p-2}(b-a)\tau _2^{p}b^{-\epsilon }, \end{aligned}$$

and

$$\begin{aligned} \text {R.H.S}&=(\zeta _1(\epsilon )-\zeta _2(\epsilon ))\tau _2^{p}b^{p-\epsilon -1}-\zeta _2(\epsilon )\tau _2^{p}a^{p-\epsilon -1}. \end{aligned}$$

If $b>a$, then

$$\begin{aligned} \text {L.H.S}=-(b-a)^{p-1}\tau _2^{p}b^{-\epsilon }\ge -\tau _2^{p}b^{p-\epsilon -1}, \end{aligned}$$

and since $\zeta _1(\epsilon )-\zeta _2(\epsilon )<-1$ and $\zeta _2(\epsilon )>0$, we have

$$\begin{aligned} \text {R.H.S}=(\zeta _1(\epsilon )-\zeta _2(\epsilon ))\tau _2^{p}b^{p-\epsilon -1}-\zeta _2(\epsilon )\tau _2^{p}a^{p-\epsilon -1}\\ <-\tau _2^{p}b^{p-\epsilon -1} \le \text {L.H.S}. \end{aligned}$$

If $b\le a$, then the L.H.S is nonnegative and the R.H.S is negative. Therefore, we have $\text {L.H.S}\ge \text {R.H.S}$.

Case 4. Let both $\tau _1,\tau _2>0$. By symmetry, we may assume that $b\ge a$. Let $t=\frac{b}{a}\ge 1$, $s=\frac{\tau _2}{\tau _1}>0$ and $\lambda =s^{p}t^{-\epsilon }$. It can be easily seen that the inequality (2.3) is equivalent to the following inequality

$$\begin{aligned} \zeta _1(\epsilon )|s t^\frac{p-\epsilon -1}{p}-1|^p\le (t-1)^{p-1}(1-\lambda )+\zeta _2(\epsilon )|s-1|^p(t^{p-\epsilon -1}+1). \end{aligned}$$

(6.2)

We first estimate the following term.

$$\begin{aligned} |st^\frac{p-\epsilon -1}{p}-1|^p&=|st^\frac{p-\epsilon -1}{p}-t^\frac{p-\epsilon -1}{p}+t^\frac{p-\epsilon -1}{p}-1|^p\\&=|(s-1)t^\frac{p-\epsilon -1}{p}+(t^\frac{p-\epsilon -1}{p}-1)|^p\\&\le 2^{p-1}|s-1|^p t^{p-\epsilon -1}+2^{p-1}|t^\frac{p-\epsilon -1}{p}-1|^p\\&=A+B, \end{aligned}$$

where

$$\begin{aligned} A=2^{p-1}|s-1|^p t^{p-\epsilon -1}\text { and }B=2^{p-1}|t^\frac{p-\epsilon -1}{p}-1|^p. \end{aligned}$$

By Lemma 6.2, we have

$$\begin{aligned} B\le \frac{2^{p-1}(t-1)^{p-1}(1-t^{-\epsilon })}{\zeta (\epsilon )}. \end{aligned}$$

As a consequence, we obtain

$$\begin{aligned} |st^\frac{p-\epsilon -1}{p}-1|^p\le 2^{p-1}|s-1|^p t^{p-\epsilon -1}+\frac{2^{p-1}(t-1)^{p-1}(1-t^{-\epsilon })}{\zeta (\epsilon )}. \end{aligned}$$

We observe that

$$\begin{aligned} 1-t^{-\epsilon }&=1-\lambda +\lambda -t^{-\epsilon } =1-\lambda +(s^p-1)t^{-\epsilon }\\&=1-\lambda +|s-1|^p t^{-\epsilon }+(s^p-1-|s-1|^p)t^{-\epsilon }. \end{aligned}$$

Therefore, we get

$$\begin{aligned} \begin{aligned}&|s t^\frac{p-\epsilon -1}{p}-1|^p \le 2^{p-1}\big (1+\frac{1}{\zeta (\epsilon )}\big )|s-1|^p t^{p-\epsilon -1}\\&\qquad +\frac{2^{p-1}}{\zeta (\epsilon )}(t-1)^{p-1}(1-\lambda )+\frac{2^{p-1}}{\zeta (\epsilon )}(t-1)^{p-1}(s^p-1-|s-1|^p)t^{-\epsilon }. \end{aligned} \end{aligned}$$

(6.3)

Next we estimate the term $T=\frac{2^{p-1}}{\zeta (\epsilon )}(t-1)^{p-1}(s^p-1-|s-1|^p)t^{-\epsilon }$ for different values of t and s.

Case (a). If $t>1$ and $s\ge 2$. Then using the fact that $s\ge 2$, it can be easily seen that there exists constant C(p) large enough such that $s^p-1-(s-1)^p\le C(p)(s-1)^p$. Therefore, we get

$$\begin{aligned} T\le \frac{C(p)}{\zeta (\epsilon )}|s-1|^p t^{p-\epsilon -1}. \end{aligned}$$

(6.4)

By inserting (6.4) into (6.3), we get

$$\begin{aligned} \begin{aligned} |s t^\frac{p-\epsilon -1}{p}-1|^p\le C(p)\big (1+\frac{1}{\zeta (\epsilon )}\big )|s-1|^p t^{p-\epsilon -1}+\frac{C(p)}{\zeta (\epsilon )}(t-1)^{p-1}(1-\lambda ). \end{aligned} \end{aligned}$$

(6.5)

Case (b). If $t=1$ or $0<s\le 1$. Then $T\le 0$. Hence, we get the estimate in (6.5).

Case (c). If $t>1$, $s\in (1,2)$. Let $r\ge p$ be the nearest integer to p. Again it follows that there exists a positive constant C(p) large enough such that $s^p-1-|s-1|^p\le C(p)|s-1|$. We have further subcases.

Case (i). If

$$\begin{aligned}t-1<\frac{r 2^{r-1}}{\epsilon }t(s-1).\end{aligned}$$

Note that we can choose C(p) large enough such that $r 2^{r-1}\le C(p)$. Hence, we have

$$\begin{aligned} T\le \frac{C(p)}{\epsilon ^{p-1}\zeta (\epsilon )}t^{p-\epsilon -1}|s-1|^p. \end{aligned}$$

(6.6)

By inserting (6.6) into (6.3), we get

$$\begin{aligned} |s t^\frac{p-\epsilon -1}{p}-1|^p \le C(p)\Big (1+\frac{1}{\zeta (\epsilon )}\Big (1+\frac{1}{\epsilon ^{p-1}}\Big )\Big )|s-1|^p t^{p-\epsilon -1}+\frac{C(p)}{\zeta (\epsilon )}(t-1)^{p-1}(1-\lambda ). \end{aligned}$$

(6.7)

Case (ii). If

$$\begin{aligned}t-1\ge \frac{r 2^{r-1}}{\epsilon }t(s-1).\end{aligned}$$

Since r is an integer, we observe that

$$\begin{aligned} s^r+s-2=(s-1)(s^{r-1}+s^{r-2}+\cdots +s+2). \end{aligned}$$

By the mean value theorem there exists $\eta \in (1,t)$ such that $t^\epsilon -1=\epsilon \eta ^{\epsilon -1}(t-1)$ and so $\epsilon =\frac{t^\epsilon -1}{\eta ^{\epsilon -1}(t-1)}$. Now, we have

$$\begin{aligned} \frac{s^r+s-2}{t-1}&=\frac{s-1}{t-1}(s^{r-1}+s^{r-2}\cdots +s+2)\\&\le \frac{\epsilon }{r 2^{r-1}t}(s^{r-1}+s^{r-2}+\cdots +s+2)\\&\le \frac{\epsilon }{t} \le \frac{t^\epsilon -1}{t\eta ^{\epsilon -1}(t-1)}, \end{aligned}$$

which gives $t\eta ^{\epsilon -1}(s^r+s-2)\le t^\epsilon -1$.

Now, the fact $\epsilon >0$ and $1<\eta <t$ gives $t\eta ^{\epsilon -1}>\eta ^\epsilon >1$. Therefore since $r\ge p$ and $s>1$, we get $s^p+s-2<s^r+s-2<t\eta ^{\epsilon -1}(s^r+s-2)\le t^\epsilon -1$. Hence, we have $s-1\le t^\epsilon -s^p=t^\epsilon (1-\lambda )$. Thus

$$\begin{aligned} T\le \frac{C(p)}{\zeta (\epsilon )}(t-1)^{p-1}(1-\lambda ). \end{aligned}$$

(6.8)

Using (6.8) into (6.3) we get

$$\begin{aligned} |s t^\frac{p-\epsilon -1}{p}-1|^p\le C(p)\big (1+\frac{1}{\zeta (\epsilon )}\big )|s-1|^p t^{p-\epsilon -1}+\frac{C(p)}{\zeta (\epsilon )}(t-1)^{p-1}(1-\lambda ). \end{aligned}$$

(6.9)

Finally from the estimates (6.5), (6.7) and (6.9), we obtain

$$\begin{aligned} \begin{aligned} |st^\frac{p-\epsilon -1}{p}-1|^p\le C(p)\Big (1+\frac{1}{\zeta (\epsilon )}(1+\frac{1}{\epsilon ^{p-1}})\Big )|s-1|^p (t^{p-\epsilon -1}+1)\\+\frac{C(p)}{\zeta (\epsilon )}(t-1)^{p-1}(1-\lambda ). \end{aligned} \end{aligned}$$

(6.10)

Multiplying $\frac{\zeta (\epsilon )}{C(p)}$ on both sides of (6.10), we obtain

$$\begin{aligned} \frac{\zeta (\epsilon )}{C(p)}|st^\frac{p-\epsilon -1}{p}-1|^p\le \Big (\zeta (\epsilon )+1+\frac{1}{\epsilon ^{p-1}}\Big )|s-1|^p(t^{p-\epsilon -1}+1)+(t-1)^{p-1}(1-\lambda ), \end{aligned}$$

which corresponds to the inequality (6.2). The lemma thus follows.