Asymptotics of the solution to the perfect conductivity problem with p-Laplacian

Abstract

We study the perfect conductivity problem with closely spaced perfect conductors embedded in a homogeneous matrix where the current-electric field relation is the power law $J=\sigma |E|^{p-2}E$. The gradient of solutions may be arbitrarily large as $\varepsilon$, the distance between inclusions, approaches to 0. To characterize this singular behavior of the gradient in the narrow region between two inclusions, we capture the leading order term of the gradient. This is the first gradient asymptotics result on the nonlinear perfect conductivity problem.

Appendix A

In the first part of the appendix, we prove Proposition 4.2. The proof essentially follows those of [26, Proposition 2.1] and [15, Lemma 5.1]. Our estimate is sharper due to a better estimate on $|Du_0|$ (Proposition 2.2).

Proof of Proposition 4.2

Similar to the proof of Theorem 2.5, for small $r \in (0,1/2)$, we take a smooth surface $\eta$ so that $\mathcal {D}_1^\varepsilon$ is surrounded by $\Gamma _{-,s}^0 \cup \eta$. See Fig. 1. We denote the surface

$$\begin{aligned} \Sigma _r^\varepsilon := \left\{ x \in \mathbb {R}^n: |x'|=r, -\frac{\varepsilon }{2} + h_2(x')< x_n < h_2(x') \right\} . \end{aligned}$$

Since $\int _{\partial \mathcal {D}_1^{0}} |D u_{0}|^{p-2} D u_{0} \cdot \nu = \mathcal {F}$ and $\int _{\partial \mathcal {D}_1^{\varepsilon }} |D u_{\varepsilon }|^{p-2} D u_{\varepsilon } \cdot \nu =0$, by integration by parts, we have

$$\begin{aligned} -\int _{\Gamma _{-,r}^0} |D u_{0}|^{p-2} D u_{0} \cdot \nu + \int _{\eta } |D u_{0}|^{p-2} D u_{0} \cdot \nu = \mathcal {F}, \end{aligned}$$

(A.1)

and

$$\begin{aligned} -\int _{\Gamma _{-,r}^\varepsilon } |D u_{\varepsilon }|^{p-2} D u_{\varepsilon } \cdot \nu + \int _{\Sigma _r^\varepsilon } |D u_{\varepsilon }|^{p-2} D u_{\varepsilon } \cdot \nu + \int _{\eta } |D u_{\varepsilon }|^{p-2} D u_{\varepsilon } \cdot \nu =0.\nonumber \\ \end{aligned}$$

(A.2)

Note that the minus signs appear because $\nu$ on $\Gamma _{-,r}^0$ and $\Gamma _{-,r}^\varepsilon$ are defined to be pointing upwards, while $\nu$ on $\eta$ and $\Sigma _r^\varepsilon$ are pointing away from $\mathcal {D}_1^\varepsilon$. By (2.5), we have $|D u_0(x)| \le C_1 e^{-\frac{C_2}{r}}$ in $\overline{\Omega _{r}^0}$, and hence

$$\begin{aligned} \left| \int _{\Gamma _{-,r}^0} |D u_{0}|^{p-2} D u_{0} \cdot \nu \right| \le C_1 e^{-\frac{C_2}{r}} \end{aligned}$$

(A.3)

for some positive $\varepsilon$-independent constants $C_1$ and $C_2$. By Theorem 2.5, we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0_+} \int _{\eta } |D u_{\varepsilon }|^{p-2} D u_{\varepsilon } \cdot \nu = \int _{\eta } |D u_{0}|^{p-2} D u_{0} \cdot \nu . \end{aligned}$$

(A.4)

By (2.7), we have $|D u_\varepsilon (x)| \le C (\varepsilon + |x'|^2)^{-1}$ in ${\Omega _{1/2}^0}$. Therefore,

$$\begin{aligned} \left| \int _{\Sigma _r^\varepsilon } |D u_{\varepsilon }|^{p-2} D u_{\varepsilon } \cdot \nu \right|&\le C \int _{|x'|=r}\int _{-\varepsilon /2+h_2(x')}^{h_2(x')} \Big ( \frac{1}{\varepsilon +|x'|^2} \Big )^{p-1} \, dx_n dS \nonumber \\&\le C\frac{\varepsilon r^{n-2}}{(\varepsilon +r^2)^{p-1}}. \end{aligned}$$

(A.5)

Finally, (4.7) follows directly from (A.1)–(A.5). Proposition 4.2 is proved. $\square$

In the following, we verify (4.12).

Lemma A.1

(4.12) holds when $p \ge (n+1)/2$.

Proof

We only give the proof for the case when $n \ge 3$. The case $n = 2$ follows similarly and is simpler. After a rotation of coordinates if necessary, we may assume that

$$\begin{aligned} D_{x'}^2 (h_1 - h_2)(0') = \text{ diag }~(\lambda _1, \ldots , \lambda _{n-1}). \end{aligned}$$

First, we replace $\delta (y)$ in the denominator with the quadratic polynomial $\varepsilon + \sum _{i=1}^{n-1} \lambda _i y_i^2/2$. By (1.8), (1.9), and the fact that $h_1,h_2$ are $C^2$, we estimate

$$\begin{aligned}&\left| \int _{|y'|< r} \Big ( \frac{\Theta (\varepsilon )}{\delta (y)}\Big )^{p-1} - \Big ( \frac{\Theta (\varepsilon )}{\varepsilon + \sum _{i=1}^{n-1} \frac{\lambda _i}{2} y_i^2}\Big )^{p-1} \, dy' \right| \\&\quad = \Theta (\varepsilon )^{p-1 } \left| \int _{|y'|< r} \frac{\Big (\varepsilon + \sum _{i=1}^{n-1} \frac{\lambda _i}{2} y_i^2\Big )^{p-1} - \delta (y)^{p-1} }{\Big [ \delta (y) \Big (\varepsilon + \sum _{i=1}^{n-1} \frac{\lambda _i}{2} y_i^2\Big ) \Big ]^{p-1}} \, dy' \right| \\&\quad \le C \Theta (\varepsilon )^{p-1} \int _{|y'|< r} \frac{h(r)|y'|^2(\varepsilon + |y'|^2)^{p-2}}{(\varepsilon + |y'|)^{2p-2}} \, dy'\\&\quad \le C h(r) \int _{|y'| < r} \Big ( \frac{\Theta (\varepsilon )}{\varepsilon + \sum _{i=1}^{n-1} \frac{\lambda _i}{2} y_i^2}\Big )^{p-1} \, dy', \end{aligned}$$

where h(r) is the modulus of continuity of $D_{x'}^2 (h_1 - h_2)$, and hence $h(r) \rightarrow 0$ as $r \rightarrow 0$, and C is some positive constant independent of $\varepsilon$ and r. Therefore, it suffices to show that for any $r > 0$,

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0_+} \int _{|y'| < r} \Big ( \frac{\Theta (\varepsilon )}{\varepsilon + \sum _{i=1}^{n-1} \frac{\lambda _i}{2} y_i^2}\Big )^{p-1} \, dy' = \frac{1}{K}. \end{aligned}$$

(A.6)

In the spherical coordinates, for $y' \in \mathbb {R}^{n-1}$, we write

$$\begin{aligned} y_1= & {} \sqrt{\frac{2}{\lambda _1}}s \cos \theta _1,\quad y_2= \sqrt{\frac{2}{\lambda _2}}s\sin \theta _1\cos \theta _2,\quad y_3=\sqrt{\frac{2}{\lambda _3}}s\sin \theta _1\sin \theta _2\cos \theta _3,\ldots , \\ y_{n-2}= & {} \sqrt{\frac{2}{\lambda _{n-2}}}s \sin \theta _1\sin \theta _2\cdots \sin \theta _{n-3}\cos \theta _{n-2}, \\ y_{n-1}= & {} \sqrt{\frac{2}{\lambda _{n-1}}}s\sin \theta _1\sin \theta _2\cdots \sin \theta _{n-3}\sin \theta _{n-2}, \end{aligned}$$

where $s \in [0, \infty )$, $\theta _1,\theta _2,\ldots ,\theta _{n-3}\in [0,\pi ]$ and $\theta _{n-2}\in [0,2\pi )$. For convenience of notation, we denote $\Sigma :=[0,\pi ]^{n-3}\times [0,2\pi )$. By this change of variables,

$$\begin{aligned}&\int _{|y'| < r} \Big ( \frac{\Theta (\varepsilon )}{\varepsilon + \sum _{i=1}^{n-1} \frac{\lambda _i}{2} y_i^2}\Big )^{p-1} \, dy' \nonumber \\&\quad = \frac{2^{\frac{n-1}{2}}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \int _0^{\frac{r}{\varphi (\theta )}}\Big ( \frac{\Theta (\varepsilon )}{\varepsilon + s^2}\Big )^{p-1} s^{n-2} J(\theta )\,dsd\theta , \end{aligned}$$

(A.7)

where

$$\begin{aligned} \varphi (\theta ) = \Big ( \frac{2}{\lambda _1}\cos ^2\theta _1 + \frac{2}{\lambda _2} \sin ^2\theta _1\cos ^2\theta _2 + \cdots + \frac{2}{\lambda _{n-1}}\sin ^2\theta _1\cdots \sin ^2\theta _{n-2} \Big )^{\frac{1}{2}}, \end{aligned}$$

and

$$\begin{aligned} J(\theta )=\sin ^{n-3}\theta _1 \sin ^{n-4}\theta _2\cdots \sin \theta _{n-3}. \end{aligned}$$

Note that

$$\begin{aligned} \sqrt{\frac{2}{\max \lambda _i}} \le \varphi (\theta ) \le \sqrt{\frac{2}{\min \lambda _i}}. \end{aligned}$$

When $p > (n+1)/2$, $\Theta (\varepsilon ) = \varepsilon ^{\frac{2p-n-1}{2(p-1)}}$. By the change of variables $t = \varepsilon ^{-1}s^2$, the right-hand side of (A.7) becomes

$$\begin{aligned} \frac{2^{\frac{n-3}{2}}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \int _0^{\frac{r^2}{\varphi ^2(\theta )\varepsilon }} \frac{t^{\frac{n-3}{2}}}{(1+t)^{p-1}} \, dt J(\theta ) d\theta . \end{aligned}$$

Since $(n-3)/2 - (p-1) = (n-2p-1)/2 < -1$, the integral converges as $\varepsilon \rightarrow 0$. Therefore,

$$\begin{aligned}&\lim _{\varepsilon \rightarrow 0_+} \int _{|y'| < r} \Big ( \frac{\Theta (\varepsilon )}{\varepsilon + \sum _{i=1}^{n-1} \frac{\lambda _i}{2} y_i^2}\Big )^{p-1} \, dy'\nonumber \\&\quad = \frac{2^{\frac{n-3}{2}}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \int _0^{\infty } \frac{t^{\frac{n-3}{2}}}{(1+t)^{p-1}} \, dt J(\theta ) d\theta \nonumber \\&\quad = \frac{2^{\frac{n-3}{2}}|\mathbb {S}^{n-2}|}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} B\Big (\frac{n-1}{2}, \frac{2p-n-1}{2} \Big ), \end{aligned}$$

(A.8)

where B is the beta function. Recalling the identities

$$\begin{aligned} |\mathbb {S}^{n-2}| = \frac{2\pi ^{\frac{n-1}{2}}}{\Gamma \big (\frac{n-1}{2} \big )}, \quad B\Big (\frac{n-1}{2}, \frac{2p-n-1}{2} \Big ) = \frac{\Gamma \big (\frac{n-1}{2} \big )\Gamma \big (p -\frac{n-1}{2} \big )}{\Gamma (p-1)}, \end{aligned}$$

(A.9)

and plugging them into (A.8), we have proved (A.6) for the case when $p > (n+1)/2$.

When $p = (n+1)/2$, $\Theta (\varepsilon ) = |\ln \varepsilon |^{-\frac{1}{p-1}}$. By the change of variables $w = \varepsilon ^{-1/2}s$, the right-hand side of (A.7) becomes

$$\begin{aligned}&\frac{2^{\frac{n-1}{2}}|\ln \varepsilon |^{-1}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \int _0^{\frac{r}{\varphi (\theta )\sqrt{\varepsilon }}}\frac{w^{n-2}}{(1+w^2)^{\frac{n-1}{2}}} \, dw J(\theta )\, d\theta \\&\quad = \frac{2^{\frac{n-1}{2}}|\ln \varepsilon |^{-1}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \int _0^{\frac{r}{\varphi (\theta )\sqrt{\varepsilon }}}\frac{w}{1+w^2} \, dw J(\theta )\, d\theta \\&\qquad + \frac{2^{\frac{n-1}{2}}|\ln \varepsilon |^{-1}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \int _0^{\frac{r}{\varphi (\theta )\sqrt{\varepsilon }}} \left( \frac{w^{n-2}}{(1+w^2)^{\frac{n-1}{2}}} - \frac{w}{1+w^2} \right) \, dw J(\theta )\, d\theta \\&\quad =: \textrm{I} + \textrm{II}. \end{aligned}$$

By direct computations,

$$\begin{aligned} \textrm{I}&= \frac{2^{\frac{n-3}{2}}|\ln \varepsilon |^{-1}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \int _0^{\frac{r}{\varphi (\theta )\sqrt{\varepsilon }}} \, d \ln (1 + w^2) J(\theta )\, d\theta \\&= \frac{2^{\frac{n-3}{2}}|\ln \varepsilon |^{-1}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} \int _\Sigma \Big [\ln \big ( \varepsilon + \frac{r^2}{\varphi ^2(\theta )} \big ) - \ln \varepsilon \Big ] J(\theta )\, d\theta . \end{aligned}$$

Therefore, by (A.9),

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0_+} \textrm{I} = \frac{2^{\frac{n-3}{2}}|\mathbb {S}^{n-2}|}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}} = \frac{(2\pi )^{\frac{n-1}{2}}}{(\lambda _1 \cdots \lambda _{n-1})^{\frac{1}{2}}\Gamma \big (\frac{n-1}{2} \big )}. \end{aligned}$$

(A.10)

To estimate $\textrm{II}$, we split the integral over $(0, \frac{r}{\varphi (\theta )\sqrt{\varepsilon }})$ into (0, 1) and $(1, \frac{r}{\varphi (\theta )\sqrt{\varepsilon }})$, and denote them by $\textrm{II}_1$ and $\textrm{II}_2$, respectively. It is easily seen that $|\textrm{II}_1| \le C|\ln \varepsilon |^{-1}$. To estimate $\textrm{II}_2$, we have

$$\begin{aligned} |\textrm{II}_2| \le C|\ln \varepsilon |^{-1} \int _\Sigma \int _1^{\frac{r}{\varphi (\theta )\sqrt{\varepsilon }}} \frac{w \Big [ w^{n-3} - (1+w^2)^{\frac{n-3}{2}} \Big ]}{(1 + w^2)^{\frac{n-1}{2}}}\, dw J(\theta )\, d\theta . \end{aligned}$$

By the mean value theorem, there exists a $\xi \in (w^2, 1+w^2)$, such that

$$\begin{aligned} w^{n-3} - (1+w^2)^{\frac{n-3}{2}} = - \frac{n-3}{2} \xi ^{\frac{n-5}{2}}. \end{aligned}$$

Note that $(w^2, 1 + w^2) \subset (w^2, 2w^2)$ when $w \ge 1$. Therefore,

$$\begin{aligned} \left| \int _1^{\frac{r}{\varphi (\theta )\sqrt{\varepsilon }}} \frac{w \Big [ w^{n-3} - (1+w^2)^{\frac{n-3}{2}} \Big ]}{(1 + w^2)^{\frac{n-1}{2}}}\, dw \right| \le C\int _1^\infty \frac{w^{n-4}}{(1 + w^2)^{\frac{n-1}{2}}} \le C, \end{aligned}$$

which implies $|\textrm{II}_2| \le C|\ln \varepsilon |^{-1}$. Hence, by (A.10) and the estimate $|\textrm{II}_1| + |\textrm{II}_2| \le C|\ln \varepsilon |^{-1}$, we have proved (A.6) for the case when $p = (n+1)/2$. $\square$