Regularity of Optimal Ship Forms Based on Michell’s Wave Resistance

Abstract

We introduce an optimal shaping problem based on Michell’s wave resistance formula in order to find the form of a ship which has an immerged hull with minimal total resistance. The problem is to find a function $u\in H^1_0(D)$, even in the z-variable, and which minimizes the functional

$$\begin{aligned} J(u)=\int _D|\nabla u(x,z)|^2dxdz+\int _D\int _Dk(x,z,x',z')u(x,z)u(x',z')dxdzdx'dz' \end{aligned}$$

with an area constraint on the set $\{(x,z)\in D\ :\ u(x,z)\not =0\}$ and with the volume constraint $\int _D u(x,z)dxdz=V$; D is a bounded open subset of $\mathbb {R}^2$, symmetric about the x-axis, and k is Michell’s kernel. We prove that u is locally $\alpha$-Hölder continuous on D for all $0<\alpha <2/5$, and locally Lipschitz continuous on $D^\star =\{(x,z)\in D\ : z\not =0\}$. The main assumption is the nonnegativity of u. We also prove that the area constraint is “saturated”. The results are first derived for a general kernel $k\in L^q(D\times D)$ with $q\in (1,+\infty ]$. A numerical simulation illustrates the theoretical result.

Appendices

Michell’s Wave Resistance Kernel

From (7.1) and (7.3), by (formally) inverting the integrals, we see that Michell’s normalized wave resistance can be written

$$\begin{aligned} J_{wave}(u)=\int _{D\times D}k_\nu (x,z,x',z')u(x,z)u(x',z')dxdzdx'dz' \end{aligned}$$

(A.1)

where

$$\begin{aligned} k_\nu (x,z,x',z')=\frac{4\nu ^4}{\pi C_F(\nu )}K(\nu (x-x'),\nu (|z|+|z'|)), \end{aligned}$$

(A.2)

and

$$\begin{aligned} K(X,Z)=\int _1^\infty e^{-\lambda ^2Z}\cos (\lambda X)\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda . \end{aligned}$$

(A.3)

This formal computation will be made rigorous below (see Corollary A.2). This expression of Michell’s resistance in terms of a kernel $k_\nu$ is well-known [28], but to the best of our knowledge, the results in Appendix A are new.

First notice that K is defined and continuous on $\mathbb {R}\times (0,+\infty )$ and

$$\begin{aligned} |K(X,Z)|\le I(Z)<+\infty \end{aligned}$$

(A.4)

for all $(X,Z)\in \mathbb {R}\times (0,+\infty )$, with

$$\begin{aligned} I(Z)=\int _1^\infty e^{-Z\lambda ^2}\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda . \end{aligned}$$

(A.5)

In particular, $k_\nu$ is continuous on $(\mathbb {R}\times \mathbb {R}^\star )^2$.

The following result is essential for the Hölder regularity of the optimal ship.

Theorem A.1

Michell’s normalized wave resistance kernel $k_\nu$ (A.2) belongs to $L^q(D\times D)$ for all $1\le q<5/4$. Moreover, if D contains an open disc centered on the x-axis, then $k_\nu$ does not belong to $L^{5/4}(D\times D)$.

Proof

It is sufficient to prove the assertion for a domain D of the form $D_l=(-l,l)\times (-l,l)$ where $l>0$ is arbitrary. Moreover, by the change of variable $(x,z,x',z')\rightarrow (\nu x,\nu z,\nu x',\nu z')$ in (A.1)–(A.2), it will suffice to consider the case $\nu =1$. We write

$$\begin{aligned} K(X,Z)=I_1(X,Z)+I_2(X,Z)+I_3(X,Z)+I_4(X,Z), \end{aligned}$$

(A.6)

with

$$\begin{aligned} I_1(X,Z)= & {} \int _1^2e^{-\lambda ^2 Z}\cos (\lambda X)\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda ,\\ I_2(X,Z)= & {} \int _2^\infty e^{-\lambda ^2Z}\cos (\lambda X)\left( \frac{\lambda ^4}{\sqrt{\lambda ^2-1}}-\lambda ^3\right) d\lambda ,\\ I_3(X,Z)= & {} -\int _0^2e^{-\lambda ^2Z}\cos (\lambda X)\lambda ^3d\lambda ,\\ I_4(X,Z)= & {} \int _0^\infty e^{-\lambda ^2Z}\cos (\lambda X)\lambda ^3d\lambda . \end{aligned}$$

By Lebesgue’s dominated convergence theorem, $I_1$ and $I_3$ are continuous on $\mathbb {R}^2$. We will prove that $(x,z,x',z')\mapsto I_2(x-x',|z|+|z'|)$ belongs to $L^{5/4}(D_l\times D_l)$ and that $(x,z,x',z')\mapsto I_4(x-x',|z|+|z'|)$ belongs to $L^q(D_l\times D_l)$ for all $1\le q<5/4$, but does not belong to $L^{5/4}(D_l\times D_l)$. The theorem will then be proved.

By the mean value theorem, there exists $C_1>0$ such that

$$\begin{aligned} 0\le \frac{1}{\sqrt{1-u}}-1\le C_1u,\quad \forall u\in [0,1/4]. \end{aligned}$$

Thus, for all $\lambda \ge 2$,

$$\begin{aligned} 0\le \frac{1}{\sqrt{1-1/\lambda ^2}}-1\le \frac{C_1}{\lambda ^2}, \end{aligned}$$

and so

$$\begin{aligned} 0\le \frac{\lambda ^4}{\sqrt{\lambda ^2-1}}-\lambda ^3\le C_1\lambda . \end{aligned}$$

We obtain

$$\begin{aligned} |I_2(X,Z)|\le C_1\int _0^\infty e^{-\lambda ^2Z}\lambda d\lambda . \end{aligned}$$

Performing the change of variable $\mu =\sqrt{Z}\lambda$, we find

$$\begin{aligned} |I_2(X,Z)|\le \frac{C_1}{Z}\int _0^\infty e^{-\mu ^2}\mu d\mu \le \frac{C_1'}{Z}. \end{aligned}$$

(A.7)

Next, we notice that for $q>1$, the integral $\int _0^l\int _0^l(z+z')^{-q}dzdz'$ is finite if and only if $q<2$. Indeed,

$$\begin{aligned} \int _0^l\int _0^l(z+z')^{-q}dzdz= & {} \frac{1}{q-1}\int _0^l[z^{1-q}-(z+l)^{1-q}]dz \end{aligned}$$

(A.8)

$$\begin{aligned}\le & {} \frac{l^{2-q}}{(q-1)(2-q)}<\infty \end{aligned}$$

(A.9)

if $q<2$, whereas the integral on the right-hand side of (A.8) is $+\infty$ if $q\ge 2$. In particular, for $q=5/4$, the function $(x,z,x',z')\mapsto 1/(|z|+|z'|)$ belongs to $L^{5/4}(D_l\times D_l)$ since

$$\begin{aligned} \int _{D_l\times D_l}\frac{1}{(|z|+|z'|)^q}dxdzdx'dz'=16l^2\int _0^l\int _0^l\frac{1}{(z+z')^q}dzdz'<\infty . \end{aligned}$$

(A.10)

By (A.7), the function $(x,z,x',z')\mapsto I_2(x-x',|z|+|z'|)$ belongs to $L^{5/4}(D_l\times D_l)$ as well.

Concerning the term $I_4$, we first perform the change of variable $\mu =\sqrt{Z}\lambda$, so that

$$\begin{aligned} I_4(X,Z)=\frac{1}{Z^2}J\left( \frac{X}{\sqrt{Z}}\right) , \end{aligned}$$

with $J(t)=\int _0^\infty e^{-\mu ^2}\cos (t\mu )\mu ^3d\mu$. By Lebesgue’s dominated convergence theorem, J is continuous on $\mathbb {R}$; in particular, J is bounded on $[-1,1]$ by a constant $C_2$. Integration by parts yields

$$\begin{aligned} J(t)=-\frac{1}{t}\int _0^\infty \sin (t\mu )\left( 3\mu ^2-2\mu ^4\right) e^{-\mu ^2}d\mu , \end{aligned}$$

so that

$$\begin{aligned} |J(t)|\le \frac{1}{|t|}\int _0^\infty \left( 3\mu ^2+2\mu ^4\right) e^{-\mu ^2}d\mu =\frac{C_3}{|t|}, \end{aligned}$$

for all $t\not =0$. Let now $q>1$. On performing the linear change of variable $X=x-x'$, $X'=x+x'$, we find that

$$\begin{aligned}&\int _{D_l\times D_l}|I_4(x-x',|z|+|z'|)|^qdxdzdx'dz'\\&\quad \le 2l\int _{-l}^l\int _{-l}^l\int _{-2l}^{2l}|I_4(X,|z|+|z'|)|^qdXdzdz'\\&\quad =8l\int _{0}^l\int _{0}^l\int _{-2l}^{2l}|I_4(X,z+z')|^qdXdzdz'\\&\quad =8l\int _{0}^l\int _{0}^l\int _{-2l}^{2l}\frac{1}{(z+z')^{2q}}\left| J\left( \frac{X}{\sqrt{z+z'}}\right) \right| ^qdXdzdz'. \end{aligned}$$

Integration with respect to X yields

$$\begin{aligned} \int _{-2l}^{2l}\left| J\left( \frac{X}{\sqrt{z+z'}}\right) \right| dX\le & {} \int _{|X|\le \sqrt{z+z'}}\left| J\left( \frac{X}{\sqrt{z+z'}}\right) \right| dX\\&+\int _{ \sqrt{z+z'}\le |X|\le 2l}\left| J\left( \frac{X}{\sqrt{z+z'}}\right) \right| dX\\\le & {} \int _{|X|\le \sqrt{z+z'}}C_2^qdX+2\int _{\sqrt{z+z'}\le X\le 2l}C_3^q\frac{\sqrt{z+z'}^q}{X^q}dX\\\le & {} 2C_2^q\sqrt{z+z'}+\frac{2C_3^q}{q-1}\sqrt{z+z'}\\\le & {} C_4\sqrt{z+z'}. \end{aligned}$$

Thus,

$$\begin{aligned} \int _{D_l\times D_l}|I_4(x-x',|z|+|z'|)|^qdxdzdx'dz'\le 8lC_4\int _{0}^l\int _0^l(z+z')^{1/2-2q}dzdz'. \end{aligned}$$

The right-hand side is finite if and only if $q<5/4$ (see (A.8)–(A.9)). This shows that $I_4$ belongs to $L^q(D_l\times D_l)$ for all $1\le q<5/4$, as claimed.

To see the optimality of this statement, first note that $J(0)>0$ and let $t_0>0$ such that $J(t)\ge J(0)/2$ for all $t\in [-t_0,t_0]$. The linear change of variable $X=x-x'$, $X'=x+x'$ maps the square $(-l,l)\times (-l,l)$ onto a square with vertices (2l, 0), (0, 2l), $(-2l,0)$ and $(0,-2l)$, which contains the square $(-l,l)\times (-l,l)$. Let $q=5/4$. We have

$$\begin{aligned}&\int _{D_l\times D_l}|I_4(x-x',|z|+|z'|)|^qdxdzdx'dz'\\&\quad \ge 4l\int _0^l\int _0^l\frac{1}{(z+z')^{2q}}\left| J\left( \frac{X}{\sqrt{z+z'}}\right) \right| ^qdXdzdz'. \end{aligned}$$

By choosing $t_0>0$ small enough so that $t_0\sqrt{2l}\le l$, we also have

$$\begin{aligned} \int _{|X|\le l}\left| J\left( \frac{X}{\sqrt{z+z'}}\right) \right| ^qdX\ge & {} \int _{|X|\le t_0\sqrt{z+z'}}\left| J\left( \frac{X}{\sqrt{z+z'}}\right) \right| ^qdX\\\ge & {} \left( \frac{J(0)}{2}\right) ^q(2t_0\sqrt{z+z'}). \end{aligned}$$

We obtain

$$\begin{aligned} \int _{D_l\times D_l}|I_4(x-x',|z|+|z'|)|^qdxdzdx'dz'\ge 8lt_0\left( \frac{J(0)}{2}\right) ^q\int _0^l\int _0^l(z+z')^{1/2-2q}dzdz'. \end{aligned}$$

The integral on the right-hand side is $+\infty$ for $q=5/4$ (see (A.8)). This concludes the proof. $\square$

As a consequence, we have:

Corollary A.2

For every $q'>5$ and for all $u\in L^{q'}(D)$, the formulations for $J_{wave}(u)$ given by (7.1)–(7.2) and (A.1)–(A.3) are equal (and finite).

Proof

Without loss of generality, we may assume that $\nu =1$ and $D=D_l=(-l,l)\times (-l,l)$ with $l>0$. Let $q'>5$, $u\in L^{q'}(D_l)$ and $q\in (1,5/4)$ such that $1/q+1/q'=1$. We use the form $S_u(\lambda )=i\lambda T_u(\lambda )$ with $T_u(\lambda )=\int _Du(x,z)e^{-i\lambda x}e^{-\lambda ^2 |z|}dxdz$ (cf. (7.3)). For any integer $N\ge 2$, we have

$$\begin{aligned} \int _1^N|T_u(\lambda )|^2\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda =\int _{D_l\times D_l}K_N(x-x',|z|+|z'|)u(x,z)u(x',z')dxdzdx'dz', \end{aligned}$$

(A.11)

where

$$\begin{aligned} K_N(X,Z)=\int _1^Ne^{-\lambda ^2Z}\cos (\lambda (x-x'))\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda . \end{aligned}$$

This is obtained by applying Fubini’s theorem for the two variables $\lambda \in (1,N)$ and $(x,z,x',z')\in D_l\times D_l$. By the monotone convergence theorem, the left-hand side of (A.11) tends to

$$\begin{aligned} \int _1^\infty |T_u(\lambda )|^2\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda \end{aligned}$$

when N tends to $+\infty$. The convergence of the right-hand side is more delicate. We will apply Lebesgue’s dominated convergence theorem. We first note that $K_N(x-x',|z|+|z'|)$ tends everywhere in $(D_l\times D_l)\cap (\mathbb {R}\times \mathbb {R}^\star )^2$ to $K(x-x',|z|+|z'|)$ (see (A.3)). By arguing as in the proof of Theorem A.1, we can show that

$$\begin{aligned} |K_N(x-x',|z|+|z'|)|\le k^\star (x,z,x',z') \text{ in } D_l\times D_l, \end{aligned}$$

(A.12)

where $k^\star \in L^q(D_l\times D_l)$ is independent of N (details are left to the reader). This implies that the right-hand side of (A.11) tends to

$$\begin{aligned} \int _{D_l\times D_l}K(x-x',|z|+|z'|)u(x,z)u(x',z')dxdzdx'dz' \end{aligned}$$

as N tends to $+\infty$, by dominated convergence. $\square$

Proposition A.3

Assume that $D\subset \{(x,z)\in \mathbb {R}^2\ :\ |z|>\delta \}$ for some $\delta >0$. Then $k_\nu$ (cf. (A.2)) belongs to $L^\infty (D\times D)$ and

$$\begin{aligned} \Vert k_\nu \Vert _{L^\infty (D\times D)}\le \frac{4\nu ^4}{\pi C_F(\nu )}e^{-\nu \delta }I(\nu \delta ), \end{aligned}$$

(A.13)

where $I:(0,+\infty )\rightarrow (0,+\infty )$ is the continuous and decreasing function defined by (A.5). In particular, if $C_F(\nu )^{-1}$ has at most a polynomial growth as $\nu$ tends to $+\infty$, then $\Vert k_\nu \Vert _{L^\infty (D\times D)}\rightarrow 0$ as $\nu \rightarrow +\infty$.

Proof

Let $(x,z,x',z')\in D\times D$, and define $X=\nu (x-x')$, $Z=\nu (|z|+|z'|)$. Then $Z\ge 2\nu \delta$, so that for $\lambda \ge 1$, we have $e^{-\lambda ^2Z}\le e^{-\nu \delta }e^{-\lambda ^2(Z-\nu \delta )}$, and integration with respect to $\lambda$ yields

$$\begin{aligned} I(Z)\le e^{-\nu \delta }I(Z-\nu \delta )\le e^{-\nu \delta }I(\nu \delta ). \end{aligned}$$

By (A.2)–(A.3), we have

$$\begin{aligned} |k_\nu (x,z,x',z')|=\frac{4\nu ^4}{\pi C_F(\nu )}|K(X,Z)|\le \frac{4\nu ^4}{\pi C_F(\nu )}I(Z). \end{aligned}$$

Putting together these two estimates yields (A.13). $\square$

The following property of $k_\nu$ will also prove useful.

Proposition A.4

For every $v\in L^1(D)$, the function $f_v$ defined by

$$\begin{aligned} f_v(x,z)=\int _Dk_\nu (x,z,x',z')v(x',z')dx'dz' \end{aligned}$$

is real analytic in $\mathbb {R}\times \mathbb {R}^\star$.

Proof

Without loss of generality, we may assume $\nu =1$ and $z>0$. Using $\cos (\lambda (x-x'))=\mathfrak {R}(e^{i\lambda (x-x')})$, we may write $f_v=(4/\pi C_F(1))\mathfrak {R}(f_v^1+f_v^2)$ with

$$\begin{aligned} f_v^1(x,z)= & {} \int _D\left( \int _1^2e^{-\lambda ^2z}e^{i\lambda x}e^{-\lambda ^2|z'|}e^{-i\lambda x'}\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda \right) v(x',z')dx'dz'\\ f_v^2(x,z)= & {} \int _D\left( \int _2^{+\infty }e^{-\lambda ^2z}e^{i\lambda x}e^{-\lambda ^2|z'|}e^{-i\lambda x'}\frac{\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda \right) v(x',z')dx'dz'. \end{aligned}$$

It is sufficient to prove that $f_v^1$ and $f_v^2$ are real analytic on $\mathbb {R}\times (0,+\infty )$ (with values in $\mathbb {C}$ considered as a vector space over $\mathbb {R}$). For this purpose, it is sufficient to show (see, e.g., [6]) that for $i=1$ or 2, $f_v^i$ is $C^\infty$ on $\mathbb {R}\times (0,+\infty )$ and that for every compact subset K of $\mathbb {R}^2$, there are positive constants $C_K$ and $M_K$ such that for all $l=(l_1,l_2)\in \mathbb {N}^2$ and for all $(x,z)\in K$,

$$\begin{aligned} \left| \frac{\partial ^{l_1+l_2}f_v^i}{\partial x^{l_1}\partial z^{l_2}}(x,z)\right| \le C_KM_K^{|l|}|l|! \end{aligned}$$

(A.14)

where $|l|=l_1+l_2$, as usual.

The function $f_v^1$ is clearly of class $C^\infty$ on $\mathbb {R}\times (0,+\infty )$, and, for any $l=(l_1,l_2)\in \mathbb {N}^2$,

$$\begin{aligned} \frac{\partial ^{l_1+l_2}f_v^1}{\partial x^{l_1}\partial z^{l_2}}(x,z)=\int _D\left( \int _1^2(-\lambda ^2)^{l_2}(i\lambda )^{l_1}e^{-\lambda ^2z}e^{i\lambda x}e^{-\lambda ^2|z'|}e^{-i\lambda x'}\frac{v(x',z')\lambda ^4}{\sqrt{\lambda ^2-1}}d\lambda \right) dx'dz. \end{aligned}$$

(A.15)

Thus, for $(x,z)\in \mathbb {R}\times (0,+\infty )$,

$$\begin{aligned} \left| \frac{\partial ^{l_1+l_2}f_v^1}{\partial x^{l_1}\partial z^{l_2}}(x,z)\right| \le 2^{l_1+2l_2+4}\Vert v\Vert _{L^1(D)}\int _1^2\frac{d\lambda }{\sqrt{\lambda ^2-1}}. \end{aligned}$$

Estimate (A.14) is satisfied with $C_K=2^4\Vert v\Vert _{L^1(D)}\int _1^2(\lambda ^2-1)^{-1/2}d\lambda$ and $M_K=4$.

Let now $\delta \in (0,1)$. Then $f_v^2$ is clearly of class $C^\infty$ on $\mathbb {R}\times (\delta ,+\infty )$; its partial derivatives are obtained on replacing the integral over (1, 2) in (A.15) by an integral over $(2,+\infty )$. This yields

$$\begin{aligned} \left| \frac{\partial ^{l_1+l_2}f_v^2}{\partial x^{l_1}\partial z^{l_2}}(x,z)\right| \le \Vert v\Vert _{L^1(D)}\int _2^{+\infty }\lambda ^{l_1+2l_2+4}e^{-\lambda ^2 z}\frac{d\lambda }{\sqrt{\lambda ^2-1}}. \end{aligned}$$

Next, we use that $(\lambda ^2-1)^{-1/2}\le C/\lambda$ on $(2,+\infty )$, for some constant C, and we perform the change of variable $\mu =\lambda ^2 z$ in the integral. We obtain

$$\begin{aligned} \left| \frac{\partial ^{l_1+l_2}f_v^2}{\partial x^{l_1}\partial z^{l_2}}(x,z)\right| \le \frac{C\Vert v\Vert _{L^1(D)}}{2z^{l_1/2+l_2+2}}\Gamma (l_1/2+l_2+2), \end{aligned}$$

where $\Gamma$ is the Gamma function. Next, we use that

$$\begin{aligned} \Gamma (l_1/2+l_2+2)\le \Gamma (l_1+l_2+2)=(l_1+l_2+2)(l_1+l_2)!. \end{aligned}$$

We find that estimate (A.14) is valid on $\mathbb {R}\times (\delta ,+\infty )$ with, e.g.,

$$\begin{aligned} C_K=C'\Vert v\Vert _{L^1(D)}/(2\delta ^2),\quad C'=C\sup _{k\in \mathbb {N}} (k+2)/2^k,\quad \text{ and } \quad M_K=2/\delta . \end{aligned}$$

$\square$

Technical Lemmas

The proof of the following lemma may be found in [9].

Lemma B.1

Let $B((x_0,z_0),r_0)$ be an open ball and $U\in C^2(B((x_0,z_0),r_0))$. Then, for all $r\in (0,r_0)$,

This remains valid for all $U\in H^1(B((x_0,z_0),r_0))$ such that $\Delta U$ is a measure satisfying

$$\begin{aligned} \int _0^rds\, s^{-1}\int _{B((x_0,z_0),s)}d|\Delta u|<\infty , \end{aligned}$$

(B.1)

and such that

(B.2)

Remark B.2

The proof shows furthermore that the condition (B.1) implies the existence of the limit in (B.2) for any $(x_0,z_0)$ whence we can take some precise representation of U defined thanks to (B.2).

The following lemma is more or less classical (see, e.g. [9, 17]).

Lemma B.3

Let $B((x_0,z_0),r_0)$ be an open ball, $r_0\le 1$, $F\in L^p(B((x_0,z_0),r_0))$, $p\in (1,2)$, $\alpha =2/p'$. Then, there exists a constant C which depends only on p and $\Vert F\Vert _{L^p(B((x_0,z_0),r_0))}$ such that, for $r\in (0,r_0)$,

• (i) if $\Delta U=F$ on $B((x_0,z_0),r_0)$, then

  $$\begin{aligned} |U|_{\alpha ,B((x_0,z_0),r/2)}\le C\left[ 1+r^{-\alpha }\Vert U\Vert _{L^\infty (B((x_0,z_0),r))}\right] , \end{aligned}$$

  (B.3)

• (ii) if $\Delta U\ge F$ and $U\ge 0$ on $B((x_0,z_0),r_0)$, then

  

  (B.4)

Proof

Recall that for the solution of

$$\begin{aligned} W\in H^1_0(B_1),\quad -\Delta W=G \text{ on } B_1, \end{aligned}$$

since $p>1$, by elliptic regularity we have

$$\begin{aligned} \Vert W\Vert _{W^{2,p}(B_1)}\le C(p)\Vert G\Vert _{L^p(B_1)}. \end{aligned}$$

We use the Sobolev imbedding $W^{2,p}(B_1)\subset C^\alpha (\overline{B_1})$ [7] and we apply this to the rescaled functions

$$\begin{aligned} \forall \xi \in B_1,\quad V(\xi ,\zeta )=U((x_0,z_0)+r(\xi ,\zeta )),\ G(\xi ,\zeta )=r^2F((x_0,z_0)+r(\xi ,\zeta )). \end{aligned}$$

We obtain

$$\begin{aligned} \Vert W\Vert _{L^\infty (B_1)}+|W|_{\alpha ,B_1}\le C'(p) r^{2-2/p}\Vert F\Vert _{L^p(B((x_0,z_0),r_0))}. \end{aligned}$$

(B.5)

For (B.3), we notice that $\Delta (V-W)=0$ on $B_1$ so that by Harnack’s inequality [17],

$$\begin{aligned} |V-W|_{\alpha ,B_{1/2}}\le \Vert \nabla (V-W)\Vert _{L^\infty (B_{1/2})}\le C\Vert V\Vert _{L^\infty (\partial B_1)}. \end{aligned}$$

Together with (B.5), this inequality gives

$$\begin{aligned} |V|_{\alpha ,B_{1/2}}\le C(p,\Vert F\Vert _{L^p(B((x_0,z_0),r_0))})\left[ r^\alpha +\Vert V\Vert _{L^\infty (\partial B_1)}\right] . \end{aligned}$$

Going back to U gives (B.3) by change of variable. For (B.4), we first notice that $-\Delta (V-W)\le 0$, so that $(V-W)(x,z)\le \int _{\partial B_1}P_{(x,z)}(x',z')V(x',z')d\sigma (x',z')$ where $P_{x,z}(\cdot )$ denotes the Poisson kernel at (x, z). Using (B.5) again and $V\ge 0$, we deduce that

The relation (B.4) follows by change of variable. $\square$

The following lemma is proved in [9].

Lemma B.4

Let $B((x_0,z_0),r_0)$ be an open ball, $r_0\le 1$, $F\in L^q(B((x_0,z_0),r_0))$, $q>2$. Then, there exists a constant $C=C(q,\Vert F\Vert _{L^q(B((x_0,z_0),r_0))})$ such that, for $r\in (0,r_0)$,

• (i) if $\Delta U=F$ on $B((x_0,z_0),r_0)$, then

  $$\begin{aligned} |U|_{1,B((x_0,z_0),r/2)}\le C\left[ 1+r^{-1}\Vert U\Vert _{L^\infty (B((x_0,z_0),r))}\right] , \end{aligned}$$

• (ii) if $\Delta U\ge F$ and $U\ge 0$ on $B((x_0,z_0),r_0)$, then