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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.

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Acknowledgements

The authors have been partially supported by the “Action Concertée Incitative: Opti-Ondes (2015–2016)” of the University of Poitiers. The authors also acknowledge the group “Phydromat”, and Germain Rousseaux in particular, for stimulating discussions. The second author is thankful to Michel Pierre for helpful discussions.

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Correspondence to Morgan Pierre.

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 \(r\in (0,r_0)\),

  1. (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)
  2. (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 (xz). 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 \(r\in (0,r_0)\),

  1. (i)

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

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

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

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Dambrine, J., Pierre, M. Regularity of Optimal Ship Forms Based on Michell’s Wave Resistance. Appl Math Optim 82, 23–62 (2020). https://doi.org/10.1007/s00245-018-9490-0

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Keywords

  • Shape optimization
  • Existence
  • Regularity
  • Dirichlet energy
  • Wave resistance

Mathematics Subject Classification

  • 49N60
  • 76B75
  • 49J40
  • 49Q10