Optimal Shape of an Underwater Moving Bottom Generating Surface Waves Ruled by a Forced Korteweg-de Vries Equation

Abstract

It is well known since Wu and Wu (in: Proceedings of the 14th symposium on naval hydrodynamics, National Academy Press, Washington, pp 103–125, 1982) that a forcing disturbance moving steadily with a transcritical velocity in shallow water can generate, continuously and periodically, a succession of solitary waves propagating ahead of the disturbance in procession. One possible new application of this phenomenon could very well be surfing competitions, where in a controlled environment, such as a pool, waves can be generated with the use of a translating bottom. In this paper, we use the forced Korteweg–de Vries equation to investigate the shape of the moving body capable of generating the highest first upstream-progressing solitary wave. To do so, we study the following optimization problem: maximizing the total energy of the system over the set of non-negative square-integrable bottoms, with uniformly bounded norms and compact supports. We establish analytically the existence of a maximizer saturating the norm constraint, derive the gradient of the functional, and then implement numerically an optimization algorithm yielding the desired optimal shape.

Appendices

Appendix A: Proofs of the Main Results

1.1 A.1 Global Well-Posedness of the fKdV Equation

Proof of Proposition 4.1

Assume $T > 0$ is fixed and $b \in L^{2}(\mathbb {R},\mathbb {R})$ is given. As a particular case of [9, Theorem 1.2], with $\sigma = -1$, $f = - \frac{\mathrm{d}b}{\mathrm{d}x}$, and initial data $u_{0} \equiv 0$, there exists a solution $u_{b} \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$ to the initial-value problem (4). Consider any other bottom $\mathfrak {b} \in L^{2}(\mathbb {R},\mathbb {R})$, with corresponding solution , then set $\delta b = \mathfrak {b} - b$ and . Clearly, one has:

$$\begin{aligned} \begin{array}{ll} \dfrac{ \partial \left( \delta u \right) }{\partial t} + \dfrac{\partial }{\partial x} \left[ \dfrac{ \left( \delta u \right) ^{2}}{2} + u_{b} \, \delta u + \dfrac{\partial ^{2} \left( \delta u \right) }{\partial x^{2}} \right] = - \dfrac{\mathrm {d}\left( \delta b \right) }{\mathrm {d}x} \in H^{-1}(\mathbb {R},\mathbb {R}), &{}\quad t \in [0,T], \\ \delta u (x,0) = 0, &{}\quad x \in \mathbb {R}. \\ \end{array} \end{aligned}$$

(19)

Although the partial derivatives in (19) have to be handled with care, since they are understood in distributional sense, we can still apply the integration-by-parts formula stated in [18, Lemma 7.3] by considering the Gelfand triple $H^{1}(\mathbb {R},\mathbb {R}) \subset L^{2}(\mathbb {R},\mathbb {R}) \subset H^{-1}(\mathbb {R},\mathbb {R})$ and the fact that we have $\delta u \in \lbrace w \in L^{2}(0,T;H^{1}(\mathbb {R},\mathbb {R})) : \partial _{t}w \in L^{2}(0,T;H^{-1}(\mathbb {R},\mathbb {R})) \rbrace$. We thus obtain for any $t \in [0,T]$:

$$\begin{aligned} \Vert \delta u \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2}= & {} \Vert \delta u ( \bullet ,0 ) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \\&+~2 \int _{0}^{t} \left\langle \partial _{t}(\delta u) ~\vert ~ \delta u \right\rangle _{H^{-1}(\mathbb {R},\mathbb {R}),H^{1}(\mathbb {R},\mathbb {R})} (\bullet ,s) \,\mathrm {d}s . \end{aligned}$$

We get $\langle \partial _{t}(\delta u) ~\vert ~ \delta u \rangle _{H^{-1}(\mathbb {R},\mathbb {R}),H^{1}(\mathbb {R},\mathbb {R})} = \langle \frac{( \delta u)^{2}}{2} + u_{b} \, \delta u + \partial _{xx}(\delta u) + \delta b ~\vert ~ \partial _{x} (\delta u) \rangle _{L^{2}(\mathbb {R},\mathbb {R}),L^{2}(\mathbb {R},\mathbb {R})}$ and $\delta u(\bullet , 0 ) = 0$ by using (19), from which it follows for any $t \in [0,T]$:

$$\begin{aligned} \Vert \delta u \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2} ~~ = ~~ 2 \int _{0}^{t} \int _{\mathbb {R}} \delta b \, \dfrac{\partial (\delta u)}{\partial x} \, \mathrm {d}x \,\mathrm {d}s - \int _{0}^{t} \int _{\mathbb {R}} \left( \delta u \right) ^2 \dfrac{\partial u_{b}}{\partial x} \, \mathrm {d}x \,\mathrm {d}s . \end{aligned}$$

We may then write, by introducing , which is a finite constant:

$$\begin{aligned} \forall t \in [0,T], ~~ \Vert \delta u \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2} \leqslant 4CT \Vert \delta b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} + C \int _{0}^{t} \Vert \delta u \left( \bullet , s \right) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} \, \mathrm {d}s. \end{aligned}$$

Since $t \in [0,T] \mapsto \Vert \delta u (\bullet ,t) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \in \mathbb {R}$ is a continuous function [18, Lemma 7.3], we can apply Grönwall’s Lemma, which yields

In particular, the uniqueness of solution to the initial-value problem (4) follows and the functional $F: b \mapsto F(b)$ given by (3) is a well-defined application from $L^{2}(\mathbb {R},\mathbb {R})$ into $\mathbb {R}$.

1.2 A.2 Explicit A Priori Estimates

Proposition A.1

Let $T > 0$ and $b \in H^{\infty }(\mathbb {R},\mathbb {R}) : = \cap _{s \geqslant 0}H^{s}(\mathbb {R},\mathbb {R})$. Then, there exist three polynomials $P_{0}$, $P_{1}$, $P_{2}$ in two variables with (non-negative) constant coefficients, such that the following estimates hold for the unique solution $u_{b} \in C^{\infty }(0,T;H^{\infty }(\mathbb {R},\mathbb {R}))$ to the initial-value problem (4):

• (i) $\displaystyle { \sup _{t \in [0,T]} \Vert u_{b} \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \leqslant P_{0} \left( T,\Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) }$,

• (ii) $\displaystyle { \sup _{t \in [0,T]} \Vert \partial _{x} u_{b} \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \leqslant P_{1} \left( T,\Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) }$,

• (iii) $\displaystyle { \sup _{t \in [0,T]} \Vert \partial _{xx} u_{b} \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \leqslant \hbox {e}^{T \left( 1 + \frac{1}{3} \Vert b \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} \right) } P_{2} \left( T,\Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) }$.

Proof

Let $T > 0$ and $b \in H^{\infty }(\mathbb {R},\mathbb {R}) : = \cap _{s \geqslant 0}H^{s}(\mathbb {R},\mathbb {R})$. Lemmas 3.1 and 3.2 imply the existence of a unique smooth solution $u_{b} \in C^{\infty }(0,T;H^{\infty }(\mathbb {R},\mathbb {R}))$ of (4). Since $b \in L^{2}(\mathbb {R},\mathbb {R})$, we may apply [9, Proposition 3.1] with final time $T+1 (> 1)$, $\sigma = -1$, $f = - \frac{\mathrm{d}b}{\mathrm{d}x}$, and homogeneous initial data $u_{0} \equiv 0$, to establish the inequality:

$$\begin{aligned} \sup _{t \in [0,T] } \Vert u_{b}(\bullet , t ) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} ~\leqslant & {} ~ \sup _{t \in [0,T+1] } \Vert u_{b}(\bullet , t ) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \\\leqslant & {} C \left( 1 + \left( 1+T \right) ^{3} \Vert \partial _{x} b \Vert _{H^{-1}(\mathbb {R},\mathbb {R})}^{3} \right) , \end{aligned}$$

for some positive constant C, which does not depend on T, b, or $u_{b}$. This proves assertion (i) with $P_{0}(x,y) := C(1+(1+x)^{3}y^{3})$, using here the fact that $\Vert \partial _{x}b \Vert _{H^{-1}(\mathbb {R},\mathbb {R})} \leqslant \Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})}$. We now exploit the Hamiltonian structure of Eq. (4) (see [22]). Although energy is not conserved, as already pointed out, an extra conserved quantity is available for the fKdV equation, which is in fact a Hamiltonian for the system. Let H be such Hamiltonian. Then,

$$\begin{aligned} \forall t \in [0,T] , \quad H(t) : = \int _{\mathbb {R}} \left[ \left( \dfrac{\partial u_{b}}{\partial x} \right) ^{2} - \dfrac{1}{3}u_b^3 - 2 b \,u_{b} \right] \, \mathrm {d}x = 0. \end{aligned}$$

(20)

The Cauchy–Schwarz inequality and $\Vert g \Vert ^{2}_{L^{\infty }(\mathbb {R},\mathbb {R})} \leqslant 2 \Vert g \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \Vert \partial _{x}g \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \leqslant \Vert g \Vert ^{2}_{H^{1}(\mathbb {R},\mathbb {R})}$ valid for any $g \in H^{1}(\mathbb {R},\mathbb {R})$ are combined with the well-known inequalities $2xy \leqslant x^{2} + y^{2}$ and $\sqrt{x+y} \leqslant \sqrt{x} + \sqrt{y}$, valid for any $x, y \geqslant 0$, so that one can deduce from (20):

$$\begin{aligned} \Vert \partial _{x} u_{b} \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \leqslant \sqrt{\dfrac{7}{5}} \left( \Vert u_{b} \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} + \Vert u_{b} \left( \bullet , t \right) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} + \Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) . \end{aligned}$$

This proves the estimate (ii) of our Proposition A.1 with $P_{1}(x,y) : = 2 [y+P_{0}(x,y)+P_{0}^{2}(x,y)]$. Finally, the same method is used to determine $P_{2}$. We first show for any $t \in [0,T]$:

$$\begin{aligned}&\dfrac{\mathrm {d}}{\mathrm {d}t} \int _{\mathbb {R}} \left[ \left( \partial _{xx} u_{b} \right) ^{2} + 2 b \, \partial _{xx}u_{b} - \dfrac{5 }{3} u_{b} \left( \partial _{x} u_{b} \right) ^{2} + \dfrac{2}{3} b \, u_{b}^{2} + \dfrac{5}{36} u_{b}^{4} \right] \,\mathrm {d}x \nonumber \\&\quad = \dfrac{2}{3} \int _{\mathbb {R}} b \, I \, \partial _{x} u_{b} \,\mathrm {d}x, \end{aligned}$$

(21)

with $I : = \partial _{xx}u_{b}+ \frac{1}{2}u_{b}^{2}+b$. Notice that both sides of (21) depend only on time t. Denote by G(t) the left-hand side of the equation. Straightforward calculations lead to:

$$\begin{aligned} G(t)= & {} \int _{\mathbb {R}} 2 \partial _{xt}u_{b} \underbrace{\left[ - \partial _{xxx}u_{b} - b_{x} \right] }_{ = \partial _{t}u_{b} + u_{b} \partial _{x} u_{b}} - \dfrac{10}{3} u_{b} \partial _{x} u_{b} \partial _{xt}u_{b} \\&- \dfrac{5}{3} \partial _{t}u_{b} \left( \partial _{x} u_{b} \right) ^{2} + \dfrac{4}{3} b u_{b} \partial _{t}u_{b} + \dfrac{5}{9} u_{b}^{3} \partial _{t} u_{b}{~} \\= & {} \int _{\mathbb {R}} \dfrac{4}{3} u_{b} \underbrace{\partial _{t}u_{b}}_{ = -I_{x}} \underbrace{\left[ \partial _{xx} u_{b} + \dfrac{u_{b}^{2}}{2} + b \right] }_{:= I} - \dfrac{1}{3} \underbrace{\partial _{t}u_{b}}_{ = -I_{x}} \left( \partial _{x} u_{b} \right) ^{2} - \dfrac{1}{9} u_{b}^{3} \underbrace{\partial _{t} u_{b}}_{ = -I_{x}} \\= & {} \int _{\mathbb {R}} \dfrac{2}{3} \partial _{x} u_{b} \, I \, b . \end{aligned}$$

Integrating equality (21) on [0, t] for any $t \in [0,T]$, following the same strategy above yields

$$\begin{aligned} \Vert \partial _{xx}u_{b} (\bullet ,t) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2}\leqslant & {} 2 \left( 1 + \frac{\Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2}}{3} \right) \int _{0}^{t} \Vert \partial _{xx} u_{b}(\bullet ,s) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} \mathrm {d}s \\&+ P_{01}\left( T,\Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) \end{aligned}$$

where we have set

$$\begin{aligned} P_{01}(x,y):= & {} 2x \left[ y^{2} \left( 1 + \dfrac{P_{1}(x,y)^{2}}{3} \right) + \dfrac{P_{0}(x,y)^{2}}{4} \left( P_{0}(x,y)^{2} + P_{1}(x,y)^{2} \right) \right] \\&+\,2 \left[ 2y^{2} + \dfrac{5 P_{1}(x,y)}{6} \left( P_{0}(x,y)^{2} + 2P_{1}(x,y)^{2} \right) \right. \\&\left. +\,\dfrac{y}{3} \left( 2 P_{0}(x,y)^{2} + P_{1}(x,y)^{2} \right) +\dfrac{5 P_{0}(x,y)^{2} }{36} \left( P_{0}(x,y)^{2} + P_{1}(x,y)^{2} \right) \right] . \end{aligned}$$

Consequently, we apply Grönwall’s Lemma to the last inequality above, from which the assertion (iii) follows by setting $P_{2}(x,y) : = 1 + P_{01}(x,y)$, and concluding the proof. $\square$

Proof of Proposition 4.2

Let $T > 0$ and $b \in L^{2}(\mathbb {R},\mathbb {R})$. First, according to Proposition 4.1, we can consider the unique solution $u_{b} \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$ of (4). Moreover, by density, a sequence $(b_{n})_{n \in \mathbb {N}}$ of smooth maps with compact support is strongly converging to b in $L^{2}(\mathbb {R},\mathbb {R})$. In addition, Proposition A.1 ensures that the sequence $(u_{b_{n}})_{n\in \mathbb {N}}$ of associated smooth maps also satisfies (4) and the a priori estimates for any $n \in \mathbb {N}$. We deduce from the quantitative estimate of Proposition 4.1 the strong convergence of $(u_{b_{n}})_{n \in \mathbb {N}}$ to $u_{b}$ in $C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$. In particular, we can correctly let $n \rightarrow + \infty$ in the inequality (i) of Proposition A.1 applied to $(u_{b_{n}},b_{n})$ in order to get $\Vert u_{b} \Vert _{C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))} \leqslant P_{0} ( T, \Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} )$.

Then, let $t \in [0,T]$ fixed. Since $(b_{n})_{n \in \mathbb {N}}$ is bounded, the sequence $(\partial _{x}u_{b_{n}}(\bullet ,t))_{n \in \mathbb {N}}$ is uniformly bounded in $H^{1}(\mathbb {R},\mathbb {R})$. Consequently, there exists a subsequence that weakly converges in $H^{1}(\mathbb {R},\mathbb {R})$ to a certain map, which has to be $\partial _{x}u_{b}(\bullet ,t)$ by considering the convergence in distributional sense. We emphasize the fact that here the subsequence depends on the time variable so it is denoted by $(\partial _{x}u_{b_{n(t)}})_{n \in \mathbb {N}}$. Considering the lower-semicontinuity of the norm with respect to the weak convergence, we obtain for any $t \in [0,T]$:

$$\begin{aligned} \Vert \partial _{x} u_{b}(\bullet , t) \Vert _{H^{1}(\mathbb {R},\mathbb {R})}\leqslant & {} \liminf _{n \in \mathbb {N}} \Vert \partial _{x} u_{b_{n(t)}}(\bullet , t) \Vert _{H^{1}(\mathbb {R},\mathbb {R})} \\\leqslant & {} \, P_{2} \left( T, \Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) \hbox {e}^{ T+\frac{T}{3} \Vert b \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} } \\&+ \, P_{1} \left( T, \Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) . \\ \end{aligned}$$

Hence, the inequality of Proposition 4.2 holds with $(b,u_{b})$, concluding the proof. $\square$

1.3 A.3 Existence of an Optimal Bottom

Proof of Proposition 4.3

The proof is very similar to the one of Proposition 4.1. Let $T > 0$, $K > 0$ and $b \in L^{2}(\mathbb {R},\mathbb {R})$ with support included in $[-K,K]$. From Proposition 4.1, the initial-value problem (4) has a unique solution $u_{b} \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$. Consider now any sequence $(b_{n})_{n \in \mathbb {N}}$ of square-integrable maps with supports all included in $[-K,K]$ that is weakly converging to b in $L^{2}(\mathbb {R},\mathbb {R})$. In particular, such a sequence is bounded so Propositions 4.1–4.2 ensure that the sequence $(u_{b_{n}})_{n\in \mathbb {N}}$ of associated maps satisfies (4) and the a priori estimate for any $n \in \mathbb {N}$. We deduce that $u_{b}$ and $(u_{b_{n}})_{n\in \mathbb {N}}$ are uniformly bounded in $C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$. First, we consider the two following compact embeddings: $H^{2}(]-K,K[,\mathbb {R}) \subset H^{1}(]-K,K[,\mathbb {R}) \subset H^{-1}(]-K,K[,\mathbb {R})$. We can apply the Aubin–Lions–Simon Lemma [17, Section 8 Corollary 4] to obtain that the following one is also compact:

$$\begin{aligned} W:= & {} \left\{ w \in L^{\infty }\left( 0,T;H^{2} \left( \left] - K, K \right[,\mathbb {R} \right) \right) : \partial _{t}w \in L^{\infty } \left( 0,T;H^{-1} \left( \left] - K, K \right[ ,\mathbb {R} \right) \right) \right\} \\&\hookrightarrow C^{0} \left( 0,T;H^{1} \left( \left] -K,K \right[,\mathbb {R} \right) \right) . \end{aligned}$$

From the foregoing and (4), $\Vert u_{b_{n}} \Vert _{L^{\infty }(0,T;H^{2}(]-K,K[,\mathbb {R}))} + \Vert \partial _{t} u_{b_{n}} \Vert _{L^{\infty }(0,T;H^{-1}(]-K,K[,\mathbb {R}))}$ is uniformly bounded, i.e. $(u_{b_{n}})_{n \in \mathbb {N}}$ is uniformly bounded in W. We deduce that there exists $u_{K} \in C^{0}(0,T;H^{1}(]-K,K[,\mathbb {R}))$ and a subsequence $(u_{b_{n'}})_{n \in \mathbb {N}}$ that is strongly converging to $u_{K}$ in $C^{0}(0,T;H^{1}(]-K,K[,\mathbb {R}))$. Then, let $n \in \mathbb {N}$. We introduce the quantities $\delta b = b_{n'} - b$ and $\delta u = u_{b_{n'}} - u_{b}$. One can check they satisfy the initial-value problem (19). We emphasize again the fact that the partial derivatives in (19) have to be handled with care since these are understood in a distributional sense. But we can still apply the integration-by-parts formula of [18, Lemma 7.3] by considering the Gelfand triple $H^{1}(\mathbb {R},\mathbb {R}) \subset L^{2}(\mathbb {R},\mathbb {R}) \subset H^{-1}(\mathbb {R},\mathbb {R})$ and the fact that $\delta u \in \lbrace w \in L^{2}(0,T;H^{1}(\mathbb {R},\mathbb {R})) : \partial _{t}w \in L^{2}(0,T;H^{-1}(\mathbb {R},\mathbb {R})) \rbrace$. Following the same calculations than we did in the proof of Proposition 4.1, we get for any $t \in [0,T]$:

$$\begin{aligned} \Vert \delta u \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2} = 2 \int _{0}^{t} \int _{\mathbb {R}} \delta b \, \dfrac{\partial (\delta u)}{\partial x} \, \mathrm {d}x \, \mathrm {d}s - \int _{0}^{t} \int _{\mathbb {R}} \left( \delta u \right) ^{2} \dfrac{\partial u_{b}}{\partial x} \, \mathrm {d}x \, \mathrm {d}s . \end{aligned}$$

Finally, we use the Cauchy–Schwarz inequality, and the fact that all the supports of the considered bottoms are included in $[-K,K]$. We thus get for any $t \in [0,T]$:

$$\begin{aligned}&\Vert \left( u_{b_{n'}} - u_{b} \right) \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2}\\&\quad \leqslant 2 \int _{0}^{T} \begin{array}{|c|} \displaystyle { \int _{-K}^{K} \left( b_{n'} - b \right) \left( x \right) \left[ \dfrac{\partial u_{b_{n'}}}{\partial x} -\dfrac{\partial u_{b}}{\partial x} \right] \left( x, s \right) \, \mathrm {d}x } \\ \end{array}~ \, \mathrm {d}s \\&\qquad +\, \Vert u_{b} \Vert _{C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))} \int _{0}^{t} \Vert \left( u_{b_{n'}} - u_{b} \right) \left( \bullet , s \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2} \, \mathrm {d}s . \\ \end{aligned}$$

Since $t \in [0,T] \mapsto \Vert \delta u (\bullet ,t) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \in \mathbb {R}$ is a continuous function [18, Lemma 7.3], we can apply Grönwall’s Lemma and we obtain:

$$\begin{aligned} \Vert u_{b_{n'}} - u_{b} \Vert _{C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))}^{2} ~~ \leqslant ~~ C \int _{0}^{T} \begin{array}{|c|} \displaystyle { \int _{-K}^{K} \left( b_{n'} - b \right) \left( x \right) \left[ \dfrac{\partial u_{b_{n'}}}{\partial x} -\dfrac{\partial u_{b}}{\partial x} \right] \left( x, s \right) \, \mathrm {d}x } \\ \end{array}\,\mathrm {d}s, \end{aligned}$$

where we have set $C : = 2\hbox {e}^{T\Vert u_{b} \Vert _{C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))}}$. Hence, it remains to prove that the right member of the above inequality converges to zero as $n \rightarrow + \infty$. For this purpose, let us now introduce the integrand $R_{n}: t \mapsto \int _{-K}^{K} \delta b(x) \, \partial _{x}(\delta u)(x,t) \, \mathrm {d} x$. Since $b_{n'}$ converges weakly to b in $L^{2}(]-K,K[,\mathbb {R})$ and $u_{b_{n'}}$ strongly to $u_{K}$ in $C^{0}(0,T;H^{1}(]-K,K[,\mathbb {R}))$, we get that $R_{n}(t)$ converges to zero for any $t \in [0,T]$. Moreover, the a priori estimate of Proposition 4.2 ensures that $R_{n}(t)$ is uniformly bounded. Hence, Lebesgue’s dominated convergence theorem applies and $\int _{0}^{T} \vert R_{n}(t) \vert \, \mathrm {d}t \rightarrow 0$ as $n \rightarrow + \infty$. One concludes from the last inequality that $(u_{b_{n'}})_{n \in \mathbb {N}}$ strongly converges to $u_{b}$ in $C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$. We also have proved the uniqueness of the limit for any other converging subsequence. Recalling that $(u_{b_{n}})_{n \in \mathbb {N}}$ is uniformly bounded, we deduce that the whole sequence converges to $u_{b}$. $\square$

Proof of Theorem 2.1

Combining Lemma 4.1 and Proposition 4.3, it is possible to extract from any maximizing sequence of (2) a subsequence $(b_{n'})_{n \in \mathbb {N}}$ that is weakly converging in $L^{2}(\mathbb {R},\mathbb {R})$ to a certain $b^{*} \in \mathcal {B}$, and such that $u_{b_{n'}} \rightarrow u_{b^{*}}$ in $C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$. From the continuity of the embedding $C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R})) \subset L^{2}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$, we deduce that $F(b^{*})= \lim _{n \rightarrow +\infty } F(b_{n'}) = \sup _{b \in \mathcal {B}} F(b)$ with $b^{*} \in \mathcal {B}$ so the supremum is a maximum and problem (2) has a global maximizer. To conclude the proof, it remains to show that such a maximizer saturates the $L^{2}$-constraint, which is proved as in Proposition B.4. Indeed, if it not the case, then choose $\theta \in ]1, (M / \Vert b^{\text {opt}} \Vert _{L^{2}(\mathbb {R},\mathbb {R})} )^{2/7}]$ and the bottom $b^{\text {opt}}_{\theta }$ of Lemma B.1 is admissible. One can check that $F(b^{\text {opt}}_{\theta })\geqslant F(b^{\text {opt}})$ and the optimality of $b^{\text {opt}}$ yields to $u_{b^{\text {opt}}} = 0$ on $[T,\theta ^{3}T]$. Considering now (4), we obtain $\partial _{x}b^{\text {opt}} = 0$ so $u_{b^{\text {opt}}} = 0$ also on [0, T] and $F(b^{\text {opt}}) = 0$, which is a contradiction. $\square$

1.4 A.4 Fréchet Differentiability of the Functional

Proof of Proposition 4.4

Formally speaking, if $v_{b}$ is a solution of (12), then the equation satisfied by $\partial _{x} v_{b}$ is already studied in [7]. However, we need to specify a bit the existence result because it is stated in terms of the so-called Bourgain space $Y_{s,\beta }$ (see [7, Sect. 2] for details). First, we apply [9, Proposition 2.1] with $\sigma = -1$, $f = - \partial _{x}b$, $u_{0} \equiv 0$, $s = \sigma + 3 = 2$ and $\beta = \varepsilon + \frac{1}{2}$, where $\varepsilon > 0$ is chosen small enough. This local existence result combined with standard global arguments [7, Proposition 5.1] establishes that the unique solution $u_{b}$ of (4) is the [0, T]-restriction of a map $U_{b} \in Y_{2,\varepsilon + 1 / 2}$.

Then, we introduce new variables $(\xi ,\tau ) :=(-x, T- t)$ to transform (12) into an initial-value problem. We set $U(\xi ,\tau ) : = U_{b}(-x,T-t)$, and we still have $U \in Y_{2,\varepsilon + 1 / 2} \subset Y_{1,\varepsilon + 1 / 2}$ but we also get $\partial _{\xi }U \in Y_{1,\varepsilon + 1 / 2}$ [7, above Theorem 5.5]. We can now apply [7, Theorem 2.6] with $s = 1$, $\beta = \varepsilon + \frac{1}{2}$, and $f = 2\partial _{\xi }U$. We deduce that there exists a unique solution $W \in Y_{1,\varepsilon + 1 / 2}$ satisfying $W(\bullet ,0) = 0$ and $\partial _{\tau } W + \partial _{\xi }\left( U\, W \right) + \partial _{\xi \xi \xi } W = 2 \partial _{\xi }U$ on $\mathbb {R} \times [0,T]$.

Finally, it remains to get back to (12). For this purpose, we consider the [0, T]-restrictions $u \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$, $w \in C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$ of the maps U and W [7, Lemma 2.3]. Using the equations they satisfy, we get $u(2-w) \in W^{1,1}(0,T;H^{-2}(\mathbb {R},\mathbb {R}))$. From standard linear semi-group theory [7, Section 1 Sect. III], there exists a unique function $v \in C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$ satisfying $v(\bullet , 0) = 0$ and the Airy equation $\partial _{\tau }v + \partial _{\xi \xi \xi }v = u(2-w)$ on $\mathbb {R} \times [0,T]$. Looking at the equation satisfied by $w - \partial _{x}v$, we deduce that $\partial _{x}v = w$. In particular, we obtain $v \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R})$ and getting back to the original variables $(x,t) := (-\xi ,T - \tau )$, one can check that the map $v_{b}(x,t): = v(\xi ,\tau )$ is a global solution of (12) in $C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$.

At last, we prove such a solution is unique. Consider two maps of $C^{0}(0,T; H^{2} (\mathbb {R},\mathbb {R}))$ solving (12), denoted $v_{1}$ and $v_{2}$, and introduce the quantity $\delta v : = v_{1} - v_{2}$. From the linearity of (12), one can check that $\delta v$ satisfies $\delta v(\bullet ,T) = 0$ and $\partial _{t}(\delta v) + u_{b} \, \partial _{x}(\delta v) + \partial _{xxx}(\delta v) = 0$. This last equality is understood in distributional sense but we can still apply the integration-by-parts formula of [18, Lemma 7.3] with the Gelfand triple $H^{1}(\mathbb {R},\mathbb {R}) \subset L^{2}(\mathbb {R},\mathbb {R}) \subset H^{-1}(\mathbb {R},\mathbb {R})$ and the fact that $\delta v \in \lbrace w \in L^{2}(0,T;H^{1}(\mathbb {R},\mathbb {R})): \partial _{t}w \in L^{2}(0,T;H^{-1}(\mathbb {R},\mathbb {R})) \rbrace$. Proceeding as below (19), we get:

$$\begin{aligned} \forall t \in [0,T], ~~ \Vert \delta v (\bullet , t) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} \leqslant \Vert \partial _{x}u_{b} \Vert _{C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))} \int _{t}^{T} \Vert \delta v (\bullet ,s) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R}))} \, \mathrm {d}s. \end{aligned}$$

It follows from the continuity of the map $t \in [0,T] \mapsto \Vert \delta v(\bullet ,t) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \in \mathbb {R}$ [18, Lemma 7.3] and Grönwall’s Lemma that $\delta v \equiv 0$ on $[0,T] \times \mathbb {R}$, i.e. $v_{1} = v_{2}$. To conclude the proof, there exists a unique global solution $v_{b} \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$ satisfying (12). $\square$

Proof of Proposition 4.5

Let $T > 0$ and $(b, h) \in L^{2}(\mathbb {R},\mathbb {R}) \times L^{2}(\mathbb {R},\mathbb {R})$. From Proposition 4.1, there exists two associated global solutions $u_{b}$ and $u_{b+h}$ in $C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$ such that $(b,u_{b})$ and $(b+h,u_{b+h})$ satisfy (4). Introducing again the quantities $\delta u := u_{b+h} - u_{b}$ and $\delta b : = (b+h)-b = h$, one can check $(\delta b, \delta u)$ satisfies (19), which has to be understood in a distributional sense. Still, we can apply the integration-by-parts formula of [18, Lemma 7.3] with the Gelfand triple $H^{1}(\mathbb {R},\mathbb {R}) \subset L^{2}(\mathbb {R},\mathbb {R}) \subset H^{-1}(\mathbb {R},\mathbb {R})$ and the fact that $\delta u$ belongs to $\lbrace w \in L^{2}(0,T;H^{1}(\mathbb {R},\mathbb {R})): \partial _{t}w \in L^{2}(0,T;H^{-1}(\mathbb {R},\mathbb {R})) \rbrace$. Proceeding as below (19), we get:

$$\begin{aligned} \forall t \in [0,T], ~~ \Vert \delta u (\bullet ,t) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} = 2 \int _{0}^{t} \int _{\mathbb {R}} \delta b \, \dfrac{\partial (\delta u) }{\partial x} \, \mathrm {d}x \, \mathrm {d}s - \int _{0}^{t} \int _{\mathbb {R}} \left( \delta u \right) ^{2} \dfrac{\partial u_{b}}{\partial x} \, \mathrm {d}x \, \mathrm {d}s . \end{aligned}$$

Using the Cauchy–Schwarz inequality, we obtain for any $t \in [0,T]$:

$$\begin{aligned} \Vert \delta u (\bullet ,t) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} \leqslant C \int _{0}^{t} \Vert \delta u (\bullet ,s) \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})} \, \mathrm {d}s + 2 T \Vert h \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \, \Vert \delta u \Vert _{C^{0}(0,T,H^{1}(\mathbb {R},\mathbb {R})}, \end{aligned}$$

where we have set $C : = \Vert u_{b} \Vert _{C^{0}(0,T,H^{2}(\mathbb {R},\mathbb {R})}$. We can now apply Grönwall’s Lemma to the map $t \in [0,T] \mapsto \Vert \delta u(\bullet ,t) \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \in \mathbb {R}$, which is continuous [18, Lemma 7.3], and it comes:

$$\begin{aligned} \Vert u_{b+h} - u_{b} \Vert _{C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))}^{2} \leqslant 2 T \hbox {e}^{CT} \Vert h \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \Vert \delta u \Vert _{C^{0}(0,T,H^{1}(\mathbb {R},\mathbb {R})} . \end{aligned}$$

(22)

Combined with the continuity of the map $b \in L^{2} \mapsto u_{b} \in C^{0}(0,T,H^{1}(\mathbb {R},\mathbb {R}))$ ensured by Corollary B.1, we can deduce from (22) that $\Vert \delta u \Vert ^{2}_{C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))} = o ( \Vert \delta b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} )$.

Then, Proposition 4.4 ensures (12) has a unique global solution $v_{b} \in C^{0}(0,T;H^{2} (\mathbb {R},\mathbb {R}))$. Hence, we can correctly compute again the integration-by-parts formula given in [18, Lemma 7.3] by considering the Gelfand triple $H^{1}(\mathbb {R},\mathbb {R}) \subset L^{2}(\mathbb {R},\mathbb {R}) \subset H^{-1}(\mathbb {R},\mathbb {R})$ combined with the fact that $(\delta u,v_{b}) \in \lbrace w \in L^{2}(0,T;H^{1}(\mathbb {R},\mathbb {R})) : \partial _{t}w \in L^{2}(0,T;H^{-1}(\mathbb {R},\mathbb {R})) \rbrace ^{2}$. Since $v_{b}(\bullet ,T) = \delta u(\bullet ,0) = 0$, we have:

$$\begin{aligned} 0 \,= & {} \, \int _{0}^{T} \left\langle \partial _{t} \left( \delta u \right) ~\vert ~ v_{b} \right\rangle _{H^{-1}(\mathbb {R},\mathbb {R}), H^{1}(\mathbb {R},\mathbb {R})} \left( \bullet , t \right) \, \mathrm {d}t\\&+ \int _{0}^{T} \left\langle \partial _{t} v_{b} ~\vert ~ \delta u \right\rangle _{H^{-1}(\mathbb {R},\mathbb {R}), H^{1}(\mathbb {R},\mathbb {R})} \left( \bullet , t \right) \, \mathrm {d}t \end{aligned}$$

Proceeding as below (19), one may obtain from the previous relation:

$$\begin{aligned} 2 \int _{0}^{T} \int _{\mathbb {R}} u_{b} \, \delta u \, \mathrm {d}x \, \mathrm {d}t = \int _{0}^{T} \int _{\mathbb {R}} \dfrac{ \left( \delta u \right) ^{2}}{2} \, \dfrac{\partial v_{b}}{\partial x} \, \mathrm {d}x \, \mathrm {d}t + \int _{0}^{T} \int _{\mathbb {R}} \delta b \, \dfrac{\partial v_{b}}{\partial x} \, \mathrm {d}x \, \mathrm {d}t. \end{aligned}$$

Recalling that $\delta b = h$ and introducing the map (3), we deduce from the last relation:

$$\begin{aligned} R_{F}(h):= & {} F(b+h) - F(b) - \int _{\mathbb {R}} h(x) \left( \int _{0}^{T} \dfrac{\partial v_{b}}{\partial x }(x,t) \, \mathrm {d}t \right) \, \mathrm {d}x \\= & {} \int _{0}^{T} \int _{\mathbb {R}} (\delta u)^{2} \left( 1 + \dfrac{1}{2} \dfrac{\partial v_{b}}{\partial x} \right) \mathrm {d}x \, \mathrm {d}t. \end{aligned}$$

Consequently, using the fact that $v_{b} \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$, we establish

$$\begin{aligned} \vert R_{F}(h) \vert \leqslant T \Vert \delta u \Vert _{C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))}^{2} \left( 1 + \dfrac{1}{2} \Vert \partial _{x}v_{b} \Vert _{C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))} \right) , \end{aligned}$$

and inserting the estimate (22) into the last one above, we get indeed $R_{F}(h) = o(\Vert h \Vert ^{2}_{L^{2}(\mathbb {R},\mathbb {R})})$. Since $h \in L^{2}(\mathbb {R},\mathbb {R}) \mapsto \int _{\mathbb {R}}h(x)[\int _{[0,T]} \partial _{x}v_{b}(x,t) \, \mathrm {d}t ] \, \mathrm {d}x \in \mathbb {R}$ is a continuous linear form, the uniqueness of the differential ensures that the functional $F_{b} : h \in L^{2}(\mathbb {R},\mathbb {R}) \mapsto F(b+h) \in \mathbb {R}$ is Fréchet differentiable at the origin, i.e. F is Fréchet differentiable at any bottom $b \in L^{2}(\mathbb {R},\mathbb {R})$ and the shape gradient is well defined by (13), concluding the proof of Proposition 4.5. $\square$

Appendix B: Other Useful Properties

1.1 B.1 Hölder Continuity of the Functional

In Sect. 4.3, given any fixed $K > 0$, we have proved the (sequential) continuity of the nonlinear map $N:b \in L^{2}(]-K,K[,\mathbb {R}) \mapsto u_{b} \in C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$ for the $L^{2}$-weak topology. Here, we first establish $N: L^{2}(\mathbb {R},\mathbb {R}) \mapsto C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$ is continuous for the $L^{2}$-strong topology. Then, by restricting N and F to any ball of $L^{2}(\mathbb {R},\mathbb {R})$, we obtain their Hölder continuity.

Proposition B.1

Let $T > 0$ and $b \in L^{2}(\mathbb {R},\mathbb {R})$. Consider any sequence $(b_{n})_{n \in \mathbb {N}}$ of square-integrable maps strongly converging to b in $L^{2}(\mathbb {R},\mathbb {R})$. Then, the sequence $(u_{b_{n}})_{n \in \mathbb {N}}$ of their associated solutions given in Proposition 4.1 strongly converges to $u_{b}$ in $C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$, where $u_{b}$ is the unique solution of Proposition 4.1 associated with b.

Proof

Let $T > 0$ and $b \in L^{2}(\mathbb {R},\mathbb {R})$. From Proposition  4.1, we can consider the unique solution $u_{b} \in C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$ satisfying (4). First, we treat the smooth case. Let $(b_{n})_{n \in \mathbb {N}}$ be a sequence of maps in $H^{\infty }(\mathbb {R},\mathbb {R})$ that is strongly converging to b for the $L^{2}$-norm. In particular, this sequence is uniformly bounded in $L^{2}(\mathbb {R},\mathbb {R})$. Applying Proposition A.1, there exists a sequence $(u_{b_{n}})_{n\in \mathbb {N}}$ of associated smooth maps satisfying (4) and the a priori estimates, from which we deduce that $(u_{b_{n}})_{n\in \mathbb {N}}$ is uniformly bounded in $C^{0}(0,T;H^{2}(\mathbb {R},\mathbb {R}))$ by a constant denoted $C > 0$. Then, applying Proposition 4.1 with b and $\mathfrak {b} = b_{n}$ for any $n \in \mathbb {N}$, we obtain that $(u_{b_{n}})_{n \in \mathbb {N}}$ strongly converges to $u_{b}$ in $C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$. We now prove that in fact the convergence occurs in $C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$. Let $(m,n) \in \mathbb {N} \times \mathbb {N}$. We set $\delta u : = u_{b_{n}} - u_{b_{m}}$ and $\delta b : = b_{n} - b_{m}$ then establish that $(u_{b_{k}})_{k \in \mathbb {N}}$ is a uniform Cauchy sequence by relating $\partial _{x}(\delta u)$ to $\delta b$ and $\delta u$. Since both $(b_{n},u_{b_{n}})$ and $(b_{m},u_{b_{m}})$ satisfy (4), we get that $(\delta u, \delta b)$ is a smooth solution to (19) (where we have replaced $u_{b}$ by $u_{b_{m}}$). We use the conservative structure of (4) and (19) by writing $\partial _{t} ( u_{b_{m}}) = -\partial _{x}I$ and $\partial _{t}(\delta u) = -\partial _{x}J$, where we set $I : = \partial _{xx}u_{b_{m}} + \frac{1}{2}(u_{b_{m}})^{2} + b_{m}$ and $J : = \partial _{xx}(\delta u) + \frac{1}{2}(\delta u)^{2} + \delta b + u_{b_{m}} \, \delta u$. We have:

$$\begin{aligned}&\dfrac{\mathrm {d}}{\mathrm {d}t} \int _{\mathbb {R}} \left[ \dfrac{\left( \delta u \right) ^{3}}{6} + u_{b_{m}} \dfrac{\left( \delta u \right) ^{2}}{2} - \dfrac{1}{2} \left( \dfrac{\partial \left( \delta u \right) }{\partial x} \right) ^{2} + \delta b \, \delta u \right] \\&\quad = \, \underbrace{2 \int _{\mathbb {R}} \partial _{t} \left( \delta u \right) \, J}_{ = \, - \int \partial _{x}( J^{2}) \, = \, 0 } + \underbrace{\frac{1}{2} \int _{\mathbb {R}} \partial _{t} (b_{m}) \left( \delta u \right) ^{2}}_{=\, - \int I \, \delta u \, \partial _{x} (\delta u)} \\&\quad = - \int _{\mathbb {R}} \delta u \, \dfrac{\partial \left( \delta u \right) }{\partial x} \left( \dfrac{\partial ^{2} u_{b_{m}}}{\partial x^{2}} + \dfrac{u_{b_{m}}^{2}}{2} + b_{m} \right) \, \mathrm {d} x. \end{aligned}$$

Proceeding as below (19) (but here the functions are regular), we obtain for any $t \in [0,T]$:

$$\begin{aligned}&\Vert \partial _{x} \left( u_{b_{n}} - u_{b_{m}} \right) \left( \bullet , t \right) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2}\nonumber \\&\quad \leqslant \dfrac{5C}{3} \Vert \left( u_{b_{n}} - u_{b_{m}} \right) (\bullet ,t) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}^{2} + 4C \Vert b_{n} - b_{m} \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \nonumber \\&\quad +\, 2TC \left( C + \dfrac{C^{2}}{2} + \sup _{k \in \mathbb {N}}\Vert b_{k} \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \right) \Vert \left( u_{b_{n}} - u_{b_{m}} \right) (\bullet , t) \Vert _{L^{2}(\mathbb {R},\mathbb {R})}. \end{aligned}$$

(23)

We deduce from (23) that $t \in [0,T] \mapsto \partial _{x}( u_{b_{n}} ) (\bullet ,t) \in L^{2}(\mathbb {R},\mathbb {R})$ is a uniform Cauchy sequence. It is thus strongly converging to a certain map in $C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$, which has to be $\partial _{x}u_{b}$ by considering the convergence in the sense of distributions. Finally, we treat the non-regular case by approximations. Let $\varepsilon > 0$ and $(b_{n})_{n \in \mathbb {N}}$ be any sequence of maps in $L^{2}(\mathbb {R},\mathbb {R})$ that is strongly converging to b. By density, for any $n \in \mathbb {N}$, there exists a sequence $(b^{k}_{n})_{k \in \mathbb {N}}$ of smooth maps with compact support that is strongly converging to $b_{n}$ in $L^{2}(\mathbb {R},\mathbb {R})$. From the foregoing, we deduce that there exists $k_{n} \in \mathbb {N}$ such that we have $\Vert u_{b_{n}^{k_{n}}}- u_{b_{n}} \Vert _{C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))} < \varepsilon$. Moreover, one can check that $(b_{n}^{k_{n}})_{n \in \mathbb {N}}$ strongly converges to b in $L^{2}(\mathbb {R},\mathbb {R})$. Again, from the foregoing, there exists $N \in \mathbb {N}$ such that for any integer $n \geqslant N$, we have $\Vert u_{b_{n}^{k_{n}}} - u_{b} \Vert _{C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))} < \varepsilon$. We deduce for any $n \geqslant N$:

$$\begin{aligned} \Vert u_{b_{n}}- u_{b} \Vert _{C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))}\leqslant & {} \Vert u_{b_{n}} - u_{b_{n}^{k_{n}}} \Vert _{C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))} \\&+ \Vert u_{b_{n}^{k_{n}}} - u_{b} \Vert _{C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))} < 2 \varepsilon . \end{aligned}$$

To conclude the proof of Proposition B.1, $(u_{b_{n}})$ converges to $u_{b}$ in $C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$. $\square$

Corollary B.1

Let $M > 0$, $T > 0$ and set $B_{M} : = \lbrace b \in L^{2}(\mathbb {R},\mathbb {R}) : \Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \leqslant M \rbrace$. Then, there exists $C(T,M) > 0$ depending only on T and M such that for any $(b, \mathfrak {b}) \in B_{M} \times B_{M}$:

In particular, the energy functional $F: \mathcal {B} \rightarrow \mathbb {R}$ given in (3) is $\frac{1}{2}$-Hölder continuous.

Proof

Let $M > 0$, $T > 0$, and set $B_{M} : = \lbrace b \in L^{2}(\mathbb {R},\mathbb {R}) : \Vert b \Vert _{L^{2}(\mathbb {R},\mathbb {R})} \leqslant M \rbrace$. First, we combine the a priori estimate of Proposition 4.2 with the fact that $(b, \mathfrak {b}) \in B_{M} \times B_{M}$. We deduce that the constant $C > 0$ appearing in the quantitative estimate of Proposition 4.1 can be bounded by one that only depends on T and M. Hence, the nonlinear map $N: b \in B_{M} \mapsto u_{b} \in C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$ is $\frac{1}{2}$-Hölder continuous. Then, the continuity of the embedding $C^{0}(0,T;L^{2}(\mathbb {R},\mathbb {R})) \subset L^{2}(0,T;L^{2}(\mathbb {R},\mathbb {R}))$ applied to b and $b_{n'}=\mathfrak {b}$ also yields to the same result for the map $F : B_{M} \mapsto \mathbb {R}$. Finally, there exists two sequences $(b_{n})_{n \in \mathbb {N}}$ and $(\mathfrak {b}_{n})_{n \in \mathbb {N}}$ of smooth maps with compact support respectively converging to b and $\mathfrak {b}$ strongly in $L^{2}(\mathbb {R},\mathbb {R})$. Proposition B.1 ensures that the associated smooth maps $(u_{b_{n}})_{n \in \mathbb {N}}$ and respectively converges to in $C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$. We can now proceed as in the proof of Proposition B.1 so (23) holds with $b_{n}$ and $b_{m} = \mathfrak {b}_{n}$. By letting $n \rightarrow + \infty$ in this inequality, we deduce from the foregoing that $N : b \in B_{M} \mapsto u_{b} \in C^{0}(0,T;H^{1}(\mathbb {R},\mathbb {R}))$ is $\frac{1}{4}$-Hölder continuous, concluding the proof of Corollary B.1. $\square$

1.2 B.2 Stability Analysis of the Numerical Scheme

Proposition B.2

The discretization (15) takes the form $L_{\varDelta x, \varDelta t}u = 0$ and approximates Eq. (6) written as $\partial _{t}u + Lu = 0$. If we assume $\varDelta t = O(\varDelta x )$, then (15) is consistent and first-order accurate: $\forall u \in C^{4} ( \mathbb {R} \times [0,+\infty [,\mathbb {R} ), \, \partial _{t}u + Lu = 0 \Rightarrow \partial _{t}u + Lu = L_{\varDelta x, \varDelta t}u + O ( \varDelta x )$.

Proof

We introduce the shift operators $s^{\pm }_{x}[(\bullet )(x,t)]: = (\bullet )(x \pm \varDelta x , t)$ in order to define $\delta ^{1}_{x} := \frac{1}{2 \varDelta x}(s^{+}_{x} - s^{-}_{x})$, $\delta ^{2}_{x} := \frac{1}{\varDelta x^2}(s^{+}_{x} - 2 + s^{-}_{x})$ and $\delta ^{3}_{x} := \delta ^{1}_{x} \delta ^{2}_{x}$. Let $u \in C^{4}(\mathbb {R} \times [0,+\infty [,\mathbb {R})$ be such that $\partial _{t}u + Lu = 0$. We get $s^{\pm }_{x}u = u \pm \varDelta x \, \partial _{x}u + \frac{\varDelta x^{2}}{2} \partial _{xx}u \pm \frac{\varDelta x ^{3}}{6} \partial _{xxx}u + O(\varDelta x^{4})$ from a Taylor expansion. We deduce $\partial _{x} u = \delta ^{1}_{x}u + O(\varDelta x^{2})$ and $\partial _{xx} u = \delta ^{2}_{x}u + O(\varDelta x^{2})$. These estimations are then combined to obtain $\partial _{xxx} u = \delta ^{3}_{x}u + O(\varDelta x)$. Therefore, we have an approximation of the linear terms of L:

$$\begin{aligned} Lu = \dfrac{3 c_{0} }{2 h_{0}} u \, \partial _{x}u + \dfrac{c_{0}h_{0}^{2}}{6} \delta ^{3}_{x}u + \dfrac{c_{0}}{2} \delta _{x}^{1} b + O(\varDelta x). \end{aligned}$$

(24)

Introducing the time operators $s^{+}_{t}[(\bullet )(x,t)]: = (\bullet )(x,t+\varDelta t)$ and $\delta _{t}^{+} : = \frac{1}{\varDelta t } (s^{+}_{t} - 1)$, similar arguments yields $\partial _{t} u = \delta _{t}^{+} u + O(\varDelta t)$ and $\partial _{t} u = s_{t}^{+} \partial _{t} u + O(\varDelta t) = -s_{t}^{+} Lu + O(\varDelta t)$. We get: $\partial _{t} u + Lu = \delta _{t}^{+} u + Lu + O(\varDelta t) = \delta _{t}^{+} u + \frac{1}{2}(Lu - \partial _{t}u) + O(\varDelta t) = \delta _{t}^{+} u + \frac{1}{2}(1+s_{t}^{+})Lu + O(\varDelta t)$. Then, we assume $\varDelta t = O(\varDelta x)$ and from $\frac{1}{2}(s_{t}^{+} +1) u^{2}= u \,s^{+}_{t}u + O(\varDelta t^{2})$, we can treat the nonlinear term of Lu: $(1+s_{t}^{+})(u \, \partial _{x}u) = \delta ^{1}_{x}[u \,s^{+}_{t}u + O(\varDelta t^{2})] + O(\varDelta x^{2}) = \delta ^{1}_{x}(u \,s^{+}_{t}u) + O(\varDelta x)$. Hence, we deduce the expected estimation $\partial _{t}u + Lu = L_{\varDelta x, \varDelta t}u + O(\varDelta x)$ with:

$$\begin{aligned} L_{\varDelta x, \varDelta t}u ~:=~ \delta ^{+}_{t} u + \dfrac{3 c_{0}}{4h_{0}} \delta ^{1}_{x} \left( u \,s^{+}_{t} u \right) + \dfrac{c_{0}h_{0}^{2}}{12} \left( 1+s^{+}_{t} \right) \left( \delta ^{3}_{x} u \right) + \dfrac{c_{0}}{2} \delta ^{1}_{x} b. \end{aligned}$$

$\square$

Proposition B.3

Consider the discretization (15) of Eq. (6) with forcing term $b = 0$. Let $\beta = \frac{3c_{0}}{2 h_{0}} \Vert \zeta \Vert _{C^{0}([-L,L] \times [0,T],\mathbb {R})}$, $\mu = \frac{c_{0}}{6} h_{0}^{2}$ and $s = \frac{\varDelta t}{\varDelta x}$. Then, Von Neumann’s stability analysis provides an amplification factor $g: [- \pi , \pi ] \rightarrow \mathbb {C}$ of the following form:

$$\begin{aligned} g(\xi ): = \dfrac{1 - i A\left( \xi \right) }{1 + i A \left( \xi \right) } \quad \mathrm {where}\quad A \left( \xi \right) : = s \left( \sin \xi \right) \left[ \dfrac{\beta }{2} + \dfrac{\mu }{\varDelta x^{2}} \left( \cos \xi - 1 \right) \right] . \end{aligned}$$

In particular, $\vert g \vert = 1$, ensuring the non-dissipative feature of the method: the scheme (15) is unconditionally stable. Moreover, the numerical dispersion $\Psi = \mathrm {arg}(g) = - \mathrm {arctan}(\frac{2A}{1 - A^{2}})$ is compared to the analytical one whose expression is given by $\Psi _{\mathrm{ref}}(\xi ) := - s \beta \xi + \frac{s \mu }{\varDelta x^{2}} \xi ^{3}$. We obtain $\Psi (\xi ) = \Psi _{\mathrm{ref}}(\xi ) + E_{\Psi }(\xi ) + O(\xi ^{7})$ where:

$$\begin{aligned} E_{\Psi }(\xi ) = \dfrac{s \beta }{6} \left( 1 + \dfrac{s^{2} \beta ^{2}}{2} \right) \xi ^{3} - \left[ \dfrac{s^{5} \beta ^{5}}{80} + \dfrac{s^{3} \beta ^{3}}{24} + \dfrac{s \beta }{120} + \dfrac{ s^{3} \beta ^{2} \mu }{4 \varDelta x^{2}} + \dfrac{s \mu }{4 \varDelta x^{2}} \right] \xi ^{5}. \end{aligned}$$

Proof

We refer to [23, (38)–(51)] for details on the proof of this result. We stress, however, a disagreement between our expression of $E_{\Psi }$ and the one provided in [23, (50)]. This seems to result from a mistake made in [23, (48)], when expanding $\Psi$ by using Taylor series. More precisely, it is wrongly stated that $\mathrm {arctan}(\frac{-2A}{1 - A^{2}}) = - 2A[1 - \frac{1}{3}A^{2} - 3 A^{4}] + O(A^7)$, as the correct expression is given by $\mathrm {arctan}(\frac{-2A}{1 - A^{2}}) = - 2A[1 - \frac{1}{3}A^{2} +\frac{1}{5} A^{4}] + O(A^7)$. $\square$

1.3 B.3 The Necessity of a $L^{2}$-Constraint

Lemma B.1

Let b(x) be a forcing function with enough regularity, as given in Lemma 3.1, and u(x, t) be the unique smooth solution of the initial-value problem (4). For any $\theta \in \mathbb {R}$, define the maps $u_{\theta }: (x,t) \mapsto \theta ^{2} u(\theta x, \theta ^{3}t)$ and $b_{\theta }: x\mapsto \theta ^{4} b(\theta x)$. Then, $u_{\theta }$ is precisely the solution of (4) with forcing function $b_{\theta }$.

Proposition B.4

Let $K > 0$ and $T > 0$. Then, the problem (11) has no global maximizer.

Proof

Assume, by contradiction, that there exists a maximizer b to (11). Then, from Lemma 3.1, we can consider its associated smooth solution $u_{b}$. Introducing the bottoms $(b_{\theta })_{\theta > 1}$ of Lemma B.1, one can check they are admissible for problem (11). Moreover, we deduce from Lemma B.1 that $F(b_{\theta }) = \int _{0}^{\theta ^{3}T} \int _{\mathbb {R}} u_{b}^{2}(x,t) \, \mathrm {d}x \, \mathrm {d}t$. Using the optimality of b, we obtain $F(b_{\theta }) = F(b)$ for any $\theta > 1$ so $u_{b} = 0$ on $[T,\theta ^{3}T]$. From (6), we get $\partial _{x}b = 0$ thus $u_{b} = 0$ also on [0, T]. Thus, $F(b) = 0$, which contradicts the optimality of b. $\square$