**2. Problem Statement**

#### *2.1. Notations and Preliminary Notes*

Let us consider geographical (geodesic) system of coordinates ( *λ*, *<sup>θ</sup>*,*<sup>r</sup>*), where *λ* ∈ [0, 2 *π*] is the geographical longitude increasing from West to East, *θ* ∈ [− *π*/2, *π*/2] is the geographical latitude increasing form South to North (*θ* = *φ* − *π*/2, *φ* ∈ [0, *<sup>π</sup>*]), *r* is the distance from the center of the Earth to a given point, the mean Earth radius is *RE*. Instead of *r* it is often convenient to introduce the coordinate *z* = *RE* − *r* of the axis *Oz*, directed along the normal to the center of the sphere *SR* of radius *RE*, along the direction of the gravity force. The unit vectors in *λ*-, *θ*- and *z*-directions are denoted by *eλ*,*eθ*,*ez*, respectively. In this case the velocity vector in the ocean is written in the form: (*<sup>u</sup>*1, *u*2, *<sup>u</sup>*3)*<sup>T</sup>* = *u*1*eλ* + *u*2*eθ* + *u*3*ez* ≡ (*u*, *<sup>w</sup>*)*<sup>T</sup>*.

Let Ω denote the connected manifold on the sphere *SR*, which is called the "reference surface" [18]. Below we consider the case when Ω does not include polar points. The ocean surface elevation is given by the equation *z* = *ξ*(*<sup>λ</sup>*, *θ*, *t*), *z* = *<sup>H</sup>*(*<sup>λ</sup>*, *θ*) is the bottom topography function, where ( *<sup>λ</sup>*,*θ*,*RE*) ∈ Ω, *t* ∈[0, *T*] is time variable ( *<sup>T</sup>*<∞), *<sup>H</sup>*(*<sup>λ</sup>*, *θ*)>0. Moreover, suppose that there exists *ε* > 0, so that *H* ≥ *ε*. Considering these notations, the total depth of the ocean is expressed as *Htot*\_*depth* = *H* − *ξ*

Below we use the following notations of differential operators:

$$\mathbf{grad}\,\Phi = \left(m\frac{\partial\Phi}{\partial\lambda}, n\frac{\partial\Phi}{\partial\theta}\right)^T, \quad \mathbf{div}\,\Phi = m\frac{\partial\Phi}{\partial\lambda} + m\frac{\partial\left[(n/m)\Phi\right]}{\partial\theta},$$

where *m* = 1/(*RE* cos *<sup>θ</sup>*), *n* = 1/*RE*. By*n* we denote the unit vector of outer normal to *∂*Ω. Let *l* denote the Coriolis parameter, *g* is the gravitational acceleration. Let us introduce the following notation for the depth-averaged horizontal velocities:

$$
\vec{U} = \frac{1}{H} \int\_{\frac{\vec{\pi}}{\xi}}^H \vec{u} \, dz.
$$

*J. Mar. Sci. Eng.* **2019**, *7*, 195

In the current study we consider the subproblem corresponding to a system of linearized shallow water equations (Step 3-c of the splitting scheme introduced above) on the time subinterval (*tj*−1, *tj*), Δ*t* = *tj* − *tj*−1, *t*0 = 0, *tJ* = *T*, *j* = 1, . . . , *J* [18]:

$$\begin{cases} \vec{\mathcal{U}}\_{l} + \begin{bmatrix} 0 & -\ell \\ \ell & 0 \end{bmatrix} \vec{\mathcal{U}} + R\_{f} \vec{\mathcal{U}} - \mathcal{g} \cdot \mathbf{grad} \, \vec{\mathcal{g}} = \vec{f}, \quad \text{in } \Omega \times (t\_{j-1}, t\_{j}), \\\\ \vec{\mathcal{J}}\_{l} - \vec{\text{div}} \left( H \vec{\mathcal{U}} \right) = 0, \quad \text{in } \Omega \times (t\_{j-1}, t\_{j}), \\ \vec{\mathcal{U}}(t\_{j-1}) = \vec{\mathcal{U}}\_{j-1}, \quad \vec{\mathcal{J}}(t\_{j-1}) = \vec{\mathcal{J}}\_{j-1}, \quad \text{in } \Omega, \end{cases} \tag{1}$$

where - *f* is a given function, *Rf* is the linear drag coefficient. Solving the problem on (*tj*−1, *tj*), we consider the functions *U* - *j*−1, *ξj*−<sup>1</sup> to be known. Detailed description of the notations and simplifications can be found, for example, in [18]. In the present work we consider the boundary condition for System (1) of the form [19]:

$$
\hbar H \overline{\Omega} \cdot \vec{n} + m\_{op} \sqrt{\text{g}} \overline{H} \overline{\xi} = m\_{op} \sqrt{\text{g}} \overline{H} d\_{\text{ss}} \qquad \text{on } \partial \Omega \times (t\_{j-1}, t\_j), \tag{2}
$$

where *mop* is the characteristic function of the "outer liquid" (open) boundary <sup>Γ</sup>*op*, i.e., *mop* = 1 if (*<sup>λ</sup>*, *θ*, *RE*) ∈ <sup>Γ</sup>*op*, *mop* = 0 otherwise. If the function *ds* is defined, Systems (1) and (2) are well posed. This boundary function will be considered below as an additional unknown.

Considering Systems (1) and (2) on (*tj*−1, *tj*), we introduce the following implicit scheme for time approximation:

$$\begin{cases} \frac{\dot{\Omega}^{j}}{\Delta t} + \begin{bmatrix} 0 & -\ell \\ \ell & 0 \end{bmatrix} \dot{\Omega}^{j} + R\_{f}\dot{\Omega}^{j} - g \cdot \mathbf{grad} \, \mathfrak{z}^{j} = \mathfrak{f}^{\mathfrak{f}} + \frac{\dot{\Omega}^{j-1}}{\Delta t}, & \text{in } \Omega, \\\ \frac{\dot{\mathfrak{z}}^{j}}{\Delta t} - \mathbf{div} \left( H\dot{\Omega}^{j} \right) = \frac{\mathfrak{z}^{j-1}}{\Delta t}, & \text{in } \Omega, \\\ H\dot{\mathcal{U}}^{j} \cdot \vec{n} + m\_{\partial P} \sqrt{gH} \xi^{j} = m\_{\partial P} \sqrt{gH} d\_{s}^{j}, & \text{on } \partial\Omega. \end{cases} \tag{3}$$

Hereafter the "semi-discrete" System (3) is the subject of the investigation, so for convenience the indices *j* will be omitted, i.e.: *U* - ≡ *U* - *j* , *ξ* ≡ *ξj*, ... We also introduce vectors ˜ - *f* = (*f j*1 + (*U*- *<sup>j</sup>*−<sup>1</sup>)1/Δ*t*, *f j*2 + (*U*- *<sup>j</sup>*−<sup>1</sup>)2/Δ*t*)*<sup>T</sup>*, ˜*f*3 = *ξj*−1/<sup>Δ</sup>*t*, ˜ - *f* ≡ ( ˜ *f*1, ˜ *f*2). Finally, System (3) can be written in the form:

$$\begin{cases} \frac{\vec{l}\vec{l}}{\Delta t} + \begin{bmatrix} 0 & -\ell \\ \ell & 0 \end{bmatrix} \left( \vec{\Omega} + R\_f \vec{\Omega} - \mathbf{g} \cdot \mathbf{grad} \, \xi = \vec{\tilde{f}}, \quad \text{in } \Omega, \\\frac{\vec{\xi}}{\Delta t} - \mathbf{div} \left( H \vec{\mathcal{U}} \right) = \vec{f}\_3, \quad \text{in } \Omega, \\\ H \vec{\mathcal{U}} \cdot \vec{\mathfrak{n}} + m\_{op} \sqrt{\mathcal{g} \mathcal{H}} \xi = m\_{op} \sqrt{\mathcal{g} \mathcal{H}} d\_{s\_{\prime}} \quad \text{on } \partial \Omega. \end{cases} \tag{4}$$

#### *2.2. Variational Data Assimilation and the Domain Decomposition Method*

We assume that the boundary *∂*Ω of the domain Ω is Lipschitz and piecewise *C*2–smooth. We also assume that Γ*in* ⊂ Ω is a hypersurface of class *C*<sup>2</sup> [28] and divides the domain Ω into two subdomains Ω1 and Ω2 without overlap, Ω ≡ Γ*in* ∪ Ω1 ∪ Ω2 (see Figure 1), Γ ¯ *in* ≡ (*∂*Ω1 ∩ *∂*Ω2), Γ*in* ⊂ *∂*Ω1 is open in *∂*Ω1, Γ*in* ⊂ *∂*Ω2 is open in *∂*Ω2. We suppose that *∂*Ω1 and *∂*Ω2 are Lipschitz and piecewise *C*2–smooth. Suppose that <sup>Γ</sup>*op* ("outer liquid", open boundary) does not intersect with Γ*in* ("inner liquid" boundary). Moreover, the essential assumption is that *ρ*(<sup>Γ</sup>*in*, <sup>Γ</sup>*op*) = inf *<sup>x</sup>*∈Γ*in*,*y*∈Γ*opx* − *<sup>y</sup><sup>R</sup>*2 *g<sup>H</sup>* · Δ*t*.

Below we use the index *i* = 1, 2 to indicate the solutions in subdomains (*i* = 1, 2). The problem (4) can be written for the functions *U* - (1) , *ξ*(1) in the subdomain Ω1, for the functions *U*- (2), *ξ*(2) in the

subdomain Ω2. Suppose *ni* is the outer normal to the boundary *∂*Ω*i* of the domain Ω*i*, *i* = 1, 2. We will require the fulfillment of the boundary conditions on the inner liquid boundary Γ*in*:

$$\mathfrak{g}^{(1)} = \mathfrak{g}^{(2)}, \quad H\mathcal{U}\_n^{(1)} = -H\mathcal{U}\_n^{(2)}, \tag{5}$$

where *U*(*i*) *n* = *U*- (*i*) · *ni*, *i* = 1, 2.

**Figure 1.** Domain with "inner" Γ*in* and "outer" <sup>Γ</sup>*op* liquid boundaries.

An additional unknown function *v* is defined as:

$$H\underline{U}\_{\text{H}}^{(1)} = \sqrt{\mathcal{g}H}v\_{\prime} \qquad \text{on } \Gamma\_{\text{im}}.$$

We obtain the following form of the boundary conditions for the problem in the subdomain Ω1:

$$H\overrightarrow{\mathcal{U}}^{(1)}\cdot\vec{n}\_1 + m\_{\mathcal{op}}\sqrt{gH}\overrightarrow{\mathcal{Y}}^{(1)} = m\_{\mathcal{op}}\sqrt{gH}d\_{\mathfrak{s}} + m\_{\text{in}}\sqrt{gH}v,\quad\text{on }\partial\Omega\_{1\prime}\tag{6}$$

and for the problem in the subdomain Ω2:

$$H\mathcal{U}\_n^{(2)} = -m\_{in}\sqrt{gH}v,\quad\text{on }\partial\Omega\_2. \tag{7}$$

Suppose there are preprocessed data of sea level anomaly measurements *ξobs* along <sup>Γ</sup>*op*. Such data can be obtained from observations or direct measurements (satellite altimetry data, "in-situ" data), or from the simulation using a global model with a coarser grid. In any cases, these data contain errors, so we cannot use *ξobs* directly in the OBC. We introduce an additional condition (closure condition):

$$
\xi = \xi\_{obs\star} \quad \text{ on } \Gamma\_{op}. \tag{8}
$$

From System (4) in each subdomain (*i* = 1, 2) we obtain:

$$\begin{cases} \frac{\partial \mathcal{I}^{(i)}}{\Delta t} + \begin{bmatrix} 0 & -\ell \\ \ell & 0 \end{bmatrix} \, \mathrm{d}^{(i)} + R\_f \mathrm{d}^{(i)} - \mathrm{g} \cdot \mathbf{grad} \, \mathfrak{E}^{(i)} = \tilde{f}, & \text{in } \Omega\_i, \\\frac{\mathcal{E}^{(i)}}{\Delta t} - \mathrm{div} \left( H \dot{\mathcal{U}}^{(i)} \right) = \bar{f}\_3, & \text{in } \Omega\_i, \\H \dot{\mathcal{U}}^{(1)} \cdot \vec{n}\_1 + m\_{op} \sqrt{gH} \xi^{(1)} = m\_{op} \sqrt{gH} d\_s + m\_{in} \sqrt{gH} v, & \text{on } \partial \Omega\_1, \\H \dot{\mathcal{U}}^{(2)} \cdot \vec{n}\_2 = -m\_{in} \sqrt{gH} v, & \text{on } \partial \Omega\_2. \end{cases} \tag{9}$$

We formulate the *inverse problem of restoring the boundary functions on the "outer" and "inner" liquid boundaries* as follows: find the vector functions of solutions in the subdomains Φ(*i*) = (*U*- (*i*), *ξ*(*i*))*<sup>T</sup>*, *i* = 1, 2, and additional unknown boundary functions *v*, *ds*, satisfying the systems of Equation (9) and additional conditions of Equations (5) and (8).

Consider the Hilbert space **H**(*i*) 0 of vector functions Φ(*i*) = (*U*- (*i*), *ξ*(*i*))*<sup>T</sup>*, *U*- (*i*) ∈ (*<sup>L</sup>*2(<sup>Ω</sup>*i*))2, *ξ*(*i*) ∈ *<sup>L</sup>*2(<sup>Ω</sup>*i*) with the scalar product:

$$\left(\Phi^{(i)}, \Phi^{(i)}\right)\_{\mathbf{H}\_0^{(i)}} = \int\_{\Omega\_i} \left[ H(\vec{\mathcal{U}}^{(i)} \cdot \hat{\mathcal{U}}^{(i)}) + \mathfrak{g}^{\mathfrak{x}^{(i)}} \mathfrak{f}^{(i)} \right] d\Omega\_{\mathbf{H}\_0}$$

By **W**(*i*) we denote a space of vector functions Φ ∈ (*<sup>L</sup>*2(<sup>Ω</sup>*i*))<sup>2</sup> × *W*12 (<sup>Ω</sup>*i*). Let *ξobs* ∈ *<sup>L</sup>*2(<sup>Γ</sup>*op*), ˜ - *f* ∈ (*<sup>L</sup>*2(Ω))2, ˜*f*3 ∈ *<sup>L</sup>*2(Ω), *H* ∈ *<sup>C</sup>*<sup>1</sup>(Ω).

The scalar product of (9) and Φ ˆ ∈ **W**(*i*) in **H**(*i*) 0 , *i* = 1, 2 is:

$$a\_1(\Phi^{(1)}, \dot{\Phi}^{(1)}) = f\_1(\dot{\Phi}^{(1)}) + b\_{op}(d\_\circ, \dot{\Phi}^{(1)}) + b\_1(v, \dot{\Phi}^{(1)}),\tag{10}$$

$$a\_2(\Phi^{(2)}, \dot{\Phi}^{(2)}) = f\_2(\dot{\Phi}^{(2)}) - b\_2(v, \dot{\Phi}^{(2)}),\tag{11}$$

where

*<sup>a</sup>*1(Φ(1), Φˆ (1)) = Ω1 *HU*- (1) ˆ*U*- (1)+*gξ*(1) ˆ*ξ*(1) Δ*t* + *Rf U*- (1) ˆ*U*- (1) + *lH*((*U*- (1))1( ˆ*U*- (1))2 − (*U*- (1))2( ˆ*U*- (1))1) *d*Ω+ + Ω1 (*gHU*- (1) **grad** ˆ*ξ*(1) − *gH* ˆ *U* - (1) **grad** *ξ*(1)) *d*Ω + <sup>Γ</sup>*op g* ˆ *ξ*(1) *gHξ*(1) *d*Γ, *<sup>a</sup>*2(Φ(2), Φˆ (2)) = Ω2 *HU*- (2) ˆ*U*- (2)+*gξ*(2) ˆ*ξ*(2) Δ*t* + *Rf U*- (2) ˆ*U*- (2) + *lH*((*U*- (2))1( ˆ*U*- (2))2 − (*U*- (2))2( ˆ*U*- (2))1) *d*Ω+ + Ω2 (*gHU*- (2) **grad** ˆ*ξ*(2) − *gH* ˆ *U* - (2) **grad** *ξ*(2)) *d*Ω, *bop*(*ds*, Φˆ (1)) = <sup>Γ</sup>*op gHdsg* ˆ*ξ*(1) *d*Γ, *bi*(*<sup>v</sup>*, Φˆ (*i*)) = Γ*in gHvg* ˆ*ξ*(*i*) *d*Γ, *i* = 1, 2, *fi*(Φ<sup>ˆ</sup> (*i*)) = [*H* ˜ - *f* · ˆ *U* - + *g* ˜ *f*3ˆ *ξ*] *d*Ω, *i* = 1, 2.

Let us introduce the Hilbert space **H***c* of vector-functions *u* = (*ds*, *<sup>v</sup>*)*<sup>T</sup>*, *ds* ∈ *<sup>L</sup>W*2 (<sup>Γ</sup>*op*), *v* ∈ *<sup>L</sup>W*2 (<sup>Γ</sup>*in*), with the norm *u***<sup>H</sup>***c* = *ds*<sup>2</sup>*LW*2 (<sup>Γ</sup>*op*) + *v*<sup>2</sup>*LW*2 (<sup>Γ</sup>*in*), where *<sup>L</sup>W*2 (Γ) is the space of functions from *<sup>L</sup>*2(Γ) with the "weighted" scalar product: (·, ·)*LW*2 (Γ) = ( *<sup>g</sup>H*·, ·)*<sup>L</sup>*2(Γ). We also introduce the space **H***ob* = *<sup>L</sup>W*2 (<sup>Γ</sup>*op*) × *<sup>L</sup>W*2 (<sup>Γ</sup>*in*).

Ω*i*

We formulate the problem in a weak form: find Φ(*i*) ∈ **<sup>W</sup>**(*i*), *u* ∈ **H***c* satisfying the conditions of Equations (10) and (11) and also the conditions of Equations (5) and (8) (in the sense of equality almost everywhere on Γ*in*, <sup>Γ</sup>*op*, respectively) ∀Φ ˆ ∈ **W**(*i*).

In order to formulate problems in operator form to study the problem theoretically the weak formulation should be modified. To execute this, we use the procedure similar to that described in [19]. From Equation (9) we receive:

$$(\vec{\mathcal{U}})\_1 = g \frac{\partial \xi}{\partial x} \frac{\theta}{\theta^2 + l^2} + g \frac{\partial \xi}{\partial y} \frac{l}{\theta^2 + l^2} + \tilde{\vec{f}}\_{1,1}$$

$$(\vec{\mathcal{U}})\_2 = g \frac{\partial \xi}{\partial y} \frac{\theta}{\theta^2 + l^2} - g \frac{\partial \xi}{\partial x} \frac{l}{\theta^2 + l^2} + \tilde{f}\_{2,1}$$

*J. Mar. Sci. Eng.* **2019**, *7*, 195

where

$$\begin{aligned} \theta &= 1/\Delta t + R\_{f'} & \bar{f}\_1 &= f\_1 \frac{\theta}{\theta^2 + l^2} + f\_2 \frac{l}{\theta^2 + l^2}, & \bar{f}\_2 &= f\_2 \frac{\theta}{\theta^2 + l^2} - f\_1 \frac{l}{\theta^2 + l^2}, \\ \text{fix } M &\text{we denote:} \end{aligned}$$

By matrix we 

$$M = \left[ \begin{array}{cc} a & b \\ -b & a \end{array} \right] \text{-} $$

where

$$a = \frac{\theta}{\theta^2 + l^2}, \quad b = \frac{l}{\theta^2 + l^2}.$$

Substituting *U* - in the weak formulation of the problem we receive:

$$\vec{a}\_1(\vec{\xi}^{(1)}, \hat{\xi}^{(1)}) = \vec{f}\_1(\hat{\xi}^{(1)}) + \vec{b}\_{op}(d\_{\mathfrak{s}\_\tau} \hat{\xi}^{(1)}) + \tilde{b}\_1(v, \hat{\xi}^{(1)}),\tag{12}$$

$$
\vec{a}\_2(\xi^{(2)}, \widehat{\xi}^{(2)}) = \vec{f}\_2(\widehat{\xi}^{(2)}) - \vec{b}\_2(v, \widehat{\xi}^{(2)}), \tag{13}
$$

.

where

*a*

˜

$$
\delta\_1^{(1)}(\xi^{(1)}, \xi^{(1)}) = \int\_{\Omega\_1} \left[ \xi^{(1)} \xi^{(1)} / \Delta t + \operatorname{gH} \cdot M \operatorname{\mathbf{grad}} \, \xi^{(1)} \cdot \operatorname{\mathbf{grad}} \, \xi^{(1)} \right] d\Omega + \int\_{\sigma\_P} \xi^{(1)} \sqrt{\operatorname{gH}} \xi^{(1)} \, d\Gamma\_{\sigma}
$$

$$
\delta\_2(\xi^{(2)}, \xi^{(2)}) = \int\_{\Omega\_2} \left[ \xi^{(2)} \xi^{(2)} / \Delta t + \operatorname{gH} \cdot M \operatorname{\mathbf{grad}} \, \xi^{(2)} \cdot \operatorname{\mathbf{grad}} \, \xi^{(2)} \right] d\Omega.
$$

$$
\tilde{b}\_{\Psi p}(d\_\sigma, \hat{\xi}^{(1)}) = \int\_{\Gamma\_{\Psi p}} \sqrt{\operatorname{gH}} d\_{\hat{\xi}} \xi^{(1)} \, d\Gamma\_{\sigma} \tag{14}
$$

$$
\Gamma^{(1)} = \int\_{\Gamma\_{\Psi p}} \xi^{(1)} \, d\Gamma\_{\Psi p} \tag{15}
$$

$$\tilde{b}\_{i}(\upsilon\_{\ast}\tilde{\mathfrak{E}}^{(i)}) = \int\_{\Gamma\_{in}} \sqrt{gH} \upsilon\_{\ast}^{\sharp(i)} \, d\Gamma\_{\ast} \quad i = 1, 2,\tag{15}$$

$$\tilde{f}\_{i}(\xi^{(i)}) = \int\_{\Omega\_{i}} [\tilde{f}\_{3}\xi - H\mathcal{M}\tilde{f} \cdot \mathbf{grad}\,\xi] \, d\Omega\_{\ast} \quad i = 1, 2.$$

Let us introduce the following notation:

˜

$$\mathcal{W}^1\_2(\Omega\_i) \equiv \mathbf{V}^{(i)}$$

The bilinear forms *<sup>a</sup>*˜*i*(*ξ*(*i*), ˆ*ξ*(*i*)), *i* = 1, 2 are bounded and positive definite for functions *ξ*(*i*), ˆ *ξ*(*i*) ∈ **V**(*i*). Equations (12) and (13) can be formulated in the following form [26]:

$$L\_1 \mathfrak{J}^{(1)} = \hat{f}\_1 + B\_{op} d\_{\mathfrak{s}} + B\_1 \upsilon\_\prime \tag{16}$$

$$L\_2 \mathfrak{F}^{(2)} = \hat{f}\_2 - B\_2 v\_\* \tag{17}$$

where the operators *Li* : **V**(*i*) → (**V**(*i*)), *Bi* : *<sup>L</sup>W*2 (<sup>Γ</sup>*in*) → (**V**(*i*)), *Bop* : *<sup>L</sup>W*2 (<sup>Γ</sup>*op*) → (**V**(*i*)) (the space (**V**(*i*)) is the dual of **V**(*i*)) are introduced using the bilinear forms *<sup>a</sup>*˜*i*(*ξ*(1), ˆ*ξ*(1)), ˜*bi*(*<sup>v</sup>*, ˆ*ξ*(*i*)), ˜*bop*(*ds*, ˆ*ξ*(*i*)), *i* = 1, 2, respectively [29], ˆ *fi* ∈ (**V**(*i*)). The adjoint operators may also be introduced, so that the following identity is satisfied (*i* = 1, 2):

$$
\tilde{a}\_i(\mathfrak{z}^{(i)}, \mathfrak{f}^{(i)}) = (L\_i \mathfrak{z}^{(i)}, \mathfrak{f}^{(i)}) = (\mathfrak{z}^{(i)}, L\_i^\* \mathfrak{f}^{(i)}).\tag{18}
$$

Note, that (*Liξ*(*i*)) ∈ (**V**(*i*)), ˆ*ξ*(*i*) ∈ **V**(*i*) and (·, ·) means their scalar product. We obtain the same equalities for the bilinear forms *bi*(*<sup>v</sup>*, ˆ*ξ*(*i*)), *bop*(*ds*, ˆ*ξ*(*i*)):

$$b\_{\hat{\imath}}(\upsilon\_{\prime}\hat{\mathfrak{f}}^{(i)}) = (B\_{\hat{\imath}}\upsilon\_{\prime}\hat{\mathfrak{f}}^{(i)}) = (\upsilon\_{\prime}B\_{\hat{\imath}}^{\*}\hat{\mathfrak{f}}^{(i)})\_{\underline{L}\_{2}^{\vert\mathbb{N}}(\Gamma\_{\hat{m}})^{\prime}} \quad b\_{\upsilon\rho}(d\_{\delta\prime}\hat{\mathfrak{f}}^{(i)}) = (B\_{\upsilon\rho}d\_{\delta\prime}\hat{\mathfrak{f}}^{(i)}) = (d\_{\delta\prime}B\_{\hat{\imath}\rho}^{\*}\hat{\mathfrak{f}}^{(i)})\_{\underline{L}\_{2}^{\vert\mathbb{N}}(\Gamma\_{\hat{\imath}\rho})}.\tag{19}$$

It could be shown that the operators *Bi*, *Bop* are bounded, there exist operators *L*−<sup>1</sup> *i* which are bounded (*i* = 1, 2) [13,29].

The closure conditions of Equations (5) and (8) are formulated in the form:

$$
\mathcal{L}\_{obs}\mathfrak{J}^{(1)} = \mathfrak{J}\_{obs\prime} \tag{20}
$$

$$\mathcal{C}\_1 \mathfrak{z}^{(1)} = \mathcal{C}\_2 \mathfrak{z}^{(2)},\tag{21}$$

where *Cobsξ* = *ξ*|<sup>Γ</sup>*op* ∀*ξ* ∈ **V**(1) is a trace operator on <sup>Γ</sup>*op*, *Cobs* : **V**(1) → *<sup>L</sup>W*2 (<sup>Γ</sup>*op*), *Ciξ*(*i*) = *ξ*(*i*)|<sup>Γ</sup>*in* ∀*ξ*(*i*) ∈ **<sup>V</sup>**(*i*), *i* = 1, 2, are trace operators on Γ*in*, *Ci* : **V**(*i*) → *<sup>L</sup>W*2 (<sup>Γ</sup>*in*).

Now the *weak formulation of the inverse problem* is equivalent to the following: find *ξ*(*i*), *ds*, *v* satisfying Equations (16), (17), (20) and (21).

The system of Equations (16), (17), (20) and (21) can be rewritten in the operator form:

$$Au = \varphi,\tag{22}$$

where

$$A = \begin{bmatrix} \mathcal{C}\_{obs} L\_1^{-1} B\_{op} & \mathcal{C}\_{obs} L\_1^{-1} B\_1 \\ \mathcal{C}\_1 L\_1^{-1} B\_{op} & \mathcal{C}\_1 L\_1^{-1} B\_1 + \mathcal{C}\_2 L\_2^{-1} B\_2 \end{bmatrix} / \cdot,$$

$$\boldsymbol{\mathcal{P}} = \begin{bmatrix} \mathcal{I}\_{obs} - \mathcal{C}\_{obs} L\_1^{-1} \hat{f}\_1 \\ \mathcal{C}\_2 L\_2^{-1} \hat{f}\_2 - \mathcal{C}\_1 L\_1^{-1} \hat{f}\_1 \end{bmatrix}.$$

Finally, the original problem is reduced to a single operator Equation (22), for which many results of the general theory of operator equations [26,27,30] are applicable. In the current study the methodology based on the methods of optimal control and adjoint equations is used [26].

#### *2.3. Optimal Control Problem*

Note that the operator *A* is bounded: *Au***<sup>H</sup>***ob* ≤ *cAu***<sup>H</sup>***c* , *cA* = *const* < ∞. However Equation (22) may be ill-posed, since the non-smooth observational data *ξobs* ∈ *<sup>L</sup>*2(<sup>Γ</sup>*op*) are used in setting the function *ϕ*, so *ϕ* ∈/ *<sup>R</sup>*(*A*). In this regard let us move on to a *generalized formulation of Equation* (22): find *u* ∈ **H***c* minimizing *J*0(*u*) = 1/2*Au* − *<sup>ϕ</sup>*<sup>2</sup>**H***ob*.

We formulate the class of optimal control problems [26]: find boundary functions *u* = (*dt*, *v*)*<sup>T</sup>* ∈ **H***<sup>c</sup>*, minimizing the functional *Jα*(*u*):

$$J\_a(u) = \frac{1}{2} a \| |u| \|\_{\mathbf{H}\_c}^2 + \frac{1}{2} \| Au - \varphi \|\_{\mathbf{H}\_{ab'}}^2 \text{ a} \ge 0. \tag{23}$$

Note that the necessary condition for the minimum of the functional *Jα*(*u*) has the form of an equation in terms of Tikhonov regularization method, where *α* is a regularization parameter. Henceforth, we will call *α* the *regularization parameter*. If *α* > 0, functional defined in Equation (23) is strictly convex, strongly convex and has a unique global minimum *J*∗ = *J*∗*α* (*u*<sup>∗</sup>(*α*)). If *ϕ* ∈ *R*(*A*) and the solution of the inverse problem is unique, *u*<sup>∗</sup>(*α*) tends to the solution when *α* → +0. Note also that for *α* = 0 the optimal control problem is equivalent to the generalized formulation of Equation (22).

*J. Mar. Sci. Eng.* **2019**, *7*, 195

#### **3. Uniqueness of the Solution**

Let the kernel of the operator *A* consist of not only a single zero element. In this case, there exists *u* = (*ds*, *v*)*<sup>T</sup>* = 0 and *Au* = 0. This condition is equivalent to the existence of a nonzero weak solution of the problem:

$$\begin{cases} \frac{\tilde{\mathcal{U}}^{(i)}}{\Delta t} + \begin{bmatrix} 0 & -\ell \\ \ell & 0 \end{bmatrix} \tilde{\mathcal{U}}^{(i)} + R\_f \tilde{\mathcal{U}}^{(i)} - \mathcal{g} \cdot \mathbf{grad}\_{\tilde{\mathcal{G}}} \tilde{\mathcal{V}}^{(i)} = 0, & \text{in } \Omega\_{i\prime} \\ \frac{\tilde{\mathcal{G}}^{(i)}}{\Delta t} - \text{div} \left( H \tilde{\mathcal{U}}^{(i)} \right) = 0, & \text{in } \Omega\_{i\prime} \\ H \tilde{\mathcal{U}}^{(1)} \cdot \vec{\mathcal{m}}\_1 + m\_{op} \sqrt{gH} \tilde{\mathcal{S}}^{(1)} = m\_{op} \sqrt{gH} d\_s + m\_{in} \sqrt{gH} v, & \text{on } \partial\Omega\_{1\prime} \\ \tilde{\mathcal{F}}^{(1)} = 0, & \text{on } \Gamma\_{op\prime} \\ H \tilde{\mathcal{U}}^{(2)} \cdot \vec{\mathcal{m}}\_2 = -m\_{in} \sqrt{gH} v, & \text{on } \partial\Omega\_{2\prime} \end{cases}$$

where *i* = 1, 2. Note that the functions ˜ *ξ* = { ˜*ξ*(*i*) in Ω*i*, *i* = 1, <sup>2</sup>}, *U* - = { *U* - (*i*) in Ω*i*, *i* = 1, 2} are the solution of a homogenous boundary-value problem in Ω with mixed boundary conditions. So, we obtain: ˜ *ξ*(*i*) = 0, ˜ *U* - (*i*) = 0 and *ds* = 0, *v* = 0. Hence, *u* = 0 and *ker*(*A*) = {0}, i.e., *the problem considered may have only a unique solution*.
