Proof. The construction of this solution is inspired by Thacker's work (Thacker Reference Thacker1981). In order to extend the Thacker solution, we first observe that

(3.8)\begin{equation} \int_{z_b}^{z_b+h} \varphi(t,x,y) \left(z-z_b - \frac{h}{2}\right) \textrm{d}z = 0, \end{equation}

for all function $\varphi (t,x,y)$. Thus, if we add a term having the form

(3.9)\begin{equation} \varphi(t,x,y) \left(z-z_b - \frac{h}{2}\right), \end{equation}

to the components of the horizontal velocity field, this does not modify a large number of the properties such as the mass conservation. However, some adjustments are necessary to meet all the requirements imposed by (2.1) and (2.2) and allowing us to define the expression of $\varphi (t,x,y)$. Asymptotically, when $\beta$ goes to zero, the proposed analytical solution converges to the one proposed by Thacker.

The proof of Proposition 3.1 is based on the verification that the proposed expressions for $h,u,v,w$ and $p$ are solutions of the equations

(3.10)\begin{gather} \frac{\partial h}{\partial t}+ \frac{\partial}{\partial x} \int_{z_b}^{z_b+h} u\,\textrm{d} z+ \frac{\partial}{\partial y} \int_{z_b}^{z_b+h} v \,\textrm{d} z=0, \end{gather}

(3.11)\begin{gather}\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+ g \frac{\partial (h+z_b)}{\partial x}=0, \end{gather}

(3.12)\begin{gather}\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}+ g \frac{\partial (h+z_b)}{\partial x}=0, \end{gather}

where (3.10) comes from an integration over the vertical axis of the divergence free condition coupled with the two kinematic boundary conditions (2.4) and (2.5).

The details of the computation are given only for (3.10). A more complete computation will be carried out for the two-dimensional case in the proof of Corollary 3.2.

From (3.2) and the definition of $f(z)$ given in (3.7), we can verify that, on the set where $h(t,x,y)>0$, we have

(3.13)\begin{equation} \frac{\partial h}{\partial t} = \frac{\gamma \omega \sin(\omega t)}{(\gamma \cos(\omega t)-1)^2}f'(), \end{equation}

with the notation $f() =f({r^2}/({\gamma \cos (\omega t)-1}))$. From (3.4), we obtain

(3.14a,b)\begin{equation} \int_{z_b}^{z_b+h} u\,\textrm{d}z =\frac{hx\omega \gamma \sin(\omega t)}{2(1-\gamma \cos(\omega t))} , \quad \int_{z_b}^{z_b+h} v \,\textrm{d}z =\frac{hy\omega \gamma \sin(\omega t)}{2(1-\gamma \cos(\omega t))}. \end{equation}

Since

(3.15)\begin{equation} \frac{\partial h}{\partial x} ={-} \frac{2 x f()}{r^4}+ \frac{2 x f'()}{r^2 (\gamma \cos(\omega t) - 1))}, \end{equation}

we have

(3.16)\begin{equation} \frac{\partial}{\partial x} \int_{z_b}^{z_b+h} u \, \textrm{d} z =\frac{\omega \gamma \sin(\omega t)}{2(1-\gamma \cos (\omega t))}\left[ \frac{f()}{r^2} -\frac{2 x^2 f()}{r^4}+\frac{2 x^2 f'()}{r^2}\right], \end{equation}

and

(3.17)\begin{equation} \frac{\partial}{\partial y} \int_{z_b}^{z_b+h} v \, \textrm{d} z =\frac{\omega \gamma \sin(\omega t)}{2(1-\gamma \cos (\omega t))}\left[ \frac{f()}{r^2} -\frac{2 y^2 f()}{r^4}+\frac{2 y^2 f'()}{r^2}\right]. \end{equation}

We deduce that

(3.18)\begin{equation} \frac{\partial}{\partial x} \int_{z_b}^{z_b+h} u \, \textrm{d} z+ \frac{\partial}{\partial y} \int_{z_b}^{z_b+h} v\, \textrm{d} z = \frac{\omega \gamma \sin(\omega t)}{(1-\gamma \cos (\omega t))^2} f'(), \end{equation}

and (3.10) is satisfied.

Proof. It is enough to verify that the expressions for $u$, $w$ and $h$ are solutions of the following equations:

(3.25)\begin{gather} \frac{\partial h}{\partial t}+ \frac{\partial}{\partial x} \int_{z_b}^{z_b+h} u(x,z) \,\textrm{d} z=0, \end{gather}

(3.26)\begin{gather}\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+w\frac{\partial u}{\partial z}+ g \frac{\partial (h+z_b)}{\partial x}=0, \end{gather}

where $w$ is then given thanks to the incompressibility condition by

(3.27)\begin{equation} w(t,x,z) ={-}\frac{\partial}{\partial x}\int_{z_b}^z u\,\textrm{d}z. \end{equation}

From (3.19) and the definition of $f(z)$ given in (3.24), we can verify that, on the set where $h(t,x)>0$, we have

(3.28)\begin{equation} \frac{\partial h}{\partial t} ={-}\gamma \cos(\omega t) f'\left(x-\frac{\gamma}{\omega} \sin(\omega t) \right). \end{equation}

From (3.20) and using (3.8) we find

(3.29)\begin{equation} \int_{z_b}^{z_b+h} u(x,z) \, \textrm{d} z = h \gamma \cos(\omega t). \end{equation}

And since

(3.30)\begin{equation} \frac{\partial h}{\partial x} =f'\left(x-\frac{\gamma}{\omega} \sin(\omega t) \right), \end{equation}

Equation (3.25) is satisfied.

In order to verify (3.26), we start from the definition of $u(t,x,z)$ leading to

(3.31)\begin{equation} \frac{\partial u}{\partial t} = \frac{\beta \gamma \cos (\omega t)}{2} f'() -\gamma \omega \sin(\omega t), \end{equation}

and

(3.32)\begin{equation} u\frac{\partial u}{\partial x} ={-}\beta \left( 4 \alpha x + \frac{1}{2} f'() \right) \left[\beta \left( z-2\alpha x^2-\frac{1}{2}f()\right)+\gamma \cos(\omega t)\right], \end{equation}

with the notation $f()= f(x-({\gamma }/{\omega }) \sin \omega t)$. Moreover, we have

(3.33)\begin{equation} w\frac{\partial u}{\partial z} = \beta\left(4 \alpha \beta x z - 8 \alpha^2 \beta x^3 - 2\alpha \beta x f() + \beta \left(\frac{z}{2}-\alpha x^2\right) f'()+4\alpha \gamma x \cos(\omega t)\right). \end{equation}

And the pressure term gives

(3.34)\begin{equation} g \frac{\partial (h + z_b)}{\partial x} = gf'()+4 g \alpha x. \end{equation}

Summing the last four expressions, all the terms without $f()$ or $f'()$ are equal to $4 g \alpha x$ whereas the terms containing $f()$ only are equal to zero. The terms containing only $f'()$ are equal to $g-({\beta \gamma }/{2})\cos \omega t$, and the terms containing $f()\cdot f'()$ equal ${\beta ^2}/{4}$. Now, we have to observe that (3.24) gives

(3.35)\begin{equation} f'\left(x-\frac{\gamma}{\omega} \sin(\omega t) \right) ={-}\frac{2 \omega(\omega x - \gamma \sin(\omega t))}{\sqrt{4 g^2 - 2 c \beta^2 - \beta^2 (\omega^2 x-2\gamma \omega \sin(\omega t)+ \gamma^2 \sin^2(\omega t))}}. \end{equation}

Since

(3.36)\begin{equation} f(\xi)\cdot f'(\xi) ={-}\frac{4 g}{\beta^2} f'()-\frac{4 \omega^2}{\beta^2}\xi, \end{equation}

the expressions for $h,u$ and $w$ inserted in (3.26) lead to $2 g\alpha x - \omega ^2 x$, and this term vanishes under the condition $\omega =\sqrt {2 g \alpha }$.

Proof. Starting from Thacker's solution with constant density (Thacker Reference Thacker1981), we have only to verify that the chosen expression for the density $\rho (t,x,y,z) = a(\eta (t,x,y)-z)$ does not modify this solution. We have $\partial _z p = - \rho (t,x,y,z) g$ due to the hydrostatic hypothesis. If we note $\eta =\eta (t,x,y) = h(t,x,y)+z_b(x,y)$, $\rho (t,x,y,z) = \phi (\eta (t,x,y)-z)$ and $p(t,x,y,\eta (t,x,y)) = p^a(t)$, for $\phi \in C^1({\mathbb {R}})$ we have

(3.38)\begin{align} \boldsymbol{\nabla}_{x,y} p &= g\boldsymbol{\nabla}_{x,y} \int_z^{\eta} \rho(t,x,y,\xi) \,\textrm{d}\xi\nonumber\\ &= g\boldsymbol{\nabla}_{x,y} \int_z^{\eta} \phi(\eta-\xi) \,\textrm{d}\xi\nonumber\\ &=g \int_z^{\eta} \boldsymbol{\nabla}_{x,y} \phi(\eta-\xi) \,\textrm{d}\xi +g\phi(0) \boldsymbol{\nabla}_{x,y} \eta \nonumber\\ &=g \int_z^{\eta} \phi'(\eta-\xi) \boldsymbol{\nabla}_{x,y} \eta \,\textrm{d}\xi +g \phi(0)\boldsymbol{\nabla}_{x,y} \eta \nonumber\\ &=g \boldsymbol{\nabla}_{x,y} \eta \int_{\eta-z}^0 \phi'(s) \,\textrm{d}s +g \phi(0) \boldsymbol{\nabla}_{x,y} \eta \quad \text{with}\ s=\eta -\xi\nonumber\\ &=g \phi(\eta-z) \boldsymbol{\nabla}_{x,y} \eta \nonumber\\ &=g \rho(t,x,y,z) \boldsymbol{\nabla}_{x,y} \eta. \end{align}

If we consider the momentum equation (2.15), using the previous evaluation of the pressure term, we can conclude that $u$ is the solution of (2.9) that corresponds to the momentum equation with constant density.

But generally, this expression for the density does not verify the conservation equation (2.14)

(3.39)\begin{equation} \partial_t \rho + \boldsymbol{\nabla}\boldsymbol{\cdot} (\rho U) =0. \end{equation}

In the proposed analytical solution, since $u$ and $w$ are independent of $x$ and $z$, then we have

(3.40)\begin{align} \partial_t \rho + \boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla}_{x,y} \rho + w \partial_z \rho &= \phi'(\eta-z) \partial_t \eta + \phi'(\eta-z) \boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla}_{x,y} \eta - \phi'(\eta-z) w \nonumber\\ &=\phi'(\eta-z) (\partial_t \eta + \boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla}_{x,y} \eta - w ) =0, \end{align}

since at the free surface the kinematic boundary condition gives ${\partial _t \eta + \boldsymbol {u}\boldsymbol {\cdot }\boldsymbol {\nabla }_{x,y} \eta - w = 0}$.

Proof. The proof is the same as for Proposition 3.3 and is not detailed here.

Proof. The proof relies on simple computations that are detailed hereafter. For the sake of simplicity, the computations are carried out in two dimensions $(x,z)$ corresponding to $\theta = 0$.

Assuming $u=\bar {u}(t,x)$, the divergence free condition (2.1) coupled with (2.5) gives

(4.3)\begin{equation} w ={-}z\frac{\partial \bar{u}}{\partial x} + \frac{\partial (z_b\bar{u})}{\partial x}, \end{equation}

and the two components of (2.2) give

(4.4)\begin{gather} \frac{\partial \bar{u}}{\partial t} + \bar{u}\frac{\partial \bar{u}}{\partial x} + \frac{\partial p}{\partial x}=0, \end{gather}

(4.5)\begin{gather}-z\left( \frac{\partial^2 \bar{u}}{\partial x\partial t} + \bar{u} \frac{\partial^2 \bar{u}}{\partial x^2} - \left(\frac{\partial \bar{u}}{\partial x}\right)^2\right) + \frac{\partial^2 (z_b\bar{u})}{\partial x\partial t} + \bar{u}\frac{\partial^2 (z_b\bar{u})}{\partial x^2} - \frac{\partial \bar{u}}{\partial x}\frac{\partial (z_b\bar{u})}{\partial x} + \frac{\partial p}{\partial z} ={-}g. \end{gather}

From (4.4), (4.5) and (2.6), we deduce that the pressure $p$ satisfies

(4.6a,b)\begin{equation} p(t,x,\eta)=p^{a,0}(t),\quad\mbox{and}\quad\frac{\partial^2 p}{\partial x\partial z} = \frac{\partial^3 p}{\partial z^3}=0, \end{equation}

therefore, the pressure $p$ has necessarily the form

(4.7)\begin{equation} \bar{p} = p^{a,0}(t) + \frac{a(t)}{2} \left( \eta^2 - z^2 \right) + b(t) \left( \eta - z \right), \end{equation}

where $a=a(t)$ and $b=b(t)$ are two functions to be determined.

Hence, the shallow water solutions of the incompressible Euler system with a free surface are characterized by

(4.8)\begin{gather} \frac{\partial h}{\partial t} + \frac{\partial (h\bar{u})}{\partial x} = 0, \end{gather}

(4.9)\begin{gather}\frac{\partial \bar{u}}{\partial t} + \bar{u}\frac{\partial \bar{u}}{\partial x} + \left(a(t)\eta + b(t)\right)\frac{\partial \eta}{\partial x} = 0, \end{gather}

(4.10)\begin{gather}\frac{\partial^2 \bar{u}}{\partial x\partial t} + \bar{u}\frac{\partial^2 \bar{u}}{\partial x^2} - \left(\frac{\partial \bar{u}}{\partial x}\right)^2 ={-}a(t), \end{gather}

(4.11)\begin{gather}\frac{\partial^2 (z_b\bar{u})}{\partial x\partial t} + \bar{u}\frac{\partial^2 (z_b\bar{u})}{\partial x^2} - \frac{\partial \bar{u}}{\partial x}\frac{\partial (z_b\bar{u})}{\partial x} = b(t)-g. \end{gather}

Now, subtracting from (4.10) the derivative of (4.9) with respect to the variable $x$ gives

(4.12)\begin{equation} 2 \left(\frac{\partial \bar{u}}{\partial x}\right)^2 = a(t) - \frac{\partial}{\partial x}\left(\left(a(t)\eta + b(t)\right)\frac{\partial \eta}{\partial x}\right). \end{equation}

Likewise, subtracting (4.10) multiplied by $z_b$ from (4.11) gives

(4.13)\begin{equation} \frac{\partial \bar{u}}{\partial t} \frac{\partial z_b}{\partial x} + \bar{u}^2\frac{\partial^2 z_b}{\partial x^2} + \bar{u}\frac{\partial \bar{u}}{\partial x}\frac{\partial z_b}{\partial x} + a(t)z_b = b(t)-g. \end{equation}

The solution $\eta =\eta (t,x)$ of (4.9) can be obtained explicitly if we assume $a(t)>0$ and then $h(t,x)$ is given by

(4.14)\begin{equation} h(t,x) ={-}z_b(x)-\frac{b(t)}{a(t)} + \frac{1}{a(t)}\sqrt{ a(t) \left( F_1(t) - \bar{u}^2 - 2\int^x \partial_t \bar{u} \,\textrm{d}s \right)}, \end{equation}

where $t\mapsto F_1(t)$ is any function such that $F_1(t)\geqslant \bar {u}^2+2\int ^x \partial _t \bar {u} \,\textrm {d}s$.

A solution of the derivative of (4.10) with respect to the variable $x$ is given by

(4.15)\begin{equation} \bar{u}(t,x)=\frac{u_1(x)}{t-t_0+t_1}. \end{equation}

Let us assume that (4.15) holds true. Inserting (4.15) into (4.8), we can write this equation formally in the form

(4.16)\begin{equation} \frac{\partial (h u_1)}{\partial x} + \frac{t-t_0+t_1}{u_1} \frac{\partial (h u_1)}{\partial t}=0, \end{equation}

then we apply the characteristic method following $t'(x) = ({t-t_0+t_1})/{u_1}$ leading to $t(x) = (t-t_0+t_1) \int ^x ({1}/{u_1(s)})\,\textrm {d}s$ and we can exhibit a formal expression for $h(x,t)$ of the form

(4.17)\begin{equation} h(t,x)=\frac{F_0\left( (t-t_0+t_1)\exp\left({-\displaystyle\int^x \dfrac{\textrm{d}s}{u_1(s)}}\right)\right)}{u_1(x)}, \end{equation}

where $\xi \mapsto F_0(\xi )$ is any function.

Likewise, inserting the expression for $\bar {u}$ given by (4.15) into (4.14) gives another expression for $h(t,x)$ with

(4.18)\begin{align} h(t,x) &={-}z_b(x)-\frac{b(t)}{a(t)}\nonumber\\ &\quad + \frac{1}{a(t)(t-t_0+t_1)}\sqrt{a(t)\left( (t-t_0+t_1)^2 F_1(t) - u_1^2(x) + 2\int^x u_1(s)\,\textrm{d}s \right)}. \end{align}

Now, inserting the expression (4.15) into (4.10) implies that necessarily

(4.19)\begin{equation} a(t) = \frac{a_0}{(t-t_0+t_1)^2}, \end{equation}

with $a_0\in \mathbb {R}$. Similarly, inserting the expression (4.15) into (4.11) allows us to obtain the expression for $b(t)$ of the form

(4.20)\begin{equation} b(t) = g+\frac{2 b_0}{(t-t_0+t_1)^2}, \end{equation}

with $b_0\in \mathbb {R}$.

Thus, (4.18) gives

(4.21)\begin{align} h(t,x) &={-}z_b(x)-\frac{b_0}{a_0} - \frac{g}{a_0}(t-t_0+t_1)^2 \nonumber\\ &\quad + \frac{1}{\sqrt{a_0}}\sqrt{(t-t_0+t_1)^2 F_1(t) - u_1^2(x) +2\int^x u_1(s)\,\textrm{d}s}. \end{align}

The two expressions for $h(t,x)$, namely (4.17) and (4.21), are compatible only if the primitive function associated with ${1}/{u_1(x)}$ is a logarithmic function of $x$, leading to

(4.22)\begin{equation} u_1(x) = \gamma x + \beta, \end{equation}

with $(\beta ,\gamma ) \in \mathbb {R}^2$. Inserting (4.22) into (4.10) gives that

(4.23)\begin{equation} \gamma^2+\gamma = a . \end{equation}

Now, from (4.18) we can set $- u_1^2(x) +2\int ^x u_1(s)\,\textrm {d}s = C$ in order to have $h(t,x)+z_b(x)$ only depending on time $t$ i.e.

(4.24)\begin{equation} \frac{\partial \eta}{\partial x} = 0. \end{equation}

This property inserted into (4.9) gives $\gamma = 1$, and using (4.23), $a_0=2$.

Finally, we have obtained that

(4.25)\begin{equation} u_1(x) = x+\beta, \end{equation}

and $h=z_b(x)+F_\eta (t)$. These two expressions inserted in (4.8) give

(4.26)\begin{equation} \left\{\begin{array}{@{}l} z_b(x) = \dfrac{c_0}{x+\beta} - b_0\\ h(t,x) = \dfrac{\alpha}{t-t_0+t_1} - \dfrac{c_0}{x+\beta} \end{array}\right. \end{equation}

These expressions are valid only if $h(t,x)\geqslant 0$. Thus, the expressions (4.25), (4.26) and (4.15) give a proof of Proposition 4.1 in the two-dimensional ($x,z$) setting (when $\theta = 0$).

Proof. The main idea of the proof is to observe that the analytical solution proposed in Proposition 4.1 satisfies

(4.29)\begin{equation} \frac{\partial {w}}{\partial {t}} = u\frac{\partial {w}}{\partial {x}} + v\frac{\partial {w}}{\partial {y}} + w\frac{\partial{w}}{\partial {z}}={-}\frac{b_0+z}{ f(t)^2}. \end{equation}

It is sufficient to reverse the direction of the components of the velocity to cancel the non-hydrostatic part of the third momentum equation. Compared to Proposition 4.1, with opposite values of $u$ and $v$, the water depth increases in time. We have then to find the pressure term allowing us to satisfy exactly the two other momentum equations.

Proof. First, we observe that

(5.2)\begin{equation} \frac{\partial h}{\partial t} = \frac{-\alpha }{(t-t_0+t_1)^2}, \end{equation}

and

(5.3a,b)\begin{equation} \frac{\partial (hu)}{\partial x} = \frac{\alpha \sin^2(\theta)}{(t-t_0+t_1)^2} , \quad \frac{\partial (h v)}{\partial y} = \frac{\alpha \cos^2(\theta)}{(t-t_0+t_1)^2} . \end{equation}

Then, taking the sum of the last three terms, the mass conservation equation is satisfied.

For the second equation, we need to verify that

(5.4)\begin{equation} \frac{\partial u}{\partial t}+\frac{\partial u^2}{\partial x}+\frac{\partial (uv)}{\partial y}+\frac{\partial (uw)}{\partial z}+ g \frac{\partial (h+z_b)}{\partial x}=0. \end{equation}

We can observe that $u = v$, moreover $h$ and $z_b$ do not depend on $x$. Then we have to check

(5.5)\begin{equation} \frac{\partial u}{\partial t}+\frac{\partial u^2}{\partial x}+\frac{\partial u^2}{\partial y}+\frac{\partial uw}{\partial z}=0, \end{equation}

with $w$ given thanks to the incompressibility condition by

(5.6)\begin{equation} w(t,x,y,z) ={-}\frac{\partial}{\partial x}\int_{z_b}^z u\,\textrm{d}z -\frac{\partial}{\partial y}\int_{z_b}^z v\,\textrm{d}z. \end{equation}

Since we have

(5.7)\begin{gather} \frac{\partial u}{\partial t} ={-}\frac{\alpha \beta}{2} f'(t) + f'(t)(x \cos^2(\theta)+y \sin^2(\theta)),\end{gather}

(5.8a,b)\begin{gather} \frac{\partial u^2}{\partial x}= 2 u f(t) \cos^2(\theta), \quad \frac{\partial u^2}{\partial y} = 2 u f(t) \sin^2(\theta), \end{gather}

and

(5.9)\begin{equation} \frac{\partial (u w)}{\partial z} ={-}u f(t) + \beta f(t)(z_b-z), \end{equation}

then, using the property $f'(t) = -f^2(t)$ and taking the sum of (5.7), (5.8a,b) and (5.9), we can easily verify that (5.5) holds true since

(5.10)\begin{align} \frac{\partial u}{\partial t}+\frac{\partial u^2}{\partial x}+\frac{\partial u^2}{\partial y}+\frac{\partial uw}{\partial z} &=f(t)\left(\frac{\alpha \beta}{2} f(t)- f(t)(x \cos^2(\theta)+y \sin^2(\theta)) -\beta(z-z_b)+u \right)\nonumber\\ &=0. \end{align}

Proof. The results are based on two observations. First, the function $f$ is such that $f'(t) = -f^2(t)$. Second, using the trigonometric formula

(5.12a,b)\begin{equation} \cos \theta = \frac{1 - \tan^2 \left(\dfrac{\theta}{2} \right)}{1 + \tan^2 \left(\dfrac{\theta}{2} \right)}, \quad \sin \theta = \frac{2 \tan \left(\dfrac{\theta}{2} \right)}{1 + \tan^2 \left(\dfrac{\theta}{2} \right)}, \end{equation}

we can observe that

(5.13)\begin{equation} \cos \theta + \sin \theta \tan \left(\frac{\theta}{2} \right) =1. \end{equation}

This formula (5.13) will be used many times in the proof.

We can easily compute that

(5.14a–c)\begin{equation} \frac{\partial u}{\partial x} = f(t) \cos \theta ,\quad \frac{\partial v}{\partial y} = f(t) \sin \theta \tan \frac{\theta}{2}, \quad \frac{\partial w}{\partial z} ={-}f(t), \end{equation}

leading to $\boldsymbol {\nabla }\boldsymbol {\cdot } U =0$ using formula (5.13). For the first momentum equation, we have to verify that the solution is such that

(5.15)\begin{equation} \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y}+ w\frac{\partial u}{\partial z} + \frac{1}{\rho_0}\frac{\partial p}{\partial x} = 0. \end{equation}

We can observe that ${\partial p}/{\partial x} = 0$ since the functions $f$, $z_b$ and $h$ do not depend on $x$. For the other terms, we have

(5.16a,b)\begin{gather} \frac{\partial u}{\partial t} ={-}(x \cos\theta +y \sin \theta) f^2(t), \quad u\frac{\partial u}{\partial x} = u f(t) \cos\theta, \end{gather}

(5.17a,b)\begin{gather} v\frac{\partial u}{\partial y} = v f(t) \sin \theta ,\quad w\frac{\partial u}{\partial z} = \gamma f(t) (z_b-z). \end{gather}

Then,

(5.18)\begin{align} u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}+ w\frac{\partial u}{\partial z} &=f(t) \left(u(2 \cos \theta + \sin \theta \tan \left( \frac{\theta}{2} \right) -1) + v \cos \theta + \gamma (z_b-z)\right)\nonumber\\ &=f(t) \left[\vphantom{\left( \frac{\theta}{2} \right)}f(t) x \cos^2 \theta + f(t) y \sin\theta \cos \theta + \gamma (z-z_b) \cos \theta\right.\nonumber\\ &\quad + \delta \cos \theta+ f(t) x \cos \theta \sin \theta \tan \left( \frac{\theta}{2} \right) +f(t) y \sin^2\theta \tan \left( \frac{\theta}{2} \right)\nonumber\\ &\quad \left.\vphantom{\left( \frac{\theta}{2} \right)} +(1-\cos \theta ) \gamma (z-z_b) - \delta \cos \theta + \gamma (z_b-z)\right]. \end{align}

We can observe that the terms depending on $\delta$ and $\gamma$ are equal to $0$. For the terms in $x$, we have

(5.19)\begin{equation} f^2(t) \left(\cos \theta \left(\cos \theta +\sin \theta \tan \left( \frac{\theta}{2} \right) \right) \right) = f^2(t) \cos \theta, \end{equation}

and for the terms in $y$

(5.20)\begin{equation} f^2(t)\sin \theta \left(\cos \theta +\sin \theta \tan \left( \frac{\theta}{2} \right) \right) = f^2(t) \sin \theta. \end{equation}

Combining all these results, we conclude that the first momentum equation is verified.

The computation for the second momentum equation is very similar. For the third momentum equation, the computation is easier since $w$ does not depend on $x$ and $y$. We have only to compute

(5.21a–c)\begin{equation} \frac{\partial w}{\partial t} ={-}f^2(z_b-z), \quad w\frac{\partial w}{\partial z} ={-} f^2(t) (z_b-z), \quad \frac{\partial p}{\partial z} ={-}g + 2 f^2(t) (z_b-z), \end{equation}

and the result is proved.

Proof. The proof relies on very simple computations since it is enough to verify that these functions are solutions of the equations of Euler given in section 2.1. The computations are very similar to those detailed in the proof of Proposition 5.3 for the mass and momentum equations. We focus here only on the transport equation (2.12) and we use mainly the property presented in (5.13). We have

1. (i) ${\partial \phi }/{\partial t} =0,$

2. (ii) $u({\partial \phi }/{\partial x}) = ({\phi _0}/{L} )\cos \theta (({z-z_b})/{h_0})f(t) (x \cos \theta + y \sin \theta ),$

3. (iii) $v({\partial \phi }/{\partial y}) = ({\phi _0}/{L}) \sin \theta (({z-z_b})/{h_0})f(t) (x \cos \theta + y \sin \theta ) \tan ( {\theta }/{2} ) ,$

4. (iv) $w({\partial \phi }/{\partial z}) = ({\phi _0}/{L})(({z-z_b})/{h_0})f(t) (x \cos \theta + y \sin \theta )$.

Then

(5.23)\begin{align} &\frac{\partial \phi}{\partial t}+ u\frac{\partial \phi}{\partial x} + v\frac{\partial \phi}{\partial y}+w\frac{\partial \phi}{\partial z} \nonumber\\ &\quad =\frac{\phi_0}{L} \frac{z-z_b}{h_0}f(t) \left( \cos \theta +\sin \theta \tan \left( \frac{\theta}{2} \right) -1 \right) (x\cos \theta+ y \sin\theta) = 0. \end{align}

In figure 11, we present a vertical profile (for $y=0$), at times $t=0$, $t=0.5$ and $t=1\text{s}$, of the analytical solution corresponding to the proposition 5.5. In figure 12, we present a horizontal profile (for $z=h_0/2$) of the same analytical solution for two given values of the parameter $\theta$.

Figure 11.

Draining of a tank (non-hydrostatic case): velocity norm and streamlines at $t=0$, 0.5, 1.0 s, in $(x,y=0,z)$ slice plane for parameters set to $\theta = 0$, $\alpha = \ \textrm {m}\ \textrm {s}$, $t_1 =1 \ \textrm {s}$, $p^{a} =0 \ \textrm {m}^{-2}\ \textrm {s}^{-2}$, $z_b =0 \ \textrm {m}$ and $L =10 \ \textrm {m}$.

Figure 12.

Draining of a tank (non-hydrostatic case): velocity norm and vectors at initial time in $(x,y,z=h_0/2)$ slice plane with $\alpha =5\ \textrm {m}\ \textrm {s}$, $t_1 =1 \ \textrm {s}$, $p^{a} =0 \ \textrm {m}^2\ \textrm {s}^{-2}$, $z_b =0 \ \textrm {m}$ and $L =10 \ \textrm {m}$.