Our aim is to study equatorial flows exhibiting general (continuous) stratification depending on depth and latitude. This work is motivated by the intriguing features observed in the Equatorial region, situated approximately 2° latitude from the Equator. In this area, strong vertical ocean stratification prevails, surpassing other ocean zones and creating a distinct thermocline—a sharp interface separating warmer, less dense surface water from colder, denser water below (see [20, 21, 22, 23, 16, 18]). Additionally, the equatorial region showcases underlying currents, including a westward drift near the surface triggered by prevailing trade winds and the presence of the Equatorial Undercurrent (EUC), an eastward pointing stream residing on the thermocline. These characteristics contribute to the complexity of large-scale ocean motions, emphasizing the importance of understanding fluid density variations in both depth and latitude, primarily influenced by temperature and, to a lesser extent, salinity (see, e.g., [16, 17, 18]).

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Our goal in this study is to establish the existence of an exact solution within a cylindrical frame to the highly nonlinear and intractable equations of geophysical fluid dynamics (GFD). In this context, we would like to mention that, recently, a number of exact solutions of the GFD governing equations were devised, cf. [14, 15, 24, 25, 20, 26, 27, 28, 29, 30, 31, 32]. The solution characterizes a flow propagating in the azimuthal direction, influenced by both gravity and surface tension, while maintaining variability in density across all directions. This form of density distribution proves sufficiently general to capture and model most geophysical fluid dynamics phenomena [19].

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The Equatorial motion has been the subject of previous studies, where has been proven the existence of exact solutions under various contexts (see, e.g., [2, 3, 13, 14, 15]). The novelty of this paper lies in deducing an existence result, considering both surface tension and density variation. While comparable results, utilizing cylindrical coordinates, have been achieved in [2] and [3], it is noteworthy that these outcomes are exclusively applicable to one of the two cases mentioned earlier.

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Considering the line of Equator ”straightened” and replaced to the z axis, and the body of the sphere, represented by a circular disc described by radius r and the angle of deflection from the equator, $\theta\in\displaystyle\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, the fluid motion can be described in cylindrical coordinates. We can provide a right-handed rotating system of coordinates (r, $\theta$, z), where r measures the distance from the center of the disc (physically, it means the distance to the center of the Earth), $\theta$ is increasing from north to south and the z axis points positively from west to east. Note that $\theta=0$ corresponds to the line of Equator. The associated unit vectors in the (r, $\theta$, z) system are ($\vec{e_{r}},\vec{e_{\theta}},\vec{e_{z}}$) and the velocity corresponding components are (u,v,w).
The equations describing the motion of an inviscid and incompressible flow are comprised in the Euler equations and, respectively, mass conservation equation. In cylindrical coordinates, they are given by

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(2.1)

$$\begin{cases}\displaystyle u_{t}+uu_{r}+\frac{v}{r}u_{\theta}+wu_{z}-\frac{v^{2}}{r}=-\frac{1}{\rho}p_{r}+F_{r}\\ ~{}\\ \displaystyle v_{t}+uv_{r}+\frac{v}{r}v_{\theta}+wvz+\frac{uv}{r}=-\frac{1}{\rho}\frac{1}{r}p_{\theta}+F_{\theta}\\ ~{}\\ \displaystyle w_{t}+uw_{r}+\frac{v}{r}+w_{\theta}+ww_{z}=-\frac{1}{\rho}p_{z}+F_{z},\end{cases}$$

and

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(2.2)

$$\frac{1}{r}\frac{\partial}{\partial r}(r\rho u)+\frac{1}{r}\frac{\partial}{\partial\theta}(\rho v)+\frac{\partial}{\partial z}(\rho w)=0,$$

respectively. We denote $\rho$ as the fluid density, $p$ as the pressure in the fluid, and ($F_{r},F_{\theta},F_{z}$) as the body-force vector, with these quantities depending on the variables $r$, $\theta$, and $z$.

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To ensure a precise examination of fluid dynamics in specific locations, it is essential to account for the Earth’s rotation effects. The additional terms to incorporate into the expression of the equations of Euler are related to the Coriolis force $2\vec{\Omega}\times\vec{u}$ and the centripetal acceleration $\vec{\Omega}\times(\vec{\Omega}\times\vec{r})$, where

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	$\displaystyle\vec{\Omega}=-\Omega((\sin\theta)\vec{e_{r}}+(\cos\theta)\vec{e_{\theta}}),$
	$\displaystyle\vec{u}=u\vec{e_{r}}+v\vec{e_{\theta}}+w\vec{e_{z}},$
	$\displaystyle\vec{r}=r\vec{e_{r}}.$

Here, $\Omega\approx 7.29\times 10^{-5}rad/s$ stands for the constant rate of rotation of the Earth.

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The two forces mentioned earlier sum up in the Euler equations. Thus, since the gravity is given by the vector $(-g,0,0)$, $2\vec{\Omega}\times\vec{u}=2\Omega(-w\cos\theta,w\sin\theta,u\cos\theta-v\sin\theta)$ and $\vec{\Omega}\times(\vec{\Omega}\times\vec{r})=r\Omega^{2}\cos\theta(-\cos\theta,\sin\theta,0),$ one has

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(2.3)

$$\begin{cases}\displaystyle u_{t}+uu_{r}+\frac{v}{r}u_{\theta}+wu_{z}-\frac{v^{2}}{r}-2\Omega w\cos\theta-r\Omega^{2}\cos^{2}\theta=-\frac{1}{\rho}p_{r}-g\\ ~{}\\ \displaystyle v_{t}+uv_{r}+\frac{v}{r}v_{\theta}+wvz+\frac{uv}{r}+2\Omega w\sin\theta+r\Omega^{2}\sin\theta\cos\theta=-\frac{1}{\rho}\frac{1}{r}p_{\theta}\\ ~{}\\ \displaystyle w_{t}+uw_{r}+\frac{v}{r}+w_{\theta}+ww_{z}+2\Omega(u\cos\theta-v\sin\theta)=-\frac{1}{\rho}p_{z}.\end{cases}$$

Besides the mass conservation and Euler equations, the water motion is also subject to boundary conditions. Considering that the fluid extends to infinity in all horizontal directions, there are two boundaries: the rigid flat bed and the free surface of the water. On the bed $r=d(\theta,z)$, we have the kinematic boundary condition

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(2.4)

$$u=wd_{z}+\displaystyle\frac{1}{r}vd_{\theta}.$$

On the free surface $r=R+h(\theta,z)$, along with the kinematic boundary condition

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(2.5)

$$u=wh_{z}+\displaystyle\frac{1}{r}vh_{\theta},$$

we also have the dynamic condition

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(2.6)

$$p=P(\theta,z)+\sigma\nabla\cdot\vec{n},$$

where $\sigma$ is the coefficient of surface tension and $\vec{n}$ is the unit normal vector, pointing outward. The function $h$ is unknown and will be determined later.

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Our aim is to determine the pressure under the additional assumption that the flow is purely azimuthal and invariant in this direction, i.e.,

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(3.1)

$$u=v=0\text{ and }w=w(r,\theta).$$

Consequently, since the dependence in the ”z” direction is eliminated, relation (2.6) has the form

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(3.2)

$$p=P(\theta)+\sigma\nabla\cdot\vec{n}.$$

Additionally, both the free surface and the bed will be characterized by $r=R+h(\theta)$ and $r=d(\theta)$, respectively.

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From (3.1), the Euler equations becomes

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(3.3)

$$\begin{cases}-2\Omega w\cos\theta-r\Omega^{2}\cos^{2}\theta=-\displaystyle\frac{1}{\rho}p_{r}-g\\ 2\Omega w\sin\theta+r\Omega^{2}\sin\theta\cos\theta=\displaystyle-\frac{1}{\rho}\frac{1}{r}p_{\theta}\\ \quad\indent\quad\indent\indent 0=-\displaystyle\frac{1}{\rho}p_{z},\end{cases}$$

or equivalently

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(3.4)

$$\begin{cases}\rho(r,\theta)\Omega\cos\theta(2w(r,\theta)+r\Omega\cos\theta)=p_{r}+g\rho(r,\theta)\\ \rho(r,\theta)r\Omega\sin\theta(2w(r,\theta)+r\Omega\cos\theta)=-p_{\theta}\\ \quad\indent\quad\indent\indent\quad\indent 0=p_{z}.\end{cases}$$

Letting $\pazocal{U}(r,\theta):=2w(r,\theta)+r\Omega\cos\theta$, if we derive the first equation with respect to $\theta$ and the second one with respect to r, we obtain

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$$\left\{\begin{aligned} \displaystyle(\rho\Omega\pazocal{U}\cos\theta)_{\theta}&=\displaystyle p_{r\theta}+g\rho_{\theta}\\ \displaystyle(\rho r\Omega\pazocal{U}\sin\theta)_{r}&=\displaystyle-p_{\theta r}\end{aligned}\right.$$

Next, we sum up the two equations to deduce

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$$(\rho\Omega\pazocal{U}\cos\theta)_{\theta}+(\rho r\Omega\pazocal{U}\sin\theta)_{r}=g\rho_{\theta},$$

which yields

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(3.5)

$$(r\sin\theta)\pazocal{Z}_{r}(r,\theta)+(\cos\theta)\pazocal{Z}_{\theta}(r,\theta)=(r\cos\theta)g\rho_{\theta},$$

where $\pazocal{Z}(r,\theta):=\rho r\Omega\pazocal{U}\cos\theta$.

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Following the method of characteristics, we look for solutions $s\rightarrow\tilde{r}(s)$ and $s\rightarrow\tilde{\theta}(s)$ such that

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(3.6)			$\displaystyle\widetilde{r}~{}^{\prime}(s)=\widetilde{r}(s)\sin\widetilde{\theta}(s),$
		$\displaystyle\widetilde{\theta}^{\prime}(s)=\cos\widetilde{\theta}(s).$

Thus, relation (3.5) is equivalent to

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(3.7)

$$\displaystyle\frac{d}{ds}\big{(}\pazocal{Z}(\widetilde{r}(s),\widetilde{\theta}(s))=(\widetilde{r}(s)\cos\widetilde{\theta}(s))g\rho_{\theta}(\widetilde{r}(s_{,}\widetilde{\theta}(s)).$$

Simple computations yield $\widetilde{r}^{\prime\prime}(s)=\widetilde{r}(s)$, leading to the solution

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(3.8)

$$\widetilde{r}(s)=c_{1}e^{s}+c_{2}e^{-s},$$

where $c_{1},c_{2}$ are some real constants. Now, since $\widetilde{r}^{\prime}(s)=c_{1}e^{s}-c_{2}e^{-s}$, we infer

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(3.9)

$$\widetilde{\theta}(s)=\arcsin\displaystyle\frac{c_{1}e^{s}-c_{2}e^{-s}}{c_{1}e^{s}+c_{2}e^{-s}}=\arcsin\displaystyle\frac{ce^{2s}-1}{ce^{2s}+1},$$

where $c=\displaystyle\frac{c_{1}}{c_{2}}$ is constant. Returning to our equation (3.5), we note that

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(3.10)

$$\displaystyle\frac{d}{ds}(\widetilde{r}(s)\cos\widetilde{\theta}(s))=0.$$

Assuming, without loss of generality, that $\widetilde{\theta}(0)=0$, we find that $c=1$, implying $c_{1}=c_{2}$. For given values of $r$ and $\theta$, we seek $s_{0}\in\mathbb{R}$ such that

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(3.11)

$$\widetilde{\theta}(s_{0})=\theta,\widetilde{r}(s_{0})=r.$$

The initial condition in (3.11) leads to $\displaystyle\frac{e^{2s_{0}}-1}{e^{2s_{0}}+1}=\sin\theta$, thus

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(3.12)

$$s_{0}(\theta)=\displaystyle\frac{1}{2}\ln\displaystyle\frac{1+\sin\theta}{1-\sin\theta},$$

while the second condition from (3.11) leads to

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$$c_{1}\bigg{(}\sqrt{\displaystyle\frac{1+\sin\theta}{1-\sin\theta}}+\sqrt{\displaystyle\frac{1-\sin\theta}{1+\sin\theta}}\bigg{)}=r,$$

which implies that

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(3.13)

$$c_{1}=\displaystyle\frac{r\cos\theta}{2}=c_{2}.$$

Consequently, relations (3.6) are given by

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(3.14)		$\displaystyle\widetilde{r}(s)=\displaystyle\frac{r\cos\theta}{2}(e^{s}+e^{-s})=r\cos\theta\cosh(s),$
(3.15)		$\displaystyle\widetilde{\theta}(s)=\arcsin\bigg{(}\displaystyle\frac{e^{2s}-1}{e^{2s}+1}\bigg{)}=\arcsin(\tanh(s)).$

From (3.10), we immediately derive

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(3.16)

$$\widetilde{r}(s)\cos\widetilde{\theta}(s)=\widetilde{r}(0)\cos\widetilde{\theta}(0)=r\cos\theta\cos(0)=r\cos\theta,$$

Whence, integrating (3.7) from $0$ to $s_{0}(\theta)$, we deduce

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$$\pazocal{Z}(r,\theta)-\pazocal{Z}(r\cos\theta,0)=gr\cos\theta\int_{0}^{s_{0}(\theta)}\rho_{\theta}(\widetilde{r}(s),\widetilde{\theta}(s))ds.$$

Now, from the definitions of $\pazocal{Z}$ and $\pazocal{U}$, we determine the expression for the velocity $w(r,\theta)$, that is,

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(3.17)

$$w(r,\theta)=-\displaystyle\frac{\Omega r\cos\theta}{2}+\frac{1}{2\rho\Omega}\left(\displaystyle\frac{F(r\cos\theta)}{r\cos\theta}+g\int_{0}^{s_{0}(\theta)}\rho_{\theta}(\widetilde{r}(s),\widetilde{\theta}(s))ds\right),$$

where $F(x):=\pazocal{Z}(x,0)$ is an arbitrarily smooth function.

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To derive the expression for pressure $p$, the first equation of (3.3) yields

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	$\displaystyle p_{r}$	$\displaystyle=-g\rho+\Omega\rho\pazocal{U}\cos\theta$
(3.18)			$\displaystyle=-g\rho+\displaystyle\frac{F(r\cos\theta)}{r}+g\cos\theta\int_{0}^{s_{0}(\theta)}\rho_{\theta}(\widetilde{r}(s),\widetilde{\theta}(s))ds,$

while from the second one we infer

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	$\displaystyle p_{\theta}$	$\displaystyle=-\Omega r\rho\pazocal{U}\sin\theta$
(3.19)			$\displaystyle=-\tan\theta\Bigg{[}F(r\cos\theta)+gr\cos\theta\int_{0}^{s_{0}(\theta)}\rho_{\theta}(\widetilde{r}(s),\widetilde{\theta}(s))ds\Bigg{]}.$

If we integrate (3) with respect to r, and use the substitution $y=r\cos\theta$, we obtain

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(3.20)

$$p(r,\theta)=-g\int_{a}^{r}\rho(\xi,\theta)d\xi+\int_{a\cos\theta}^{r\cos\theta}\Bigg{[}\displaystyle\frac{F(y)}{y}+H(y,\theta)\Bigg{]}dy+\widetilde{C}(\theta),$$

where $a$ is a constant, $\theta\rightarrow\widetilde{C}(\theta)$ is a smooth function and

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(3.21)

$$H(y,\theta)=\int_{0}^{s_{0}(\theta)}g\rho_{\theta}(y\cosh(s),\widetilde{\theta}(s))ds.$$

In equation (3.20), differentiating with respect to $\theta$ yields the expression for $p_{\theta}$,

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	$\displaystyle p_{\theta}$	$\displaystyle=-g\int_{a}^{r}\rho_{\theta}(\xi,\theta)d\xi-\tan\theta\left(F(r\cos\theta)-F(a\cos\theta)\right)$
		$\displaystyle\quad+\int_{a\cos\theta}^{r\cos\theta}H_{\theta}(y,\theta)dy-\sin\theta\left(r\,H(r\cos\theta,\theta)-aH(a\cos\theta,\theta)\right)+\widetilde{C}^{\prime}(\theta).$

Since

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(3.22)

$$H_{\theta}(y,\theta)=g\rho_{\theta}(y\cosh(s_{0}(\theta)),\widetilde{\theta}(s_{0}(\theta)))\cdot\frac{1}{\cos\theta},$$

and

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$$\cosh(s_{0})=\displaystyle\frac{\widetilde{r}(s_{0})}{r\cos\theta}=\displaystyle\frac{1}{\cos\theta},$$

from (3.11) we obtain

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(3.23)

$$H_{\theta}(y,\theta)=g\rho_{\theta}\bigg{(}\displaystyle\frac{y}{\cos\theta},\theta\bigg{)}\cdot s_{0}^{\prime}(\theta)=\displaystyle\frac{g}{\cos\theta}\rho_{\theta}\bigg{(}\displaystyle\frac{y}{\cos\theta},\theta\bigg{)}.$$

Consequently,

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(3.24)

$$\int_{a\cos\theta}^{r\cos\theta}H_{\theta}(y,\theta)dy=\int_{a\cos\theta}^{r\cos\theta}\displaystyle\frac{g}{\cos\theta}\rho_{\theta}\bigg{(}\frac{y}{\cos\theta},\theta\bigg{)}dy=\int_{a}^{r}g\rho_{\theta}(\widetilde{r},\theta)d\widetilde{r}.$$

Hence, we derive the expression for $p_{\theta}$ given by

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	$\displaystyle p_{\theta}$	$\displaystyle=-F(r\cos\theta)\tan\theta+\tan\theta\,F(a\cos\theta)-r\sin\theta\,H(r\cos\theta,\theta)$
(3.25)			$\displaystyle\quad+a\,\sin\theta\,H(a\cos\theta,\theta)+\widetilde{C}^{\prime}(\theta).$

From the formulas (3.25) and (3), we deduce

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$$\widetilde{C}^{\prime}(\theta)+\tan\theta\,F(a\cos\theta)+a\sin\theta\,H(a\cos\theta,\theta)=0,$$

which gives,

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(3.26)

$$\widetilde{C}(\theta)=A-\int_{0}^{\theta}\tan\xi\,F(a\cos\xi)d\xi-a\int_{0}^{\theta}\sin\xi\,H(a\cos\xi,\xi)d\xi.$$

From all the previous results, we derive the formula for the pressure,

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(3.27)		$\displaystyle p(r,\theta)$	$\displaystyle=A-g\int_{a}^{r}\rho(\xi,\theta)d\xi+\int_{a\cos\theta}^{r\cos\theta}\Bigg{[}\displaystyle\frac{F(y)}{y}+H(y,\theta)\Bigg{]}dy$
		$\displaystyle\quad-\int_{0}^{\theta}\tan\xi\,F(a\cos\xi)d\xi-a\int_{0}^{\theta}\sin\xi\,H(a\cos\xi,\xi)d\xi.$