2.1. Governing equations

We consider the unsteady motion of a circular cylinder of radius $\varepsilon$ under the free surface of an inviscid incompressible fluid of infinite depth and constant density $\rho$. The two-dimensional flow is considered in Cartesian coordinate system $Oxy$ with $x$-axis lying on the undisturbed horizontal free surface, which is subject to a constant atmospheric pressure, and $y$-axis directed upwards, as shown in figure 1. The fluid and the cylinder are initially at rest. The cylinder starts moving impulsively from an initial position $(0,-h)$ ($h>\varepsilon$) with constant acceleration $a_0$, remaining totally submerged under the free surface $y = \eta (x,t)$ at all times $t$. We will distinguish between two cases of the accelerated cylinder motion: with ($u_0 \neq 0, a_0 \neq 0$) and without ($u_0 = 0, a_0 \neq 0$) initial impulsive velocity. The case of a cylinder moving with constant impulsive velocity without acceleration ($u_0 \neq 0, a_0 = 0$) was investigated previously by Tyvand & Miloh (Reference Tyvand and Miloh1995).

Figure 1. Scheme of the problem.

The physical variables are given in dimensionless form by use of $h$, $\rho$ and $u_0$ as a dimensionally independent set, i.e.

(2.1a–e)\begin{equation} \bar{x} = \frac{x}{h}, \quad \bar{y} = \frac{y}{h}, \quad \bar{\varepsilon} = \frac{\varepsilon}{h}, \quad \bar{t} = \frac{tu_0}{h}, \quad \bar{a}_0 = \frac{a_0 h}{u_0^2}. \end{equation}

Quantities $\rho h^2$, $\rho u_0^2$, $\rho h u_0^2$ and $\rho h u_0^2$ denote the units of mass, pressure, force and momentum, respectively. The important non-dimensional physical parameter $\lambda = gh/u_0^2$ is the square of inverse Froude number. Special attention needs to be given to the case of the vanishing initial velocity ($u_0 = 0$). In this case, the velocity scale is calculated as $\sqrt {a_0h}$, with constant acceleration $a_0$ taken as a unit, and the inverse Froude number reduces to the ratio $\lambda = g/a_0$. For consistency, all equations, input parameters and resulting figures will be given in dimensionless form with the bars dropped.

Since the fluid is assumed incompressible and the flow is irrotational, the fluid velocity potential $\varPhi (x,y)$ satisfies the Laplace's equation in the flow domain,

(2.2)\begin{equation} \nabla^2 \varPhi = 0, \end{equation}

nonlinear boundary kinematic and dynamic conditions on the free surface,

(2.3)$$\begin{gather} \frac{\partial \eta}{\partial t} + \frac{\partial \varPhi}{\partial x} \frac{\partial \eta}{\partial x} = \frac{\partial \varPhi}{\partial y}, \quad y=\eta(x, t), \end{gather}$$

(2.4)$$\begin{gather}\frac{\partial \varPhi}{\partial t} + \frac12 |\boldsymbol{\nabla} \varPhi|^2 + \lambda \eta = 0, \quad y = \eta(x,t) \end{gather}$$

and nonlinear kinematic condition at the cylinder surface,

(2.5)\begin{equation} (\boldsymbol{r} - \boldsymbol{R}) \boldsymbol{\cdot} \left(\boldsymbol{\nabla} \varPhi - \frac{{\rm d}\boldsymbol{R}}{{\rm d}t} \right) = 0, \quad \vert \boldsymbol{r} - \boldsymbol{R} \vert = \varepsilon. \end{equation}

Vectors $\boldsymbol {r}$ and $\boldsymbol {R}(t)$ are position vectors of the fluid particles and the cylinder centre, respectively. Under the assumption of a semi-infinite fluid domain, the far-field condition implies

(2.6a–c)\begin{equation} \boldsymbol{\nabla} \varPhi \to 0, \quad x \to \pm\infty, \quad y \to -\infty. \end{equation}

Tyvand & Miloh (Reference Tyvand and Miloh1995) pointed out that initiation of the flow in a reversible system requires an algorithmic preparation, which is the only way a time arrow can emerge in a system without entropy production. The impulsive start of a submerged body initiates a process evolving with time. Mathematically we must allow a temporal singularity at $t=0$. An impulsively started velocity gives a Dirac delta function singularity for the flow, while the velocity potential starts as a Heaviside unit step function. It is important to consider negative times where the entire system is at rest, obeying Newton's laws of the balance of forces. Once the cylinder is put into motion, Newton's third law requires that the surrounding fluid responds to the induced motion by a net pressure force opposite to the force applied on the cylinder. We prescribe the motion, so that the induced forces are caused by the motion and not the other way round. The static state at rest means that the potential $\varPhi$ is identical to zero for all $t<0$. It also means that all time derivatives of $\varPhi$ are zero. This is worth noting because any initiation of the forced cylinder motion will lead to a Heaviside singularity in the leading velocity potential. Due to the Bernoulli equation, the time singularity for the pressure will be more severe than that of the velocity potential. As we will see in § 5, the leading-order force for constant impulsive velocity will represent a Dirac delta function in time.

It is assumed that the cylinder is set into motion during an infinitesimal time interval $0 < t < 0^+$ by the impulsive force. The initial velocity is finite, the fluid is initially at rest and the free surface is flat. Due to the zero horizontal force during the impulsive start, initial horizontal velocity of the fluid is zero on the free surface at $t = 0^+$. Hence, the initial conditions for the forced free surface flow are given as

(2.7a,b)\begin{equation} \varPhi(x,0,0^+) = 0,\quad \eta(x, 0^+) = 0. \end{equation}

In this study, we consider the forced motion of the cylinder, which is given as a series

(2.8)\begin{equation} \boldsymbol{R}(t) = \left(\begin{array}{@{}c@{}} 0 \\ - 1\end{array}\right) + H(t) (\boldsymbol{R}_1 t + \boldsymbol{R}_2 t^2 + \boldsymbol{R}_3 t^3 + \dots). \end{equation}

The Heaviside unit step function $H(t)$ makes the expansion (2.8) applicable to both negative and positive times. At negative times $t<0$ the fluid is at rest with horizontal free surface, and the cylinder is kept at rest with zero net force. It is useful to include negative times for consistent treatment of the temporal singularity of an impulsive force. Equations (2.2)–(2.8) form the complete formulation of the nonlinear initial boundary-value problem of unsteady motion of a cylinder under the free surface in dimensionless variables.

The instantaneous position vectors of the forced cylinder motion can be expressed using the angles $\alpha$ and $\beta$ between the vectors of the forced velocity and acceleration and the horizontal axis, respectively (see figure 2),

(2.9a,b)\begin{equation} \boldsymbol{R}_1(\alpha) = \delta \left(\begin{array}{@{}c@{}} \cos\alpha \\ \sin\alpha \end{array}\right), \quad \boldsymbol{R}_2(\beta) = \frac{a}{2}\left(\begin{array}{@{}c@{}} \cos\beta \\ \sin\beta \end{array}\right)= \left\{\!\!\begin{array}{ll} \dfrac {\delta a}{2}\, \boldsymbol{R}_1(\beta), & \delta \neq 0, \\[8pt] \dfrac{a}{2}\,\boldsymbol{R}_1(\beta), & \delta = 0. \end{array}\right. \end{equation}

Here, the dimensionless parameter $\delta$ is equal to unity, when the cylinder accelerates with non-zero initial velocity, and is equal to zero otherwise. The meaning of this parameter is that it exposes the explicit dependencies between velocity and acceleration magnitudes to each order in the small-time expansion that will follow. For the development of the higher-order theory, it is convenient to include the factor $\delta$ explicitly in the derived expressions, and discard terms with $\delta$ if initial velocity is zero. Here and in subsequent derivations, we agree upon using the dimensionless acceleration parameter $a_0$ without zero subscript. It should be noted that direction angles $\alpha$ and $\beta$ are independent from each other.

Figure 2. Angles of the initial velocity and acceleration at the initial position of the cylinder.

2.2. The small-time expansion

We seek a solution to the initial boundary-value problem (2.2)–(2.8) in the form of a power series with respect to the time variable $t$, given as

(2.10)$$\begin{gather} \varPhi(x,t) = H(t)( \varPhi_0 + \varPhi_1 t + \varPhi_2 t^2 + \dots), \end{gather}$$

(2.11)$$\begin{gather}\eta(x,t) = H(t)( \eta_1 t +\eta_2t^2 + \eta_3 t^3 + \dots). \end{gather}$$

It follows immediately that Laplace's equation (2.2) and the far-field condition (2.6a–c) are satisfied to each order in the small-time expansion (2.11),

(2.12)$$\begin{gather} \nabla^2 \varPhi_n = 0, \quad x^2 + (y-1)^2 > \varepsilon, \quad n = 1,2,\dots \end{gather}$$

(2.13)$$\begin{gather}|\boldsymbol{\nabla} \varPhi_n| \to 0, \quad x \to \pm \infty, \quad y \to -\infty \quad n = 1,2,\dots \, . \end{gather}$$

Successive application of the operator of total time differentiation at the moving free surface

(2.14)\begin{equation} \left.\frac{{\rm d}}{{\rm d}t}\right|_{\eta} = \frac{\partial}{\partial t} + \frac{\partial\eta}{\partial t} \frac{\partial}{\partial y} \end{equation}

to free-surface kinematic condition (2.3) and putting $t=0$ in the resulting small-time expansions leads to the set of general kinematic conditions up to the third order,

(2.15)$$\begin{gather} \eta_1 = \frac{\partial\varPhi_0}{\partial y}, \quad y = 0, \end{gather}$$

(2.16)$$\begin{gather}2 \eta_2 = \frac{\partial\varPhi_1}{\partial y}, \quad y = 0, \end{gather}$$

(2.17)$$\begin{gather}6\eta_3 = 2\frac{\partial\varPhi_2}{\partial y} + 2\frac{\partial\varPhi_0}{\partial y}\frac{\partial^2\varPhi_1}{\partial y^2} + \left(\frac{\partial\varPhi_0}{\partial y}\right)^2\frac{\partial^3\varPhi_0}{\partial y^3}\nonumber\\ \quad - 2\frac{\partial^2\varPhi_0}{\partial x\partial y}\frac{\partial\varPhi_1}{\partial x} - 2\left(\frac{\partial^2\varPhi_0}{\partial x\partial y}\right)^2\frac{\partial\varPhi_0}{\partial y}, \quad y = 0, \end{gather}$$

and simplified kinematic condition to the fourth order,

(2.18)\begin{equation} 4 \eta_{4a} = \frac{\partial\varPhi_3}{\partial y},\quad y = 0; \end{equation}

valid for the case of constant acceleration without initial impulsive velocity. We have added a subscript $a$ to the fourth-order free surface elevation to indicate that this contribution represents the pure impulsive acceleration. It should be noted that initially the free surface is undisturbed and its impulsive horizontal velocity is zero, which implies the identities

(2.19a,b)\begin{equation} \frac{\partial\varPhi_0}{\partial x} = 0, \quad \frac{\partial^2\varPhi_0}{\partial x^2} = 0. \end{equation}

Thereby, the free surface elevation of $n$th order is induced by the potential up to the $(n-1)$th order and can be calculated recurrently.

The corresponding dynamic conditions on the free surface to each order are derived in a similar way. Keeping in mind (2.15)–(2.16), we obtain the set of general dynamic conditions up to the second order

(2.20)$$\begin{gather} \varPhi_0 = 0, \quad y = 0, \end{gather}$$

(2.21)$$\begin{gather}\varPhi_1 ={-}\tfrac{1}{2}\eta_1^2, \quad y=0, \end{gather}$$

(2.22)$$\begin{gather}2\varPhi_2 ={-}4\eta_1\eta_2 - \lambda\eta_1, \quad y=0, \end{gather}$$

and simplified dynamic condition to the fourth order

(2.23)\begin{equation} 3 \varPhi_3 ={-} 4\eta_2^2 - \lambda\eta_2, \quad (y=0); \end{equation}

valid only for the case of constant acceleration without initial impulsive velocity. It should be noted that, in this case, the first-order free surface elevation $\eta _1$ and consequently the first-order potential $\varPhi _1$ should necessarily vanish.

In order to construct the expansion of the boundary conditions on the cylinder surface (2.5), we need to introduce the polar coordinates $(R,\varTheta )$ with origin in the initial cylinder centre, defined in a way suitable for the bipolar coordinates to be introduced below,

(2.24a,b)\begin{equation} x = R\sin\varTheta, \quad y ={-}1 + R\cos\varTheta. \end{equation}

Here, coordinate $\varTheta$ measures an angle between the unit vector $\boldsymbol {i}_R$ extending from the cylinder centre and the vertical axis, and varies in the interval $(-{\rm \pi},{\rm \pi} )$, see figure 2. The kinematic conditions on the cylinder to each order are derived in a similar way from the non-leaking condition (2.5) by applying successively the operator of total time differentiation,

(2.25)\begin{equation} \left.\frac{{\rm d}}{{\rm d}t}\right|_{S_\varepsilon} = \frac{\partial}{\partial t} + \frac{{\rm d}\boldsymbol{R}}{{\rm d}t} \boldsymbol{\cdot}\boldsymbol{\nabla}, \end{equation}

tracking the cylinder in its forced motion. The first factor $\boldsymbol {r} - \boldsymbol {R}$ in the scalar product in (2.5) has no effect on time differentiation and can be replaced by the parallel unit vector $\boldsymbol {i}_R$. Evaluation of each time derivative (2.25) of (2.5) at $t = 0$ results in the succession of the exact kinematic conditions at the cylinder contour up to the second order,

(2.26)$$\begin{gather} \boldsymbol{i}_R \boldsymbol{\cdot}\left(\boldsymbol{\nabla} \varPhi_0 - \boldsymbol{R}_1 \right) = 0, \quad R = \varepsilon, \end{gather}$$

(2.27)$$\begin{gather}\boldsymbol{i}_R \boldsymbol{\cdot} \left(\boldsymbol{\nabla}\varPhi_1 - 2 \boldsymbol{R}_2) + \boldsymbol{R}_1 \boldsymbol{\cdot} \boldsymbol{\nabla} (\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla} \varPhi_0\right) = 0, \quad R = \varepsilon, \end{gather}$$

(2.28)\begin{gather} \boldsymbol{i}_R \boldsymbol{\cdot} (2 \boldsymbol{\nabla} \varPhi_2 - 6 \boldsymbol{R}_3) + 2 \boldsymbol{R}_2 \boldsymbol{\cdot} \boldsymbol{\nabla}(\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla}\varPhi_0) \nonumber\\ \quad + 2 \boldsymbol{R}_1 \boldsymbol{\cdot}\boldsymbol{\nabla}(\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla} \varPhi_1) + \boldsymbol{R}_1 \boldsymbol{\cdot}\boldsymbol{\nabla} (\boldsymbol{R}_1 \boldsymbol{\cdot} \boldsymbol{\nabla}(\boldsymbol{i}_R \boldsymbol{\cdot} \boldsymbol{\nabla} \varPhi_0 ))= 0, \quad R = \varepsilon, \end{gather}

(2.29)\begin{gather} 3\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla} \varPhi_3 + 2 \boldsymbol{R}_2 \boldsymbol{\cdot}\boldsymbol{\nabla} (\boldsymbol{i}_R \boldsymbol{\cdot} \boldsymbol{\nabla}\varPhi_1)= 0, \quad R = \varepsilon, \end{gather}

where the last equation is valid only for the case of constant acceleration.

2.3. Bipolar coordinates

We will follow the approach proposed by Tyvand & Miloh (Reference Tyvand and Miloh1995) and will solve the exact boundary value problems to each order by means of bipolar coordinates. Figure 3 gives a sketch of the computational domain as represented by bipolar coordinates $\zeta$ and $\theta$. In bipolar coordinates, the curves of constant $\zeta$ correspond to non-concentric circles, concurring with the contour of the cylinder at time $t=0$ and in the limit $\zeta \to 0$ coincide with undisturbed free surface $y = 0$. In the new coordinate system, the doubly connected semi-infinite flow domain is mapped into a rectangular $\{(\theta,\zeta ): -{\rm \pi} < \theta < {\rm \pi},\,0 < \zeta < \zeta _0\}$. Another important feature of bipolar coordinates is that Laplace's equation (2.2) in two dimensions separates to

(2.30)\begin{equation} \left(\frac{\partial^2}{\partial \zeta^2} + \frac{\partial^2}{\partial \theta^2} \right)\varPhi = 0. \end{equation}

Figure 3. Coordinate lines of bipolar coordinate system $(\zeta,\theta )$.

The transformation equations between Cartesian and bipolar coordinate systems is given by

(2.31a,b)\begin{equation} x = \frac{r\sin \theta}{\cosh \zeta + \cos \theta}, \quad y ={-} \frac{r\sinh \zeta}{\cosh \zeta + \cos \theta}. \end{equation}

The scale factors for $\zeta$ and $\theta$ are equal and given by

(2.32)\begin{equation} h_\zeta = h_\theta = \frac{r}{\cosh \zeta + \cos \theta}, \end{equation}

where $r$ is a dimensionless length, representing the centres $(0, \pm r)$ of the bipolar coordinate system and is related to the radius via

(2.33)\begin{equation} r = \sqrt{1 - \varepsilon^2}. \end{equation}

The expressions of the bipolar coordinate $\zeta _0$, appearing in the mathematical derivations of subsequent sections, can be represented through the dimensionless radius of the cylinder,

(2.34a–c)\begin{equation} \sinh\zeta_0 = \frac{r}{\varepsilon}, \quad \cosh\zeta_0 = \frac{1}{\varepsilon}, \quad \tanh\zeta_0 = r. \end{equation}

For simplicity of mathematical expressions, that will follow, we introduce a series of constants determined through hyperbolic functions by analogy with constants $r$ and $\varepsilon$ from the above,

(2.35a–c)\begin{equation} \sinh n\zeta_0 = \frac{r_n}{\varepsilon_n}, \quad \cosh n\zeta_0 = \frac1{\varepsilon_n}, \quad \tanh n\zeta_0 = r_n, \quad n = 2,3,\dots \, . \end{equation}

The angle transformations between polar and bipolar coordinate systems will become useful when integrating along the cylinder contour,

(2.36a–c)\begin{equation} \sin \varTheta = \frac{r\sin \theta}{1+ \varepsilon\cos \theta}, \quad \cos \varTheta = \frac{\varepsilon + \cos \theta}{1 + \varepsilon\cos \theta}, \quad \frac{{\rm d}\varTheta}{{\rm d}\theta} = \frac{r}{1+\varepsilon\cos\theta}. \end{equation}

Note, that expressions (2.36a–c) are only valid along the cylinder contour. In what follows, we will utilize bipolar coordinates to represent the solutions. The derivatives in the equations presented above in Cartesian coordinates will be transformed to bipolar coordinates by the formula

(2.37)\begin{equation} \left.\frac{\partial }{\partial y}\right|_{y=0} = \left.-\frac{2\cos^2(\theta/2)}r\frac{\partial }{\partial \zeta}\right|_{\zeta=0} . \end{equation}

3.1. Zeroth-order potential and first-order free surface elevation

It follows from (2.20) and (2.26) that the zeroth-order potential $\varPhi _0$ includes only the term of $\psi$-type,

(3.1)\begin{equation} \varPhi_0 = \delta \psi_0, \end{equation}

which satisfies the boundary conditions

(3.2a,b)\begin{equation} \psi_0 = 0 \quad (\zeta = 0), \quad \frac{\partial\psi_0}{\partial \zeta} = \delta \,\boldsymbol{R}_1(\alpha)\boldsymbol{\cdot}\boldsymbol{i}_R \quad (\zeta = \zeta_0). \end{equation}

Tyvand & Miloh (Reference Tyvand and Miloh1995), in the problem of the cylinder impulsive motion without acceleration, found function $\psi _0$ by use of the Fourier expansion technique in the form of an infinite series,

(3.3)\begin{equation} \psi_0(\theta,\zeta) = 2 r \sum_{n=1}^\infty \frac{k_n}n \sin (n \theta + \alpha)\sinh (n \zeta), \end{equation}

where coefficients

(3.4)\begin{equation} k_n = \frac{({-}1)^n n e^{{-}n\zeta_0}}{\cosh n\zeta_0}, \quad n = 1, 2, \dots, \end{equation}

are introduced here for brevity. In the case of the cylinder acceleration without initial impulsive velocity ($\delta = 0$), the zeroth-order potential $\psi _0$ gives no contribution to the general solution. The expression for the gradient $\boldsymbol {\nabla }\psi _0$ at the cylinder contour,

(3.5)\begin{equation} \boldsymbol{\nabla} \psi_0(\theta,\zeta_0) = \frac{2(1+\varepsilon\cos\theta)}\varepsilon\sum_{n=1}^\infty \frac{k_n}{\varepsilon_n} \left(\begin{array}{@{}c@{}} r_n\cos(n\theta + \alpha) \\ \sin(n\theta+\alpha) \end{array}\!\! \right), \end{equation}

will be useful for calculation of the hydrodynamic force in § 5.

The first-order free surface elevation is derived from (2.15) with the use of coordinate transformation (2.37), and attains the form

(3.6)\begin{equation} \eta_1(\theta; \alpha) ={-} 4 \delta \cos^2(\theta/2)\sum_{n=1}^{\infty}k_n\sin(n\theta + \alpha), \end{equation}

or, alternatively, the linear form with respect to the trigonometric functions as

(3.7)\begin{equation} \eta_1(\theta;\alpha) ={-}\delta \sum_{i={-}1}^1\sum_{n=1}^\infty C^2_{i+1}k_n\sin([n+i]\theta + \alpha), \end{equation}

where $C^2_{i+1}$ is the binomial coefficient, equal to the number of $(i+1)$-combinations of a set with two elements. The expression (3.7) will be sometimes preferable for the purpose of the higher-order analysis of the solution. Hereafter, the free surface elevation can be expressed in Cartesian coordinates by using the inverse transformation formula,

(3.8)\begin{equation} \theta(x) = 2\arctan\left(x/r\right). \end{equation}

In the following subsections, we will apply the same approach to calculate next-order terms of the flow potential taking into account the influence of impulsive acceleration and its interaction with impulsive velocity.

3.2. First-order potential and second-order free surface elevation

The first-order potential $\varPhi _1$ has two contributors: from the self-interaction of initial velocity proportional to $\delta ^2$; and from the initial acceleration, proportional to $a$,

(3.9)\begin{equation} \varPhi_1 = \delta^2(\hat{\phi}_{1} + \hat{\psi}_1) + a\psi_{1}, \end{equation}

which means that acceleration of the cylinder has an effect on the flow starting from the first order. This follows from substitution of the functions $\eta _1$ and $\psi _0$ obtained above into the boundary conditions (2.21) and (2.27). Thus, we can formulate the conditions satisfied by harmonic functions $\hat {\phi }_1$, $\hat {\psi }_1$ and $\psi _1$ on both boundaries of the flow domain,

(3.10a,b) $$\begin{gather} \hat{\phi}_1={-}\frac12\frac{\eta_1^2}{\delta^{2}} \quad (\zeta = 0), \quad \frac{\partial\hat{\phi}_1}{\partial\zeta} = 0 \quad (\zeta = \zeta_0), \end{gather}$$

(3.11a,b)$$\begin{gather}\hat{\psi}_1=0 \quad (\zeta = 0), \quad \frac{\partial\hat\psi_1}{\partial \zeta} = - \frac{\boldsymbol{R}_1(\alpha)}\delta \boldsymbol{\cdot} \nabla (\boldsymbol{i}_R \boldsymbol{\cdot} \nabla \psi_0) \quad (\zeta = \zeta_0), \end{gather}$$

(3.12a,b)$$\begin{gather}\psi_1 = 0, \quad (\zeta = 0), \quad \frac{\partial\psi_1}{\partial\zeta} = \frac{2\boldsymbol{R}_2(\beta)}a\boldsymbol{\cdot}\boldsymbol{i}_R \quad (\zeta = \zeta_0). \end{gather}$$

Potentials $\hat {\phi }_1$ and $\hat {\psi }_1$, were found previously by Tyvand & Miloh (Reference Tyvand and Miloh1995) for the problem of the cylinder moving with constant impulsive speed without acceleration, and we refer to the solution given therein, reformulated in our notation,

(3.13)\begin{align} \hat{\phi}_{1}(\theta,\zeta) &= \frac14 \sum_{n,m=1}^\infty\sum_{i,j,\kappa={-}1}^1 \kappa\, C^2_{i+1}C^2_{j+1}k_nk_m \cos([n + \kappa m + i +\kappa j]\theta + [\kappa+1]\alpha) \nonumber\\ &\quad \times \frac{\cosh(n+\kappa m+i+\kappa j)(\zeta-\zeta_0)}{\cosh(n+\kappa m + i + \kappa j)\zeta_0}, \end{align}

(3.14)\begin{align} \hat{\psi}_1(\theta,\zeta) &={-}\frac{\boldsymbol{R}_1(\alpha)}\delta\boldsymbol{\nabla} \psi_0 - \sin\alpha\sum_{i={-}1}^1\sum_{n=1}^{\infty}C^2_{i+1}k_n\sin([n+i]\theta+\alpha)\nonumber\\ &\quad \times \frac{\cosh(n+i)(\zeta-\zeta_0)}{\cosh(n+i)\zeta_0}. \end{align}

The respective derivatives from $\hat {\phi }_1$ and $\hat {\psi }_1$, taken at the undisturbed free surface ($\zeta = 0$),

(3.15)\begin{align} \frac{\partial\hat{\phi}_1}{\partial y}(\theta,0)& = \frac{\cos^2(\theta/2)}{2r} \sum_{n,m=1}^\infty\sum_{i,j,\kappa={-}1}^1 \kappa(n+\kappa m+i+\kappa j) C^2_{i+1}C^2_{j+1}k_nk_m r_{n+\kappa m + i + \kappa j} \nonumber\\ &\quad \times \cos([n + \kappa m + i +\kappa j]\theta + [\kappa+1]\alpha), \end{align}

(3.16)\begin{align} \frac{\partial\hat{\psi}_1}{\partial y}(\theta,0) & ={-}\frac{4\cos\alpha}{x^2+r}\sum_{n=1}^{\infty}k_n(\sin\theta\sin(n\theta+\alpha) - 2n\cos^2(\theta/2)\cos(n\theta+\alpha)) \nonumber\\ &\quad -\frac{2\sin\alpha}r \cos^2(\theta/2)\sum_{i={-}1}^1\sum_{n=1}^{\infty}C^2_{i+1}k_nr_{n+i}(n+i)\sin([n+i]\theta+\alpha), \end{align}

will be used below in the calculation of the higher-order terms in the expansion of the free surface elevation. The value of the second function at the cylinder contour will be needed in the calculation of the forces in § 5:

(3.17)\begin{align} \hat{\psi}_1(\theta,\zeta_0) &= \frac{2r\sin\theta}\varepsilon \sum_{n=1}^\infty \frac{k_n}{\varepsilon_n}\left( \cos\alpha\sin(n\theta+\alpha) + r_n \sin\alpha\cos(n\theta+\alpha)\right) \nonumber\\ &\quad + \frac{2(\varepsilon+\cos\theta)}\varepsilon \sum_{n=1}^\infty \frac{k_n}{\varepsilon_n} \left( \sin\alpha\sin(n\theta+\alpha) - r_n \cos\alpha\cos(n\theta+\alpha) \right) \nonumber\\ &\quad -\sin\alpha \sum_{i={-}1}^1 \sum_{n=1}^\infty C_{i+1}^2 k_n\varepsilon_{n+i} \sin([n+i]\theta+\alpha). \end{align}

The first-order potential $\psi _{1}$ due to impulsive acceleration can be related to the zeroth-order potential $\psi _0$ due to impulsive velocity by making the substitution $\alpha \to \beta$ in (3.2a,b) and using transformation (2.9a,b), as follows:

(3.18)\begin{equation} \psi_1(\theta,\zeta;\beta) = \psi_0(\theta,\zeta;\beta) = 2r \sum_{n=1}^\infty \frac{k_n}n\sin (n \theta + \beta)\sinh(n\zeta). \end{equation}

Applying expansion (3.9) to free surface boundary condition (2.16) gives the corresponding expansion for the second-order free surface elevation, expressed in terms of the first-order potentials,

(3.19)\begin{equation} \eta_2(x) = \frac{\delta^2}2\left(\frac{\partial\hat{\phi}_1}{\partial y}(x,0)+\frac{\partial\hat{\psi}_1}{\partial y}(x,0)\right)+ \frac{a}2 \frac{\partial\psi_1}{\partial y}(x,0). \end{equation}

Replacing the respective derivatives of the potentials with the expressions provided above, we obtain the explicit formula for the calculation of the second-order free surface elevation, induced by both impulsive velocity and impulsive acceleration,

(3.20)\begin{align} \eta_2(\theta;\alpha,\beta) &= \frac{\delta^2\cos^2(\theta/2)}{4r} \sum_{n,m=1}^\infty\sum_{i,j,\kappa={-}1}^1 \kappa(n+\kappa m+i+\kappa j) C^2_{i+1}C^2_{j+1}k_nk_m r_{n+\kappa m + i + \kappa j} \nonumber\\ &\quad \times \cos([n + \kappa m + i +\kappa j]\theta + [\kappa+1]\alpha) \nonumber\\ &\quad - \frac{2\delta^2\cos\alpha}{x^2+r} \sum_{n=1}^{\infty}k_n(\sin\theta\sin(n\theta+\alpha) - 2n \cos^2(\theta/2)\cos(n\theta+\alpha)) \nonumber\\ &\quad-\frac{\delta^2\sin\alpha\cos^2(\theta/2)}r \sum_{i={-}1}^1\sum_{n=1}^{\infty}C^2_{i+1}k_nr_{n+i}(n+i)\sin([n+i]\theta+\alpha) \nonumber\\ &\quad + \eta_{2a}(\theta;\beta). \end{align}

Here the term $\eta _{2a}$ is responsible for the contribution from the acceleration alone and is directly related to the first-order solution $\eta _1$ via the transformation

(3.21)\begin{equation} \eta_{2a}(\theta, \beta) = \frac{a}{2}\frac{\partial\psi_1}{\partial y}(x,0;\beta) = \frac{a}{2}\frac{\partial\psi_0}{\partial y}(x,0;\beta) = \frac{a}{2\delta}\eta_1(\theta;\beta). \end{equation}

It is remarkable that the effect of the forced acceleration on the free surface flow is inherited from that of the forced velocity and is explicitly described by the formula

(3.22)\begin{equation} \eta_{2a}(\theta, \beta) ={-}2a \cos^2(\theta/2)\sum_{n=1}^{\infty}k_n\sin(n\theta + \beta). \end{equation}

The analysis of the general solution for arbitrary angles of impulsive velocity and acceleration can be complicated. Therefore, we limit the presentation of our results to the cases where both vectors of impulsive velocity and acceleration are aligned and the cylinder motion is rectilinear. Figure 4 illustrates the effects of reciprocity between initial impulsive velocity and acceleration on the second-order free surface elevation, predicted by (3.20), for the cylinders of two representative radii starting to move in three primary directions. The plots in figure 4 were obtained by truncating the sums in (3.20) to 20 terms. For upward motion of the cylinder, initial impulsive velocity and acceleration reinforce each other creating a larger cumulative swell above the body. For vertical submersion, two factors work in opposite directions resulting in the mitigation of the free surface disturbance. As it is observed in figure 4, acceleration is responsible for the wave, which travels with the horizontally moving cylinder, while velocity creates the wave, which is symmetric with respect to the vertical axis. This can be explained by the fact that the respective terms of the solution due to velocity and acceleration are represented by even and odd functions, respectively.

Figure 4. Second-order free surface elevation $\eta _2(x)$ for the cylinder of non-dimensional radius (a–c) $\varepsilon = 0.4$ and (d–f) $\varepsilon = 0.8$ moving in different directions: (a,d) upward ($\alpha = \beta = {\rm \pi}/2$), (b,e) downward ($\alpha = \beta = -{\rm \pi} /2$), (c, f) horizontal ($\alpha = \beta = 0$).

3.3. Second-order potential and third-order free surface elevation

As will be appreciated from the description below, the second-order potential $\varPhi _2$ has three contributions, among which are the effect of the gravitational force and the leading nonlinear interaction between impulsive velocity and impulsive acceleration with arbitrary respective angles $\alpha$ and $\beta$ of initial motion,

(3.23)\begin{equation} \varPhi_2 = \delta\phi_{2g} + a\delta (\phi_2 + \psi_2)+ \delta^3(\bar{\phi}_2 + \bar{\psi}_2). \end{equation}

The term with $\delta ^3$, originating from triple self-interaction of the impulsive velocity, will be disregarded in subsequent considerations. It should be noted that $\varPhi _2$ gives no contribution to the solution when initial impulsive velocity is zero ($\delta = 0$) and includes no self-interaction of impulsive acceleration. Using the solutions to the previous orders given above, the kinematic and dynamic boundary conditions (2.22) and (2.28) can be broken down into three distinct boundary-value problems for each of the harmonic functions in expansion (3.23):

(3.24a,b)$$\begin{gather} \phi_{2g}={-}\frac\lambda{2\delta}\eta_1 \quad (\zeta = 0), \quad \frac{\partial\phi_{2g}}{\partial\zeta} = 0 \quad (\zeta = \zeta_0), \end{gather}$$

(3.25a,b)$$\begin{gather}\phi_2={-}2\frac{\eta_1}{\delta}\,\frac{\eta_{2a}}{a} \quad (\zeta = 0), \quad \frac{\partial\phi_2}{\partial\zeta} = 0 \quad (\zeta = \zeta_0) , \end{gather}$$

(3.26a,b)$$\begin{gather}\psi_2=0 \quad (\zeta = 0), \quad \frac{\partial \psi_2}{\partial\zeta} ={-}\frac{\boldsymbol{R}_2(\beta)}{a} \boldsymbol{\cdot} \boldsymbol{\nabla}(\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla} \psi_0) - \frac{\boldsymbol{R}_1(\alpha)}{\delta} \boldsymbol{\cdot}\boldsymbol{\nabla}(\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla} \psi_1) \quad (\zeta = \zeta_0). \end{gather}$$

Equations (3.24a,b) and (3.25a,b) can be solved immediately by analytical continuation of the functions in the right-hand sides into the half-plane $\zeta > 0$. Here, it is convenient to use (3.7) to expand these into the linear series with respect to trigonometric functions. The unknown potentials can be found, as follows:

(3.27)\begin{align} \phi_{2g}(\theta,\zeta;\alpha) &= \frac\lambda2 \sum_{i={-}1}^1 \sum_{n=1}^{\infty} C^2_{i+1}k_n \sin([n+i]\theta + \alpha)\frac{\cosh (n+i)(\zeta - \zeta_0)}{\cosh (n+i)\zeta_0}, \end{align}

(3.28)\begin{align} \phi_2(\theta,\zeta;\alpha,\beta) &= \frac12\sum_{i,j,\kappa={-}1}^1 \sum_{n,m=1}^{\infty} \kappa C^2_{i+1}C^2_{j+1} k_n k_m \nonumber\\ &\quad \times \cos([n+\kappa m+i+\kappa j]\theta +\alpha +\kappa\beta)\frac{\cosh(n+\kappa m+i+\kappa j)(\zeta-\zeta_0)}{\cosh (n+\kappa m+i+\kappa j)\zeta_0} . \end{align}

With transformations (2.9a,b) and (3.18) between the leading-order terms of the two modes of motion, the condition (3.26b) at the cylinder contour attains the form

(3.29)\begin{equation} \frac{\partial \psi_2}{\partial\zeta} ={-}\frac{\boldsymbol{R}_1(\beta)}{2\delta} \boldsymbol{\cdot} \boldsymbol{\nabla}(\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla} \psi_0(\alpha)) - \frac{\boldsymbol{R}_1(\alpha)}{\delta} \boldsymbol{\cdot}\boldsymbol{\nabla} (\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla}\psi_0(\beta)) \quad (\zeta = \zeta_0). \end{equation}

Hence the function $\psi _2$ can be broken down into two similar components differing by a constant and positioning of the arguments $\alpha$ and $\beta$:

(3.30)\begin{equation} \psi_2(\alpha,\beta) = \tfrac{1}{2}\tilde{\psi}_2(\beta,\alpha) + \tilde{\psi}_2(\alpha,\beta). \end{equation}

Then we look for harmonic function $\tilde {\psi }_2$, which satisfies the following boundary conditions:

(3.31a,b)\begin{equation} \tilde{\psi}_2 = 0 \quad (\zeta = 0), \quad \frac{\partial\tilde{\psi}_2}{\partial \zeta} = - \frac{\boldsymbol{R}_1(\alpha)}{\delta} \boldsymbol{\cdot} \nabla (\mathbf {i}_R \boldsymbol{\cdot} \nabla \psi_0(\beta)) \quad (\zeta = \zeta_0). \end{equation}

From this point, it is sufficient to work with the vector of impulsive velocity $\boldsymbol {R}_1$ and zeroth-order potential $\psi _0$, which refer to the case of constant impulsive velocity only. Thus, function $\tilde {\psi }_2$ expresses the effect of finite penetration of constant velocity into the flow field established by the constant acceleration. By mere substitution we can calculate the effect of finite penetration of constant acceleration into the flow field established by constant velocity. It should not be surprising that these two effects are linked by the factor $\frac 12$, indicating that the second effect appears as the smallest of the two. In view of the fact that (3.11a,b) and (3.31a,b) share the same structure, we can express the solution $\psi _2$ based on the previous results as

(3.32)\begin{align} \psi_{2}(\theta,\zeta;\alpha,\beta) &={-}\frac{\boldsymbol{R}_1(\alpha)}{\delta}\boldsymbol{\cdot}\boldsymbol{\nabla}\psi_0(\beta)- \frac{\boldsymbol{R}_1(\beta)}{2\delta}\boldsymbol{\cdot}\boldsymbol{\nabla}\psi_0(\alpha)\nonumber\\ &\quad - \sum_{i={-}1}^1\sum_{n=1}^{\infty}C_{i+1}^2 k_n\left(\sin\alpha\sin([n+i]\theta + \beta)+\frac12\sin\beta\sin([n+i]\theta+\alpha)\right)\nonumber\\ &\quad \times \frac{\cosh (n+i)(\zeta - \zeta_0)}{\cosh(n+i)\zeta_0}, \end{align}

where the gradient operator refers to the Cartesian coordinate system, to which all the physical quantities belong. The value of this potential at the cylinder contour will be used in the calculation of the forces in § 5:

(3.33)\begin{align} \psi_{2}(\theta,\zeta_0) & ={-}\frac{\boldsymbol{R}_1(\alpha)}{\delta} \boldsymbol{\cdot}\boldsymbol{\nabla}\psi_0(\beta)- \frac{\boldsymbol{R}_1(\beta)}{2\delta} \boldsymbol{\cdot}\boldsymbol{\nabla}\psi_0(\alpha) \nonumber\\ &\quad - \sum_{i={-}1}^1\sum_{n=1}^{\infty}C_{i+1}^2 k_n\varepsilon_{n+i}\left(\!\sin\alpha\sin([n+i]\theta + \beta)+\frac12\sin\beta\sin([n+i]\theta+\alpha)\!\right). \end{align}

Substituting the expansions (3.1), (3.9) and (3.23) into (2.17) and neglecting the terms proportional to $\delta ^3$, we find the following representation of the third-order free surface elevation through derivatives of the functions obtained above:

(3.34)\begin{align} \eta_3(x) &=\frac{\delta}3\frac{\partial\phi_{2g}}{\partial y}(x,0) \nonumber\\ &\quad +\frac{a\delta}3\left( \frac{\partial\phi_2}{\partial y}(x,0) + \frac{\partial\psi_2}{\partial y}(x,0) + \frac{\partial \phi_0}{\partial y}\frac{\partial^2 \psi_1}{\partial y^2}(x,0) - \frac{\partial^2 \phi_0}{\partial x\partial y}\frac{\partial \psi_1}{\partial x}(x,0)\right). \end{align}

The equation (3.34) can be further simplified by use of the fact that the derivatives of $\psi _1$ in the last two terms of the formula vanish. Based on previous results of this section, we find the explicit formula for the third-order free surface elevation in bipolar coordinates:

(3.35)\begin{align} &\eta_3(\theta;\alpha,\beta)\nonumber\\ &\quad = \eta_{3g}(\theta;\alpha) + \frac{a\delta \cos^2(\theta/2)}{3r}\nonumber\\ &\qquad \times\sum_{n,m=1}^{\infty}\sum_{i,j,\kappa={-}1}^1 C^2_{i+1}C^2_{j+1}\kappa(n+\kappa m+i+\kappa j)k_nk_m r_{n+\kappa m+i+\kappa j}\nonumber\\ &\qquad \times \cos([n+\kappa m+i+\kappa j]\theta+\alpha+\kappa\beta)\nonumber\\ &\qquad -\frac{4a\delta\sin\theta}{3(x^2+r)}\sum_{n=1}^\infty k_n\left(\sin(n\theta+\beta)\cos\alpha + \frac12\sin(n\theta+\alpha)\cos\beta\right) \nonumber\\ &\qquad + \frac{4a\delta(1+\cos\theta)}{3(x^2+r)}\sum_{n=1}^\infty nk_n\left(\cos(n\theta+\beta)\cos\alpha + \frac12\cos(n\theta+\alpha)\cos\beta\right)\nonumber\\ &\qquad - \frac{a\delta(1+\cos\theta)}{3r} \sum_{i={-}1}^1\sum_{n=1}^{\infty} C_{i+1}^2k_nr_{n+i}(n+i)\nonumber\\ &\qquad \times \left(\sin\alpha\sin([n+i]\theta+\beta)+\frac12\sin\beta\sin([n+i]\theta+\alpha)\right), \end{align}

where the term

(3.36)\begin{equation} \eta_{3g}(\theta;\alpha) = \frac{\delta\lambda\cos^2(\theta/2)}{3r}\sum_{n=1}^{\infty}\sum_{i={-}1}^1C^2_{i+1}(n+i)k_n r_{n+i} \sin([n+i]\theta+\alpha) \end{equation}

represents the contribution of the gravitational force.

The way the third-order free surface elevation $\eta _3$ depends on $\alpha$ and $\beta$ has additional complexities compared with the second-order term $\eta _2$. It contains the products of trigonometric functions depending on the two direction angles $\alpha$ and $\beta$, representing the forced velocity and the forced acceleration, respectively. This dependency demonstrates that the interplay between the velocity and acceleration grows stronger with time.

Figure 5 shows the third-order free surface elevation predicted by (3.35), with the sums truncated to 20 terms, for different non-dimensional radii of the cylinder and directions of motion. Without acceleration the gravity is the only driving force for the fluid motion to this order. By prescribing alternately $a = 0$ and $a = 1$ in (3.35), we can isolate the terms responsible for the gravitational force and nonlinear interaction between impulsive velocity and acceleration. Figure 5 demonstrates that impulsive acceleration intensifies the flow field, induced by gravitational force, when the cylinder moves vertically, and violates the symmetry of the wave profile when the cylinder moves horizontally. Nonlinear terms contribute to the splitting of the splash jet formed by impulsive submersion of the cylinder observed for the cylinders of bigger radius.

Figure 5. Third-order free surface elevation $\eta _3(x)$ for $\lambda = 6$ and the cylinder of non-dimensional radius (a–c) $\varepsilon = 0.4$ and (d–f) $\varepsilon = 0.8$ moving in different directions: (a,d) upward ($\alpha = \beta = {\rm \pi}/2$), (b,e) downward ($\alpha = \beta = -{\rm \pi} /2$), (c, f) horizontal ($\alpha = \beta = 0$).

3.4. Third-order potential and fourth-order free surface elevation

So far, we have included the impulsive velocity and impulsive acceleration in our development of the solution in the form of the Taylor expansion in time. The impulsive start of the cylinder was accompanied by simultaneous velocity and acceleration directed arbitrarily. In this paragraph, we sum up the results for the case of pure acceleration ($\delta = 0, a=1$), and calculate the third-order potential and the fourth-order free surface elevation. In this case, the zeroth- and second-order potentials $\varPhi _0$ and $\varPhi _2$ give no contributions to the general solution. The only non-zero component of the first-order potential $\psi _1$ is related to the zeroth-order potential $\psi _0$ by (3.18) and the leading-order free surface elevation $\eta _{2a}$ is given by (3.22).

According to boundary conditions (2.23) and (2.29) the third-order velocity potential $\varPhi _3$ can be decomposed into three harmonic components

(3.37)\begin{equation} \varPhi_3 = \phi_{3g} + \phi_3 + \psi_3,\end{equation}

satisfying the following boundary conditions:

(3.38a,b)$$\begin{gather} \phi_{3g}={-}\frac{\lambda}{3}\eta_{2a} \quad (\zeta = 0), \quad \frac{\partial\phi_{3g}}{\partial\zeta} = 0 \quad (\zeta = \zeta_0), \end{gather}$$

(3.39a,b)$$\begin{gather}\phi_3={-}\frac{4\eta_{2a}^2}3 \quad (\zeta = 0), \quad \frac{\partial\phi_3}{\partial\zeta} = 0 \quad (\zeta = \zeta_0) , \end{gather}$$

(3.40a,b)$$\begin{gather}\psi_3=0 \quad (\zeta = 0), \quad \frac{\partial \psi_3}{\partial\zeta} ={-}\frac{2\boldsymbol{R}_2}{3} \boldsymbol{\cdot}\boldsymbol{\nabla}(\boldsymbol{i}_R \boldsymbol{\cdot}\boldsymbol{\nabla} \psi_{1}) \quad (\zeta = \zeta_0) . \end{gather}$$

Since the dimensionless acceleration parameter $a$ equals unity in this case, the different powers of $a$ are not relevant in expansion (3.37). In the absence of the forced impulsive velocity, the dimensionless parameter $\lambda$ in (3.38a,b) represents the ratio between the gravitational acceleration and the forced acceleration of the cylinder: $\lambda = g/a_0$.

Conditions (3.38a,b) and (3.39a,b) can be brought into correspondence with conditions (3.24a,b) and (3.10a,b), respectively, by use of (3.21) relating the second-order free surface elevation $\eta _{2a}$ due to impulsive acceleration with the first-order free surface elevation $\eta _1$ due to impulsive velocity. Hence, combining the results of the previous paragraphs, the first two components of the third-order potentials can be related to the previous-order contributions:

(3.41a,b)\begin{equation} \phi_{3g}(\theta,\zeta;\beta) = \tfrac{1}{3} \phi_{2g}(\theta,\zeta;\beta), \quad \phi_3(\theta,\zeta;\beta) = \tfrac{2}{3} \hat{\phi}_1(\theta,\zeta;\beta). \end{equation}

The similarity between two sets of (3.11a,b) and (3.40a,b) allows us to establish the direct correspondence between their solutions by use of transformation (2.9a,b), relating the vectors of impulsive velocity and acceleration. By analogy, we can obtain

(3.42)\begin{equation} \psi_3(\theta,\zeta;\beta) = \tfrac{1}{3}\hat{\psi}_1(\theta,\zeta;\beta). \end{equation}

As implied by (2.18), in the case of pure acceleration, the only contribution to the fourth-order surface elevation $\eta _4$ comes from the third-order potential $\varPhi _3$. It is worth mentioning that the second-order surface elevation $\eta _2$ has been obtained previously by a similar formula (see (2.16)). Fourth-order free surface elevation can be then found as

(3.43)\begin{equation} \eta_{4a} = \frac{1}{12}\,\frac{\partial \phi_{2g}}{\partial y}(x,0) +\frac{1}{6}\,\frac{\partial \hat{\phi}_1}{\partial y}(x,0)+ \frac{1}{12}\,\frac{\partial\hat{\psi}_1}{\partial y}(x,0). \end{equation}

Comparing (3.34) and (3.43), we can point out that gravitational third- and fourth-order terms of free surface elevation are directly related through the constant $1/4$.

We have now completed the nonlinear theory up to the fourth order in time for the free surface elevation caused by impulsive start of a circular cylinder of arbitrary radius from a state of rest with constant acceleration.

4.1. The non-gravitational linearized elevation of the free surface

We will now study the instantaneous elevation of the free surface caused by a finite displacement of the cylinder $\Delta \boldsymbol {R}$. We neglect gravitational effects since we want to discuss the relationship between the finite displacement of the cylinder and the resulting finite elevations of the initially horizontal free surface.

In the general case, the displacement of the cylinder, put into the forced motion with initial velocity and initial acceleration during the time interval $\Delta t$, is given by the vector sum

(4.1)\begin{equation} \Delta \boldsymbol{R} = \boldsymbol{R}_1 \Delta t + \boldsymbol{R}_2 (\Delta t)^2, \end{equation}

and in the Cartesian coordinates as

(4.2a,b)\begin{equation} \Delta X = \delta\Delta t \cos \alpha + \frac{a}{2}(\Delta t)^2 \cos \beta, \quad \Delta Y = \delta\Delta t \sin \alpha + \frac{a}{2} (\Delta t)^2 \sin \beta, \end{equation}

where $\Delta X$ and $\Delta Y$ are the horizontal and vertical components of the cylinder displacement, respectively.

To check that the superposition principle holds for the net effect of the two added motions, we calculate the total free surface elevation according to the linear theory:

(4.3)\begin{equation} \eta(\theta, \Delta t) = \eta_1(\theta;\alpha) \Delta t + \eta_{2a}(\theta;\beta) (\Delta t)^2. \end{equation}

From (3.21), after inserting $\delta = 1$, we have

(4.4)\begin{equation} \eta_{2a}(\theta;\beta) = \frac{a}{2}\eta_1(\theta;\beta). \end{equation}

The analytical verification of the superposition principle for the total free surface elevation due to the general motions with different direction angles for the initial velocity ($\alpha$) and initial acceleration ($\beta$) can be carried out numerically. The focus of the present paper is on the superposition of leading-order elevations for the case when two direction angles are equal ($\alpha = \beta$). In this case, (4.3) reduces to

(4.5)\begin{equation} \eta(\theta, \Delta t) = \eta_1(\theta;\alpha) \Delta t \left(1 + \frac{a}{2}\Delta t \right), \end{equation}

and the horizontal and vertical displacements are related as

(4.6)\begin{equation} \frac{\Delta X}{\cos \alpha} = \frac{\Delta Y}{\sin \alpha} = \Delta t \left(1 + \frac a2 \Delta t \right). \end{equation}

The last two equations show that initial acceleration aligned with the initial velocity increases the displacement components by a factor equal to the relative increment of the free surface elevation. Thus, we have confirmed the validity of the superposition principle for a rectilinear motion of the cylinder, by which the total free surface elevation sums up the contributions from the initially forced velocity and acceleration.

4.2. On the leading nonlinear free surface elevation

In our further analysis we will compare the free surface elevations predicted by the leading-order nonlinear theory, treating constant velocity and constant acceleration separately. In that case, when both direction angles are equal, the radial displacement of the cylinder during the time interval $\Delta t$ is given by

(4.7)\begin{equation} \Delta R = \delta\Delta t + \frac{a}{2}(\Delta t)^2. \end{equation}

If to consider the two cases of cylinder motion separately, the time interval needed to achieve a radial displacement $\Delta R$ is given by two options:

(4.8)\begin{equation} \Delta t = \left\{\!\!\!\begin{array}{ll} \Delta R, & \delta = 1, \ a=0, \\ \sqrt{2\Delta R}, & \delta = 0, \ a = 1. \end{array}\right. \end{equation}

We consider the exposition of a total nonlinear solution, found in § 3, as a small-parameter expansion in terms of the dimensionless radial displacement $\Delta R$ instead of a small-time expansion, using the substitution of variables (4.8). The two versions of the free surface elevation, corresponding to the cases of constant impulsive velocity and constant impulsive acceleration, can be written in the form

(4.9)\begin{equation} \eta(x,\Delta R) = \left\{\!\!\!\begin{array}{ll} \eta_1(x)\Delta R + \left( \dfrac12 \dfrac{\partial\hat{\phi}_1}{\partial y}(x,0) + \dfrac12 \dfrac{\partial\hat{\psi}_1}{\partial y} (x,0)\right)\Delta R^2 + \dots, & \delta = 1, \ a=0,\\[12pt] 2\eta_{2a}(x)\Delta R + \left( \dfrac23 \dfrac{\partial\hat{\phi}_1}{\partial y}(x,0) + \dfrac13 \dfrac{\partial\hat{\psi}_1}{\partial y} (x,0)\right)\Delta R^2 + \dots, & \delta = 0, \ a=1. \end{array}\right. \end{equation}

Note, that contribution of the gravitational force has been neglected in (4.9).

Earlier, we have established that free surface elevations to the leading order, $\eta _1$ and $\eta _{2a}$, are related by factor $\frac 12$, see (4.4). Here, we elaborate a similar comparison for the leading nonlinear contributions to the free surface elevation induced by the forced velocity and the forced acceleration. In (4.9), the nonlinear terms consist of two parts, responsible for the free-surface nonlinearity, $\partial \hat {\phi }_1/\partial y$, and for the nonlinearity due to the finite displacement of the cylinder, $\partial \hat {\psi }_1/\partial y$, with different weights. The case of constant acceleration increases the elevation due to the dynamic pressure at the free surface by factor $\frac 43$, compared with the case of constant impulsive velocity. The finite penetration of the cylinder into the fluid which has the opposite effect of decreasing the elevation by factor $\frac 23$. Since the relative contributions of $\partial \hat {\phi }_1/\partial y$ and $\partial \hat {\psi }_1/\partial y$ depend on parameters $\varepsilon$ and $\alpha$, we cannot produce the general conclusion whether the leading nonlinear effects on the surface elevation are enhanced or reduced for constant acceleration, as compared with constant velocity leading to the same displacement of the cylinder.

4.3. Some comments on the accuracy of the expansions

We have now compared the leading-order free surface elevations induced by a cylinder with constant acceleration and by a cylinder with constant impulsive velocity. With time chosen to ensure equal displacements of the cylinders, the nonlinear free surface elevations for these two cases will always differ, but their orders of magnitude will remain equal.

It would be a challenge to check the accuracy of our expansion in comparison with the full nonlinear problem. We will not attempt it here and refer to one recent work where such detailed comparisons have been carried out instead. Tyvand, Mulstad & Bestehorn (Reference Tyvand, Mulstad and Bestehorn2021a) solved a nonlinear Cauchy–Poisson problem of impulsive start with a small-time expansion with respect to the free surface elevation. The analysis was similar to the present one, but was implemented one order further, providing the exact solution to the third order in time. This expansion in Eulerian variables was appropriately followed up by a second-order expansion in terms of the Lagrangian description of motion (Tyvand, Mulstad & Bestehorn Reference Tyvand, Mulstad and Bestehorn2021b).

Since the Lagrangian description is expected to treat the free surface nonlinearity more efficiently than the Eulerian, these two truncated expansions could be considered as sufficient approximations to the exact nonlinear problem as far as they were sufficiently close to one another. This plausible argument was confirmed by a third approach of a fully nonlinear numerical simulation, described in Tyvand et al. (Reference Tyvand, Mulstad and Bestehorn2021a) and carried out in Tyvand et al. (Reference Tyvand, Mulstad and Bestehorn2021b).

These publications reveal how the breakdown of asymptotic solutions can be visualized when the time limit for convergence is reached for strongly nonlinear flows. The Lagrangian solution develops a visible mass defect, where there is a mismatch between the water above and below the zero level of the initial free surface. The Eulerian solution obeys exact mass balance, but it develops undulations of the surface, which are unphysical and absent in the Lagrangian solution. Such analyses are outside the reach of our present work, but illustrates how Lagrangian and Eulerian approaches to the same problem may provide mutual corrections for the respective asymptotic theories.

The experiments by Greenhow & Lin (Reference Greenhow and Lin1983) showed that a large cylinder in rapid upward motion tends to surround itself with an almost uniformly thick fluid layer in the radial direction around the cylinder. Tyvand & Miloh (Reference Tyvand and Miloh1995) confirmed this effect for the case of impulsive velocity. Equation (4.9) shows that this trend of a radially uniform fluid layer above the cylinder will be preserved for any rapid upward motion starting from rest with undisturbed free surface.

5.1. The zeroth-order force

The total steady hydrodynamic force arises as a result of two simultaneous processes: the emerging force induced by impulsive velocity and the immediate reaction force due to impulsive acceleration. The instantaneous force is evaluated as a Fourier series in bipolar coordinates as

(5.8)\begin{equation} \boldsymbol{F}_{{-}1} ={-}\delta \int_{S_\varepsilon}\psi_0\,{\rm d}\boldsymbol{S} = 4{\rm \pi} \delta r^2 \sum_{n=1}^\infty \frac{r_nk_n^2}{n\varepsilon_n^2} \left(\begin{array}{@{}c@{}} \cos \alpha \\ \sin\alpha \end{array}\right). \end{equation}

Following the expansions (3.1) and (3.9), the zeroth-order force (5.5) can be decomposed into four integral components:

(5.9)\begin{equation} \boldsymbol{F}_0 ={-}\delta^2\int_{S_\varepsilon} \hat{\phi}_1 \,{\rm d}\boldsymbol{S} - \delta^2 \int_{S_\varepsilon} \hat{\psi}_1 \,{\rm d}\boldsymbol{S} - \frac{\delta^2}{2} \int_{S_\varepsilon} (\boldsymbol{\nabla}\psi_0)^2 \, {\rm d}\boldsymbol{S} -a\int_{S_\varepsilon} \psi_1\,{\rm d}\boldsymbol{S}. \end{equation}

Here, the first integral represents the steady force due to the dynamic pressure set up by the initial flow surrounding the cylinder, the second integral is due to the finite geometric displacement of the cylinder, the third integral is the steady force set up from the free surface due to its leading-order nonlinearity, and the fourth integral is responsible for the force influenced by acceleration. Integrating the functions obtained in the previous section over the cylinder surface leads to

(5.10) \begin{align} \int_{S_\varepsilon} \hat{\phi}_1 \,{\rm d}\boldsymbol{S} &= {-}\frac{{\rm \pi} r}{2} \sum_{i,j,\kappa={-}1}^1\sum_{n,m=1}^\infty \kappa C_{i+1}^2C_{j+1}^2k_n k_m k_{n+\kappa m+i+\kappa j} \left(\begin{array}{@{}c@{}} \sin(\kappa+1)\alpha \\ \cos(\kappa+1)\alpha \end{array}\right), \end{align}

(5.11) \begin{align} \int_{S_\varepsilon}\hat \psi_1 {\rm d}\boldsymbol{S} &= \Bigl[- 4{\rm \pi} r\sum_{n=1}^\infty\sum_{i=0}^1 \frac{k_nk_{n+i}}{\varepsilon^i\varepsilon_n\varepsilon_{n+i}} + \frac{2{\rm \pi} r^2}\varepsilon \sum_{n=1}^\infty \frac{k_nk_{n+1}(r_{n+1}-r_n)}{\varepsilon_n\varepsilon_{n+1}}\Bigr]\left( \begin{array}{@{}c@{}} \sin\alpha\cos\alpha \\ \sin^2\alpha\end{array} \right)\nonumber\\ &\quad + \Bigl[4{\rm \pi} r\sum_{n=1}^\infty \frac{k_n^2r_n}{\varepsilon_n^2} + \frac{2{\rm \pi} r}\varepsilon \sum_{n=1}^\infty \frac{k_nk_{n+1}(r_{n+1}+r_n)}{\varepsilon_n\varepsilon_{n+1}}\Bigr]\left( \begin{array}{@{}c@{}} -\sin\alpha\cos\alpha \\ \cos^2\alpha\end{array} \right)\nonumber\\ &\quad + 2{\rm \pi} r\sin\alpha\sum_{n=1}^\infty\sum_{i=-1}^1 C_{i+1}^2 k_nk_{n+i} \left( \begin{array}{@{}c@{}} \cos\alpha \\ \sin\alpha \end{array} \right), \end{align}

(5.12)\begin{align} \int_{S_\varepsilon} (\boldsymbol{\nabla} \psi_0)^2 \,{\rm d}\boldsymbol{S} &= 4{\rm \pi} r \sum_{n=1}^\infty\sum_{i=0}^1 \frac{k_nk_{n+i}}{\varepsilon_n\varepsilon_{n+i}}\frac{(1+r_nr_{n+i})}{\varepsilon^i} \left( \begin{array}{@{}c@{}} 0 \\ 1\end{array}\right). \end{align}

The added momentum due to the impulsive acceleration can be found directly from the instantaneous force impulse, obtained above, by substitution $\alpha \to \beta$:

(5.13)\begin{equation} \int_{S_\varepsilon} \psi_1\,{\rm d}\boldsymbol{S} = \int_{S_\varepsilon}\psi_0(\beta)\,{\rm d}\boldsymbol{S}. \end{equation}

In this zeroth-order force, only the impulsive acceleration is taken into account ($\delta =0$). It is independent from the zeroth-order force due to nonlinear effects for the case of constant velocity ($\delta =1$), which we find in Tyvand & Miloh (Reference Tyvand and Miloh1995).

5.2. The first-order force

Following the expansions (3.1), (3.9) and (3.23), the first-order hydrodynamic force (5.6) can be decomposed into four integral components:

(5.14)\begin{equation} \boldsymbol{F}_1 ={-}2\delta\int_{S_\varepsilon} \phi_{2g} \,{\rm d}\boldsymbol{S} - 2a\delta\int_{S_\varepsilon} \phi_2 \,{\rm d}\boldsymbol{S} -2a\delta\int_{S_\varepsilon} \psi_2 \,{\rm d}\boldsymbol{S}- a\delta\int_{S_\varepsilon} \boldsymbol{\nabla}\psi_0\boldsymbol{\nabla} \psi_1 \,{\rm d}\boldsymbol{S}. \end{equation}

Here, the first integral represents the leading-order gravitational force due to the impulsive acceleration, the second integral is due to the force from the leading nonlinear free-surface interaction of impulsive velocity and impulsive acceleration, the third integral is the force from the leading nonlinear effects of finite geometric displacement of the cylinder contour, which combines the impulsive velocity and impulsive acceleration, and the fourth integral is responsible for the force from the leading nonlinear effects of dynamic pressure on the cylinder, which combines the impulsive velocity and impulsive acceleration.

Only first-order forces which are influenced by impulsive acceleration are taken into account. Thus, the first-order force due to triple nonlinear self-interactions of the forced velocity and its induced surface motion will be disregarded in this paper. The contribution from the gravitational force due to the forced velocity is governed by the linear theory and is given by

(5.15)\begin{equation} \int_{S_\varepsilon} \phi_{2g} \,{\rm d}\boldsymbol{S} ={-}\lambda{\rm \pi} r\sum_{i={-}1}^1\sum_{n=1}^\infty C^2_{i+1}k_n k_{n+i} \left(\begin{array}{@{}c@{}}\cos\alpha \\ \sin\alpha\end{array}\right) . \end{equation}

It should be noted that below there will be calculated the second-order counterpart of the gravitational force due to forced acceleration,

(5.16)\begin{align} \int_{S_\varepsilon} \phi_2 \,{\rm d}\boldsymbol{S} &={-}{\rm \pi} r\sum_{i,j,\kappa={-}1}^1 \sum_{n,m=1}^{\infty}\kappa C^2_{i+1}C^2_{j+1} k_nk_mk_{n+\kappa m+i+\kappa j} \left(\begin{array}{@{}c@{}} -\sin\alpha \\ \cos\alpha\end{array}\right), \end{align}

(5.17)\begin{align} \int_{S_\varepsilon} \psi_2 \,{\rm d}\boldsymbol{S}& ={-}2{\rm \pi} r \sum_{n=1}^\infty\sum_{i=0}^1 \frac{k_nk_{n+i}}{\varepsilon_n\varepsilon_{n+i}\varepsilon^i} \left(\begin{array}{@{}c@{}}(2+r_n)\sin\alpha\cos\beta + (1+2r_n)\cos\alpha\sin\beta \\ 3\sin\alpha\sin\beta + 3r_n\cos\alpha\cos\beta \end{array}\right) \nonumber\\ &\quad +{\rm \pi} r \sum_{n=1}^\infty \sum_{i=-1}^1 C_{i+1}^{2}k_nk_{n+i}\left(\begin{array}{@{}c@{}} 2\sin\alpha \cos\beta + \cos\alpha\sin\beta\\ 3\sin\alpha \sin\beta\end{array}\right). \end{align}

The only other first-order force that we will calculate is the force due to the leading-order nonlinear mutual interaction between impulsive velocity and impulsive acceleration. The calculation must be done carefully for the general case with two different direction angles: $\alpha$ for the forced velocity; and $\beta$ for the forced acceleration of the cylinder. Using the transformation (3.18) the expression for the force can be written in the form

(5.18)\begin{equation} \int\nolimits_{S_\varepsilon} \boldsymbol{\nabla} \psi_0 \boldsymbol{\cdot} \boldsymbol{\nabla} \psi_1\,{\rm d}\boldsymbol{S} = \int\nolimits_{S_\varepsilon} \boldsymbol{\nabla} \psi_0(\alpha) \boldsymbol{\cdot}\boldsymbol{\nabla} \psi_0(\beta)\,{\rm d}\boldsymbol{S}, \end{equation}

and we insert the zeroth-order potential (3.3) to obtain the first-order pressure along the cylinder contour. In bipolar coordinates it is given by evaluating the integrals of the triple products of trigonometric functions, giving

(5.19)\begin{equation} \int\nolimits_{S_\varepsilon} \boldsymbol{\nabla} \psi_0 \boldsymbol{\cdot} \boldsymbol{\nabla} \psi_1\,{\rm d}\boldsymbol{S} = 4{\rm \pi} r \sum_{n=1}^\infty\sum_{i=0}^1 \frac{k_nk_{n+i}}{\varepsilon_n\varepsilon_{n+i}}\frac{(1+r_nr_{n+i})}{\varepsilon^i} \left( \begin{array}{@{}c@{}} 0 \\ \cos(\alpha-\beta)\end{array}\right). \end{equation}

Equation (5.19) demonstrates that the first-order interaction force has only the vertical component and it is determined by the mutual difference of the directional angles. The interaction force vanishes when the directions of the forced velocity and acceleration are orthogonal.

5.3. The second-order force

In this paragraph, we provide the second-order hydrodynamic force for the special case of the cylinder starting without initial impulsive velocity ($\delta = 0, a=1$). Substituting the expansions (3.9) and (3.37) into the integral equation (5.7) we can decompose the second-order force into four principal constituents:

(5.20)\begin{equation} \boldsymbol{F}_2 ={-}3\int_{S_\varepsilon} \phi_{3g} \,{\rm d}\boldsymbol{S}-3\int_{S_\varepsilon} \phi_3 \,{\rm d}\boldsymbol{S} -3\int_{S_\varepsilon} \psi_3 \,{\rm d}\boldsymbol{S} -\tfrac{1}{2}\int_{S_\varepsilon} |\boldsymbol{\nabla}\psi_1|^2 \,{\rm d}\boldsymbol{S}. \end{equation}

Having obtained the solution up to third order, we can relate directly the integral components of the fourth-order hydrodynamic force to those of the zeroth- and the first-order forces. Application of relations (3.18), (3.41a,b) and (3.42) in (5.20) gives alternative integral representation of the force,

(5.21)\begin{equation} \boldsymbol{F}_2 ={-}\int_{S_\varepsilon} \phi_{2g} \,{\rm d}\boldsymbol{S} -2\int_{S_\varepsilon} \hat{\phi}_1 \,{\rm d}\boldsymbol{S} -\int_{S_\varepsilon}\hat{\psi}_1 \,{\rm d}\boldsymbol{S} -\tfrac{1}{2}\int_{S_\varepsilon} |\boldsymbol{\nabla}\psi_0(\beta)|^2 \,{\rm d}\boldsymbol{S}, \end{equation}

which can be evaluated based on the results of the previous paragraphs.