3.1 Velocity potential due to a single source: the Green function

The Green function $G(x,z;x_{0},z_{0})$ is defined as the velocity potential at point $p(x,z)$ due to a source at $q(x_{0},z_{0})$ . It then satisfies the following equation

(3.1) $$\begin{eqnarray}\unicode[STIX]{x1D6FB}^{2}G=2\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FF}(x-x_{0})\unicode[STIX]{x1D6FF}(z-z_{0}),\end{eqnarray}$$

and the same boundary conditions as those in (2.3), (2.4), (2.6) and (2.7). Here, $\unicode[STIX]{x1D6FF}(x)$ is the Dirac delta function. In order to derive $G$ , we shall use the Fourier transform

(3.2) $$\begin{eqnarray}\tilde{G}=\frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{+\infty }G\text{e}^{-\text{i}kx}\,\text{d}x.\end{eqnarray}$$

We may replace $\unicode[STIX]{x1D714}$ in (2.1) with $\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D714}-\text{i}\unicode[STIX]{x1D716}$ , where $\unicode[STIX]{x1D716}\rightarrow 0^{+}$ (Lighthill Reference Lighthill1978). For a non-zero $\unicode[STIX]{x1D716}$ , $G\rightarrow 0$ when $|x|\rightarrow \infty$ , and the radiation condition in (2.7) will be satisfied when $\unicode[STIX]{x1D716}$ is taken to be zero. Applying (3.2) to (3.1), we obtain

(3.3) $$\begin{eqnarray}-k^{2}\tilde{G}+\frac{\unicode[STIX]{x2202}^{2}\tilde{G}}{\unicode[STIX]{x2202}z^{2}}=\unicode[STIX]{x1D6FF}(z-z_{0})\text{e}^{-\text{i}kx_{0}}.\end{eqnarray}$$

Here we introduce

(3.4a,b ) $$\begin{eqnarray}z_{{>}}=\max (z,z_{0})\quad \text{and}\quad z_{{<}}=\min (z,z_{0}).\end{eqnarray}$$

Then the solution for (3.3) with the boundary condition in (2.6) can be written in the following form

(3.5) $$\begin{eqnarray}\tilde{G}=\frac{\text{e}^{-\text{i}kx_{0}}}{k}[\unicode[STIX]{x1D6FC}k\cosh (kz_{{>}})+\unicode[STIX]{x1D6FD}\sinh (kz_{{>}})]Z(z_{{<}}),\end{eqnarray}$$

where $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D6FD}$ are two unknown constants to be solved,

(3.6) $$\begin{eqnarray}Z(z)=\cosh [k(z+H)].\end{eqnarray}$$

Integrating (3.3) with respect to $z$ from $z_{0}^{-}$ to $z_{0}^{+}$ , we have

(3.7) $$\begin{eqnarray}\left(\frac{\unicode[STIX]{x2202}\tilde{G}}{\unicode[STIX]{x2202}z}\right)_{z=z_{0}^{-}}^{z=z_{0}^{+}}=\text{e}^{-\text{i}kx_{0}}.\end{eqnarray}$$

Substituting (3.5) into (3.7), we obtain

(3.8) $$\begin{eqnarray}-\unicode[STIX]{x1D6FC}k\sinh (kH)+\unicode[STIX]{x1D6FD}\cosh (kH)=1.\end{eqnarray}$$

Fourier transform can also be applied to the boundary condition on the ice sheet in (2.3). We have

(3.9) $$\begin{eqnarray}\tilde{U} +(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})\frac{\unicode[STIX]{x2202}\tilde{G}}{\unicode[STIX]{x2202}z}-\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}\tilde{G}=0,\end{eqnarray}$$

where

(3.10) $$\begin{eqnarray}\tilde{U} =\frac{L}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{+\infty }\frac{\unicode[STIX]{x2202}^{5}G}{\unicode[STIX]{x2202}x^{4}\unicode[STIX]{x2202}z}\text{e}^{-\text{i}kx}\,\text{d}x.\end{eqnarray}$$

Through integrating by parts and using the condition of $G\rightarrow 0$ at $|x|\rightarrow \infty$ when $\unicode[STIX]{x1D716}$ is non-zero, (3.10) can be given as

(3.11) $$\begin{eqnarray}\tilde{U} =\frac{L}{2\unicode[STIX]{x03C0}}\left[\left(\frac{\unicode[STIX]{x2202}^{4}G}{\unicode[STIX]{x2202}x^{3}\unicode[STIX]{x2202}z}+\text{i}k\frac{\unicode[STIX]{x2202}^{3}G}{\unicode[STIX]{x2202}x^{2}\unicode[STIX]{x2202}z}-k^{2}\frac{\unicode[STIX]{x2202}^{2}G}{\unicode[STIX]{x2202}x\unicode[STIX]{x2202}z}-\text{i}k^{3}\frac{\unicode[STIX]{x2202}G}{\unicode[STIX]{x2202}z}\right)\text{e}^{-\text{i}kx}\right]_{x=0^{+}}^{x=0^{-}}+Lk^{4}\frac{\unicode[STIX]{x2202}\tilde{G}}{\unicode[STIX]{x2202}z}.\end{eqnarray}$$

At the ice crack, we have condition (2.4). It is expected that the displacements and the slopes on both sides of the crack are discontinuous. Thus we may write

(3.12) $$\begin{eqnarray}\displaystyle & \displaystyle \left(\frac{\unicode[STIX]{x2202}G}{\unicode[STIX]{x2202}z}\right)_{x=0^{+}}^{x=0^{-}}=A,\quad (z=0), & \displaystyle\end{eqnarray}$$

(3.13) $$\begin{eqnarray}\displaystyle & \displaystyle \left(\frac{\unicode[STIX]{x2202}^{2}G}{\unicode[STIX]{x2202}x\unicode[STIX]{x2202}z}\right)_{x=0^{+}}^{x=0^{-}}=B,\quad (z=0). & \displaystyle\end{eqnarray}$$

Then (3.11) can be written as

(3.14) $$\begin{eqnarray}\tilde{U} =-\frac{L}{2\unicode[STIX]{x03C0}}(\text{i}k^{3}A+k^{2}B)+Lk^{4}\frac{\unicode[STIX]{x2202}\tilde{G}}{\unicode[STIX]{x2202}z}.\end{eqnarray}$$

It may be noticed that for crack conditions different from those in (3.12) and (3.13), a similar procedure may be followed, leading to a different form of (3.14). Substituting (3.5) together with (3.14) into (3.9) and invoking (3.8), we obtain

(3.15) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FC}=\frac{L}{2\unicode[STIX]{x03C0}}\frac{\text{i}k^{3}A+k^{2}B}{K(\unicode[STIX]{x1D6FE},k)Z(z_{0})}\text{e}^{\text{i}kx_{0}}-\frac{Lk^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2}}{K(\unicode[STIX]{x1D6FE},k)Z(0)}, & \displaystyle\end{eqnarray}$$

(3.16) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6FD}=\frac{L}{2\unicode[STIX]{x03C0}}\frac{(\text{i}k^{3}A+k^{2}B)Z^{\prime }(0)}{K(\unicode[STIX]{x1D6FE},k)Z(z_{0})Z(0)}\text{e}^{\text{i}kx_{0}}-\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}{K(\unicode[STIX]{x1D6FE},k)Z(0)}, & \displaystyle\end{eqnarray}$$

where the prime denotes partial derivative with respect to $z$ . Substituting (3.15) and (3.16) into (3.5), we have

(3.17) $$\begin{eqnarray}\displaystyle \tilde{G} & = & \displaystyle -\frac{\text{e}^{-\text{i}kx_{0}}}{k}\frac{(Lk^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})k\cosh (kz_{{>}})+\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}\sinh (kz_{{>}})}{K(\unicode[STIX]{x1D6FE},k)Z(0)}Z(z_{{<}})\nonumber\\ \displaystyle & & \displaystyle +\,\frac{L}{2\unicode[STIX]{x03C0}}\frac{(\text{i}k^{3}A+k^{2}B)[\cosh (kz_{{>}})+\tanh (kH)\sinh (kz_{{>}})]}{K(\unicode[STIX]{x1D6FE},k)Z(z_{0})}Z(z_{{<}}).\end{eqnarray}$$

Using

(3.18) $$\begin{eqnarray}\displaystyle & \displaystyle \ln \left(\frac{r_{1}}{H}\right)=\int _{0}^{+\infty }\frac{\text{e}^{-kH}-\text{e}^{-k(z_{{>}}-z_{{<}})}\cos [k(x-x_{0})]}{k}\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.19) $$\begin{eqnarray}\displaystyle & \displaystyle \ln \left(\frac{r_{2}}{H}\right)=\int _{0}^{+\infty }\frac{\text{e}^{-kH}-\text{e}^{-k(z_{{>}}+z_{{<}}+2H)}\cos [k(x-x_{0})]}{k}\,\text{d}k, & \displaystyle\end{eqnarray}$$

where $r_{1}$ is the distance between $p$ and $q$ , and $r_{2}$ is the distance between $p$ and the mirror image of $q$ about the flat seabed $z=-H$ , and then performing the inverse Fourier transform of $\tilde{G}$ , the solution due to a single source can be written as

(3.20) $$\begin{eqnarray}G=G_{ice}(x,z;x_{0},z_{0})+A\times \frac{\unicode[STIX]{x2202}G_{1}(x,z)}{\unicode[STIX]{x2202}x}+B\times G_{1}(x,z),\end{eqnarray}$$

where

(3.21) $$\begin{eqnarray}G_{ice}=\ln \left(\frac{r_{1}}{H}\right)+\ln \left(\frac{r_{2}}{H}\right)-2\int _{0}^{+\infty }\frac{\text{e}^{-kH}}{k}\{P(k)Z(z)Z(z_{0})\cos [k(x-x_{0})]+1\}\,\text{d}k,\end{eqnarray}$$

with

(3.22) $$\begin{eqnarray}P(k)=\frac{(Lk^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})k+\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}{K(\unicode[STIX]{x1D6FE},k)Z(0)},\end{eqnarray}$$

and

(3.23) $$\begin{eqnarray}G_{1}=\frac{L}{\unicode[STIX]{x03C0}}\int _{0}^{+\infty }\frac{Z(z)}{K(\unicode[STIX]{x1D6FE},k)Z(0)}k^{2}\cos (kx)\,\text{d}k.\end{eqnarray}$$

We may notice that $G_{ice}$ is in fact the Green function without the crack (Evans & Porter Reference Evans and Porter2003). We may also notice that $G_{1}$ satisfies (2.3) and

(3.24) $$\begin{eqnarray}G_{1}(x,z)=\frac{L}{2\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}\left(\frac{\unicode[STIX]{x2202}^{3}G_{ice}(x_{0},z_{0};x,z)}{\unicode[STIX]{x2202}z_{0}\unicode[STIX]{x2202}x_{0}^{2}}\right)_{x_{0}=0,z_{0}=0}.\end{eqnarray}$$

For the current problem with a crack, there are still two unknown coefficients, $A$ and $B$ in (3.20), which can be determined by the condition in (2.4). The zero bending moment and shear force conditions respectively give

(3.25) $$\begin{eqnarray}\displaystyle & \displaystyle \lim _{x,z\rightarrow 0}G_{zxx}=\frac{2\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}{L}G_{1}(x_{0},z_{0})+A\lim _{x,z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{4}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{3}}+B\lim _{x,z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{3}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{2}}=0,\qquad & \displaystyle\end{eqnarray}$$

(3.26) $$\begin{eqnarray}\displaystyle & \displaystyle \lim _{x,z\rightarrow 0}G_{zxxx}=-\frac{2\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}{L}\frac{\unicode[STIX]{x2202}G_{1}(x_{0},z_{0})}{\unicode[STIX]{x2202}x_{0}}+A\lim _{x,z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{5}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{4}}+B\lim _{x,z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{4}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{3}}=0,\qquad & \displaystyle\end{eqnarray}$$

where (3.24) has been used. We then notice that when $x=r\sin \unicode[STIX]{x1D6E9}$ and $z=-r\cos \unicode[STIX]{x1D6E9}$ we have at $z<0$

(3.27) $$\begin{eqnarray}\frac{\text{e}^{\text{i}(m+1)\unicode[STIX]{x1D6E9}}}{r^{m+1}}=\frac{1}{m!}\int _{0}^{+\infty }\text{e}^{kz}k^{m}\text{e}^{\text{i}kx}\,\text{d}k.\end{eqnarray}$$

This means that when $\unicode[STIX]{x1D6E9}\rightarrow \unicode[STIX]{x03C0}/2$ , or $z\rightarrow 0$ , we have

(3.28) $$\begin{eqnarray}\lim _{z\rightarrow 0}\int _{0}^{+\infty }\text{e}^{kz}k^{m}\cos (kx)\,\text{d}k=0\quad \text{for an even }m,\end{eqnarray}$$

and

(3.29) $$\begin{eqnarray}\lim _{z\rightarrow 0}\int _{0}^{+\infty }\text{e}^{kz}k^{m}\sin (kx)\,\text{d}k=0\quad \text{for an odd }m.\end{eqnarray}$$

Equations (3.28) and (3.29) further provide

(3.30) $$\begin{eqnarray}\lim _{z\rightarrow 0}\int _{0}^{+\infty }\frac{Z^{\prime }(z)}{Z^{\prime }(0)}k^{m}\cos (kx)\,\text{d}k=0\quad \text{for an even }m,\end{eqnarray}$$

and

(3.31) $$\begin{eqnarray}\lim _{z\rightarrow 0}\int _{0}^{+\infty }\frac{Z^{\prime }(z)}{Z^{\prime }(0)}k^{m}\sin (kx)\,\text{d}k=0\quad \text{for an odd }m.\end{eqnarray}$$

Using these two equations, we have

(3.32) $$\begin{eqnarray}\displaystyle & \displaystyle \lim _{z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{3}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{2}}=-\frac{1}{\unicode[STIX]{x03C0}}\int _{0}^{+\infty }\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}-(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})k\tanh (kH)}{K(\unicode[STIX]{x1D6FE},k)}\cos (kx)\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.33) $$\begin{eqnarray}\displaystyle & \displaystyle \lim _{z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{4}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{3}}=\frac{1}{\unicode[STIX]{x03C0}}\int _{0}^{+\infty }\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}-(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})k\tanh (kH)}{K(\unicode[STIX]{x1D6FE},k)}k\sin (kx)\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.34) $$\begin{eqnarray}\displaystyle & \displaystyle \lim _{z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{5}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{4}}=\frac{1}{\unicode[STIX]{x03C0}}\int _{0}^{+\infty }\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}-(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})k\tanh (kH)}{K(\unicode[STIX]{x1D6FE},k)}k^{2}\cos (kx)\,\text{d}k. & \displaystyle\end{eqnarray}$$

Through these results, (3.26) and (3.25) then respectively give

(3.35) $$\begin{eqnarray}A=\frac{2\unicode[STIX]{x03C0}^{2}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}{L{\mathcal{I}}_{1}}\frac{\unicode[STIX]{x2202}G_{1}(x_{0},z_{0})}{\unicode[STIX]{x2202}x_{0}},\end{eqnarray}$$

and

(3.36) $$\begin{eqnarray}B=\frac{2\unicode[STIX]{x03C0}^{2}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}{L{\mathcal{I}}_{2}}G_{1}(x_{0},z_{0}),\end{eqnarray}$$

where

(3.37a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{I}}_{1}=\unicode[STIX]{x03C0}\lim _{x,z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{5}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{4}}=\int _{0}^{+\infty }\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}-(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})k\tanh (kH)}{K(\unicode[STIX]{x1D6FE},k)}k^{2}\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.37b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{I}}_{2}=-\unicode[STIX]{x03C0}\lim _{x,z\rightarrow 0}\frac{\unicode[STIX]{x2202}^{3}G_{1}(x,z)}{\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{2}}=\int _{0}^{+\infty }\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}-(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D6FE}^{2})k\tanh (kH)}{K(\unicode[STIX]{x1D6FE},k)}\,\text{d}k. & \displaystyle\end{eqnarray}$$

(3.38) $$\begin{eqnarray}G=G_{ice}(x,z;x_{0},z_{0})+\frac{2\unicode[STIX]{x03C0}^{2}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FE}^{2}}{L}\left[\frac{1}{{\mathcal{I}}_{1}}\frac{\unicode[STIX]{x2202}G_{1}(x,z)}{\unicode[STIX]{x2202}x}\frac{\unicode[STIX]{x2202}G_{1}(x_{0},z_{0})}{\unicode[STIX]{x2202}x_{0}}+\frac{G_{1}(x,z)G_{1}(x_{0},z_{0})}{{\mathcal{I}}_{2}}\right].\end{eqnarray}$$

The replacement of $\unicode[STIX]{x1D714}$ with $\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D714}-\text{i}\unicode[STIX]{x1D716}$ in (2.8) during the derivation moves the pole $k=\unicode[STIX]{x1D706}$ off the real axis of $k$ . Letting $\unicode[STIX]{x1D716}\rightarrow 0^{+}$ , the Taylor expansion of the dispersion equation $K(\unicode[STIX]{x1D6FE},k)$ with respect to $\unicode[STIX]{x1D716}$ provides the root at

(3.39) $$\begin{eqnarray}k=\unicode[STIX]{x1D706}-\text{i}\unicode[STIX]{x1D716}^{\prime },\end{eqnarray}$$

where

(3.40) $$\begin{eqnarray}\unicode[STIX]{x1D716}^{\prime }=\frac{\unicode[STIX]{x1D716}}{\unicode[STIX]{x1D714}}\frac{L\unicode[STIX]{x1D706}^{4}+\unicode[STIX]{x1D70C}g}{L\unicode[STIX]{x1D706}^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D714}^{2}}\left/\left[\frac{1}{2\unicode[STIX]{x1D706}}+\frac{H}{\sinh (2\unicode[STIX]{x1D706}H)}+\frac{2L\unicode[STIX]{x1D706}^{3}}{L\unicode[STIX]{x1D706}^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D714}^{2}}\right]\right..\end{eqnarray}$$

Thus when $\unicode[STIX]{x1D716}\rightarrow 0^{+}$ , the pole will approach the real axis of $k$ from below. Therefore, the integration route ${\mathcal{D}}$ in (3.38) from $0$ to $+\infty$ should pass over the pole at $k=\unicode[STIX]{x1D706}$ , which is consistent with the radiation condition, i.e. the wave will propagate away from the source into infinity. With such a definition of ${\mathcal{D}}$ , $\unicode[STIX]{x1D6FE}$ in (3.38) can be replaced by $\unicode[STIX]{x1D714}$ . We may notice that (3.38) is a summation of the Green function $G_{ice}$ for the ice sheet without the crack and two additional terms equivalent to the loads applied at the crack to ensure that the shear force and bending moment there are zero. We may also notice that the integrand of $G$ in (3.38) has only a simple pole at $k=\unicode[STIX]{x1D706}$ where $K(\unicode[STIX]{x1D714},k)=0$ . The integration may be calculated by using the Gauss–Legendre principal value integration procedure (Noble Reference Noble2000). This may also be done by following the numerical procedure in Wu & Eatock Taylor (Reference Wu and Eatock Taylor1987), which effectively removes the singularity.

3.2 Multipole expansion

Potentials due to multipoles or singularities of higher order may be found by differentiating (3.38) with respect to the position of the source point. In such a way, none of the boundary conditions will be affected and will be still satisfied. As shown in figure 1, we have $x-x_{0}=r\sin \unicode[STIX]{x1D703}$ and $z-z_{0}=r\cos \unicode[STIX]{x1D703}$ . Similar to Wu (Reference Wu1998), we may apply the operator

(3.41) $$\begin{eqnarray}(D_{\pm })^{n}=-\frac{1}{2^{n-1}(n-1)!}\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}z_{0}}\pm \text{i}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x_{0}}\right)^{n}\end{eqnarray}$$

to (3.38). Using the result

(3.42) $$\begin{eqnarray}(D_{\pm })^{n}\ln (r_{1})=\frac{\text{e}^{\pm \text{i}n\unicode[STIX]{x1D703}}}{r^{n}},\end{eqnarray}$$

and noticing $(D_{+})\exp (\pm kz_{0}\pm \text{i}kx_{0})=0$ and $(D_{-})\exp (\pm kz_{0}\mp \text{i}kx_{0})=0$ , we have

(3.43) $$\begin{eqnarray}\displaystyle f_{n} & = & \displaystyle (D_{+})^{n}G\nonumber\\ \displaystyle & = & \displaystyle \frac{\text{e}^{+\text{i}n\unicode[STIX]{x1D703}}}{r^{n}}+\frac{(-1)^{n}}{(n-1)!}\int _{0}^{+\infty }\text{e}^{-k(z+z_{0}+2H)-\text{i}k(x-x_{0})}k^{n-1}\,\text{d}k\nonumber\\ \displaystyle & & \displaystyle +\,\frac{2\text{i}^{n}}{(n-1)!}\int _{{\mathcal{D}}}Z(z)\cosh [k(z_{0}+H)+\text{i}k(x-x_{0})-\text{i}n\unicode[STIX]{x03C0}/2]\text{e}^{-kH}P(k)k^{n-1}\,\text{d}k\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}}{(n-1)!}\left[\frac{\text{i}{\mathcal{G}}_{1}(n)}{{\mathcal{I}}_{1}}\frac{\unicode[STIX]{x2202}G_{1}(x,z)}{\unicode[STIX]{x2202}x}-\frac{{\mathcal{G}}_{1}(n-1)G_{1}(x,z)}{{\mathcal{I}}_{2}}\right],\end{eqnarray}$$

and

(3.44) $$\begin{eqnarray}\displaystyle g_{n} & = & \displaystyle (D_{-})^{n}G\nonumber\\ \displaystyle & = & \displaystyle \frac{\text{e}^{-\text{i}n\unicode[STIX]{x1D703}}}{r^{n}}+\frac{(-1)^{n}}{(n-1)!}\int _{0}^{+\infty }\text{e}^{-k(z+z_{0}+2H)+\text{i}k(x-x_{0})}k^{n-1}\,\text{d}k\nonumber\\ \displaystyle & & \displaystyle +\,\frac{2\text{i}^{n}}{(n-1)!}\int _{{\mathcal{D}}}Z(z)\cosh [k(z_{0}+H)-\text{i}k(x-x_{0})-\text{i}n\unicode[STIX]{x03C0}/2]\text{e}^{-kH}P(k)k^{n-1}\,\text{d}k\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}}{(n-1)!}\left[\frac{\text{i}{\mathcal{G}}_{2}(n)}{{\mathcal{I}}_{1}}\frac{\unicode[STIX]{x2202}G_{1}(x,z)}{\unicode[STIX]{x2202}x}+\frac{{\mathcal{G}}_{2}(n-1)G_{1}(x,z)}{{\mathcal{I}}_{2}}\right],\end{eqnarray}$$

where

(3.45a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{G}}_{1}(n)=\int _{{\mathcal{D}}}\frac{(-1)^{n}\text{e}^{-k(z_{0}+H)+\text{i}kx_{0}}-\text{e}^{k(z_{0}+H)-\text{i}kx_{0}}}{K(\unicode[STIX]{x1D714},k)Z(0)}k^{n+3}\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.45b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{G}}_{2}(n)=\int _{{\mathcal{D}}}\frac{(-1)^{n}\text{e}^{-k(z_{0}+H)-\text{i}kx_{0}}-\text{e}^{k(z_{0}+H)+\text{i}kx_{0}}}{K(\unicode[STIX]{x1D714},k)Z(0)}k^{n+3}\,\text{d}k. & \displaystyle\end{eqnarray}$$

$\text{i}$

$-\text{i}$

$f_{n}$

$g_{n}$

3.3 Solution for the radiation problem

Here we consider the wave radiation due to the body undergoing sinusoidal oscillation with frequency $\unicode[STIX]{x1D714}$ . By using (3.43) and (3.44), we can write the potentials $\unicode[STIX]{x1D719}_{j}$ in terms of multipole expansions in the polar coordinate system as

(3.46) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{j}=\mathop{\sum }_{n=1}^{\infty }a^{n}(A_{n}^{j}\,f_{n}+B_{n}^{j}g_{n}).\end{eqnarray}$$

$\unicode[STIX]{x1D719}_{j}$ now satisfies the governing equation and all the boundary conditions apart from that on the body surface, which is to be used to obtain the coefficients $A_{n}^{j}$ and $B_{n}^{j}$ . When $j=1,2$ , we respectively have

(3.47a,b ) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{1}}{\unicode[STIX]{x2202}r}=-\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{1}}{\unicode[STIX]{x2202}n}=\sin \unicode[STIX]{x1D703}\quad \text{and}\quad \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{2}}{\unicode[STIX]{x2202}r}=-\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{2}}{\unicode[STIX]{x2202}n}=\cos \unicode[STIX]{x1D703},\quad (r=a),\end{eqnarray}$$

where the minus sign is due to the fact that $\boldsymbol{n}$ points into the body. Since

(3.48) $$\begin{eqnarray}\text{e}^{kz\pm \text{i}kx}=\exp (kz_{0}\pm \text{i}kx_{0}+kr\text{e}^{\pm \text{i}\unicode[STIX]{x1D703}})=\text{e}^{kz_{0}\pm \text{i}kx_{0}}\mathop{\sum }_{l=0}^{\infty }\frac{k^{l}r^{l}\text{e}^{\pm \text{i}l\unicode[STIX]{x1D703}}}{l!},\end{eqnarray}$$

$\unicode[STIX]{x1D719}_{j}$ can be expressed in the polar coordinate system as

(3.49) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D719}_{j} & = & \displaystyle \mathop{\sum }_{n=1}^{\infty }A_{n}^{j}a^{n}\left\{\frac{\text{e}^{+\text{i}n\unicode[STIX]{x1D703}}}{r^{n}}+\frac{1}{(n-1)!}\mathop{\sum }_{l=0}^{\infty }\frac{r^{l}}{l!}[\text{e}^{+\text{i}l\unicode[STIX]{x1D703}}C(n,l)+\text{e}^{-\text{i}l\unicode[STIX]{x1D703}}D(n,l)]\right\}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{n=1}^{\infty }B_{n}^{j}a^{n}\left\{\frac{\text{e}^{-\text{i}n\unicode[STIX]{x1D703}}}{r^{n}}+\frac{1}{(n-1)!}\mathop{\sum }_{l=0}^{\infty }\frac{r^{l}}{l!}[\text{e}^{-\text{i}l\unicode[STIX]{x1D703}}C(l,n)+\text{e}^{+\text{i}l\unicode[STIX]{x1D703}}E(n,l)]\right\},\end{eqnarray}$$

where

(3.50a ) $$\begin{eqnarray}\displaystyle & \displaystyle C(n,l)=(-1)^{n+l}{\mathcal{J}}_{1}(n+l)+\frac{{\mathcal{J}}_{3}(n+l)}{2}-\frac{L\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}}{4}\left[\frac{{\mathcal{G}}_{1}(n){\mathcal{G}}_{2}(l)}{{\mathcal{I}}_{1}}+\frac{{\mathcal{G}}_{1}(n-1){\mathcal{G}}_{2}(l-1)}{{\mathcal{I}}_{2}}\right], & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.50b ) $$\begin{eqnarray}\displaystyle & \displaystyle D(n,l)=[(-1)^{l}+(-1)^{n}]\frac{{\mathcal{J}}_{2}(n+l)}{2}+\frac{L\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}}{4}\left[\frac{{\mathcal{G}}_{1}(n){\mathcal{G}}_{1}(l)}{{\mathcal{I}}_{1}}-\frac{{\mathcal{G}}_{1}(n-1){\mathcal{G}}_{1}(l-1)}{{\mathcal{I}}_{2}}\right], & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.50c ) $$\begin{eqnarray}\displaystyle & \displaystyle E(n,l)=[(-1)^{l}+(-1)^{n}]\frac{{\mathcal{J}}_{2}(n+l)}{2}+\frac{L\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}}{4}\left[\frac{{\mathcal{G}}_{2}(n){\mathcal{G}}_{2}(l)}{{\mathcal{I}}_{1}}-\frac{{\mathcal{G}}_{2}(n-1){\mathcal{G}}_{2}(l-1)}{{\mathcal{I}}_{2}}\right], & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(3.51a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{J}}_{1}(n)=\unicode[STIX]{x1D6E4}(n)/[2(z_{0}+H)]^{n}, & \displaystyle\end{eqnarray}$$

(3.51b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{J}}_{2}(n)=\int _{{\mathcal{D}}}\text{e}^{-kH}P(k)k^{n-1}\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.51c ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{J}}_{3}(n)=\int _{{\mathcal{D}}}[(-1)^{n}\text{e}^{-2k(z_{0}+H)}+\text{e}^{2k(z_{0}+H)}]\text{e}^{-kH}P(k)k^{n-1}\,\text{d}k. & \displaystyle\end{eqnarray}$$

$\unicode[STIX]{x1D6E4}(n)$

(3.52) $$\begin{eqnarray}-l\frac{A_{l}^{j}}{a}+\mathop{\sum }_{n=1}^{\infty }\frac{A_{n}^{j}a^{n+l-1}C(n,l)}{(n-1)!(l-1)!}+\mathop{\sum }_{n=1}^{\infty }\frac{B_{n}^{j}a^{n+l-1}E(n,l)}{(n-1)!(l-1)!}=u_{l}(j),\end{eqnarray}$$

and

(3.53) $$\begin{eqnarray}-l\frac{B_{l}^{j}}{a}+\mathop{\sum }_{n=1}^{\infty }\frac{B_{n}^{j}a^{n+l-1}C(l,n)}{(n-1)!(l-1)!}+\mathop{\sum }_{n=1}^{\infty }\frac{A_{n}^{j}a^{n+l-1}D(n,l)}{(n-1)!(l-1)!}=v_{l}(j),\end{eqnarray}$$

for $l=1,2,\ldots \,$ , where

(3.54a,b ) $$\begin{eqnarray}u_{l}(1)={\textstyle \frac{1}{2\text{i}}}\unicode[STIX]{x1D6FF}_{l1},\quad v_{l}(1)=-u_{l}(1),\end{eqnarray}$$

and

(3.55a,b ) $$\begin{eqnarray}u_{l}(2)={\textstyle \frac{1}{2}}\unicode[STIX]{x1D6FF}_{l1},\quad v_{l}(2)=u_{l}(2).\end{eqnarray}$$

Here $\unicode[STIX]{x1D6FF}_{l1}$ is the Kronecker delta function, and $j=1,2$ in (3.52) and (3.53) correspond to $\unicode[STIX]{x1D719}_{1}$ and $\unicode[STIX]{x1D719}_{2}$ respectively. For each $l$ , (3.52) and (3.53) provide two linear equations for $A_{n}^{j}$ and $B_{n}^{j}$ . Thus when the infinite summation is truncated at a finite number, we can obtain a matrix equation, from which $A_{n}^{j}$ and $B_{n}^{j}$ can be solved. When the solutions for $\unicode[STIX]{x1D719}_{1}$ and $\unicode[STIX]{x1D719}_{2}$ have been obtained, the added mass $\unicode[STIX]{x1D707}_{js}$ and damping coefficients $\unicode[STIX]{x1D706}_{js}$ can be computed through (Newman Reference Newman1977)

(3.56) $$\begin{eqnarray}\unicode[STIX]{x1D70F}_{js}=\unicode[STIX]{x1D707}_{js}-\text{i}\frac{\unicode[STIX]{x1D706}_{js}}{\unicode[STIX]{x1D714}}=\unicode[STIX]{x1D70C}\int _{S_{0}}\unicode[STIX]{x1D719}_{s}n_{j}\,\text{d}S,\end{eqnarray}$$

for $j,s=1,2$ , where $S_{0}$ represents the body surface. Multiplying both sides of (3.52) and (3.53) respectively by $\text{e}^{+\text{i}l\unicode[STIX]{x1D703}}$ and $\text{e}^{-\text{i}l\unicode[STIX]{x1D703}}$ , then taking summation with respect to $l$ from $1$ to $+\infty$ , substituting the obtained results into (3.49), and letting $r=a$ , we have (Wu Reference Wu1993)

(3.57) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{s}=2\mathop{\sum }_{n=1}^{\infty }[A_{n}^{s}\text{e}^{+\text{i}n\unicode[STIX]{x1D703}}+B_{n}^{s}\text{e}^{-\text{i}n\unicode[STIX]{x1D703}}]+a\mathop{\sum }_{n=1}^{\infty }\frac{1}{n}[\text{e}^{+\text{i}n\unicode[STIX]{x1D703}}u_{n}(s)+\text{e}^{-\text{i}n\unicode[STIX]{x1D703}}v_{n}(s)]+\unicode[STIX]{x1D719}_{c}^{s},\end{eqnarray}$$

where

(3.58) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{c}^{s}=\mathop{\sum }_{n=1}^{\infty }\frac{A_{n}^{s}a^{n}}{(n-1)!}[C(n,0)+D(n,0)]+\mathop{\sum }_{n=1}^{\infty }\frac{B_{n}^{s}a^{n}}{(n-1)!}[C(0,n)+E(n,0)].\end{eqnarray}$$

Substituting (3.57) into (3.56), we obtain

(3.59) $$\begin{eqnarray}\unicode[STIX]{x1D70F}_{1s}=-\text{i}2\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a(A_{1}^{s}-B_{1}^{s})-\unicode[STIX]{x1D6FF}_{1s}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2},\end{eqnarray}$$

and

(3.60) $$\begin{eqnarray}\unicode[STIX]{x1D70F}_{2s}=-2\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a(A_{1}^{s}+B_{1}^{s})-\unicode[STIX]{x1D6FF}_{2s}\unicode[STIX]{x03C0}\unicode[STIX]{x1D70C}a^{2}.\end{eqnarray}$$

3.4 Solution for the diffraction problem

Wave diffraction potential consists of two parts and the total diffraction potential $\unicode[STIX]{x1D719}_{D}$ can be written as

(3.61) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{D}=\unicode[STIX]{x1D719}_{D}^{1}+\unicode[STIX]{x1D719}_{D}^{2},\end{eqnarray}$$

where $\unicode[STIX]{x1D719}_{D}^{1}$ is the diffraction potential by the crack to $\unicode[STIX]{x1D719}_{I}$ , and $\unicode[STIX]{x1D719}_{D}^{2}$ is the diffraction potential by the cylinder to $\unicode[STIX]{x1D711}=\unicode[STIX]{x1D719}_{I}+\unicode[STIX]{x1D719}_{D}^{1}$ . For a periodic wave from $x=-\infty$ to $x=+\infty$ , $\unicode[STIX]{x1D719}_{I}$ can be given as

(3.62) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{I}=I\text{e}^{-\text{i}\unicode[STIX]{x1D706}x}\unicode[STIX]{x1D713}(z),\end{eqnarray}$$

with

(3.63) $$\begin{eqnarray}\unicode[STIX]{x1D713}(z)=\cosh [\unicode[STIX]{x1D706}(H+z)]/\text{cosh}(\unicode[STIX]{x1D706}H),\end{eqnarray}$$

and $I=g/\text{i}\unicode[STIX]{x1D714}$ . Applying the Green’s second identity to $\unicode[STIX]{x1D719}_{D}^{1}$ and $G$ in (3.38) over the fluid boundary $S$ , we have

(3.64) $$\begin{eqnarray}2\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}_{D}^{1}(x,z)=\int _{S}\left[\unicode[STIX]{x1D719}_{D}^{1}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D701})\frac{\unicode[STIX]{x2202}G(x,z;\unicode[STIX]{x1D709},\unicode[STIX]{x1D701})}{\unicode[STIX]{x2202}n}-G(x,z;\unicode[STIX]{x1D709},\unicode[STIX]{x1D701})\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{D}^{1}(\unicode[STIX]{x1D709},\unicode[STIX]{x1D701})}{\unicode[STIX]{x2202}n}\right]\,\text{d}S,\end{eqnarray}$$

where the derivative and integration are with respect to $(\unicode[STIX]{x1D709},\unicode[STIX]{x1D701})$ . Invoking the boundary conditions for $\unicode[STIX]{x1D719}_{D}^{1}$ and $G$ , only the ice sheet will remain on the right-hand side of (3.64). Using condition (2.3) and through integrating by parts, this equation can be further written as

(3.65) $$\begin{eqnarray}2\unicode[STIX]{x03C0}\unicode[STIX]{x1D719}_{D}^{1}=\frac{L}{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}}\left[\left(\frac{\unicode[STIX]{x2202}^{4}\unicode[STIX]{x1D719}_{D}^{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{3}\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}\frac{\unicode[STIX]{x2202}G}{\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}-\frac{\unicode[STIX]{x2202}^{3}\unicode[STIX]{x1D719}_{D}^{1}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}^{2}\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}\frac{\unicode[STIX]{x2202}^{2}G}{\unicode[STIX]{x2202}\unicode[STIX]{x1D709}\unicode[STIX]{x2202}\unicode[STIX]{x1D701}}\right)_{\unicode[STIX]{x1D709}=0^{+}}^{\unicode[STIX]{x1D709}=0^{-}}\right]_{\unicode[STIX]{x1D701}=0}.\end{eqnarray}$$

Noticing that $\unicode[STIX]{x1D711}=\unicode[STIX]{x1D719}_{I}+\unicode[STIX]{x1D719}_{D}^{1}$ should satisfy the crack condition in (2.4), then $\unicode[STIX]{x1D719}_{D}^{1}(x,z)$ can be obtained directly through the Green function as

(3.66) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{D}^{1}=-\unicode[STIX]{x03C0}I\unicode[STIX]{x1D706}^{3}\tanh (\unicode[STIX]{x1D706}H)\left[\frac{\text{i}\unicode[STIX]{x1D706}}{{\mathcal{I}}_{1}}\frac{\unicode[STIX]{x2202}G_{1}(x,z)}{\unicode[STIX]{x2202}x}+\frac{G_{1}(x,z)}{{\mathcal{I}}_{2}}\right].\end{eqnarray}$$

To obtain the solution for $\unicode[STIX]{x1D719}_{D}^{2}$ , $\unicode[STIX]{x1D711}=\unicode[STIX]{x1D719}_{I}+\unicode[STIX]{x1D719}_{D}^{1}$ can be expressed in the polar coordinate system through using (3.48)

(3.67) $$\begin{eqnarray}\unicode[STIX]{x1D711}=I\mathop{\sum }_{n=0}^{\infty }\frac{r^{n}}{4n!}[\text{e}^{+\text{i}n\unicode[STIX]{x1D703}}{\mathcal{F}}_{1}(n)+\text{e}^{-\text{i}n\unicode[STIX]{x1D703}}{\mathcal{F}}_{2}(n)],\end{eqnarray}$$

where

(3.68a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{F}}_{1}(n)=\frac{2(-\unicode[STIX]{x1D706})^{n}\text{e}^{-\unicode[STIX]{x1D706}(z_{0}+H)-\text{i}\unicode[STIX]{x1D706}x_{0}}}{\cosh (\unicode[STIX]{x1D706}H)}-L\unicode[STIX]{x1D706}^{3}\tanh (\unicode[STIX]{x1D706}H)\left[\frac{\unicode[STIX]{x1D706}{\mathcal{G}}_{2}(n)}{{\mathcal{I}}_{1}}-\frac{{\mathcal{G}}_{2}(n-1)}{{\mathcal{I}}_{2}}\right], & \displaystyle\end{eqnarray}$$

(3.68b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{F}}_{2}(n)=\frac{2\unicode[STIX]{x1D706}^{n}\text{e}^{\unicode[STIX]{x1D706}(z_{0}+H)-\text{i}\unicode[STIX]{x1D706}x_{0}}}{\cosh (\unicode[STIX]{x1D706}H)}+L\unicode[STIX]{x1D706}^{3}\tanh (\unicode[STIX]{x1D706}H)\left[\frac{\unicode[STIX]{x1D706}{\mathcal{G}}_{1}(n)}{{\mathcal{I}}_{1}}+\frac{{\mathcal{G}}_{1}(n-1)}{{\mathcal{I}}_{2}}\right]. & \displaystyle\end{eqnarray}$$

$\unicode[STIX]{x1D719}_{D}^{2}$

(3.69) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{D}^{2}}{\unicode[STIX]{x2202}n}=-\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D711}}{\unicode[STIX]{x2202}n},\end{eqnarray}$$

we have

(3.70a,b ) $$\begin{eqnarray}u_{l}(0)=-\frac{Ia^{l-1}{\mathcal{F}}_{1}(l)}{4(l-1)!},\quad v_{l}(0)=-\frac{Ia^{l-1}{\mathcal{F}}_{2}(l)}{4(l-1)!},\end{eqnarray}$$

for $j=0$ in (3.52) and (3.53) respectively. Invoking (3.57) we further have at $r=a$

(3.71) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{D}^{2}=2\mathop{\sum }_{n=1}^{\infty }[A_{n}^{0}\text{e}^{+\text{i}n\unicode[STIX]{x1D703}}+B_{n}^{0}\text{e}^{-\text{i}n\unicode[STIX]{x1D703}}]+a\mathop{\sum }_{n=1}^{\infty }\frac{1}{n}[\text{e}^{+\text{i}n\unicode[STIX]{x1D703}}u_{n}(0)+\text{e}^{-\text{i}n\unicode[STIX]{x1D703}}v_{n}(0)]+\unicode[STIX]{x1D719}_{c}^{0}.\end{eqnarray}$$

When the solution for $\unicode[STIX]{x1D719}_{D}^{2}$ has been obtained, the wave exciting force can be computed through (Newman Reference Newman1977)

(3.72) $$\begin{eqnarray}f_{E,j}=-\text{i}\unicode[STIX]{x1D714}\unicode[STIX]{x1D70C}\int _{S_{0}}\unicode[STIX]{x1D719}_{0}n_{j}\,\text{d}S.\end{eqnarray}$$

Substituting (3.71) together with (3.67) into (3.72), we have

(3.73) $$\begin{eqnarray}\displaystyle & \displaystyle f_{E,1}=-2\unicode[STIX]{x03C0}a\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}(A_{1}^{0}-B_{1}^{0}), & \displaystyle\end{eqnarray}$$

(3.74) $$\begin{eqnarray}\displaystyle & \displaystyle f_{E,2}=\text{i}2\unicode[STIX]{x03C0}a\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}(A_{1}^{0}+B_{1}^{0}). & \displaystyle\end{eqnarray}$$

We may notice that $f_{E,j}$ in (3.72) can be also computed through $\unicode[STIX]{x1D711}$ and $\unicode[STIX]{x1D719}_{j}$ by using the Haskind relation (C 5).

3.5 Some special cases

3.5.1 Simplified solution procedure at large $|x_{0}/a|$

When the circular cylinder is located far away from the crack, or $|x_{0}/a|\gg 1$ , the problems in §§ 3.3 and 3.4 can be solved by using the wide spacing approximation, based on the assumption that the mutual effects of evanescent waves between the crack and the body can be neglected. Without losing generality, here we may assume $x_{0}>0$ and the incoming wave is from $x=-\infty$ . We denote the radiation and scattering potentials of the body, with its centre at $x=0$ and submerged below the ice sheet without the crack, as $\unicode[STIX]{x1D713}_{j}^{r}$ and $\unicode[STIX]{x1D713}_{0}$ respectively. The solutions of these potentials can be obtained by dropping the terms of ${\mathcal{I}}_{1}$ and ${\mathcal{I}}_{2}$ in (3.43) and (3.44), and the procedure will be much simpler. We further define the velocity potentials $\unicode[STIX]{x1D713}_{L}$ and $\unicode[STIX]{x1D713}_{R}$ for ice sheet with a crack but without the body. They correspond to incoming waves from $x=-\infty$ and $x=+\infty$ respectively, and contain both the incident and diffracted potentials. These two potentials can be obtained directly from (3.62) and (3.66). Using these results the solution for the current problem of a body submerged beneath an ice sheet with a crack can be easily constructed for $|x_{0}/a|\gg 1$ , following the method used for multiple bodies with a large gap (Srokosz & Evans Reference Srokosz and Evans1979) and a floating body in a wide polynya (Li et al. Reference Li, Shi and Wu2017a ).

We may notice that the transmission coefficient $T_{c}$ and reflection coefficient $R_{c}$ corresponding to $\unicode[STIX]{x1D713}_{L}$ and $\unicode[STIX]{x1D713}_{R}$ are the same, which can be easily confirmed through reversing the direction of the $x$ axis. At infinity, the evanescent waves will tend to zero and the velocity potentials defined above will have the following asymptotic forms

(3.75a ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D713}_{j}^{r}=A_{j}^{\pm }\text{e}^{\mp \text{i}\unicode[STIX]{x1D706}x}\unicode[STIX]{x1D713}(z)\text{ as }x\rightarrow \pm \infty ,\quad (j=1,2), & \displaystyle\end{eqnarray}$$

(3.75b ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D713}_{0}=t_{0}\text{e}^{-\text{i}\unicode[STIX]{x1D706}x}\unicode[STIX]{x1D713}(z)\text{ as }x\rightarrow +\infty , & \displaystyle\end{eqnarray}$$

(3.75c ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D713}_{0}=(\text{e}^{-\text{i}\unicode[STIX]{x1D706}x}+r_{0}\text{e}^{+\text{i}\unicode[STIX]{x1D706}x})\unicode[STIX]{x1D713}(z)\text{ as }x\rightarrow -\infty , & \displaystyle\end{eqnarray}$$

(3.75d ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D713}_{L}=T_{c}\text{e}^{-\text{i}\unicode[STIX]{x1D706}x}\unicode[STIX]{x1D713}(z)\text{ as }x\rightarrow +\infty , & \displaystyle\end{eqnarray}$$

(3.75e ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D713}_{L}=(\text{e}^{-\text{i}\unicode[STIX]{x1D706}x}+R_{c}\text{e}^{+\text{i}\unicode[STIX]{x1D706}x})\unicode[STIX]{x1D713}(z)\text{ as }x\rightarrow -\infty , & \displaystyle\end{eqnarray}$$

(3.75f ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D713}_{R}=(\text{e}^{+\text{i}\unicode[STIX]{x1D706}x}+R_{c}\text{e}^{-\text{i}\unicode[STIX]{x1D706}x})\unicode[STIX]{x1D713}(z)\text{ as }x\rightarrow +\infty , & \displaystyle\end{eqnarray}$$

(3.75g ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D713}_{R}=T_{c}\text{e}^{+\text{i}\unicode[STIX]{x1D706}x}\unicode[STIX]{x1D713}(z)\text{ as }x\rightarrow -\infty . & \displaystyle\end{eqnarray}$$

For the submerged circular cylinder undergoing oscillation in the $j$ th mode, by following the procedure in Srokosz & Evans (Reference Srokosz and Evans1979), the velocity potential near the body can be written as a summation of the potential without a crack and the potential due to the wave reflected by the crack

(3.76) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{j}(x,z)=\unicode[STIX]{x1D713}_{j}^{r}(x-x_{0},z)+\unicode[STIX]{x1D716}_{j}\unicode[STIX]{x1D713}_{0}(x-x_{0},z).\end{eqnarray}$$

Near the crack we may write the velocity potential as

(3.77) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{j}(x,z)=\unicode[STIX]{x1D702}_{j}\unicode[STIX]{x1D713}_{R}(x,z),\end{eqnarray}$$

which is due to the incoming wave to the crack generated by the body motion and its reflection by the crack. $\unicode[STIX]{x1D716}_{j}$ and $\unicode[STIX]{x1D702}_{j}$ in the two equations are two unknown coefficients to be determined. $x-x_{0}$ in (3.76) reflects the fact that the solutions for $\unicode[STIX]{x1D713}_{j}^{r}$ and $\unicode[STIX]{x1D713}_{0}$ correspond to the cylinder with centre located at $(0,z_{0})$ , which in the current problem is at $(x_{0},z_{0})$ .

The solution has now been constructed based on results which could be obtained much more easily. The only task is to find the two unknown coefficients. We may consider (3.76) and (3.77) at a vertical line at $x=x_{I}$ between the crack and cylinder. On this vertical interface, the continuity conditions of pressure and normal velocity may be enforced, i.e.

(3.78a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{j}(x_{I-},z)=\unicode[STIX]{x1D719}_{j}(x_{I+},z),\quad \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{j}(x_{I-},z)}{\unicode[STIX]{x2202}x}=\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{j}(x_{I+},z)}{\unicode[STIX]{x2202}x},\end{eqnarray}$$

where the subscripts $+$ and $-$ mean that the potentials should be taken from the solutions on the right- and left-hand sides of the vertical line, or (3.76) and (3.77), respectively. On the basis $0\ll x_{I}\ll x_{0}$ , (3.75a )–(3.75g ) become applicable at $x=x_{I}$ . Substituting (3.76) and (3.77), and the corresponding asymptotic expansions into (3.78), we can obtain

(3.79) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D716}_{j}=\frac{A_{j}^{-}R_{c}}{\text{e}^{2\text{i}\unicode[STIX]{x1D706}x_{0}}-R_{c}r_{0}}, & \displaystyle\end{eqnarray}$$

(3.80) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D702}_{j}=\frac{A_{j}^{-}\text{e}^{\text{i}\unicode[STIX]{x1D706}x_{0}}}{\text{e}^{2\text{i}\unicode[STIX]{x1D706}x_{0}}-R_{c}r_{0}}. & \displaystyle\end{eqnarray}$$

Substituting (3.76) into (3.56), we can obtain the added mass and damping coefficient as

(3.81) $$\begin{eqnarray}\unicode[STIX]{x1D70F}_{js}=\unicode[STIX]{x1D70F}_{js}^{o}-\unicode[STIX]{x1D716}_{s}f_{E,j}^{o}/g,\end{eqnarray}$$

where the superscript $o$ means that the results are from the ice sheet without crack.

Similarly, for the diffraction problem, near the body we may write the potential as

(3.82) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{0}(x,z)=\unicode[STIX]{x1D716}_{0}\unicode[STIX]{x1D713}_{0}(x-x_{0},z),\end{eqnarray}$$

and near the crack we may write the potential as

(3.83) $$\begin{eqnarray}\unicode[STIX]{x1D719}_{0}(x,z)=\unicode[STIX]{x1D713}_{L}(x,z)+\unicode[STIX]{x1D702}_{0}\unicode[STIX]{x1D713}_{R}(x,z).\end{eqnarray}$$

Using the asymptotic expansion and imposing the continuity condition (3.78) at $x=x_{I}$ , we have

(3.84) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D716}_{0}=\frac{T_{c}\text{e}^{\text{i}\unicode[STIX]{x1D706}x_{0}}}{\text{e}^{2\text{i}\unicode[STIX]{x1D706}x_{0}}-R_{c}r_{0}}, & \displaystyle\end{eqnarray}$$

(3.85) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D702}_{0}=\frac{T_{c}r_{0}}{\text{e}^{2\text{i}\unicode[STIX]{x1D706}x_{0}}-R_{c}r_{0}}. & \displaystyle\end{eqnarray}$$

Substituting (3.82) into (3.72), we can obtain the wave exciting force on the cylinder below the ice sheet with a crack from the result without the crack

(3.86) $$\begin{eqnarray}f_{E,j}=\unicode[STIX]{x1D716}_{0}f_{E,j}^{o}.\end{eqnarray}$$

We should notice that (3.81) and (3.86) are not limited to a circular cylinder and they are applicable for a body of arbitrary shape.

3.5.2 Solution for infinite water depth

When the water depth tends to infinity or $H\rightarrow +\infty$ , $G_{ice}$ and $G_{1}$ in the Green function (3.38) respectively become

(3.87) $$\begin{eqnarray}G_{ice}=\ln (r_{1})-\ln (r_{1}^{\prime })-2\int _{{\mathcal{D}}}\frac{Lk^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D714}^{2}}{K(\unicode[STIX]{x1D714},k)}\text{e}^{k(z+z_{0})}\cos [k(x-x_{0})]\,\text{d}k,\end{eqnarray}$$

and

(3.88) $$\begin{eqnarray}G_{1}=\frac{L}{\unicode[STIX]{x03C0}}\int _{{\mathcal{D}}}\frac{\text{e}^{kz}}{K(\unicode[STIX]{x1D714},k)}k^{2}\cos (kx)\,\text{d}k,\end{eqnarray}$$

together with ${\mathcal{I}}_{1}$ and ${\mathcal{I}}_{2}$ simplified as

(3.89a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{I}}_{1}=\int _{{\mathcal{D}}}\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}-(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D714}^{2})k}{K(\unicode[STIX]{x1D714},k)}k^{2}\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.89b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{I}}_{2}=\int _{{\mathcal{D}}}\frac{\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}-(\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D714}^{2})k}{K(\unicode[STIX]{x1D714},k)}\,\text{d}k, & \displaystyle\end{eqnarray}$$

where $r_{1}^{\prime }$ is the distance between $p$ and the mirror image of $q$ about the mean upper surface $z=0$ , and

(3.90) $$\begin{eqnarray}K(\unicode[STIX]{x1D714},k)=(Lk^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D714}^{2})k-\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}.\end{eqnarray}$$

For the multipole expansions, (3.45a ) and (3.45b ) become

(3.91a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{G}}_{1}(n)=-2\int _{{\mathcal{D}}}\frac{\text{e}^{kz_{0}-\text{i}kx_{0}}}{K(\unicode[STIX]{x1D714},k)}k^{n+3}\,\text{d}k, & \displaystyle\end{eqnarray}$$

(3.91b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{G}}_{2}(n)=-2\int _{{\mathcal{D}}}\frac{\text{e}^{kz_{0}+\text{i}kx_{0}}}{K(\unicode[STIX]{x1D714},k)}k^{n+3}\,\text{d}k, & \displaystyle\end{eqnarray}$$

and (3.51a )–(3.51c ) are simplified as

(3.92a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{J}}_{1}(n)=0, & \displaystyle\end{eqnarray}$$

(3.92b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{J}}_{2}(n)=0, & \displaystyle\end{eqnarray}$$

(3.92c ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{J}}_{3}(n)=2\int _{{\mathcal{D}}}\left[1+\frac{2\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}}{K(\unicode[STIX]{x1D714},k)}\right]\text{e}^{2kz_{0}}k^{n-1}\,\text{d}k. & \displaystyle\end{eqnarray}$$

For the problem of wave propagation across the crack, ${\mathcal{F}}_{1}(n)$ and ${\mathcal{F}}_{2}(n)$ in (3.67) respectively become

(3.93a ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{F}}_{1}(n)=-L\unicode[STIX]{x1D706}^{3}\left[\frac{\unicode[STIX]{x1D706}{\mathcal{G}}_{2}(n)}{{\mathcal{I}}_{1}}-\frac{{\mathcal{G}}_{2}(n-1)}{{\mathcal{I}}_{2}}\right], & \displaystyle\end{eqnarray}$$

(3.93b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{F}}_{2}(n)=4\unicode[STIX]{x1D706}^{4}\text{e}^{\unicode[STIX]{x1D706}z_{0}-\text{i}\unicode[STIX]{x1D706}x_{0}}+L\unicode[STIX]{x1D706}^{3}\left[\frac{\unicode[STIX]{x1D706}{\mathcal{G}}_{1}(n)}{{\mathcal{I}}_{1}}+\frac{{\mathcal{G}}_{1}(n-1)}{{\mathcal{I}}_{2}}\right]. & \displaystyle\end{eqnarray}$$

The simplified solution in § 3.5.1 can be also extended to infinite water depth. This is obtained by setting $\unicode[STIX]{x1D713}(z)$ in (3.75a–g ) as $\text{e}^{\unicode[STIX]{x1D706}z}$ , while the other equations remain the same.

When there is no ice crack, i.e. removing the terms ${\mathcal{I}}_{1}$ and ${\mathcal{I}}_{2}$ , ${\mathcal{F}}_{1}(n)$ in (3.93a ) and $E(n,l)$ in (3.50c ) become zero. Then (3.52) provides $A_{l}^{0}=0$ . By letting $H\rightarrow +\infty$ in (3.44), we obtain $\lim _{x\rightarrow -\infty }g_{n}=0$ . Thus according to (3.46), we can conclude that there is no wave reflection by the circular cylinder. If the ice thickness is taken to be zero, this becomes the well-known case for a free surface, as has been shown by Dean (Reference Dean1948) and Ursell (Reference Ursell1950). When the ice sheet has a crack, this conclusion is not true in general. However (3.91b ) provides $\lim _{x_{0}\rightarrow +\infty }{\mathcal{G}}_{2}(n)=0$ and we can also obtain $A_{l}^{0}=0$ . Thus there will be no reflection in such a case. This is of course quite obvious as the wave due to $\unicode[STIX]{x1D711}=\unicode[STIX]{x1D719}_{I}+\unicode[STIX]{x1D719}_{D}^{1}$ becomes the normal incident wave when $x_{0}\rightarrow +\infty$ , or the cylinder is infinitely far away from the crack.

4.1 Wave scattering by an ice crack

We may use the problem of wave propagation beneath an ice sheet with a crack to first verify the present formulation. The case has been considered by Evans & Porter (Reference Evans and Porter2003). In that work, the problem was first divided into the symmetric and antisymmetric parts, and then Green’s identity was applied to each part together with the Green function satisfying the condition on the ice sheet without a crack. Thus the result for each part turned out to be a single series multiplied by an unknown constant which was further determined by the condition at the crack. Here, by taking the advantage of the Green function in (3.38), which itself already satisfies the ice crack condition, the solution can be obtained directly through the Green identity, as shown in (3.66). If we convert the integration in (3.66) into series form, it can be found that the solution is identical to that of Evans & Porter (Reference Evans and Porter2003). Taking $x\leqslant 0$ as an example, we first extend the route of integrations in (3.66) to $k\in (-\infty ,+\infty )$ , and add an integration along a semicircle of infinite radius in the lower half-complex plane of $k$ . The latter does not affect the result as it is in fact a zero term, which follows Jordan’s lemma. Then using residue theorem and noticing the poles at $-k_{n}$ , $n=-2,-1,0,1,\ldots$ in the lower half-plane, the potential $\unicode[STIX]{x1D711}$ can be expressed in the series form as

(4.2) $$\begin{eqnarray}\unicode[STIX]{x1D711}=\frac{I}{Z_{0}(0)}\unicode[STIX]{x1D711}_{0},\end{eqnarray}$$

where

(4.3) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D711}_{0} & = & \displaystyle \text{e}^{\text{i}k_{0}x}Z_{0}(z)\nonumber\\ \displaystyle & & \displaystyle +\,\left[\frac{k_{0}^{3}Z_{0}^{\prime }(0)}{\displaystyle \mathop{\sum }_{n=-2}^{\infty }\frac{k_{n}^{6}Z_{n}^{\prime }(0)}{{\mathcal{X}}^{\prime }(k_{n})}}\mathop{\sum }_{n=-2}^{\infty }\frac{k_{n}^{3}Z_{n}(z)\text{e}^{-\text{i}k_{n}x}}{{\mathcal{X}}^{\prime }(k_{n})}-\frac{k_{0}^{2}Z_{0}^{\prime }(0)}{\displaystyle \mathop{\sum }_{n=-2}^{\infty }\frac{k_{n}^{4}Z_{n}^{\prime }(0)}{{\mathcal{X}}^{\prime }(k_{n})}}\mathop{\sum }_{n=-2}^{\infty }\frac{k_{n}^{2}Z_{n}(z)\text{e}^{-\text{i}k_{n}x}}{{\mathcal{X}}^{\prime }(k_{n})}\right],\qquad\end{eqnarray}$$

and

(4.4) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{X}}(k)=(L\unicode[STIX]{x1D706}^{4}+\unicode[STIX]{x1D70C}g-m\unicode[STIX]{x1D714}^{2})k\sinh (kH)-\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}^{2}\cosh (kH), & \displaystyle\end{eqnarray}$$

(4.5) $$\begin{eqnarray}\displaystyle & \displaystyle Z_{n}(z)=\cosh [k_{n}(z+H)]. & \displaystyle\end{eqnarray}$$

Here, $k_{0}=-\unicode[STIX]{x1D706}$ , $k_{n}$ with $n=1,2,\ldots$ are the purely positive imaginary roots of (2.8), $k_{-1}$ and $k_{-2}$ are the corresponding two complex roots with positive imaginary parts. It can be seen that (4.3) is identical to (2.17) in Evans & Porter (Reference Evans and Porter2003) for the normal incident case, by noticing that the time factor adopted by them is $\text{e}^{-\text{i}\unicode[STIX]{x1D714}t}$ , while $\text{e}^{+\text{i}\unicode[STIX]{x1D714}t}$ is used in (2.1) here.

At $x\rightarrow \pm \infty$ , by applying the Fourier integrals (13.16) in Wehausen & Laitone (Reference Wehausen and Laitone1960) to $\unicode[STIX]{x1D719}_{D}^{1}$ in (3.66), the potential $\unicode[STIX]{x1D711}$ can be given as

(4.6) $$\begin{eqnarray}\unicode[STIX]{x1D711}=\left\{\begin{array}{@{}ll@{}}\displaystyle I\times T_{c}\text{e}^{-\text{i}\unicode[STIX]{x1D706}x}\unicode[STIX]{x1D713}(z) & \text{at }x\rightarrow +\infty \\ \displaystyle I(\text{e}^{-\text{i}\unicode[STIX]{x1D706}x}+R_{c}\text{e}^{+\text{i}\unicode[STIX]{x1D706}x})\unicode[STIX]{x1D713}(z) & \text{at }x\rightarrow -\infty ,\end{array}\right.\end{eqnarray}$$

with transmission coefficient $T_{c}$ and reflection coefficient $R_{c}$ given in (B 7) and (B 8) respectively. It may be interesting to compare these results for wave propagation below an ice sheet with a crack and those through a polynya confined between two semi-infinite ice sheets. When the polynya width $b$ tends to zero, the result is expected to tend to that of an ice sheet with a crack. Comparison of the reflection coefficients is shown in figure 2, where $\unicode[STIX]{x1D70E}=\unicode[STIX]{x1D714}^{2}h/g$ , and $|R_{p}|$ for polynya with $b=0.05$ is computed by MEE (Ren et al. Reference Ren, Wu and Thomas2016). From the figure, it can be seen that $|R_{p}|$ and $|R_{c}|$ agree with each other very well within the frequency range calculated.

Figure 2.

Comparison of reflection coefficients. Solid line: $R_{c}$ for a crack; dashed line: $R_{p}$ for a polynya with $b=0.05$ . ( $H=100$ , $h=1$ , $m=0.9$ , $L=45582$ .)

4.2 The radiation problem

We consider the radiation problem of a submerged circular cylinder, with its radius $a$ taken as the characteristic length. The convergence study with $N$ is first carried out through the hydrodynamic coefficients against $\unicode[STIX]{x1D70E}=\unicode[STIX]{x1D714}^{2}a/g$ , as presented in figure 3. It can be seen that the numerical results converge very rapidly, and $N=5$ and $N=7$ give graphically indistinguishable results in the figure. The far field formula in the appendix has been used for calculating the damping coefficient and excellent agreement with the near field result has been found. For the results obtained subsequently we have taken $N=5$ .

Figure 3.

Hydrodynamic coefficients of a submerged circular cylinder against $\unicode[STIX]{x1D70E}$ with different $N$ . Solid lines: $N=7$ ; dashed lines: $N=5$ ; dash-dotted lines: $N=3$ ; dotted lines: $N=1$ . ( $a=1$ , $x_{0}=20$ , $z_{0}=-1.2$ , $H=20$ , $h=0.2$ , $m=0.18$ , $L=72.932$ .)

In figure 4 we show the hydrodynamic coefficients at different $x_{0}$ . As can be observed in this figure, when $\unicode[STIX]{x1D70E}\rightarrow 0$ , the results with the crack will tend to those without the crack. This is due to the fact that at very small $\unicode[STIX]{x1D70E}$ , the condition in (2.3) tends to $\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{j}/\unicode[STIX]{x2202}z=0$ on $z=0$ , which means $\unicode[STIX]{x2202}^{3}\unicode[STIX]{x1D719}_{j}/\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{2}\rightarrow 0$ and $\unicode[STIX]{x2202}^{4}\unicode[STIX]{x1D719}_{j}/\unicode[STIX]{x2202}z\unicode[STIX]{x2202}x^{3}\rightarrow 0$ on the whole ice sheet. Therefore, the conditions at the crack will be automatically satisfied. Also as $\unicode[STIX]{x1D70E}\rightarrow 0$ , the effect of the position of $x_{0}$ diminishes. When $x_{0}=0$ , we have ${\mathcal{G}}_{1}(n)={\mathcal{G}}_{2}(n)$ in (3.45a ) and (3.45b ), which means in (3.50a )–(3.50c ) $C(n,l)=C(l,n)$ and $D(n,l)=E(n,l)$ . Then (3.52) and (3.53) give

(4.7a,b ) $$\begin{eqnarray}A_{l}^{1}+B_{l}^{1}=0,\quad A_{l}^{2}-B_{l}^{2}=0.\end{eqnarray}$$

This suggests that $\unicode[STIX]{x1D719}_{1}$ and $\unicode[STIX]{x1D719}_{2}$ in (3.46) are antisymmetric and symmetric about the body centre respectively, as fully expected. Therefore, we have $\unicode[STIX]{x1D70F}_{js}$ equal to zero when $j\neq s$ .

Figure 4.

Hydrodynamic coefficients of a submerged circular cylinder against $\unicode[STIX]{x1D70E}$ at different $x_{0}$ . Solid lines: ice sheet without crack; dashed lines: $x_{0}=0$ ; dash-dotted lines: $x_{0}=10$ . ( $a=1$ , $z_{0}=-1.2$ , $H=20$ , $h=0.2$ , $m=0.18$ , $L=72.932$ .)

Figure 5.

Hydrodynamic coefficients of a submerged circular cylinder against $\unicode[STIX]{x1D6EC}$ in heave mode. In (a) and (d), $\unicode[STIX]{x1D70E}=0.50$ ; in (b) and (e), $\unicode[STIX]{x1D70E}=3.00$ ; in (c) and (f), $\unicode[STIX]{x1D70E}=6.00$ . Solid line is for the exact results, while dashed line denotes the results from wide spacing approximation. ( $a=1$ , $z_{0}=-1.2$ , $H=20$ , $h=0.2$ , $m=0.18$ , $L=72.932$ .)

We may notice in figure 4 that when $x_{0}\neq 0$ , the results are oscillatory with the frequency. Obviously, this is due to the fact that the wave generated by the body motion will arrive at the crack and then be reflected back to the body. This process is very much frequency dependent. Also we may infer that if we move the body position from $x_{0}$ to $x_{0}+\ell /2$ , where $\ell =2\unicode[STIX]{x03C0}/\unicode[STIX]{x1D706}$ is the wavelength of the travelling wave generated by the body, the effect of the crack may be almost the same. To verify that, in figure 5 we plot $\unicode[STIX]{x1D707}_{22}$ and $\unicode[STIX]{x1D706}_{22}$ against $\unicode[STIX]{x1D6EC}=\unicode[STIX]{x1D706}x_{0}$ at various $\unicode[STIX]{x1D70E}$ , or a fixed $\unicode[STIX]{x1D706}$ when $x_{0}$ varies. It can be seen that the results become virtually periodic with $x_{0}$ when $\unicode[STIX]{x1D6EC}>4$ and they also become graphically indistinguishable to those from wide approximation method. To provide further insight into what has been observed in figure 5, we may follow the procedure in Li et al. (Reference Li, Shi and Wu2017a ) and use the equations in § 3.5.1.

Similar to (B 4), $f_{E,j}^{o}$ for the ice sheet without the crack can be expressed through the far field formula as

(4.8) $$\begin{eqnarray}f_{E,j}^{o}=2I\unicode[STIX]{x1D70C}\unicode[STIX]{x1D714}Q_{0}C_{g}A_{j}^{-},\end{eqnarray}$$

where $Q_{0}$ and $C_{g}$ are given in (A 7) and (A 8) respectively. Then invoking (3.81), we have

(4.9) $$\begin{eqnarray}\unicode[STIX]{x1D70F}_{js}=\unicode[STIX]{x1D70F}_{js}^{o}+2\text{i}\unicode[STIX]{x1D70C}Q_{0}C_{g}A_{js},\end{eqnarray}$$

where

(4.10) $$\begin{eqnarray}A_{js}=\frac{A_{j}^{-}A_{s}^{-}R_{c}}{\text{e}^{2\text{i}\unicode[STIX]{x1D6EC}}-R_{c}r_{0}}.\end{eqnarray}$$

For a given $\unicode[STIX]{x1D70E}$ , (4.10) shows that $A_{js}$ is periodic against $x_{0}$ with period of $\ell /2$ . This means that $\unicode[STIX]{x1D707}_{js}$ and $\unicode[STIX]{x1D706}_{js}$ both will respectively have the same values at $x_{0}+n\ell /2$ , where $n$ includes all integers to ensure that $x_{0}$ is positive. From (4.10) it can be also seen that $A_{js}$ maps the unit circle $\text{e}^{2\text{i}\unicode[STIX]{x1D6EC}}$ to a circle with centre at

(4.11) $$\begin{eqnarray}y_{c,r}=x_{c,r}+\text{i}z_{c,r}=-|R_{c}r_{0}|r_{r}\text{e}^{\text{i}\unicode[STIX]{x1D717}_{r}},\end{eqnarray}$$

and radius as

(4.12) $$\begin{eqnarray}r_{r}=\frac{|A_{j}^{-}A_{s}^{-}R_{c}|}{1-|R_{c}r_{0}|^{2}},\end{eqnarray}$$

where

(4.13) $$\begin{eqnarray}\text{e}^{\text{i}\unicode[STIX]{x1D717}_{r}}=\frac{A_{j}^{-}A_{s}^{-}\bar{r_{0}}}{|A_{j}^{-}A_{s}^{-}\bar{r_{0}}|}.\end{eqnarray}$$

From (4.11) and (4.12), (4.10) can be rewritten as

(4.14) $$\begin{eqnarray}A_{js}=y_{c,r}+r_{r}\text{e}^{\text{i}\unicode[STIX]{x1D70F}_{r}}.\end{eqnarray}$$

This indicates that the peak ( $+$ ) and trough ( $-$ ) of $\unicode[STIX]{x1D707}_{js}$ respectively equal

(4.15) $$\begin{eqnarray}\unicode[STIX]{x1D707}_{js}^{\pm }=\unicode[STIX]{x1D707}_{js}^{o}-2\unicode[STIX]{x1D70C}Q_{0}C_{g}(z_{c,r}\mp r_{r}),\end{eqnarray}$$

and they occur respectively when $\unicode[STIX]{x1D70F}_{r}$ equals $-\unicode[STIX]{x03C0}/2$ or $\unicode[STIX]{x03C0}/2$ . Similarly, the peak ( $+$ ) and trough ( $-$ ) of $\unicode[STIX]{x1D706}_{js}$ respectively equal

(4.16) $$\begin{eqnarray}\unicode[STIX]{x1D706}_{js}^{\pm }=\unicode[STIX]{x1D706}_{js}^{o}-2\unicode[STIX]{x1D714}\unicode[STIX]{x1D70C}Q_{0}C_{g}(x_{c,r}\mp r_{r}),\end{eqnarray}$$

and they occur respectively when $\unicode[STIX]{x1D70F}_{r}$ equals $\unicode[STIX]{x03C0}$ or $0$ . We may notice that the corresponding $\unicode[STIX]{x1D6EC}$ for $\unicode[STIX]{x1D707}_{js}^{\pm }$ and $\unicode[STIX]{x1D706}_{js}^{\pm }$ can be further obtained from (4.10) and (4.14).

The above discussions are valid for a body of general shape. For the circular cylinder, (4.10) may be further simplified. As shown in § 3.5.2, $|r_{0}|$ is zero for infinite water depth. For the finite water depth neither $\unicode[STIX]{x1D707}_{l}(0)$ nor $E(n,l)$ is zero, and $|r_{0}|$ will not exactly be zero. However, $|r_{0}|$ is still much smaller than $|t_{0}|$ . In fact it has been found that $|r_{0}|<0.01$ over the range of $\unicode[STIX]{x1D70E}$ in figures 4 and 5. Thus (4.10) can be approximated as

(4.17) $$\begin{eqnarray}A_{js}\approx A_{j}^{-}A_{s}^{-}R_{c}\text{e}^{-2\text{i}\unicode[STIX]{x1D6EC}},\end{eqnarray}$$

which gives $y_{c,r}\approx 0$ and $r_{r}\approx |A_{j}^{-}A_{s}^{-}R_{c}|$ . These indicate that the local extrema of $\unicode[STIX]{x1D707}_{js}$ and $\unicode[STIX]{x1D706}_{js}$ will appear when $A_{js}$ respectively becomes a purely imaginary or real number, and if one of them reaches a local extremum, the other will reach the value for ice sheet without the crack.

Figure 6.

Hydrodynamic coefficients of a submerged circular cylinder against $H$ . Solid line represents the results by the method for finite water depth, while dashed line represents those by the method for infinite water depth. ( $a=1$ , $(x_{0},z_{0})=(5.0,-1.2)$ , $\unicode[STIX]{x1D70E}=3.00$ , $h=0.2$ , $m=0.18$ , $L=72.932$ .)

In figure 6, we show the hydrodynamic coefficients against $H$ for a given $\unicode[STIX]{x1D70E}$ . As can be expected, the results tend to those for infinite water depth as $H$ becomes larger. However for the heave mode, or $j=2$ , it can be seen that this trend is very slow.

For the problem of a free surface with infinite water depth, Ogilvie (Reference Ogilvie1963) found that the wave generated by the forced body oscillation would propagate only in one direction if the path of the circular cylinder centre follows a circular orbit. When the water surface is covered by an ice sheet, a similar conclusion may be drawn. Removing the terms related to ${\mathcal{I}}_{1}$ and ${\mathcal{I}}_{2}$ in (3.50a )–(3.50c ) and noticing (3.92a )–(3.92c ), invoking (3.52) we have

(4.18) $$\begin{eqnarray}A_{l}^{2}=\text{i}A_{l}^{1}.\end{eqnarray}$$

We may take the clockwise rotation as an example. This is achieved by taking the motion amplitudes $|\unicode[STIX]{x1D6FC}_{1}|=|\unicode[STIX]{x1D6FC}_{2}|$ in (2.1), and their phases $\unicode[STIX]{x1D717}_{2}=\unicode[STIX]{x1D717}_{1}+\unicode[STIX]{x03C0}/2$ . Invoking (3.46) and (4.18) we obtain

(4.19) $$\begin{eqnarray}\unicode[STIX]{x1D6F7}(x,z,t)=\text{Re}\left[\text{i}\unicode[STIX]{x1D714}\text{e}^{\text{i}\unicode[STIX]{x1D714}t}\unicode[STIX]{x1D6FC}_{1}\mathop{\sum }_{n=1}^{\infty }a^{n}(B_{n}^{1}\text{e}^{\text{i}\unicode[STIX]{x1D717}_{1}}+\text{i}B_{n}^{2}\text{e}^{\text{i}\unicode[STIX]{x1D717}_{1}})g_{n}\right].\end{eqnarray}$$

Since $\lim _{x\rightarrow -\infty }g_{n}=0$ , we can conclude that clockwise oscillation will generate the wave propagating to $x=+\infty$ only. Similarly, the counterclockwise oscillation will generate the wave propagating to $x=-\infty$ only due to $\lim _{x\rightarrow +\infty }f_{n}=0$ . When there is a crack in the ice sheet, since the equations for $A_{l}^{j}$ and $B_{l}^{j}$ are coupled, (4.18) is not valid in general. However, when $x_{0}\rightarrow +\infty$ , we have ${\mathcal{G}}_{2}(n)\rightarrow 0$ in (3.91b ), which gives $E(n,l)\rightarrow 0$ in (3.50c ). Substituting this into (3.52), we can obtain (4.18) again, i.e. no wave will propagate to $x=-\infty$ for clockwise oscillation.

Figure 7.

Wave exciting force on a submerged circular cylinder against $\unicode[STIX]{x1D70E}$ at different $x_{0}$ . Solid lines: $x_{0}=0$ ; dashed lines: $x_{0}=10$ ; dash-dotted lines: $x_{0}=20$ ; dotted lines: ice sheet without crack. $\times$ and $+$ are the wide spacing solutions for $x_{0}=10$ and $x_{0}=20$ , respectively. ( $a=1$ , $z_{0}=-1.2$ , $H=20$ , $h=0.2$ , $m=0.18$ , $L=72.932$ .)

4.3 The diffraction problem

We then consider the problem of wave diffraction by the submerged circular cylinder. Figure 7 presents the results for the excitation force at different $x_{0}$ against $\unicode[STIX]{x1D70E}$ , together with those for the ice sheet without crack. From each figure, it can be seen that the results tend to the same value when $\unicode[STIX]{x1D70E}\rightarrow 0$ , for the reason discussed in § 4.2. It can also be seen that when $x_{0}$ is sufficiently large, the exact results are hardly different from those from the wide spacing method in § 3.5.1, even for very long waves, although the approximation is based on the assumption that $x_{0}$ is much larger than the wavelength. The reason for that is when $\unicode[STIX]{x1D70E}\rightarrow 0$ , $\unicode[STIX]{x1D716}_{0}\rightarrow 1$ . Then invoking (3.86) we have that $f_{E,j}$ tends to $f_{E,j}^{o}$ , or it tends to the exact result. This reveals that the wide spacing approximation can work well for both small and large $\unicode[STIX]{x1D70E}$ .

We may use the wide spacing approximation in (3.86) to give some insights into the features of the wave exciting force. Substituting (3.84) into (3.86) and invoking (4.8), we obtain

(4.20) $$\begin{eqnarray}f_{E,j}=-2\text{i}\unicode[STIX]{x1D70C}gQ_{0}C_{g}(A_{j}^{-}\unicode[STIX]{x1D716}_{0}\text{e}^{\text{i}\unicode[STIX]{x1D6EC}})\text{e}^{-\text{i}\unicode[STIX]{x1D6EC}}.\end{eqnarray}$$

Here

(4.21) $$\begin{eqnarray}A_{j}^{-}\unicode[STIX]{x1D716}_{0}\text{e}^{\text{i}\unicode[STIX]{x1D6EC}}=\frac{A_{j}^{-}T_{c}\text{e}^{2\text{i}\unicode[STIX]{x1D6EC}}}{\text{e}^{2\text{i}\unicode[STIX]{x1D6EC}}-R_{c}r_{0}}.\end{eqnarray}$$

It should be noticed that the oscillatory term $A_{j}^{-}\unicode[STIX]{x1D716}_{0}\text{e}^{\text{i}\unicode[STIX]{x1D6EC}}$ is due to the interaction between the cylinder and the crack, while the oscillatory term $\text{e}^{-\text{i}\unicode[STIX]{x1D6EC}}$ is due to the initial phase of the incident potential $\unicode[STIX]{x1D719}_{I}$ relative to the body at $x_{0}$ . From (4.20), we can see that at a fixed $\unicode[STIX]{x1D70E}$ , $f_{E,j}$ is periodic against $\unicode[STIX]{x1D6EC}$ or it varies periodically with $x_{0}$ . The period for $\text{Re}(f_{E,j})$ and $\text{Im}(f_{E,j})$ equals $2\unicode[STIX]{x03C0}$ , as can be observed in figure 8.

Figure 8.

Wave exciting force on a submerged circular cylinder against $\unicode[STIX]{x1D6EC}$ . Solid line is for the exact results, while dashed line denotes the results from wide spacing approximation. ( $a=1$ , $z_{0}=-1.2$ , $\unicode[STIX]{x1D70E}=3.00$ , $H=20$ , $h=0.2$ , $m=0.18$ , $L=72.932$ .)

From (4.21) it can be seen that $A_{j}^{-}\unicode[STIX]{x1D716}_{0}\text{e}^{\text{i}\unicode[STIX]{x1D6EC}}$ maps the unit circle $\text{e}^{2\text{i}\unicode[STIX]{x1D6EC}}$ to a circle with centre at

(4.22) $$\begin{eqnarray}y_{c,d}=x_{c,d}+\text{i}z_{c,d}=\frac{|1-2|R_{c}r_{0}|^{2}|}{|R_{c}r_{0}|}r_{d}\text{e}^{\text{i}\unicode[STIX]{x1D717}_{d}},\end{eqnarray}$$

and radius as

(4.23) $$\begin{eqnarray}r_{d}=\frac{|A_{j}^{-}T_{L}R_{c}r_{0}|}{1-|R_{c}r_{0}|^{2}},\end{eqnarray}$$

where

(4.24) $$\begin{eqnarray}\text{e}^{\text{i}\unicode[STIX]{x1D717}_{d}}=\frac{(1-2|R_{c}r_{0}|^{2})A_{j}^{-}T_{c}}{|(1-2|R_{c}r_{0}|^{2})A_{j}^{-}T_{c}|}.\end{eqnarray}$$

From (4.22) and (4.23), (4.21) can be rewritten as

(4.25) $$\begin{eqnarray}A_{j}^{-}\unicode[STIX]{x1D716}_{0}\text{e}^{\text{i}\unicode[STIX]{x1D6EC}}=y_{c,d}+r_{d}\text{e}^{\text{i}\unicode[STIX]{x1D70F}_{d}}.\end{eqnarray}$$

This indicates that the peak ( $+$ ) and trough ( $-$ ) of $|f_{E,j}|$ respectively equal

(4.26) $$\begin{eqnarray}|f_{E,j}^{\pm }|=2\unicode[STIX]{x1D70C}gQ_{0}C_{g}||y_{c,d}|\pm r_{d}|,\end{eqnarray}$$

and they occur respectively when $\unicode[STIX]{x1D70F}_{d}$ equals $\unicode[STIX]{x1D717}_{d}$ or $\unicode[STIX]{x1D717}_{d}+\unicode[STIX]{x03C0}$ , and the corresponding $\unicode[STIX]{x1D6EC}$ can be further obtained according to (4.21) and (4.25). For the special case $1-2|R_{c}r_{0}|^{2}=0$ , we have $y_{c,d}=0$ and $|f_{E,j}^{+}|=|f_{E,j}^{-}|$ , which means that the location of the cylinder no longer has an effect on the magnitude of the force and the magnitude of the force is no longer oscillatory with respect to $x_{0}$ .

We may notice that the above discussions are valid for a body of generalized shape. For the present case of a circular cylinder, we may use the fact that $r_{0}\approx 0$ . Equation (4.21) becomes

(4.27) $$\begin{eqnarray}A_{j}^{-}\unicode[STIX]{x1D716}_{0}\text{e}^{\text{i}\unicode[STIX]{x1D6EC}}\approx A_{j}^{-}T_{c}.\end{eqnarray}$$

Substituting (4.27) into (4.20), it can be found that the oscillations of $\text{Re}(f_{E,j})$ and $\text{Im}(f_{E,j})$ in figure 8 are purely due to the fact that the cylinder is placed at a different horizontal position which causes the relative phase change of the incident wave. The magnitude of the exciting force becomes $|f_{E,j}|=2\unicode[STIX]{x1D70C}gQ_{0}C_{g}|A_{j}^{-}T_{c}|$ , which is not affected by $x_{0}$ , as can be seen in figure 8. We may notice that although $|f_{E,j}|$ tends to a constant as $x_{0}$ increases, the value of the constant is different from the force on the body without a crack. This is because $\unicode[STIX]{x1D719}_{D}^{1}$ in (3.61) at the body does not tend to zero even when $x_{0}\rightarrow +\infty$ . In fact, we have $|f_{E,j}|<|f_{E,j}^{o}|$ because $|T_{c}|<1$ is always valid based on the energy conservation equation (Barrett & Squire Reference Barrett and Squire1996).