On explicit geometric solutions of some inviscid flows with free boundary.

Suppose that the equation $F(t,\vec{x})=F(t,x,y,z)=0$ describes the moving free boundary surface. For $\vec{x}\in\partial\Omega(t)$ at time $t\geq 0$, let $\{\vec{X}(s;t,\vec{x})\}_{0\leq s\leq t}$ be the backward characteristic curve emanating from $(t,\vec{x})$. Then $F(s,\vec{X}(s;t,\vec{x}))=0$ for all $0\leq s\leq t$. Putting the ansatz (3.1) into the differential equation (1.4), one has

For simplicity, denote $X^{\sigma}_{0}\equiv X^{\sigma}(0;t,\vec{x})$ for $\sigma=x,\,y,\,z$. Then solving this, one has

Suppose that at $s=0$, the initial configuration for the free boundary is determined by a given function $\vec{x}\to f(\vec{x})\in\mathbb{R}$, i.e. $F(0,\vec{x})=f(\vec{x})=0$, Then it follows that

Thus the general equations for the free surface takes the form (3.2), with $\vec{x}\to f(\vec{x})$ being function independent of the variables $(t,x,y,z)$.

Substituting the ansatz (3.1) into the Bernoulli’s principle (1.6c), one has

Rewriting the above equally in accordance to (3.2), we have

Two equations (3.2) and (3.3) must be consistent, which means that we must have

for some fixed constants $c_{1},\,c_{2}\in\mathbb{R}$. The first equation of the above can be rewritten as $6g(t)=\ln(2c_{1}\frac{4+A^{\prime}}{2-A^{\prime}})$. Taking derivative, and noting that $g(t)=\int_{0}^{t}\frac{2}{A}{\mathrm{d}}s$, one gets

Multiplying the above with $A^{\prime}$ we have that $\frac{A^{\prime\prime}A^{\prime}}{(4+A^{\prime})(2-A^{\prime})}=\frac{2A^{\prime}}{A}$. Applying the partial fraction decomposition on the left hand side of this equation, we have

Multiplying the above with $3$, then integrating the resultant equation we obtain that

with $(A_{0},A^{\prime}_{0})\equiv(A(0),A^{\prime}(0))$. Next, substituting $6g(t)=\ln(2c_{1}\frac{4+A^{\prime}}{2-A^{\prime}})$ into the second equation of (3.4), then using (3.6), we also obtain that for $c_{0}\equiv c(0)$,

The root for (3.8) is given by one of the three possible solutions of Viète’s trigonometric formula (3.11):

for $k=0,\,1,\,2$. The relevant integer $k$ must be chosen so that solution satisfies the initial condition $A_{0}^{\prime}+2=V_{k}(1)$. By the definition of $\beta(\cdot)$ in (3.12), it can be verified that $\arccos(1-\tfrac{2}{\beta(A_{0}^{\prime})})\in[0,\pi]$ for $A_{0}^{\prime}\in[-6,2]$. Using this, we have

Therefore, depending on the initial data $A_{0}^{\prime}$, $a(t)$ solves the differential equation:

Integrating the above in $s\in[0,t]$, and by the fact that $a(0)=1$, we have for $t\geq 0$,

Let $T>0$ be the maximal time for which $a(t)\mapsto\mathcal{I}_{k}(a(t))$ is invertible. Then

In this case, the one physical real root for (3.8) is given by the Cardano’s formula (3.9):

Let $T\vcentcolon=\sup\{t>0\vcentcolon|a(t)|^{6}>\beta(A_{0}^{\prime})\}$, then $a(t)$ satisfies the differential equation

Integrating the above in $s\in[0,t]$, and using $a(0)=1$ we have that for $t\in[0,T)$,

Using the integral expressions (3.15) and (3.19), one can numerically plot the curve $a\mapsto\mathcal{I}_{0}(a),\mathcal{I}_{1}(a),\mathcal{I}_{2}(a),\mathcal{I}_{H}(a)$, which is then analysed to determine the convergence of $(A(t),A^{\prime}(t))$ as $t\to\pm\infty$. Without loss of generality, we restrict our analysis to the case $A_{0}>0$. The case $A_{0}<0$ can be obtained by the transformation: $(t,A)\mapsto(-t,-A)$, which preserves the differential equations (3.5)–(3.6). We consider initial data $(A_{0},A_{0}^{\prime})$ in $5$ distinct regimes. The statements and figures listed below, describing the behaviour of $(A,A^{\prime})$ as $t\to\infty$ are based on the numerical integration of (3.15) and (3.19).

In this case, $1\geq\beta(A_{0}^{\prime})$ and the differential equation is given by $A^{\prime}(t)=W(\frac{A(t)}{A_{0}})-2$. Moreover, (3.6) implies that $\mu>0$ and $t\mapsto A^{\prime}(t)$ is monotone decreasing due to $A>0$ and (3.5). From the fact that $\mu>0$ and (3.6), it follows that as $t$ increases from $0$, one has the limit $A\to 0^{+}$ and $A^{\prime}\to-\infty$. On the other hand, as $t$ decreases from $0$, one has the limit $A\to A_{\ast}^{-}$ and $A^{\prime}\to-6^{-}$ where $A_{\ast}\vcentcolon=(\frac{\mu}{(2-A^{\prime})(4+A^{\prime})^{2}})^{\frac{1}{6}}|_{A^{\prime}=-6}$. Thus, we aim to find the times: $T_{+}>0$ for which $A^{\prime}\to-\infty$, and $T_{-}<0$ for which $A^{\prime}\to-6$. By the numerical simulation, we have

1. fnum@enumiitem (i)(i)

   For the case of increasing $t>0$, the integral (3.19) converges to a positive finite time: $\mathcal{I}_{H}(a)\to\frac{T_{+}}{A_{0}}>0$ as $a=\frac{A}{A_{0}}\to 0$. Thus $(A,A^{\prime})\to(0,-\infty)$ in finite time as $t\to T_{+}$.

2. fnum@enumiitem (ii)(ii)

   For the case of decreasing $t<0$, the integral (3.19) converges to a negative finite time: $\mathcal{I}_{H}(a)\to\frac{T_{-}}{A_{0}}<0$ as $a=\frac{A}{A_{0}}\to\frac{A_{\ast}}{A_{0}}$. Thus $(A,A^{\prime})\to(A_{\ast},-6)$ in finite time as $t\to T_{-}$. In this scenario, we take $(A,A^{\prime})|_{t=T_{-}}=(A_{\ast},-6)$ as an initial data starting in Regime 2(ii) then concatenate the resulting solution with $A(t)|_{T_{-}\leq t\leq 0}$.

   

   Regime 2: $A_{0}^{\prime}\in[-6,-4)$.

   In this case, $1<\beta(A_{0}^{\prime})$ and $A$ solves $A^{\prime}(t)=V_{2}(\frac{A(t)}{A_{0}})-2$ according to (3.14). Moreover, (3.6) implies that $\mu>0$, and $t\mapsto A^{\prime}(t)$ is monotone decreasing since $A>0$ and (3.5). From the fact that $\mu>0$ and (3.6), it follows that as $t$ increases from $0$, one has the limit $A\to A_{\ast}^{+}$ and $A^{\prime}\to-6^{+}$, where $A_{\ast}\vcentcolon=(\frac{\mu}{(2-A^{\prime})(4+A^{\prime})^{2}})^{\frac{1}{6}}|_{A^{\prime}=-6}$. On the other hand, as $t$ decreases from $0$, one has the limit $A\to\infty$ and $A^{\prime}\to-4^{-}$. Thus, we aim to find the times $T_{+}>0$ for which $A^{\prime}\to-6$, and $T_{-}<0$ for which $A^{\prime}\to-4^{-}$. According to the numerical simulation, we have

   1. fnum@enumiiitem (i)(i)

      For the case of increasing $t>0$, the integral (3.15) converges to a positive finite time: $\mathcal{I}_{2}(a)\to\frac{T_{+}}{A_{0}}>0$ as $a=\frac{A}{A_{0}}\to\frac{A_{\ast}}{A_{0}}$. Thus $(A,A^{\prime})\to(A_{\ast},-6)$ in finite time as $t\to T_{+}$. In this scenario, we take $(A,A^{\prime})|_{t=T_{+}}=(A_{\ast},-6)$ as an initial data starting in Regime 1(i) then concatenate the resulting solution with $A(t)|_{0\leq t\leq T_{+}}$.

   2. fnum@enumiiitem (ii)(ii)

      For the case of decreasing $t<0$, it can be verified that $\frac{1}{V_{2}(a)-2}\to-\frac{1}{4}$ as $a\to\infty$. Hence the integral (3.15) diverges to negative infinity: $\mathcal{I}_{2}(a)\to-\infty$ as $a=\frac{A}{A_{0}}\to\infty$, which implies that $T_{-}=-\infty$. Thus $(A,A^{\prime})\to(\infty,-4)$ in infinite time as $t\to-\infty$.

      

      

      Regime 3: $A_{0}^{\prime}\in(-4,0]$.

      In this case, $1<\beta(A_{0}^{\prime})$ and the differential equation is given by $A^{\prime}(t)=V_{1}(\frac{A(t)}{A_{0}})-2$ according to (3.14). Moreover, (3.6) implies that $\mu>0$, and $t\mapsto A^{\prime}(t)$ is monotone increasing since $A>0$ and (3.5). From the fact that $\mu>0$ and (3.6), it follows that as $t$ increases from $0$, one has the limit $A\to A_{\ast}^{+}$ and $A^{\prime}\to 0^{-}$, where $A_{\ast}\vcentcolon=(\frac{\mu}{(2-A^{\prime})(4+A^{\prime})^{2}})^{\frac{1}{6}}|_{A^{\prime}=0}$. On the other hand, as $t$ decreases from $0$, one has the limit $A\to\infty$ and $A^{\prime}\to-4^{+}$. Thus, we aim to find the times $T_{+}>0$ for which $A^{\prime}\to 0$, and $T_{-}<0$ for which $A^{\prime}\to-4$. According to the numerical simulation, we have

      1. fnum@enumiiiitem (i)(i)

         For the case of increasing $t>0$, the integral (3.15) converges to a positive finite time: $\mathcal{I}_{1}(a)\to\frac{T_{+}}{A_{0}}>0$ as $a=\frac{A}{A_{0}}\to\frac{A_{\ast}}{A_{0}}$. Thus $(A,A^{\prime})\to(A_{\ast},0)$ in finite time as $t\to T_{+}$. In this scenario, we take $(A,A^{\prime})|_{t=T_{+}}=(A_{\ast},0)$ as an initial data starting in Regime 4(i) then concatenate the resulting solution with $A(t)|_{0\leq t\leq T_{+}}$.

      2. fnum@enumiiiitem (ii)(ii)

         For the case of decreasing $t<0$, it can be verified that $\frac{1}{V_{1}(a)-2}\to-\frac{1}{4}$ as $a\to\infty$. Hence the integral (3.15) diverges to negative infinity: $\mathcal{I}_{1}(a)\to-\infty$ as $a=\frac{A}{A_{0}}\to\infty$, which implies that $T_{-}=-\infty$. Thus $(A,A^{\prime})\to(\infty,-4)$ as $t\to-\infty$.

         Regime 4: $A_{0}^{\prime}\in[0,2)$.

         In this case, $1<\beta(A_{0}^{\prime})$ and the differential equation is given by $A^{\prime}(t)=V_{0}(\frac{A(t)}{A_{0}})-2$ according to (3.14). Moreover, (3.6) implies that $\mu>0$, and $t\mapsto A^{\prime}(t)$ is monotone increasing since $A>0$ and (3.5). From the fact that $\mu>0$ and (3.6), it follows that as $t$ increases from $0$, one has the limit $A\to\infty$ and $A^{\prime}\to 2^{-}$. On the other hand, as $t$ decreases from $0$, one has the limit $A\to A_{\ast}^{+}$ and $A^{\prime}\to 0^{+}$, where $A_{\ast}\vcentcolon=(\frac{\mu}{(2-A^{\prime})(4+A^{\prime})^{2}})^{\frac{1}{6}}|_{A^{\prime}=0}$. Thus, we aim to find the times $T_{+}>0$ for which $A^{\prime}\to 2^{-}$, and $T_{-}<0$ for which $A^{\prime}\to 0^{+}$. By the numerical simulation, we have

         1. fnum@enumivitem (i)(i)

            For the case of increasing $t>0$, it can be verified that $\frac{1}{V_{0}(a)-2}\to\frac{1}{2}$ as $a\to\infty$. Hence the integral (3.15) diverges to positive infinity: $\mathcal{I}_{0}(a)\to\infty$ as $a=\frac{A}{A_{0}}\to\infty$, which implies that $T_{+}=\infty$. Thus $(A,A^{\prime})\to(\infty,2)$ in infinite time as $t\to\infty$.

         2. fnum@enumivitem (ii)(ii)

            For the case of decreasing $t<0$, the integral (3.15) converges to a negative finite time: $\mathcal{I}_{0}(a)\to\frac{T_{-}}{A_{0}}<0$ as $a=\frac{A}{A_{0}}\to\frac{A_{\ast}}{A_{0}}$. Thus $(A,A^{\prime})\to(A_{\ast},0)$ in finite time as $t\to T_{-}$. In this scenario, we take $(A,A^{\prime})|_{t=T_{-}}=(A_{\ast},0)$ as an initial data starting in Regime 3(ii) then concatenate the resulting solution with $A(t)|_{T_{-}\leq t\leq 0}$.

            

            Regime 5: $A_{0}^{\prime}\in(2,\infty)$.

            In this case, $1>\beta(A_{0}^{\prime})$ and the differential equation is given by $A^{\prime}(t)=W(\frac{A(t)}{A_{0}})-2$. Moreover, (3.6) implies that $\mu<0$, and $t\mapsto A^{\prime}(t)$ is monotone decreasing since $A>0$ and (3.5). From the fact that $\mu<0$ and (3.6), it follows that as $t$ increases from $0$, one has the limit $A\to\infty$ and $A^{\prime}\to 2^{+}$. On the other hand, as $t$ decreases from $0$, one has the limit $A\to 0^{+}$ and $A^{\prime}\to\infty$. Thus, we aim to find the time $T_{+}$ for which $A^{\prime}\to 2^{+}$. From (3.6), we see that $A\to\infty$ as $A^{\prime}\to 2^{+}$, and $T_{-}<0$ for which $A^{\prime}\to\infty$. By the numerical simulation, we have

            1. fnum@enumv(i)(i)

               For the case of increasing $t>0$, it can be verified that $\frac{1}{W(a)-2}\to\frac{1}{2}$ as $a\to\infty$. Hence the integral (3.19) diverges to positive infinity: $\mathcal{I}_{H}(a)\to\infty$ as $a=\frac{A}{A_{0}}\to\infty$, which implies that $T_{+}=\infty$. Thus $(A,A^{\prime})\to(\infty,2)$ in infinite time as $t\to\infty$.

            2. fnum@enumv(ii)(ii)

               For the case of decreasing $t<0$, the integral (3.19) converges to a negative finite time: $\mathcal{I}_{0}(a)\to\frac{T_{-}}{A_{0}}<0$ as $a=\frac{A}{A_{0}}\to 0$. Thus $(A,A^{\prime})\to(0,\infty)$ in finite time as $t\to T_{-}$.

               

               3.3 Free boundary surface and phase space

               The free boundary surface of the flow is determined by (3.3), (3.6), and (3.7). This is stated as:

               $$(2-A^{\prime})(4+A^{\prime})(x^{2}+y^{2})+2(4+A^{\prime})^{2}z^{2}=\dfrac{\mu\nu}{A^{4}},$$

               (3.20)

               with $\mu=A_{0}^{6}(2-A_{0}^{\prime})(4+A_{0}^{\prime})^{2}$ and $\nu=\frac{c_{0}}{(2-A_{0}^{\prime})(4+A_{0}^{\prime})}$. From this, we see that

               • •

                 If $A^{\prime}<-4$ or $A^{\prime}>2$, then the surface is either an one-sheeted or two-sheeted hyperboloid, or a cone;

               • •

                 If $-4<A^{\prime}<2$, then the surface is an ellipsoid.

               For a given initial data $A_{0}=A(0)$ and $A_{0}^{\prime}=A^{\prime}(0)$, the solution to (3.5)–(3.6) exists up to the maximum time of existence $T\neq 0$ and $A(t)\in\mathcal{C}^{2}\big{(}[0,T)\big{)}$ or $A(t)\in\mathcal{C}^{2}\big{(}(T,0]\big{)}$ due to the construction given in the previous section. In what follows, we study the evolution of its corresponding free surface. Once again, due to the time symmetry of the differential equations (3.5)–(3.6) for $(t,A)\leftrightarrow(-t,-A)$, we restrict our analysis to the case $A_{0}>0$. We split the initial data into 9 cases:

               Case I(a): $A_{0}^{\prime}<-4$ and $c_{0}>0$. The initial free boundary is an One-Sheeted Hyperboloid.

               In this case, $\mu>0$ and $\nu<0$. Rewriting (3.20) using (3.6), we have

               $$x^{2}+y^{2}+\dfrac{2(4+A^{\prime})}{2-A^{\prime}}z^{2}=\nu\sqrt[\scriptstyle 3]{\dfrac{\mu(4+A^{\prime})}{2-A^{\prime}}}.$$

               (3.21)

               By the analysis of Regime 1 and Regime 2 in the previous section, we have

               1. fnum@enumvi(i)(i)

                  For increasing $t>0$, there exists a finite time $T_{+}>0$ such that $A(t)\to 0$, $A^{\prime}(t)\to-\infty$ as $t\to T_{+}$. Thus, as $t\to T_{+}$, the free surface converges to an One-Sheeted Hyperboloid given by the following equation:

                  $$x^{2}+y^{2}-2z^{2}=-\nu\sqrt[\scriptstyle 3]{\mu}>0.$$

               2. fnum@enumvi(ii)(ii)

                  For decreasing $t<0$, we have the limit $A(t)\to\infty$, $A^{\prime}(t)\to-4$ as $t\to-\infty$. Thus, as $t\to-\infty$, the free surface collapses to the $z$-axis described by the equation $x^{2}+y^{2}=0$.

                  Case I(b): $A_{0}^{\prime}<-4$ and $c_{0}<0$. The initial free boundary is a Two-Sheeted Hyperboloid.

                  In this case, $\mu>0$ and $\nu>0$. The evolution of surface is described by the same equation as (3.21). According to Regime 1 and Regime 2, we have

                  1. fnum@enumvii()()

                     For increasing $t>0$, there exists a finite time $T_{+}>0$ such that $A(t)\to 0$, $A^{\prime}(t)\to-\infty$ as $t\to T_{+}$. Thus, as $t\to T_{+}$, the free surface converges to a Two-Sheeted Hyperboloid given by the following equation

                     $$x^{2}+y^{2}-2z^{2}=-\nu\sqrt[\scriptstyle 3]{\mu}<0.$$

                  2. fnum@enumvii()()

                     For decreasing $t<0$, we have the limit $A(t)\to\infty$, $A^{\prime}(t)\to-4$ as $t\to-\infty$. Thus, as $t\to-\infty$, the free surface collapses to the $z$-axis described by the equation $x^{2}+y^{2}=0$.

                     Case I(c): $A_{0}^{\prime}<-4$ and $c_{0}=0$. The initial free boundary is a Cone.

                     In this case, $\mu>0$ and $\nu=0$. The evolution of surface is described by the same equation as (3.21). By the analysis of Regime 1 and Regime 2 in the previous section, we have

                     1. fnum@enumviii()()

                        For increasing $t>0$, there exists a finite time $T_{+}>0$ such that $A(t)\to 0$, $A^{\prime}(t)\to-\infty$ as $t\to T_{+}$. Thus, as $t\to T_{+}$, the free surface converges to a Cone with slope $2$ given by the equation

                        $$x^{2}+y^{2}-2z^{2}=0.$$

                     2. fnum@enumviii()()

                        For decreasing $t<0$, we have the limit $A(t)\to\infty$, $A^{\prime}(t)\to-4$ as $t\to-\infty$. Thus, as $t\to-\infty$, the free surface collapses to the $z$-axis described by the equation $x^{2}+y^{2}=0$.

                        Case II: $-4<A_{0}^{\prime}<2$ and $c_{0}\geq 0$. The initial free boundary is an Ellipsoid or a point.

                        The free surface is described by (3.20) and solution $A(t)$ as

                        $$(2-A^{\prime})(4+A^{\prime})(x^{2}+y^{2})+2(4+A^{\prime})^{2}z^{2}=\dfrac{\mu\nu}{A^{4}}.$$

                        If $c_{0}=0$ then (3.20) reduces to the equation $(2-A^{\prime})(x^{2}+y^{2})+2(4+A^{\prime})z^{2}=0$ which describes a point at the origin. In addition, we remark that $c_{0}<0$ is not possible in this case. By the analysis of Regime 3 and Regime 4 in the previous section, we have

                        1. fnum@enumix()()

                           For increasing $t>0$, we have $A(t)\to\infty$, $A^{\prime}(t)\to 2$ as $t\to\infty$. Therefore in the limit $t\to\infty$, the equation (3.20) reduces to $z^{2}=0$, which means that the free surface collapses into the $(x,y)$-Plane.

                        2. fnum@enumix()()

                           For decreasing $t<0$, we have the limit $A(t)\to\infty$, $A^{\prime}(t)\to-4$ as $t\to-\infty$. Thus, as $t\to-\infty$, the free surface collapses to the $z$-Axis described by the equation $x^{2}+y^{2}=0$.

                           Case III(a): $A_{0}^{\prime}>2$ and $c_{0}>0$. The initial free boundary is a Two-Sheeted Hyperboloid.

                           In this case, $\mu<0$ and $\nu<0$. The free surface is described by the same equation as (3.21). By the analysis of Regime 5 in the previous section, we have

                           1. fnum@enumx()()

                              For increasing $t>0$, we have $A(t)\to\infty$, $A^{\prime}(t)\to 2^{+}$ as $t\to\infty$. Taking this limit in (3.20), the equation reduces to $z^{2}=0$ as $t\to\infty$. Thus, the free surface collapses to the (x,y)-Plane in the limit $t\to\infty$.

                           2. fnum@enumx()()

                              For decreasing $t<0$, there exists a finite time $-\infty<T_{-}<0$ such that $A(t)\to 0$, $A^{\prime}(t)\to\infty$ as $t\to T_{-}$. Thus, as $t\to T_{-}$, the free surface converges to a Two-Sheeted Hyperboloid described by the equation:

                              $$x^{2}+y^{2}-2z^{2}=-\nu\sqrt[\scriptstyle 3]{\mu}<0$$

                              Case III(b): $A_{0}^{\prime}>2$ and $c_{0}<0$. The initial free boundary is an One-Sheeted Hyperboloid.

                              In this case, $\mu<0$ and $\nu>0$. The free surface is described by the same equation as (3.21). According to Regime 5 in the previous section, we have

                              1. fnum@enumxi()()

                                 For increasing $t>0$, we have $A(t)\to\infty$, $A^{\prime}(t)\to 2^{+}$ as $t\to\infty$. Taking this limit in (3.20), the equation reduces to $z^{2}=0$ as $t\to\infty$. Thus, the free surface collapses to the (x,y)-Plane in the limit $t\to\infty$.

                              2. fnum@enumxi()()

                                 For decreasing $t<0$, there exists a finite time $-\infty<T_{-}<0$ such that $A(t)\to 0$, $A^{\prime}(t)\to\infty$ as $t\to T_{-}$. Thus, as $t\to T_{-}$, the free surface converges to an One-Sheeted Hyperboloid described by the equation:

                                 $$x^{2}+y^{2}-2z^{2}=-\nu\sqrt[\scriptstyle 3]{\mu}>0$$

                                 Case III(c): $A_{0}^{\prime}>2$ and $c_{0}=0$. The initial free boundary is a Cone.

                                 In this case, $\mu<0$ and $\nu=0$. The free surface is an evolving cone described by the same equation as (3.21). By the analysis in Regime 5, we have

                                 1. fnum@enumxii()()

                                    For increasing $t>0$, we have $A(t)\to\infty$, $A^{\prime}(t)\to 2^{+}$ as $t\to\infty$. Taking this limit in (3.20), the equation reduces to $z^{2}=0$ as $t\to\infty$. Thus, the free surface collapses to the (x,y)-Plane in the limit $t\to\infty$.

                                 2. fnum@enumxii()()

                                    For decreasing $t<0$, there exists a finite time $-\infty<T_{-}<0$ such that $A(t)\to 0$, $A^{\prime}(t)\to\infty$ as $t\to T_{-}$. Thus, as $t\to T_{-}$, the free surface converges to the Cone with slope $2$ described by the equation:

                                    $$x^{2}+y^{2}-2z^{2}=0$$

                                    Case IV: $A_{0}\in\mathbb{R}$ and $A_{0}^{\prime}=-4$. The initial free boundary is the $z$-axis.

                                    For this initial data, $\mu\nu=0$ and the solution is explicitly given by $A(t)=-4t+A_{0}$. The equation (3.20) for the surface remains as the $z$-Axis given by the equation $x^{2}+y^{2}=0$.

                                    Case V: $A_{0}\in\mathbb{R}$ and $A_{0}^{\prime}=2$. The initial free boundary is a Two-Sheeted Plane perpendicular to the $z$-axis.

                                    For this initial data, $\mu=0$ and $\mu\nu=6c_{0}A_{0}^{6}$. The solution is explicitly given by $A(t)=2t+A_{0}$, and the equation (3.20) for the surface is:

                                    $$z^{2}=\dfrac{c_{0}|A(t)|^{6}}{9}=\dfrac{c_{0}}{9}|2t+A_{0}|^{2}.$$

                                    This describes a two-sheeted plane perpendicular to the $z$-axis moving along the $z$-direction with velocities $\pm\frac{c_{0}(2t+A_{0})}{9}$. Note that in this case $c_{0}<0$ is impossible.

                                    

                                    Remark 3.1.

                                    The point $(A_{0}^{\prime},A_{0})$ for which $A_{0}^{\prime}(t)\notin\{-4,2\}$ and $A_{0}=0$ cannot exist as a solution to (3.6). Thus the set $\{(q,p)\in\mathbb{R}^{2}\vcentcolon q\in(-\infty,-4)\cup(-4,2)\cup(2,\infty)\text{ and }p=0\}$ is excluded from the phase space of $(A^{\prime},A)$.

                                    3.4 Time asymptotic of $(A,A^{\prime})$

                                    In this section, we study the time asymptotic behaviour of $A(t)$ and $A^{\prime}(t)$. As before, we restrict our study to the case $A_{0}>0$ since the solution for $A_{0}<0$ can be obtained by the transformation: $(t,A)\mapsto(-t,-A)$.

                                    Time asymptotic for Case I.

                                    The initial data in this case satisfies $A_{0}^{\prime}\in(-\infty,-4)$ and $c_{0}\in\mathbb{R}$. According to Regime 1(i) and Regime 2(i), if we set

                                    $$T_{+}\vcentcolon=\mathcal{I}_{H}(0)=A_{0}\int_{1}^{0}\dfrac{{\mathrm{d}}s}{W(s)-2}>0,$$

                                    then $(A(t),A^{\prime}(t))\to(0,-\infty)$ as $t\nearrow T_{+}$. It follows from (3.19) that

                                    $$A_{0}\int_{0}^{A/A_{0}}\dfrac{{\mathrm{d}}s}{W(s)-2}=t-T_{+}.$$

                                    (3.22)

                                    Since $A_{0}^{\prime}<-4$, (3.12) implies that $\beta(A_{0}^{\prime})<0$. By the expression (3.17), the following series expansion holds:

                                    $$\dfrac{1}{W(s)-2}=\Big{(}\dfrac{|\beta(A_{0}^{\prime})|}{16}\Big{)}^{\frac{1}{3}}s^{2}+\mathcal{O}(s^{4})\quad\text{as }\ s\to 0^{+}.$$

                                    Substituting the above into (3.22), it follows that as $A\to 0^{+}$,

                                    $$t-T_{+}=A_{0}\int_{0}^{A/A_{0}}\!\!\Big{\{}\Big{(}\dfrac{|\beta(A_{0}^{\prime})|}{16}\Big{)}^{\frac{1}{3}}s^{2}+\mathcal{O}(s^{4})\Big{\}}\,{\mathrm{d}}s=\dfrac{1}{3A_{0}^{2}}\Big{(}\dfrac{|\beta(A_{0}^{\prime})|}{16}\Big{)}^{\frac{1}{3}}A^{3}+\mathcal{O}(A^{5}).$$

                                    Therefore we obtain the convergence rate

                                    $$A(t)=\mathcal{O}\big{(}\sqrt[\scriptstyle 3]{T_{+}-t}\big{)}\qquad\text{as }\ t\nearrow T_{+}.$$

                                    Combining the above convergence rate with (3.17), we conclude that

                                    $$A^{\prime}(t)=W\big{(}\dfrac{A(t)}{A_{0}}\big{)}-2=\mathcal{O}\Big{(}(T_{+}-t)^{-2/3}\Big{)}\qquad\text{as }\ t\nearrow T_{+}.$$

                                    Next, we consider the asymptotic of $(A,A^{\prime})$ in the limit $t\to-\infty$. According to the analysis given in Regime 1(ii) and Regime 2(ii), as $t\to-\infty$, $A(t)$ takes the form:

                                    	$$\displaystyle A(t)=A_{0}\mathcal{I}_{2}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)},\quad\text{ with }\ \mathcal{I}_{2}(x)\vcentcolon=\int_{1}^{x}\dfrac{{\mathrm{d}}s}{V_{2}(s)-2},$$
                                    	$$\displaystyle\text{where }\ V_{2}(s)=4\cos\Big{\{}\dfrac{1}{3}\arccos\Big{(}1-\dfrac{2}{s^{6}\beta(A_{0}^{\prime})}\Big{)}-\dfrac{4\pi}{3}\Big{\}}.$$		(3.23)

                                    It can be verified that $\lim_{s\to\infty}V_{2}(s)=-2$. This implies $\lim_{x\to\infty}\mathcal{I}_{2}(x)=-\infty$. Hence its inverse satisfies $\lim_{t\to-\infty}\mathcal{I}_{2}^{-1}(t)=\infty$. In addition, there exists $N\in\mathbb{N}$ such that for all $x\geq N$, $-5\leq V_{2}(x)-2\leq-3$, hence

                                    $$|\mathcal{I}_{2}(x)|\geq\int_{N}^{x}\dfrac{{\mathrm{d}}s}{|V_{2}(s)-2|}-|\mathcal{I}_{2}(N)|\geq\dfrac{x}{5}-\dfrac{N}{5}-|\mathcal{I}_{2}(N)|,\quad\text{for all }\ x\geq N.$$

                                    Thus we have $\liminf_{x\to\infty}\frac{|\mathcal{I}_{2}(x)|}{x}\geq\frac{1}{5}$. For inverse function $\mathcal{I}_{2}^{-1}(t)$, we have that

                                    $$\limsup\limits_{t\to-\infty}\dfrac{|\mathcal{I}_{2}^{-1}(t)|}{|t|}=\limsup\limits_{x\to\infty}\dfrac{x}{|\mathcal{I}_{2}(x)|}=\Big{(}\liminf\limits_{x\to\infty}\dfrac{|\mathcal{I}_{2}(x)|}{x}\Big{)}^{-1}\leq 5.$$

                                    Therefore we conclude

                                    $$A(t)=A_{0}\mathcal{I}_{2}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)}=\mathcal{O}(t)\qquad\text{as }\ t\to-\infty.$$

                                    Since $\lim_{x\to\infty}\mathcal{I}_{2}^{\prime}(x)=-\frac{1}{4}$, by inverse function theorem,

                                    $$\lim\limits_{t\to-\infty}A^{\prime}(t)=\lim\limits_{t\to-\infty}\dfrac{1}{\mathcal{I}_{2}^{\prime}\big{(}\mathcal{I}_{2}^{-1}(t/A_{0})\big{)}}=-4,\ \text{ which implies }\ A^{\prime}(t)+4=o(1).$$

                                    The following series expansion holds:

                                    	$\displaystyle\arccos(1-x^{-6})=\sqrt{2}x^{-3}+\dfrac{1}{6\sqrt{2}}x^{-9}+\mathcal{O}(x^{-15})$	$\displaystyle\text{as }\ x\to\infty,$
                                    	$\displaystyle 4\cos\big{(}\dfrac{x}{3}-\dfrac{4\pi}{3}\big{)}+2=-\dfrac{2}{\sqrt{3}}x+\dfrac{1}{9}x^{2}+\mathcal{O}(x^{3})$	$\displaystyle\text{as }\ x\to 0.$

                                    Using these in the expression (3.23), we obtain the series expansion:

                                    $$V_{2}(s)+2=-\Big{(}\dfrac{16}{3\beta(A_{0}^{\prime})}\Big{)}^{\frac{1}{2}}s^{-3}+\dfrac{4}{9\beta(A_{0}^{\prime})}s^{-6}+\mathcal{O}(s^{-9})=\mathcal{O}(s^{-3})\qquad\text{as }\ s\to\infty.$$

                                    Applying the fact that $\mathcal{I}_{2}^{-1}(t)=\mathcal{O}(t)$ as $t\to-\infty$, we conclude that

                                    $\displaystyle A^{\prime}(t)+4=V_{2}\big{(}\mathcal{I}_{2}^{-1}(\tfrac{t}{A_{0}})\big{)}+2=\mathcal{O}\big{(}|\mathcal{I}_{2}^{-1}(\tfrac{t}{A_{0}})|^{-3}\big{)}=\mathcal{O}(t^{-3})\qquad\text{as }\ t\to-\infty.$

                                    In summary if $A_{0}^{\prime}\in(-\infty,-4)$ and $c_{0}\in\mathbb{R}$, then the corresponding solution $A(t)$ satisfies the following asymptotic behaviour: there exists $T_{+}=T_{+}(A_{0},A_{0}^{\prime})\in(0,\infty)$ such that

                                    	$\displaystyle A(t)=\mathcal{O}\big{(}\sqrt[\scriptstyle 3]{T_{+}-t}\big{)},$	$\displaystyle\quad\text{and}\quad A^{\prime}(t)=\mathcal{O}\Big{(}(T_{+}-t)^{-2/3}\Big{)}\qquad$	$\displaystyle\text{as }\ t\nearrow T_{+},$
                                    	$\displaystyle A(t)=\mathcal{O}(t),$	$\displaystyle\quad\text{and}\quad A^{\prime}(t)=-4+\mathcal{O}(t^{-3})$	$\displaystyle\text{as }\ t\to-\infty.$

                                    Time asymptotic for Case II.

                                    The initial data in this case satisfies $A_{0}^{\prime}\in(-4,2)$ and $c_{0}\geq 0$. According to the analysis given in Regime 3(i) and Regime 4(i), in the limit $t\to\infty$, the solution $A(t)$ is takes the form:

                                    	$$\displaystyle A(t)=A_{0}\mathcal{I}_{0}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)},\quad\text{ with }\ \mathcal{I}_{0}(x)\vcentcolon=\int_{1}^{x}\dfrac{{\mathrm{d}}s}{V_{0}(s)-2},$$
                                    	$$\displaystyle\text{where }\ V_{0}(s)=4\cos\Big{\{}\dfrac{1}{3}\arccos\Big{(}1-\dfrac{2}{s^{6}\beta(A_{0}^{\prime})}\Big{)}\Big{\}}.$$		(3.24)

                                    It can be verified that $\lim_{s\to\infty}V_{0}(s)=4$. This implies $\lim_{x\to\infty}\mathcal{I}_{0}(x)=\infty$. Hence its inverse satisfies $\lim_{t\to\infty}\mathcal{I}_{0}^{-1}(t)=\infty$. In addition, there exists $N\in\mathbb{N}$ such that for all $x\geq N$, $1\leq V_{0}(x)-2\leq 3$, hence

                                    $$|\mathcal{I}_{0}(x)|\geq\int_{N}^{x}\dfrac{{\mathrm{d}}s}{|V_{0}(s)-2|}-|\mathcal{I}_{0}(N)|\geq\dfrac{x}{3}-\dfrac{N}{3}-|\mathcal{I}_{0}(N)|,\quad\text{for all }\ x\geq N.$$

                                    Thus we have $\liminf_{x\to\infty}\frac{|\mathcal{I}_{0}(x)|}{x}\geq\frac{1}{3}$. For inverse function $\mathcal{I}_{0}^{-1}(t)$, we have that

                                    $$\limsup\limits_{t\to\infty}\dfrac{|\mathcal{I}_{0}^{-1}(t)|}{t}=\limsup\limits_{x\to\infty}\dfrac{x}{|\mathcal{I}_{0}(x)|}=\Big{(}\liminf\limits_{x\to\infty}\dfrac{|\mathcal{I}_{0}(x)|}{x}\Big{)}^{-1}\leq 3.$$

                                    Therefore we conclude

                                    $$A(t)=A_{0}\mathcal{I}_{0}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)}=\mathcal{O}(t)\qquad\text{as }\ t\to\infty.$$

                                    Since $\lim_{x\to\infty}\mathcal{I}_{0}^{\prime}(x)=\frac{1}{2}$, by inverse function theorem,

                                    $$\lim\limits_{t\to\infty}A^{\prime}(t)=\lim\limits_{t\to\infty}\dfrac{1}{\mathcal{I}_{0}^{\prime}\big{(}\mathcal{I}_{0}^{-1}(t/A_{0})\big{)}}=2,\ \text{ which implies }\ A^{\prime}(t)-2=o(1).$$

                                    The following series expansion holds:

                                    	$\displaystyle\arccos(1-x^{-6})=\sqrt{2}x^{-3}+\dfrac{1}{6\sqrt{2}}x^{-9}+\mathcal{O}(x^{-15})$	$\displaystyle\text{as }\ x\to\infty,$
                                    	$\displaystyle 4\cos\big{(}\dfrac{x}{3}\big{)}-4=-\dfrac{2}{9}x^{2}+\dfrac{1}{486}x^{4}+\mathcal{O}(x^{5})$	$\displaystyle\text{as }\ x\to 0.$

                                    Using these in the expression (3.24), we obtain the series expansion:

                                    $$V_{0}(s)-4=-\dfrac{8}{9\beta(A_{0}^{\prime})}s^{-6}+\dfrac{8}{243\beta(A_{0}^{\prime})}s^{-12}+\mathcal{O}(s^{-15})=\mathcal{O}(s^{-6})\qquad\text{as }\ s\to\infty.$$

                                    Applying the fact that $\mathcal{I}_{0}^{-1}(t)=\mathcal{O}(t)$ as $t\to\infty$, we conclude that

                                    $\displaystyle A^{\prime}(t)-2=V_{0}\big{(}\mathcal{I}_{0}^{-1}(\tfrac{t}{A_{0}})\big{)}-4=\mathcal{O}\big{(}|\mathcal{I}_{0}^{-1}(\tfrac{t}{A_{0}})|^{-6}\big{)}=\mathcal{O}(t^{-6})\qquad\text{as }\ t\to\infty.$

                                    Next we consider the asymptotic behaviour of $A(t)$, $A^{\prime}(t)$ in the limit $t\to-\infty$. In this case, according to Regime 3(ii) and Regime 4(ii), the solution $A(t)$ is given by

                                    	$$\displaystyle A(t)=A_{0}\mathcal{I}_{1}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)},\quad\text{ with }\ \mathcal{I}_{1}(x)\vcentcolon=\int_{1}^{x}\dfrac{{\mathrm{d}}s}{V_{1}(s)-2},$$
                                    	$$\displaystyle\text{where }\ V_{1}(s)=4\cos\Big{\{}\dfrac{1}{3}\arccos\Big{(}1-\dfrac{2}{s^{6}\beta(A_{0}^{\prime})}\Big{)}-\dfrac{2\pi}{3}\Big{\}}.$$		(3.25)

                                    It can be verified that $\lim_{s\to\infty}V_{1}(s)=-2$. This implies $\lim_{x\to\infty}\mathcal{I}_{1}(x)=-\infty$. Hence its inverse satisfies $\lim_{t\to-\infty}\mathcal{I}_{1}^{-1}(t)=\infty$. In addition, there exists $N\in\mathbb{N}$ such that for all $x\geq N$, $-5\leq V_{0}(x)-2\leq-3$, hence

                                    $$|\mathcal{I}_{1}(x)|\geq\int_{N}^{x}\dfrac{{\mathrm{d}}s}{|V_{1}(s)-2|}-|\mathcal{I}_{1}(N)|\geq\dfrac{x}{5}-\dfrac{N}{5}-|\mathcal{I}_{1}(N)|,\quad\text{for all }\ x\geq N.$$

                                    Thus we have $\liminf_{x\to\infty}\frac{|\mathcal{I}_{1}(x)|}{x}\geq\frac{1}{5}$. For inverse function $\mathcal{I}_{1}^{-1}(t)$, we have that

                                    $$\limsup\limits_{t\to-\infty}\dfrac{|\mathcal{I}_{1}^{-1}(t)|}{|t|}=\limsup\limits_{x\to\infty}\dfrac{x}{|\mathcal{I}_{1}(x)|}=\Big{(}\liminf\limits_{x\to\infty}\dfrac{|\mathcal{I}_{1}(x)|}{x}\Big{)}^{-1}\leq 5.$$

                                    Therefore we conclude

                                    $$A(t)=A_{0}\mathcal{I}_{1}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)}=\mathcal{O}(t)\qquad\text{as }\ t\to-\infty.$$

                                    Since $\lim_{x\to\infty}\mathcal{I}_{1}^{\prime}(x)=-\frac{1}{4}$, by inverse function theorem,

                                    $$\lim\limits_{t\to-\infty}A^{\prime}(t)=\lim\limits_{t\to-\infty}\dfrac{1}{\mathcal{I}_{1}^{\prime}\big{(}\mathcal{I}_{1}^{-1}(t/A_{0})\big{)}}=-4,\ \text{ which implies }\ A^{\prime}(t)+4=o(1).$$

                                    The following series expansion holds:

                                    	$\displaystyle\arccos(1-x^{-6})=\sqrt{2}x^{-3}+\dfrac{1}{6\sqrt{2}}x^{-9}+\mathcal{O}(x^{-15})$	$\displaystyle\text{as }\ x\to\infty,$
                                    	$\displaystyle 4\cos\big{(}\dfrac{x}{3}-\dfrac{2\pi}{3}\big{)}+2=\dfrac{2}{\sqrt{3}}x+\dfrac{1}{9}x^{2}+\mathcal{O}(x^{3})$	$\displaystyle\text{as }\ x\to 0.$

                                    Using these in the expression (3.25), we obtain the series expansion:

                                    $$V_{1}(s)+2=\Big{(}\dfrac{16}{3\beta(A_{0}^{\prime})}\Big{)}^{\frac{1}{2}}s^{-3}+\dfrac{4}{9\beta(A_{0}^{\prime})}s^{-6}+\mathcal{O}(s^{-9})=\mathcal{O}(s^{-3})\qquad\text{as }\ s\to\infty.$$

                                    Applying the fact that $\mathcal{I}_{1}^{-1}(t)=\mathcal{O}(t)$ as $t\to-\infty$, we conclude that

                                    $\displaystyle A^{\prime}(t)+4=V_{1}\big{(}\mathcal{I}_{1}^{-1}(\tfrac{t}{A_{0}})\big{)}+2=\mathcal{O}\big{(}|\mathcal{I}_{1}^{-1}(\tfrac{t}{A_{0}})|^{-3}\big{)}=\mathcal{O}(t^{-3})\qquad\text{as }\ t\to-\infty.$

                                    In summary, if $A_{0}^{\prime}\in(-4,2)$ and $c_{0}\geq 0$, then the corresponding solution $A(t)$ satisfies the following time asymptotic behaviour:

                                    	$\displaystyle A(t)=\mathcal{O}(t),\quad\text{and}\quad A^{\prime}(t)=2+\mathcal{O}(t^{-6})$	$\displaystyle\text{as }\ t\to\infty,$
                                    	$\displaystyle A(t)=\mathcal{O}(t),\quad\text{and}\quad A^{\prime}(t)=-4+\mathcal{O}(t^{-3})$	$\displaystyle\text{as }\ t\to-\infty.$

                                    Time asymptotic for Case III.

                                    The initial data in this case satisfies $A_{0}^{\prime}\in(2,\infty)$ and $c_{0}\in\mathbb{R}$. According to Regime 5(i), as $t\to\infty$, the solution $A(t)$ takes the form:

                                    $$A(t)=A_{0}\mathcal{I}_{H}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)},\quad\text{ with }\ \mathcal{I}_{H}(x)\vcentcolon=\int_{1}^{x}\dfrac{{\mathrm{d}}s}{W(s)-2},$$

                                    where $W(s)$ is defined in (3.17). It can be verified that $\lim_{s\to\infty}W(s)=4$. This implies $\lim_{x\to\infty}\mathcal{I}_{H}(x)=\infty$. Hence its inverse satisfies $\lim_{t\to\infty}\mathcal{I}_{H}^{-1}(t)=\infty$. In addition, there exists $N\in\mathbb{N}$ such that for all $x\geq N$, $1\leq W(x)-2\leq 3$, hence

                                    $$|\mathcal{I}_{H}(x)|\geq\int_{N}^{x}\dfrac{{\mathrm{d}}s}{|W(s)-2|}-|\mathcal{I}_{H}(N)|\geq\dfrac{x}{3}-\dfrac{N}{3}-|\mathcal{I}_{H}(N)|,\quad\text{for all }\ x\geq N.$$

                                    Thus we have $\liminf_{x\to\infty}\frac{|\mathcal{I}_{H}(x)|}{x}\geq\frac{1}{3}$. For inverse function $\mathcal{I}_{H}^{-1}(t)$, we have that

                                    $$\limsup\limits_{t\to\infty}\dfrac{|\mathcal{I}_{H}^{-1}(t)|}{t}=\limsup\limits_{x\to\infty}\dfrac{x}{|\mathcal{I}_{H}(x)|}=\Big{(}\liminf\limits_{x\to\infty}\dfrac{|\mathcal{I}_{H}(x)|}{x}\Big{)}^{-1}\leq 3.$$

                                    Therefore we conclude

                                    $$A(t)=A_{0}\mathcal{I}_{H}^{-1}\big{(}\dfrac{t}{A_{0}}\big{)}=\mathcal{O}(t)\qquad\text{as }\ t\to\infty.$$

                                    Since $\lim_{x\to\infty}\mathcal{I}_{H}^{\prime}(x)=\frac{1}{2}$, by inverse function theorem,

                                    $$\lim\limits_{t\to\infty}A^{\prime}(t)=\lim\limits_{t\to\infty}\dfrac{1}{\mathcal{I}_{H}^{\prime}\big{(}\mathcal{I}_{H}^{-1}(t/A_{0})\big{)}}=2,\ \text{ which implies }\ A^{\prime}(t)-2=o(1).$$

                                    From (3.17), one has the following series expansion:

                                    $$W(s)-4=-\dfrac{8(3|\beta(A_{0}^{\prime})|^{2}-2)}{9(\beta(A_{0}^{\prime}))^{3}}s^{-6}+\mathcal{O}(s^{-9})=\mathcal{O}(s^{-6})\qquad\text{as }\ s\to\infty.$$

                                    Applying the fact that $\mathcal{I}_{H}^{-1}(t)=\mathcal{O}(t)$ as $t\to\infty$, we conclude that

                                    $\displaystyle A^{\prime}(t)-2=W\big{(}\mathcal{I}_{H}^{-1}(\tfrac{t}{A_{0}})\big{)}-4=\mathcal{O}\big{(}|\mathcal{I}_{H}^{-1}(\tfrac{t}{A_{0}})|^{-6}\big{)}=\mathcal{O}(t^{-6})\qquad\text{as }\ t\to\infty.$

                                    Next, we consider the asymptotic of $(A,A^{\prime})$ as $t$ decreases. According to the analysis given in Regime 5(ii), if we set

                                    $$T_{-}\vcentcolon=\mathcal{I}_{H}(0)=A_{0}\int_{1}^{0}\dfrac{{\mathrm{d}}s}{W(s)-2}<0,$$

                                    then $(A(t),A^{\prime}(t))\to(0,\infty)$ as $t\searrow T_{-}$. It follows from (3.19) that

                                    $$A_{0}\int_{0}^{A/A_{0}}\dfrac{{\mathrm{d}}s}{W(s)-2}=t-T_{-}.$$

                                    (3.26)

                                    Since $A_{0}^{\prime}>2$, (3.12) implies that $\beta(A_{0}^{\prime})>0$. By the expression (3.17), the following series expansion holds:

                                    $$\dfrac{1}{W(s)-2}=-\Big{(}\dfrac{|\beta(A_{0}^{\prime})|}{16}\Big{)}^{\frac{1}{3}}s^{2}+\mathcal{O}(s^{4})\quad\text{as }\ s\to 0^{+}.$$

                                    Substituting the above into (3.26), it follows that as $A\to 0^{+}$,

                                    $$t-T_{-}=A_{0}\int_{0}^{A/A_{0}}\!\!\Big{\{}-\Big{(}\dfrac{|\beta(A_{0}^{\prime})|}{16}\Big{)}^{\frac{1}{3}}s^{2}+\mathcal{O}(s^{4})\Big{\}}\,{\mathrm{d}}s=-\dfrac{1}{3A_{0}^{2}}\Big{(}\dfrac{|\beta(A_{0}^{\prime})|}{16}\Big{)}^{\frac{1}{3}}A^{3}+\mathcal{O}(A^{5}).$$

                                    Therefore we obtain the convergence rate

                                    $$A(t)=\mathcal{O}\big{(}\sqrt[\scriptstyle 3]{t-T_{-}}\big{)}\qquad\text{as }\ t\searrow T_{-}.$$

                                    Combining the above convergence rate with (3.17), we conclude that

                                    $$A^{\prime}(t)=W\big{(}\dfrac{A(t)}{A_{0}}\big{)}-2=\mathcal{O}\Big{(}(t-T_{-})^{-2/3}\Big{)}\qquad\text{as }\ t\searrow T_{-}.$$

                                    In summary if $A_{0}^{\prime}\in(2,\infty)$ and $c_{0}\in\mathbb{R}$, then the corresponding solution $A(t)$ satisfies the following asymptotic behaviour: there exists $T_{-}=T_{-}(A_{0},A_{0}^{\prime})\in(-\infty,0)$ such that

                                    	$\displaystyle A(t)=\mathcal{O}(t),$	$\displaystyle\quad\text{and}\quad A^{\prime}(t)=2+\mathcal{O}(t^{-6})\qquad$	$\displaystyle\text{as }\ t\to\infty,$
                                    	$\displaystyle A(t)=\mathcal{O}\big{(}\sqrt[\scriptstyle 3]{t-T_{-}}\big{)},$	$\displaystyle\quad\text{and}\quad A^{\prime}(t)=\mathcal{O}\Big{(}(t-T_{-})^{-2/3}\Big{)}\qquad$	$\displaystyle\text{as }\ t\searrow T_{-}.$

                                    4 Flow of Parabola with Drift

                                    Suppose that $A(t)$ is a solution to (3.5)–(3.6). We consider the following ansatz:

                                    $$\phi(t,x,y,z)=\dfrac{x^{2}+y^{2}-2z^{2}}{A(t)}+B(t)z$$

                                    (4.1)

                                    For each point $\vec{x}\in\partial\Omega(t)$ at time $t$, let $\vec{X}(s;t,\vec{x})=(X^{x},X^{y},X^{z})(s;t,\vec{x})$ be the backward characteristic curve emanating from $(t,\vec{x})$, which satisfies $\vec{X}(t;t,x)=\vec{x}$ and the differential equations

                                    $$\dfrac{\partial X^{x}}{\partial s}=\dfrac{2X^{x}}{A(s)},\qquad\dfrac{\partial X^{y}}{\partial s}=\dfrac{2X^{y}}{A(s)},\qquad\dfrac{\partial X^{z}}{\partial s}=B(s)-\dfrac{4X^{z}}{A(s)}.$$

                                    For simplicity denote $X^{\sigma}_{0}\equiv X^{\sigma}(0;t,\vec{x})$ for $\sigma=x,\,y,\,z$. Solving these, one has

                                    	$$\displaystyle X^{x}_{0}=xe^{-g(t)},\quad X^{y}_{0}=ye^{-g(t)},\ \quad X^{z}_{0}=ze^{2g(t)}-h(t),$$		(4.2)
                                    	$$\displaystyle\text{where }\quad h(t)\vcentcolon=\int_{0}^{t}\!\!B(s)e^{2g(s)}\,{\mathrm{d}}s,\quad\text{ and }\quad g(t)=\int_{0}^{t}\!\!\dfrac{2}{A(s)}\,{\mathrm{d}}s.$$

                                    Therefore, the equation describing the free surface must satisfies the form:

                                    $$f\big{(}X^{x}_{0},X^{y}_{0},X^{z}_{0}\big{)}=f\big{(}xe^{-g(t)},ye^{-g(t)},ze^{2g(t)}-h(t)\big{)}=0,$$

                                    (4.3)

                                    for some function $f(\cdot,\cdot,\cdot)$ independent of $(t,x,y,z)$. On the other hand, substituting (4.1) into (1.6c), we have

                                    $$\Big{\{}\dfrac{2-A^{\prime}}{A^{2}}(x^{2}+y^{2})+\dfrac{2(4+A^{\prime})}{A^{2}}z^{2}+\big{(}B^{\prime}-\dfrac{4B}{A}\big{)}z\Big{\}}\Big{|}_{F(t,\vec{x})=0}=c(t)-\dfrac{B^{2}}{2}.$$

                                    (4.4)

                                    Rewriting the above in terms of $\big{(}X^{x}_{0},X^{y}_{0},X^{z}_{0}\big{)}$ given in (4.2), one obtains

                                    	$\displaystyle 0=$	$\displaystyle\dfrac{2-A^{\prime}}{A^{2}}e^{2g}|X^{x}_{0}|^{2}+\dfrac{2-A^{\prime}}{A^{2}}e^{2g}|X^{y}_{0}|^{2}+\dfrac{2(4+A^{\prime})}{A^{2}}e^{-4g}|X^{z}_{0}|^{2}$
                                    		$\displaystyle+e^{-2g}\Big{(}\dfrac{4(4+A^{\prime})}{A^{2}}he^{-2g}+B^{\prime}-\dfrac{4B}{A}\Big{)}X^{z}_{0}$
                                    		$\displaystyle+he^{-2g}\Big{(}B^{\prime}-\dfrac{4B}{A}+\dfrac{2(4+A^{\prime})}{A^{2}}he^{-2g}\Big{)}+\dfrac{B^{2}}{2}-c(t).$

                                    Since the above equation must be consistent with (4.3), it follows that there exists constants $c_{1}$, $c_{2}$, $c_{3}\in\mathbb{R}$ such that

                                    	$$\displaystyle\dfrac{2-A^{\prime}}{A^{2}}e^{2g}=c_{1}\dfrac{2(4+A^{\prime})}{A^{2}}e^{-4g},$$		(4.5a)
                                    	$$\displaystyle\dfrac{2-A^{\prime}}{A^{2}}e^{2g}=c_{2}e^{-2g}\Big{(}\dfrac{4(4+A^{\prime})}{A^{2}}he^{-2g}+B^{\prime}-\dfrac{4B}{A}\Big{)},$$		(4.5b)
                                    	$$\displaystyle\dfrac{2-A^{\prime}}{A^{2}}e^{2g}=c_{3}\Big{\{}\big{(}B^{\prime}-\dfrac{4B}{A}\big{)}he^{-2g}+\dfrac{2(4+A^{\prime})}{A^{2}}h^{2}e^{-4g}+\dfrac{B^{2}}{2}-c(t)\Big{\}}.$$		(4.5c)

                                    From (4.5a), we have the identity $\frac{1}{2c_{1}}=\frac{4+A^{\prime}}{2-A^{\prime}}e^{-6g}$. Substituting this into (4.5b),

                                    $\displaystyle 1=\dfrac{2c_{2}}{c_{1}}h+c_{2}\Big{(}B^{\prime}A-4B\Big{)}\dfrac{A}{2-A^{\prime}}e^{-4g}\ \iff\ h=\dfrac{c_{1}}{2}\Big{\{}\dfrac{1}{c_{2}}+\dfrac{4AB-B^{\prime}A^{2}}{2-A^{\prime}}e^{-4g}\Big{\}}.$

                                    Taking derivative on the above equation, and using the definition $h(t)=\int_{0}^{t}\!\!B(s)e^{2g(s)}\,{\mathrm{d}}s$, we obtain

                                    $$0=\dfrac{2B}{c_{1}}+\Big{\{}B^{\prime\prime}A+4(A^{\prime}-1)B^{\prime}-\dfrac{12A^{\prime}}{A}B\Big{\}}\dfrac{A}{2-A^{\prime}}e^{-6g}.$$

                                    Using the identity $\frac{1}{2c_{1}}=\frac{4+A^{\prime}}{2-A^{\prime}}e^{-6g}$ on the above equation once again, then rearranging terms, we have

                                    $$B^{\prime\prime}+\dfrac{4(A^{\prime}-1)}{A}B^{\prime}+\dfrac{8(2-A^{\prime})}{A^{2}}B=0.$$

                                    (4.6)

                                    Assuming further that $A(t)=-4t+4a_{0}$ where $a_{0}=\frac{A_{0}}{4}\neq 0$, then one has the equation

                                    	$$\displaystyle B^{\prime\prime}+\dfrac{5B^{\prime}}{t-a_{0}}+\dfrac{3B}{(t-a_{0})^{2}}=0,\quad\text{ which implies }\quad B(t)=\dfrac{k_{1}}{t-a_{0}}+\dfrac{k_{2}}{(t-a_{0})^{3}},$$
                                    	$$\displaystyle\text{where }\quad k_{1}\vcentcolon=\frac{a_{0}}{2}(a_{0}B_{0}^{\prime}-3B_{0}),\quad k_{2}\vcentcolon=\frac{a_{0}^{3}}{2}(B_{0}-a_{0}B_{0}^{\prime}),\quad(B_{0},B_{0}^{\prime})\vcentcolon=(B(0),B^{\prime}(0)).$$

                                    In addition, using the expression (4.5c), one can calculate to get that c(t)-B22 = ( c_0 - B022 )a03(a0-t)3. where $c_{0}\equiv c(0)$. Substituting these expression of $A$ and $B$ into (4.4), we conclude that the free surface is a Paraboloid described by the equation

                                    $$z=\dfrac{3(t-a_{0})^{2}}{16k_{2}}(x^{2}+y^{2})+\dfrac{a_{0}^{3}}{2k_{2}}\Big{(}c_{0}-\dfrac{B_{0}^{2}}{2}\Big{)}(t-a_{0}).$$

                                    (4.7)

                                    As $|t|\to\infty$, this paraboloid will collapse to $z$-Axis described by $x^{2}+y^{2}=0$.

                                    5 Flow of Ellipsoid with Constant Vorticity

                                    In this section we construct an explicit solution to (1.6b)–(1.6c) with vorticity, i.e. when the condition (1.6a) is not satisfied. More precisely, the solution has constant vorticity $\textnormal{{curl}}\vec{\mathrm{v}}=-2\varepsilon\hat{z}$ for some constant $\varepsilon>0$. We remark that such solution was first studied by Ovsjannikov in [Ovsyannikov].

                                    Let $\vec{X}(t)=(x(t),y(t),z(t))$ be the position vector of the free surface at time $t\geq 0$, and denote $\vec{X}_{0}\vcentcolon=\vec{X}(0)=\vcentcolon(x_{0},y_{0},z_{0})$. We impose the following ansatz:

                                    $$\vec{X}(t)=A(t)\cdot\vec{X}_{0}\qquad\text{where}\qquad A(t)\vcentcolon=\begin{pmatrix}\alpha&\beta&0\\ -\beta&\alpha&0\\ 0&0&\gamma\end{pmatrix}.$$

                                    (5.1)

                                    Here $\alpha(t),\,\beta(t),\,\gamma(t)$ are functions of $t$ such that $A(0)=I$ with $I$ being the $3$-by-$3$ identity matrix. Let $P=P(t,\vec{X})$ be the pressure. By chain rule one has,

                                    $$\nabla_{\vec{X}_{0}}\big{(}P(t,\vec{X})\big{)}=\nabla_{\vec{X}_{0}}\big{(}P(t,A\cdot\vec{X}_{0})\big{)}=A^{\top}\cdot\nabla_{\vec{X}}P\big{|}_{\vec{X}=A\vec{X}_{0}}.$$

                                    Then Newton’s second law yields

                                    $$\vec{X}^{\prime\prime}(t)=-\nabla_{\vec{X}}P\big{|}_{\vec{X}=A\vec{X}_{0}}=-(A^{\top})^{-1}\cdot\nabla_{\vec{X}_{0}}\big{(}P(t,\vec{X})\big{)}.$$

                                    Substituting the ansatz $\vec{X}(t)=A(t)\vec{X}_{0}$ into the above equation, we obtain that,

                                    $$A^{\top}(t)\cdot A^{\prime\prime}(t)\cdot\vec{X}_{0}=-\nabla_{\vec{X}_{0}}\big{(}P(t,\vec{X})\big{)}.$$

                                    Integrating in $\vec{X}_{0}$ and noting that on the free surface, pressure is constant for given time $t$, we get that

                                    $$\frac{1}{2}\vec{X}_{0}^{\top}\cdot A^{\top}(t)\cdot A^{\prime\prime}(t)\cdot\vec{X}_{0}=c(t).$$

                                    In addition, we also impose the following similarity relations:

                                    $$A^{\top}A^{\prime\prime}=a(t)I,\qquad c(t)=a(t)\dfrac{|\vec{X}_{0}|^{2}}{2},$$

                                    (5.2)

                                    Our aim is then to find $\alpha,\beta,\gamma,$ and $a$ that solve the problem with initial conditions:

                                    $$\alpha(0)=1,\quad\beta(0)=0,\quad\gamma(0)=1,\quad a(0)=a_{0}.$$

                                    (5.3)

                                    Since $\vec{X}(t)$ is also the particle trajectory, if we set $\vec{\mathrm{v}}(t,\vec{X})$ as the velocity, then

                                    $$\vec{\mathrm{v}}\big{(}t,\vec{X}(t)\big{)}=\dfrac{{\mathrm{d}}\vec{X}}{{\mathrm{d}}t}(t)=A^{\prime}(t)\vec{X}_{0}=A^{\prime}A^{-1}\vec{X}(t).$$

                                    From the incompressibility $\textnormal{{div}}\,\vec{\mathrm{v}}=0$, and Jacobi’s formula for determinant, one has

                                    $$0=\textnormal{{div}}_{\vec{X}}\big{(}\vec{\mathrm{v}}(t,\vec{X})\big{)}=\textrm{tr}(A^{\prime}A^{-1})=\dfrac{{\mathrm{d}}}{{\mathrm{d}}t}\ln(\det A),$$

                                    It follows that $\det A(t)=\det A(0)=1$, hence by (5.1),

                                    $$\alpha^{2}+\beta^{2}=\frac{1}{\gamma}.$$

                                    (5.4)

                                    Next, by a direct computation one has

                                    $$A^{\top}A^{\prime\prime}=\begin{pmatrix}\alpha&-\beta&0\\ \beta&\alpha&0\\ 0&0&\gamma\end{pmatrix}\begin{pmatrix}\alpha^{\prime\prime}&\beta^{\prime\prime}&0\\ -\beta^{\prime\prime}&\alpha^{\prime\prime}&0\\ 0&0&\gamma^{\prime\prime}\end{pmatrix}=\begin{pmatrix}\alpha\alpha^{\prime\prime}+\beta\beta^{\prime\prime}&\alpha\beta^{\prime\prime}-\beta\alpha^{\prime\prime}&0\\ \beta\alpha^{\prime\prime}-\alpha\beta^{\prime\prime}&\beta\beta^{\prime\prime}+\alpha\alpha^{\prime\prime}&0\\ 0&0&\gamma\gamma^{\prime\prime}\end{pmatrix}.$$

                                    Then the similarity assumption (5.2) yields

                                    $$\left\{\begin{array}[]{lll}\alpha\beta^{\prime\prime}-\alpha^{\prime\prime}\beta=0,\\ \alpha\alpha^{\prime\prime}+\beta\beta^{\prime\prime}=\gamma\gamma^{\prime\prime}.\end{array}\right.$$

                                    (5.5)

                                    From here, we wish to reduce the system into a single ODE for $\gamma$ and solve it. Using (5.5), we get that $(\alpha\beta^{\prime}-\alpha^{\prime}\beta)^{\prime}=0,$ hence after integration we have

                                    $$\alpha\beta^{\prime}-\alpha^{\prime}\beta=\epsilon=\mbox{const}.$$

                                    Moreover, differentiating (5.4), we have $\alpha\alpha^{\prime}+\beta\beta=-\frac{\gamma^{\prime}}{2\gamma^{2}}$. Putting these equations together, it forms the system:

                                    	$\displaystyle\alpha\beta^{\prime}-\alpha^{\prime}\beta=\epsilon,$		(5.6)
                                    	$\displaystyle\alpha\alpha^{\prime}+\beta\beta=-\frac{\gamma^{\prime}}{2\gamma^{2}}.$		(5.7)

                                    Multiplying the first equation by $\alpha$, and the second one by $\beta$ and adding up we get $\alpha^{2}\beta^{\prime}-\alpha\alpha^{\prime}\beta+\alpha\alpha^{\prime}\beta+\beta^{2}\beta^{\prime}=\alpha\epsilon-\frac{\gamma^{\prime}\beta}{2\gamma^{2}}$, or equivalently

                                    $$(\alpha^{2}+\beta^{2})\beta^{\prime}=\alpha\epsilon-\frac{\gamma^{\prime}\beta}{2\gamma^{2}}.$$

                                    Utilizing (5.4) we obtain

                                    $$\beta^{\prime}=\alpha\epsilon\gamma-\frac{\gamma^{\prime}\beta}{2\gamma}.$$

                                    Repeating the same calculation but with $\alpha$ and $\beta$ exchanged, we also obtain

                                    $$\alpha^{\prime}=-\beta\epsilon\gamma-\frac{\gamma^{\prime}\alpha}{2\gamma}.$$

                                    Using the last two equations and (5.4) one more time we get that

                                    $$(\alpha^{\prime})^{2}+(\beta^{\prime})^{2}=\alpha^{2}\epsilon^{2}\gamma^{2}+\frac{(\gamma^{\prime})^{2}\beta^{2}}{4\gamma^{2}}+\beta^{2}\epsilon^{2}\gamma^{2}+\frac{(\gamma^{\prime})^{2}\alpha^{2}}{4\gamma^{2}}=\epsilon^{2}\gamma+\frac{(\gamma^{\prime})^{2}}{4\gamma^{3}}.$$

                                    (5.8)

                                    Differentiating (5.7) yields

                                    $$(\alpha^{\prime})^{2}+\alpha\alpha^{\prime\prime}+(\beta^{\prime})^{2}+\beta\beta^{\prime\prime}=-\frac{\gamma^{\prime\prime}}{2\gamma^{2}}+\frac{(\gamma^{\prime})^{2}}{\gamma^{3}}.$$

                                    Substituting (5.5) and (5.8) into the above, we obtain

                                    $$\epsilon^{2}\gamma+\frac{(\gamma^{\prime})^{2}}{4\gamma^{3}}+\gamma\gamma^{\prime\prime}=-\frac{\gamma^{\prime\prime}}{2\gamma^{2}}+\frac{(\gamma^{\prime})^{2}}{\gamma^{3}}.$$

                                    Hence we get the desired ODE for $\gamma$

                                    $$(\gamma^{3}+\frac{1}{2})\gamma\gamma^{\prime\prime}-\frac{3}{4}(\gamma^{\prime})^{2}+\epsilon^{2}\gamma^{4}=0,\qquad\gamma(0)=1.$$

                                    (5.9)

                                    In order to solve (5.9) we make the substitution $2\psi(\gamma)=(\gamma^{\prime})^{2}$, for a new unknown $\psi$. Then $\psi$ solves the following first order ODE

                                    $$(\gamma^{3}+\frac{1}{2})\gamma\psi^{\prime}(\gamma)-\frac{3}{4}\psi(\gamma)+\epsilon^{2}\gamma^{4}=0.$$

                                    Note that

                                    $$\frac{d}{d\gamma}\left(\frac{(\gamma^{3}+\frac{1}{2})\psi^{\prime}(\gamma)}{\gamma^{3}}\right)=\frac{d}{d\gamma}\left((1+\frac{1}{2\gamma^{3}})\psi(\gamma)\right)=-\frac{3}{2\gamma^{4}}+\frac{\gamma^{3}+\frac{1}{2}}{\gamma^{3}}\psi^{\prime}(\gamma).$$

                                    This yields,

                                    $$\frac{d}{d\gamma}\left(\frac{(\gamma^{3}+\frac{1}{2})\psi(\gamma)}{\gamma^{3}}+\epsilon^{2}\gamma\right)=0\quad\text{ or }\quad\frac{(\gamma^{3}+\frac{1}{2})\psi(\gamma)}{\gamma^{3}}+\epsilon^{2}\gamma=C.$$

                                    Recalling the definition of $\psi$, we finally get

                                    $$(\gamma^{\prime})^{2}=\frac{2\gamma^{3}(C-\epsilon^{2}\gamma)}{\gamma^{3}+\frac{1}{2}}.$$

                                    Denote $\gamma^{*}=C/\epsilon^{2}$, then

                                    $$(\gamma^{\prime})^{2}=\frac{2\epsilon^{2}\gamma^{3}(\gamma^{*}-\gamma)}{\gamma^{3}+\frac{1}{2}}.$$

                                    (5.10)

                                    We immediately see that $\gamma\leq\gamma^{*}$. Using the initial condition $\gamma(0)=1$, we get that

                                    $$\gamma^{\prime}(0)=2|\epsilon|\sqrt{\frac{\gamma^{*}-1}{3}},\quad\gamma^{*}\geq 1.$$

                                    For this case, if $t>0$ and close to $t=0$, $\gamma(t)$ must be increasing. Thus integrating (5.10), we have

                                    $$\int_{1}^{\gamma}\gamma^{-\frac{3}{2}}\sqrt{2\gamma^{3}+1}\frac{d\gamma}{\sqrt{\gamma^{*}-\gamma}}=2|\epsilon|t.$$

                                    As $\gamma$ increases, it attains the value $\gamma^{*}$ at $t=t^{*}$, where $t^{*}$ is determined from

                                    $$\int_{1}^{\gamma^{*}}\gamma^{-\frac{3}{2}}\sqrt{2\gamma^{3}+1}\frac{d\gamma}{\sqrt{\gamma^{*}-\gamma}}=2|\epsilon|t^{*}.$$

                                    From (5.9) and (5.10) it follows that $\gamma$ has a local maximum at $t=t^{*}$, and $\gamma(t)$ decrease for $t>t^{*}$. For these values of $t$ we have

                                    $$\int^{\gamma^{*}}_{\gamma}\gamma^{-\frac{3}{2}}\sqrt{2\gamma^{3}+1}\frac{d\gamma}{\sqrt{\gamma^{*}-\gamma}}=2|\epsilon|(t-t^{*}).$$

                                    From here it follows that as $t\to\infty$ then $\gamma\to 0$. The graph of $\gamma$ is shown in Figure 5.6.

                                    

                                    The functions $\alpha,\beta$ can be calculated as functions of $\gamma$ as follows: in light of the incompressibility condition (5.4), and constant vorticity condition (5.6), we let

                                    $$\alpha=\frac{1}{\sqrt{\gamma}}\cos\theta,\quad\beta=\frac{1}{\sqrt{\gamma}}\sin\theta.$$

                                    To determine $\theta$, we use (5.6) to obtain that

                                    $$\alpha=\frac{1}{\sqrt{\gamma}}\cos\left(\epsilon\int_{0}^{t}\gamma dt\right),\quad\beta=\frac{1}{\sqrt{\gamma}}\sin\left(\epsilon\int_{0}^{t}\gamma dt\right).$$

                                    Now let us discuss the geometric picture for this flow. The pressure can be assumed zero on the sphere $\xi^{2}+\eta^{2}+\zeta^{2}=1$, and the mapping given by the matrix $A$ implies

                                    $$\gamma(x^{2}+y^{2})+\frac{1}{\gamma^{2}}z^{2}=1.$$

                                    We see that for $0<t<t^{*}$ the sphere $\Gamma_{0}$ at $t=0$ turns into an ellipsoid of revolution with axis $z$ until the moment when its major semiaxis becomes equal to $\gamma^{*}$. After that the ellipsoid reverses its motion back to the unit sphere, and consequently flattens out to the plane $z=0$, as $t\to\infty$ (see Figure 5.6). The 3D animation of this flow can be viewed here; https://www.maths.ed.ac.uk/~aram/mov.html .

                                    To clarify the mechanical phenomenon, let us look at the velocity $\vec{\mathrm{v}}=A^{\prime}(0)\vec{X}_{0}$ at initial data. Since $\beta^{\prime}(0)=\epsilon,\gamma^{\prime}(0)=2|\epsilon|\sqrt{\frac{\gamma^{*}-1}{3}}$, and $\alpha^{\prime}(0)=-\frac{1}{2}\gamma^{\prime}(0)$, we have

                                    $$A^{\prime}(0)=\begin{pmatrix}\alpha^{\prime}(0)&\epsilon&0\\ -\epsilon&\alpha^{\prime}(0)&0\\ 0&0&-2\alpha^{\prime}(0)\end{pmatrix}.$$

                                    Observe that the matrix $A^{\prime}(0)$ is not symmetric, which indicates that the motion is not irrotational. In fact $\textnormal{{curl}}\vec{\mathrm{v}}=-2\epsilon\hat{z}$. Therefore the initial state of the motion, at $t=0$, corresponds to uniform rotation of the sphere $\Gamma_{0}$ about $z$ axis with angular velocity $\epsilon.$ If $\epsilon=0$ then we get a potential flow discussed earlier, for which the needle stretches to a line as $t\to\infty$. However, for $\epsilon\not=0$, the stretching into line is impossible and the flow collapses to a hyperplane, as explained above.

                                    The velocity at every point can be written in the form $\nabla\phi+\theta$, where $\phi$ is a potential function and $\theta$ is the rotation component about the $z$-axis. The kinematic energy of the system is the sum of kinetic energies

                                    $$K=K_{\nabla\phi}+K_{\theta}.$$

                                    During the flow $K_{\nabla\phi}$ transfers to $K_{\theta}$ up to the point $t=t^{*}$, where $K_{\nabla\phi}(t^{*})=0$, after that the transfer reverses its direction from $K_{\theta}$ to $K_{\nabla\phi}$ as $t\to\infty$, until it purges and gives $K_{\theta}(\infty)=0$ and $K(\infty)=K_{\nabla\phi}(\infty)$.

                                    References