3.1 Methodology

In the soliton boundary approach, one looks for periodic solutions that satisfy the boundary conditions

(3.1)\begin{equation} n\rightarrow1,\quad u\rightarrow0,\quad \phi\rightarrow0,\quad {\rm when}\ x\rightarrow\infty. \end{equation}

Now, right at the onset, it should be noted that non-trivial (i.e. non-constant) periodic functions have no asymptotic limit. As such, the very notion of introducing a limit when $x$ or $\xi$ approaches infinity is deeply flawed. Despite this obvious mathematical error, we proceed by introducing the usual reductive perturbation expansions

(3.2)\begin{gather} n=1+\varepsilon n_{1}+\varepsilon^{2}n_{2}+\cdots, \end{gather}

(3.3)\begin{gather}u=\varepsilon u_{1}+\varepsilon^{2}u_{2}+\cdots, \end{gather}

(3.4)\begin{gather}\phi=\varepsilon\phi_{1}+\varepsilon^{2}\phi_{2}+\cdots, \end{gather}

along with the stretched coordinates

(3.5a,b)\begin{equation} \xi=\varepsilon^{{1}/{2}}\left(x-Mt\right),\quad \tau=\varepsilon^{{3}/{2}}t.\end{equation}

Substitution of the expansions (3.2), (3.3), (3.4), along with the stretched coordinates (3.5a,b) into the continuity equation (2.1) leads to an equation that yields terms of differing orders of $\varepsilon$. At the lowest order $O(\varepsilon ^{{3}/{2}})$, one obtains

(3.6)\begin{equation} {-}M\frac{\partial n_{1}}{\partial\xi}+\frac{\partial u_{1}}{\partial\xi}=0,\end{equation}

while the higher order $O(\varepsilon ^{{5}/{2}})$ yields

(3.7)\begin{equation} {-}M\frac{\partial n_{2}}{\partial\xi}+\frac{\partial n_{1}}{\partial\tau}+ \frac{\partial u_{2}}{\partial\xi}+\frac{\partial}{\partial\xi}\left(n_{1}u_{1}\right)=0.\end{equation}

Similarly, the momentum equation (2.2) gives

(3.8)\begin{equation} {-}M\frac{\partial u_{1}}{\partial\xi}={-}\frac{\partial\phi_{1}}{\partial\xi},\end{equation}

at order $O(\varepsilon ^{{3}/{2}})$, and

(3.9)\begin{equation} {-}M\frac{\partial u_{2}}{\partial\xi}+\frac{\partial u_{1}}{\partial\tau}+u_{1} \frac{\partial u_{1}}{\partial\xi}={-}\frac{\partial\phi_{2}}{\partial\xi},\end{equation}

at order $O(\varepsilon ^{{5}/{2}})$. Finally, Poisson's equation gives

(3.10)\begin{equation} 0=\phi_{1}-n_{1},\end{equation}

at order $O(\varepsilon )$, and

(3.11)\begin{equation} \frac{\partial^{2}\phi_{1}}{\partial\xi^{2}}=\phi_{2}+\frac{1}{2}\phi_{1}^{2}-n_{2},\end{equation}

at order $O(\varepsilon ^{2}).$ Following the usual reductive perturbation approach, we start by solving the lowest-order equations (3.6), (3.8) and (3.10), and use the resulting solutions to solve the higher-order equations (3.7), (3.9) and (3.11).

One can solve (3.6) by integrating with respect to $\xi,$ resulting in the equation

(3.12)\begin{equation} {-}Mn_{1}+u_{1}=C_{1},\end{equation}

where $C_{1}$ is an integration constant. To find $C_{1}$, we use the ill-posed boundary conditions, namely that $n\rightarrow 1$ and $u\rightarrow 0$ when $x\rightarrow \infty$. It then follows that $n_{1}\rightarrow 0$ and $u_{1}\rightarrow 0$ when $\xi \rightarrow \infty$, so that $C_{1}=0$ and

(3.13)\begin{equation} u_{1}=Mn_{1}.\end{equation}

We now proceed to solve the lowest-order equation associated with the momentum equation (3.8). Once again, a straightforward integration leads to

(3.14)\begin{equation} {-}Mu_{1}+\phi_{1}=C_{2},\end{equation}

where $C_{2}$ is an integration constant. As before, the ill-posed boundary conditions $u\rightarrow 0$ and $\phi \rightarrow 0$ yield that $C_{2}=0$. From (3.13) it follows that

(3.15)\begin{equation} {-}Mu_{1}+\phi_{1}=0,\end{equation}

so that

(3.16)\begin{equation} n_{1}=\frac{1}{M^{2}}\phi_{1}.\end{equation}

Substituting this expression of $n_{1}$ into the lowest-order equation derived from Poisson's equation (3.10), one obtains the compatibility condition

(3.17)\begin{equation} \left(1-\frac{1}{M^{2}}\right)\phi_{1}=0.\end{equation}

Hence, it follows that $M=\pm 1$. Since we are interested in waves propagating in the direction of the magnetic field, we choose $M=1$. From (3.13) and (3.16) it then follows that $n_{1}=u_{1}=\phi _{1}$.

We now proceed to the higher-order equations. Substituting $M=1$, $n_{1}=\phi _{1}$ and $u_{1}=\phi _{1}$ into (3.7) gives the following partial differential equation:

(3.18)\begin{equation} \frac{\partial u_{2}}{\partial\xi}= \frac{\partial n_{2}}{\partial\xi}-\frac{\partial\phi_{1}}{\partial\tau}- \frac{\partial}{\partial\xi}\left(\phi_{1}^{2}\right).\end{equation}

Similarly, (3.9) gives

(3.19)\begin{equation} {-}\frac{\partial u_{2}}{\partial\xi}+\frac{\partial\phi_{1}}{\partial\tau}+ \phi_{1}\frac{\partial\phi_{1}}{\partial\xi}+\frac{\partial\phi_{2}}{\partial\xi}=0.\end{equation}

Substituting (3.18) into (3.19) then leads to

(3.20)\begin{equation} \frac{\partial n_{2}}{\partial\xi}=2\frac{\partial\phi_{1}}{\partial\tau}+3\phi_{1}\frac{\partial\phi_{1}}{\partial\xi}+\frac{\partial\phi_{2}}{\partial\xi}.\end{equation}

Finally, the KdV equation is obtained by differentiating (3.11) with respect to $\xi$, and substituting (3.20) into the resulting equation, giving

(3.21)\begin{equation} \frac{\partial\phi_{1}}{\partial\tau}+\frac{1}{2} \frac{\partial^{3}\phi_{1}}{\partial\xi^{3}}+\phi_{1} \frac{\partial\phi_{1}}{\partial\xi}=0.\end{equation}

In order to derive the cnoidal wave solutions for the KdV equation (3.21), it is customary to look for solutions that remain constant in the moving frame that propagate slightly faster than the acoustic speed. We therefore introduce the variable

(3.22)\begin{equation} \eta=\xi-v\tau.\end{equation}

In terms of the original coordinates, this is given by

(3.23)\begin{equation} \eta=\varepsilon^{{1}/{2}}\left[x-(1+\varepsilon v)t\right].\end{equation}

By substituting the variable (3.22) into the KdV equation (3.21), one reduces the KdV equation to the following third-order ordinary differential equation (ODE):

(3.24)\begin{equation} {-}v\frac{{\rm d}\phi_{1}}{{\rm d}\eta}+\frac{1}{2}\frac{{\rm d}^{3}\phi_{1}}{{\rm d}\eta^{3}}+\phi_{1} \frac{{\rm d}\phi_{1}}{{\rm d}\eta}=0.\end{equation}

Integrating this equation once with respect to $\eta$ gives

(3.25)\begin{equation} {-}v\phi_{1}+\frac{1}{2}\frac{{\rm d}^{2}\phi_{1}}{{\rm d}\eta^{2}}+ \frac{1}{2}\phi_{1}^{2}=C_{3},\end{equation}

where $C_{3}$ is a constant of integration. The usual approach is to choose $C_{3}=0$, based on the argument that if $\phi \rightarrow 0$ in the limit when $\eta \rightarrow \infty$, then ${{\rm d}^{2}\phi _{1}}/{{\rm d}\eta ^{2}}\rightarrow 0$ in the same limit. The resulting second-order ODE is then given by

(3.26)\begin{equation} {-}v\phi_{1}+\frac{1}{2}\frac{{\rm d}^{2}\phi_{1}}{{\rm d}\eta^{2}}+\frac{1}{2}\phi_{1}^{2}=0.\end{equation}

In order to perform another integration, (3.25) is then multiplied with ${{\rm d}\phi _{1}}/{{\rm d}\eta }$ and integrated to get

(3.27)\begin{equation} \frac{1}{2}\left(\frac{{\rm d}\phi_{1}}{{\rm d}\eta}\right)^{2}+ \frac{1}{3}\phi_{1}^{3}-v\phi_{1}^{2}=C_{4},\end{equation}

where $C_{4}$ is a constant of integration. If one were to stay mathematically consistent, one would note that, since $\phi _{1}\rightarrow 0$ when $\eta \rightarrow \infty$, one must then have that ${{\rm d}\phi _{1}}/{{\rm d}\eta }\rightarrow 0$ when $\eta \rightarrow \infty$, similarly to the argument used to obtain $C_3=0$. In this case, the constant $C_{4}=0$, so that the equilibrium is recovered, and no periodic solutions are found. However, here, the soliton boundary approach contradicts itself by choosing $C_{4}$ to be an arbitrary constant. This implies that ${{\rm d}\phi _{1}}/{{\rm d}\eta }$ approaches a non-zero value in the limit $\eta \rightarrow \infty$. Not only does this contradict the use of the previous limits, but it clearly also implies that the solutions are unbounded rather than periodic. Nevertheless, by following this reasoning one obtains the following equation:

(3.28)\begin{equation} \frac{1}{2}\left(\frac{{\rm d}\phi_{1}}{{\rm d}\eta}\right)^{2}+ \frac{1}{3}\phi_{1}^{3}-v\phi_{1}^{2}-C_{4}=0.\end{equation}

The differential equation (3.28) has cnoidal wave solutions of the form

(3.29)\begin{equation} \phi_{1}(\eta)=\mu_{2}+\left(\mu_{3}-\mu_{2}\right)\text{cn}^{2} \left[\sqrt{\frac{\mu_{3}-\mu_{1}}{6}}\eta,m\right],\end{equation}

where $\mu _{1}<\mu _{2}<\mu _{3}$ are the real roots of the ‘Sagdeev potential’, i.e. the cubic polynomial

(3.30)\begin{equation} \tfrac{1}{3}\phi_{1}^{3}-v\phi_{1}^{2}-C_{4}=0,\end{equation}

and

(3.31)\begin{equation} m=\frac{\mu_{3}-\mu_{2}}{\mu_{3}-\mu_{1}}.\end{equation}

The solution (3.29) is periodic with period $L=12K(m)/(\mu _{3}-\mu _{1})$, where $K(m)$ is the complete elliptic integral of the first kind.

Clearly, a necessary condition for the existence of cnoidal wave solutions is that the equation (3.30) must have three distinct real roots. This depends on the choices of $v$ and $C_{4}$. Indeed, for a fixed choice of $v$, one can show that the polynomial (3.30) has three real roots provided that

(3.32)\begin{equation} 0< C_4<\tfrac{4}{3}v^{3}. \end{equation}

As an example, let us consider cnoidal wave solutions associated with the soliton boundary approach corresponding to periodic waves propagating with a velocity of $M=1.01$. To achieve this, we set $\varepsilon =0.01$, $v=1$, and use different choices of ${C_{4}\in (0,4/3)}$. To this end, we let $C_{4}=\frac {4}{3}a,$ where $0< a<1$. The solutions are shown in figure 1. Here, panel (a) shows the phase portraits for six different choices of $a$. It is worth pointing out that all these solutions are inconsistent with the original set of boundary conditions. Clearly all choices of $a$ result in solutions that remain positive for all values of $\eta$. Indeed, if $\phi _{1}$ remains positive, it can never approach $0$ in any limit, and therefore violates the original set of boundary conditions. Moreover, these results are inconsistent with the conservation of number density. To show this, consider figure 1(b), where the number density is plotted in the original coordinates at $t=0$. The blue curve corresponds to $a=0.2$, and the red curve corresponds to $a=0.99$. In addition, the black dotted line shows the equilibrium number density. It is clear that neither of these solutions conserves the number density due to the fact that the perturbed periodic wave oscillates above the equilibrium $n=1$ level. As such, it is clear that

(3.33)\begin{equation} \frac{1}{L}\int_0^L (n-1)\,{{\rm d}x}>0, \end{equation}

where $L$ is the spatial period of the solution, so that the conservation of number density is violated.

Figure 1.

Cnoidal wave solutions obtained by means of the soliton boundary approach. In the left panel, a phase portrait is shown for different choices of $a$, as shown on the outside of each curve. In (b), the solution of $n$ is shown in the original spatial coordinate $x$ at $t=0.$ The blue curve corresponds to $a=0.2$ and the red curve corresponds to $a=0.99$. The dashed line shows the equilibrium number density.

3.2 Deconstruction of the method

As we mentioned above, there are a number of mathematical flaws and contradictions in the derivation of the cnoidal wave solutions. In particular, the boundary conditions for the number density $n$, fluid velocity $u$ and electrostatic potential $\phi$ are all inconsistent with periodic solutions. In addition, the integration constant $C_{4}$ is constructed in a way that implies that the derivative ${{\rm d}\phi _{1}}{{\rm d}\eta }$ approaches a non-zero value in the limit $\eta \rightarrow \infty$, clearly contradicting the boundary condition stating that $\phi _{1}\rightarrow 0$ when $\eta \rightarrow \infty$. Despite these fundamental flaws, one obtains cnoidal wave solutions.

As it turns out, the calculation of the integration constants can instead be obtained through the application of a set of initial conditions. The key idea here is to notice that the integration constants are determined at a specific choice of $\xi$ or $\eta$, rather than at the boundaries.

To simplify the deconstruction, we look to express the conditions that allow us to determine the integration constants as an initial condition. To do so, we set $t=0$, so that $\xi =\varepsilon ^{{1}/{2}}x$ and $\eta =\varepsilon ^{{1}/{2}}x$. Let us start with the integration constant $C_{1}$. By setting $C_{1}=0$, it follows that there exists a $\xi _{0}$ such that $n_{1}(\xi _{0})=0$ and $u_{1}(\xi _{0})=0$. By setting $t=0$, it follows that $n(x_{0},0)=1$ and $u(x_{0},0)=0$, where $x_{0}=\varepsilon ^{-{1}/{2}}\xi _{0}$. Similarly, the second integration constant can be obtained by setting $\phi (x_{0},0)=0$. This implies that the function $\phi (x,0)$ has a position $x=x_0$ where the solution satisfies the equilibrium conditions.

The next integration constant was chosen in a way that $C_{3}=0$. At $x=x_{0}$ and $t=0$, this implies that

(3.34)\begin{equation} \frac{\partial^{2}\phi}{\partial x^{2}}\left(x_{0},0\right)=0. \end{equation}

In other words, this boundary condition shows that the function $\phi _{1}(x,0)$ has a spatial inflection point at $x=x_{0}$ and $t=0$.

For the final integration constant $C_{4}$, it follows that $({{\rm d}\phi _{1}}/{{\rm d}\eta })(\eta _{0})=\pm \sqrt {2C_{4}}$. By choosing the point where the slope reaches its maximum, we can use the positive $+$ sign. If we rewrite this in the original coordinates at $t=0$, it follows that

(3.35)\begin{equation} \frac{\partial\phi}{\partial x}\left(x_{0},0\right)= \varepsilon^{{3}/{2}}\sqrt{2C_{4}}.\end{equation}

In other words, at $t=0$ and $x_{0}=\varepsilon ^{-{1}/{2}}\xi _{0}$, the function $\phi (x,0)$ satisfies three conditions, namely (i) the function reaches its equilibrium value, (ii) the function has a non-zero slope and (iii) the function has an inflection point. The latter implies that the function reaches its maximum slope at $x_{0}$.

When expressed in the form of an initial condition problem, the soliton limit approach is mathematically sound. However, the problem of number density conservation still remains. This is due to the fact that the solutions are forced to reach their equilibrium state at the inflection points. In the next section, we derive a new formalism by relaxing this condition.

5.1 Necessary condition for existence of solutions

The analysis of the cnoidal wave solution is complicated by the fact that the solution has three free variables, namely $v$, $\alpha ^{2}$ and $\phi _{10}$. Here, $v$ is directly related to the propagation speed of the periodic wave. On the other hand, $\alpha ^{2}$ is related to the slope of the periodic solution at the inflection point $\xi =0$. In addition, $\phi _{10}$ is chosen in a way that ensures that the number density is conserved. For the cnoidal wave solution, larger values of $\alpha ^{2}$ correspond to larger fluctuations.

It is clear from the solution (4.21) that finding three real roots to the polynomial (4.22) is crucial in order to analyse the periodic waves. For our analysis, we treat $v$ as a fixed parameter, while $\alpha ^{2}$ is varied within this choice of $v$, and $\phi _{10}$ depends on the choice of $\alpha$.

It is well known that the nature of the roots of a generalized cubic polynomial of the form

(5.1)\begin{equation} ax^{3}+bx^{2}+cx+d=0,\end{equation}

depends on the discriminant

(5.2)\begin{equation} \varDelta=18abcd-4b^{3}d+b^{2}c^{2}-4ac^{3}-27a^{2}d^{2},\end{equation}

see for example Abramowitz & Stegun (Reference Abramowitz and Stegun1965). In particular, if $\varDelta >0$, the polynomial has three distinct real roots. This is a necessary condition for the existence of cnoidal wave solutions. For the polynomial (4.22) we have $a=\frac {1}{3}$, $b=v$, $c=-\left(2v\phi_{10}+\phi_{10}^2\right)$ and $d=-(\alpha ^{2}-\frac {2}{3}\phi _{10}^{3}-v\phi _{10}^{2})$.

In the following, we assume that $v$, $\alpha$ and $\phi _{10}$ are chosen in a way that ensures that the discriminant $\varDelta$ is strictly positive. The relationship between $\alpha ^{2}$ and $\phi _{10}$ will be considered in more detail after the final solution has been derived.

5.2 Analytical expressions for the roots $\mu _{1}$, $\mu _{2}$ and $\mu _{3}$

The three real roots of the cubic polynomial (4.22) satisfying $\varDelta >0$ can be expressed in terms of trigonometric functions as follows:

(5.3)\begin{equation} r_{j}=t_{j}-\frac{b}{3a}\end{equation}

for $j=0,1,2$, where

(5.4)\begin{gather} t_{j}=2\sqrt{-\frac{p}{3}}\cos\left[\frac{1}{3}\arccos\left(\frac{3q}{2p} \sqrt{-\frac{3}{p}}\right)+\frac{2{\rm \pi}}{3}j\right], \end{gather}

(5.5)\begin{gather}p=\frac{3ac-b^{2}}{3a^{2}}, \end{gather}

and

(5.6)\begin{equation} q=\frac{2b^{3}-9abc+27a^{2}d}{27a^{3}}.\end{equation}

For the polynomial (4.22), one then obtains the following roots:

(5.7)\begin{gather} r_{0}=2\left(v+\phi_{10}\right)\cos\frac{\theta}{3}-v, \end{gather}

(5.8)\begin{gather}r_{1}=2\left(v+\phi_{10}\right)\cos\frac{\theta+2{\rm \pi}}{3}-v, \end{gather}

and

(5.9)\begin{equation} r_{2}=2\left(v+\phi_{10}\right)\cos\frac{\theta+4{\rm \pi}}{3}-v,\end{equation}

where

(5.10)\begin{equation} \theta=\arccos\left[\frac{3\alpha^{2}-2\left(v+\phi_{10}\right)^{3}}{2\left(v+\phi_{10}\right)^{3}}\right].\end{equation}

Since $0\leq \theta \leq {\rm \pi}$, one can easily confirm by plotting the functions $\cos ({\theta }/{3})$, $\cos ({(\theta +2{\rm \pi} )}/{3})$ and $\cos ({(\theta +4{\rm \pi} )}/{3})$ that

(5.11)\begin{equation} r_{1}< r_{2}< r_{0},\end{equation}

whenever the polynomial has three real roots. As such, we have that,

(5.12a–c)\begin{equation} \mu_{1}=r_{1},\quad \mu_{2}=r_{2},\quad \mu_{3}=r_{0}.\end{equation}

By substituting these expressions into (4.21), it follows that the solution can be expressed as

(5.13)\begin{equation} \phi_{1}(\xi,\tau)=\beta_{0}+\beta_{1}\text{cn}^{2}\left(\beta_{2}(\xi+v\tau),\beta_{3}\right),\end{equation}

where

(5.14)\begin{gather} \beta_{0}=2\left(v+\phi_{10}\right)\cos\frac{\theta+4{\rm \pi}}{3}-v, \end{gather}

(5.15)\begin{gather}\beta_{1}=\sqrt{12}\left(v+\phi_{10}\right)\cos\frac{2\theta+{\rm \pi}}{6}, \end{gather}

(5.16)\begin{gather}\beta_{2}=3^{{-}1/4}\sqrt{\left(v+\phi_{10}\right)\cos\frac{2\theta-{\rm \pi}}{6}}, \end{gather}

and

(5.17)\begin{equation} \beta_{3}=\frac{\cos\dfrac{2\theta+{\rm \pi}}{6}}{\cos\dfrac{2\theta-{\rm \pi}}{6}}.\end{equation}

5.3 Choice of $\phi _{10}$

The general solution (5.13) depends on three free parameters, namely $\alpha$, $v$ and $\phi _{10}$. In order to construct solutions, we consider a fixed choice of the velocity $v$. For the resulting propagation speed we then construct periodic solutions with different amplitudes. These solutions are obtained by varying $\alpha$. Finally, for each choice of $\alpha$ we look for the value of $\phi _{10}$ such that the conservation of number density (4.10) is satisfied. That is, we look for a choice of $\phi _{10}$ such that the following identity holds:

(5.18)\begin{equation} \int_{0}^{L}\phi_{1}(\xi,\tau)\,{\rm d}\xi=0\end{equation}

for all $\tau$. Here, $L$ is the wavelength, given by

(5.19)\begin{equation} L=\frac{2K\left(\beta_{3}\right)}{\beta_{2}},\end{equation}

where $K(\beta _{3})$ is the complete elliptic integral of the first kind.

The integral (5.18) can be expressed analytically, thanks to the following indefinite integral:

(5.20)\begin{equation} \int\text{cn}^{2}(z,m)\,{\rm d} z=z-\frac{z}{m}+ \frac{E\left[\left.\text{am}\left(z\,|\,m\right)\right|m\right] \left[\text{cn}^{2}\left(z\,|\,m\right)+\dfrac{1}{m}-1\right]}{\text{dn}(z\,|\,m) \sqrt{1-m\text{sn}^{2}(z\,|\,m)}}+C.\end{equation}

Here, the functions $\text {sn}(z\,|\,m)$, $\text {cn}(z\,|\,m)$ and $\text {dn}(z\,|\,m)$ are Jacobi elliptic functions, $\text {am}(z\,|\,m)$ is the Jacobi elliptic amplitude function and $E(z\,|\,m)$ is the incomplete elliptic integral of the second kind.

By using the incomplete integral (5.20) along with the identities

(5.21)\begin{equation} \text{sn}(0\,|\,m)=\text{sn}\left[2K(m)\,|\,m\right]=\text{am}(0\,|\,m)=E(0\,|\,m)=0,\end{equation}

and

(5.22a–c)\begin{equation} \text{cn}\left[2K(m)\,|\,m\right]={-}1,\quad \text{dn}\left[2K(m)\,|\,m\right]=1,\quad \text{am}\left[2K(m)\,|\,m\right]={\rm \pi},\end{equation}

it follows that

(5.23)\begin{equation} \int_{0}^{L}\phi_{1}(\xi,\tau)\,{\rm d}\xi=\frac{\left(\beta_{0}+\beta_{1}\right) \beta_{3}-\beta_{1}}{\beta_{3}}L+\frac{\beta_{1}}{\beta_{2}\beta_{3}}E \left({\rm \pi}\,|\,\beta_{3}\right).\end{equation}

Notice that, for a fixed choice of $v$ and $\alpha$, this integral depends on $\phi _{10}$, as is obvious from (5.10) and (5.14)–(5.17). As a result, finding $\phi _{10}$ becomes a root-finding problem for the function $I(\phi _{10})$, given by the right-hand side of (5.23).

In order to show the relationship between $\alpha$ and $\phi _{10}$, consider the special case where $v=1$. The dependence of $\phi _{10}$ on $\alpha$ is shown in figure 2. The figure clearly shows that $\phi _{10}$ increases with $\alpha$. In terms of the solution, this means that the vertical position of $\phi _{1}$, the point where the spatial slope reaches its maximum, must increase in order to compensate for the increased asymmetry associated with larger amplitudes. These aspects are explored in more detail in § 6.

Figure 2.

Dependence of $\phi _{10}$ on $\alpha$ for $v=1$.