Wilton Ripples with High-Order Resonances in Weakly Nonlinear Models

Abstract

Resonant periodic traveling waves (Wilton ripples) are examined asymptotically for a family of weakly nonlinear partial differential equations. Wilton ripple resonances can occur between pairs of wavenumbers, here labeled $k=1$ and N. Typical studies consider $N=2,$ the triad resonance, but this work examines $N>2,$ denoted “high-order resonance.” We present explicit formulas for the coefficients in the asymptotic series expansions and answer previously unaddressed questions including what modes are present at each order in the asymptotic expansion and at what order we can expect a non-zero Wilton ripple. The character of the solutions is presented using the example of the Kawahara equation. Finally, we comment on the factors which are indicative of convergence for the asymptotic expansions, and present an example where the series degenerates in the Benjamin equation.

Appendices

Appendix A: Simplified Formulas

This section presents the formulas for $s_{n-1},$ $b_{n-2},$ and $\beta _{n,j}$ when $N \ge 4$ once Theorems 1 and 2 have been applied. The original formulas are given by (41), (43), and (38), respectively. The new formulas will depend on the parity of n and N (and j in the case of $\beta _{n,j}$), and thus separate formulas must be given for each parity. Fortunately, for many combinations of parities, the coefficients are always 0.

We begin with the formula for $s_{n-1}.$ If n is even (and thus $n-1$ is odd), then $s_{n-1}$ will be zero. If n is odd, then the formula depends on the parity of N. For even N,  we find

$$\begin{aligned} s_{n-1}= & {} -\Biggl (\beta _{n-1,2} + \sum _{l=1}^{\frac{n-3}{2}}\Bigl (b_{2l}(\beta _{n-2l,N-1}+\beta _{n-2l,N+1})\nonumber \\{} & {} +\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne N/2 \end{array}}^{K_N(2l)/2}\bigl (\beta _{2l,2k}((1-\delta _{k,1})\beta _{n-2l,2k-1}+\beta _{n-2l,2k+1})\bigr )+\frac{1}{2}\sum _{k=1}^{(K_N(2l+1)-1)/2}\nonumber \\{} & {} \bigl (\beta _{2l+1,2k+1}((1-\delta _{2k,N})\beta _{n-2l-1,2k}+(1-\delta _{2k+2,N})\beta _{n-2l-1,2k+2})\bigr )\Bigr )\Biggr ) \end{aligned}$$

(50)

and for odd N,  we find

$$\begin{aligned} s_{n-1}= & {} -\Biggl (\beta _{n-1,2} + \sum _{l=1}^{\frac{n-3}{2}}\Bigl (b_{2l+1}(\beta _{n-2l-1,N-1}+\beta _{n-2l-1,N+1})\nonumber \\{} & {} +\frac{1}{2}\sum _{k=1}^{K_N(2l)/2}\bigl (\beta _{2l,2k}((1-\delta _{2k,N+1})\beta _{n-2l,2k-1}+(1-\delta _{2k,N-1})\beta _{n-2l,2k+1})\bigr )\nonumber \\{} & {} +\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne (N-1)/2 \end{array}}^{(K_N(2l+1)-1)/2}\bigl (\beta _{2l+1,2k+1}(\beta _{n-2l-1,2k}+\beta _{n-2l-1,2k+2})\bigr )\Bigr )\Biggr ). \end{aligned}$$

(51)

For $b_{n-2},$ the result is zero if N and n do not have the same parity. For even N and n,  the formula reduces to

$$\begin{aligned} \Omega (N)b_{n-2}= & {} {\overline{\beta }}_{n-1,N-1}+{\overline{\beta }}_{n-1,N+1} + \sum _{l=1}^{\frac{n-2}{2}}\Bigl ((1-\delta _{2l,n-2})b_{2l}(s_{n-2l}+\beta _{n-2l,2N}) \nonumber \\{} & {} +\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne N/2 \end{array}}^{K_N(2l)/2}\bigl (\beta _{2l,2k}(\beta _{n-2l,|N-2k|}+\beta _{n-2l,N+2k})\bigr )\nonumber \\{} & {} +(1-\delta _{2l,n-2})\frac{1}{2}\sum _{k=1}^{(K_N(2l+1)-1)/2}\nonumber \\{} & {} \bigl (\beta _{2l+1,2k+1}((1-\delta _{|N-2k-1|,1})\beta _{n-2l-1,|N-2k-1|}+\beta _{n-2l-1,N+2k+1})\bigr )\Bigr )\nonumber \\ \end{aligned}$$

(52)

and for odd N and n,  the formula reduces to

$$\begin{aligned} \Omega (N)b_{n-2}= & {} {\overline{\beta }}_{n-1,N-1}+{\overline{\beta }}_{n-1,N+1} + \sum _{l=1}^{\frac{n-3}{2}}\Bigl ((1-\delta _{2l+1,n-2})b_{2l+1}(s_{n-2l-1}+\beta _{n-2l-1,2N}) \nonumber \\{} & {} +\frac{1}{2}\sum _{k=1}^{K_N(2l)/2}\bigl (\beta _{2l,2k}((1-\delta _{|N-2k|,1})\beta _{n-2l,|N-2k|}+\beta _{n-2l,N+2k})\bigr )\nonumber \\{} & {} +\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne (N-1)/2 \end{array}}^{(K_N(2l+1)-1)/2}\bigl (\beta _{2l+1,2k+1}(\beta _{n-2l-1,|N-2k-1|}+\beta _{n-2l,N+2k+1})\bigr )\Bigr ). \end{aligned}$$

(53)

Finally, for $\beta _{n,j},$ the result is zero whenever n and j have different parity. One must also consider that $\beta _{n,N\pm 1}$ will depend on $b_{n-1},$ which is unknown. In practice, one can ignore the terms dependent on $b_{n-1}$ at first, as this returns ${\overline{\beta }},$ which is all that is needed to eventually determine $b_{n-1}.$ Once $b_{n-1}$ is known, the previously ignored terms can be added to ${\overline{\beta }}$ to return the true value of the coefficient. We begin by presenting the formulas for even N. If n and j are both even, then

$$\begin{aligned} \beta _{n,j}= & {} \frac{1}{\widehat{{\mathcal {R}}}(j)}\biggl ((1-\delta _{j,2})\beta _{n-1,j-1}+\beta _{n-1,j+1}+\sum _{l=1}^{\frac{n-2}{2}}\Bigl (s_{n-2l}\beta _{2l,j}\nonumber \\{} & {} +b_{2l}((1-\delta _{j,2N})\beta _{n-2l,|N-j|}+\beta _{n-2l,N+j})+\frac{1}{2}\delta _{j,2N}b_{2l}b_{n-2l}\nonumber \\{} & {} +\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne N/2 \end{array}}^{K_N(2l)/2}\bigl (\beta _{2l,2k}((1-\delta _{2k,j}-\delta _{|2k-j|,N})\beta _{n-2l,|2k-j|}+(1-\delta _{2k+j,N})\beta _{n-2l,2k+j})\bigr )\nonumber \\{} & {} +\frac{1}{2}(1-\delta _{2l,n-2})\sum _{k=1}^{(K_N(2l+1)-1)/2}\bigl (\beta _{2l+1,2k+1}((1-\delta _{|2k+1-j|,1})\beta _{n-2l-1,|2k+1-j|}\nonumber \\{} & {} +\beta _{n-2l-1,2k+1+j})\bigr )\Bigr )\biggr ) \end{aligned}$$

(54)

and for n and j both odd

$$\begin{aligned} \beta _{n,j}= & {} \frac{1}{\widehat{{\mathcal {R}}}(j)}\biggl ((\delta _{j,N-1}+\delta _{j,N+1})b_{n-1}+(1-\delta _{j,N+1})\beta _{n-1,j-1}+(1-\delta _{j,N-1})\beta _{n-1,j+1}\nonumber \\{} & {} +\sum _{l=1}^{\frac{n-3}{2}}\Bigl (s_{n-2l-1}\beta _{2l+1,j}+b_{2l}((1-\delta _{|N-j|,1})\beta _{n-2l,|N-j|}+\beta _{n-2l,N+j})\nonumber \\{} & {} +\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne N/2 \end{array}}^{K_N(2l)/2}\bigl (\beta _{2l,2k}((1-\delta _{|2k-j|,1})\beta _{n-2l,|2k-j|}+\beta _{n-2l,2k+j})\bigr )\nonumber \\{} & {} +\frac{1}{2}\sum _{k=1}^{(K_N(2l+1)-1)/2}\bigl (\beta _{2l+1,2k+1}((1-\delta _{2k+1,j}-\delta _{|2k+1-j|,N})\beta _{n-2l-1,|2k+1-j|}\nonumber \\{} & {} +(1-\delta _{2k+1+j,N})\beta _{n-2l-1,2k+1+j})\bigr )\Bigr )\biggr ). \end{aligned}$$

(55)

For N odd, if n and j are both even,

$$\begin{aligned} \beta _{n,j}= & {} \frac{1}{\widehat{{\mathcal {R}}}(j)}\biggl ((\delta _{j,N-1}+\delta _{j,N+1})b_{n-1}+(1-\delta _{j,2}-\delta _{j,N+1})\beta _{n-1,j-1}+(1-\delta _{j,N-1})\nonumber \\{} & {} \beta _{n-1,j+1}+\sum _{l=1}^{\frac{n-2}{2}}\left( s_{n-2l}\beta _{2l,j}+\frac{1}{2}\sum _{k=1}^{K_N(2l)/2}\bigl (\beta _{2l,2k}((1-\delta _{2k,j})\beta _{n-2l,|2k-j|}+\beta _{n-2l,2k+j})\bigr )\right. \nonumber \\{} & {} +(1-\delta _{2l,n-2})\Bigl [b_{2l+1}((1-\delta _{|N-j|,1}-\delta _{j,2N})\beta _{n-2l-1,|N-j|}+\beta _{n-2l-1,N+j})\nonumber \\{} & {} + \frac{1}{2}\delta _{j,2N}b_{2l+1}b_{n-2l-1}+\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne (N-1)/2 \end{array}}^{(K_N(2l+1)-1)/2}\bigl (\beta _{2l+1,2k+1}((1-\delta _{|2k+1-j|,1}-\delta _{|2k+1-j|,N})\nonumber \\{} & {} \left. \beta _{n-2l-1,|2k+1-j|}+(1-\delta _{2k+1+j,N})\beta _{n-2l-1,2k+1+j})\bigr )\Bigr ]\Bigr )\right) , \end{aligned}$$

(56)

and if n and j both odd

$$\begin{aligned} \beta _{n,j}= & {} \frac{1}{\widehat{{\mathcal {R}}}(j)}\left( \beta _{n-1,j-1}+\beta _{n-1,j+1}+\sum _{l=1}^{\frac{n-3}{2}}\Bigl (s_{n-2l-1}\beta _{2l+1,j} + b_{2l+1}(\beta _{n-2l-1,|j-N|}+\beta _{n-2l-1,j+N})\right. \nonumber \\{} & {} +\frac{1}{2}\sum _{k=1}^{K_N(2l)/2}\bigl (\beta _{2l,2k}((1-\delta _{|2k-j|,1}-\delta _{|2k-j|,N})\beta _{n-2l,|2k-j|}+(1-\delta _{2k+j,N})\beta _{n-2l,2k+j})\bigr )\nonumber \\{} & {} \left. +\frac{1}{2}\sum _{\begin{array}{c} k=1\\ k\ne (N-1)/2 \end{array}}^{(K_N(2l+1)-1)/2}\bigl (\beta _{2l+1,2k+1}((1-\delta _{2k+1,j})\beta _{n-2l-1,|2k+1-j|}+\beta _{n-2l-1,2k+1+j})\bigr )\Bigr ]\Bigr )\right) .\nonumber \\ \end{aligned}$$

(57)

Maple code which finds the solution by executing the algorithm presented in Sect. 2 and python code which finds the solution through the recursive relations presented above can both be found at https://github.com/raylanger/WiltonRipples.

Appendix B: Proof of Theorems

1.1 B.1 Proof of Theorem 1

Proof

For the proofs of Theorems 1 and 2, it is more convenient to use notation which is consistent for the coefficients of the resonant mode, wave speed correction, and the non-resonant modes. That is, instead of $b_n,$ $s_n,$ and $\beta _{n,k},$ we use $\gamma _{n,k},$ where in general $\gamma _{n,k} = \beta _{n,k},$ except for $\gamma _{n,N} = b_n$ and $\gamma _{n,0} = s_n.$ Using this notation, the expansion of the solution in the small parameter $\epsilon$ is

$$\begin{aligned} f&= \sum _{n=1}^{\infty } \epsilon ^n \sum _{k=1}^{\infty } \gamma _{n,k} \phi _k = \epsilon f_1 + \sum _{n=2}^{\infty } \epsilon ^n \sum _{k=2}^{\infty } \gamma _{n,k} \phi _k\end{aligned}$$

(58)

$$\begin{aligned} c&= c_0 + s = c_0 + \sum _{n=1}^{\infty } \epsilon ^n s_n = c_0 + \sum _{n=1}^{\infty } \epsilon ^n \gamma _{n,0} \phi _0. \end{aligned}$$

(59)

Note that for all $n > 1,$ we have excluded the first mode, equivalent to setting $\gamma _{n,1} = 0,$ and that we have used the fact that the wave speed is a constant term to express it as the coefficient of the zeroth mode, since $\phi _0 = 1.$ Substituting this into (13) allows it to be rewritten as

$$\begin{aligned} \sum _{l=1}^{\infty }\sum _{m=1}^{\infty } \epsilon ^{l+m} \sum _{a=1}^{\infty } \sum _{b=0}^{\infty } \gamma _{l,a} \gamma _{m,b} \phi _a \phi _b = \sum _{n=1}^{\infty } \sum _{i=1}^{\infty } \epsilon ^n \left( \hat{{\mathcal {L}}}(i) - c_0 \right) \gamma _{n,i} \phi _i, \end{aligned}$$

where the sums over l and a produce f,  and the sum over m and b produce $s+f,$ hence the addition of the zeroth mode. The right-hand side denotes ${\mathcal {R}}[f].$ We proceed by isolating terms at each order by ensuring $l+m = n,$ or $l = n-m,$ and using identity (12), the above becomes

$$\begin{aligned} \sum _{m=1}^{n-1}\sum _{a=1}^{\infty } \sum _{b=0}^{\infty } \gamma _{m,a} \gamma _{n-m,b} \phi _a \phi _b&= \sum _{i=1}^{\infty } \left( \hat{{\mathcal {L}}}(i) - c_0 \right) \gamma _{n,i} \phi _i \nonumber \\ \frac{1}{2}\sum _{m=1}^{n-1}\sum _{a=1}^{\infty } \sum _{b=0}^{\infty } \gamma _{m,a} \gamma _{n-m,b} \left( \phi _{a+b} + \phi _{|a-b|} \right)&= \sum _{i=1}^{\infty } \left( \hat{{\mathcal {L}}}(i) - c_0 \right) \gamma _{n,i} \phi _i, \end{aligned}$$

(60)

where it is now clear that the right-hand side contains all n-th order coefficients, and the left-hand side contains all lower order coefficients.

The following is a proof by induction. We begin with the base case obtained by solving up to the third order, shown from the beginning of Sect. 4 up to the solution of Eqs. (35) and (36). For $N \ge 4,$ this produces $\gamma _{1,N} = 0,$ $\gamma _{1,0} = 0,$ $\gamma _{2,0} = -1/2{\mathcal {R}}(2),$ and $\gamma _{2,2} = 1/2{\mathcal {R}}(2).$ All other coefficients up to second order are known to be zero, except $\gamma _{2,N}.$ The non-resonant coefficients at third order would also be known, though that is irrelevant to the following proof. As expected, all coefficients satisfy Theorem 1. After the proof, we will remark on the special case of $N = 3.$

Like any induction proof, we assume that Theorem 1 is true up to some order $n-1,$ and wish to show that it continues to be true after solving at order n. As before, the unknowns at order n are $\gamma _{n,k}$ for $k \ne 0,$ 1,  or N,  $\gamma _{n-1,0},$ and $\gamma _{n-2,N}.$ The reasoning behind this will be discussed shortly.

To solve (60), we equate the coefficients of $\phi _i.$ On the left-hand side of (60), this is done by setting $a+b = i \rightarrow b = i - a,$ which is valid for $a \le i,$ and $|a-b| = i \rightarrow b = a+i$ (always valid) and $b = a-i$ (valid for $a \ge i$). Thus, we can rewrite the above to simply obtain the coefficients of $\phi _i$ as

$$\begin{aligned} \frac{1}{2}\sum _{m=1}^{n-1}\sum _{a=1}^\infty \gamma _{m,a}\left( \gamma _{n-m,a+i} + (1+\delta _{a,i})\gamma _{n-m,|a-i|}\right) =\left( \hat{{\mathcal {L}}}(i) - c_0 \right) \gamma _{n,i}, \end{aligned}$$

(61)

where $\delta _{a,i}$ is used because for $a = i,$ both $b = i-a$ and $b = a-i$ are valid, in which case $b = 0.$ In general, this is analogous to (19), though in the case of $i = 1$ or $i = N,$ we find the two solvability conditions, (32) and (33), respectively.

For $i \ne 1$ or N,  the right-hand side of (61) is non-zero, and so $\gamma _{n,i}$ is the unknown in this equation. The key observation is that m and $n-m$ will have the same parities if n is even, but different parities if n is odd. Let us first consider n to be even, and solely the term $\gamma _{m,a}\gamma _{n-m,a+i}.$ Since the orders m and $n-m$ have the same parity, the modes a and $a + i$ must also share that parity, or one of the coefficients will necessarily be zero, according to Theorem 1. This can only happen if i is even. Similar logic may be applied to $\gamma _{m,a}\gamma _{n-m,|a-i|},$ which means that if i is odd, the left-hand side of (61) will be zero, proving that $\gamma _{n,i}$ is zero for n even, i odd. The argument is nearly identical when n is odd, with the exception that now that m and $n-m$ have different parity, so too must a and $a+i$ (or a and $|a-i|$), requiring instead that i be odd, now proving that $\gamma _{n,i}$ is zero for n odd, i even.

Recall that the above argument is only sufficient for $i \ne 1$ or N. At these values of i,  $\widehat{{\mathcal {L}}}(i) - c_0 = 0,$ meaning that $\gamma _{n,i}$ does not appear on the right-hand side. Thus, we assume that $\gamma _{n,1} = 0,$ and use these equations to solve for the resonant coefficient and wave speed correction at previous orders. Furthermore, we must consider the resonant coefficient from two orders previous, as in (61) $\gamma _{n-1,N}$ is paired with $\gamma _{1,N\pm 1}$ for $i = 1$ and $\gamma _{1,N \pm N}$ for $i = N,$ all of which are zero according to the base case.

Once this is known, the argument is largely the same. It is still true that if n and i have different parities, all terms on the left-hand side of (61) will be zero, outside of those containing the unknowns $\gamma _{n-1,0}$ and $\gamma _{n-2,N}.$ The wave speed correction is solved for using $i = 1,$ shown in (41). As i is odd, we will find $\gamma _{n-1,0} = 0$ whenever n is even, which is when $n-1$ is odd. Likewise, the resonant coefficient is solved for using $i = N,$ as seen in (43), and thus we find $\gamma _{n-2,N} = 0$ whenever n and N have different parity, which coincides with when $n-2$ and N have different parity.

Thus, we have shown that if Theorem 1 is true up to some order $n-1,$ it will continue to be true at order n. Alongside the base case, this proves that Theorem 1 holds for all orders. $\square$

Remark

The above proof was specific to $N \ge 4.$ For $N = 3,$ the base case is changed which affects how the special cases $i = 1$ and $i = N = 3$ are treated. Solving up to third order for $N = 3$ produces non-zero $\gamma _{1,3},$ $\gamma _{2,4}$ and $\gamma _{2,6},$ in addition to the coefficients which were non-zero in the base case for $N = 4,$ as seen at the beginning of Sect. 3. The result is that the solvability conditions obtained from the special cases produce coupled equations for the wave speed correction and resonant coefficient, as given by the matrix in (28). As 1 and 3 are both odd, we find that both $s_{n-1}$ and $b_{n-2}$ are zero whenever n is even, in accordance with Theorem 1.

1.2 B.2 Proof of Theorem 2

Proof

For this proof, it is useful to consider an equivalent definition of Theorem 2: If n satisfies $k(N-2)\le n < (k+1)(N-2)$ for some non-negative integer k,  then the highest mode $K_N(n)$ found at ${\mathcal {O}}(\epsilon ^n)$ is $n + 2k.$ Each increase of 2 in $K_N(n)$ relative to n will henceforth be referred to as a boost, and the goal of the proof is to show that the number of boosts at each order is equal to $\left\lfloor n/(N-2)\right\rfloor .$

The highest mode solved for at ${\mathcal {O}}(\epsilon ^n)$ will be the highest mode $\phi _{a+b}$ in (61) for which $\gamma _{m,a}$ and $\gamma _{n-m,b}$ are both non-zero, for all choices of $m < n.$ To produce the largest possible sum, a and b must necessarily be the highest modes at their respective orders, and so we may define the following formula

$$\begin{aligned} K^*_N(n) = \max {(\{K_N(m) + K_N(n-m) \quad \forall \ 1 \le m \le n-1\})}, \end{aligned}$$

(62)

where $K^*_N(n)$ denotes the highest mode solved for at ${\mathcal {O}}(\epsilon ^n),$ as opposed to $K_N(n),$ which denotes the highest mode found. There is an important distinction between solved for at ${\mathcal {O}}(\epsilon ^n)$ and found at ${\mathcal {O}}(\epsilon ^n).$ Typically, they are equivalent, as for most modes we solve for their coefficients at the current order. So, if a mode is solved for at some order, it is also found at that order. For only one mode is this not true, and that is the resonant mode $\phi _N,$ as we solve for $\gamma _{n-2,N}$ at ${\mathcal {O}}(\epsilon ^n),$ as discussed in Theorem 1. $K^*_N(n)$ is still very useful as it always provides the highest known mode at ${\mathcal {O}}(\epsilon ^n)$ after solving at that order.

Because the resonant mode is a special case, we must split the proof into two parts. The first is determining at what order the resonant mode first appears. Once this is known, the solution up to that order serves as a base case for the second part of the proof, an induction proof showing that Theorem 2 holds for all higher orders.

We will show that the resonant mode first appears at ${\mathcal {O}}(\epsilon ^{N-2}),$ which is also the first instance of a boost. This requires that we solve up to ${\mathcal {O}}(\epsilon ^{N}),$ and in doing so, we will also show that for all $n<N-2,$ the highest mode present at ${\mathcal {O}}(\epsilon ^n)$ is $\phi _n.$ The proof of this follows and serves as a nice introduction to the rest of the proof. As with Theorem 1, we provide the proof for $N \ge 4,$ and remark on the case of $N = 3$ at the end.

The initial condition $f_1 = \phi _1$ serves as the base case for an induction proof showing that $K_N(n) = n$ for $n < N-2.$ From (62), it is clear that if $K_N(m) = m \ \forall \ m<n,$ then $K^*_N(n) = m+(n-m) = n.$ Alongside the initial condition, this shows that up to ${\mathcal {O}}(\epsilon ^{N-3}),$ the highest mode is the same as the order.

We proceed carefully to ${\mathcal {O}}(\epsilon ^{N-2}).$ It is also true that $K_N^*(N-2) = N-2,$ and thus after solving at that order the highest known mode is $\phi _{N-2}.$ However, $\gamma _{N-2,N}$ remains an unknown. Thus, we tentatively set $K_N(N-2) = N-2$ to be used for the subsequent orders, but note that this may be subject to change should $\gamma _{N-2,N}$ be non-zero. Using $K_N(N-2) = N-2,$ we find $K_N^*(N-1) = N-1,$ but similarly $\gamma _{N-1,N+1}$ is still unknown as it depends on $\gamma _{N-2,N},$ as seen in (40). We again set $K_N(N-1) = N-1$ while acknowledging that this is not yet certain. Proceeding to ${\mathcal {O}}(\epsilon ^N),$ we find $K^*_N(N) = N,$ meaning that at this order we solve for the resonant mode. This produces a non-zero $\gamma _{N-2,N},$ the first instance of a boost, as well as a non-zero $\gamma _{N-1,N+1}.$ While this means that we must retroactively change $K_N(N-2)$ to N,  $K_N(N-1)$ to $N+1,$ and $K_N(N)$ to $N+2,$ it is still true that ${\mathcal {O}}(\epsilon ^N)$ is the first order at which we solve for the resonant mode, which is the important part of this proof.

Now that we have shown that the resonant mode always first appears at ${\mathcal {O}}(\epsilon ^{N-2}),$ we may proceed to the rest of the proof of Theorem 2. Recall that we have also shown that for all orders $m < N-2,$ $K_N(m) = m,$ in accordance with Theorem 2. We aim to show that the number of boosts only increases at orders which are multiples of $N-2,$ and otherwise the highest mode increases by only 1 with the order. To do so, we show that all orders of the form ${\mathcal {O}}(\epsilon ^{k(N-2)+p}),$ with k and $p \in {\mathbb {Z}}$ and $0\le p<N-2,$ have k boosts.

Now that we are only considering orders $n > N - 2,$ $K^*_N(n)$ will always be greater than N,  and thus we may use $K^*_N(n) = K_N(n).$ To simplify, we let m in (62) take on the form

$$\begin{aligned} m = j(N-2)+q,\quad j \in \{0,1,2\ldots k\},\ q\in \{0,1,2\ldots N-3\}, \end{aligned}$$

(63)

such that the corresponding order always has j boosts. Substituting (63) and $n = k(N-2)+p$ into $n-m,$ we obtain

$$\begin{aligned} k(N-2)-m+p = (k-j)(N-2)+p-q. \end{aligned}$$

(64)

The number of boosts at this order is less clear, as it depends on the values of p and q. Substituting (63) and (64) into (44) produces

$$\begin{aligned} K_N(m) = j(N-2)+q + 2\left\lfloor \frac{j(N-2)+q}{N-2}\right\rfloor = j(N-2)+q + 2j \end{aligned}$$

(65)

and

$$\begin{aligned} K_N\bigl (k(N-2)-m+p\bigr ) = (k-j)(N-2)+p-q + 2\left\lfloor \frac{(k-j)(N-2)+p-q}{N-2}\right\rfloor . \nonumber \\ \end{aligned}$$

(66)

Finally, substitution of (65) and (66) into (62) produces

$$\begin{aligned} K_N^*\bigl (k(N-2)+p\bigr ) = \max {\Biggl (k(N-2)+p+2j+2\left\lfloor \frac{(k-j)(N-2)+p-q}{N-2}\right\rfloor \Biggr )}, \nonumber \\ \end{aligned}$$

(67)

with the maximum being taken over all values of j and q. The j terms will cancel once a value of q is chosen, so we simply must maximize the value of the floor function, which is achieved by choosing $q\le p.$ Thus, we may express the highest mode

$$\begin{aligned} K_N^*\bigl (k(N-2)+p\bigr ) = K_N\bigl (k(N-2)+p\bigr ) = k(N-2)+p+2k, \end{aligned}$$

(68)

which evidently has k boosts.

Note that if $p = 0,$ that is, if the order is a multiple of $N-2,$ then only $q = 0$ satisfies $q \le p.$ As p increases, the number of q which satisfy that inequality increases, but this still only leads to k boosts, and as such increasing p by 1 only increases the highest mode by 1. Additionally, if we let $p = -1,$ then no values of q satisfy $q \le p,$ and thus that order will have $k-1$ boosts, showing that the number of boosts does increment at orders which are exactly multiples of $N-2.$ This completes the proof of Theorem 2. $\square$

Remark

For $N = 3,$ we begin with $f_1 = \phi _1 + b_1 \phi _3,$ meaning that the preliminary step of determining when the resonant mode first appears is unnecessary. Furthermore, as $N - 2 = 1,$ we expect the number of boosts to increment at every order, and thus the highest mode should increase by 3 every time. This can simply be shown through an induction proof, using $K_3(1) = 3$ as a base case. If we assume that $K_3(m) = 3m\ \forall \ m < n,$ then (62) returns $K^*_3(n) = K_3(n) = 3n.$ This is sufficient to show that for $N = 3,$ the highest mode is always 3 times the order, in accordance with Theorem 2.