Global Solutions and Stability Properties of the 5th Order Gardner Equation

Abstract

In this work, we deal with the initial value problem of the 5th-order Gardner equation in ${\mathbb {R}}$, presenting the local well-posedness result in $H^2({\mathbb {R}})$. As a consequence of the local result, in addition to $H^2$-energy conservation law, we are able to prove the global well-posedness result in $H^2({\mathbb {R}})$. Finally as a direct application, we prove that some globally defined functions, e.g. breather solutions of 5th order Gardner equation, are $H^2({\mathbb {R}})$ stable.

Appendices

Appendix A. Proof of Theorem 1.4

The aim of this section is to prove the weak ill-posedness of (1.1) for $s > 0$, which, in addition to the first author’s recent work [4], completely justifies that the 5th Gardner Eq. (1.1) is a quasilinear equation in the sense that the flow map from data to solutions is not (locally) uniformly continuous for all regularities, see Corollary 1.5. Since the weak-illposedness phenomenon occurs due to the strong high-low interaction in the quadratic nonlinearity with three derivatives, Theorem 1.2 in [49] seems to guarantee the lack of uniform continuity of the flow map associated to (1.1) for $s > 0$. This section contributes to prove that the Eq. (1.1) is indeed weakly ill-posed for $s > 0$.

The proof basically follows the argument used in [49], initially introduced by Koch-Tzvetkov [45]. Since the (weak) ill-posedness phenomenon arises from the strong high-low quadratic nonlinearity (high frequency waves with low frequency perturbations), the main part of the proof is identical to the argument in [49]. Thus, we, here, provide an additional estimate to be needed for the other nonlinearities.

In view of the argument presented in Sect. 2.6, it suffices to show the ill-posedness of (2.1) with small initial data.

1.1 A.1 Setting

We first define the approximate solution, which is an ansatz to cause the (weak) ill-posedness phenomenon. Let $\phi , \widetilde{\phi } \in C_0^{\infty }({\mathbb {R}})$ be smooth bump functions satisfying

$$\begin{aligned}\phi \equiv 1, \quad |x| < 1, \quad \text{ and } \quad \phi \equiv 0, \quad |x| > 2\end{aligned}$$

and

$$\begin{aligned}\widetilde{\phi } \equiv 1, \quad x \in \text{ supp } (\phi ) \quad \text{ and } \quad \widetilde{\phi }\phi \equiv \phi ,\end{aligned}$$

respectively. For $N \ge 1$ and $0< \delta < 1$, set

$$\begin{aligned}\phi _{N}(x) := \phi \left( \frac{x}{N^{4+\delta }}\right) , \quad \widetilde{\phi }_{N}(x) := \widetilde{\phi }\left( \frac{x}{N^{4+\delta }}\right) .\end{aligned}$$

Let $\epsilon > 0$ be a sufficiently small for the initial data to satisfy (2.37). Let

$$\begin{aligned}u_{0,l}^{\pm }(x) := \pm \epsilon N^{-3}\widetilde{\phi }_{N}(x)\end{aligned}$$

and $u_l^{\pm }(t,x)$ be the solution to (2.1) with the initial data $u_{0,l}^{\pm }(x)$. Let $\Phi _N(t) := (N^5-10\mu ^2\lambda ^2N^3)t$ and

$$\begin{aligned} u_h^{\pm }(t,x) := N^{-\frac{4+\delta }{2}-s}\phi _{N}(x) \cos \left( N x -\Phi _N(t) {\mp } t\right) \end{aligned}$$

(A.1)

be a high frequency part of the approximate solution, and thus define the approximate solution as

$$\begin{aligned}u_{ap}^{\pm }(t,x):=u_l^{\pm }(t,x) + u_h^{\pm }(t,x).\end{aligned}$$

Then the main task is to prove the following proposition:

Proposition A.1

(Proposition 6.2 in [49]) Let $\max (0,2-2s)< \delta < 1$. Let $u_{N}^{\pm }$ be the unique solution to (2.1) with initial data

$$\begin{aligned}u_{N}^{\pm }(0,x) = \pm \epsilon N^{-3}\widetilde{\phi }_{N}(x) + N^{-\frac{4+\delta }{2}-s}\phi _{N}(x) \cos \left( N x \right) .\end{aligned}$$

Then, we have

$$\begin{aligned} \left\Vert u_{N}^{\pm } - u_{ap}^{\pm }\right\Vert _{H^s} = o(1), \end{aligned}$$

(A.2)

for $s > 0$ and $|t| <1$, as $N \rightarrow \infty$.

Once (A.2) holds true, one conclude that

$$\begin{aligned}\begin{aligned} \left\Vert u_{N}^+ - u_{N}^-\right\Vert _{H^s} =&~{} N^{-\frac{4+\delta }{2}-s}\\&\quad \left\Vert \phi _{N}(x)\left( \cos \left( N x -\Phi _N(t) + t\right) -\cos \left( N x -\Phi _N(t) - t\right) \right) \right\Vert _{H^s} + o(1)\\ =&~{} 2N^{-\frac{4+\delta }{2}-s}\left\Vert \phi _{N}(x)\sin \left( N x -\Phi _N(t)\right) \right\Vert _{H^s}|\sin t| + o(1), \end{aligned}\end{aligned}$$

which, in addition to Lemma A.2 below, implies

$$\begin{aligned}\lim _{N \rightarrow \infty }\left\Vert u_{N}^+ - u_{N}^-\right\Vert _{H^s} \ge c|\sin t| \sim c|t|,\end{aligned}$$

for $|t| < 1$. This completes the proof of Theorem 1.4.

We recall from [45, 49] the following useful lemmas to prove Proposition A.1.

Lemma A.2

(Lemma 2.3 in [45]) Let $s \ge 0$, $\delta > 0$ and $\gamma \in {\mathbb {R}}$. Then,

$$\begin{aligned}\lim _{N \rightarrow \infty } N^{-\frac{4+\delta }{2}-s}\left\Vert \phi _{N}(x)\sin \left( N x + \gamma \right) \right\Vert _{H^s} = c_0 \left\Vert \phi \right\Vert _{L^2},\end{aligned}$$

for some $c_0 >0$.

Lemma A.3

(Lemma 6.3 in [49]) Let K be a positive integer and $K-2-s \ge k \ge 0$. Then, we have

$$\begin{aligned} \left\Vert \partial _x^ku_l^{\pm }(t,\cdot )\right\Vert _{L^2} \lesssim _K N^{-\frac{2-\delta }{2}-k(4+\delta )} \end{aligned}$$

(A.3)

$$\begin{aligned} \left\Vert \partial _x^ku_l^{\pm }(t,\cdot )\right\Vert _{L^{\infty }} \lesssim _K N^{-3-k(4+\delta )} \end{aligned}$$

(A.4)

$$\begin{aligned} \left\Vert u_l^{\pm }(t,\cdot )-u_{0,l}^{\pm }(\cdot )\right\Vert _{L^2} \lesssim _K N^{-15-3\delta } \end{aligned}$$

(A.5)

Proof

The proof of (A.3) and (A.4) follows from a direct computation and Theorem 1.2, in particular, a priori bound (2.38). Moreover, the proof of (A.5) follows from a direct calculation in (2.1) and (A.3)–(A.4). The proof is almost identical to the proof of Lemma 6.3 in [49], thus we omit the details. $\square$

Lemma A.4

Let

$$\begin{aligned} {\mathcal {P}}^{\pm }(t,x):=u_{ap, t}^{\pm }+u_{ap, 5x}^{\pm }+10\mu ^2\lambda ^2u_{ap, 3x}^{\pm } +{\mathcal {N}}_2(u_{ap}^{\pm }) + {\mathcal {N}}_3(u_{ap}^{\pm }) + {{\mathcal {S}}}{{\mathcal {N}}}(u_{ap}^{\pm }), \end{aligned}$$

(A.6)

where ${\mathcal {N}}_2(\cdot )$, ${\mathcal {N}}_3(\cdot )$ and ${{\mathcal {S}}}{{\mathcal {N}}}(\cdot )$ are defined as in (2.2)–(2.3), respectively. Let $s > 0$, $0<\delta <2$ and $|t| \le 1$. Then, we have

$$\begin{aligned} \left\Vert {\mathcal {P}}^{\pm }(t,\cdot )\right\Vert _{L^2} \lesssim N^{-s-\delta } + N^{\frac{2-\delta }{2} - 2s} + N^{-1-\delta -3s} + N^{1-\frac{3(4+\delta )}{2}-4s} +N^{1-2(4+\delta )-5s}. \end{aligned}$$

(A.7)

Moreover, if $\sigma > 0$, we have

$$\begin{aligned} \left\Vert {\mathcal {P}}^{\pm }(t,\cdot )\right\Vert _{H^{\sigma }}\lesssim & {} N^{-s-\delta +\sigma } + N^{\frac{2-\delta }{2} - 2s + \sigma } + N^{-1-\delta -3s + \sigma } \nonumber \\&+ N^{1-\frac{3(4+\delta )}{2}-4s + \sigma } +N^{1-2(4+\delta )-5s + \sigma }. \end{aligned}$$

(A.8)

Proof

It suffices to consider ${\mathcal {P}}^{+}$, since an identical argument holds true for ${\mathcal {P}}^{-}$. We drop the super-index $+$. We decompose ${\mathcal {P}}$ into ${\mathcal {P}}_1+{\mathcal {P}}_2$, where ${\mathcal {P}}_2= {\mathcal {N}}_3(u_{ap}) -{\mathcal {N}}_3(u_l) + {{\mathcal {S}}}{{\mathcal {N}}}(u_{ap}) - {{\mathcal {S}}}{{\mathcal {N}}}(u_l)$ and ${\mathcal {P}}_1 = {\mathcal {P}} -{\mathcal {P}}_2$. Lemma 6.4 in [49] exactly shows (A.7) and (A.8) for ${\mathcal {P}}_1$¹³. Our setting of $\phi$, $\widetilde{\phi }$ and $u_l$ is essential to deal with

$$\begin{aligned}\Lambda := \epsilon N^{-\frac{4+\delta }{2}-s}\phi _N(x)(\partial _t+\partial _x^5+10\mu ^2\lambda ^2\partial _x^3 + \epsilon ^{-1}u_l\partial _x^3)\cos \left( N x -\Phi _N(t) - t\right) \end{aligned}$$

contained in ${\mathcal {P}}_1$ (compared to $F_4$ in the proof of Lemma 6.4 in [49]). Indeed, a direct calculation in addition to $u_{0,l}(x) := \epsilon N^{-3}\widetilde{\phi }_{N}(x)$ and $\phi \widetilde{\phi } = \phi$ gives

$$\begin{aligned}\begin{aligned} \Lambda =&~{} N^{-\frac{4+\delta }{2}-s}\phi _N(x)\left( u_lN^3 - \epsilon \right) \sin \left( N x -\Phi _N(t) - t\right) \\ =&~{} N^{-\frac{4+\delta }{2}-s}N^3\phi _N(x)\left( u_l - u_{0,l} \right) \sin \left( N x -\Phi _N(t) - t\right) , \end{aligned}\end{aligned}$$

which is handled by using (A.5). Thus, it suffices to show (A.7) and (A.8) for ${\mathcal {P}}_2$. Putting first $u_{ap} = u_l + u_h$ into $10u_{ap}^2u_{ap, 3x} - 10 u_l^2u_{l, x}$ in ${\mathcal {P}}_2$, one has

$$\begin{aligned} 10u_l^2u_{h, xxx} + 20u_lu_hu_{l, xxx} + 20u_lu_hu_{h, xxx} + 10u_h^2u_{l, xxx} + 10u_h^2u_{h, xxx}. \end{aligned}$$

(A.9)

Note that

$$\begin{aligned}\begin{aligned} u_{h,x} =&~{} N^{-\frac{4+\delta }{2}-s}\left( \partial _x\phi _{N}(x) \cos \left( N x -\Phi _N(t) - t\right) +\phi _{N}(x) \partial _x\cos \left( N x -\Phi _N(t) - t\right) \right) \\ =&~{}N^{-\frac{4+\delta }{2}-s}\left( N^{-(4+\delta )}\phi _{N,x}(x)\cos \left( N x -\Phi _N(t) - t\right) \right. \\&\quad \left. - N \phi _{N}(x)\sin \left( N x -\Phi _N(t) - t\right) \right) . \end{aligned}\end{aligned}$$

Thus, one can see that the worst term arises from the case when the derivative acts on $\cos \left( N x -\Phi _N(t) - t\right)$. Using Lemmas A.2 and A.3 , one estimates

$$\begin{aligned} \left\Vert (A.9)\right\Vert _{L^2} \lesssim N^{-3-s} + N^{-\frac{4+\delta }{2}-2s} + N^{-1-\delta -3s}.\end{aligned}$$

An analogous argument yield

$$\begin{aligned} \left\Vert u_{ap, x}^3 - u_{l, x}^3\right\Vert _{L^2}\lesssim & {} N^{-5-2(4+\delta )-s} + N^{-1-\frac{3(4+\delta )}{2}-2s} + N^{-1-\delta -3s},\\ \left\Vert u_{ap}u_{ap, x}u_{ap, xx} - u_{l}u_{l, x}u_{l, xx}\right\Vert _{L^2}\lesssim & {} N^{-8-\delta -s} + N^{-\frac{4+\delta }{2}-2s} + N^{-1-\delta -3s},\\ \left\Vert u_{ap}^4u_{ap,x} - u_{l}^4u_{l,x}\right\Vert _{L^2}\lesssim & {} N^{-11-s} + N^{-8-\frac{4+\delta }{2}-2s} + N^{-5-(4+\delta )-3s} \\&+ N^{-2-\frac{3(4+\delta )}{2}-4s}+ N^{1-2(4+\delta )-5s},\\ \left\Vert u_{ap}u_{ap, x} - u_{l}u_{l, x}\right\Vert _{L^2}\lesssim & {} N^{-2-s} + N^{1-\frac{4+\delta }{2}-2s}\end{aligned}$$

and

$$\begin{aligned}\left\Vert u_{ap}^3u_{ap, x} - u_{l}^3u_{l, x}\right\Vert _{L^2} \lesssim N^{-8-s} + N^{-5-\frac{4+\delta }{2}-2s} +N^{-2-(4+\delta )-3s}+N^{1-\frac{3(4+\delta )}{2}-4s}.\end{aligned}$$

Collecting all, we completes the proof of (A.7). Moreover, the fractional Leibniz rule ensure at least $\left\Vert {\mathcal {P}}\right\Vert _{{\dot{H}}^{\sigma }} \lesssim _{\sigma } N^{\sigma } \left\Vert {\mathcal {P}}\right\Vert _{L^2}$, which in addition to (A.7) implies (A.8), since $u_{l, t}+u_{l, 5x}+10\mu ^2\lambda ^2u_{l, 3x} + 30\mu ^4 \lambda ^4 u_{l, x} +{\mathcal {N}}_2(u_l) + {\mathcal {N}}_3(u_l) + {{\mathcal {S}}}{{\mathcal {N}}}(u_l) = 0$ and the others contains at least one $u_h$. We complete the proof. $\square$

1.2 A.2. Proof of Proposition A.1

Let $w^{\pm } := u_N^{\pm } - u_{ap}^{\pm }$. We only show $\left\Vert w^+\right\Vert _{H^s} = o(1)$ as $N \rightarrow \infty$ and drop the super-index $+$. For $s \ge 2$, the local well-posedness theory is available. A direct calculation gives

$$\begin{aligned}\Gamma w + {\mathcal {N}}_2(u_N) - {\mathcal {N}}_2(u_{ap}) + {\mathcal {N}}_3(u_N) - {\mathcal {N}}_3(u_{ap}) + {{\mathcal {S}}}{{\mathcal {N}}}(u_N) - {{\mathcal {S}}}{{\mathcal {N}}}(u_{ap}) + {\mathcal {P}} =0,\end{aligned}$$

where $\Gamma := \partial _t+\partial _x^5 +10\mu ^2\lambda ^2 \partial _x^3$ and ${\mathcal {P}}$ is as in (A.6). For $2 \le \sigma$, the local well-posedness, in particular (2.38), ensures

$$\begin{aligned} \left\Vert u_N\right\Vert _{C_TH^\sigma } + \left\Vert u_N\right\Vert _{F^\sigma (T)} \lesssim \left\Vert u_N(0)\right\Vert _{H^{\sigma }} \lesssim N^{\sigma -s}. \end{aligned}$$

(A.10)

Moreover, a direct calculation and the local theory (for $u_l$) gives

$$\begin{aligned} \left\Vert u_{ap}\right\Vert _{C_TH^\sigma } + \left\Vert u_{ap}\right\Vert _{F^\sigma (T)} \lesssim N^{-\frac{2-\delta }{2}} + N^{\sigma -s}. \end{aligned}$$

(A.11)

Using Propositions 2.15, Propositions 2.7, 2.8, 2.9 and 2.13 , and (A.7) under (A.10) and (A.11), one concludes

$$\begin{aligned}\left\Vert w\right\Vert _{F^0(T)} \lesssim \left\Vert {\mathcal {P}}\right\Vert _{L_T^1L_x^2} = O(N^{-s -\beta }),\end{aligned}$$

for $\beta = \min (\delta , -\frac{2-\delta }{2}+s) > 0$, which, in addition to Proposition 2.14, implies

$$\begin{aligned} \left\Vert w\right\Vert _{L_T^{\infty }L_x^2} = O(N^{-s -\beta }). \end{aligned}$$

(A.12)

Furthermore, an analogous argument (but using (A.8) instead of (A.7)) in addition to

$$\begin{aligned}\left\Vert u_{ap}\right\Vert _{F^2s(T)}\left\Vert w\right\Vert _{F^0(T)} = O(N^{s}N^{-s-\beta }) = O(N^{-\beta }),\end{aligned}$$

ensures $\left\Vert w\right\Vert _{F^s(T)} = O(N^{-\beta })$, which concludes (A.2) as $N \rightarrow \infty$ for $s \ge 2$.

To fill the regularity range $0< s < 2$, we use the conservation law and the interpolation theorem. $H^2$ conservation law (2.39) and a direct calculation yield

$$\begin{aligned}\left\Vert u_N\right\Vert _{H^2} \lesssim N^{2-s} \quad \text{ and } \quad \left\Vert u_{ap}\right\Vert _{H^2} \lesssim N^{2-s},\end{aligned}$$

respectively, which concludes

$$\begin{aligned} \left\Vert w\right\Vert _{H^2} \lesssim N^{2-s}. \end{aligned}$$

(A.13)

The interpolation between (A.12) and (A.13) ensures

$$\begin{aligned}\left\Vert w\right\Vert _{H^s} \lesssim \left\Vert w\right\Vert _{L^2}^{1-\frac{s}{2}}\left\Vert w\right\Vert _{H^3}^{\frac{s}{2}} \lesssim N^{-\frac{\beta (2-s)}{2}},\end{aligned}$$

which proves (A.2) as $N \rightarrow \infty$ for $0< s < 2$.

Appendix B. Proof of Lemma 3.2.

We are going to prove the identity (3.2)

$$\begin{aligned} {{\tilde{B}}}_{\mu ,t} = (\alpha ^2 + \beta ^2)^2 B_\mu + 2\big (\alpha ^2 - \beta ^2 - 5\mu ^2\big )\big (B_{\mu ,xx} + 2B_\mu ^3 + 6\mu B_\mu ^2\big ). \end{aligned}$$

Firstly and for the sake of simplicity, we will use the following notation:

$$\begin{aligned}&A_1:= (\alpha ^2+\beta ^2)^2 ,\quad A_2:= 2(\alpha ^2 - \beta ^2 - 5\mu ^2),\\&\Delta =\alpha ^2+\beta ^2-4\mu ^2,\quad e^{z} = \cosh (z) + \sinh (z),\\&D:= f^2 + g^2,\quad \text {where}\quad f, g~~\text {and its derivatives are given by:} \end{aligned}$$

$$\begin{aligned} \begin{aligned}&f=\cosh (\beta y_2)-\frac{2\beta \mu }{\alpha \sqrt{\alpha ^2+\beta ^2}\sqrt{\Delta }}(\alpha \cos (\alpha y_1) -\beta \sin (\alpha y_1) ),\\&f_1:=f_x=\beta \sinh (\beta y_2)+\frac{2\beta \mu }{\sqrt{\alpha ^2+\beta ^2}\sqrt{\Delta }}(\beta \cos (\alpha y_1) +\alpha \sin (\alpha y_1)), \\&f_2:=f_t=\beta \gamma _5\sinh (\beta y_2)+\frac{2\beta \delta _5\mu }{\sqrt{\alpha ^2+\beta ^2}\sqrt{\Delta }}(\beta \cos (\alpha y_1) +\alpha \sin (\alpha y_1) ),\\&f_3:=f_{xx}=\beta ^2\cosh (\beta y_2)+\frac{2\alpha \beta \mu }{\sqrt{\alpha ^2+\beta ^2}\sqrt{\Delta }}(-\alpha \cos (\alpha y_1) +\beta \sin (\alpha y_1) ),\\&f_4:=f_{xxx}=\beta ^3\sinh (\beta y_2)-\frac{2\alpha ^2\beta \mu }{\sqrt{\alpha ^2+\beta ^2}\sqrt{\Delta }}(\beta \cos (\alpha y_1) +\alpha \sin (\alpha y_1) ) \end{aligned} \end{aligned}$$

(B.1)

and

$$\begin{aligned} \begin{aligned}&g=\frac{\beta \sqrt{\alpha ^2+\beta ^2}}{\alpha \sqrt{\Delta }}\sin (\alpha y_1) - \frac{2\beta \mu e^{\beta y_2}}{\Delta },\\&g_1:=g_x=\frac{\beta \sqrt{\alpha ^2+\beta ^2}}{\sqrt{\Delta }}\cos (\alpha y_1) - \frac{2\beta ^2\mu e^{\beta y_2}}{\Delta },\\&g_2:=g_t=\frac{\beta \delta _5\sqrt{\alpha ^2+\beta ^2}}{\sqrt{\Delta }}\cos (\alpha y_1) - \frac{2\beta ^2\gamma _5\mu e^{\beta y_2}}{\Delta },\\&g_3:=g_{xx}=-\frac{\alpha \beta \sqrt{\alpha ^2+\beta ^2}}{\sqrt{\Delta }}\sin (\alpha y_1) - \frac{2\beta ^3\mu e^{\beta y_2}}{\Delta },\\&g_4:=g_{xxx}=-\frac{\alpha ^2\beta \sqrt{\alpha ^2+\beta ^2}}{\sqrt{\Delta }}\cos (\alpha y_1) - \frac{2\beta ^4\mu e^{\beta y_2}}{\Delta }, \end{aligned} \end{aligned}$$

(B.2)

where velocities $(\gamma _5,\delta _5)$ are given in (1.7). From the explicit expression of the breather solution (1.7) but now written in terms of the above derivatives (B.1)–(B.2), we obtain that:

$$\begin{aligned} B_\mu = 2\frac{g_1f-f_1g}{D} \qquad \text {and}\qquad {{\tilde{B}}}_{\mu ,t} = 2\frac{g_2f-f_2g}{D}. \end{aligned}$$

(B.3)

Moreover we get

$$\begin{aligned} B_\mu ^2 = 4\left( \frac{g_1f-f_1g}{D}\right) ^2\qquad \text {and}\qquad B_\mu ^3 = 8\left( \frac{g_1f-f_1g}{D}\right) ^3. \end{aligned}$$

(B.4)

Now, we compute $B_{\mu ,xx}$. First we get

$$\begin{aligned}B_{\mu ,x} = -\frac{2}{D^2}\left( f^3 g_3-f^2 (2 f_1 g_1+f_3 g)+f g \left( 2 f_1^2+g g_3-2 g_1^2\right) +g^2 (2 f_1 g_1-f_3 g)\right) ,\end{aligned}$$

and then

$$\begin{aligned} B_{\mu ,xx} = 2\frac{M_1}{D^3}, \end{aligned}$$

(B.5)

where

$$\begin{aligned} \begin{aligned} M_1:=&\Big (f^5 g_4-f^4 (3 f_1 g_3+3 f_3 g_1+f_4 g)\\&+2 f^3 \left( 3 f_1^2 g_1+3 f_1 f_3 g+g^2 g_4-3 g g_1 g_3-g_1^3\right) \\&-2 f^2 g \left( 3 f_1^3-9 f_1 g_1^2+f_4 g^2\right) \\&+f g^2 \left( -18 f_1^2 g_1+6 f_1 f_3 g+g^2 g_4-6 g g_1 g_3+6 g_1^3\right) \\&+g^3 \left( 2 f_1^3+f_1 \left( 3 g g_3-6 g_1^2\right) +g (3 f_3 g_1-f_4 g)\right) \Big ), \end{aligned} \end{aligned}$$

(B.6)

and therefore from (B.3), (B.4), (B.5) and (B.6), we get

$$\begin{aligned} A_1B_\mu + A_2(B_{\mu ,xx} + 2B_\mu ^3 + 6\mu B_\mu ^2) = \frac{M_2}{D^3}, \end{aligned}$$

(B.7)

where

$$\begin{aligned} M_2:=2\Big (A_1D^2(fg_1-f_1g) + A_2(8(fg_1-f_1g)^3 + 12\mu D(f_1g-fg_1)^2 + M_1)\Big ),\nonumber \\ \end{aligned}$$

(B.8)

Now, we verify that, after expanding $f's$ and $g's$ terms (B.1)–(B.2) and lengthy rearrangements, the above term (B.8) simplifies as follows:

$$\begin{aligned}M_2 = 2D^2(g_2f-gf_2).\end{aligned}$$

Finally, remembering (B.7), we have that

$$\begin{aligned}A_1B_\mu + A_2(B_{\mu ,xx} + 2B_\mu ^3 + 6\mu B_\mu ^2) = \frac{M_2}{D^3} = \frac{2D^2(g_2f-gf_2)}{D^3} = {\tilde{B}}_{\mu ,t},\end{aligned}$$

and we conclude.