Periodic Solutions to Nonlinear Wave Equation with X-Dependent Coefficients Under the General Boundary Conditions

Abstract

In this paper we consider a class of nonlinear wave equations with x-dependent coefficients and prove existence of families of time-periodic solutions under the general boundary conditions. Such a model arises from the forced vibrations of a nonhomogeneous string and the propagation of seismic waves in nonisotropic media. The proof is based on a Lyapunov–Schmidt reduction together with a differentiable Nash–Moser iteration scheme.

Appendix

1.1 Proof of Remark 2.2

(i) Decomposing uv as

$$\begin{aligned} uv=\sum _{l\in \mathbf {Z}}\left( \sum _{k\in \mathbf {Z}}u_{l-k}v_{k}\right) e^{\mathrm {i}lt},\quad \forall u,v\in H^s \end{aligned}$$

and using the Cauchy inequality may yields

$$\begin{aligned} \Vert uv\Vert ^2_{s}&=\sum \limits _{l}(1+l^{2s})\Vert \sum \limits _{k}u_{l-k}v_{k}\Vert ^2_{H^1}\le \sum \limits _{l}\left( \sum \limits _{k}(1+l^{2s})^{\frac{1}{2}} c_{lk}\Vert u_{l-k}v_{k}\Vert _{H^1}\frac{1}{c_{lk}}\right) ^2\\&{\le }\sum \limits _{l}\left( \sum \limits _{k}\frac{1}{c^2_{lk}}\right) \left( \sum \limits _{k}\Vert u_{l-k}\Vert ^2_{H^1}(1+|l-k|^{2s})\Vert v_{k}\Vert ^2_{H^1}(1+k^{2s})\right) , \end{aligned}$$

where

$$\begin{aligned} c_{lk}:=\sqrt{\frac{(1+k^{2s})(1+|l-k|^{2s})}{1+l^{2s}}}. \end{aligned}$$

A simple process yields

$$\begin{aligned} 1+l^{2s}&\le 1+(|k|+|l-k|)^{2s}\le 1+2^{2s-1}(k^{2s}+|l-k|^{2s})\\&<2^{2s-1}(1+k^{2s}+1+|l-k|^{2s}), \end{aligned}$$

which leads to

$$\begin{aligned} \sum \limits _{k\in \mathbf {Z}}\frac{1}{c^2_{lk}}<2^{2s-1}\left( \sum \limits _{k\in \mathbf {Z}}\frac{1}{1+k^{2s}}+ \sum \limits _{k\in \mathbf {Z}}\frac{1}{1+|l-k|^{2s}}\right) =2^{2s}\sum \limits _{k\in \mathbf {Z}}\frac{1}{1+k^{2s}}{\mathop {=:}\limits ^{s>1/2}}C(s)^2. \end{aligned}$$

Hence $\Vert uv\Vert _{s}$ may be bounded from above by $C(s)\Vert u\Vert _s\Vert v\Vert _s$.

(ii) It follows from the Cauchy inequality that

$$\begin{aligned} \sum \limits _{l\in \mathbf {Z}}\Vert u_l\Vert _{L^{\infty }}\le \mathrm {c}\sum \limits _{l\in \mathbf {Z}}\Vert u_l\Vert _{H^1}\le \mathrm {c}\left( \sum \limits _{l\in \mathbf {Z}}\Vert u_l\Vert _{H^1}(1+l^{2s})\right) ^{\frac{1}{2}}\left( \sum \limits _{l\in \mathbf {Z}}\frac{1}{1+l^{2s}}\right) ^{\frac{1}{2}}\le C(s)\Vert u\Vert _{s}. \end{aligned}$$

1.2 Preliminaries

By the definitions of $H^{s}$, for completeness, we list Lemmas 5.1–5.3 and the proof can be found in [8].

Lemma 5.1

(Moser–Nirenberg) For all $u_1,u_2\in H^{s'}\cap H^s$ with $s'\ge 0$ and $s>\frac{1}{2}$, we have

$$\begin{aligned} \Vert u_1u_2\Vert _{{s'}}&\le C(s')\left( \Vert u_1\Vert _{L^{\infty }(\mathbf {T},H^1(0,\pi ))}\Vert u_2\Vert _{{s'}}+\Vert u_1\Vert _{{s'}}\Vert u_2\Vert _{L^{\infty }(\mathbf {T},H^1(0,\pi ))}\right) \end{aligned}$$

(5.1)

$$\begin{aligned}&\le C(s')\left( \Vert u_1\Vert _{s}\Vert u_2\Vert _{{s'}}+\Vert u_1\Vert _{{s'}}\Vert u_2\Vert _{s}\right) . \end{aligned}$$

(5.2)

Lemma 5.2

(Logarithmic convexity) Let $0\le \mathfrak {a}'\le \mathfrak {a}\le \mathfrak {b}\le \mathfrak {b}'$ satisfy $\mathfrak {a}+\mathfrak {b}=\mathfrak {a}'+\mathfrak {b}'$. Taking $\mathfrak {p}:=\frac{\mathfrak {b}'-\mathfrak {a}}{\mathfrak {b}'-\mathfrak {a}'}$, one has

$$\begin{aligned} \Vert u_1\Vert _{\mathfrak {a}}\Vert u_2\Vert _{\mathfrak {b}}\le \mathfrak {p}\Vert u_1\Vert _{\mathfrak {a}'}\Vert u_2\Vert _{\mathfrak {b}'} +(1-\mathfrak {p})\Vert u_2\Vert _{\mathfrak {a}'}\Vert u_1\Vert _{\mathfrak {b}'},\quad \forall u_1,u_2\in \mathcal {H}^{\mathfrak {b}'}. \end{aligned}$$

In particular

$$\begin{aligned} \Vert u\Vert _{\mathfrak {a}}\Vert u\Vert _{\mathfrak {b}}\le \Vert u\Vert _{\mathfrak {a}'}\Vert u\Vert _{\mathfrak {b}'}, \quad \forall u\in \mathcal {H}^{\mathfrak {b}'}. \end{aligned}$$

(5.3)

Let $\mathscr {C}_k$ denote the following space composed by the space-independent functions:

$$\begin{aligned} \mathscr {C}_k:=\left\{ f\in C([0,\pi ]\times \mathbf {R};\mathbf {R}):~u\mapsto f(\cdot ,u)~{\mathrm {belongs}}~{\mathrm { to}}~C^{k}(\mathbf {R};H^{1}(0,\pi ))\right\} . \end{aligned}$$

Lemma 5.3

Let $f\in \mathscr {C}_1$, $\mathfrak {C}:=\Vert u\Vert _{L^{\infty }(0,\pi )}$. Then the composition operator $u(x)\mapsto f(x,u(x))$ belongs to $C(H^1(0,\pi );H^1(0,\pi ))$ with

$$\begin{aligned} \Vert f(x,u(x))\Vert _{H^1}\le C\Big (\max _{u\in [-\mathfrak {C},\mathfrak {C}]}\Vert f(\cdot ,u)\Vert _{H^1}+\max _{u\in [-\mathfrak {C},\mathfrak {C}]}\Vert \partial _uf(\cdot ,u)\Vert _{H^1}\Vert u\Vert _{H^1}\Big ). \end{aligned}$$

In particular one has

$$\begin{aligned} \Vert f(x,0)\Vert _{H^1}\le C. \end{aligned}$$

With the help of Lemmas 5.1–5.3, the following lemma can be obtained.

Lemma 5.4

Let $f\in \mathcal {C}_{k}$ with $k\ge 1$. Then, for all $s>\frac{1}{2},0\le s'\le k-1$, the composition operator $u(t,x)\mapsto f(t,x,u(t,x))$ is in $C(H^{s}\cap {H}^{s'};{H}^{s'})$ with

$$\begin{aligned} \Vert f(t,x,u)\Vert _{{s'}}\le C(s',\Vert u\Vert _{s})(1+\Vert u\Vert _{{s'}}). \end{aligned}$$

(5.4)

Proof

If $s'=\mathrm {p}$ is an integer, for all $\mathrm {p}\in \mathbf {N}$ with $\mathrm {p}\le k-1$, $u\in H^{s}\cap H^{\mathrm {p}}$, we have to prove

$$\begin{aligned} \Vert f(t,x,u)\Vert _{\mathrm {p}}\le C(\mathrm {p},\Vert u\Vert _{s})(1+\Vert u\Vert _{\mathrm {p}}) \end{aligned}$$

(5.5)

and that

$$\begin{aligned} f(t,x,u_n)\rightarrow f(t,x,u) \quad \text {as } u_n\rightarrow u ~\text {in}~ {H}^{s}\cap {H}^{\mathrm {p}}. \end{aligned}$$

(5.6)

Let us verify (5.5)–(5.6) by a recursive argument. For all $g\in \mathscr {C}_1$, it is clear that

$$\begin{aligned} \Vert g(x,u(x))\Vert _{H^1}{\mathop {\le }\limits ^{\text {Lemma }5.3 }} C'(1+\Vert u\Vert _{H^1}). \end{aligned}$$

(5.7)

Firstly, for $\mathrm {p}=0$, using (5.7) and Remark 2.2$\mathrm {(ii)}$ yields

$$\begin{aligned} \Vert f(t,x,u)\Vert _{0}&\le C\max _{t\in \mathbf {T}}\Vert f(t,\cdot ,u(t,\cdot ))\Vert _{H^1(0,\pi )}{\le }C'(1+\max _{t\in \mathbf {T}}\Vert u(t,\cdot )\Vert _{H^1(0,\pi )})\nonumber \\&\le C''(1+\Vert u\Vert _{s})=:C(\Vert u\Vert _{s}). \end{aligned}$$

(5.8)

If $k\ge 2$, then a similar argument as above can yield

$$\begin{aligned} \Vert \partial _{t}f(t,x,u)\Vert _{0}\le C(\Vert u\Vert _{s}),\quad \max _{t\in \mathbf {T}}\Vert \partial _{u}f(t,\cdot ,u(t,\cdot ))\Vert _{H^1(0,\pi )}\le C(\Vert u\Vert _{s}). \end{aligned}$$

(5.9)

Moreover Remark 2.2$\mathrm {(ii)}$ leads to

$$\begin{aligned} \max _{t\in \mathbf {T}}\Vert u_{n}(t,\cdot )-u(t,\cdot )\Vert _{H^1(0,\pi )}\rightarrow 0 \quad \text{ as } u_n\rightarrow u \text{ in } H^{s}\cap H^{0}. \end{aligned}$$

Then it follows from the continuity property in Lemma 5.3 and the compactness of $\mathbf {T}$ that

$$\begin{aligned} \Vert f(t,x,u_{n})-f(t,x,u)\Vert _{0}\le C\max _{t\in \mathbf {T}}\Vert f(t,\cdot ,u_{n}(t,\cdot ))-f(t,\cdot ,u(t,\cdot ))\Vert _{H^1(0,\pi )}\rightarrow 0 \end{aligned}$$

as $u_n\rightarrow u$ in $H^{s}\cap H^{0}$.

Assume that (5.5) holds for $\mathrm {p}=\mathfrak {k}$ with $\mathfrak {k}\in \mathbf N^+$, then we have to verify that it holds for $\mathrm {p}=\mathfrak {k}+1$ with $\mathfrak {k}+1 \le k-1$.

Since $\partial _{t}f,\partial _{u}f\in \mathcal {C}_{k-1}$, by the above assumption for $\mathrm {p}=\mathfrak {k}$, we get that for $u\in H^s\cap H^{\mathfrak {k}+1}$,

$$\begin{aligned} \Vert \partial _{t}f(t,x,u)\Vert _{\mathfrak {k}}\le C(\mathfrak {k},\Vert u\Vert _{s})(1+\Vert u\Vert _{\mathfrak {k}}),\quad \Vert \partial _{u}f(t,x,u)\Vert _{\mathfrak {k}}\le C(\mathfrak {k},\Vert u\Vert _{s})(1+\Vert u\Vert _{\mathfrak {k}}). \end{aligned}$$

(5.10)

Let $\mathfrak {q}(t,x):=f(t,x,u(t,x))$. We write $\mathfrak {q}$ as the form

$$\begin{aligned} \mathfrak {q}(t,x)=\sum _{j\in \mathbf {Z}}\mathfrak {q}_{j}(x)e^{\mathrm{i}jt}. \end{aligned}$$

It is obvious that $\mathfrak {q}_{t}(t,x)=\sum _{j\in \mathbf {Z}}\mathrm{i}j\mathfrak {q}_{j}(x)e^{\mathrm{i}jt}$. By the definition of s-norm, we obtain

$$\begin{aligned} \Vert \mathfrak {q}(t,x)\Vert ^2_{{\mathfrak {k}+1}}&=\sum \limits _{j\in \mathbf {Z}}\left( 1+j^{2(\mathfrak {k}+1)}\right) \Vert \mathfrak {q}_{j}\Vert ^2_{H^1}= \sum _{j\in \mathbf {Z}}\Vert \mathfrak {q}_{j}\Vert ^2_{H^1}+\sum _{j\in \mathbf {Z}}j^{2\mathfrak {k}}\Vert \mathrm{i}j\mathfrak {q}_j\Vert ^2_{H^1}\\&\le \Vert \mathfrak {q}(t,x)\Vert ^2_{0}+\Vert \mathfrak {q}_{t}(t,x)\Vert ^2_{\mathfrak {k}} \le \left( \Vert \mathfrak {q}(t,x)\Vert _{0}+\Vert \mathfrak {q}_{t}(t,x)\Vert _{\mathfrak {k}}\right) ^2. \end{aligned}$$

As a consequence

$$\begin{aligned} \Vert f(t,x,u)\Vert _{{\mathfrak {k}+1}}\le \Vert f(t,x,u)\Vert _{0}+\Vert \partial _{t}f(t,x,u)\Vert _{\mathfrak {k}}+\Vert \partial _{u}f(t,x,u)\partial _{t}u\Vert _{\mathfrak {k}}. \end{aligned}$$

(5.11)

For $\mathfrak {k}=0$ ($\mathrm {p}=1$), formulae (5.8)–(5.9) and (5.11) carry out

$$\begin{aligned} \Vert f(t,x,u)\Vert _{1}&\le \Vert f(t,x,u)\Vert _{0}+\Vert \partial _{t}f(t,x,u)\Vert _{0}+C\max _{t\in \mathbf {T}}\Vert \partial _{u}f(t,\cdot ,u(t,\cdot ))\Vert _{H^1(0,\pi )}\Vert \partial _xu\Vert _{0}\\&\le ~2C(\Vert u\Vert _{s})+C'(\Vert u\Vert _{s})\Vert u\Vert _{1}\le C(1,\Vert u\Vert _{s})(1+\Vert u\Vert _{1}), \end{aligned}$$

where $C(1,\Vert u\Vert _{s}):=\max \{2C(\Vert u\Vert _{s}),C'(\Vert u\Vert _{s})\}$. Letting $s_{1}\in (1/2,\min (1,s))$, it is straightforward that

$$\begin{aligned} {\left\{ \begin{array}{ll} s_1<\mathfrak {k}<s_1+1<\mathfrak {k+1},\quad \mathfrak {k}=1,\\ s_1<s_1+1<\mathfrak {k}<\mathfrak {k}+1,\quad \forall \mathfrak {k}\ge 2. \end{array}\right. } \end{aligned}$$

Then using (5.3) gives rise to

$$\begin{aligned} \Vert u\Vert _{\mathfrak {k}}\Vert u\Vert _{{s_1+1}}{\le }\Vert u\Vert _{{\mathfrak {k}+1}}\Vert u\Vert _{{s_1}}\le \Vert u\Vert _{{\mathfrak {k}+1}}\Vert u\Vert _{s}. \end{aligned}$$

(5.12)

Thus, by combining (5.1), (5.10)–(5.12), Remark 2.2$\mathrm {(ii)}$ with the above assumption for $\mathrm {p}=\mathfrak {k}$, we get

$$\begin{aligned} \Vert f(t,x,u)\Vert _{{\mathfrak {k}+1}}&\le C(\Vert u\Vert _{s})+C(\mathfrak {k},\Vert u\Vert _{s})(1+\Vert u\Vert _{\mathfrak {k}})\\&\quad +C(\mathfrak {k})\Vert \partial _{u}f(t,x,u)\Vert _{\mathfrak {k}}\Vert \partial _{t}u\Vert _{L^{\infty }(\mathbf {T},H^1(0,\pi ))}\\&\quad +C(\mathfrak {k}) \Vert \partial _{u}f(t,x,u)\Vert _{L^{\infty }(\mathbf {T},H^1(0,\pi ))}\Vert u\Vert _{{\mathfrak {k}+1}}\\&\le C(\Vert u\Vert _{s})+C(\mathfrak {k},\Vert u\Vert _{s})(1+\Vert u\Vert _{\mathfrak {k}})+C(\mathfrak {k})C(\mathfrak {k},\Vert u\Vert _{s})(1+\Vert u\Vert _{\mathfrak {k}})\Vert u\Vert _{{s_1+1}}\\&\quad +\,C(\mathfrak {k})C(\Vert u\Vert _{s})\Vert u\Vert _{{\mathfrak {k}+1}}\\&\le C(\mathfrak {k}+1,\Vert u\Vert _{s})(1+\Vert u\Vert _{{\mathfrak {k}+1}}), \end{aligned}$$

where $C(\mathfrak {k}+1,\Vert u\Vert _{s})=4\max \{C(\Vert u\Vert _{s}),C(\mathfrak {k},\Vert u\Vert _{s}),C(\mathfrak {k})C(\mathfrak {k},\Vert u\Vert _{s})(1+\Vert u\Vert _{s}),C(\mathfrak {k})C(\Vert u\Vert _{s})\} .$ This implies that (5.5) is satisfied for $\mathrm {p}=\mathfrak {k}+1$.

Finally, we assume that (5.6) holds for $\mathrm {p}=\mathfrak {k}$. Using inequality (5.11) yields that the continuity property of f with respect to u also holds for $\mathrm {p}=\mathfrak {k}+1$ with $\mathfrak {k}+1\le k-1$.

When $s'$ is not an integer, we can obtain the result by the Fourier dyadic decomposition. The argument is similar to the proof of the Lemma A.1 in [20]. $\square$

Lemma 5.5

Letting $f\in \mathcal {C}_{k}$ with $k\ge 3$, for all $0\le s'\le k-3$, a map F is defined as

$$\begin{aligned} F:\quad H^{s}\cap H^{s'}&\rightarrow {H}^{s'},u\mapsto f(t,x,u). \end{aligned}$$

Then F is a $C^2$ map with respect to u and

$$\begin{aligned} {\mathrm{D}}_{u}F(u)[h]=\partial _{u}f(t,x,u)h, \quad \mathrm{D}^2_{u}G(u)[h,h]=\partial ^2_{u}f(t,x,u)h^2,\quad \forall h\in H^s\cap H^{s'}. \end{aligned}$$

Moreover one has

$$\begin{aligned}&\Vert \partial _{u}f(t,x,u)\Vert _{{s'}}\le C(s',\Vert u\Vert _{s})(1+\Vert u\Vert _{{s'}}),\quad \Vert \partial ^2_{u}f(t,x,u)\Vert _{{s'}}\le C(s',\Vert u\Vert _{s})(1+\Vert u\Vert _{{s'}}). \end{aligned}$$

(5.13)

Proof

It is straightforward that $\partial _{u}f,\partial ^2_{u}f$ are in $\mathcal {C}_{k-1},\mathcal {C}_{k-2}$. Hence it follows from Lemma 5.4 that the maps $u\mapsto \partial _{u}f(t,x,u)$, $u\mapsto \partial ^2_{u}f(t,x,u)$ are continuous and that (5.13) holds. Let us check that F is $C^2$ respect to u. Using the continuity property of $u\mapsto \partial _{u}f(t,x,u)$, we deduce

$$\begin{aligned}&\Vert f(t,x,u+h)-f(t,x,u)-\partial _{u}f(t,x,u)h\Vert _{{s'}}\\&\quad =\Vert h\int _0^1(\partial _{u}f(t,x,u+\mathfrak {v} h) -\partial _{u}f(t,x,u))\mathrm {d}\mathfrak {v}\Vert _{{s'}}\\&\quad \le C(s')\Vert h\Vert _{{\max {\{s,s'\}}}}\max _{\mathfrak {v}\in [0,1]}\Vert \partial _{u}f(t,x,u+\mathfrak {v} h)-\partial _{u}f(t,x,u)\Vert _{{\max {\{s,s'\}}}}\\&\quad =o(\Vert h\Vert _{{\max {\{s,s'\}}}}), \end{aligned}$$

which carries out

$$\begin{aligned} \mathrm{D}_{u}F(u)[h]=\partial _{u}f(t,x,u)h,\quad \forall h\in H^s\cap H^{s'} \end{aligned}$$

and that $u\mapsto \mathrm{D}_uF(u)$ is continuous. In addition

$$\begin{aligned}&\partial _{u}f(t,x,u+\sigma h)h-\partial _{u}f(t,x,u)h-\partial ^2_{u}f(t,x,u)h^2\\&\quad =h^2\int _0^1(\partial ^2_{u}f(t,x,u+\mathfrak {v} h)-\partial ^2_{u}f(t,x,u))~\mathrm {d}\mathfrak {v}. \end{aligned}$$

The same discussion as above yields that F is twice differentiable with respect to u and that $u\mapsto \mathrm{D}^2_uF(u)$ is continuous. $\square$

1.3 Proof of Lemma 4.14

Proof

By (3.10), let $\vartheta _i(\xi )=c^2 \frac{d_i(\psi )}{\rho (\psi )}(\xi ),i=1,2$. Define

$$\begin{aligned} T_1 u:=\frac{\mathrm {d}^{2}}{\mathrm {d}{\xi ^2}}u+\vartheta _1(\xi )u, \quad T_2 u:=\frac{\mathrm {d}^{2}}{\mathrm {d}{\xi ^2}}u+\vartheta _2(\xi )u. \end{aligned}$$

It follows from (3.2), Lemma 5.3, $m\in H^1$ and $\rho \in H^3$ that

$$\begin{aligned} \Vert \vartheta _iu\Vert _{L^2}\le \Vert \vartheta _i\Vert _{L^{\infty }}\Vert u\Vert _{L^2}=\left\| c^2{d_i(\psi )}/{\rho (\psi )}\right\| _{{H}^1(0,\pi )}\Vert u\Vert _{L^2}\le \tilde{\mathfrak {C}}\Vert u\Vert _{L^2}. \end{aligned}$$

This indicates that $\vartheta _i\in \mathcal {B}(L^2,L^2)$. It is obvious that $T_1,T_2$ are self-adjoint using Theorem 4.13. By means of Theorem 4.13, Lemma 5.5 and the inverse Liouville substitution of (3.3), for all $(\epsilon ,w)\in (\epsilon _1,\epsilon _2)\times \{W\cap H^s:\Vert w\Vert _{s}< r\}$, we derive

$$\begin{aligned} |\lambda _j(d_2)-\lambda _j(d_1)|&\le \frac{1}{c^2}|\mu _{j}(\vartheta _2)-\mu _j(\vartheta _1)| \le \frac{1}{c^2}\Vert \vartheta _2-\vartheta _1\Vert _{\mathcal {B}(L^2,L^2)} \le \frac{1}{c^2}\Vert \vartheta _2-\vartheta _1\Vert _{L^{\infty }}\\&\le \kappa _0\Vert d_2-d_1\Vert _{{H}^1(0,\pi )}\\&\le \kappa (|\epsilon -\epsilon '|+\Vert w-w'\Vert _s). \end{aligned}$$

$\square$

1.4 Proof of Formula (4.76)

Proof

If $j>\max {\{\hat{N},{4 c\mathfrak {M}}\}}$, $\forall \epsilon \in (\epsilon _1,\epsilon _2)$ ,$\forall \Vert w\Vert _s< r$, then formula (4.23) shows that either

$$\begin{aligned} \inf \left| \sqrt{\lambda _{j+1}(\epsilon ,w)}-\sqrt{\lambda _j(\epsilon ,w)}\right|&\ge \frac{1}{c}-\left| \sqrt{\lambda _{j+1}(\epsilon ,w})-\frac{j+1}{c}\right| -\left| \sqrt{\lambda _{j}(\epsilon ,w)}-\frac{j}{c}\right| \\&\ge \frac{1}{c}-\frac{2\mathfrak {M}}{j}>\frac{1}{2c} \end{aligned}$$

or

$$\begin{aligned} \inf \left| \sqrt{\lambda _{j+1}(\epsilon ,w)}-\sqrt{\lambda _j(\epsilon ,w)}\right|&\ge \frac{1}{c}-\left| \sqrt{\lambda _{j+1}(\epsilon ,w)}-\frac{j+1+1/2}{c}\right| \\&\quad -\,\left| \sqrt{\lambda _{j}(\epsilon ,w)}-\frac{j+1/2}{c}\right| \\&>\frac{1}{2c} \end{aligned}$$

holds. For $0\le j\le \max {\{\hat{N},{4c\mathfrak {M}}\}}$, we may obtain

$$\begin{aligned} \mathfrak {w}_j:=\inf _{{\mathop {w\in \{W\cap H^s:\Vert w\Vert _{s}< r\}}\limits ^{\epsilon \in (\epsilon _1,\epsilon _2)}}}\left| \sqrt{\lambda _{j+1}(\epsilon ,w)}-\sqrt{\lambda _j(\epsilon ,w)}\right| . \end{aligned}$$

Note that $\hat{N}$ is seen in Lemma 3.8. $\square$