Response Solutions for Wave Equations with Variable Wave Speed and Periodic Forcing

Abstract

We consider a model of nonlinear wave equations with periodically varying wave speed and periodic external forcing. By imposing non-resonance conditions on the frequency, we establish the existence of response solutions (i.e., periodic solutions with the same frequency as the forcing) for such a model in a Cantor set of asymptotically full measure. The proof relies on a Lyapunov–Schmidt reduction together with the Nash–Moser iteration.

Appendix

In the Appendix, we will supplement some results used in the proof of Theorem 1.2 for completeness. First, we need to give the proof of Remark 1.1.

Proof

(Proof of Remark 1.1) (i) For all $u,v\in H^s$, we decompose $uv=\sum _{j} (\sum _{k}u_{j-k}v_{k})e^{\mathrm {i}jx}$. Then using the Cauchy inequality yields that

$$\begin{aligned} \Vert uv\Vert ^2_{s}&=\textstyle \sum \limits _{j}(1+j^{2s})\Vert \sum \limits _{k}u_{j-k}v_{k}\Vert ^2_{H^1} \le \sum \limits _{j}(\sum _{k}(1+j^{2s})^{\frac{1}{2}}c_{jk}\Vert u_{j-k}v_{k}\Vert _{H^1}\frac{1}{c_{jk}})^2\\&\le \textstyle \sum \limits _{j}(\sum _{k}\frac{1}{c^2_{jk}})( \sum \limits _{k}\Vert u_{j-k}\Vert ^2_{H^1}(1+(j-k)^{2s})\Vert v_{k}\Vert ^2_{H^1}(1+k^{2s})), \end{aligned}$$

where $c^2_{jk}=\frac{(1+k^{2s})(1+(j-k)^{2s})}{1+j^{2s}}$. Moreover, it is clear that

$$\begin{aligned} 1+j^{2s}&\le 1+(k+j-k)^{2s}\le 1+2^{2s-1}(k^{2s}+(j-k)^{2s})\\&\le 2^{2s-1}(1+k^{2s})+2^{2s-1}(1+(j-k)^{2s}). \end{aligned}$$

This leads to

$$\begin{aligned} \textstyle \sum \limits _{k}\frac{1}{c^2_{jk}}\le 2^{2s-1}(\sum \limits _{k}\frac{1}{1+k^{2s}}+\sum \limits _{k}\frac{1}{1+(j-k)^{2s}}) =2^{2s}\sum \limits _{k}\frac{1}{1+k^{2s}}{\mathop {=:}\limits ^{s>1/2}}(C(s))^2. \end{aligned}$$

Hence we get the first conclusion in Remark 1.1.

(ii) From the Cauchy inequality, we get

$$\begin{aligned} \textstyle \max _{x\in {\mathbb {T}}}\Vert u(\cdot ,x)\Vert _{{H}^1({\mathbb {T}})}&\le C\sum \limits _{j}\Vert u_j\Vert _{H^1}\le C (\sum \limits _{j}\Vert u_j\Vert ^2_{H^1}(1+j^{2s}))^{\frac{1}{2}}(\sum \limits _{j}\frac{1}{1+j^{2s}})^{\frac{1}{2}}\\&\le C(s)\Vert u\Vert _{s}. \end{aligned}$$

Thus we arrive at the other one in Remark 1.1. $\square$

The next thing is to recall Lemmas 5.1–5.3 introduced in [4, cf. Lemmas 2.1–2.3].

Lemma 5.1

(Moser–Nirenberg) Let $s'\ge 0$ and $s>1/2$. For all $u_1,u_2\in H^{s'}\cap H^s$, one has

$$\begin{aligned} \Vert u_1u_2\Vert _{s'}&\le \textstyle C(s')(\Vert u_1\Vert _{L^{\infty }({\mathbb {T}};H^1({\mathbb {T}}))}\Vert u_2\Vert _{{s'}}+\Vert u_1\Vert _{{s'}}\Vert u_2\Vert _{L^{\infty }({\mathbb {T}};H^1({\mathbb {T}}))}) \end{aligned}$$

(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 $0\le \mathrm {a}\le \mathrm {a}'\le \mathrm {b}'\le \mathrm {b}$ with $\mathrm {a}+\mathrm {b}=\mathrm {a}'+\mathrm {b}'$, and ${\mathfrak {v}}:=\frac{\mathrm {b}-\mathrm {a}'}{\mathrm {b}-\mathrm {a}}$. Then

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

In particular, one has

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

(5.3)

Lemma 5.3

Let $y\longmapsto f(\cdot ,y)$ be in $C^1({\mathbb {R}};H^{1}({\mathbb {T}}))$ and $m:=\Vert y\Vert _{L^{\infty }({\mathbb {T}})}$. Then the composition operator $y(t)\longmapsto f(t,y(t))$ belongs to $C(H^1({\mathbb {T}});H^1({\mathbb {T}}))$ satisfying

$$\begin{aligned} \textstyle \Vert f(\cdot ,y)\Vert _{H^1}\le C(\max _{y\in [-m,m]}\Vert f(\cdot ,y)\Vert _{H^1}+\max _{y\in [-m,m]}\Vert \partial _yf(\cdot ,y)\Vert _{H^1}\Vert y\Vert _{H^1}). \end{aligned}$$

Based on Lemmas 5.1–5.3, we have the following lemma.

Lemma 5.4

Let $(x,u)\longmapsto f(\cdot ,x,u)$ belong to $C^\ell ({\mathbb {T}}\times {\mathbb {R}};H^1({\mathbb {T}}))$ with $\ell \ge 1$. For all $s>{1}/{2}$ and $0\le s'\le \ell -1$, the composition operator $u(t,x)\longmapsto f(t,x,u(t,x))$ is in $C(H^{s}\cap H^{s'};H^{s'})$ with

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

Proof

If $s'=\mathrm {p}\in {\mathbb {N}}$ with $\mathrm {p}\le \ell -1$, then we prove that for all $u\in H^{s}\cap H^{\mathrm {p}}$,

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

(5.4)

and that

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

(5.5)

Let us verify (5.4)–(5.5) by a recursive argument.

For $\mathrm {p}=0$, it follows from Lemma 5.3 and Remark 1.1 that

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

(5.6)

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

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

(5.7)

Moreover, it can be shown from Remark 1.1 that

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

Then using the continuity property in Lemma 5.3 and the compactness of ${\mathbb {T}}$ yields that

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

if $u_n$ tends to u in $H^{s}\cap H^{0}$.

Suppose that (5.4) could hold for $\mathrm {p}=k$, with $k\in {\mathbb {N}}^+$. We will show that it follows for $\mathrm {p}=k+1$, where $k+1 \le \ell -1$. Then the assumption for $\mathrm {p}=k$ implies that for all $u\in H^s\cap H^{k+1}$,

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

(5.8)

If we set $\rho (t,x):=f(t,x,u(t,x))$, then $\rho (t,x)=\sum _{j}\rho _{j}(t)e^{\mathrm{i}j x}$. Observe that $\partial _{x}\rho (t,x)=\sum _{j}\mathrm {i}j\rho _{j}(t)e^{\mathrm{i}j x}$. Then the definition of s–norm (recall (1.6)) yields that

$$\begin{aligned} \textstyle \Vert \rho \Vert ^2_{k+1}&=\textstyle \sum \limits _{j}(1+j^{2(k+1)})\Vert \rho _{j}\Vert ^2_{H^1}= \sum \limits _{j}\Vert \rho _{j}\Vert ^2_{H^1}+\sum \limits _{j}j^{2k}j^2\Vert \rho _j\Vert ^2_{H^1}\\&=\textstyle \sum \limits _{j}\Vert \rho _{j}\Vert ^2_{H^1}+\sum \limits _{j}j^{2k}\Vert (\mathrm {i}j)\rho _j\Vert ^2_{H^1} \le (\Vert \rho \Vert _{0}+\Vert \partial _{x}\rho \Vert _{k})^2. \end{aligned}$$

This leads to

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

(5.9)

If $\mathrm {p}=1$ ($k=0$), then it follows from (5.6)–(5.7) and (5.9) that

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

In addition, denote ${\mathfrak {s}}\in (\frac{1}{2},\min (1,s))$. It is easy to see that

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

Combining this with (5.3) shows that

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

(5.10)

As a consequence, by (5.1), (5.6)–(5.10), we deduce

$$\begin{aligned} \Vert f(t,x,u)\Vert _{k+1}&\le C(\Vert u\Vert _{s})+C(k,\Vert u\Vert _{s})(1+\Vert u\Vert _{k})+C(k)\Vert \partial _{u}f(t,x,u)\Vert _{k} \Vert u_{x}\Vert _{L^{\infty }({\mathbb {T}};H^1({\mathbb {T}}))}\\&\quad +C(k) \Vert \partial _{u}f(t,x,u)\Vert _{L^{\infty }({\mathbb {T}};H^1({\mathbb {T}}))}\Vert u\Vert _{k+1}\\&\le C(\Vert u\Vert _{s})+C(k,\Vert u\Vert _{s})(1+\Vert u\Vert _{k})+C(k)C(k,\Vert u\Vert _{s})(1+\Vert u\Vert _{k})\Vert u\Vert _{{{\mathfrak {s}}+1}}\\&\quad +C(k)C(\Vert u\Vert _{s})\Vert u\Vert _{k+1}\\&\le C(k+1,\Vert u\Vert _{s})(1+\Vert u\Vert _{k+1}). \end{aligned}$$

Hence (5.4) follows for $\mathrm {p}=k+1$.

On the other hand, suppose that (5.5) could hold for $\mathrm {p}=k$. From formula (5.9), we get the continuity property of f with respect to u for $\mathrm {p}=k+1$ with $k+1\le \ell -1$.

If $s'$ is not an integer, then the conclusion follows from the Fourier dyadic decomposition. The argument is similar to the proof of Lemma A.1 in [14]. $\square$

Lemma 5.5

Let $(x,u)\longmapsto f(\cdot ,x,u)$ be in $C^\ell ({\mathbb {T}}\times {\mathbb {R}};H^1({\mathbb {T}}))$ with $\ell \ge 3$. For all $0\le s'\le \ell -3$, define the mapping

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

Then F is a $C^2$ mapping with respect to u and satisfies that for all $h\in H^s\cap H^{s'}$,

$$\begin{aligned} \mathrm{D}F(u)[h]=\partial _{u}f(t,x,u)h, \quad \mathrm{D}^2F(u)[h,h]=\partial ^2_{u}f(t,x,u)h^2, \end{aligned}$$

where

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

(5.11)

Proof

For $\ell \ge 3$, observe that

$$\begin{aligned} (x,u)&\longmapsto \partial _{u}f(\cdot ,x,u)\in C^{\ell -1}({\mathbb {T}}\times {\mathbb {R}};H^1({\mathbb {T}})),\\ (x,u)&\longmapsto \partial ^2_{u}f(\cdot ,x,u)\in C^{\ell -2}({\mathbb {T}}\times {\mathbb {R}};H^1({\mathbb {T}})). \end{aligned}$$

In view of Lemma 5.4, the mappings $u\longmapsto \partial _{u}f(t,x,u)$, $u\longmapsto \partial ^2_{u}f(t,x,u)$ are continuous and satisfy two estimates in (5.11).

It remains to verify that F is $C^2$ with respect to u. Using the continuity property of $u\longmapsto \partial _{u}f(t,x,u)$ yields that

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

This implies that $\mathrm{D}F(u)[h]=\partial _{u}f(t,x,u)h$ and that $u\longmapsto \mathrm{D}F(u)$ is continuous. In addition,

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

By the similar argument as above, F is twice differentiable with respect to u and that $u\longmapsto \mathrm{D}^2_uF(u)$ is continuous.

Thus this completes the proof. $\square$