On the Cauchy problem for a class of semilinear second order evolution equations with fractional Laplacian and damping

Abstract

In the present paper, we prove time decay estimates of solutions in weighted Sobolev spaces to the second order evolution equation with fractional Laplacian and damping for data in Besov spaces. Our estimates generalize the estimates obtained in the previous studies (Karch in Stud Math 143:175–197, 2000; Ikeda et al. in Nonlinear Differ. Equ. Appl. 24:10, 2017). The second aim of this article is to apply these estimates to prove small data global well-posedness for the Cauchy problem of the equation with power nonlinearities. Especially, the estimates obtained in this paper enable us to treat more general conditions on the nonlinearities and the spatial dimension than the results in the papers (Chen et al. in Electron. J. Differ. Equ. 2015:1–14, 2015; Ikeda et al. in Nonlinear Differ. Equ. Appl. 24:10, 2017).

Appendices

Proof of a nonlinear estimate

Lemma 8.1

Let $z=(z_j)_{j=-1}^1, w=(w_j)_{j=-1}^1 \subset \mathbb R$. For any real sequence $(a_j)_{j=-1}^1$, let $\Delta ^{(2)}(a) = a_{1}+a_{-1} - 2 a_{0}$. Then

$$\begin{aligned}&| \Delta ^{(2)}(\mathcal M (z)) - \Delta ^{(2)}(\mathcal M (w))|\\&\quad \le C \max _j |z_j|^{\rho _0} | \Delta ^{(2)} (z-w)|\\&\quad \quad + C \max _j (|z_j|+|w_j|)^{(\rho _0-2)_+} \max _j |z_j-w_j|^{\min (\rho _0-1,1)} | \Delta ^{(2)} (w)|\\&\quad \quad + C \max _j |z_j|^{(\rho _0-2)_+} (|z_1-z_0| + |z_0 - z_{-1}|)^{\min (\rho _0-1,1)} |z_0 - w_0 - z_{-1} + w_{-1}|\\&\quad \quad + C \max _j (|z_j|+|w_j|)^{(\rho _0-2)_+} |w_0 - w_{-1}|\\&\quad \quad \cdot \min ( (|z_1-z_0| + |z_0 - z_{-1}| + |w_1-w_0| + |w_0 - w_{-1}|), \max _j|z_j - w_j| )^{\min (\rho _0-1,1)} \end{aligned}$$

with $\mathcal M$ of Lemma 6.3.

Proof

Let

$$\begin{aligned} z_\theta = {\left\{ \begin{array}{ll} \theta z_1 + (1-\theta ) z_0 &{}\mathrm {if} \quad \theta \in [0,1],\\ |\theta | z_{-1} + (1-|\theta |) z_0 &{}\mathrm {if} \quad \theta \in [-1,0]. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned}&\Delta ^{(2)}(\mathcal M(z))\\&\quad = \int _0^1 \mathcal M' (z_\theta ) d \theta (z_1-z_0) - \int _{-1}^0 \mathcal M' (z_\theta ) d \theta (z_0-z_{-1})\\&\quad = \int _0^1 \mathcal M' (z_\theta ) d \theta \Delta ^{(2)}(z) + \int _0^{1} (\mathcal M'(z_\theta ) - \mathcal M' (z_{-\theta })) d \theta (z_{0}- z_{-1}). \end{aligned}$$

Therefore

$$\begin{aligned}&\Delta ^{(2)}(\mathcal M(z) - \mathcal M(w))\\&\quad = \int _0^1 \mathcal M' (z_\theta ) d \theta \Delta ^{(2)}(z-w) + \int _0^1 (\mathcal M' (z_\theta ) - \mathcal M'(w_\theta )) d \theta \Delta ^{(2)}(w)\\&\quad \quad + \int _0^{1} (\mathcal M'(z_\theta ) - \mathcal M' (z_{-\theta })) d \theta (z_{0} - w_0 - z_{-1} + w_{-1})\\&\quad \quad + \int _0^{1} (\mathcal M'(z_\theta ) - \mathcal M' (z_{-\theta }) - \mathcal M'(w_\theta ) + \mathcal M' (w_{-\theta }) ) d \theta (w_0 - w_{-1}). \end{aligned}$$

Since $z_\theta - z_{-\theta } = \theta (z_1 - z_{-1})$, the assertion is obtained. $\square$

Proof of Lemma 4.2

Proof of Lemma 4.2

It is sufficient for us to show

$$\begin{aligned} \Vert \phi _j *f \sum _{k \le j-2} \phi _k *g \Vert _{L^{p_0}}&\le C \Vert \phi _j *f \Vert _{L^{p_1}} \Vert g \Vert _{L^{p_2}}, \end{aligned}$$

(9.1)

$$\begin{aligned} \bigg \Vert 2^{sj} \Vert \sum _{k \ge 0} (\phi _{j+k} *f) (\phi _{j+k} *g) \Vert _{L^{p_0}} \bigg \Vert _{\ell ^2}&\le C \Vert f \Vert _{\dot{B}_{p_1,2}^s} \Vert g \Vert _{L^{p_2}}. \end{aligned}$$

(9.2)

(9.1) follows from the Young estimate and the fact that

$$\begin{aligned} \Vert \sum _{k \le j-2} \phi _k \Vert _{L^1} = \Vert \sum _{k \le 0} \phi _k \Vert _{L^1} \end{aligned}$$

holds for any $j \in \mathbb Z$. (9.2) also holds because we have

$$\begin{aligned} \bigg \Vert 2^{sj} \Vert \sum _{k \ge 0} (\phi _{j+k} *f) (\phi _{j+k} *g) \Vert _{L^{p_0}} \bigg \Vert _{\ell _j^2}&\le \Vert g \Vert _{L^{p_2}} \sum _{k \ge 0} 2^{-sk} \Vert 2^{s(j+k)} \Vert \phi _{j+k} *f \Vert _{L^{p_1}} \Vert _{\ell _j^2}\\&\le C \Vert g \Vert _{L^{p_2}} \Vert f \Vert _{\dot{B}_{p_1,2}^s}. \end{aligned}$$

$\square$