Fixed analytic radius lower bound for the dissipative KdV equation on the real line

Abstract

We study the global analyticity for the dissipative KdV equation with an analytic initial data on the real line. We show that the analytic radius of the solution has a fixed positive lower bound uniformly for all time. This reflects the dissipative effect to some extent, since the best known analytic radius of the KdV equation may decay polynomially as time goes to infinity.

Appendix: Analyticity in the case \(\alpha \ge 1\)

In [26], the spatial analyticity is proved for the critical Burgers equation on the circle ${\mathbb {T}}$. Now we adapt the approach to show the spatial analyticity of the dissipative KdV equation, with $\alpha \ge 1$, on the real line. According to [28], for every data $u_0\in L^2({\mathbb {R}})$, the dissipative KdV equation (1.4) has a global solution

$$\begin{aligned} u\in C([0,\infty );L^2({\mathbb {R}}))\bigcap C([0,\infty ); H^\infty _x({\mathbb {R}})). \end{aligned}$$

The solution is in fact uniformly analytic with respect to the spatial variable. The precise statement is as follows.

Proposition 4.1

Let $\alpha \ge 1$ and $u_0\in L^2({\mathbb {R}})$. Then there exists a constant $\sigma _0, C>0$ depending only on $\Vert u_0\Vert _{L^2({\mathbb {R}})}$ so that when u solves (1.4)

$$\begin{aligned} \Vert u(t)\Vert _{G^{\sigma _0}}\le C, \quad \text{ for } \text{ all } t\ge 1. \end{aligned}$$

Proof

The following calculation is carried out formally, which can be rigorous by a Galerkin approximation as in [26].

Acting the operator $e^{\frac{t}{2}|D_x|}$ on both sides of (1.4) we obtain

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \Big ( e^{\frac{t}{2}|D_x|}u \Big )+e^{\frac{t}{2}|D_x|}u_{xxx}+e^{\frac{t}{2}|D_x|}\big ( uu_x\big ) + (|D_x|^{\alpha }-\frac{1}{2}|D_x|)e^{\frac{t}{2}|D_x|}u=0. \end{aligned}$$

(4.1)

Taking the inner product of (4.1) with $(1-\partial _x^4)e^{\frac{t}{2}|D_x|}u$, and using the cancelation property

$$\begin{aligned} \Big (e^{\frac{t}{2}|D_x|}u_{xxx}+e^{\frac{t}{2}|D_x|}ue^{\frac{t}{2}|D_x|}u_x,(1-\partial _x^4)e^{\frac{t}{2}|D_x|}u\Big )=0 \end{aligned}$$

we infer that

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} Z&=2\Big (e^{\frac{t}{2}|D_x|}ue^{\frac{t}{2}|D_x|}u_x-e^{\frac{t}{2}|D_x|}\big ( uu_x\big ), (1-\partial _x^4)e^{\frac{t}{2}|D_x|}u\Big )\nonumber \\&\quad + 2\Big ((\frac{1}{2}|D_x|-|D_x|^{\alpha })e^{\frac{t}{2}|D_x|}u, (1-\partial _x^4)e^{\frac{t}{2}|D_x|}u\Big )\nonumber \\&:=I_1+I_2, \end{aligned}$$

(4.2)

where $Z=\int _{\mathbb {R}}|e^{\frac{t}{2}|D_x|}u|^2+|\partial _x^2e^{\frac{t}{2}|D_x|}u|^2\mathrm{d}x$.

Now we are going to estimate $I_1, I_2$ separately. For $I_1$, using Parseval identity we can rewrite

$$\begin{aligned} I_1 = \int _{\xi +\xi _1+\xi _2=0}i\xi (1+\xi ^4)e^{\frac{t}{2}|\xi |}\Big (e^{\frac{t}{2}|\xi _1|}e^{\frac{t}{2}|\xi _2|}-e^{\frac{t}{2}|\xi |} \Big ){\widehat{u}}(\xi ){\widehat{u}}(\xi _1){\widehat{u}}(\xi _2)\mathrm{d}\xi _1 \mathrm{d}\xi . \end{aligned}$$

(4.3)

One can check that, see [35, Lemma 12], if $\xi +\xi _1+\xi _2=0$, then

$$\begin{aligned} 0\le e^{\frac{t}{2}|\xi _1|}e^{\frac{t}{2}|\xi _2|}-e^{\frac{t}{2}|\xi |}\le t\min \{|\xi _1|,|\xi _2|\}e^{\frac{t}{2}|\xi _1|}e^{\frac{t}{2}|\xi _2|}. \end{aligned}$$

Plugging this into (4.3), using the symmetry of $|\xi _1|$ and $|\xi _2|$, we find

$$\begin{aligned} I_1 \le 2\int _{\xi +\xi _1+\xi _2=0, |\xi _1|\le |\xi _2|} t|\xi _1\Vert \xi |(1+\xi ^4)e^{\frac{t}{2}|\xi |}|{\widehat{u}}(\xi ) |e^{\frac{t}{2}|\xi _1|}|{\widehat{u}}(\xi _1)|e^{\frac{t}{2}|\xi _2|}|{\widehat{u}}(\xi _2)|\mathrm{d}\xi _1 \mathrm{d}\xi . \end{aligned}$$

(4.4)

Noting $|\xi |\le |\xi _1|+|\xi _2|\le 2|\xi _2|$ in the integration, we deduce from (4.4) that

$$\begin{aligned} I_1&\lesssim \int _{\xi +\xi _1+\xi _2=0 \atop |\xi _1|\le |\xi _2|} t|\xi _1\Vert \xi _2|^{\frac{1}{2}}(1+\xi _2^2)|\xi |^{\frac{1}{2}}(1+\xi ^2)e^{\frac{t}{2}|\xi |}|{\widehat{u}}(\xi ) |e^{\frac{t}{2}|\xi _1|}|{\widehat{u}}(\xi _1)|e^{\frac{t}{2}|\xi _2|}|{\widehat{u}}(\xi _2)|\mathrm{d}\xi _1 \mathrm{d}\xi \nonumber \\&\lesssim t\int _{{\mathbb {R}}} |\xi _1|e^{\frac{t}{2}|\xi _1|}|{\widehat{u}}(\xi _1)|\mathrm{d}\xi _1\int _{{\mathbb {R}}} |\xi |(1+\xi ^2)^2e^{t|\xi |}|{\widehat{u}}(\xi )|\mathrm{d}\xi \nonumber \\&\lesssim t Z^{\frac{1}{2}} \int _{{\mathbb {R}}} |\xi |(1+\xi ^4)e^{t|\xi |}|{\widehat{u}}(\xi )|^2\mathrm{d}\xi , \end{aligned}$$

(4.5)

where in the last two steps we used the Cauchy-Schwarz inequality.

For $I_2$, by Parseval and split the integral $\int _{\mathbb {R}}=\int _{|\xi |\le 1}+\int _{|\xi |>1}$, we have

$$\begin{aligned} I_2&=\int _{\mathbb {R}}(|\xi |-2|\xi |^{\alpha })(1+\xi ^4)e^{t|\xi |}|{\widehat{u}}(\xi )|^2\mathrm{d}\xi \nonumber \\&\le 2e^t\Vert u(t)\Vert ^2_{L^2({\mathbb {R}})}-\int _{\mathbb {R}}|\xi |(1+\xi ^4)e^{t|\xi |}|{\widehat{u}}(\xi )|^2\mathrm{d}\xi . \end{aligned}$$

(4.6)

It follows from (4.2), (4.5)-(4.6) that, noting $\Vert u(t)\Vert _{L^2({\mathbb {R}})}\le \Vert u_0\Vert _{L^2({\mathbb {R}})}$,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}Z\le 2e^t\Vert u_0\Vert ^2_{L^2({\mathbb {R}})}+(Ct Z^{\frac{1}{2}}-1)\int _{\mathbb {R}}|\xi |(1+\xi ^4)e^{t|\xi |}|{\widehat{u}}(\xi )|^2\mathrm{d}\xi . \end{aligned}$$

(4.7)

We analyze (4.7) as follows. If for some $t_0\in (0,1]$,

$$\begin{aligned} Ct Z^{\frac{1}{2}}-1\le 0, \quad t\in [0,t_0], \end{aligned}$$

(4.8)

then we deduce from (4.7) that

$$\begin{aligned} Z(t)\le 2e\Vert u_0\Vert ^2_{L^2({\mathbb {R}})}+Z(0), \quad t\in [0,t_0]. \end{aligned}$$

(4.9)

Note that $Z(0)\lesssim \Vert u_0\Vert _{H^2({\mathbb {R}})}$, if we take

$$\begin{aligned} t_0=\frac{1}{1+C\sqrt{2e\Vert u_0\Vert ^2_{L^2({\mathbb {R}})}+Z(0)}} \end{aligned}$$

then (4.8)-(4.9) holds. In particular, this implies that if $u_0\in H^2({\mathbb {R}})$, then we have $u(t_0)\in G^{t_0}$ for some $t_0>0$.

Since the dissipative KdV equation has a smoothing effect, namely the solution $u(t)\in H^2({\mathbb {R}})$ for every $t>0$, we conclude the analyticity of u(t) for all $t\ge 1$. This completes the proof. $\square$