On the well-posedness of dispersive–dissipative one dimensional equations with non decaying initial data

Abstract

In this paper, we study the well-posedness of the Cauchy problem for a dissipative version of dispersive one dimensional equations of Korteweg de Vries type without any assumption on the decay of its initial data. We consider purely dispersion operators between BO and KdV combined with purely dissipation operators. We show in particular the local well-posedness of our equations in Zhidkov spaces $Z^s$ with $s>1/2$. Besides, we prove the convergence of the perturbed solution to that of the purely dispersive KdV type equation as the dissipation factor tends to 0.

Appendix

We present a concise proof of Lemma 3.2. We mainly follow [21] where a more general statement is proven.

1.1 Proof of Lemma 3.2

Knowing that our goal is to prove

$$\begin{aligned} \Vert g_{\xi } \Vert _{H^{1}({ I\!\!R})} \lesssim \Vert f\Vert _{L^2({ I\!\!R})} \quad \forall \xi \in { I\!\!R}, \end{aligned}$$

we recall (3.3) and proceed by direct calculations on $g_{\xi }(t)$:

$$\begin{aligned} \displaystyle g_{\xi }(t)= & {} 1\hspace{-1mm} \textrm{I}_{{ I\!\!R}_+} (t) \ \eta (t)\int _0^t e^{-\nu |t-t'| |\xi |^{2\beta }} f(t') \ dt' \\ \displaystyle= & {} 1\hspace{-1mm} \textrm{I}_{{ I\!\!R}_+} (t) \ \eta (t) \ e^{-\nu |t||\xi |^{2 \beta }} \int _0^t e^{\nu t' |\xi |^{2\beta }} f(t') \ dt' \\ \displaystyle= & {} 1\hspace{-1mm} \textrm{I}_{{ I\!\!R}_+} (t) \ \eta (t) \ e^{-\nu |t||\xi |^{2 \beta }} \int _0^t e^{\nu t' |\xi |^{2\beta }} \bigg ( \int _{\mathbb {R}} e^{i t' \tau } \hat{f}(\tau ) \ d\tau \bigg ) \ dt' \\ \displaystyle= & {} 1\hspace{-1mm} \textrm{I}_{{ I\!\!R}_+} (t) \ \eta (t) \ e^{-\nu |t||\xi |^{2 \beta }} \int _{\mathbb {R}} \hat{f}(\tau )\int _0^t e^{t'(i \tau +\nu |\xi |^{2\beta })} \ dt' \ d\tau \\ \displaystyle= & {} 1\hspace{-1mm} \textrm{I}_{{ I\!\!R}_+} (t) \ \eta (t) \ e^{-\nu |t||\xi |^{2 \beta }} \int _{\mathbb {R}} \hat{f}(\tau )\ \frac{e^{(i t \tau + \nu t |\xi |^{2\beta })} - e^0}{i \tau +\nu |\xi |^{2\beta }} \ d\tau \\ \displaystyle= & {} 1\hspace{-1mm} \textrm{I}_{{ I\!\!R}_+} (t) \ \eta (t) \int _{\mathbb {R}} \hat{f}(\tau )\ \frac{e^{i t \tau } - e^{-\nu |t||\xi |^{2 \beta }}}{i \tau +\nu |\xi |^{2\beta }} \ d\tau \\ \displaystyle:= & {} 1\hspace{-1mm} \textrm{I}_{{ I\!\!R}_+} (t) \ k(t), \end{aligned}$$

where k(t) is a function defined on $\mathbb {R}$ that can be seen as $k(t)=k_1(t) + k_2(t)$ with

$$\begin{aligned} k_1(t):= \eta (t) \int _{\mathbb {R}} \hat{f}(\tau )\ \frac{1 - e^{-\nu |t||\xi |^{2 \beta }}}{i \tau + \nu |\xi |^{2\beta }} \ d\tau , \end{aligned}$$

and

$$\begin{aligned} k_2(t):= \eta (t) \int _{\mathbb {R}} \hat{f}(\tau )\ \frac{e^{i t \tau } - 1}{i \tau + \nu |\xi |^{2\beta }} \ d\tau . \end{aligned}$$

We remark that since $f\in H^1({ I\!\!R})$, the Lebesgue convergence theorem ensures that k is continuous with $k(0)=0$. Therefore, proving

$$\begin{aligned} ||k||_{H^1} \lesssim \ ||f||_{L^2} \end{aligned}$$

(6.1)

is sufficient to terminate this proof. We recall the following product estimate for $s\ge 0$,

$$\begin{aligned} ||u \ v||_{H^s} \le ||u||_{H^s} \ ||v||_{L^{\infty }} + ||u||_{L^{\infty }} \ ||v||_{\dot{H^s}} \end{aligned}$$

(6.2)

and the fact that for any $\lambda >0$ we have

$$\begin{aligned} ||f(\lambda t)||_{\dot{H^s}} = {\lambda }^{s-\frac{1}{2}} \ ||f(t)||_{\dot{H^s}}, \end{aligned}$$

(6.3)

both of which will be frequently used throughout this proof. Starting with $k_1$, we split the estimation of $||k_1||_{H^1}$ into two frequency domains:

• Over $\{|\xi | \ge 1 \}$: Using (6.2), we get

  $$\begin{aligned} \displaystyle ||k_1||_{H^1}\le & {} \bigg | \int _{\mathbb {R}} \frac{\hat{f}(\tau )}{i \tau + \nu |\xi |^{2\beta }} d\tau \bigg | \ ||\eta (t) (1 - e^{-\nu |t||\xi |^{2 \beta }})||_{H^1} \\ \displaystyle\le & {} \bigg | \int _{\mathbb {R}} \frac{\hat{f}(\tau )}{i \tau +\nu |\xi |^{2\beta }} d\tau \bigg | \ (||\eta ||_{H^1} + ||\eta \ e^{-\nu |t||\xi |^{2 \beta }}||_{H^1} ) \\ \displaystyle\le & {} \bigg | \int _{\mathbb {R}} \frac{\hat{f}(\tau )}{i \tau +\nu |\xi |^{2\beta }} d\tau \bigg | ( ||\eta ||_{H^1} + ||e^{-\nu |t||\xi |^{2 \beta }}||_{L^{\infty }}\\{} & {} + ||\eta (t)||_{L^{\infty }} + ||e^{-\nu |t||\xi |^{2 \beta }}||_{\dot{H^1}} ). \end{aligned}$$

  At this point, we use Cauchy–Schwarz inequality followed by a change of variable ($\tau := \theta \, \nu |\xi |^{2 \beta }$) to get

  $$\begin{aligned} \bigg | \int _{\mathbb {R}} \frac{\hat{f}(\tau )}{i \tau + \nu |\xi |^{2\beta }} d\tau \bigg | \le ||f||_{L^2} \bigg ( \int _{\mathbb {R}} \frac{d\tau }{|i \tau + \nu |\xi |^{2\beta }|^2} \bigg )^{\frac{1}{2}} \le \nu ^{-1/2} \ |\xi |^{-\beta } \ ||f||_{L^2}, \end{aligned}$$

  along with (6.3) on

  $$\begin{aligned} ||e^{-|t||\xi |^{2 \beta }}||_{\dot{H^1}} \lesssim \nu ^{1/2} \ |\xi |^{\beta }. \end{aligned}$$

  We hence obtain

  $$\begin{aligned} ||k_1||_{H^1} \lesssim ||f||_{L^2}\quad \quad \forall \ |\xi |\ge 1. \end{aligned}$$

• Over $\{|\xi | \le 1 \}$: Similar to some steps in the previous case, we write

  $$\begin{aligned} \displaystyle ||k_1||_{H^1}\le & {} \bigg | \int _{\mathbb {R}} \frac{\hat{f}(\tau )}{i \tau +\nu |\xi |^{2\beta }} d\tau \bigg | \ ||\eta \ (1- e^{-\nu |t||\xi |^{2 \beta }})||_{H^1} \nonumber \\ \displaystyle\le & {} \nu ^{-1/2} \ |\xi |^{-\beta } \ ||f||_{L^2} \ ||\eta \ (1- e^{-\nu |t||\xi |^{2 \beta }})||_{H^1}. \end{aligned}$$

  But, for the right term, we follow [17], and use the fact that for $n\ge 1$, we have $|| \ |t^n| \ \eta (t)||_{H^1} \lesssim 2^n$, to get

  $$\begin{aligned} \displaystyle ||\eta (t) \ (1- e^{-\nu |t||\xi |^{2 \beta }})||_{H^1}\lesssim & {} \sum _{n\ge 1} \bigg | \bigg | \frac{|t^n| \ \eta (t) \ | \nu ^n | \ |\xi |^{2\beta n }}{n!} \bigg | \bigg | _{H^1} \\ \displaystyle\lesssim & {} \sum _{n \ge 1} \frac{\nu ^n |\xi |^{2 \beta n}}{n !} \ || \ |t^n| \ \eta (t)||_{H^1} \\ \displaystyle\lesssim & {} \nu |\xi |^{2 \beta } \sum _{n\ge 1} \frac{2^n}{n!} \lesssim \nu |\xi |^{2 \beta } \;. \end{aligned}$$

  And as $|\xi | \le 1$, we obtain

  $$\begin{aligned} ||k_1||_{H^1} \lesssim \sqrt{\nu } \, |\xi |^{\beta } \ ||f||_{L^2} \lesssim ||f||_{L^2}. \end{aligned}$$

Finally, covering both frequency domains, we get

$$\begin{aligned} ||k_1||_{H^1} \lesssim ||f||_{L^2} \quad \quad \forall \ \xi \in \mathbb {R}. \end{aligned}$$

(6.4)

Now, we move to estimate $||k_2||_{H^1}$ which in turn will be split in the following sense

$$\begin{aligned} \displaystyle k_2(t)= & {} \eta (t) \int _{|\tau | \le 1} \hat{f}(\tau )\ \frac{e^{i t \tau } - 1}{i \tau + \nu |\xi |^{2\beta }} \ d\tau + \eta (t) \int _{|\tau | \ge 1} \hat{f}(\tau )\ \frac{e^{i t \tau } - 1}{i \tau +\nu |\xi |^{2\beta }} \ d\tau \\ \displaystyle:= & {} k_{2,l}(t) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + \quad \quad \quad \quad \quad \quad k_{2,h}(t). \end{aligned}$$

We proceed with estimating $||k_{2,l}||_{H^1}$. Knowing that

$$\begin{aligned} \sum _{n \ge 1} \frac{x^n}{n !} = e^x - 1, \end{aligned}$$

and using the fact that $|\tau | \le 1$ followed by Cauchy–Schwarz inequality, we write

$$\begin{aligned} \displaystyle || k_{2,l} ||_{H^1}= & {} \bigg | \bigg | \eta (t) \int _{|\tau | \le 1} \sum _{n \ge 1} \frac{(i t \tau )^n}{n !} \ \frac{\hat{f}(\tau )}{i \tau +\nu |\xi |^{2\beta }} d\tau \bigg | \bigg |_{H^1} \\ \displaystyle\lesssim & {} \sum _{n \ge 1} \frac{|| \ |t^n| \ \eta (t)||_{H^1}}{n !} \ \int _{|\tau |\le 1} \bigg | \frac{(i \tau )^n \ \hat{f}(\tau )}{i \tau + \nu |\xi |^{2\beta }} \bigg | d \tau \\ \displaystyle\lesssim & {} \bigg ( \sum _{n \ge 1} \frac{1}{(n-1)!} \bigg ) \ ||f||_{L^2}\; \sup _{n\ge 1} \bigg ( \int _{|\tau |\le 1} \frac{|\tau |^{2n} \, d\tau }{\tau ^2 + \nu ^2 |\xi |^{4 \beta }} \bigg ) ^{\frac{1}{2}} \\ \displaystyle\lesssim & {} \ ||f||_{L^2} . \end{aligned}$$

Hence,

$$\begin{aligned} ||k_{2,l}||_{H^1} \lesssim ||f||_{L^2} \quad \quad \forall \ \xi \in \mathbb {R}. \end{aligned}$$

(6.5)

On the other hand, benefiting from the continuity of the Fourier Transform operator from $L^1$ to $L^{\infty }$, Cauchy–Schwarz inequality and Parseval-Plancherel identity, along with (6.2), we get

$$\begin{aligned} \displaystyle || k_{2,h} ||_{H^1}= & {} \bigg | \bigg | \eta (t) \overline{{\mathcal {F}}}_t \bigg ( \frac{\hat{f}(\tau ) \ {1\hspace{-1mm} \textrm{I}_{|\tau | \ge 1} }}{i \tau +\nu |\xi |^{2\beta }} \bigg )(t) \bigg | \bigg |_{H^1_t} \nonumber \\ \displaystyle\lesssim & {} \bigg | \bigg |\overline{{\mathcal {F}}}_t\bigg ( \frac{\hat{f}(\tau ) \ 1\hspace{-1mm} \textrm{I}_{|\tau | \ge 1} }{i \tau +\nu |\xi |^{2\beta }}\bigg ) \bigg | \bigg |_{L^{\infty }} \ ||\eta ||_{H^1} + ||\eta ||_{L^\infty } \ \bigg | \bigg |\overline{{\mathcal {F}}}_t \bigg ( \frac{\hat{f}(\tau ) \ 1\hspace{-1mm} \textrm{I}_{|\tau | \ge 1} }{i \tau +\nu |\xi |^{2\beta }} \bigg ) \bigg | \bigg |_{\dot{H}^1} \nonumber \\ \displaystyle\lesssim & {} \bigg | \bigg |\frac{\hat{f}(\tau ) \ 1\hspace{-1mm} \textrm{I}_{|\tau | \ge 1} }{i \tau +\nu |\xi |^{2\beta }} \bigg | \bigg |_{L^1} + \bigg | \bigg |\overline{{\mathcal {F}}}_t \bigg ( \frac{\hat{f}(\tau ) \ 1\hspace{-1mm} \textrm{I}_{|\tau | \ge 1} }{i \tau + \nu |\xi |^{2\beta }} \bigg ) \bigg | \bigg |_{\dot{H}^1} \nonumber \\ \displaystyle\lesssim & {} ||f||_{L^2} \ \bigg ( \int _{|\tau | \ge 1} \frac{d \tau }{|\tau |^2 + \nu ^2 |\xi |^{4 \beta }} \bigg )^{\frac{1}{2}} + ||f||_{L^2} \ \sup _{|\tau | \ge 1} \bigg ( \frac{|\tau |^2}{|\tau |^2 + \nu ^2 |\xi |^{4 \beta }} \bigg )^{\frac{1}{2}} \nonumber \\ \displaystyle\lesssim & {} \ ||f||_{L^2} . \end{aligned}$$

(6.6)

Ultimately, gathering (6.4), (6.5), and (6.6), we end up with

$$\begin{aligned} ||k||_{H^1} \lesssim ||f||_{L^2} \quad \forall \ \xi \in \mathbb {R} \end{aligned}$$

and this terminates the proof. $\square$