Well-posedness in Sobolev spaces for semi-linear 3-evolution equations

Abstract

We prove local in time well-posedness of the Cauchy problem in Sobolev spaces for semi-linear 3-evolution equations of the first order. We require real principal part, but complex valued coefficients for the lower order terms. Therefore decay conditions on the imaginary parts are needed, as $x\rightarrow \infty$.

1 Introduction and main result

Let us consider the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} P(t,x,u(t,x),D_t,D_x)u(t,x)=f(t,x),&{}(t,x)\in [0,T]\times {\mathbb {R}}\\ u(0,x)=u_0(x),\quad x\in {\mathbb {R}}\end{array}\right. \qquad \end{aligned}$$

(1.1)

for the semi-linear operator

$$\begin{aligned} P:=&D_t +a_3(t)D_x^3+a_2(t,x,u)D_x^2+a_1(t,x,u)D_x+a_0(t,x,u),\qquad \end{aligned}$$

(1.2)

where $D:=\frac{1}{i}\partial , a_3\in C([0,T];{\mathbb {R}}), a_j\in C([0,T]; C^\infty ({\mathbb {R}}\times {\mathbb {C}}))$ and $x\mapsto a_j(t,x,w)\in {\mathcal B}^\infty ({\mathbb {R}})$ (here ${\mathcal B}^\infty ({\mathbb {R}})$ is the space of complex valued functions which are bounded on ${\mathbb {R}}$ together with all their derivatives), for $j=0,1,2$.

We are so dealing with a semi-linear non-kowalewskian 3-evolution equation $Pu=f$ with the real characteristic (in the sense of Petrowski) $\tau =-a_3(t)\xi ^3$. In the case $a_3(t)\equiv -1, a_2\equiv a_0\equiv 0, a_1(t,x,u)=-6u$, we recover the Korteweg-de Vries equation.

The aim of this paper is to give suitable decay conditions on the coefficients in order that the Cauchy problem (1.1) is locally in time well-posed in $H^s$ with $s$ great enough, and in $H^\infty$.

The well-posedness result will be achieved by developing the linear technique of [5] (coming from the examples in [7, 8] and used also in [3, 4]), and applying then a fixed point argument, following the ideas of [1, 2, 9].

We consider here $x\in {\mathbb {R}}$ only for simplicity’s sake; $x\in {\mathbb {R}}^n, n\ge 2$ could be considered with only technical changes in our proofs, see [10, 13].

The assumption $a_3(t)\in {\mathbb {R}}$ is due to the necessary condition of the Lax-Mizohata Theorem (cf. [15]), while the assumptions $a_j(t,x,w)\in {\mathbb {C}}$ for $0\le j\le 2$ imply some decay conditions on the coefficients because of the necessary condition of Ichinose (cf. [12]). We shall thus assume, in the following, that there exists a constant $C_3>0$ such that

$$\begin{aligned} a_3(t)\ge C_3\quad \forall t\in [0,T], \end{aligned}$$

(1.3)

and that there exist constants $C,\varepsilon >0$ and a function $h:\ {\mathbb {C}}\rightarrow {\mathbb {R}}^+$ bounded on compact sets (for instance, $h$ continuous) such that for all $(t,x,w)\in [0,T]\times {\mathbb {R}}\times {\mathbb {C}}$:

$$\begin{aligned} |\mathrm{Im\,}a_2(t,x,w)|&\le \frac{C}{\langle x\rangle ^{1+\varepsilon }}h(w) \end{aligned}$$

(1.4)

$$\begin{aligned} |\mathrm{Im\,}a_1(t,x,w)|&\le \frac{C}{\langle x\rangle ^{1/2}}h(w) \end{aligned}$$

(1.5)

$$\begin{aligned} |\mathrm{Re\,}a_2(t,x,w)|&\le Ch(w) \end{aligned}$$

(1.6)

$$\begin{aligned} |\partial _x\mathrm{Re\,}a_2(t,x,w)|&\le \frac{C}{\langle x\rangle ^{1/2}}h(w) \end{aligned}$$

(1.7)

$$\begin{aligned} |\partial _w a_2(t,x,w)|&\le \frac{C}{\langle x\rangle ^{1/2}}h(w), \end{aligned}$$

(1.8)

with the notation $\langle x\rangle :=\sqrt{1+x^2}$.

Under the assumptions above we prove the following result:

Theorem 1.1

Let $P$ be as in (1.2) satisfying (1.3)–(1.8). Then the Cauchy problem (1.1) is locally in time well-posed in $H^\infty$. More precisely, for every given $s>5/2$ and for all $f\in C([0,T];H^s({\mathbb {R}}))$ and $u_0\in H^s({\mathbb {R}})$, there exists $0<T^*\le T$ and a unique solution $u\in C([0,T^*]; H^{s}({\mathbb {R}}))$ of (1.1) satisfying the following inequality:

$$\begin{aligned} \Vert u(t,\cdot )\Vert ^2_s\le e^{\sigma t} \left( \Vert u_0\Vert ^2_s+\int \limits _0^t\Vert f(\tau ,\cdot )\Vert ^2_sd\tau \right) \qquad \forall t\in [0,T^*], \end{aligned}$$

(1.9)

for some positive constant $\sigma$ depending on $s$.

Remark 1.2

Estimate (1.9) gives local in time well-posedness of the Cauchy problem (1.1) in $H^s, s>5/2$. By the same estimate we gain also $H^\infty$ well-posedness: if the Cauchy data are $f\in C([0,T];H^\infty ({\mathbb {R}}))$ and $u_0\in H^\infty ({\mathbb {R}})$, then the solution $u\in C([0,T]; H^s)$ for every $s>5/2$, and then by Sobolev’s embeddings we immediately get $u\in C([0,T]; H^\infty )$.

Example 1.3

Let us consider the non-linear equation

$$\begin{aligned} P(t,x,u,D_t,D_x)=D_t u+a_3(t)D_x^3+a_2(x,u)D_x^2u=f(t,x) \end{aligned}$$

with

$$\begin{aligned} f&\in C([0,T];H^s({\mathbb {R}})),\quad \ s\ge 5/2\\ a_3(t)&\in C([0,T];{\mathbb {R}}),\quad a_3(t)\ge C_3>0\ \forall t\in [0,T]\\ a_2(x,w)&= i\frac{\sin x^\alpha }{(1+x^2)^{\frac{1+\varepsilon }{2}}}\frac{1}{1+w^2}, \qquad \alpha ,\varepsilon >0. \end{aligned}$$

Then

$$\begin{aligned}&|\mathrm{Im\,}a_2|\le \frac{1}{\langle x\rangle ^{1+\varepsilon }}\\&|\partial _w a_2|=\left| -i\frac{\sin x^\alpha }{\langle x\rangle ^{1+\varepsilon }} \frac{2w}{(1+w^2)^2}\right| \le \frac{2}{\langle x\rangle ^{1+\varepsilon }} \le \frac{2}{\langle x\rangle ^{1/2}}\,. \end{aligned}$$

Therefore Theorem 1.1 can be applied to get, for some $0<T^*\le T$, a unique solution $u\in C([0,T^*];H^{s}({\mathbb {R}}))$ of the Cauchy problem

$$\begin{aligned} \left\{ \!\begin{array}{l} P(t,x,u,D_t,D_x)u(t,x)=f(t,x)\quad (t,x)\in [0,T^*]\times {\mathbb {R}}\\ u(0,x)=u_0(x)\in H^s({\mathbb {R}}), \quad x\in {\mathbb {R}}. \end{array}\right. \end{aligned}$$

The same result holds if, more in general, we take

$$\begin{aligned} a_2(t,x,w)=i a'_2(t,x)a''_2(w) \end{aligned}$$

for some real valued functions $a'_2\in C([0,T];{\mathcal B}^\infty ({\mathbb {R}}))$ satisfying (1.4) and $a''_2\in C([0,T];C^\infty ({\mathbb {R}}))$ with bounded derivative $\partial _w a''_2$.

Example 1.4

By simple computations it is easy to check that Example 1.3 works also considering, for example, $a_2(t,x,w)={\displaystyle }\frac{ia_2'(t,x)}{\langle x+w\rangle ^{1+\varepsilon }}$, or $a_2(t,x,w)={\displaystyle }\frac{ia_2'(t,x)}{\langle x\rangle ^{1+\varepsilon }+w^2}$, with a real valued function $a'_2\in C([0,T];{\mathcal B}^\infty ({\mathbb {R}}))$ satisfying (1.4).

2 Notation and main tools

The proof of Theorem 1.1 is based on the pseudo-differential calculus. In this paper we denote by $S^m:=S^m({\mathbb {R}}^2)$ the space of symbols $a(x,\xi )$ such that for every $\alpha ,\beta \in {\mathbb {N}}$

$$\begin{aligned} \sup _{x,\xi \in {\mathbb {R}}}\vert \partial _\xi ^\alpha D_x^\beta a(x,\xi )\vert \langle \xi \rangle _h^{-m+|\alpha |}<\infty , \end{aligned}$$

where $\langle \xi \rangle _h:=\sqrt{h^2+\xi ^2}, h\ge 1$ fixed. Our symbols will be of the form $a(x,w,\xi )$, depending smoothly on a parameter $w\in {\mathbb {C}}$.

The idea of the proof is to fix $u\in B_r$,

$$\begin{aligned} B_r:=\{u\in C([0,T];H^s):\ \sup _{t\in [0,T]}\Vert u(t,\cdot )\Vert _s\le r\}, \end{aligned}$$

with $r>0$ to be determined later on, to solve the linear Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{l} P(t,x,u,D_t,D_x)v=f\\ v(0,x)=u_0(x) \end{array}\right. \end{aligned}$$

(2.1)

in the unknown $v(t,x)$ following [5], and then use a fixed point argument to find the solution of the non-linear Cauchy problem (1.1).

For this reason we recall now some definitions and results from [5]. According to [5], formula (2.4) and Remark 3.1], we define

$$\begin{aligned}&\lambda _2(x,\xi ):=M_2\int \limits _0^x\langle y\rangle ^{-1-\varepsilon } \psi \left( \frac{\langle y\rangle }{\langle \xi \rangle _h^2}\right) dy\end{aligned}$$

(2.2)

$$\begin{aligned}&\lambda _1(x,\xi ):=M_1\int \limits _0^x\langle y\rangle ^{-\frac{1}{2}} \psi \left( \frac{\langle y\rangle }{\langle \xi \rangle _h^2}\right) dy \cdot \langle \xi \rangle _h^{-1} \end{aligned}$$

(2.3)

where the constants $M_1,M_2>0$ have to be chosen in the sequel, $\psi \in C^\infty _0({\mathbb {R}})$ satisfies $0\le \psi \le 1$ and

$$\begin{aligned} \psi (y)=\left\{ \begin{array}{l@{\quad }l} 1&{}|y|\le \frac{1}{2}\\ 0&{}|y|\ge 1. \end{array}\right. \end{aligned}$$

Then

$$\begin{aligned}&|\lambda _2(x,\xi )|\le M_2\int \limits _0^{\langle x\rangle } \langle y\rangle ^{-1-\varepsilon }dy\le C_2\\&|\lambda _1(x,\xi )|\le CM_1\langle x\rangle ^{\frac{1}{2}} \langle \xi \rangle _h^{-1}\chi _{\mathrm{supp\,}\psi }(x)\le C_1M_1, \end{aligned}$$

for some $C_2,C_1>0$, where $\chi _{\mathrm{supp\,}\psi }$ is the characteristic function of the support of $\psi (\langle x\rangle / \langle \xi \rangle _h^2)$.

Therefore, for $\Lambda (x,\xi ):=\lambda _1(x,\xi )+\lambda _2(x,\xi )$, we have that

$$\begin{aligned} |\Lambda (x,\xi )|\le C'_2 \end{aligned}$$

(2.4)

for some $C'_2>0$; moreover, from [5, Lemma 2.1] (with $\delta =0$):

$$\begin{aligned}&|\partial _\xi ^\alpha D_x^\beta \Lambda (x,\xi )|\le \delta _{\alpha ,\beta } \langle \xi \rangle _h^{-\alpha }\qquad \forall \alpha ,\beta \in {\mathbb {N}}, \end{aligned}$$

(2.5)

for some $\delta _{\alpha ,\beta }>0$.

This proves that the pseudo-differential operator $e^{\Lambda (x,D_x)}$ has symbol $e^{\Lambda (x,\xi )}\in S^0$, and then we can apply the following:

Lemma 2.1

(see Lemma 2.3, [5]) Let $\Lambda (x,\xi )$ satisfy (2.5). There exists a constant $h_0\ge 1$ such that for $h\ge h_0$ the operator $e^\Lambda$ is invertible and

$$\begin{aligned}&(e^\Lambda )^{-1}=e^{-\Lambda }(I+R), \end{aligned}$$

(2.6)

where $I$ is the identity operator and $R$ is an operator of the form $R=\sum _{n=1}^{+\infty }r^n$ with principal symbol

$$\begin{aligned} \tilde{r}(x,\xi )=\partial _\xi \Lambda (x,\xi ) D_x\Lambda (x,\xi ). \end{aligned}$$

(2.7)

We conclude this section by recalling two results that will be crucial in determining the minimal assumptions needed on the coefficients $a_j$ in (1.2) to get the well-posedness result here presented:

Theorem 2.2

(Sharp-Gårding inequality, [14]) Let $a(x,D_x)$ be a pseudo-differential operator with symbol $a(x,\xi )\in S^m$ suche that $\mathrm{Re\,}a(x,\xi )\ge 0$. Then there exists $c>0$ such that

$$\begin{aligned} \mathrm{Re\,}\langle a(x,D_x)u,u\rangle \ge -c\Vert u\Vert _{(m-1)/2}^2. \end{aligned}$$

(2.8)

Theorem 2.3

(Fefferman–Phong inequality, [11]) Let $a(x,\xi )\in S^m$ with $a(x,\xi )\ge 0$. Then there exists $c>0$ such that

$$\begin{aligned} \mathrm{Re\,}\langle a(x,D_x)u,u\rangle \ge -c\Vert u\Vert _{(m-2)/2}^2. \end{aligned}$$

(2.9)

3 Proof of Theorem 1.1

To start with the proof we fix $s>5/2, f,u\in C([0,T];H^s)$ and $u_0\in H^s({\mathbb {R}})$ , and consider the linear Cauchy problem (2.1). A direct application of [5, Theorem 1.1 and Remark 1.5] immediately gives the existence of a unique solution $v\in C([0,T];H^s)$ of problem (2.1) such that

$$\begin{aligned} \Vert v(t,\cdot )\Vert _s^2\le C_s(u)\left( \Vert u_0\Vert _s^2+\int \limits _0^t \Vert f(\tau ,\cdot )\Vert _s^2d\tau \right) \qquad \forall t\in [0,T]\qquad \end{aligned}$$

(3.1)

for some $C_s(u)>0$, since assumption (1.4) gives no loss of derivatives ($\sigma =2\delta =0$ in [5, Theorem 1.1]). This is not enough for our purposes, since to proceed with the proof and apply a fixed point scheme we need to know precisely the constant $C_s(u)$. We thus quickly retrace in what follows the proof of Theorem 1.1 in [5], taking care of the dependence of the constants on the fixed function $u$, and taking advantage of the choice of $p=3$.

We write

$$\begin{aligned} iP(t,x,u,D_t,D_x)=\partial _t+A(t,x,u,D_x) \end{aligned}$$

with

$$\begin{aligned} A(t,x,u,D_x):=ia_3(t)D_x^3+ia_2(t,x,u)D_x^2+ia_1(t,x,u)D_x +ia_0(t,x,u) \end{aligned}$$

and compute the symbol of the pseudo-differential operator $(e^\Lambda )^{-1}Ae^\Lambda$.

We have:

$$\begin{aligned} \sigma (Ae^\Lambda )&= (ia_3\xi ^3+ia_2\xi ^2+ia_1\xi )e^\Lambda +(3ia_3\xi ^2+2ia_2\xi )D_xe^\Lambda \nonumber \\&+\frac{1}{2}(6ia_3\xi )D_x^2e^\Lambda +\tilde{A} e^\Lambda \end{aligned}$$

(3.2)

for some $\tilde{A}\in S^0$.

To compute then $\sigma ((e^\Lambda )^{-1}Ae^\Lambda )$ we need to write down the symbol of $(e^\Lambda )^{-1}$ by means of (2.6) and (2.7).

In the sequel it will be useful to estimate, from (2.2) and (2.3):

$$\begin{aligned} |\partial _\xi \lambda _2(x,\xi )|&\le M_2\left| \int \limits _0^x \langle y\rangle ^{\!-1\!-\varepsilon }\langle y\rangle \left( \partial _\xi \frac{1}{\langle \xi \rangle _h^2}\right) \psi '\left( \frac{\langle y\rangle }{\langle \xi \rangle _h^2}\right) dy\right| \!\le \! C'_2M_2\frac{\langle x\rangle ^{1-\varepsilon }}{\langle \xi \rangle _h^3} \chi _{\mathrm{supp\,}\psi '}\end{aligned}$$

(3.3)

$$\begin{aligned} |\partial _x\lambda _2(x,\xi )||&\le M_2\langle x\rangle ^{-1-\varepsilon } \psi \left( \frac{\langle x\rangle }{\langle \xi \rangle _h^2}\right) \le C'_2M_2\langle x\rangle ^{-1-\varepsilon }\end{aligned}$$

(3.4)

$$\begin{aligned} |\partial _\xi \lambda _1(x,\xi )|&\le M_1\left| \int \limits _0^x\langle y\rangle ^{-\frac{1}{2}}\langle y\rangle \left( \partial _\xi \frac{1}{\langle \xi \rangle _h^2}\right) \psi '\left( \frac{\langle y\rangle }{\langle \xi \rangle _h^2}\right) dy\right| \langle \xi \rangle _h^{-1}\nonumber \\&+\,M_1\left| \int \limits _0^x\langle y\rangle ^{-\frac{1}{2}} \psi \left( \frac{\langle y\rangle }{\langle \xi \rangle _h^2}\right) dy\cdot \left( \partial _\xi \frac{1}{\langle \xi \rangle _h}\right) \right| \nonumber \\&\le C'_1M_1\left( \frac{\langle x\rangle ^{3/2}}{\langle \xi \rangle _h^4} \chi _{\mathrm{supp\,}\psi '}+\frac{\langle x\rangle ^{1/2}}{\langle \xi \rangle _h^2} \right) \end{aligned}$$

(3.5)

$$\begin{aligned} |\partial _x\lambda _1(x,\xi )|&\le M_1\langle x\rangle ^{-\frac{1}{2}} \psi \left( \frac{\langle x\rangle }{\langle \xi \rangle _h^2}\right) \langle \xi \rangle _h^{-1} \le C'_1M_1\frac{\langle x\rangle ^{-\frac{1}{2}}}{\langle \xi \rangle _h} \end{aligned}$$

(3.6)

for some $C'_2,C'_1>0$, where $\chi _{\mathrm{supp\,}\psi '}$ is the characteristic function of

$$\begin{aligned} \mathrm{supp\,}\psi '\left( \frac{\langle x\rangle }{\langle \xi \rangle _h^2}\right) \subseteq \left\{ x\in {\mathbb {R}}:\ \frac{1}{2}\langle \xi \rangle _h^2\le \langle x\rangle \le \langle \xi \rangle _h^2\right\} . \end{aligned}$$

Therefore

$$\begin{aligned} |\tilde{r}(x,\xi )|=|\partial _\xi \Lambda (x,\xi ) \cdot D_x\Lambda (x,\xi )|\le C_{M_1,M_2} \langle x\rangle ^{-\frac{1}{2}-\varepsilon }\langle \xi \rangle _h^{-2}; \end{aligned}$$

by simple computations we get that $\tilde{r}(x,\xi )\in S^{-2}$ and, by (2.6) and (2.7):

$$\begin{aligned} (e^\Lambda )^{-1}=e^{-\Lambda }(I+ \tilde{r}+R_{-3}) \end{aligned}$$

with $\tilde{r}(x,D)$ a pseudo-differential operator with symbol $\tilde{r}(x,\xi )$ and $R_{-3}$ an operator of order $-3$.

Then, from (3.2):

$$\begin{aligned} \sigma ((e^\Lambda )^{-1}Ae^\Lambda )&= (e^{-\Lambda }+e^{-\Lambda } \tilde{r}) \left( ia_3\xi ^3+ia_2\xi ^2+ia_1\xi \right) e^\Lambda \\&+\,(e^{-\Lambda }+e^{-\Lambda } \tilde{r})\left( 3ia_3\xi ^2+2ia_2\xi \right) (D_x\Lambda )e^\Lambda \\&+\,(e^{-\Lambda }+e^{-\Lambda } \tilde{r})(3ia_3\xi ) \left( D_x^2\Lambda +(D_x\Lambda )^2\right) e^\Lambda \\&-\,(\partial _\xi \Lambda )\left( iD_xa_2\xi ^2\right) -(\partial _\xi \Lambda ) \left( ia_3\xi ^3+ia_2\xi ^2\right) (D_x\Lambda )\\&-\,(\partial _\xi \Lambda )\left( 3ia_3\xi ^2\right) \left( D_x^2\Lambda +(D_x\Lambda )^2\right) \\&+\,\frac{1}{2}\left( \partial _\xi ^2\Lambda +(\partial _\xi \Lambda )^2\right) \left( ia_3\xi ^3\right) \left( D_x^2\Lambda +(D_x\Lambda )^2\right) +A'_0\\&= ia_3\xi ^3+ia_2\xi ^2+ia_1\xi + \tilde{r}(x,\xi )\left( ia_3\xi ^3\right) +\left( 3ia_3\xi ^2\right) (D_x\Lambda ) \\&+\,(2ia_2\xi )(D_x\Lambda )+(3ia_3\xi )\left( D_x^2\Lambda +(D_x\Lambda )^2\right) \\&-\,(\partial _\xi \Lambda ) \left( iD_xa_2\xi ^2\right) -\tilde{r}(x,\xi )\left( ia_3\xi ^3\right) +A_0''\\&= ia_3\xi ^3 +\left[ ia_2\xi ^2+\left( 3ia_3\xi ^2\right) (D_x\lambda _2)\right] \\&+\,\left[ ia_1\xi +\left( 3ia_3\xi ^2\right) (D_x\lambda _1) +(2ia_2\xi )(D_x\lambda _2)\right. \\&\left. +\,(3ia_3\xi )\left( D_x^2\lambda _2+\left( D_x\lambda _2\right) ^2\right) -(\partial _\xi \lambda _2)\left( iD_xa_2\xi ^2\right) \right] +A_0 \end{aligned}$$

for some $A'_0, A''_0, A_0\in S^0$, since $a_3=a_3(t)$ and because of (3.3) and (3.5).

Therefore

$$\begin{aligned} \sigma ((e^\Lambda )^{-1}Ae^\Lambda )=A_3+A_2+A_1+A_0, \end{aligned}$$

with $A_j\in S^j$ defined by:

$$\begin{aligned} A_3(t,\xi )&:= ia_3\xi ^3\\ A_2(t,x,u,\xi )&:= ia_2\xi ^2+(3ia_3\xi ^2)(D_x\lambda _2)\\ A_1(t,x,u,\xi )&:= ia_1\xi +(3ia_3\xi ^2)(D_x\lambda _1) +(2ia_2\xi )(D_x\lambda _2)\\&+\,(3ia_3\xi )(D_x^2\lambda _2+(D_x\lambda _2)^2) -(\partial _\xi \lambda _2)(iD_xa_2\xi ^2). \end{aligned}$$

Note that assumption (1.3) implies

$$\begin{aligned} \mathrm{Re\,}A_3(t,\xi )=0. \end{aligned}$$

(3.7)

As in the proof of Theorem 1.1 of [5], we look first for $M_2>0$ great enough to apply the Fefferman–Phong inequality (2.9) to

$$\begin{aligned} \mathrm{Re\,}A_2=-\mathrm{Im\,}a_2\xi ^2+3a_3\xi ^2\partial _x\lambda _2. \end{aligned}$$

(3.8)

By (1.4)

$$\begin{aligned} |\mathrm{Im\,}a_2(t,x,u)\xi ^2|\le \frac{C}{\langle x\rangle ^{1+\varepsilon }} h(u)\langle \xi \rangle _h^2, \end{aligned}$$

(3.9)

while, by (1.3) and (3.4), for $|\xi |\ge h$ we have

$$\begin{aligned} 3a_3(t)\xi ^2\partial _x\lambda _2(x,\xi )&= 3M_2a_3(t)\xi ^2\langle x\rangle ^{-1-\varepsilon } \psi \left( \frac{\langle x\rangle }{\langle \xi \rangle _h^2}\right) \nonumber \\&\ge 3M_2C_3\psi \langle x\rangle ^{-1-\varepsilon }|\xi |^2\nonumber \\&\ge \frac{3}{\sqrt{2}}M_2C_3\psi \langle x\rangle ^{-1-\varepsilon } \langle \xi \rangle _h^2. \end{aligned}$$

(3.10)

Substituting (3.9) and (3.10) in (3.8):

$$\begin{aligned} \mathrm{Re\,}A_2&\ge \frac{3}{\sqrt{2}}M_2C_3\psi \frac{\langle \xi \rangle _h^2}{\langle x\rangle ^{1+\varepsilon }}- \frac{C}{\langle x\rangle ^{1+\varepsilon }}h(u)\langle \xi \rangle _h^2\\&= \psi \left( \frac{3}{\sqrt{2}}C_3M_2-Ch(u)\right) \frac{\langle \xi \rangle _h^2}{\langle x\rangle ^{1+\varepsilon }} -Ch(u)\frac{\langle \xi \rangle _h^2}{\langle x\rangle ^{1+\varepsilon }} (1-\psi )\\&\ge \psi \left( \frac{3}{\sqrt{2}}C_3M_2-Ch(u)\right) \frac{\langle \xi \rangle _h^2}{\langle x\rangle ^{1+\varepsilon }} -2Ch(u) \end{aligned}$$

since $\langle \xi \rangle _h^2\le 2\langle x\rangle$ on $\mathrm{supp\,}\left( 1- \psi \left( \frac{\langle x\rangle }{\langle \xi \rangle _h^2}\right) \right)$.

We thus choose $M_2>\sqrt{2}Cc_r/3C_3$, where

$$\begin{aligned} c_r:={\displaystyle }\sup _{\genfrac{}{}{0.0pt}1{(t,x)\in [0,T]\times {\mathbb {R}}}{u\in B_r}}h(u) \end{aligned}$$

is a positive constant because $h$ maps compact sets into bounded sets by assumption and $\sup _{(t,x)\in [0,T]\times {\mathbb {R}}}|u(t,x)|\le C_s\sup _{t\in [0,T]}\Vert u(t,\cdot )\Vert _s$ since $s>\frac{5}{2}>\frac{1}{2}$ by Sobolev embedding Theorem.

Then

$$\begin{aligned} \mathrm{Re\,}A_2(t,x,u,\xi )\ge -2Cc_r \end{aligned}$$

and, applying the Fefferman–Phong inequality (2.9) to the operator $\mathrm{Re\,}A_2(t,x,u,\xi )+2Cc_r$, we have that

$$\begin{aligned} \mathrm{Re\,}\langle \mathrm{Re\,}A_2z,z\rangle \ge -c(1+c_r) \Vert z\Vert _0^2 \end{aligned}$$

(3.11)

for some fixed constant $c>0$.

On the other hand, we can write the operator $\mathrm{Im\,}A_2(t,x,u,D_x)=i\mathrm{Re\,}a_2(t,x,u)D_x^2$ as

$$\begin{aligned} \mathrm{Im\,}A_2=\frac{\mathrm{Im\,}A_2+(\mathrm{Im\,}A_2)^*}{2}+ \frac{\mathrm{Im\,}A_2-(\mathrm{Im\,}A_2)^*}{2} \end{aligned}$$

(3.12)

with

$$\begin{aligned} \mathrm{Re\,}\left\langle \frac{\mathrm{Im\,}A_2-(\mathrm{Im\,}A_2)^*}{2} z,z\right\rangle&= \frac{1}{2}\mathrm{Re\,}\langle \mathrm{Im\,}A_2 z,z\rangle -\frac{1}{2}\mathrm{Re\,}\langle z, \mathrm{Im\,}A_2 z\rangle \nonumber \\&= \frac{1}{2}\mathrm{Re\,}\langle \mathrm{Im\,}A_2 z,z\rangle -\frac{1}{2}\mathrm{Re\,}\overline{\langle \mathrm{Im\,}A_2 z,z\rangle }=0\qquad \qquad \quad \end{aligned}$$

(3.13)

and $\frac{\mathrm{Im\,}A_2+(\mathrm{Im\,}A_2)^*}{2}$ of order 1 since

$$\begin{aligned} \sigma (\mathrm{Im\,}A_2)+\sigma ((\mathrm{Im\,}A_2)^*)&\sim \sigma (\mathrm{Im\,}A_2)+\sum _{\alpha \ge 0}\frac{1}{\alpha !} \partial _\xi ^\alpha D_x^\alpha \overline{\sigma (\mathrm{Im\,}A_2)}\\&= i\mathrm{Re\,}a_2\xi ^2-i\mathrm{Re\,}a_2\xi ^2+\partial _\xi D_x(-i\mathrm{Re\,}a_2\xi ^2)+B_0\\&= -\,2\partial _x\mathrm{Re\,}a_2\xi +B_0 \end{aligned}$$

for some $B_0\in S^0$.

Let us now choose $M_1>0$ in order to apply the sharp-Gårding inequality (2.8) to

$$\begin{aligned} \tilde{A}_1(t,x,u,D_x):=A_1(t,x,u,D_x)-2(\partial _x\mathrm{Re\,}a_2)D_x \end{aligned}$$

with symbol

$$\begin{aligned} \tilde{A}_1(t,x,u,\xi )&= ia_1\xi +(3ia_3\xi ^2)(D_x\lambda _1) +(2ia_2\xi )(D_x\lambda _2)\nonumber \\&\!\!\!+\,(3ia_3\xi )(D_x^2\lambda _2+(D_x\lambda _2)^2) -(\partial _\xi \lambda _2)(iD_x a_2\xi ^2) -2(\partial _x\mathrm{Re\,}a_2)\xi .\nonumber \\ \end{aligned}$$

(3.14)

By (1.3) and (2.3):

$$\begin{aligned} \mathrm{Re\,}(3ia_3\xi ^2 D_x\lambda _1)&= 3a_3\xi ^2\partial _x\lambda _1\nonumber \\&\ge 3c|\xi |^2M_1\langle x\rangle ^{-1/2} \psi \left( \frac{\langle x\rangle }{\langle \xi \rangle _h^2}\right) \langle \xi \rangle _h^{-1}\nonumber \\&\ge \frac{3}{\sqrt{2}}C_3M_1\psi \frac{\langle \xi \rangle _h}{\langle x\rangle ^{1/2}} \end{aligned}$$

(3.15)

if $|\xi |\ge h$.

On the other hand, by (1.5):

$$\begin{aligned} |\mathrm{Re\,}(ia_1\xi )|=|\mathrm{Im\,}a_1|\cdot |\xi |\le \frac{C}{\langle x\rangle ^{1/2}}h(u)\langle \xi \rangle _h. \end{aligned}$$

(3.16)

By (1.6) and (3.4):

$$\begin{aligned} |\mathrm{Re\,}[(2ia_2\xi )(D_x\lambda _2)]&= |2\mathrm{Re\,}a_2\xi \partial _x\lambda _2|\nonumber \\&\le 2|\mathrm{Re\,}a_2|\langle \xi \rangle _hM_2\langle x\rangle ^{-1-\varepsilon } \psi \left( \frac{\langle x\rangle }{\langle \xi \rangle _h^2}\right) \nonumber \\&\le 2CM_2h(u)\psi \frac{\langle \xi \rangle _h}{\langle x\rangle ^{1/2}}. \end{aligned}$$

(3.17)

By (1.3):

$$\begin{aligned} \mathrm{Re\,}[(3ia_3\xi )(D_x^2\lambda _2+(D_x\lambda _2)^2)=0. \end{aligned}$$

(3.18)

By (1.7), (1.8) and (3.3):

$$\begin{aligned} |\mathrm{Re\,}[(\partial _\xi \lambda _2)(iD_xa_2\xi ^2)]|&= |\partial _\xi \lambda _2|\cdot |\mathrm{Re\,}\partial _x\big (a_2(t,x,u)\big )|\cdot |\xi |^2\nonumber \\&\le cM_2\frac{\langle x\rangle ^{1-\varepsilon }}{\langle \xi \rangle _h^3} \chi _{\mathrm{supp\,}\psi '} \cdot |\mathrm{Re\,}(\partial _x a_2)+\mathrm{Re\,}(\partial _w a_2)(\partial _x u)| \langle \xi \rangle _h^2\nonumber \\&\le cM_2\frac{1}{\langle \xi \rangle _h^2} \chi _{\mathrm{supp\,}\psi '}\cdot \frac{C}{\langle x\rangle ^{1/2}}h(u)(1+|\partial _xu|) \langle \xi \rangle _h^2\nonumber \\&\le cCM_2h(u)(1+|\partial _x u|) \frac{\langle \xi \rangle _h}{\langle x\rangle ^{1/2}} \end{aligned}$$

(3.19)

for some $c>0$.

By (1.7) and (1.8):

$$\begin{aligned} \left| \partial _x\mathrm{Re\,}\big (a_2(t,x,u(t,x))\big )\xi \right|&= |\partial _x(\mathrm{Re\,}a_2)+\mathrm{Re\,}(\partial _w a_2)(\partial _x u)|\cdot |\xi |\nonumber \\&\le \frac{C}{\langle x\rangle ^{1/2}}h(u)(1+|\partial _x u|) \langle \xi \rangle _h. \end{aligned}$$

(3.20)

Substituting (3.15)–(3.20) in (3.14) and taking into account that $\langle x\rangle ^{-1/2}\langle \xi \rangle _h\le 2$ on $\mathrm{supp\,}(1-\psi )$, we finally find a constant $c>0$, which depends also on the already chosen $M_2$, such that

$$\begin{aligned} \mathrm{Re\,}\tilde{A}_1&\ge \bigg (\frac{3C_3}{\sqrt{2}}M_1\psi -Ch(u)-2CM_2\psi h(u)\\&-\,cCM_2h(u)(1+|\partial _xu|) -Ch(u)(1+|\partial _xu|)\bigg ) \frac{\langle \xi \rangle _h}{\langle x\rangle ^{1/2}}\\&= \psi \left( \frac{3C_3}{\sqrt{2}}M_1-C(M_2)h(u)(1+|\partial _xu|)\right) \frac{\langle \xi \rangle _h}{\langle x\rangle ^{1/2}}\\&-\,(1-\psi )C(M_2)h(u)(1+|\partial _xu|) \frac{\langle \xi \rangle _h}{\langle x\rangle ^{1/2}}\\&\ge -\,2C(M_2)C_r \end{aligned}$$

for some constant $C(M_2)>0$ which depends on the already chosen $M_2$, and for $M_1\ge \frac{\sqrt{2}C(M_2)}{3C_3}C_r$ with

$$\begin{aligned} C_r:=\sup _{\genfrac{}{}{0.0pt}1{(t,x)\in [0,T]\times {\mathbb {R}}}{u\in B_r}} h(u)(1+|\partial _xu|)\ge c_r. \end{aligned}$$

Applying the sharp-Gårding inequality (2.8) to $\tilde{A}_1+2C(M_2)C_r$ we obtain that

$$\begin{aligned} \mathrm{Re\,}\langle \tilde{A}_1(t,x,u,D_x)z,z\rangle \ge -c(1+2C(M_2)C_r) \Vert z\Vert _0^2 \end{aligned}$$

(3.21)

for some fixed constant $c>0$.

Summing up, we have chosen $M_1,M_2>0$ sufficiently large so that $A_\Lambda :=(e^\Lambda )^{-1}Ae^\Lambda$ satisfies:

$$\begin{aligned} \mathrm{Re\,}\langle (e^\Lambda )^{-1}Ae^\Lambda z,z\rangle \ge -\tilde{C}(1+C_r)\Vert z\Vert _0^2 \end{aligned}$$

(3.22)

for some fixed constant $\tilde{C}>0$, because of (3.7), (3.11), (3.13) and (3.21).

Now, for every $z\in C([0,T];H^3)\cap C^1([0,T];L^2)$, from the identity $iP_\Lambda =\partial _t+A_\Lambda$, where $P_\Lambda :=(e^\Lambda )^{-1}Pe^\Lambda , A_\Lambda :=(e^\Lambda )^{-1}Ae^\Lambda$, we have:

$$\begin{aligned} \frac{d}{dt}\Vert z\Vert _0^2&= 2\mathrm{Re\,}\langle \partial _t z,z\rangle =2\mathrm{Re\,}\langle iP_\Lambda z,z\rangle -2\mathrm{Re\,}\langle A_\Lambda z,z\rangle \\&\le 2(\Vert P_\Lambda z\Vert _0^2+\Vert z\Vert _0^2)+\tilde{C}(1+C_r)\Vert z\Vert _0^2\\&= 2\Vert P_\Lambda z\Vert _0^2+(2+\tilde{C}(1+C_r))\Vert z\Vert _0^2. \end{aligned}$$

By Gronwall’s Lemma:

$$\begin{aligned} \Vert z\Vert _0^2\le e^{(2+\tilde{C}(1+C_r))t}\left( \Vert z(0,\cdot )\Vert _0^2 +\int \limits _0^t2\Vert P_\Lambda z(\tau ,\cdot )\Vert _0^2d\tau \right) \!. \end{aligned}$$

By usual arguments we get also, for $s\ge 5/2$:

$$\begin{aligned} \Vert z\Vert _s^2\le e^{(3+\tilde{C}(1+C_r))t}\left( \Vert z(0,\cdot )\Vert _s^2 +\int \limits _0^t\Vert P_\Lambda z(\tau ,\cdot )\Vert _s^2d\tau \right) \!. \end{aligned}$$

(3.23)

The a-priori estimate (3.23) gives existence and uniqueness of a solution $z\in C([0,T];H^s)$ of the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{l} P_\Lambda (t,x,u,D_t,D_x)z(t,x)=f_\Lambda (t,x)\\ z(0,x)=(u_0)_\Lambda (x) \end{array}\right. \end{aligned}$$

(3.24)

equivalent to (1.1) for $f_\Lambda :=(e^\Lambda )^{-1}f, (u_0)_\Lambda :=(e^\Lambda )^{-1}u_0$; moreover the solution satisfies the following energy estimate:

$$\begin{aligned} \Vert z\Vert _s^2\le e^{(3+\tilde{C}(1+C_r))t}\left( \Vert (u_0)_\Lambda \Vert _s^2 +\int \limits _0^t\Vert f_\Lambda (\tau ,\cdot )\Vert _s^2d\tau \right) \!. \end{aligned}$$

(3.25)

Remark now that $z$ is a solution of (3.24) if and only if $v=e^\Lambda z$ is a solution of (2.1). Since $e^\Lambda \in S^0$, from (3.25) we thus have that the solution $v$ of the Cauchy problem (2.1) satisfies:

$$\begin{aligned} \Vert v\Vert _s^2&\le c_1\Vert z\Vert _s^2\le c_1e^{(3+\tilde{C}(1+C_r))t} \left( \Vert (u_0)_\Lambda \Vert _s^2 +\int \limits _0^t\Vert f_\Lambda (\tau ,\cdot )\Vert _s^2d\tau \right) \nonumber \\&\le c_2e^{(3+\tilde{C}(1+C_r))t}\left( \Vert u_0\Vert _s^2 +\int \limits _0^t\Vert f(\tau ,\cdot )\Vert _s^2d\tau \right) \!, \end{aligned}$$

(3.26)

for some fixed constants $c_1,c_2>0$. Note that (3.26) implies (3.1) for $C_s(u):={c_2}e^{(3+\tilde{C}(1+C_r))T}$.

It is then defined a map

$$\begin{aligned} S:\ B_r&\rightarrow C([0,T];H^s)\\ u&\mapsto v \end{aligned}$$

which associates, to every fixed $u\in B_r$, the unique solution $v\in C([0,T];H^s)$ of the Cauchy problem (2.1), satisfying

$$\begin{aligned} \Vert v(t,\cdot )\Vert _s\le \sqrt{c_2}e^{\frac{1}{2}(3+\tilde{C}(1+C_r))t}(\Vert u_0\Vert _s+ \sqrt{t}\Vert f(t,\cdot )\Vert _s)\qquad \forall t\in [0,T].\qquad \qquad \end{aligned}$$

(3.27)

We now choose $r>2e\sqrt{c_2}\max \{\Vert u_0\Vert _s,\sup _{t\in [0,T]}\Vert f(t,\cdot )\Vert _s\}$. Then

$$\begin{aligned} \Vert v(t,\cdot )\Vert _s\le \frac{r}{2}(1+\sqrt{t})e^{\frac{1}{2}(3+\tilde{C}(1+C_r))t-1}<r \end{aligned}$$

if $t\in [0,T_0]$ for $T_0$ sufficiently small.

For such a choice of $T_0$ we thus have that, for

$$\begin{aligned} u\in B_r^0:=\{u\in C([0,T_0];H^s):\ \sup _{t\in [0,T_0]}\Vert u(t,\cdot )\Vert _s\le r\}, \end{aligned}$$

the Cauchy problem (2.1) admits a unique solution $v\in B_r^0$, i.e.

$$\begin{aligned} S:\ B_r^0\rightarrow B_r^0. \end{aligned}$$

We are now ready to use a fixed point argument. Fix $u,\tilde{u}\in B_r^0$, let $v=S(u)$ and $\tilde{v}=S(\tilde{u})$ the corresponding solutions of (2.1) and set $w=v-\tilde{v}$.

From

$$\begin{aligned}&D_tv+a_3(t)D_x^3v+a_2(t,x,u)D_x^2v+a_1(t,x,u)D_xv+a_0(t,x,u)=f(t,x)\\&D_t\tilde{v}+a_3(t)D_x^3\tilde{v}+a_2(t,x,\tilde{u})D_x^2\tilde{v}+ a_1(t,x,\tilde{u})D_x \tilde{v}+a_0(t,x,\tilde{u})=f(t,x) \end{aligned}$$

we have that

$$\begin{aligned}&D_tw+a_3(t)D_x^3w+a_2(t,x,u)D_x^2v-a_2(t,x,\tilde{u})D_x^2\tilde{v}\\&\quad +\,a_1(t,x,u)D_xv-a_1(t,x,\tilde{u})D_x\tilde{v}+a_0(t,x,u) -a_0(t,x,\tilde{u})=0, \end{aligned}$$

i.e.

$$\begin{aligned}&D_tw+a_3(t)D_x^3w+a_2(t,x,u)D_x^2w+a_1(t,x,u)D_xw\\&\quad +\,[a_2(t,x,u)-a_2(t,x,\tilde{u})]D_x^2\tilde{v} +[a_1(t,x,u)-a_1(t,x,\tilde{u})]D_x\tilde{v}\\&\quad \quad +[a_0(t,x,u)-a_0(t,x,\tilde{u})]=0. \end{aligned}$$

This means that $w$ is a solution of

$$\begin{aligned} \tilde{P}(t,x,u,D_t,D_x)w(t,x)=\tilde{f}(t,x,u,\tilde{u}, \tilde{v}), \end{aligned}$$

where $\tilde{P}(t,x,u,D_t,D_x):= P(t,x,u,D_t,D_x)-a_0(t,x,u)$ and

$$\begin{aligned} \tilde{f}(t,x,u,\tilde{u}, \tilde{v})&:= [a_2(t,x,u) -a_2(t,x,\tilde{u})]D_x^2\tilde{v}\\&\!\!+\,[a_1(t,x,u)-a_1(t,x,\tilde{u})]D_x\tilde{v} +[a_0(t,x,u)-a_0(t,x,\tilde{u})]. \end{aligned}$$

Since $u,\tilde{u}, \tilde{v}\in C([0,T_0];H^s)$ we have that $\tilde{f}\in C([0,T_0];H^{s-2})$ and, from (3.27) and $w(0,x)=0$:

$$\begin{aligned} \Vert w(t,\cdot )\Vert _{s-2}\le \sqrt{c_2}e^{\frac{1}{2}(3+\tilde{C}(1+C_r))T_0}\sqrt{T_0} \sup _{t\in [0,T_0]}\Vert \tilde{f}\Vert _{s-2} \end{aligned}$$

(3.28)

with

$$\begin{aligned} \Vert \tilde{f}\Vert _{s-2}&\le \Vert (a_2(t,x,u) \!-\!a_2(t,x,\tilde{u}))D_x^2\tilde{v}\Vert _{s-2} \!+\!\Vert (a_1(t,x,u) \!-\!a_1(t,x,\tilde{u}))D_x\tilde{v}\Vert _{s-2}\\&\!\!+\Vert a_0(t,x,u) -a_0(t,x,\tilde{u})\Vert _{s-2}. \end{aligned}$$

Since $s-2>1/2$ by assumption, then $H^{s-2}({\mathbb {R}})$ is an algebra and

$$\begin{aligned} \Vert (a_2(t,x,u) -a_2(t,x,\tilde{u}))D_x^2\tilde{v}\Vert _{s-2}&\le C_s\Vert a_2(t,x,u) -a_2(t,x,\tilde{u})\Vert _{s-2}\Vert D_x^2\tilde{v}\Vert _{s-2}\nonumber \\&\le C_{s,r}\Vert u-\tilde{u}\Vert _{s-2} \end{aligned}$$

(3.29)

where $C_{s,r}$ is a positive constant depending on $s$ and $r$, and more precisely

$$\begin{aligned} C_{s,r}=C_s'\left( \sum _{\alpha +\beta \le [s]-1}\sup _{\genfrac{}{}{0.0pt}1{(t,x)\in [0,T_0]\times {\mathbb {R}}}{|w|\le C_sr}} |D_x^\alpha D_w^{\beta +1}a_2(t,x,w)|\right) \Vert \tilde{v}\Vert _s \end{aligned}$$

for some $C'_s>0$.

Analogously, up to changing the constant $C_{s,r}$,

$$\begin{aligned} \Vert (a_1(t,x,u) -a_1(t,x,\tilde{u}))D_x\tilde{v}\Vert _{s-2}&\le C_{s,r}\Vert u-\tilde{u}\Vert _{s-2}\end{aligned}$$

(3.30)

$$\begin{aligned} \Vert a_0(t,x,u) -a_0(t,x,\tilde{u})\Vert _{s-2}&\le C_{s,r}\Vert u-\tilde{u}\Vert _{s-2}. \end{aligned}$$

(3.31)

Substituting (3.29), (3.30) and (3.31) in (3.28) we have that

$$\begin{aligned} \Vert w\Vert _{s-2}\le 3C_{s,r}\sqrt{c_2}e^{\frac{1}{2}(3+\tilde{C}(1+C_r))T_0}\sqrt{T_0} \sup _{t\in [0,T_0]}\Vert u-\tilde{u}\Vert _{s-2}. \end{aligned}$$

(3.32)

We now choose $T^*\le T_0$ sufficiently small so that

$$\begin{aligned} L:=3C_{s,r}\sqrt{c_2}e^{\frac{1}{2}(3+\tilde{C}(1+C_r))T^*}\sqrt{T^*}<1, \end{aligned}$$

and define

$$\begin{aligned} |||u|||_s&:= \sup _{t\in [0,T^*]}\Vert u(t,\cdot )\Vert _s\,,\\ B_r^*&:= \{u\in C([0,T^*];H^{s}):\ |||u(t,\cdot )|||_{s}\le r\}. \end{aligned}$$

Then (3.32) implies that $S:\ B_r^*\rightarrow B_r^*$ is a contraction with the $|||\cdot |||_{s-2}$ norm:

$$\begin{aligned} |||S(u)-S(\tilde{u})|||_{s-2}\le L|||u-\tilde{u}|||_{s-2},\qquad 0<L<1. \end{aligned}$$

(3.33)

Define now recursively

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} u_1=S(u_0)\\ u_{n+1}=S(u_n), &{}n\ge 1. \end{array}\right. \end{aligned}$$

From (3.33):

$$\begin{aligned} |||u_{n+1}-u_n|||_{s-2}&= |||S(u_n)-S(u_{n-1})|||_{s-2}\le L|||u_n-u_{n-1}|||_{s-2}\\&= L|||S(u_{n-1})-S(u_{n-2})|||_{s-2}\le L^2|||u_{n-1} -u_{n-2}|||_{s-2}\\&\le \ldots \le L^n|||u_1-u_0|||_{s-2}. \end{aligned}$$

Therefore,

$$\begin{aligned} |||u_{n+p}-u_n|||_{s-2}&\le |||u_{n+p}-u_{n+p-1}|||_{s-2}+|||u_{n+p-1}-u_{n+p-2}|||_{s-2} \\&\!\!+\ldots +|||u_{n+1}-u_n|||_{s-2}\\&\le L^n(1+L+\ldots +L^{p-1})|||u_1-u_0|||_{s-2}\\&\le \frac{L^n}{1-L}|||u_1-u_0|||_{s-2}, \end{aligned}$$

so that $\{u_n\}_{n\in {\mathbb {N}}}$ is a Cauchy sequence in $C([0,T^*];H^{s-2})$ and hence converges in $C([0,T^*];H^{s-2})$ to some $u\in C([0,T^*];H^{s-2})$. In particular, for every fixed $t\in [0,T^*]$,

$$\begin{aligned} u_n(t,\cdot )\rightarrow u(t,\cdot )\qquad \text{ in }\ H^{s-2}. \end{aligned}$$

(3.34)

At the same time, since $H^s({\mathbb {R}})$ is a reflexive space and $\Vert u_n(t,\cdot )\Vert _s\le r$, by Kakutani’s Theorem we have that there exists a subsequence $\{u_{n_h}\}_{h\in {\mathbb {N}}}$ which weakly converges in $H^s$ to some $\tilde{u}\in H^s({\mathbb {R}})$:

$$\begin{aligned} u_{n_h}(t,\cdot )\rightharpoonup \tilde{u}(t,\cdot )\qquad \text{ in }\ H^{s} \end{aligned}$$

(3.35)

and hence

$$\begin{aligned} \Vert \tilde{u}(t,\cdot )\Vert _s\le \liminf _{h\rightarrow +\infty }\Vert u_{n_h}(t,\cdot )\Vert _s. \end{aligned}$$

(3.36)

From (3.34) and (3.35) we have that $u(t,\cdot )= \tilde{u}(t,\cdot )\in H^s({\mathbb {R}})$.

Moreover, by (3.33):

$$\begin{aligned} |||S(u_n)-S(u)|||_{s-2}\le L|||u_n-u|||_{s-2}\rightarrow 0. \end{aligned}$$

Therefore, as $n\rightarrow +\infty$:

$$\begin{aligned} u\leftarrow u_{n+1}=S(u_n)\rightarrow S(u) \qquad \text{ in }\ C([0,T^*];H^{s-2}), \end{aligned}$$

so that $S(u)=u\in C([0,T^*];H^{s})$ and we have thus found a solution $u\in C([0,T^*];H^{s})$ of the Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{ll} P(t,x,u,D_t,D_x)u(t,x)=f(t,x),&{}(t,x)\in [0,T^*]\times {\mathbb {R}}\\ u(0,x)=u_0(x),&{}x\in {\mathbb {R}}. \end{array}\right. \end{aligned}$$

Since (3.26) is satisfied with $v(t,\cdot )=u_{n_h}(t,\cdot )$, for $t\in [0,T^*]$, from (3.36) we have that

$$\begin{aligned} \Vert u(t,\cdot )\Vert ^2_s\le {c_2}e^{(3+\tilde{C}(1+C_r))t} \left( \Vert u_0\Vert ^2_s+\int \limits _0^t\Vert f(\tau ,\cdot )\Vert ^2_sd\tau \right) \qquad \forall t\in [0,T^*] \end{aligned}$$

which gives (1.9).

Uniqueness follows from (3.33).

The proof is thus complete. $\square$