Show that this map is a contraction (PDE) I was asked a question by my research advisor and I really don't know how to think about it. The goal is to prove the existence of a function $u(x,t):\mathbb{R}\times[0,T]\to\mathbb{R}$ satisfying the following PDE:
$$\begin{cases}\partial_t^2 u-\partial_x^2 u=(\partial_t u)(\partial_x u)\\u(-,0)=f\in H^s(\mathbb{R})\\\partial_tu(-,0)=g\in H^{s-1}(\mathbb{R}),\end{cases}$$
where $H^s(\mathbb R)$ is the Sobolev space
$$H^s(\mathbb{R})=\{f:\lVert f\rVert_{s,2}=\lVert\hat f(\xi)(1+\lvert\xi\rvert^2)^{s/2}\rVert_2<\infty\},$$
and $s$ is sufficiently large. To do this, I was told to define the norm
$$\lVert u\rVert_X=\sup_{t\in[0,T]}(\lVert u\rVert_{s,2}+\lVert\partial_t u\rVert_{s-1,2})$$
on the space
$$X=\{u\in C([0,T];H^s(\mathbb R))\cap C^1([0,T];H^{s-1}(\mathbb R)) :\lVert u\rVert_X\leq c(\lVert f\rVert_{s,2}+\lVert g\rVert_{s-1,2})\}.$$
Now let $w_0$ satisfy the wave equation $\partial_t^2 w_0-\partial_x^2 w_0=0$ with initial data $w_0(-,0)=f$ and $\partial_t w_0(-,0)=g$, and for $n>0$ let $w_n$ satisfy
$$\begin{cases}\partial_t^2 w_n-\partial_x^2 w_n=(\partial_t w_{n-1})(\partial_x w_{n-1})\\w_n(-,0)=f\\\partial_t w_n(-,0)=g.\end{cases}$$
If we can show $\Phi=(w_n\mapsto w_{n+1}):X\to X$ is a contraction mapping and that $X$ is complete, then we are done by the Picard iteration theorem.
I do not know how to prove $\Phi$ is a contraction. I understand all of the definitions, I've tried to prove through direct computation, and I've tried to apply any of the relevant theorems I know, but I have had no luck. I also don't know how to show that $\Phi(u)$ even exists for general $u$.
If anyone has an idea of how to proceed with this, I would very much appreciate hearing their thoughts. Thank you.
 A: Sketch: 
First, observe we have the following integral formulation
\begin{align}
u(t, x) = \cos(t\sqrt{-\Delta})f(x)+\frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}g(x) -\int^t_0 \frac{\sin((t-t')\sqrt{-\Delta})}{\sqrt{-\Delta}}[\partial_t u(t', \cdot) \partial_x u(t', \cdot)]\ dt',
\end{align}
then we see that
\begin{align}
\|u(t)\|_{s,2} \leq&\ \left\| \cos(t\sqrt{-\Delta}) f\right\|_{s,2} + \left\| \frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}g\right\|_{s,2}\\
&\ + \left\|\int^t_0 \frac{\sin((t-t')\sqrt{-\Delta})}{\sqrt{-\Delta}}[\partial_t u(t', \cdot) \partial_x u(t', \cdot)]\ dt' \right\|_{s, 2}.
\end{align}
Next, note that
\begin{align}
\left\| \cos(t\sqrt{-\Delta}) f\right\|_{s,2}^2 =&\ \int d\xi\ (1+|\xi|^2)^s \cos^2(t|\xi|)|\hat f(\xi)|^2\\
\leq&\  \int d\xi\ (1+|\xi|^2)^s|\hat f(\xi)|^2 = \| f\|_{s, 2}^2
\end{align}
and
\begin{align}
\left\| \frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}g\right\|_{s,2}^2=&\ \int d\xi\ \frac{\sin^2(t|\xi|)}{|\xi|^2}(1+|\xi|^2)^s|\hat g(\xi)|^2 \\
=&\ \int_{|\xi|\leq 1}d\xi\ \frac{t^2\sin^2(t|\xi|)}{t^2|\xi|^2}(1+|\xi|^2)^s|\hat g(\xi)|^2 + \int_{|\xi|\geq 1} d\xi\ \frac{\sin^2(t|\xi|)}{|\xi|^2}(1+|\xi|^2)^s|\hat g(\xi)|^2 \\
\leq&\ 2t^2 \int_{|\xi|\leq 1} d\xi\  (1+|\xi|^2)^{s-1}|\hat g(\xi)|^2\\
& + \int_{|\xi|\geq 1} d\xi\  (1+|\xi|^2)^{s-1}|\hat g(\xi)|^2\\
\leq&\ (1+2T^2)\|g\|_{s-1, 2}^2.
\end{align}
For the nonlinear term, we see that
\begin{align}
&\left\|\int^t_0 \frac{\sin((t-t')\sqrt{-\Delta})}{\sqrt{-\Delta}}[\partial_t u(t', \cdot) \partial_x u(t', \cdot)]\ dt' \right\|_{s, 2}\\
&\leq \sqrt{1+2T^2}\int^T_0\left\| \partial_t u(t', \cdot) \partial_x u(t', \cdot) \right\|_{s-1, 2}\ dt'\\
&=\sqrt{1+2T^2}\int^T_0\left\| \partial_t u(t', \cdot) \partial_x u(t', \cdot) \right\|_{s-1, 2}\ dt'.
\end{align}
Using the Sobolev product estimate and $L^\infty$-Sobolev embedding, we get
\begin{align}
\left\| \partial_t u(t', \cdot) \partial_x u(t', \cdot) \right\|_{s-1, 2} \leq&\ \|\partial_t u\|_{s-1, 2}\|\partial_x u\|_{0, \infty}+\|\partial_t u\|_{0, \infty}\|\partial_x u\|_{s-1, 2}\\
\leq&\ C\|\partial_t u\|_{s-1, 2}\|u\|_{s, 2}
\end{align}
provided $s>\frac{3}{2}$. 
Finally, we see that
\begin{align}
\sqrt{1+2T^2}\int^T_0\left\| \partial_t u(t', \cdot) \partial_x u(t', \cdot) \right\|_{s-1, 2}\ dt'\leq C\sqrt{1+2T^2}T \sup_{t \in [0, T]}\|\partial_t u(t)\|_{s-1, 2}\sup_{t \in [0, T]}\|u(t)\|_{s, 2}.
\end{align}
Hence it follows
\begin{align}
\|u(t)\|_{s,2}  \leq C(T)\left(\|f\|_{s, 2}+\|g\|_{s-1, 2}+T\|u\|_X^2\right).
\end{align}
I will let you show that
\begin{align}
\|\partial_t u(t)\|_{s-1,2}  \leq C(T)\left(\|f\|_{s, 2}+\|g\|_{s-1, 2}+T\|u\|_X^2\right).
\end{align}
Lastly, we see that
\begin{align}
\|u\|_X \leq C(T)\left(\|f\|_{s, 2}+\|g\|_{s-1, 2}+T\|u\|_X^2\right).
\end{align}
Hence for $T$  sufficiently small we have that there exists $C>0$ such that
\begin{align}
\|u\|_X \leq C(\|f\|_{s, 2}+\|g\|_{s-1, 2}).
\end{align}
Thus, if we set up the iteration scheme, we will get
\begin{align}
\|u_n\|_X \leq C(\|f\|_{s, 2}+\|g\|_{s-1, 2})
\end{align}
for all $n$. 
I will let you fill in the remaining detail.
One more hint: To show you have a contraction, it suffices to show
\begin{align}
\int^T_0\|\partial_t u(t, \cdot)\partial_x u(t, \cdot)-\partial_t v(t, \cdot)\partial_x v(t, \cdot)\|_{s-1, 2} dt \leq L\|u-v\|_X
\end{align}
for some constant $L$ depending on $T$. 
