Help Implicit funtions theorem question

Question

Let

\begin{cases} u_t(x,t) + u(x,t)u_x(x,t) = 0, & \text{for } x\in\mathbb{R},\; t > 0 \\ u(x,0) = h(x), & \text{for } x\in\mathbb{R}. \end{cases}

Suppose that $h\in \mathcal C^2 (\mathbb{R})$ and that $h' (x) > 0$ for all $x\in \mathbb{R}$. Check, with the help of implicit function theorem, for $(x, t)$ sufficiently close to $(x_0, 0)$ (where $x_0\in\mathbb{R}$ is a fixed number), that the equation $u = h(x - tu)$ defines a solution to the Cauchy problem (for $x$ sufficiently close to $x_0$ and $t$ close to zero).

Answer 1

Let $$ F=F(x,t,u)=u-h(x-tu) $$ Then $F: \mathbb R^2\times\mathbb R^1\to\mathbb R^1$ is $C^2$, and $F_u\in C^1$ and $$ F_u=1+th'(x+tu)>0\quad $$ is $x$ $t$ near $0$ and $x$ near $x_0$. Hence, IFT provides the existence of $u(x,t)\in C^1$, for an open set of $(x_0,0)$.

Note. If fact, this is Burger's equation, and if $h'>0$, then this IVP possesses a unique global solution ($t\ge 0$). But if $h'<0$, then it develops shocks in finite time.