Quasilinear first order scalar

Characteristics

Let $a, b : \R^3\to\R$ be two smooth functions and consider the PDE
\begin{equation} u_t + a(x, t, u)u_x = b(x, t, u). \tag{Q} \end{equation}
Let $O\subset\R^2$ be an open subset, and assume that $u\in C^1(O)$ is a solution of (Q).
A characteristic of the PDE (Q) is a curve in $\R^3$ given by $x=\bx (t)$, $u=\bu (t)$, where the functions $\bx (t), \bu (t)$ satisfy
\begin{equation} \frac{d\bx} {dt} = a(\bx, t, \bu),\qquad \frac{d\bu} {dt} = b(\bx, t, \bu) . \tag{Ch} \end{equation}
The existence and uniqueness theorem for ordinary differential equations implies that for any point $(x_0, t_0, u_0)$ there is a unique characteristic passing through that point.
Let $u:O\to\R$ be a $C^1$ function, and let $(x_0, t_0)\in O$ be any point. Write $\bx (t)$ for the solution to the ODE \[ \frac{d\bx } {dt} = a\bigl(\bx (t), t, u(\bx (t), t)\bigr), \qquad \bx (t_0) = x_0. \] If $u$ is a solution of the pde (Q), then we have \begin{align*} \frac{d} {dt} u(\bx (t), t) &= u_x(\bx (t), t) \frac{d\bx } {dt} + u_t(\bx (t),t)\\ &= u_x(\bx (t), t)\; a\bigl(\bx (t), t, u(\bx (t), t)\bigr) + u_t(\bx (t),t)\\[4pt] &= b\bigl(\bx (t), t, u(\bx (t), t)\bigr) \end{align*} In other words $(\bx (t), \bu (t))$ (where $\bu (t) = u(\bx (t), t)$) is a solution to the characteristic equation (Ch).
Theorem. If $u:O\to\R$ is a $C^1$ solution of (Q), then the graph of $u$, \[ \Gamma = \bigl\{\bigl(x, t, u(x, t)\bigr)\in\R^3 : (x, t)\in O \bigr\} \] is a union of characteristics.

Local existence for Burger’s equation

The initial value problem for Burger’s equation is \begin{equation} u_t + u\, u_x = 0, \qquad u(x, 0) = g(x), \tag{Bu} \end{equation} where $g:I\to\R$ is a $C^1$ function defined on some open interval $I\subset \R$.
The characteristics are very simple, namely they are straight lines: the characteristic equations \[ \frac{d\bx}{dt} = \bu, \qquad \frac{d\bu}{dt}=0, \] imply that $\bu (t) = \bu (0)$ is constant along characteristics, while $\bx (t) = \bx (0) + \bu (0)t$. It follows that a solution $u(x, t)$ must satisfy \[ u(\bx (t), t) = \bu (t) \] for any characteristic $(\bx (t), \bu (t))$. If $(\bx (t), \bu (t))$ is the characteristic starting at $x=a$ for $t=0$, then $\bu (t) = \bu (0) = u(a, 0) = g(a)$ and $\bx (t) = \bx (0) + \bu (0) t = a + g(a)t$. Thus we have \[ u\bigl(a + g(a)t, t\bigr) = g(a). \] To recover the solution $u(x, t)$ as a function of $(x,t)$ we must solve $x=a+g(a)t$ for $a$. When $t=0$ we have $x=a$ so no need to solve. By the Implicit Function Theorem there is a unique solution $a = A(x, t)$ of the equation $a+tg(a) = x$, provided $t$ is small. We then find that the solution, if it exists, is given by \[ u(x, t) = g\bigl(A(x, t)\bigr), \qquad \text{ where }a=A(x,t)\iff a+tg(a) = x. \]
Theorem. If $g\in C^1(\R)$ satisfies $M=\sup_{x\in\R}|g'(x)|\lt\infty$, then the initial value problem (Bu) for Burger’s equation has a $C^1$ solution $u(x, t)$ that is defined for $x\in\R$, $|t|\lt \varepsilon$, where $\varepsilon=1/M$.
Proof. The solution is given by the characteristics solution above, i.e. $u(x, t)=g(A(x, t))$, where $a=A(x, t)$ is the solution to $a+tg(a)=x$. If $|t|\lt 1/M$ then the function $\varphi(a) = a+tg(a)$ satisfies \[ 1-tM \le \varphi'(a) \le 1+tM \] for all $a\in\R$. Since $1-tM\gt0$ this implies that $a\mapsto \varphi(a)$ is a strictly increasing function with $\lim_{a\to\pm\infty}\varphi(a) = \pm\infty$. Thus $\varphi(a) = x$ has a unique solution $a=A(x, t)$ for any $x\in\R$, and we see that the solution $u(x, t) = g(A(x,t))$ is indeed defined for $(x,t)\in\R\times(-\varepsilon,\varepsilon)$.    ////