[example:BurgersSquared]
Consider the conservation law problem on $\Omega = \{ (x_1, x_2) \in \mathbb{R}^2 : x_1 > 0 \}$ (where $x_1$ represents time $t$ and $x_2$ represents space):
\begin{align*}
\frac{\partial u}{\partial x_1} + u^2 \frac{\partial u}{\partial x_2} = 0, \quad u(0, x_2) = x_2.
\end{align*}
Here, the [function](/page/Function) $F: \mathbb{R}^2 \times \mathbb{R} \times \mathbb{R}^2 \to \mathbb{R}$ is defined by:
\begin{align*}
F(a, b, c) = a_1 + b^2 a_2
\end{align*}
**(1) The ODE System**
We compute the necessary partial [derivatives](/page/Derivative) of $F$:
\begin{align*}
\frac{\partial F}{\partial a_1} = 1, \quad \frac{\partial F}{\partial a_2} = b^2, \quad \frac{\partial F}{\partial b} = 2b a_2, \quad \frac{\partial F}{\partial c_i} = 0.
\end{align*}
The characteristic system for the trajectory $T(s) = (G(s), U(s), X(s))$ becomes:
\begin{align*}
\dot{X}_1 &= 1 \\
\dot{X}_2 &= U^2 \\
\dot{U} &= G_1(1) + G_2(U^2) = a \cdot \nabla_a F = 0 \quad (\text{since } F=0 \text{ implies } a \cdot \nabla_a F = 0 \text{ for homogeneous parts})
\end{align*}
**(2) The Flow Map $\Psi$**
We parametrize the boundary $\Gamma$ (the $x_2$-axis at $x_1=0$) using the identity chart, so $y' \in \mathbb{R}$ corresponds to the initial position $x_2$.
The initial conditions at $s=0$ are:
\begin{align*}
X(0, y') &= (0, y') \\
U(0, y') &= g(X(0, y')) = y'
\end{align*}
Solving the ODEs:
* $X_1(s, y') = s$
* $U(s, y') = y'$ (constant along characteristic)
* $X_2(s, y') = X_2(0) + \int_0^s U^2 d\tau = y' + (y')^2 s$
Thus, the Characteristic Flow Map is:
\begin{align*}
\Psi(s, y') = \begin{pmatrix} s \\ y' + s(y')^2 \end{pmatrix}
\end{align*}
**(3) The Envelope**
We compute the Jacobian matrix $D\Psi$ with respect to the parameters $(s, y')$:
\begin{align*}
D\Psi = \begin{pmatrix}
\frac{\partial X_1}{\partial s} & \frac{\partial X_1}{\partial y'} \\
\frac{\partial X_2}{\partial s} & \frac{\partial X_2}{\partial y'}
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\
(y')^2 & 1 + 2s y'
\end{pmatrix}
\end{align*}
The determinant is:
\begin{align*}
\det(D\Psi) = 1(1 + 2s y') - 0 = 1 + 2s y'
\end{align*}
The envelope occurs where $\det(D\Psi) = 0$:
\begin{align*}
1 + 2s y' = 0 \implies y' = -\frac{1}{2s}
\end{align*}
Substituting this critical parameter back into the map for $X_2$:
\begin{align*}
X_2 &= \left(-\frac{1}{2s}\right) + s \left(-\frac{1}{2s}\right)^2 = -\frac{1}{2s} + \frac{1}{4s} = -\frac{1}{4s}.
\end{align*}
Since $X_1 = s$, the curve in physical space is the hyperbola $x_2 = -\frac{1}{4x_1}$.
[/example]