Kruzkov theorem

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Revision as of 06:32, 15 September 2005

The scalar Cauchy problem

$u_t + f(u)_x = 0, \ \ \ f \in C^1(\mathbb{R})$

with initial condition

$u(0,x) = u_o(x), \ \ \ u_o \in L^\infty(\mathbb{R})$

has a unique entropy solution

$u \in L^\infty(\mathbb{R}_+ \times \mathbb{R})$

which fulfills (important for numerics)

$|| u(t, \cdot)||_{L^\infty} \le || u_o ||_{L^\infty}, \mbox{ a.e. in } t \in \mathbb{R}_+$

$u(t, \cdot) \ge v(t, \cdot) \ \ \ \mbox{ a.e. in } \mathbb{R}, \mbox{ a.e. in } t \in \mathbb{R}_+$

$u(t, \cdot) \in BV(\mathbb{R}) \ \ \ \textrm{ and } \ \ \ TV(u(t, \cdot)) \le TV(u_o)$

$\int_{\mathbb{R}} u(t,x) d x = \int_\mathbb{R} u_o(x) d x, \ \ \ \mbox{ a.e. in } t \in \mathbb{R}_+$

Finite domain of dependence: If $u, v$ are two entropy solutions, $u_o, v_o \in L^\infty$ and

$M = \max\{ | f^\prime(\phi) | : | \phi | \le \max( ||u_o||_{L^\infty}, ||v_o||_{L^\infty}) \}$

then

$\int_{|x| \le R} | u(t,x) - v(t,x) | d x \le \int_{|x| \le R + Mt} | u_o(x) - v_o(x) | d x$