Locally Integrable f(x+ct) is Weak Solution to Transport Equation

Theorem
Consider the transport equation:


 * $\dfrac {\partial u} {\partial t} - c \dfrac {\partial u} {\partial x} = 0$

with the initial condition:


 * $\map u {x, 0} = \map f x$

where $c \in \R$.

Then it has a weak solution of the form:


 * $\map u {x, t} := \map f {x + ct}$

where $f \in \map {L^1_{loc} } \R$ is a locally integrable function.

Proof
Let $\map u {x, t} = \map f {x + ct}$ be a locally integrable function.

We have that a locally integrable function defines a distribution.

Let $T_u \in \map {\DD'} {\R^2}$ be a distribution associated with $u$.

Let $\phi \in \map \DD {\R^2}$ be a test function.

Then:

Introduce real variables $\xi, \eta$ such that:


 * $\xi = x + ct$


 * $\eta = t$

Then:


 * $x = \xi - c \eta$


 * $t = \eta$

Let the expressions above define the mapping $\Psi$ between these variables:


 * $\tuple {\xi, \eta} \stackrel {\Psi} {\mapsto} \tuple {x, t}$

Suppose $F : \R^2 \to \R$ is a real function.

Then:

where $\circ$ denotes the composition of mappings, and $\mathbf J_\Psi$ is the Jacobian matrix:

Hence:


 * $\map \det { {\mathbf J_\Psi} } = 1$.

Suppose:


 * $\map F {x, t} = \map f {x + ct} \paren {c \map {\dfrac {\partial \phi}{\partial x} } {x, t} - \map {\dfrac {\partial \phi}{\partial t} } {x, t} }$

Then:

Hence:

Therefore, $\map u {x, t} := \map f {x + c t}$ is a weak solution to the transport equation.