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

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 Schwartz distribution associated with $u$.
Let $\phi \in \map \DD {\R^2}$ be a test function.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial} {\partial t} \map {T_u} \phi - c \dfrac {\partial} {\partial x} \map {T_u} \phi
      | r = - \map {T_u} {\dfrac {\partial \phi}{\partial t} } + \map {T_u} {c \dfrac {\partial \phi}{\partial x} }
      | c = {{Defof|Distributional Partial Derivative}}
}} 
{{eqn | r = - \iint_{\R^2} \map f {x + ct} \dfrac {\partial \phi}{\partial t} \rd x \rd t + \iint_{\R^2} \map f {x + ct} c \dfrac {\partial \phi}{\partial x} \rd x \rd t
      | c = {{Defof|Schwartz Distribution}}
}}
{{eqn | r = \iint_{\R^2} \map f {x + ct} \paren {c \dfrac {\partial \phi}{\partial x} - \dfrac {\partial \phi}{\partial t} } \rd x \rd t
}}
{{end-eqn}}
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.
By Change of Variables Theorem (Multivariable Calculus):
{{begin-eqn}}
{{eqn | l = \iint_{\R^2} \map F {x, t} \rd x \rd t
      | r = \iint_{\R^2} \map {\paren {F \circ \Psi} } {\xi, \eta} \map \det { {\mathbf J_\Psi} } \rd \xi \rd \eta
}}
{{end-eqn}}
where $\circ$ denotes the composition of mappings, and $\mathbf J_\Psi$ is the Jacobian matrix:
{{begin-eqn}}
{{eqn | l = \mathbf J_\Psi 
      | r = \begin{pmatrix}
 \map {\dfrac {\partial x} {\partial \xi} } {x, t}  & \map {\dfrac {\partial x} {\partial \eta} } {x, t} \\
 \map {\dfrac {\partial t} {\partial \xi} } {x, t}  & \map {\dfrac {\partial t} {\partial \eta} } {x, t}
\end{pmatrix}
}}
{{eqn | r = \begin{pmatrix}
 1  & -c \\
 0 & 1
\end{pmatrix}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map {\paren {F \circ \Psi} } {\xi, \eta}
      | r = \map F {\map \Psi {\xi, \eta} }
}}
{{eqn | r = \map F {\xi - c \eta, \eta}
}}
{{eqn | r = \map f {\paren {\xi - c \eta} + c \eta} \paren {c \map {\dfrac {\partial \phi}{\partial x} }{\xi - c \eta, \eta} - \map {\dfrac {\partial \phi}{\partial t} }{\xi - c \eta, \eta} }
}}
{{eqn | r = \map f \xi \paren {- \dfrac {\partial \paren {\xi - c \eta} }{\partial \eta} \map {\dfrac {\partial \phi}{\partial x} }{\xi - c \eta, \eta} - \dfrac {\partial \eta}{\partial \eta} \map {\dfrac {\partial \phi}{\partial t} }{\xi - c \eta, \eta} }
}}
{{eqn | r = - \map f \xi \paren { \map {\dfrac {\partial \phi}{\partial x} }{\xi - c \eta, \eta} \dfrac {\partial \paren {\xi - c \eta} }{\partial \eta} + \map {\dfrac {\partial \phi}{\partial t} }{\xi - c \eta, \eta} \dfrac {\partial \eta}{\partial \eta} }
}}
{{eqn | r = - \map f \xi \dfrac {\partial}{\partial \eta} \map \phi {\xi - c \eta, \eta}
}}
{{eqn | r = - \map f \xi \map {\dfrac {\partial \paren {\phi \circ \Psi} }{\partial \eta} } {\xi, \eta}
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map {\paren {\dfrac {\partial} {\partial t} T_u - c \dfrac {\partial} {\partial x} T_u} } \phi
      | r = \iint_{\R^2} \map f {x + ct} \paren {c \map {\dfrac {\partial \phi}{\partial x} } {x, t} - \map {\dfrac {\partial \phi}{\partial t} } {x, t} } \rd x \rd t
}}
{{eqn | r = \iint_{\R^2} - \map f \xi \map {\dfrac {\partial \paren {\phi \circ \Psi} }{\partial \eta} } {\xi, \eta} \size {1} \rd \xi \rd \eta
}}
{{eqn | r = - \int_{-\infty}^\infty \map f \xi \paren {\int_{-\infty}^\infty \map {\dfrac {\partial \paren {\phi \circ \Psi} }{\partial \eta} } {\xi, \eta} \rd \eta} \rd \xi
}}
{{eqn | r = -\int_{-\infty}^\infty \map f \xi \paren {\bigintlimits {\map {\paren {\phi \circ \Psi} } {\xi,\eta} } {\eta \mathop = -\infty} {\eta \mathop = \infty} } \rd \xi
      | c = Fundamental Theorem of Calculus
}}
{{eqn | r = -\int_{-\infty}^\infty \map f \xi \paren {\lim_{\eta \mathop \to \infty} \map {\paren {\phi \circ \Psi} } {\xi,\eta} - \lim_{\eta \mathop \to -\infty} \map {\paren {\phi \circ \Psi} } {\xi,\eta} } \rd \xi
}}
{{eqn | r = -\int_{-\infty}^\infty \map f \xi \paren {\lim_{\eta \mathop \to \infty} \map \phi {\xi - c \eta,\eta} - \lim_{\eta \mathop \to -\infty} \map \phi {\xi - c \eta,\eta} } \rd \xi
}}
{{eqn | r = -\int_{-\infty}^\infty \map f \xi \paren {0 - 0} \rd \xi
      | c = {{Defof|Test Function}}
}}
{{eqn | r = -\int_{-\infty}^\infty \map f \xi 0 \rd \xi
}}
{{eqn | r = 0
}}
{{end-eqn}}
Therefore, $\map u {x, t} := \map f {x + c t}$ is a weak solution to the transport equation.
{{qed}}