Determine the solution of this EDP using the characteristic method (parametric form).
$\begin{equation} \frac{\partial w}{\partial t}-\frac{\partial{w}}{\partial x}=0 \\ w(x,0)=e^{2x} \,\,\,\, \text{for $(x,0)$ in $\gamma$ } \end{equation}$
I don't understand much of this method... I'm stuck in some parts, i hope you can help me to understand this (:
My attempt
Note $\gamma$ is a function such that his point are of the form $(e^{2x},0)$.
We need find a parametric solution for the superfice $S$ defined by the graph of $z=w(x,0)$.
Then
$(x(r,s),t(r,s),w(x(r,s),t(r,s))$ where $w(x(r,s),t(r,s))=w(r,s).$
If $s=0$ we have the parametrization of the condition $w(x,0)=e^{2x}$.
Then
$(x(r,0),t(r,0),w(x,0))=(e^{2x},0,e^{2x})$
Here i'm stuck. Can someone help me? I don't understand very well this.
$$ \begin{align}\begin{cases} \frac{\partial w}{\partial t} - \frac{\partial w}{\partial x} = 0 \\ w(x,0) = e^{2x} \end{cases} \end{align} \tag{1}$$
We parameterize the curve by $r$ such that $\gamma = \{ (r,0)\}$ the characteristics are given by
$$ \frac{dt}{ds}(r,s) = 1 \\ \frac{dx}{ds}(r,s) = -1 \\ \frac{dz}{ds}(r,s) = 0 \tag{2}$$
with the initial condition given as
$$ t(r,0) = 0 \\ x(r,0) = 0 \\ z(r,0) = e^{2r} \tag{3}$$
when we solve the system
$$ t(r,s) = s+c_{1}(r) \\ x(r,s) = c_{2}(r)-s \\ z(r,s) = c_{3}(r) \tag{4}$$ imposing the initial condition $$ t(r,s) = s \\ x(r,s) = -s\\ z(r,s) = e^{2r} \tag{5}$$
now
$$ r(x,t) =x \\ s(x,t) = t \tag{6} $$
$$ z(r(x,t),s(x,t)) = \phi(x) = e^{2x} \tag{7} $$
note that the solution for the characteristic is given by
$$ w(x,t) = e^{2(x-t)} \tag{8} $$