Determine the solution of this PDE using the characteristic method (parametric form). 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

$$ w(x,t) = e^{2(x-t)} \tag{8} $$