11

Below is an image of my lecture notes explaining the idea behind the method of characteristics for quasilinear first order PDEs.

However I don't understand how the curve $C_s$ is defined and how it translates into the graph $Gr_u$. How does $C_s$ 'start' on the initial curve if there is no $f(s)$ or $g(s)$ in the equation for $C_s$? What does $x(t;s)$ mean, does that mean (say for the $x$ term) that for fixed $s$ one starts at $x(s-\epsilon_s)$ and finishes at $x(s+\epsilon_s)$? If so what is this 'function' $x$?

Thanks, I would really appreciate someone explaining this to me! :

doraemonpaul
  • 16,178
  • 3
  • 31
  • 75
user53076
  • 1,594

2 Answers2

0

As i understand it the main idea behind the Characteristic Method is to "transform" your PDE in a system of (infinite) ODE's.

However I think your lecture notes are missing the starting conditions: $x(0,s) = f(s)$ and $y(0,s) = g(s)$ and $z(0,s) = h(s)$.

So each $s \in (\alpha,\beta)$ gives you one point on the initial curve $\gamma$:

$(x(0,s),y(0,s)) $

and thus $(x(t,s),y(t,s))$ is a curve which goes through that point in $\gamma$.

The key part is now that later one, you will discover that under certain assumptions $x(t,s)$, $y(t,s)$ and $z(t,s)$ solve a system of ODE's which is called the $\textit{Characterstic System}$. And in simple cases one can first solve the Characterstic System and then transform the solution to an $\textit{explicit}$ solution of the PDE.

Jahi02
  • 107
-3

To clarify what your notes imply, the curve ${\cal C}_s$ contains points $(x(t;s),y(t;s),z(t;s))$ defined by solving the ODE $\frac{d}{dt}(x,y,z)=(a(x,y,z),b(x,y,z),c(x,y,z))$ for $-\varepsilon_s<t<\varepsilon_s$, and condition $(x(0;s),y(0;s),z(0;s))=(f(s),g(s),h(s))$ on $\Gamma$ at $t=0$.

Aimé
  • 11