4

So I am trying to understand the non-algorithmic part of the method of characteristics for solving a first order quasilinear PDE:

$ a(x,y,u)u_{x}+b(x,y,u)u_{y}=c(x,y,u) \hspace{1cm } $ (1)

I understand why the characteristic curves are solutions to the equation:

An integral surface $ z=u(x,y) $ has normal vector $(u_{x},u_{y},-1)$, so (1) is equal to the condition that the normal of the integral surface is perpendicular to the vector $(a(x,y,z),b(x,y,z),c(x,y,z))$.

So the characteristic curves are defined as the curves with tangent vector (a,b,c) for every point of the curve. I also understand how to obtain them solving the ODE's system.

The problem is that, in my notes, given the particular Cauchy problem:

$ \left\{\begin{matrix} a(x,y)u_{x}+b(x,y)u_{y}=c(x,y) \\ u(f(s),g(s))=h(s) \end{matrix}\right.$

It is stated that the Cauchy problem has a solution if the curve $ \{f(s),g(s)\} $is not characteristic. I can't get why, Is this for the sole purpose of the change of variables to work well? I can neither understand why the characteristic curves are said to be "constant" solutions of the PDE.

D1X
  • 2,222

1 Answers1

6

The idea of the method of characteristics is to change co-ordinates so as to rewrite the differential operator $(a \partial_x + b \partial_y)$ (the operator that is applied to $u$ on the left-hand side of your (1)) as a single derivative with respect to a certain variable. Let this variable be $t$, and this then gives you an ODE, $$ \frac{d}{dt}u(x(t),y(t)) = c(x(t),y(t),u). \tag{2} $$ In particular, this equation holds on the characteristic curves, which are defined as you specify.

Provided that the characteristics only have point-like intersections with the initial data curve $(f(\xi),g(\xi))$, the inverse function theorem allows us to produce a parametrisation of the characteristics by where they intersect the initial data curve.

Further, because (2) is an ODE on a characteristic, the values of $u$ on the characteristic only depend on the initial value of $u$ on the characteristic (i.e. the intersection of it with the $(f,g)$-curve) (as well as the values of $c$ along the characteristic), so this means that each point's initial data propagates along only one characteristic. If $c\equiv 0$, the data does not change along a characteristic, so the function is constant on each characteristic.

If you can do this, you can label every point where the initial data specifies values of $u$ by two co-ordinates: the characteristic variable $t$ (distance along a characteristic) and the characteristic it is on (given by intersection with the initial data at a specific $\xi$). With this change of coordinates, you re-express the function as a function of $(t,\xi)$, which gives you the system $$ \partial_t u(t,\xi) = c(t,\xi,u) \\ u(0,\xi) = h(\xi), $$ which is easy to solve.

Chappers
  • 67,606
  • So, if I understand well, we want to use the characteristic curves because they depend only on a single parameter, which allows us to solve the PDE along that curve as an ODE. Is $ \partial_s u(t,\xi) = c(t,\xi,u) $ correct on your post? – D1X Jun 06 '15 at 21:47
  • I've put a partial derivative there because we have to ignore $\xi$ (i.e. stay on a particular characteristic). The previous ODE only applies on a characteristic, but $u$ was still expressed as a function of $x$ and $y$ there. – Chappers Jun 06 '15 at 22:00
  • I've made it more specific. – Chappers Jun 06 '15 at 22:01
  • Sorry, I still don't get why is it you do $\partial_s u(t,\xi) = c(t,\xi,u)$ instead of $\partial_t u(t,\xi) = c(t,\xi,u)$ and why the initial condition needs to be a non characteristic curve. – D1X Jun 07 '15 at 10:27
  • I understand you cannot use the Inverse Function Theorem then, but I can't get why the initial condition cannot be made along a characteristic with a constant value. – D1X Jun 07 '15 at 10:42
  • Oh, typo. Whoops. I wrote with $s$, as is commonly used, but then saw that's what you'd used for the initial curve. – Chappers Jun 07 '15 at 11:33