# First order linear PDE-the idea of characteristic curves

1. Sep 19, 2009

### kingwinner

"Consider a first order linear PDE. (e.g. y ux + x uy = 0)
If u(x,y) is constant along the curves y2 - x2 = c, then this implies that the general solution to the PDE is u(x,y) = f(y2 - x2) where f is an arbitrary differentiable funciton of one variable. We call the curves along which u(x,y) is constant the characteristic curves."
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I don't understand the implication above highlighted in red. The idea of characteristic curves seems to be very important in solving first order linear PDEs, but I am never able to completely understand the idea of it. Why would finding the characteristic curves help us find the general solution to the PDE?

Thanks for explaining!

Last edited: Sep 19, 2009
2. Sep 20, 2009

### LCKurtz

3. Sep 20, 2009

### kingwinner

For that link, they seem to be using a different approach. In particular, the characteristic equation is different and not matching that in my textbook...so I'm getting more confused(?) :(

4. Sep 21, 2009

### LCKurtz

I admit it's a little tricky to see. You understand that for a solution u(x,y), the level curves satisfy dy/dx = x/y in your example and these curves are of the form y2-x2=c.

These happen to be hyperbolas in your example.

Now any function that has those level curves will be a solution to the PDE because the surface is made up of its level curves. Think of a mountain being the aggregate of its level curves (contour lines). Of course u(x,y) = y2-x2 is such a solution. But that is just one.

What about u(x,y) = exp(y2-x2)? It has the same type of level curves. Or sin(y2-x2).

In fact you can take any differentiable function of (y2-x2):

u(x,y) = f(y2-x2)

To verify this take

ux = f'(y2-x2)(-2x)
uy = f'(y2-x2))(2y)

dy/dx = - ux/uy = x/y

so u(x,y) = f(y2-x2) is a solution for any f.

Hope this helps. Sack time here.