# Homework Help: Hyperbolic PDE, Cauchy-type problem

1. Mar 22, 2013

### thelema418

1. The problem statement, all variables and given/known data
Consider the equation
$4y^2u_{xx} + 2(1-y^2)u_{xy} - u_{yy} - \frac{2y}{1+y^2} (2u_x - u_y) = 0$

Find the solution $u(x,y)$ which satisfies $u(x,0) = g(x)$, and $u_y(x,0) = f(x)$, where $f, g \in \mathcal{C}^2(\mathbb{R})$ are arbitrary functions.

2. Relevant equations

I used the discriminant method to show that this PDE can be categorized as hyperbolic. I found that the characteristics are $s= - \frac{2}{3}y^3 +x$ and $t = 2y + x$. I know the general solution is the form $u(x,y) = F( - \frac{2}{3}y^3 +x) + G(2y + x)$. (Depending on algebra of finding the characteristics, this may be stated in a different way.)

3. The attempt at a solution

There are other problems I have done without such abstract conditions, so I followed the general solution style I have been using. NAMELY:

Find $u_y$: $-2y^2 F'( - \frac{2}{3}y^3 +x) + 2G'(2y+x)$. Then, look at the initial values.

$u(x,0) = F(x) + G(x) = g(x)$
$u_y (x,0) = 0 + 2 G'(x) = f(x)$

Since $u_y$ has already isolated one of the general functions, I work with it. I integrate with respect to $x$. Ultimately, I get

$G(x) = \frac{1}{2} \int f(x) \; dx$

Solving for $F(x)$ with this information, I get

$F(x) = g(x) - \frac{1}{2} \int f(x) \; dx$

At this point I've reached a "SO WHAT?" stage. In the previous problems, the function was simply a multiplication factor that was easy to manipulate. I found the multiplication factor and then I multiplied the original information by it to get a $u(x,y)$.

I feel that I should calculate:
$F(-\frac{2}{3} y^3 + x) = g(-\frac{2}{3} y^3 + x) + \frac{1}{2} \int -\frac{2}{3} y^3 + x \; d(-\frac{2}{3} y^3 + x)$
$G(2y+x) = \frac{1}{2} \int 2y + x \; d(2y + x)$

Yet, I feel like this is going to get me nowhere.