# 2nd order linear hyperbolic PDE?

1. Oct 5, 2009

### kingwinner

uxx - x2 uyy = 0 (assume x>0)

Is there any systematic method (e.g. change of variables) to solve this hyperbolic equation?

dy/dx = [B + sqrt(B2 - AC)]/A
=> dy/dx = x
=> 2y -x2 = c

dy/dx = [B - sqrt(B2 - AC)]/A
=> dy/dx = -x
=> 2y + x2 = k

So the characteristic curves are 2y -x2 = c and 2y + x2 = k

Now does this imply that the general solution is u = f(2y-x2)+g(2y+x2) ?

Thanks!

2. Oct 5, 2009

### matematikawan

u = f(2y-x2)+g(2y+x2)

Is uxx - x2 uyy identically zero? Then it must be a general solution

3. Oct 6, 2009

### kingwinner

Hi, I tried to verify it by computing a bunch of tedious second derivatives, but somehow I didn't get 0... I may have made a mistake in my calculations...

4. Oct 6, 2009

### kosovtsov

The most direct way to solve this PDE is the Fourier method, that is, supposing that solution has the form

u(x,y)=\int_{-\infty}^{\infty}F(x,k)exp(iyk) dk

we obtain ODE for F

F_{xx}+x^2k^2F=0 ,

which solution is as follows

F(x,k)=x^{1/2}[F1(k)J_{1/4}(x^2k/2)+F2(k)K_{1/4}(x^2k/2)] ,

where J_{1/4} and K_{1/4} are modified Bessel functions; F1 and F2 are arbitrary functions.

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