# Solving S.O. PDE by transforming to normal form

1. Jun 24, 2010

### prhzn

1. The problem statement, all variables and given/known data

Transform to normal form and solve:
1) $$u_{xx}+u_{xy}-2u_{yy} = 0$$

2. Relevant equations

Normal form: $$Au_{xx}+2Bu_{xy}+Cu_{yy}$$, hence, $$A = 1, B = \frac{1}{2}, C = -2$$.
Since $$AC-B^2 = -2.25 < 0$$ this is a hyperbolic equation.
Want to transform it by setting

$$v = \Phi(x,y), z = \Psi(x,y)$$ where $$\Phi = const$$ and $$\Psi = const$$ are solutions $$y=y(x)$$ of

2) $$A(y')^2-2By'+C = 0$$

3. The attempt at a solution

From 2) I find

$$(y'-2)(y'+1) = 0$$, giving me $$y_1 = 2x+C, y_2 = -x+C$$. From this I choose choose $$v = 2x-y, z = x + y$$ as the transformations for the variables.

Finding the new expressions:

$$u_{xx} = (u_x)_x = ..... = 4u_{vv}+4u_{zv}+u_{zz}$$
$$u_{xy} = (u_x)_y = .... = -2u_{vv}+u_{zv}+u_{zz}$$
$$u_{yy} = (u_y)_y = .... = u_{vv}-2u_{zv}+u_{zz}$$

Inserting these into 1) I get that

$$9u_{zv} = 0 \leftrightarrow u_{zv} = 0$$, yielding $$u(v,z) = f_1(v)+f_2(z) \rightarrow u(x,y) = f_1(2x-y)+f_2(x+y)$$.

I'm not sure if this is correct, hence, I hope someone could let me know if I'm onto something or way off :)

Last edited: Jun 24, 2010
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