I'm given the following PDE
$$u_{xx} – 2\sin x u_{xy} – (3 + \cos^2x)u_{yy} + u_x + (2 – \sin x – \cos x)u_y = 0$$
with conditions
$$u(x, \cos x) = 0, \quad u_y(x, \cos x) = e^{-x/2}\cos x.$$
It's a hyperbolic equation. I used substitutions
$\xi = y – \cos x – 2x, \eta = y – \cos x + 2x$ and
$v(\xi(x, y), \eta(x,y)) = u(x, y)$ and got the canonical form
$$-16v_{\xi\eta} + 4v_\eta = 0.$$
Then I should integrate it with respect to $\eta$ and $\xi$ and get arbitrary functions $\phi(\xi)$ and $\psi(\eta)$, but I don't understand how to do it. In examples I've seen the canonical form was
$u_{\xi\eta} = 0$ which gives $u(\xi, \eta) = \phi(\xi) + \psi(\eta)$. What is the result of integration for my problem?
The next step would be getting a system of two equations using given conditions and solving it and for this problem it should be simple.
Best Answer
Hint.
$$ -16v_{\xi\eta} + 4v_\eta = 0\Rightarrow\partial_{\eta}\left(-16\partial_{\xi}+4\right)v =0 $$
so calling $V = \left(-16\partial_{\xi}+4\right)v$ we solve first
$$ \partial_{\eta}V = 0\Rightarrow V(\xi,\eta) = \phi(\xi) $$
following with
$$ \left(-16\partial_{\xi}+4\right)v = \phi(\xi) \Rightarrow v(\xi,\eta) = e^{\xi /4} \left(\psi (\eta )-\frac{1}{16}\int_1^{\xi } e^{-\zeta /4} \phi (\zeta ) \, d\zeta \right) $$
or
$$ v(\xi,\eta) = e^{\xi /4} \left(\psi (\eta )+\omega(\xi)\right) $$