Q:

Trouble in Checking PDE Boundary Solution Problem

I'm doing this problem with some other students, but it seems that our solution doesn't work?
We have the partial differential equation $\dfrac{\partial^2 u}{\partial x^2} -2 \dfrac{\partial^2 u}{\partial x \, \partial y} - 3\dfrac{\partial^2 u}{\partial y^2} = 0$. 
We found the general solution to be $u(x, y) = F_1(x - y) + F_2(3x + y)$, where $F_1$ and $F_2$ are arbitrary functions.
We want a solution to the equation with the general boundary conditions $u(x,0) = g_0(x)$ and $u_y(x,0) = g_1(x)$.
$$u(x,0) = g_0(x): u(x, 0) = F_1(x) + F_2(3x) = g_0(x)$$
$$u_y(x,0) = g_1(x): u_y(x, 0) = -F_1'(x) + F_2'(3x) = g_1(x)$$
Our work is as follows:
So we have $F_1(x) + F_2(3x) = g_0(x)$ and $-F_1'(x) + F_2'(3x) = g_1(x)$.
Integration of $-F_1'(x) + F_2'(3x) = g_1(x)$:
$-F_1(x) + \frac13 F_2(3x) = \int g_1(x)dx+c_1$
Adding this to $F_1(x) + F_2(3x) = g_0(x)$:
$F_2(3x) =\frac34 \left(g_0(x)+\int g_1(x)dx +c_1\right)$
And since $F_1(x) + F_2(3x) = g_0(x) \rightarrow F_1(x) = g_0(x) - F_2(3x)$:
$$F_1(x) = g_0(x) - \frac{3}{4} \int g_1(x) \ dx - \frac{3}{4} g_0(x)$$
And so we check our work:
$$u(x,y) = F_1(x - y) + F_2(3x + y) = \\ = g_0(x - y) - \frac{3}{4} \int g_1 (x - y) dx - \frac{3}{4} g_0(x - y) + \frac{3}{4} \int g_1 (x - y) dx + \frac{3}{4} g_0 (x - y)$$
$$= g_0(x - y)$$
$$u_y(x, y) = - g_0(x - y) $$
$$u_y(x, 0) = - g_0(x) \not= g_1(x)$$
If I am not mistaken, we should have $u_y(x, 0) = g_1(x)$ here?
I would greatly appreciate it if people could please take the time to clear this up for us.

A:

Your problemms come from the integration of the equation $-F'_1(x)+F_2(3\,x)=g_1(x)$. You should write
$$
-F_1(x)+\frac13\,F_2(3\,x)=\int_0^xg_1(t)\,dt.
$$
The choice of $0$ as the lower limit of integration is arbitrary. Then
\begin{align}
F_1(x)&=\frac14\,g_0(x)-\frac34\int_0^xg_1(t)\,dt\\
F_2(x)&=\frac34\,g_0\Bigl(\frac x3\Bigr)+\frac34\int_0^{x/3}g_1(t)\,dt
\end{align}

