Stuck with boundary value PDE problem

Question

The last time I posted this I got many down votes and I don't know why. Maybe because I forgot to include my work?

I got the PDE $\dfrac{\partial^2 u}{\partial x^2} -2 \dfrac{\partial^2 u}{\partial x \, \partial y} - 3\dfrac{\partial^2 u}{\partial y^2} = 0$

General solution found to be $u(x, y) = F_1(x - y) + F_2(3x + y)$ these are arbitrary functions.

I want to get a solution to the equation that satisfies the boundary conditions $u(x,0) = x$, $u_y(x,0) = 0$.

$$u(x, 0) = x: u(x, 0) = F_1(x) + F_2(3x) = x$$

$$u_y(x, y) = - F_1'(x - y) + F_2'(3x + y)$$

$$u_y(x,0) = 0: u_y(x, 0) = - F_1'(x) + F_2'(3x) = 0$$

And also get a solution to the equation with the general boundary conditions $u(x,0) = g_0(x)$ and $u_y(x,0) = g_1(x)$.

$$u(x,0) = g_0(x): u(x, 0) = F_1(x) + F_2(3x) = g_0(x)$$

$$u_y(x,0) = g_1(x): u_y(x, 0) = -F_1'(x) + F_2'(3x) = g_1(x)$$

I am stuck. Please show me how this is done. Thank you.

Answer 1

You have $F_1(x) + F_2(3x) = x$, differentiating by $x$ we'll get $F_1'(x)+3F_2'(3x)=1$. From the last boundary condition we have $-F_1'(x)+F_2'(3x)=0$. Adding two equations together, we get $4F_2'(3x)=1$. Now we can integrate and find $F_2$: $4\int{F_2'(3x)}dx=\int{dx}=x+c$; $F_2(x)=\frac{x}{12}+c$. Now substitute $F_2$ back in the initial condition and find $F_1$: $F_1(x)+\frac{3x}{12}+c=x$; $F_1=...$

Answer 2

$$ F_1(x) + F_2(3x) = g_0(x)$$

$$ -F_1'(x) + F_2'(3x) = g_1(x)$$

Integrating the second equation :

$$ -F_1(x) + \frac13 F_2(3x) = \int g_1(x)dx+c_1$$

Adding to the first equation : $$F_2(3x) =\frac34 \left(g_0(x)+\int g_1(x)dx +c_1\right)$$

$$F_1(x)=g_0(x)-\frac34 \left(g_0(x)+\int g_1(x)dx +c_1\right)$$ $$\begin{cases} F_2(3x) =\frac34 g_0(x)+\frac34\int g_1(x)dx +c'_1 \\ F_1(x)=\frac14 g_0(x)-\frac34 \int g_1(x)dx -c'_1 \end{cases}$$ $F_2(x) =\frac34 g_0(\frac{x}{3})+\frac14\int g_1(\frac{x}{3})dx +c'_1$