Methods of characteristics for second order PDE $u_{xx}-3u_{xy}+2u_{yy}=0$ Given $$u_{xx}-3u_{xy}+2u_{yy}=0$$
Can you apply the methods of characteristics to this problem? I was required to but don't know how to do it for second order PDEs.
with the boundary condition $$u(x,0)=-x^2, \frac{\partial u}{\partial y}(x,0)=0$$
Also in which range are these $x$ and $y$ valid?
EDIT:
Based on the comment we got $$c_1=y+x$$ and $$c_2=y+2x$$
The general solution is $$u(x,y)=f(y+x)+g(y+2x)$$
Now given the boundary condition $u(x,0)=-x^2$ and $u_y(x,0)=0$ I need to find the special solution.
So obviously $$u(x,0)=f(x)+g(2x)=-x^2$$
$$u_y(x,0)=f_y(x)+g_y(2x)=0\Rightarrow f(x,0)+g(x,0)=C=-x^2$$
But i don't know how to proceed..
 A: Yes, you can use the method of characteristic to tackle these types of problems. The trick is you need to factor the differential operator. In this case we can write $$\left(\partial_{xx} + 2\partial_{yy} - 3\partial_{xy}\right)u = \left(\partial_x-\partial_y\right)\left(\partial_x -2\partial_y\right)u$$
You can now use the method of characteristics to solve each piece separately.
Update:
Here's how to find the solution. Following the above, we solve each first order hyperbolic pde separately. Let's take a look at the first one. We want to solve
$$u_x - u_y = 0$$
Notice that this equation can be rewritten in the form
$$\nabla u \cdot \begin{bmatrix} -1 \\ 1\end{bmatrix} = 0$$
implying that long lines with slope $\frac{dx}{dy} = -1$ the solution $u$ is constant. Solving this ODE for $x$ (i.e. using the method of characteristics) we find that $x(y) = -y + x_0$ where $x_0$ is the initial condition. Since $u$ is constant along these lines, we have
$$u(x(y),y) = u(x_0,0) = -x_0^2$$
Since $x_0 = x(y)+y$, this tells us that the first solution is $u_1 = C_1(x+y)^2$. Using the same kind of argument for the second equation, we find that the second solution is $u_2 = C_2(x+\frac{y}{2})^2$. Since each of these two solutions solve the equation, the general solution is
$$u_{\text{gen}} = C_1(x+y)^2 + C_2(x+ \frac{y}{2})^2$$
To get the particular solution, we use the fact that $u(x,0) = -x^2$ and $\frac{\partial u}{\partial y}(x,0) = 0$. Using these conditions we get the following system of equations for the two constants
$$C_1 +C_2 = 0$$
$$2C_1 + C_2 = 0$$
Solving this system, we find that $C_1 = 1$ and $C_2 = -2$. Therefore, the particular solution is
$$u = (x+y)^2 - 2(x + \frac{y}{2})^2.$$
