Solve the following PDE:
$$f_{xxx}=a_1f_{xx}+b_1f_{x} \ \ (1)$$
$$f_{yyy}=a_2f_{yy}+b_2f_{y} \ \ \ (2)$$
$f(x,y)$ is a smooth function. $a_1,b_1,a_2,b_2$ are constatns.
For an ODE, $g'''=ag''+bg'$, the solution is not hard by the method of characteristics. Here is my solution:
If there are three zero roots, then $f$ is quadratic
If there are two zero roots, then $f=C_1x+C_2e^{C_3x}$
If there is only one zero root, then either $f=C_1e^{C_2x}+C_3e^{C_4x}$ or $f=(C_1x+C_2)e^{C_3x}.$
I am not so used to the Laplace transformation but I will continue with my approach. First we consider the case where the characteristic equation of both (1) and (2) have two zero roots, then the solution to (1) would be:
$$f=C_1(y)+C_2(y)e^{C_3(y)x}$$ The solution to (2): $$f=C_4(x)+C_5(x)e^{C_6(x)y}$$
Those two solutions must equal (?) so,
$f=C_7+C_8e^{C_3x+C_6y+C_7xy}+C_9e^{C_{10}x}+C_{11}e^{C_{12}x}$+...
By repeating the steps for other cases of the characteristic equation, we can get all the solutions. Does it sound correct?
I am still trying to come-up with possible solutions. One comment suggests that the functional form might be $f=g(x)+h(y)$ by using the Laplace transformation. I agree with the approach but I also observe that $f=g(x)h(y)$ might also work?