I tried to solve the following PDE:
$$au_x+bu_y+cu_z=0$$
where $a,b,c \in \mathbb{C} $.
The results I obtained are the follwing:
$$ \begin{cases} \frac{1}{a}dx=\frac{1}{b}dy\\ \frac{1}{b}dy=\frac{1}{c}dz \end{cases} $$
Solving this system
$$ \begin{cases} bdx=ady\\ cdy=bdz \end{cases} $$
Integrating, yields
$$ \begin{cases} b(x+K_1)=a(y+K_2)\\ c(y+K_3)=b(z+K_4) \end{cases} $$
Therefore
$$ \begin{cases} \zeta_1:= aK_2-bK_1 = bx-ay\\ \zeta_2:= bK_4-cK_3 = cy-bz\\ \end{cases} $$
So
$$u(x,y,z)=\Phi(\zeta_1,\zeta_2)=\Phi(bx-ay,cy-bz)$$
At this point, I needed to verify the solution, and I started like this:
Let $p=f(x,y,z)=bx-ay \quad$ and $\quad q=g(x,y,z)=cy-bz$
So, using the chain rule, I wrote:
$$u_x = f_x\Phi_p + g_x\Phi_q = b\Phi_p$$ $$u_y = f_y\Phi_p + g_y\Phi_q = -a\Phi_p + c\Phi_q$$ $$u_z = f_z\Phi_p + g_z\Phi_q = -b\Phi_q$$
By substitution of these values in the PDE, I am not able to conclude. Surely, I am getting wrong in some passages. Can anyone tell me how to do the correct ones?
Thanks in advance.
Your characteristic equation is correct and your solution is also correct.
Substituting this into the PDE you will obtain:
$$aF_x+bF_y+cF_z=a\left[\dfrac{\partial F}{\partial (bx-ay)}\dfrac{\partial (bx-ay)}{\partial x} +\dfrac{\partial F}{\partial (cy-bz)}\dfrac{\partial (cy-bz)}{\partial x}\right]$$ $$+b\left[\dfrac{\partial F}{\partial (bx-ay)}\dfrac{\partial (bx-ay)}{\partial y}+\dfrac{\partial F}{\partial (cy-bz)}\dfrac{\partial (cy-bz)}{\partial y}\right]$$ $$+c\left[\dfrac{\partial F}{\partial (bx-ay)}\dfrac{\partial (bx-ay)}{\partial z}+\dfrac{\partial F}{\partial (cy-bz)}\dfrac{\partial (cy-bz)}{\partial z}\right]$$
$$=a\left[\dfrac{\partial F}{\partial (bx-ay)}b +\dfrac{\partial F}{\partial (cy-bz)}\cdot 0\right]$$ $$+b\left[\dfrac{\partial F}{\partial (bx-ay)}(-a)+\dfrac{\partial F}{\partial (cy-bz)}c\right]$$ $$+c\left[\dfrac{\partial F}{\partial (bx-ay)}\cdot 0+\dfrac{\partial F}{\partial (cy-bz)}(-b)\right]=0$$ Can you complete it from here?