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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.

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  • $\begingroup$ It might be convenient to write it as $(a,b,c)\cdot\nabla u=0$. $\endgroup$ – timur Sep 26 '17 at 13:40
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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?

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  • $\begingroup$ Yes! :) $$ a\left(b\frac{\partial\Phi(bx-ay,cy-bz)}{\partial(bx-ay)}\right)+ $$ $$ +b\left(-a\frac{\partial\Phi(bx-ay,cy-bz)}{\partial(bx-ay)}+ c\frac{\partial\Phi(bx-ay,cy-bz)}{\partial(cy-bz)}\right)+$$ $$+c\left(-b\frac{\partial\Phi(bx-ay,cy-bz)}{\partial(cy-bz)}\right)=0$$ That's perfect! $\endgroup$ – Clyde A. Jansen Sep 26 '17 at 15:09

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