# Clarification on Method of Characteristics

I'm trying to solve the PDE: $-2u_x +6u_y=0$ using the method of characteristics.

My solution: Let $\xi=x$, $\eta = Ax+By$.

$\implies u_x = u_\xi + Au_\eta$ and $u_y=Bu_\eta$.

Then $-2(u\xi +Au_eta) + 6Bu_\eta = 0$. Let $B=1$ and $A=3$. Then $-2u_\xi = 0 \implies u_\xi = 0$.

The solution is then ANY equation of the form $u=f(3x+y)$.

Am I done? Is that all there is to this?

And also, what if we added the initial condition $u(4y,y)=2y+1$?

I'm thinking $\implies u(4y,y) = f(12y+y)=f(13y) = 2y+1$, so we just need any solution to $f(13y)=2y+1$, like for instance $f(z) = \frac 2{13}z+1$.

So then the general solution should be $u(x,y) = f(3x+y) = \frac 2{13}(3x+y)+1$.

Is that correct? -- I'm new to this and my textbook doesn't explain it all that well.

• For the first part, your general solution is correct (you can check by differentiation). For the second part, I'm not used to seeing the notation $x \to 4y \implies u(x,y) = u(4y,y)$ but it seems to be fine and still satisfies the PDE. – Mattos Mar 9 '15 at 13:36
• OK, good. Thanks. – user222183 Mar 9 '15 at 13:37
• I would say this is the method of change of variable. The method of characteristics is more general. The characteristics (or the coordinate change in your language) can be curvilinear; it is generally solved from the PDE, rather than guessed by trial and error. – Fan Zheng Mar 10 '15 at 1:41