I am asked to solve $$au_x+bu_y+cu=0$$ I am tempted to first solve $au_x+au_y=0$ which has characteristic lines $C=ay-bx$ and thus a solution to this is given by $$u(x,y)=f(ay-bx)$$ where $f$ is an arbitrary function. Then substituting back into the original equation yields $$au_x+bu_y+cu=0+cu=cf(ay-bx)=0$$ implying that I have merely found the trivial solution $u(x,y)=f(ay-bx)=0$.

So far the book I am using has only explained the method of characteristic equations and I have solved various difficult ones like $\sqrt{1-x^2}u_x+u_y=0$. So I am guessing that I should be able to solve $au_x+bu_y+cu=0$ using this method combined with maybe some clever thinking. I might be able to use the fact that the directional derivative of $u$ along the lines $C=ay-bx$ is $-cu$ and so maybe along these lines $u=e^{-cf(ay-bx)}$ or something. If anyone has any suggestions I would be thankful.

  • $\begingroup$ I think I just found a solution $$u=Ke^{-c\left(\frac{1}{2}\left(x+y\right)\right)}$$ where $K$ is an arbitrary constant. This was sort of trial and error though, so anyone who has an interesting way to arrive here speak up please :)! $\endgroup$ – Slugger Feb 14 '13 at 20:50
  • $\begingroup$ Your solution doesn't hold for arbitrary $a$ and $b$. You can check by simple substitution. It's valid only for $a = b = \frac 12$ $\endgroup$ – Kaster Feb 14 '13 at 23:59

Make a change of variables $$ t = bx + ay \\ p = bx - ay $$ So $$ u_x = b(u_t + u_p) \\ u_y = a(u_t - u_p) $$ After substituting in PDE $$ ab(u_t + u_p) + ab(u_t - u_p) + cu = 2ab\ u_t + cu = 0 $$ It can be easily integrated $$ \frac {u_t}u = -\frac c{2ab} \\ \ln u = -\frac c{2ab}t+f(p) \\ u = F(p)e^{-\frac c{2ab}t} $$ or, in initial variables $$ u = F(bx-ay)e^{-\frac c{2ab}(bx+ay)} $$ where $F(x) = e^{f(x)}$


If you use $$ t = ax + by \\ p = bx - ay \\ $$ so equation is $$ (a^2+b^2)u_t + cu = 0 \\ \ln u = -\frac c{a^2+b^2}t \\ u = F(p)e^{-\frac {ct}{a^2+b^2}} \\ u = F(bx - ay)e^{-\frac c{a^2+b^2} (ax + by)} $$ which is also a solution.

  • $\begingroup$ Sorry, I made a mistake. Fixed it though. $\endgroup$ – Kaster Feb 14 '13 at 23:26
  • $\begingroup$ Very clear answer, thanks! $\endgroup$ – Slugger Feb 16 '13 at 14:35

let $v(x,y)=e^{cx}u(x,y)$ and compute $v_x,v_y$ then we have$$v_x+v_y=0$$ by geometric method we have$$ v(x,y)=e^{\frac{-c}{a}}f(y-x)$$ easily conclude $$u(x,y)=e^{\frac{-c}{a}}f(\frac{y}{b}-\frac{x}{a})$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.