Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 :)! –  Slugger Feb 14 '13 at 20:50
    
Your solution doesn't hold for arbitrary $a$ and $b$. You can check by simple substitution. It's valid only for $a = b = \frac 12$ –  Kaster Feb 14 '13 at 23:59
add comment

2 Answers

up vote 2 down vote accepted

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 $F(x) = e^{f(x)}$

Update

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.

share|improve this answer
    
Sorry, I made a mistake. Fixed it though. –  Kaster Feb 14 '13 at 23:26
    
Very clear answer, thanks! –  Slugger Feb 16 '13 at 14:35
add comment

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 $$u(x,y)=e^{\frac{-c}{a}}f(\frac{y}{b}-\frac{x}{a})$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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