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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.

doraemonpaul
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Slugger
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  • 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

3 Answers3

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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.

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

M.H
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Generally, your idea is correct. However, some implication is need. We need to parametrize the characteristic curve with parameter r. Change the original equation to $\frac{du}{dr}+cu=0$ with $$y=br$$ $$x=ar+c_0$$ where $c_0$ denote which characteristic line we are focus on. Second, use separation of variable to solve it. $$\frac{du}{u}=-cdr$$ $$ln|u|=-cr+c_1$$ $$u=\pm e^{c_1}e^{-cr}$$ Actually, $c_1$ depends on which characteristic line we are studying. Therefore, $e^{c_1}$ can be seen as a constant depending on $c_0=by-ax$, written as $F(by-ax)$. Substitute $r$ with $r=\frac{y}{b}$ we got the solution $u=e^{-\frac{c}{b}y}F(bx-ay)$. PS:It may differ with other solutions by some function of $bx-ay$. Say $u=e^{-\frac{c}{b}y}f(bx-ay)\cdot e^{-\frac{bx-ay}{2ab}}=e^{-\frac{c}{2ab}(bx+ay)}$

Brian
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