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Well I have given the following PDE:

$y\cdot\partial_xu(x,y)+\partial_yu(x,y)=0$

Now I have to solve it by using the method of characteristics.

The coefficient are $(a,b,c)=(y,1,0)$. Then I have to solve the following differential equations: $\dot x(t) = y$ and $\dot y=1$. $\Longrightarrow y(t)=t+c$ and $x(t)=\frac {t^2}2+tc +d$ with any constant numbers $c,d$.

But now I don't know how to continue. How do I get $u(x,y)$ and is my ansatz correct?

Thanks for helping!!

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1 Answer 1

up vote 2 down vote accepted

Yes your ansatz is correct. Now you only have to use the fact that $u$ remains constant along a characteristic which is parameterized by $$(x,y) = (\tfrac12 t^2 +t c +d,t+c).$$ Eliminating, $t$ we have the explicit equation $$y^2-2x =c^2 -2d.$$ Setting $u=f(c^2-2d)$ along the characteristic, we have $$u(x,y)= f(c^2-2d)= f(y^2-2x)$$ as the solution of the differential equation with $f(x)$ an arbitrary function.

Edit:

We eliminate $t$ by writing down the expression of $y^2$ in terms of $t$ and reexpress it in terms of $x$, $$y^2 = (t+c)^2 = t^2 +2 t c + c^2 = 2x + c^2 -d.$$

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thanks for your help!! But how did you get $y^2-2x=c^2-2d$ ? I just don't get by eliminating t :( –  sheldoor Apr 19 '12 at 16:56
    
@sheldoor: I had a typo in my last reply (I had $x$ and $y$ interchanged). Most likely this was the reason why you couldn't get my result. I added a line trying to explain the elimination. –  Fabian Apr 19 '12 at 19:41
    
THANK YOU VERY MUCH! now I get everything! thx! –  sheldoor Apr 19 '12 at 19:43

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