I am asked to solve a PDE in the sense of distributions, but I'm kinda confused by the definition. So I am trying to solve the PDE: $$au_x+bu_y=0$$ Now, let us define $f(bx-ay) = u(x,y)$, where $f$ is only locally integrable. I'm supposed to show that $u(x,y)$ solves this PDE in the sense of distributions. The hint is to make a change of variable so the characteristic directions are the coordinate axes.

Now, I began to solve it: $$\left< au_x+bu_y ,\phi \right> = \iint_{\mathbb R^2} -u(a\phi_x+b\phi_y) dx dy$$ Where $\phi\in C^{\infty}_c (\mathbb R^2)$. But how do I proceed from here? I'm kinda confused by what the question is asking.

  • $\begingroup$ Replace $u(x,y)$ with $f(bx-ay)$ and make a variable change to $\xi=bx-ay$ and $\eta=ax+by$. $\endgroup$ – md2perpe Nov 10 '17 at 18:14
  • $\begingroup$ How do you go about solving this PDE with that change in the sense of distributions? I understand it changes the PDE to $2ab u_{\eta}=0$, and integrating you will get the desired result, but I have done nothing in the sense of distributions. How does this come into play? $\endgroup$ – Felicio Grande Nov 10 '17 at 20:42
  • $\begingroup$ What exactly is the goal of the exercise? Is it to find a general solution or to show that $(x,y) \mapsto f(bx-ay)$ solves the PDE in the sense of distributions? $\endgroup$ – md2perpe Nov 10 '17 at 21:14
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    $\begingroup$ Now you got it! $\endgroup$ – md2perpe Nov 11 '17 at 20:36
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    $\begingroup$ You are right. There will also be a factor $1/(a^2+b^2)$ because of the coordinate change from $dx \, dy$ to $d\xi \, d\eta.$ $\endgroup$ – md2perpe Nov 13 '17 at 14:10

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