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I am trying to solve the PDE $xu_x+yu_y=0$. I reach a point where, by setting $\frac{\partial{y}}{\partial{x}}=\frac{y}{x}$ which ends up implying that y=Ax, for some constant A. From there, I am trying to show that , since the derivative of any general solution $u(x,y)$ is $0$, then $u(x,Dx)=u(0,0)=u(x,y)$. From here I don't where to go to have a general solution. Any suggestions? Thanks in advance!

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The equation $$ \begin{align} 0 &=x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\\ &=(x,y)\cdot\nabla u \end{align} $$ means that the gradient of $u$ is perpendicular to $(x,y)$. This means that in any radial direction, $u$ is constant. In other words, $u$ is a function of $\arg(x,y)$. This is a bit more general than $u$ being a function of $\frac yx$ since we can have $u(x,y)\ne u(-x,-y)$. For example, $u(x,y)=\sin(\arg(x,y))$.

Since $u(x,y)=\frac12\log\left(x^2+y^2\right)$ is a particular solution of $xu_x+yu_y=1$, the general solution is $$ u(x,y)=\frac12\log\left(x^2+y^2\right)+f(\arg(x,y)) $$ where $f$ should be $2\pi$-periodic if we want $u$ to be smooth away from $(0,0)$.

Note that in polar coordinates, this becomes $$ u(r,\theta)=\log(r)+f(\theta) $$

Example $$ u(x,y)=\log\left(\sqrt{x^2+y^2}\right)+\frac y{\sqrt{x^2+y^2}} $$ Is an example of a solution which is smooth away from $(0,0)$ and cannot be written as $\log|x|+f\left(\frac yx\right)$ since $$ u(3,4)=\log(5)+\frac35\ne u(-3,-4)=\log(5)-\frac35 $$ but no matter what $f$ is $$ \log|3|+f\left(\frac43\right)=\log|{-}3|+f\left(\frac{-4}{-3}\right) $$

robjohn
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  • $f\left(\arg(x,y)\right) = f\left(\arctan\left(\frac{y}{x}\right)\right)=F(\frac{y}{x})$ Is it more general or the same ? – JJacquelin Sep 04 '15 at 10:09
  • It is more general since functions of $\frac yx$ are $\pi$-periodic functions of $\arg(x,y)$, and we can use any $2\pi$-periodic functions of $\arg(x,y)$. Take for example, $\log(r)+\sin(\theta)$. $\sin(\theta)$ cannot be written as a function of $\frac yx$. – robjohn Sep 04 '15 at 10:13
  • $\sin(\theta)=\sin\left(\arctan\left(\frac{y}{x}\right)\right)=F\left( \frac{y}{x} \right)$ – JJacquelin Sep 04 '15 at 10:17
  • Note that $\sin\left(\arctan\left(\frac yx\right)\right)$ is not smooth on the line $x=0$. Nor is it equal to $\sin(\theta)$ when $x\lt0$. – robjohn Sep 04 '15 at 10:18
  • What is the problem ? The function of function is smooth on the line $x=0$ – JJacquelin Sep 04 '15 at 10:21
  • here is a plot of $\sin\left(\arctan\left(\frac yx\right)\right)$. – robjohn Sep 04 '15 at 10:28
  • I would agree with you if the wording of the question was with an ODE in polar coordinates. But, if we want to strictly answer to the question, i.e; give a function $u(x,y)$ ,not on the form $u(r,\theta)$, then of course it is possible to use the function of function $f\left(\arg(x,y)\right)$. Do you think that $\arg(x,y)$ is not a function of $\frac{y}{x}$ ? . – JJacquelin Sep 04 '15 at 12:17
  • $\arg(1,1)=\frac\pi4$ and $\arg(-1,-1)=-\frac{3\pi}4$. $\frac yx=1$ in both cases, but $\arg(x,y)$ is different. Therefore, $\arg(x,y)$ cannot possibly be a function of $\frac yx$. – robjohn Sep 04 '15 at 13:05
  • It doesn't matter whether we specify the question in polar or rectangular coordinates. Using only functions of $\frac yx$ misses solutions that are valid in either case. – robjohn Sep 04 '15 at 13:08
  • @robjohn : Indeed, the solution I gave is incomplete. I guess it is because of the assumption I made when changing to variables $\xi, \eta$ which implies that only $x,y > 0$ can be written in such way. – Rogelio Molina Sep 07 '15 at 03:21
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$$xu_x+yu_y=1$$ First, we look for one (any one) particular solution. The simplest way is to look for a solution on the form $u=f(x)$ for example : $$xf'(x)=1$$ $$f(x)=\ln|x|$$ Then let $u(x,y)=\ln|x[+U(x,y)$

Bringing it back into $xu_x+yu_y=1$ leads to : $$xU_x+xU_y=0$$ which general solution is easy to find thanks to the change of variables $X=\ln[x[$ and $Y=\ln|y|$. The resulting PDE is $U_X+U_Y=0$ which general solution is $U(X,Y)=F(X-Y)$, any derivable function $F$.

$U(x,y)=F(\ln|x|-\ln|y|)$ $$u(x,y)=\ln|x|+F(\ln|x|-\ln|y|)$$ where $F$ is any derivable function.

This is the same as $u(x,y)=\ln|y|+G(\ln|x|-\ln|y|)$ where $G$ is any derivable function.

This general solution can be written on many different equivalent forms, for example : $u(x,y)=\ln|x|+\Phi\left(\frac{y}{x}\right)=\ln|y|+\phi\left(\frac{x}{y}\right)=\ln|y|+\varphi\left(\frac{y}{x}\right)=...$ where $\Phi$, $\phi$, $\varphi$ are any derivable functions.

Another way consists to transform the PDE given in Cartesian coordinates to a PDE in polar coordinates. This can be donne very easily as robjohn smartly did. Below, a more hard work is shown, just for the fun :

enter image description here

It is interresting to compare the solutions found in polar and in Cartesian coordinates and try to answer to the question : in which conditions are they equivalent or not ?

enter image description here

JJacquelin
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  • If $\varphi$ is differentiable, then $\log|y|+\varphi\left(\frac yx\right)$ is not differentiable near $y=0$. – robjohn Sep 05 '15 at 00:26
  • The PDE involves $yu_y$ NOT $u_y$ alone. For $y$ tending to 0, then $y\frac{d\ln(y)}{y}$ tends to $1$. – JJacquelin Sep 05 '15 at 05:29
  • I'm not claiming the function does not satisfy the equation where the derivative exists, just that the solution is not smooth, so the derivative does not exists at $y=0$. – robjohn Sep 05 '15 at 12:24
  • I agree. Of course, you are right. But even true, this gives an incomplete picture in some cases. – JJacquelin Sep 05 '15 at 14:59
  • For example, look at the behaviour in a range close to $x=0$ and $y>0$ of the piecewise function $f(x,y)=\sin\left(\arctan \left( \frac{y}{x}\right)+n:\pi\right)$ with determinations $n=0$ in $x>0$ and $n=1$ in $x<0$. For $x$ tending to $0$ both sides, the arctan tends to $+\infty$ and $f(x,y)$ tends to $1$ Moreover $\frac{\partial f}{\partial x}$ tends to $0$. So the partial derivative is well defined at limit and equal to $0$ – JJacquelin Sep 05 '15 at 15:00
  • This corresponds exactly to the derivative of $\sin(\theta)$ at $\theta=\frac{\pi}{2}$. As well, the piecewise function is smooth at the junction between the two pieces and is derivable at this point. – JJacquelin Sep 05 '15 at 15:03
  • $\sin\left(\arctan\left(\frac yx\right)+n\pi\right)$, as you have defined it, is $\sin(\arg(x,y))$. This is the same as $\frac y{\sqrt{x^2+y^2}}$, which I mention in the example from my answer. It is indeed infinitely differentiable away from $(0,0)$. – robjohn Sep 05 '15 at 15:22