What is the geometric interpretation of the solution to PDE $xu_x+yu_y=0$

Question

I have the following PDE

$$xu_x+yu_y=0$$

for which I get the characteristic function $$y=cx$$ along which the u(x,y) is constant. The general solution is $$u(x,y)=f(\frac{y}{x})$$.

I understand that the characteristic functions is a fan of lines in the x,y plane going through (0,0) point. It is unclear to me how the general solution could be presented.

What is the geometric solution/interpretation to the PDE in the x,y,u dimensions?

Answer 1

$$xu_x+yu_y=0$$ FIRST PART, Solving with the method of characteristics :

From $\quad xu_x+yu_y=0 u \quad$ the set of characteristic equations is : $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{0}$$ because the coefficient of $u_x$ is $x$ , the coefficient of $u_y$ is $y$ and the coefficient of $u$ is $0$ .

From $\frac{dx}{x}=\frac{dy}{y}$ the equation of the set of characteristic curves is $\ln|y|-\ln|x|=$constant, or : $$\frac{y}{x}=c_1$$ To be finite $\frac{du}{0}$ implies $u=$constant. So, the equation of the set of characteristic curves is : $$u=c_2$$ The general solution of the PDE expressed on implicit form is : $$F\left(\frac{y}{x} \:,\: u\right)=0$$ where $F$ is any differentiable function of two variables.

Solving this implicit equation for the second variable leads to the explicit form : $$u=f\left(\frac{y}{x}\right)$$ where $f$ is any differentiable function.

Note : Since $f$ is any function, the solution includes $u=\phi(\tan^{-1}\left(\frac{y}{x}\right))$ where $\phi$ is any function. This is equivalent to $u=\phi(\theta)$ where $\theta=\tan^{-1}\left(\frac{y}{x}\right)$ and with the function $g\equiv f(\tan^{-1})$.

SECOND PART

Answer to the question of graphical interpretation :

Your graph represents the set of characteristic curves $$y=c_1x$$ drawn with various values of $c_1$

On one characteristic curve corresponding to $c_1$ : $$u=f\left(\frac{y}{x}\right)=f(c_1)$$ since $c_1$ is a constant, $f(c_1)$ is constant, so $u$ is constant. $$u=f(c_1)=c_2$$

Thus $u(x,y)$ is constant all along the characteristic curve considered.

But $u(x,y)$ isn't constant if the point $(x,y)$ goes from one curve $(c_1)$ to another curve $(c'_1)$, because $u(x,y)$ varies from $f(c_1)=c_2$ to $f(c'_1)=c'_2$ which are different.

If you draw in 3D the function of two variables $\frac{y}{x}$ you obtain the representation of a particular solution of the PDE : $u(x,y)=f\left(\frac{y}{x}\right)=\frac{y}{x}$ in the case of $f(X)=X$. Of course, this isn't a representation of the characteristic curves.

Answer 2

First of all, the "general solution" $u(x,y)=f(\frac{y}{x})$ is not general. All functions of this form have symmetry $u(-x,-y) = u(x,y)$, which need not be satisfied by solutions of the PDE: for example, $u(x,y)=y/\sqrt{x^2+y^2}$ is a solution that is not of the above form.

The correct form of general solution is $$u(x,y) = f(\theta)$$ where $\theta$ is the polar angle and $f$ is a differentiable $2\pi$-periodic function.

The characteristic curves of the PDE are not lines; they are half-lines. The geometry of the solution is that it's constant on each such line, so it depends on the polar angle only.

Answer 3

It has been shown here by several people that the solutions $(x,y)\mapsto u(x,y)$ have to be constant on rays emanating from $(0,0)$. It follows that either $u$ is globally constant, or undefined at $(0,0)$. This means that the solutions are of one of the following forms: Either $$ u(x,y)\equiv c \in{\mathbb R}\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ ,$$ or there is a differentiable $2\pi$-periodic function $f$ such that $$u(x,y)=f\bigl({\rm arg}(x,y)\bigr)\qquad\bigl((x,y)\in\dot{\mathbb R}^2\bigr)\ .\tag{2}$$ It remains to prove that the functions of type $(2)$ are indeed solutions. To this end we write ${\rm arg}(x,y)=:\phi$ and compute $$u_x x+u_y y=f'(\phi)\phi_x\,x+f'(\phi)\phi_y\,y=f'(\phi)\left({-y\,x\over x^2+y^2}+{x\,y\over x^2+y^2}\right)\equiv0\ .$$ Some of the OP's difficulties stem from the fact that the variables $x$ and $y$ were not treated in an equal-rights way.

Answer 4

Rewrite it as

$$xu_x+yu_y=\left(x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y}\right) u =0$$

Now recall that

$$r^2=x^2+y^2$$

so, taking the full derivative

$$rdr=xdx+ydy$$

Dualising this

$$r\frac{\partial}{\partial r}=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$$

So, our equation is

$$r\frac{\partial}{\partial r} u = 0 \Rightarrow \frac{\partial}{\partial r} u=0$$

That is, $u$ is constant when moving along $r$. Introducing polar coordinates we see that a solution depends only on the angle $\theta$.