Solution of the Cauchy problem $xu_x + yu_y = 0$, $u(x,y) = x$ on $x^2 +y^2 =1$

Question

The Cauchy problem $xu_x + yu_y = 0$, $u(x,y) = x$ on $x^2 +y^2 =1$ has

1. a solution for $x ,y \in \Bbb R$

2. a unique bounded solution in $\{x ,y \mid (x,y) \neq (0, 0)\}$

3. an unique solution in $\{x ,y \mid (x,y) \neq (0, 0)\}$ but not bounded.

I have tried to solve this problem. Can anyone please check and tell me how to proceed further and which option(s) will be right?

Answer 1

From your notes, it's not quite clear (to me) what you have actually done, but what the PDE says is that $u$ must be constant on rays (half-lines) from the origin. Together with the given values on the unit circle, that should tell you anything you need to know about $u$.

Answer 2

Taking $$ xu_{x}+yu_{y}=0\text{ with } u(x,y)=x \text{ on }x^{2}+y^{2}=1 $$

Due to the contour conditions we will consider the change of variables

\begin{align} x & = r\cos\theta\\ y & = r\sin\theta \end{align}

with

\begin{align} dx & = dr\cos\theta-r\sin\theta d\theta\\ dy & = dr\sin\theta+r\cos\theta d\theta \end{align}

we have \begin{align} u_{x} & = u_{r}\frac{dr}{dx}+u_{\theta}\frac{d\theta}{dx}\\ u_{y} & = u_{r}\frac{dr}{dy}+u_{\theta}\frac{d\theta}{dy} \end{align}

Here from the characteristic curves

$$ \frac{dx}{x}=\frac{dy}{y}\implies\frac{dy}{dx}=\frac{y}{x}=\tan\theta $$

so we obtain

$$ xu_{x}+yu_{y}=0\iff2ru_{r}=0\implies u(r,\theta)=C+\Phi(\theta) $$

Note that

\begin{align} \frac{dr}{dx} & = \frac{1}{\cos\theta}\\ \frac{dr}{dy} & = \frac{1}{\sin\theta}\\ \frac{d\theta}{dx} & = 0\\ \frac{d\theta}{dy} & = 0 \end{align}

Now with the boundary conditions

$$ u(1,\theta)=\cos\theta\implies\Phi(\theta)=\cos\theta\text{ and }C=0 $$

and finally

$$ u(r,\theta)=\cos\theta $$

or in $(x,y)$ coordinates, $$du(x,y)=\frac{x}{\sqrt{x^{2}+y^{2}}}$$

Answer 3

Let $$v(r,\theta) = u(r\cos \theta, r\sin \theta)$$

Then $$u_x = v_r\cos\theta - v_\theta r \sin \theta \\ u_y = v_r\sin\theta + v_\theta r \cos \theta $$ $$xu_x + yu_y = v_r r\cos^2\theta - v_\theta r^2 \sin \theta \cos\theta \\ + v_r r\sin^2\theta + v_\theta r^2 \cos \theta \sin \theta$$ $$ 0 = xu_x + yu_y = r v_r $$

Therefore $v_r = 0$ everywhere except possibly at the origin, and $$ v(r, \theta) = f(\theta) \\ u(x,y) = f(\mbox{atan2 } (y/x) ) $$ where the atan2 function is the usual quadrant-aware arctangent (atan2 is in the third quadrant if $x$ and $y$ are both negative, etc).

Now if we impose the boundary conditions, the unique solution is $$ u(x,y) = f(\cos\left(\mbox{atan2 } (y/x) \right) $$

This is a bounded function, but it is not defined at $(x,y) = (0,0)$ because on an arbitrarily small circle around the origin, ($u(x,y)$ takes on all values in $[-1,1]$.

So the answer is 2 - bounded everywhere but domain does not include the origin.