Cauchy Problem and Region of Validity

Question

I have the Cauchy Problem

$$ 2xu_x+(x+y)u_y=2u $$ with data

$$ u(x,-x)=\sqrt{x},x>0$$

Omitting details, my answer is

$$u(x,y) =\sqrt{x}\left ( \frac{y}{2x}-\frac{1}{2} \right)^{-\frac{1}{3}} $$

or

$$u(x,y) =\frac{2^{\frac{1}{3}}x^{\frac{5}{6}}}{\left ( y-x \right)^{\frac{1}{3}} } $$

If you solve the Cauchy problem, do you arrive at the same $u(x,y)$? What about the region of validity?

It seems this becomes invalid when $y=x$ and when $x<0$, so I propose that my region of validity is $x>0$, $y<x$

Answer 1

The solution of the PDE is given by:

$$\color{blue}{u(x,y) = f\left(\frac{y-x}{\sqrt{x}} \right) \, x, \quad x > 0} \tag{1}$$

for some arbitrary function $f$.

In the other hand, the solution of the Cauchy problem is:

$$\color{blue}{ u(x,y) = 2x \, \frac{\sqrt{x}}{x-y}, \quad y \neq x, \ \ x>0} \tag{2} $$

What happens when $x=0$ is left to the reader.

Hope this helps. Cheers!

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Edit:

Here's a contour plot of the solution where the region $x=y$ is highlighted:

Answer 2

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=2x$ , letting $x(0)=1$ , we have $x=e^{2t}$

$\dfrac{dy}{dt}=x+y=e^{2t}+y$ , letting $y(0)=y_0+1$ , we have $y=e^{2t}+y_0e^t=x+y_0\sqrt x$

$\dfrac{du}{ds}=2u$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)e^{2t}=f\left(\dfrac{y-x}{\sqrt x}\right)x$

$u(x,-x)=\sqrt x$ :

$f(-2\sqrt x)x=\sqrt x$

$f(-2\sqrt x)=\dfrac{1}{\sqrt x}$

$f(-2x)=\dfrac{1}{x}$

$f(x)=-\dfrac{2}{x}$

$\therefore u(x,y)=-\dfrac{2}{\dfrac{y-x}{\sqrt x}}\times x=\dfrac{2x^\frac{3}{2}}{x-y}$