PDE with no solution in neighborhood.

[renegade05]

QUESTION:

Show that:

$\displaystyle (y-x)u_x - (x+y)u_y = 0 \qquad u(x,0)=sin(x)$

has no solution that exists in a neighborhood of the origin (x, y) = (0, 0).

ATTEMPT AT SOLUTION:

$\displaystyle \frac{dx}{y-x}=\frac{dy}{x+y}$

$\displaystyle \frac{dx}{dx}=\frac{x+y}{y-x}$

Using WOLFRAM I got an answer of:

$\displaystyle y(x)=x(\frac{\pm \sqrt{2c_1^4x^4-c_1^2x^2}}{c_1^2x^2}+1)=x \pm \sqrt{2x^2-c}$

Now solving for c:

$\displaystyle c=2x^2-(y-x)^2$ So these are our base curves.

So:

$\displaystyle u(x,y)=f(2x^2-(y-x)^2)$

Now using: $\displaystyle u(x,0)=sin(x)$ We get that:

$\displaystyle u(x,y)=\sin{\pm \sqrt{2x^2-(y-x)^2}}$

I think this is right? But now how can I SHOW that there are no solutions around the origin.

Thanks!

 

[Ishuda]

Given that that is the solution, what is u_x and u_y at the origin?

 

[renegade05]

Given that that is the solution, what is u_x and u_y at the origin?

Something nasty.

$\displaystyle u_x=\frac{STUFF}{\sqrt{-x^2+2yx-y^2}}$

(similar for u_y)

So at (0,0) there is a singularity, is this the answer? Seems trivial almost.....

 

[Ishuda]

Something nasty.

\(\displaystyle u_x=\frac{STUFF}{\sqrt{-x^2+2yx-y^2}}\)

(similar for u_y)

So at (0,0) there is a singularity, is this the answer? Seems trivial almost.....

I would think so. Note also that there is no solution along the lines $\displaystyle y = (1 \pm \sqrt{2}) x$