renegade05
Full Member
- Joined
- Sep 10, 2010
- Messages
- 260
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!
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!