1. ## PDE-Characteristic Curves

Let $\displaystyle f(x,y)$ be the soloution of $\displaystyle xu_x +yu_y = u^4$ that is defined in the whole plane. Prove that $\displaystyle f = 0$ .
Hint: Think of the characteristic curves of this PDE.

HOPE You'll be able to help me

2. Originally Posted by WannaBe
Let $\displaystyle f(x,y)$ be the soloution of $\displaystyle xu_x +yu_y = u^4$ that is defined in the whole plane. Prove that $\displaystyle f = 0$ .
Hint: Think of the characteristic curves of this PDE.

HOPE You'll be able to help me

Solve the characteristic system $\displaystyle dx/x=dy/y=du/u^4$ and assure that any nonzero solution on the whole plane will develop a discontinuity at the origin.

3. I've already done this...We'll get:
$\displaystyle y=c_1 \cdot x$ and: $\displaystyle u^3 = -\frac{1}{3ln(x) +3k(y)}$ where is k is a function of y ...
We'll have a discontinuity every time that $\displaystyle ln(x) = -k(y)$ and every time that x=0...
How did you deduced the concloution that the discontinuities are at the origin?

Thanks !

4. From y=c_1x. Thus there will always be a discontinuity at the origin, but the expression lnx+k(y) may be nonzero for a suitable choice of initial data.

5. $\displaystyle y=c_1 \cdot x$ is a line segmant in $\displaystyle R^2$ ...Why there is a discontinuity at the origin?

6. It is one of the characteristics though. The constant c_1 is the slope and will vary along each characteristic, while being (explicitly) determined by c_1=y/x. So at x=0 (for which y=0) the slope will become infinite and the solution will develop a shock. check your notes.

7. actually, at x=0, the expression ln(x) has a discontinuity... So we can deduce the result from this (I think...)...

Thanks...I think I understand it