# PDE-Characteristic Curves

• Nov 2nd 2010, 02:29 AM
WannaBe
PDE-Characteristic Curves
Let $f(x,y)$ be the soloution of $xu_x +yu_y = u^4$ that is defined in the whole plane. Prove that $f = 0$ .
Hint: Think of the characteristic curves of this PDE.

HOPE You'll be able to help me
• Nov 4th 2010, 03:20 AM
Rebesques
Quote:

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

HOPE You'll be able to help me

Solve the characteristic system $dx/x=dy/y=du/u^4$ and assure that any nonzero solution on the whole plane will develop a discontinuity at the origin.
• Nov 4th 2010, 07:17 AM
WannaBe
$y=c_1 \cdot x$ and: $u^3 = -\frac{1}{3ln(x) +3k(y)}$ where is k is a function of y ...
We'll have a discontinuity every time that $ln(x) = -k(y)$ and every time that x=0...
How did you deduced the concloution that the discontinuities are at the origin?

Thanks !
• Nov 4th 2010, 09:27 AM
Rebesques
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.
• Nov 4th 2010, 10:42 AM
WannaBe
$y=c_1 \cdot x$ is a line segmant in $R^2$ ...Why there is a discontinuity at the origin?
• Nov 4th 2010, 12:07 PM
Rebesques
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.
• Nov 4th 2010, 12:57 PM
WannaBe
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