# Method of Characteristics: Prove the u and v is a solution

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Oct 25th 2010, 06:46 AM
Ackbeet
Correct. So your argument for $ku$ being a solution is complete, because all the steps are reversible.

Quote:

$c(x,y)u^2+c(x,y)v^2$
Actually, you'd get $c(x,y)(u^{2}+2uv+v^{2})$. So the proof for the $u+v$ case won't end up with zero, because you can't get zero out of this messier expression. Make sense?
• Oct 25th 2010, 07:25 AM
bugatti79
Quote:

Originally Posted by Ackbeet
Correct. So your argument for $ku$ being a solution is complete, because all the steps are reversible.

Actually, you'd get $c(x,y)(u^{2}+2uv+v^{2})$. So the proof for the $u+v$ case won't end up with zero, because you can't get zero out of this messier expression. Make sense?

O yes as it would be (u+v)^2. I guess both unknown functions u and v would have to be 0 for that term to go to 0. Thanks Ackbeet!
• Oct 25th 2010, 10:04 AM
Ackbeet
Right. You're welcome. Have a good one!
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12