# Quasi-Linear PDE xuu_x+y^2u_y=u^2

Printable View

• May 6th 2011, 11:43 AM
bugatti79
Quasi-Linear PDE xuu_x+y^2u_y=u^2
Hi folks,

What does one do when the particular solution is defined implicitly and it has to be differentiated an put back into original DE to check it?

See attached pdf...thanks!
• May 6th 2011, 12:38 PM
Jester
You're missing a 2 in your answer. As for showing it satisfies the PDE you can either (1) try and solve for $u$ explicitly, then sub into the PDE or (2) differentiate your solution implicitly and solve for $u_x$ and $u_y$ and sub into your PDE.
• May 6th 2011, 01:49 PM
bugatti79
Quote:

Originally Posted by Danny
You're missing a 2 in your answer. As for showing it satisfies the PDE you can either (1) try and solve for $u$ explicitly, then sub into the PDE or (2) differentiate your solution implicitly and solve for $u_x$ and $u_y$ and sub into your PDE.

Ok but I first need to find where im missing the 2....I have gone over it numerous times! :-)

Ok I spotted the 2
• May 7th 2011, 07:52 AM
bugatti79
Quote:

Originally Posted by Danny
You're missing a 2 in your answer. As for showing it satisfies the PDE you can either (1) try and solve for $u$ explicitly, then sub into the PDE or (2) differentiate your solution implicitly and solve for $u_x$ and $u_y$ and sub into your PDE.

I managed to get u(x,y) explicitly and got u_x and u_y.

In terms of your method (1), you say solve for u explicitly and then sub into PDE, but we have to get u_x and u_y beore we sub in right?

For your (2), this looks interesting but wouldnt it be difficult to get u_x and u_y if its implicitly defined? otherwise (1) is the exact same (2)?

Thanks
• May 7th 2011, 08:55 AM
Jester
Difficult - not really. From your solution

$\dfrac{u}{x} = G\left(\dfrac{1}{u} - \dfrac{1}{y}\right)$

then

$\dfrac{u_x}{x} - \dfrac{u}{x^2} = G'\left(\dfrac{1}{u} - \dfrac{1}{y}\right)\left(- \dfrac{u_x}{u^2}\right)$

and solving for $u_x$ gives

$u_x = \dfrac{u^3}{x(u^2 + xG')}$

Similarly

$u_y = \dfrac{xu^2G'}{y^2(u^2+xG')}$.

Substituting into your PDE and simplifying shows it's identically satisfied!
• May 7th 2011, 11:37 AM
bugatti79
Quote:

Originally Posted by Danny
Difficult - not really. From your solution

$\dfrac{u}{x} = G\left(\dfrac{1}{u} - \dfrac{1}{y}\right)$

then

$\dfrac{u_x}{x} - \dfrac{u}{x^2} = G'\left(\dfrac{1}{u} - \dfrac{1}{y}\right)\left(- \dfrac{u_x}{u^2}\right)$

and solving for $u_x$ gives

$u_x = \dfrac{u^3}{x(u^2 + xG')}$

Similarly

$u_y = \dfrac{xu^2G'}{y^2(u^2+xG')}$.

Substituting into your PDE and simplifying shows it's identically satisfied!

I see...thats cool. Never saw that before :-)

Cheers for that!