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

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May 6th 2011, 10:43 AM
bugatti79

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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, 11:38 AM
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.
May 6th 2011, 12:49 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, 06:52 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, 07:55 AM
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!
May 7th 2011, 10:37 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!