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

  1. #1
    Senior Member bugatti79's Avatar
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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!
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  2. #2
    MHF Contributor
    Jester's Avatar
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    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.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    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
    Last edited by bugatti79; May 6th 2011 at 01:06 PM. Reason: Correction
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    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
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  5. #5
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    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!
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    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!
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