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Math Help - Quasi Linear PDE's

  1. #1
    Senior Member bugatti79's Avatar
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    Quasi Linear PDE's

    Hi All,

    I have 2 problems that I am stuck on (see attached), I dont know how to go any further. Any tips or advice on what Im doing wrong?

    Thanks
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  2. #2
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    You're not doing anything wrong - just not going far enough.

    (1) You have

    \dfrac{du}{dx} = \dfrac{-2xu}{x^2 + 3B^2u^2+3u^3}

    If you write it as

    \dfrac{dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-2xu}. It's Bernoulli in x.

    (2) You have

    \dfrac{d(x+y)}{y-x} = \dfrac{du}{x-y} which integrates giving x+y+u = c_1.

    Then

    \dfrac{d(x-y)}{x+ y - 2u} = \dfrac{du}{x-y}

    or

    (x-y)d(x-y) = (c_1 - 3u)du .

    This also integrates.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    You're not doing anything wrong - just not going far enough.

    (1) You have

    \dfrac{du}{dx} = \dfrac{-2xu}{x^2 + 3B^2u^2+3u^3}

    If you write it as

    \dfrac{dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-2xu}. It's Bernoulli in x.

    (2) You have

    \dfrac{d(x+y)}{y-x} = \dfrac{du}{x-y} which integrates giving x+y+u = c_1.

    Then

    \dfrac{d(x-y)}{x+ y - 2u} = \dfrac{du}{x-y}

    or

    (x-y)d(x-y) = (c_1 - 3u)du .

    This also integrates.

    Hi Danny,

    1) I am not sure how to convert this expression into Bernoulli's. Have I got the correct form as attached?

    2) I derived an implicit solution given the initial conditions. Not sure if it is right?.....

    Thanks
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  4. #4
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    In part (1), multiply by 2x and let X = x^2 (it's linear now)

    In (2) - looks good to me :-)
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    In part (1), multiply by 2x and let X = x^2 (it's linear now)

    In (2) - looks good to me :-)
    Lovely, I will inform you on how I get on in a few days.

    Thanks
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  6. #6
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    Quote Originally Posted by bugatti79 View Post
    Lovely, I will inform you on how I get on in a few days.

    Thanks
    Getting stuck on this one. How does it become linear? I multiplied accros by 2x on both sides and let X=x^2.
    Its bernoulli in x but arent we solving for u? See attached my attempt. THanks
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  7. #7
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    Multiplying by 2x gives

    \dfrac{2x dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-u}

    and letting X = x^2 gives

    \dfrac{dX}{du} = \dfrac{X + 3B^2u^2+3u^3}{-u}

    or

    \dfrac{dX}{du}  + \dfrac{X}{u} = - 3B^2u-3u^2. This is now linear!
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Multiplying by 2x gives

    \dfrac{2x dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-u}

    and letting X = x^2 gives

    \dfrac{dX}{du} = \dfrac{X + 3B^2u^2+3u^3}{-u}

    or

    \dfrac{dX}{du}  + \dfrac{X}{u} = - 3B^2u-3u^2. This is now linear!

    I have brought this further based on Danny's tip. See attached. I am stuck now getting the particular solution for the given IC's
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  9. #9
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    First replace your B with B = \dfrac{y}{u} so your solution is


    \dfrac{u}{y} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)

    or

    x^2u+y^2u + \frac{3}{4}u^4\right = f\left(\dfrac{u}{y}\right).

    Are you sure about the IC b/c with u(x,2x) = 0, the only solution is u \equiv 0.
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  10. #10
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    First replace your B with B = \dfrac{y}{u} so your solution is


    \dfrac{u}{y} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)

    or

    x^2u+y^2u + \frac{3}{4}u^4\right = f\left(\dfrac{u}{y}\right).

    Are you sure about the IC b/c with u(x,2x) = 0, the only solution is u \equiv 0.
    You are right Danny...once again! The IC was for another similar question and this question is to find general solution only.

    However shouldnt the GS be
    \dfrac{y}{u} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)

    or

    x^2u+y^2u + \frac{3}{4}u^4\right = f\left(\dfrac{y}{u}\right)

    ie, you had u/y

    Assuming I am correct, how do you know that the only soltion is u \equiv 0
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  11. #11
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    As for your first question, it doesn't matter. Define

    f(\lambda) = \bar{f}\left(\dfrac{1}{\lambda}\right) then drop the bar.

    For your second question, let go back when we introduced new coordiantes such that

    u_s = u_x x_s + u_y y_s gave your PDE by choosing

    x_s=x^2 + 3y^2 + 3u^3
    y_s =-2xy
    u_s = -2xu

    with the IC as x = r, y = 2r, u = 0 when s = 0. From the second two we find

    y u_s - u y_s = 0 which integrates gives \dfrac{u}{y} = a(r). Imposing the IC gives a(r) = 0 giving u = 0.
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  12. #12
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    \dfrac{u}{y} = a(r). Imposing the IC gives a(r) = 0 giving u = 0.
    I dont understand this last bit. How doyou impose the IC?

    Thanks
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  13. #13
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    Quote Originally Posted by bugatti79 View Post
    I dont understand this last bit. How doyou impose the IC?

    Thanks
    If \dfrac{u}{y} = a(r). Imposing the IC x = r, y = 2r, u = 0 gives \dfrac{0}{2r} = a(r) gives a(r) = 0 giving u = 0.
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  14. #14
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    If \dfrac{u}{y} = a(r). Imposing the IC x = r, y = 2r, u = 0 gives \dfrac{0}{2r} = a(r) gives a(r) = 0 giving u = 0.
    Ok, that makes sense.

    However, refering to f(\lambda) = \bar{f}\left(\dfrac{1}{\lambda}\right)

    Ie you say it doesnt matter whether the LHS is y/u or u/y but if you have y/u and u=o then the LHS goes to infinity...
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  15. #15
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    For the general solution, it doesn't matter but for the IC u(x,2x) = 0, then it would matter.
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