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Thread: 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

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

    If you write it as

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

    (2) You have

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

    Then

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

    or

    $\displaystyle (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

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

    If you write it as

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

    (2) You have

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

    Then

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

    or

    $\displaystyle (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 $\displaystyle 2x$ gives

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

    and letting $\displaystyle X = x^2$ gives

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

    or

    $\displaystyle \dfrac{dX}{du} + \dfrac{X}{u} = - 3B^2u-3u^2$. 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 $\displaystyle 2x$ gives

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

    and letting $\displaystyle X = x^2$ gives

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

    or

    $\displaystyle \dfrac{dX}{du} + \dfrac{X}{u} = - 3B^2u-3u^2$. 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 $\displaystyle B$ with $\displaystyle B = \dfrac{y}{u}$ so your solution is


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

    or

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

    Are you sure about the IC b/c with $\displaystyle u(x,2x) = 0$, the only solution is $\displaystyle u \equiv 0$.
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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 $\displaystyle B$ with $\displaystyle B = \dfrac{y}{u}$ so your solution is


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

    or

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

    Are you sure about the IC b/c with $\displaystyle u(x,2x) = 0$, the only solution is $\displaystyle u \equiv 0$.
    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
    $\displaystyle \dfrac{y}{u} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)$

    or

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

    ie, you had u/y

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

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

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

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

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

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

    $\displaystyle y u_s - u y_s = 0$ which integrates gives $\displaystyle \dfrac{u}{y} = a(r)$. Imposing the IC gives $\displaystyle a(r) = 0$ giving $\displaystyle u = 0$.
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  12. #12
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    $\displaystyle \dfrac{u}{y} = a(r)$. Imposing the IC gives $\displaystyle a(r) = 0$ giving $\displaystyle u = 0$.
    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 $\displaystyle \dfrac{u}{y} = a(r)$. Imposing the IC $\displaystyle x = r, y = 2r, u = 0$ gives $\displaystyle \dfrac{0}{2r} = a(r)$ gives $\displaystyle a(r) = 0$ giving $\displaystyle u = 0$.
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  14. #14
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    Quote Originally Posted by Danny View Post
    If $\displaystyle \dfrac{u}{y} = a(r)$. Imposing the IC $\displaystyle x = r, y = 2r, u = 0$ gives $\displaystyle \dfrac{0}{2r} = a(r)$ gives $\displaystyle a(r) = 0$ giving $\displaystyle u = 0$.
    Ok, that makes sense.

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

    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 $\displaystyle u(x,2x) = 0$, then it would matter.
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