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Insanely hard differential equation

  1. Aug 6, 2013 #1
    I'm trying to find the equilibrium size of a planetesimal which is made of a material whose density is pressure dependent. (quite a mouthful)

    I have to solve this differential equation:

    [tex]y(x^2y''+2xy')+2y^2+bxy'=ax^2y^3+(xy')^2+by[/tex]

    where a and b are constants.
     
  2. jcsd
  3. Aug 6, 2013 #2
    I have no idea where to start. :(
     
  4. Aug 6, 2013 #3

    mfb

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    Hmm, the y3 is an issue.

    Here is what I would do (after testing some functions like polynomials, or polynomial*exponential and so on): get everything on one side, see what y=1, y=x, y=1/x, y=x2 and so on produce as result and see if you can combine them to get a zero somehow.
     
  5. Aug 6, 2013 #4

    MathematicalPhysicist

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    $$(1) y(x^2 y''+2xy)-(xy')^2 = y(x^2 y')' - (x^2 y')y' = (x^2 y'/y)' y^2 $$

    $$(2) bxy' -by = b (y/x)' x^2$$

    $$(3) 2y^2 -ax^2 y^3 = y^2 (2- ax^2 y) $$

    Your equation is: $$ (1)+(2)+(3) = 0$$

    So you get:

    $$ (x^2 y'/y)' /x^2 + b(y/x)' /y^2 + (2/x^2 - ay) = 0$$

    I wish I could translate all this equation into a derivative of some expression, but I don't see exactly how to do it here. Anyone?
     
  6. Aug 6, 2013 #5

    phyzguy

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    It should be straightforward to integrate numerically. Is a numerical solution acceptable, or do you want an analytic expression?
     
  7. Aug 6, 2013 #6
    An analytic expression would be best.
     
  8. Aug 6, 2013 #7
    Absolutely an analytic expression. I mean just any old body can numerically integrate it. This is what I came up with early this morning:

    Let [itex]u=x^2y'[/itex] so that:

    [tex]ydu+2y^2+bxy'=ax^2 y^3+y'u+by[/tex]

    then I went to bed. Sadly, that's the best I got.

    . . . hummmm, wonder if anyone else is lookin' at this problem? :)

    Edit: Oh yeah, I say we let a=1 and b=1 or whatever neat numbers that might make it possible to solve for a particular case for starters.
     
    Last edited: Aug 6, 2013
  9. Aug 11, 2013 #8
    Well you could use Laplace transformation. It's good for this type of equations.
     
  10. Aug 14, 2013 #9

    HallsofIvy

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    NO, it's not. The Laplace transform only works for linear equations (and a few non-linear equations that can be easily linearized).
     
  11. Aug 14, 2013 #10
    I was intrigued by this equation when you first posted it. This equation is very nonlinear, and I tried to do an approximate solution using an expansion, but even that is difficult because of a couple of discrete convolutions. It may be best (If you truly cannot obtain an analytic solution) to substitute a polynomial of some specified degree (This depends on how accurate you want to be) and match terms.
     
  12. Aug 14, 2013 #11
    I think an "analytical" solution, if all else fails, can be computed via power series. I realize we'll have to cube a power series and will have other messy terms, but I am optimistic we can obtain one, however I suspect it will have a small radius of convergence. Still, approaching it via power series would be quite an interesting approach.
     
    Last edited: Aug 14, 2013
  13. Aug 15, 2013 #12

    Stephen Tashi

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    Since this equation is from physics, how do the physical units work out? Does the equation make sense as a physical law in that respect?
     
  14. Aug 15, 2013 #13
    May I ask if there is any way to check if an algebraic function is a solution to this equation? The reason I ask is that I've noticed sometimes they are solutions to highly-nonlinear equations. Take for example a problem solved in here earlier:

    [tex]1+p^2-yp'p-2y(1+p^2)^{3/2}=0[/tex]

    Well, it turns out that the algebraic function, [itex]p(y)[/itex] defined implicitly by:

    [tex]f(y,p)=(y^2-2c_1 y^2-y^2+c_1^2)+(y^4-2y^2+1)p^2=0[/tex]

    is a solution. So, suppose we start with the equation:

    [tex]1+p^2-yp'p-2y(1+p^2)^{3/2}=0[/tex]

    is there a way to determine if the function:

    [tex]f(y,p)=a_0(y)+a_2(y) p^2=0[/tex]
    where [itex]a_i(y)[/itex] are polynomials in y, is a solution to this equation? And if so, is there a way likewise to determine if

    [tex]f(x,y)=a_0(x)+a_1(x)y+\cdots+a_n(x)y^n=0[/tex]

    is a solution to:

    [tex]y(x^2 y ′′ +2xy ′ )+2y^2 +bxy ′ =ax^2 y^3 +(xy ′ )^ 2 +by [/tex]

    Suppose that is the assignment:

    Given [itex]g(x,y,y',y'')=0[/itex], determine if there are solutions [itex]y(x)[/itex] that can be written in the form:
    [tex]f(x,y)=a_0(x)+a_1(x)y+\cdots+a_n(x)y^n=0[/tex]
    I think that's an interesting problem don't you guys think?
     
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