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2nd Order Elliptic Equation

  1. Jul 12, 2015 #1
    I've been searching for exact solution of d2y/dx2 = C/y, where C is some constant, such equation take place when deriving equation of motion in gravitationnal field, I'm more interested in how to solve it, yet I only managed to express it as power series using taylor's theorem at x = 0, just pick y(0) = c1, y'(0) = c2, so y''= C/c1, y'''(x) = C*c2/c12 and so on, until y ≈ c1 + c2*x + C/2c1 * x2+ C*c2/6c12 *x3 + ......., Is there's any special function that solves the equation or any other approximisation ?, I want to hear some thoughts !,
     
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  3. Jul 12, 2015 #2

    fzero

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    If we let ##v = dy/dx##, then

    $$\frac{C}{y} = \frac{d^2y}{dx^2} = \frac{dv}{dx} = \frac{dy}{dx} \frac{dv}{dy} = v \frac{dv}{dy}.$$

    We can integrate this for ##v(y)## (it's the square root of log). Solving ##v = dy/dx## for ##y## in closed form is probably not possible.
     
  4. Jul 12, 2015 #3

    phyzguy

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    There is a trick for solving second order equations of this type, where the independent variable x is missing. Using the notation y' = dy/dx, write d2y/dx2 as follows:
    [tex] \frac{d^2 y}{dx^2} = \frac{dy'}{dx} = \frac{dy'}{dy} \frac{dy}{dx} = y' \frac{dy'}{dy}[/tex]

    Now you have eliminated x from the equation and turned it into a first order equation in y and y'. For your equation:

    [tex] y' \frac{dy'}{dy} = \frac{C}{y} [/tex]

    or:

    [tex] y' dy'= \frac{C}{y} dy[/tex]

    Now you can integrate both sides, and solve for y' in terms of y. Then you replace y' with dy/dx and integrate a second time. So you should be able to solve your equation analytically without needing power series.
     
  5. Jul 13, 2015 #4
    Thanks for quick replies, as far as I tried this seems "Damn" hard to solve but can get other approximisation from it , thanks :p
     
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