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Nonlinear first order

  1. Feb 14, 2006 #1
    Hi,
    As far as I know this is a first order, nonlinear diff eqn with both dependant and independant variables...so it is not solvable??

    y'+ay^2 = bx

    If anybody knows if there is a method to solving it, please let me know.

    Thanks,
    danmag
    :confused:
     
  2. jcsd
  3. Feb 14, 2006 #2

    Haelfix

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    I think this is a Ricatti diff eqn. Its like the Bernouilli equation, but with nonhomogenous parts (its a bit of a pain if I recall right, you need to do some guessing).

    But anyway make the substitution v = 1/y and go from there.
     
  4. Feb 15, 2006 #3

    saltydog

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    As Haelfix mentioned, it's a Riccati equation of the form:

    [tex]y^{'}+Q(x)y+R(x)y^2=P(x)[/tex]

    So use the standard approach:

    [tex]y=\frac{u^{'}}{Ru}[/tex]

    to convert it to a second-order ODE with variable coefficients. This then can be solved via power series or just bust it up directly with Airy functions (equal . . . nevermind).:rolleyes:
     
  5. Feb 19, 2006 #4

    saltydog

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    Know what, this turned out to be a quagmire. But very interesting nevertheless. For example:

    1. What is the relationship between the power-series solution of the reduced Riccati equation:

    [tex]u^{''}-abxu=0[/tex]

    and the solution expressed in terms of Airy functions?

    (not easy)

    2. Using the solutions above, how then does one proceed to solve an IVP for the original Riccati equation:

    [tex]y^{'}+ay^{2}=bx,\quad y(0)=a[/tex]

    (just barely)

    3. What is the relationship between the Airy function expressed in terms of a hypergeometric series and the integral expression for the Airy function:

    [tex]Ai(x)=\frac{1}{3^{2/3}\Gamma(2/3)} F_1(2/3,z^3/9)-\frac{z}{3^{1/3}\Gamma(1/3)}F_1(4/3,z^3/9)[/tex]

    [tex]Ai(x)=\frac{1}{\pi}\int_0^{\infty}Cos[t^3/3+xt]dt[/tex]

    (not even close)

    See . . . quagmire.
     
    Last edited: Feb 20, 2006
  6. Feb 20, 2006 #5

    saltydog

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    More complex contours:

    So I reduce the equation down to:

    [tex]u^{''}-axy=0[/tex]

    Now, consider the Fourier Transform of u(x):

    [tex]g(\xi)=\mathcal{F}\left\{u\right\}=
    \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x)e^{-i\xi x}dx[/tex]

    So I'll take the Fourier transform of the DE and noting:

    [tex]\mathcal{F}\left\{xu(x)\right\}=i\frac{dg}{d\xi}[/tex]

    I obtain:

    [tex]-\xi^2 g-ia\frac{dg}{d\xi}=0[/tex]

    That's a piece of cake to solve and yields:

    [tex]g(\xi)=Ce^{i(\xi^3/(3ai)}[/tex]

    Now, inverting the transform:

    [tex]u(x)=\frac{C}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{i(\xi^3/(3a)+\xi x)}d\xi[/tex]

    That's close. :smile: Looks like I can just split up the integral via Euler's formula to get a Cos and iSin. That might be all there is to it but need to check . . .
     
    Last edited: Feb 20, 2006
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