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Solution of a first order ODE.

  1. Feb 17, 2006 #1
    We have the first order ODE

    [tex] y'=4t \sqrt y,~y(0)=1, [/tex]

    for which i have found the exact solution, namely a fourth order polynomial.

    I want a numerical method to solve the problem exactly. This method has to be a fourth order method, since this implies that the local error vanishes.

    Now we change the problem so it becomes

    [tex] y'=4t \sqrt y - \lambda(y-(1+t^2)^2),~y(0)=a, [/tex]

    and the question is: for which values of [itex]\lambda[/itex] and [itex]a[/itex] does a method that has the above mentioned property solve the new problem exactly.

    Of course, the obvious case is [itex]\lambda=0[/itex] and [itex]a=1[/itex], because in this case the new problem reduces to the first problem.

    My idea is that the solution must be a fourth order polynomial, since a fourth order numerical method has to solve the new problem exactly.

    Although I want your view on this and a strategy to find the values of [itex]\lambda[/itex] and [itex]a[/itex] for which the new problem is solved exactly by a fourth order numerical method.
     
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  3. Feb 17, 2006 #2

    Tide

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    By definition, a numerical solution is not exact.
     
  4. Feb 17, 2006 #3

    arildno

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    What are you talking about, tide?
    A numerical method is at times perfectly capable of yielding the exact solution's function values at the grid points.

    That is what is ordinarily meant with "solving exactly" in CFD, for example.
     
    Last edited: Feb 17, 2006
  5. Feb 17, 2006 #4

    Tide

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    Any numerical solution is limited (at best) to machine precision - ergo it is not exact. :)

    You surely are not suggesting that an exact analytic solution is mathematically equivalent to a numerical approximation? As a practical matter and when making engineering use of an exact analytic solution one is similarly limited by machine precision. Nevertheless, the (actual) real solution has many advantages over an approximate solution.
     
  6. Feb 18, 2006 #5

    arildno

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    I'm not talking about finite-precision arithmetic (and the errors induced by that), but the discretization techniques used on the original diff. eq It is common to call that a numerical METHOD (as OP said). If the exact solution of the difference equation equals the exact solution of the differential equation (restricted to the grid points), then the discretization scheme (i.e, numerical method) is said to reproduce the diff eq.solution.

    The term "numerical SOLUTION" is often restricted to the solution as given by a machine, but neither OP or me used that term. Only you did that.
     
    Last edited: Feb 18, 2006
  7. Feb 18, 2006 #6
    By "I want a numerical method to solve the problem exactly", I mean a method for which the local error, in theory, vanishes. However, that is not my problem, because I know that with a fourth order numerical method this is achieved.

    On the contrary, I want some help with the the second problem I mentioned in the original post.
     
  8. Feb 18, 2006 #7

    arildno

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    Note that insofar as y is a polynomial, then the term [tex]t\sqrt{y}[/tex] must be a polynomial as well.
    Try first to see what conditions on [itex],\lambda[/itex] you'd get (if any) on a fourth-order solution on the form [tex]y=(t^{2}+kt+\sqrt{a})^{2}[/tex]
    where k is some constant.

    Due to the square root term, we have of course a>=0
    We get, by inserting it in the diff. eq.:
    [tex]4t^{3}+6kt^{2}+2(k^{2}+2\sqrt{a})t+2k\sqrt{a}=4t^{3}+4kt^{2}+4\sqrt{a}t-\lambda(2kt^{3}+(k^{2}+2\sqrt{a}-2)t^{2}+2k\sqrt{a}t+(a-1))[/tex]
    Setting each coefficient of different powers zero, you'll get the conditions on your constants.

    You'll get that k is necessarily 0, for example..
     
    Last edited: Feb 18, 2006
  9. Feb 12, 2007 #8
    What is the exact fourth order polynomial solution given y(0) = 1 ?

    I can easily go from y' = 4t*sqrt(y) to
    y = t^4

    But I am not able to perform the integration when using the initial value y(0) = 1.
     
    Last edited: Feb 12, 2007
  10. Feb 16, 2007 #9
    Found a away to get the solution y = (t^2 + C)^2
    where C equals the initial value for y(0) = C
     
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