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How to solve this nonlinear ODE?

  1. Dec 8, 2008 #1
    Hello,

    How to solve the equation as follows:
    F''+F'*F'-k*F=0, where k is a constant and k>0

    Is there any analytical solution?
     
  2. jcsd
  3. Dec 8, 2008 #2

    HallsofIvy

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    Certainly, there exist an analytic function that satisfies the equation. If by "analytical solution" you mean, rather, that there is a calculus method for getting an exact solution, I would suspect "no" simply because "almost all" non-linear differential equations cannot be solved that way.
     
  4. Dec 8, 2008 #3
    Your equation is equivalent to a first-order system, if we let [tex]F = u[/tex] and [tex]F' = v[/tex]:
    [tex] \begin{cases}\dot{u} = v \\ \dot{v} = ku-v^2\end{cases}[/tex]

    From there you can quite easily find a first integral and then separate variables in the second equation:

    [tex] \begin{cases}\dot{u} = v \\ \dot{v}^2 = v^4 + kv^2-kE\end{cases}[/tex]

    Where E is a constant determined by the initial conditions.

    edit: the function I found is not a first integral :/
     
    Last edited: Dec 9, 2008
  5. Dec 8, 2008 #4
    Thanks for your reply.
    While I cannot see how you obtain the fourth equation: [tex]\dot{v} ^2 = v^4 + kv^2-kE\[/tex]
    Would you please support more details?
     
  6. Dec 9, 2008 #5
    I assumed you were familiar with the definition of first integral. A first integral for the system [tex]y' = g(y), g : D \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^n[/tex] is a [tex]C^1[/tex] scalar function [tex]E : D \rightarrow \mathbb{R}[/tex] constant on every solution of the system. In other words, if [tex]\phi : I \rightarrow D[/tex] is a solution of the system, [tex]E(\phi(t)) = \mathrm{const.} \quad \forall t \in I[/tex].

    It follows from the definition that the gradient of E is everywhere normal to the field g:
    [tex]\nabla E(y) \cdot g(y) = 0 \quad \forall y \in D[/tex].

    In your equation, [tex]g = (v,ku-v^2)[/tex]. A field normal to that is [tex]f = (-ku+v^2,v)[/tex]. The problem is, the equation I wrote yesterday is wrong because it is not so obvious to find a primitive of a conservative field parallel to this field f. I was too much in a hurry to check, sorry ;)
     
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