Can Nonlinear ODEs Like F''+F'*F'-k*F=0 Be Solved Analytically?

[bobls86]

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?

 

[HallsofIvy]

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.

 

[Matthaeus]

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 :/

 

[bobls86]

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.

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?

 

[Matthaeus]

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 ;)