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 F = u and F' = v:
\begin{cases}\dot{u} = v \\ \dot{v} = ku-v^2\end{cases}

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

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

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 F = u and F' = v:
\begin{cases}\dot{u} = v \\ \dot{v} = ku-v^2\end{cases}

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

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

Where E is a constant determined by the initial conditions.

Thanks for your reply.
While I cannot see how you obtain the fourth equation: \dot{v} ^2 = v^4 + kv^2-kE\
Would you please support more details?

 

[Matthaeus]

I assumed you were familiar with the definition of first integral. A first integral for the system y' = g(y), g : D \subseteq \mathbb{R}^n \rightarrow \mathbb{R}^n is a C^1 scalar function E : D \rightarrow \mathbb{R} constant on every solution of the system. In other words, if \phi : I \rightarrow D is a solution of the system, E(\phi(t)) = \mathrm{const.} \quad \forall t \in I.

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

In your equation, g = (v,ku-v^2). A field normal to that is f = (-ku+v^2,v). 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 ;)