I'd like to know if there is an analytical solution of the following Ordinary Differential Equation :

$\displaystyle{\dfrac{d}{dt}x_i(t) = D_i + \sum_{j=1}^{n}L_{ij}x_j(t) + \sum_{j=1}^{n}\sum_{k=1}^{n}C_{ijk}x_j(t)x_k(t)}$

where $\forall i,j,k=1,...,n$ we have $x_i(t) \in M_{T\times 1}(\mathbb{R})$ and $D_i,L_{ij},C_{ijk} \in \mathbb{R}$

The linear case :

$\displaystyle{\dfrac{d}{dt}x_i(t) = D_i + \sum_{j=1}^{n}L_{ij}x_j(t)}$

is straightforward, but i don't see how to solve this ODE with the quadratic terms.

  • 1
    $\begingroup$ Also posted to m.se, math.stackexchange.com/questions/694444/… $\endgroup$ – Gerry Myerson Feb 28 '14 at 22:34
  • $\begingroup$ While Alexandre's discussion and example below are important ones to have, one should bear in mind that the answer to your question really does depend on the constants $C$, $L$, and $D$. If the vector field $$X = \left(D^i + L^i_jx^j+C^i_{jk} x^j x^k\right)\frac{\partial}{\partial x^i}$$ happens to lie in a finite dimensional Lie algebra of a known transitive action of a Lie group on some completion of $\mathbb{R}^n$ (as it always does when $C=0$), then there will be an explicit elementary solution of the above ODE. $\endgroup$ – Robert Bryant May 4 '15 at 8:35

Everything depends on the meaning of the word "analytical".

In the common usage there is a word "analytic", which means "can be expanded in a convergent power series". Then there is a general theorem: if you have a differential equation $y'=F(y)$, where $y$ is a function with values in $R^n$, and $F$ is an analytic function (in your case, $F$ is a polynomial of degree $2$), then for every $t_0$ and every $y_0$, there exists a solution with the property $y(t_0)=y_0$, which is analytic in some neighborhood of $t_0$. The neighborhood depends on $F$ and on $y_0$.

However I suspect that your word "analytical" means some kind of "explicit formula". Of course, this has to be defined exactly, but if you are looking for solutions in terms of functions that contain algebraic functions, exponentials, logarithms, any sums, products, ratios or superpositions of the above and integrals of the above, then the answer is no.

Example: $$y_1'=y_1^2+y_2,\quad y_2^\prime=1.$$ This is equivalent to the single equation $y'=y^2+t+C,$ and J. Liouville proved in XIX century that it has no solutions of the type described above.

  • $\begingroup$ There are, however, solutions of Alexandre's example in terms of Airy functions and their derivatives: $$y \left( t \right) ={\frac {{{\rm Ai}^{(1)}\left(-C-t\right)}{k }+{{\rm Bi}^{(1)}\left(-C-t\right)}}{{{\rm Ai}\left(-C-t\right)}{\it k}+{{\rm Bi}\left(-C-t\right)}}}$$ where $k$ is an arbitrary constant. $\endgroup$ – Robert Israel May 4 '15 at 6:59

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