Impossible Kinds Of Differential Equations

Question

Today my professor told me that there are some differential equations that cannot be solved. Is this true? If it is true, why can they not be solved? How complex would that kind of differential equation have to be?

Perhaps a more interesting version of this question would ask how we can prove that some some differential equations cannot be solved analytically. I have wondered that myself. — Bobson Dugnutt, Jun 26 '17 at 22:28
Yes. I should have phrased my question that way. I hear that it is impossible, but I'd like to see more information behind it. — AB408, Jun 26 '17 at 22:30
If you are interested in integrals that are not analytically solvable: https://math.stackexchange.com/questions/265780/how-to-determine-with-certainty-that-a-function-has-no-elementary-antiderivative/265884#265884 — miracle173, Jun 26 '17 at 22:56

score 2 · Answer 1 · answered Jun 26 '17 at 22:43

Ordinary Differential Equations generally admit solutions in the event that the defining functions are "reasonable", i.e. possessed of Lipschitz continuity. Differentiable functions are generally locally Lipschitz, so we know that equations of the form

$\dot {\vec x} = \vec f(\vec x, t) \tag{1}$

where $\vec x \in \Bbb R^n$, with differentiable $\vec f$, have unique solutions when the initial data

$\vec x(t_0) = \vec x_0 \tag{2}$

is specified. Most equations of practical interest, say in the sciences or engineering, fall into this category so there's really no problem in the applications.

Partial Differential Equations, on the other hand, yield us no such good fortune. There are even very simple, first order PDEs which admit no solutions whatsoever. See, for instance, Lewy's example. Lewy showed that even such a simple PDE as

$\dfrac{\partial u}{\partial \bar z} - iz \dfrac{\partial u}{\partial t} = \phi'(t) \tag{3}$

admits no local solutions near $0$ on $\Bbb R \times \Bbb C$ when $\phi$ is not analytic.

Impossible Kinds Of Differential Equations

1 Answers1