Differential Equations without Analytical Solutions

In many talks, I have heard people say that the differential equation they are interested in has no analytical solution. Do they really mean that? That is:

Can you prove a differential equation has no analytical solution?

I suspect what they mean is that no one has been able to derive one, but I could be wrong. I also have a question related to the former case.

What are some simple examples of differential equations with no known analytical solution?

The differential equations courses at my university are method based (identify the DE and use the method provided) which is completely fine. However, I'd like to have some examples which look easy (or look similar to ones for which the given methods will work) in order to show students that not all differential equations are so easily solved.

Added later: Taking the comments into account, I suppose the type of differential equations I am looking for in the second question are ones which, at this point in time, can only be solved using numerical methods (which, as Emmad Kareem points out, would be good motivation for learning such methods).

The kind of thing I'm looking for: I was talking to my friend who does Fluid Mechanics and he suggested the Blasius equation $$f''' + \frac{1}{2}ff'' = 0.$$ Apart from $f(x) = ax + b$, there are no known (as far as he knows) analytical solutions.

• It very much depends on what you mean by closed-form/analytical solutions. For example, for suitably restrictive definitions of closed form, Bessel's differential equations do not admit closed form solutions, but most ODE users would consider Bessel functions well known and perhaps part of their "elementary" toolkit. – Willie Wong Oct 10 '12 at 7:12
• The cynical side of me wants to say: if you can't express the solution of a "simple looking" ODE in analytical terms (allowing the use of special functions), then you might as well define a new special function that refers to the solution of that ODE.... – Willie Wong Oct 10 '12 at 7:23
• @WillieWong: I did consider your first point and I suppose it depends on what you mean by a special function. I know that differential Galois theory comes into the picture here as there is some problem with trying to capture all of the special functions. As J.M.'s answer to the question you linked to indicates, there isn't a universally accepted definition for 'closed form', but I would consider Bessel's functions to be analytical solutions. I'm not sure whether I can change my second question so that I can avoid this issue. – Michael Albanese Oct 10 '12 at 7:33
• You'd imagine that every book on numerical methods (and its directly related subjects) should start with an answer to your question. However, this is not the case. – NoChance Oct 10 '12 at 8:02

There's something worse than having no analytical solution. Pour-El and Richards found an ordinary differential equation $\phi'(t)=F(t,\phi(t))$ with $F$ computable and no computable solution. A reference is Marian Boykan Pour-El and Ian Richards, A computable ordinary differential equation which possesses no computable solution, Ann. Math. Logic 17 (1979), no. 1-2, 61–90, MR0552416 (81k:03064).

• Thanks for this reference. However, I was hoping for an explicit DE as the students I had in mind would not be familiar with the concept of computability. – Michael Albanese Dec 17 '12 at 23:17

Take the initial value problem $$y'=\cases{x\bigl(1+2\log|x|\bigr)\quad &(x\ne0) \cr 0&(x=0)\cr}\ ,\qquad y(0)=0\ .$$ This example obviously fulfills the assumptions of the existence and uniqueness theorem, so there is exactly one solution. As is easily checked this solution is given by $$y(x)=\cases{x^2\>\log|x|\quad&(x\ne0)\cr 0&(x=0)\cr}\ .$$ This function is not analytic in any neighborhood of $x=0$.

Yes, it can be shown that differential equations do not have analytic solutions, using differential galois theory. An example is the second order linear ode $y''+ x y' = 0$ this should be a good example for your purposes as it is only slightly different from the second order linear odes with constant coefficients which are easily solved.

PS: I assume you understand the difference between the ode not having an analytical solution and it not having a solution at all.

• can you link to a proof to that equation having no analytic solution? – Lolman Aug 26 '15 at 14:21
• because it doesn't seem to be true. – Lolman Aug 26 '15 at 15:52
• y=0 seams to be a solution and last time I checked, this function was analytical..... – Börge Oct 21 '16 at 8:59
• Furthermore the solution seams to be the errorfunction (if I am not mistaken). The errorfunction is the antiderivative of the gaussian function, which is analytic, I think every solution of that equation should be analytic.... – Börge Oct 21 '16 at 9:05

Hint: I have posted an example here in MO for a differential Equation which has no analytical solution that is :$\displaystyle \ f'={e}^{{f}^{-1}}$ . with nice answer for Non -existence of analytic solution of it by Christian Remling

For an example, why not start with the first differential equation ever, Newton's DE for falling objects, derived in high school physics class, you remember, $$A = \frac{G M_{earth}}{R^{2}}.$$ It's the universities' little secret that most differential equations describing the real (non-linear) world are not solvable. So, they don't tell you in high school that the equation is not solvable. They don't tell you in your university physics class (I've checked the text used by my local U), or even the classical mechanics class (this text notes that a modified equation is solvable and can be 'inverted' to get a solution, but they don't mention that the 'inversion' is not closed, i.e. is a procedure for approximation).

In introductory physics, the equation of motion for a pendulum is given as $$ml^2\frac{d^2\theta}{dt^2}=-mgl\theta$$. This gives you an easy solution in sines and cosines. That uses the "small angle approximation". The true equation is $$ml^2\frac{d^2\theta}{dt^2}=-mgl\sin\theta$$, admitting no "analytical/closed form solutions", depending on how you define that.