y(x) = x^3/9 - 6/x, const = - 28/3

Sum of two known solutions is solution too.

Should the solution you look for be nonzero? If not, y(x)=0 (identically zero) satisfies your equation and is bounded.

Artur: Yes, it should be nonzero. I hadn't thought of that possibility, though, so thanks. I'll modify the problem statement to reflect that.

Another solution is y = - 1/9 x^3, with constant -2/3. Not bounded, though.

Jorg: Nice!

I just noticed a sign error in the equation as originally posted. The correct equation is [y''' - (y')^2 + xy]' = 0. I've changed this in the question statement. Jorg's solution becomes y = 1/9 x^3 with const. = 2/3.

y(x) = x^3/9 - 6/x, const = - 28/3

Sum of two known solutions is solution too.

Gennadiy: Not for a nonlinear equation.

R. Bradley: For a nonlinear equation sum of solutions is not a solution in general case. But in this particular case sum of solutions found by you and Jorg is solution.

Gennadiy: You're right! That's awesome. It is very surprising but true that the two solutions add to give a new one. I wonder if we could get a deeper understanding of how that happens. Sadly, though, the new solution is not bounded either.

Inspired by Gennadiy's discovery, I just posted a new question in which I ask "When do two nontrivial solutions to a nonlinear ODE add to give a new nontrivial solution?" Maybe we can get some insight into this strange phenomenon in a more general context.

In order to find out, whether

y''' + y'^2 + x y = c (1)

has solutions that are bounded on the whole real domain, it is first useful to do a local analysis at infinity, to see whether and which bounded behaviours are possible there.

I find three dominant balances for x>>1

a) x y ~ c, this gives y ~ c/x, i.e. bounded behaviour at infinity; for c=-6, the subdominant terms cancel each other, so we have an exact solution (which is however not bounded for all finite x)

b) y'^2 ~ x y, which gives y ~ x^3/9, i.e. unbounded behaviour at infinity; for c=2/3, the subdominant terms cancel, so the solution becomes exact

c) y''' ~ -x y, which leads to y ~ exp(-(3/4) (-x)^(4/3)); if we admit one of the complex third roots of -x, this leads to oscillatory unbounded behaviour, if we restrict ourselves to the real root, we have a dominant balance only for c=0, the solution is bounded for large x, but there is no reason to believe that the same solution that is bounded at x>>1 is also bounded at -x>>1

So only behaviour a) has some likelyhood of allowing for globally bounded solutions. Consider now equation (1) without the nonlinear term and formally take the Fourier transform

g(k) = \int_{-\infty}^{\infty} dx exp(-i k x) y(x) (2)

This produces a first-order differential equation for g(k):

g'(k) - k^3 g(k) = - 2 \pi i c delta(k) (3)

(delta(k) is the Dirac distribution). The general solution of this is

g(k) = - 2\pi i c exp(k^4/4) Theta(k) + d exp(k^4/4), (4)

where Theta(k) is the Heaviside function and d a constant of integration. Unfortunately, the inverse Fourier transform of g(k) is not convergent, so the solution is not helpful. My interpretation is that equation (1) *without* the nonlinear term does not have bounded solutions. Note that if the sign of the x y term was changed, we would have exp(-k^4/4) instead of exp(k^4/4) in (4), which would produce a nice globally bounded solution of the linearized equation with the asymptotic behaviour y ~ -c/x.

So if equation (1) is to have bounded solutions, the nonlinearity must care for the boundedness. This seems unlikely, because it can compensate only divergences of one sign (as it is always positive). We may also Fourier tranform the full equation, which leads to

g'(k) - k^3 g(k) + 1/(2\pi i) \int{-\infty}^{\infty} dk' (k-k') g(k-k') k' g(k') = - 2 \pi i c delta(k) , (5)

which may be recast as

d/dk exp(-k^4/4) g(k) = - 2 \pi i c delta(k) - 1/(2\pi i) exp(-k^4/4) \int{-\infty}^{\infty} dk' (k-k') g(k-k') k' g(k')

or

g(k) = - 2\pi i c exp(k^4/4) Theta(k) + d exp(k^4/4)

- 1/(2\pi i) \int_{-\infty}^k dk'

\int{-\infty}^{\infty} dk'' exp((k^4-k'^4)/4) (k'-k'') g(k'-k'') k'' g(k''). (6)

This is an integral equation for g(k). But as long as there are additive terms \propto exp(k^4/4), this is not likely to have a valid inverse Fourier transform. If we drop the term with the integration constant d and iterate the equation once, we obtain

g(k) = - 2\pi i c exp(k^4/4) Theta(k)

- 2\pi i c^2 exp(k^4/4) \int_{-\infty}^k dk' exp(-k'^4/4) \int_0^k' dk'' (k'-k'') k'' exp((k'-k'')^4/4) exp(k''^4/4) + multiple integrals involving g (7)

If the integral term could be supposed to become smaller on iteration, this might converge to a solution, but things do not look like that. The last integral written out in (7) contains exp(k''^4/2) as a factor, i.e. the prefactor of the positive fourth power of the wave vector becomes larger on iteration. Therefore, it seems that g(k) cannot tend to something integrable (after multiplication with exp(i k x)). This suggests strongly that (1) does not have bounded solutions.

If the sign of the x y term and hence the powers of k^4, k'^4 etc. in the exponentials, were reversed, then iteration might work and convergence would presumably be good. So it seems likely that the equation with the x y term replaced by -x y does have bounded solutions.

At this point, my analysis stops, because I do not know how to proceed further with the integral equation (6).

Sorry, equation (1) should of course have a minus sign in front of the nonlinear term and the nonlinearity is therefore always negative. But I have not analysed the wrong equation, it is just a typo (otherwise the local behaviour b would be different). Also when I am saying there are "no bounded solutions", it is to be understood as "no nontrivial bounded solutions". The solution y=0 for c=0 is of course bounded, but it may be the only globally bounded solution.