I'd think there isn't such a solution. Since the y values would be bounded, y''' would be bounded and therefore (some interpolation) y' and y'' would be bounded. Therefore the trajectory in R^3 would be bounded and the omega-limit set would be bounded. If the limit is y=1, the omega limit set would be contained in y=1, so that y'=y''=0. But this is only one point, and it is not an equilibrium. So y=1 cannot be a limit.

Not sure I can help enormously, but found myself interested to consider associated problems. In the meantime, discovered how cool WolframAlpha is.

My train of thought was to consider problems of the same form as yours but with lower powers and derivatives, starting with y' + y = 0, then y'' + y^2 = 0. While the n=1 case is clearly trivial, there are already nasty things arising for n = 2. WolframAlpha told me that the solution involves a Weierstrass Elliptic function, which I'd not heard of before.

I also had a quick look at power series approaches, to see if anything popped out, and also functions involving exp(-x/3), which seemed like a sensible thing to try but wasn't fruitful.

I'm guessing you're aware of all of this stuff, if you're actually researching these types of problem, so I'm sure my answer is mostly useless.

Thanks for posting a nice teaser, something to take me away from my main research.

Neil, Thanks for the input. y = \sqrt{105/8} x^{-3/2} is a solution, but it doesn't satisfy the boundary conditions.

Boundary conditions are the principal point in this problem, isn't it? Without them the trivial solution y=0 will be feasible :)

Mikhail: You're right.

I'd think there isn't such a solution. Since the y values would be bounded, y''' would be bounded and therefore (some interpolation) y' and y'' would be bounded. Therefore the trajectory in R^3 would be bounded and the omega-limit set would be bounded. If the limit is y=1, the omega limit set would be contained in y=1, so that y'=y''=0. But this is only one point, and it is not an equilibrium. So y=1 cannot be a limit.

Wow! Thanks, Arnd.

Please see my newer post for a refined and corrected version of this problem.

The good old method is to search the solution as infinite series see the attachment

Andrey: I don't see the attachment.

My e-mail is [email protected]

Try attach the preposition

If you posit that there is a solution of the form y = r x ^(-a) for constants r and a (after trying other simple forms for y and noticing that the solution space is not preserved under scalar multiplication because of the y^3 term), you can show rather directly that a = 3/2 and r = sqrt(6) / 4 gives a solution of the equation, but not the boundary conditions

now i think existence and uniqueness theorems can be used to show that there is no solution for the provided boundary conditions

@ Christopher: I pointed out the existence of a solution of this form in a post earlier. We differ on the prefactor though: I get r = sqrt(105/8). Arnd Scheel already pointed out that a solution satisfying the boundary conditions doesn't exist. I posted a corrected question a few days ago that I believe does have a solution.

i just tried to read andrey's attachment - please everybody use PDF, since i (and i presume many others) do not have any direct access to MS word - but there was part of the file missing (it looked like an OLE connection without both files present)

There exists two parametric family of an analytic solutions to this equations being represented in form: y(x)=[(ax+b)sqrt(cx+d)]^(-1). (for example: a=2/105,c=210,b=d) But this solution tends to 0 as x tends to plus or minus infinity and isn't defined for all x

Hi Professor,

I do not think that such method exists, since the family of exact solutions can be written as : c_1 exp(-x) +c_2 exp(x/2)*sin(sqrt(3)*x/2) +c_3 exp(x/2)*cos(sqrt(3)*x/2)

where c_1, c_2 and c_3 are constants, so it is impossible to get a bounded solution as x tends to plus/minus infinity

Cordially

No, this is a solution to y'''+y=0 but not to the equation y'''+y^3=0 !

Sorry r.bradley, i did not see all of the earlier answers; and yes, my algebra failed me, i do get the same answer as you once i fix it - what i wonder is how do we decide to try out that particular form?

Thanks everyone for the very helpful comments and suggestions!

Again, what about the Handbuch of Erich Kamke

I think such equations can be solved by first integral method....

The solution will be

y=(〖2x/∛105+1)〗^((-3)/2)

if we ignore given conditions

This equation has not a solution that satisfies given conditions, because these conditions, as asymptotic solutions, should also satisfy in the equation. Hence, Given condition are not correct. y=0 when x goes to plus or minus infinity.

Dr. Savari --- You're right. Several people observed this earlier. I went back to the physics and developed a more accurate ODE, which I posted in the question "How can I solve a^2 y = y^3 - y''' - y'?" There has been quite a bit of subsequent discussion on that equation. Thanks for your comment.

I would say it it worth trying maple software . It gives analytical solution almost for any equation it has one.

Dear Dr Bradley

I am not sure that a is constant or not. If it is not constant then this equation can have several solutions. For example, Y=a=SinX or Y=a=CosX,...

I will be pleased if you tell me where did you get such equations. Since we are preparing a nonlinear ODEs book these equation can be helpful. Regards

Dear Dr Bradley

Sorry Dr I forgot to say if a is constant the solution will be Y=a.

Analytical solution for equations of the type y'''=Ay^a exist only for

a=-7/2 , -5/3 , -4/3 , -7/6 , -2 , -1/2 , -5/4

Source:

Handbook of exact solution for ordinary differential equations

A. D. Polyanin, V. F. Zaitsev

CRC Press, 1995, chapter 3.2.2

Analytical solution for equations of the type y'''=Ay^a exist for

a=-7/2 , -5/3 , -4/3 , -7/6 , -2 , -1/2 , -5/4, that's right, but not "exists only for". Try a=0,1,...etc

Is correct. I assumed trivial cases were out.

Marta, the independent variable is not y !

@Marta: Your approach would be correct if we had to solve y''' = -x^3 rather than y''' = -y^3.

Ok

I think that an analytical solutions for y^3 + y''' = 0 is very easy:

y= sqrt( 105 / 8 ) * x^( -3 / 2 )

My Answer is in the file attachment. Please, answer my. Your sincerely, Anna Tomova.

@Mehdi Delkhosh: your solution is correct (and elegant), but I regret to say that it doesn't satisfy the requested boundary conditions, namely "y tends to plus or minus one as x tends to plus or minus infinity". Instead in either case they tend to zero.

Moreover, it is requested that "[t]he solution should also be bounded *everywhere*", and your solution tends to infinity as x tends to zero from the right, and it tends to plus imaginary infinity, as x tends to zero from the left. Having said that, I don't know if Mark Bradley's request may tolerate one unbounded point, provided that for all other values the solution is bounded.

Dear Arturo Ortiz Tapia,

I have read Your answer to @Mehdi Delkhosh. In My Answer 5.doc (see above) I have summarized some individual solutions of ODE of the type y^(n)+Ay^m=0, n in N, m in Q, A - const. Of course, these solutions don't satisfy the Mark Bradley's request. In this regard I have added my question:

"What would be the formula about the integral general for ODE y'''+y^3=0 or y^(n)+Ay^m=0, n in N, m in Q, A - const?.", because between the individual solutions of the integral general we could find the solution of the Mark Bradley's request (if it exists) or between any special integrals. I add the solution of Wolframalpha.com too.Please, answer me. Your sincerely, Anna Tomova.

Hello everybody---

Thanks for all of your help with this ODE. Your input showed that the correct ODE to look at was somewhat different than the one I asked about here. The paper that came out of this line of research is attached. Hopefully you'll find it interesting.

Best regards, 

Mark Bradley

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