# Fourth Order Nonlinear ODE

I was looking at an ode $w^{(4)} + w^3 = 0$ with initial conditions $[w'''(0),w''(0),w'(0),w(0)]=[1,0,0,0]$. I can see via maple that there is a blowup around 3.7. I was wondering if there was a way to show there is a blowup without using a computer, or if someone could direct me to some material on how to prove blowups. I'm having trouble finding literature on it. Thanks so much for your consideration.

Generally speaking, if your ODE is amenable to finding a solution using series, then the radius of convergence (ROC) of the series tells you there's a singularity at a distance of the ROC from the initial time. A singularity could be a pole -- a "blow up" -- or it could be an essential singularity, such as increasingly wild oscillation on shorter and shorter time scales. If the ROC is infinite, there are no singularities, and we expect a solution for all time.

For a series solution, we guess at the form of the answer, substitute it into the ODE, and equate coefficients of like order.

For this ODE, we might guess $$w(t) = t^{-2} \sum_0^\infty a_n t^n$$ because the indicial-like equation gives $r=-2$. (I say "indicial-like" because this equation is not linear.) The integer root (non-fractional) means using a Laurent series. OTOH, the finite initial conditions mean two negative-order Laurent coeffiecients are zero, and we have a plain old Taylor series. Plug and play from there: substitute this form in the ODE and ICs, and equate like terms.

For the ICs you specified, you'll wind up with most of the terms being $0$, and a form like $$w(t) = t^3 \sum_0^\infty b_n t^{10n}$$. From there the algebra gets hairy because of the cube term. The Cauchy product means the recurrence relation among coefficients involves terms of all order. I'm not sure it's possible to find a closed-form expression for the coefficients in terms of $n$. But if you could find such an expression, you could use the ratio test or root test to find the ROC of the series. The first few terms are $$w(t) = \frac{1}{6}t^3 - \frac{1}{3706560}t^{13} + \frac{1}{9452617574400}t^{23} - \frac{1}{21722750401872199680}t^{33} + \frac{661}{37638496964414476201623552000}t^{43} - \frac{405469}{66922504728527550129991809682636800000}t^{53} + \ldots$$

I wasn't able to find such an expression myself, but I believe $b_n$ is asymptotically $Cn\lambda^n$. I did some experimenting on the first few hundred or so coefficients, and they fit that pattern very closely. A fit gives $\lambda \approx \exp(15.068)$. That would mean the radius of convergence of the series is $\exp(1.5068)\approx 4.512$.

I wouldn't be surprised if there's a trick to this problem, an elegant way of transforming it to show the nature of the growth. I also suspect the equation is integrable, but only could find an exact form for the third derivative.

BTW, just curious: where did this problem come from?

The solution is $$w(t)=1 + \frac{t^4}{24} + \frac{t^8}{13440} + \frac{73 t^{12}}{159667200} + \frac{491 t^{16}}{232475443200}+\cdots$$

Maybe you could estimate the decay of the coefficients of the series and then estimate its ratio of convergence.

• Careful, $w(0) = 0$ while $w'''(0) 1$, I think you input it wrongly on Maple.
– D G
Jan 9 '14 at 13:28

There is an explicit solution that blows up in finite time. It is given by $w=\frac{2 i \sqrt{30}}{(t-t_0)^2}$, where $t_0$ is a free constant.

• In the present form of your solution, $w(0)\neq 0$ and so does not satisfy the initial conditions in the OP. Oct 10 '14 at 2:22

This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0506.pdf

By integrating, we obtain

$$2w'w'''-(w'')^2=-\dfrac{w^4}{2}+c$$

Substitute $$w'''(0)=1$$ , $$w''(0)=0$$ , $$w'(0)=0$$ , $$w(0)=0$$ :

$$c=0$$

$$\therefore2w'w'''-(w'')^2=-\dfrac{w^4}{2}$$

Let $$z=(w')^\frac{3}{2}$$ ,

Then $$\dfrac{d^2z}{dw^2}=-\dfrac{3w^4}{8z^\frac{5}{3}}$$

Which reduces to an Emden–Fowler equation.