# An interesting ODE

I've been thinking about approaches to solve

$$\frac{\textrm{d}^{3}y}{\textrm{d}x^{3}}=\textrm{e}^{-y(x)}.$$

My initial thought was to set $$y(x)=\ln(z(x))$$ in order to obtain an ODE relating $$z$$ to $$x$$. However, this in fact makes things much worse (or at least much messier).

Any insights much appreciated.

Thanks

• Why do you think it has a closed-form solution? – Moo Feb 23 at 15:25
• Fair point. I'm not sure it does. There's a neat solution to $y''(x)=\textrm{e}^{-y(x)}$, so that made me wonder. Not that the second-order problem has any bearing on this problem, just curious. – Juggler Feb 23 at 15:36

Your equation is autonomous. The standard method is to let $$u(y)=y'(x)$$, then carefully writing out the equation we have

$$u(u')^2+u^2 u''=e^{-y}$$

Now $$y$$ is the independent variable. A particular solution is $$u_p=\left( \frac{9}{2} \right)^{1/3}e^{-y/3}$$, letting $$u=u_p+v$$ just produces another awful nonlinear equation for $$v$$. At least we have one solution to the original equation: $$y_p(x)=3 \ln(x-c_1)+\ln(6)$$. In the absence of some clever integrating factor trick, we can look at perturbative solutions

$$y'''=\varepsilon e^{- y}$$

When $$\varepsilon=0$$, we have $$y_0(x)=Ax^2+Bx+C$$. With the ansatz

$$y(x)=\sum\limits_{n=0}^\infty y_n(x)\varepsilon^n$$

Choosing $$y(0)=1, \ y'(0)=-1, \ y''(0)=0$$, we have $$y_0=1-x$$, and we can find the next term by matching powers of $$\varepsilon^\dagger$$

$$y_1'''=e^{-y_0}=e^{-1+x}$$

Which is easily integrated

$$y_1(x)=-\frac{1}{2e}(2-2e^x +2x+x^2)$$

Here is a plot of the numerical integration and the two term series $$y_0+y_1$$:

$$\dagger$$ To get the higher order terms, you'd want a systematic way of getting down the right coefficients of $$\varepsilon$$ from $$\sum_n y_n \varepsilon^n$$ that's in the exponential. For example, at the next order, when equating $$\varepsilon^2$$ terms, we have on the right

$$\varepsilon e^{-y_0-\varepsilon y_1 - \dots}=\varepsilon e^{-y_0} e^{-\varepsilon y_1} \dots =\varepsilon e^{-y_0}(1-\varepsilon y_1) \dots$$

So that the $$\varepsilon^2$$ coefficient on the right is $$-y_1 e^{-y_0}$$, and the next term in our perturbation series is given by solving

$$y_2'''=-y_1 e^{-y_0}$$

Update: while I do not have a closed form for the coefficients of $$\varepsilon^n$$, they can be found diagrammatically, as they are in correspondence with the integer partitions of $$n$$, or equivalently the Young diagrams of $$n$$ blocks. Consider

$$e^{-\sum \varepsilon^n y_n}=e^{-y_0} e^{-\varepsilon y_1} e^{-\varepsilon^2 y_2} e^{-\varepsilon^3 y_3} \dots \\ =e^{-y_0}\left(1-\varepsilon y_1 + \frac{\varepsilon^2 y_1^2}{2} \dots \right) \left(1-\varepsilon^2 y_2 + \frac{\varepsilon^4 y_2^2}{2} \dots \right)\left(1-\varepsilon^3 y_3 + \frac{\varepsilon^6 y_3^2}{2} \dots \right)\dots$$

The coefficient of $$\varepsilon^n$$ (from the exponential, excluding the single factor outside the exponential) will be given by all the ways in which expanding the parentheses out can produce $$\varepsilon^n$$. For example, the coefficients of $$\varepsilon^4$$ would be represented by the diagrams of four blocks:

Where I'm letting a column of length $$n$$ = one factor of $$y_n$$. Thus the fifth order equation reads

$$y_5'''=e^{-y_0} \left( \frac{(-y_1)^4}{4!} + (-y_2) \frac{(-y_1)^2}{2!} +\frac{(-y_2)^2}{2!} + (-y_3)(-y_1) +(-y_4) \right)$$

In this way you may generate the solution to any order. The expressions become too cumbersome to type here, but Mathematica has no trouble solving them. Here is a plot:

Since this is a perturbation series, the most we can hope for is that series is asymptotic. If you wished to take this further, note that after you have the $$N$$th order solution $$y_N$$, you have a sum of finitely many known terms

$$y=\sum\limits_{n=0}^N \varepsilon^n y_n$$

Then you could consider various methods of re-summing this series in $$\varepsilon$$. You could also investigate slightly different perturbation problems, eg: $$y'''=e^{-\varepsilon y}$$. I've tried the perturbation ansatz on the differential equation for $$u(y)$$, but that leads to more intractable differential equations at orders beyond first.

• Thanks for the answer. I had tried a similar approach with $u(y)=[y'(x)]^{2}$, although that didn't really get me anywhere. Unless I'm missing something shouldn't the ODE become $$u[(u')^{2}+uu'']=\textrm{e}^{-y}.$$ ? – Juggler Feb 23 at 16:51
• Whoops! You are correct. – Sal Feb 23 at 16:57
• Having looked at your answers, it seems that you enjoy approximations as much as I do ! Cheers :-) – Claude Leibovici Feb 24 at 10:08
• @ClaudeLeibovici Absolutely! Cheers! – Sal Feb 24 at 14:06
• Amazing stuff! Thanks for the insight, turns out it's actually a really interesting problem! – Juggler Feb 26 at 16:07