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I am looking to solve the following third order nonlinear ODE:

$$\frac{\textrm{d}^{3}y}{\textrm{d}x^{3}}+\biggl(\frac{\textrm{d}y}{\textrm{d}x}\biggr)^{2}-y\frac{\textrm{d}^{2}y}{\textrm{d}x^{2}}=0,$$

subject to

$$y(x=0)=0,\qquad\frac{\textrm{d}y}{\textrm{d}x}(x=0)=-1,\qquad\frac{\textrm{d}y}{\textrm{d}x}(x\to\infty)\to0.$$

From inspection I can see that the solution is $y(x)=e^{-x}-1$. However, I would like to be able to derive this solution for myself. I've made a couple of attempts that so far have proved unsuccessful. For example, if I set $z=\textrm{d}y/\textrm{d}x$ then

\begin{align*} \frac{\textrm{d}^{2}y}{\textrm{d}x^{2}}&=z\frac{\textrm{d}z}{\textrm{d}y}, \\ \frac{\textrm{d}^{3}y}{\textrm{d}x^{3}}&=z\biggl(\frac{\textrm{d}z}{\textrm{d}y}\biggr)^{2}+z^{2}\frac{\textrm{d}^{2}z}{\textrm{d}y^{2}}. \end{align*}

So that

$$z\biggl(\frac{\textrm{d}z}{\textrm{d}y}\biggr)^{2}+z^{2}\frac{\textrm{d}^{2}z}{\textrm{d}y^{2}}+z^{2}-yz\frac{\textrm{d}z}{\textrm{d}y}=0.$$

The above could be rewritten like so

$$\frac{\textrm{d}}{\textrm{d}y}\biggl(z\frac{\textrm{d}z}{\textrm{d}y}\biggr)+z-y\frac{\textrm{d}z}{\textrm{d}y}=0,$$

which is equivalent to

$$\frac{\textrm{d}}{\textrm{d}y}\biggl(z\frac{\textrm{d}z}{\textrm{d}y}\biggr)+z^{2}\frac{\textrm{d}}{\textrm{d}y}\biggl(\frac{y}{z}\biggr)=0.$$

Any suggestions as to where to go from here or am I barking up the wrong tree?

Thanks

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  • $\begingroup$ $$y'''=yy''-(y')^2$$ Divide everything by $y^2$. Then by the identity $$ \left(\dfrac{y'}y\right)'=\dfrac{y''y-(y')^2}{y^2} $$ we have $$\frac{y'''}{y^2}=\left(\dfrac{y'}y\right)'$$ I'm not sure what to do with the LHS after this (I could integrate by parts, but that doesn't seem to make the problem easier). $\endgroup$ – Axion004 Jan 31 '20 at 14:19
  • $\begingroup$ Yours ODE can be reduced to Abel equations of the First Kind,but closed-form probably not exist. Only with numerics can be solved. Solution looks like: $y(x)\approx 14.9669\, -15.9711 \text{erf}\left(1.3159\, +0.3067 x-0.0184 x^2\right)$ $\endgroup$ – Mariusz Iwaniuk Feb 2 '20 at 11:00
  • $\begingroup$ This looks like the chazy equation and this question has a solution, there is a difference in the coefficients so they are not exactly equals but should be similar I think at least but I barely get the idea about about how is solved, there is also a comment with a link to a pdf that looks useful too. $\endgroup$ – Daniel D. Feb 2 '20 at 13:08
  • $\begingroup$ @ManojKumar as far as I can see $y(x)=e^{-x}-1$, is a solution of the ODE. It seems strange, to me at least, that it is so difficult to derive such a simple solution. $\endgroup$ – Juggler Feb 3 '20 at 12:03
  • $\begingroup$ @Juggler Yes you are right, I made a silly mistake in substitution. $\endgroup$ – Manoj Kumar Feb 3 '20 at 12:55
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I found a solution to the ODE, but it does not fit all initial conditions.

Ignoring initial conditions, a solution to this ODE is $$y=\frac{-6c}{cx+d}. $$

How I found this solution: We have $y'''=yy''-(y')^2$. By induction, we can prove $$y^{(2n)}=\sum_{r=0}^{n}b_r^{2n}y^{(r)}y^{(2n-1-r)}$$ and $$y^{(2n+1)}=\sum_{r=0}^{n}b_r^{2n+1}y^{(r)}y^{(2n-r)}.$$ Now let $n=1$, (I know that it is not true, but it helps to find a solution.) we have $$y''=b_0^{2}yy'+b_1^{2}yy'=(b_0^{2}+b_1^{2})yy'=\frac{b_0^{2}+b_1^{2}}{2}(y^2)'$$ which has the solution of the form $$y=\frac{a}{cx+d}. $$ Plugging this $y$ in the ODE, shows that $a=-6c$.

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  • $\begingroup$ +1 You could rewrite it as $y=-\frac{6}{x+a}$. Pity it doesn't fit the initial conditions;) $\endgroup$ – pisoir Feb 7 '20 at 13:20
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THIS NOTE MAY HELP

We want to solve $$ y'''+(y')^2-yy''=0\tag 1 $$ Instead of (1) I will solve $$ y'''-(y')^2-yy''=0\tag 2 $$ We have $$ y'''-(y')^2-yy''=0\Leftrightarrow y'''-(yy')'=0\Leftrightarrow y''-yy'=-C_1\Leftrightarrow $$ $$ (y'-\frac{y^2}{2})'=(-C_1x)'\Leftrightarrow y'-y^2/2=-C_1x-C_2 $$ If we set $y=-2u'/u$ we arrive to $$ u''=\frac{1}{2}(C_1x+C_2)u $$ The last equation is solvable with Airy $\textrm{Ai}(x)$,$\textrm{Bi}(x)$ functions see Wikipedia. $$ y(x)=-2^{2/3}C_1^{1/3}\frac{\textrm{Bi}'\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)+\textrm{Ai}'\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)C_3}{\textrm{Bi}\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)+\textrm{Ai}\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)C_3} $$ For the conditions $y(0)=0$, $y'(0)=-1$, $y'(\infty)=0$, we easily get $C_1=1/2$,$C_2=1$,$C_3=-\textrm{Bi}'(2^{1/3})/\textrm{Ai}'(2^{1/3})$.

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    $\begingroup$ Unfortunately the sign in (1) makes huge difference... $\endgroup$ – pisoir Feb 6 '20 at 8:05
  • $\begingroup$ Could you explain further or give another hint about how you got that solution? $\endgroup$ – Daniel D. Feb 7 '20 at 2:03
  • $\begingroup$ Thanks I got it now, +1 $\endgroup$ – Daniel D. Feb 7 '20 at 10:06
  • $\begingroup$ In this case, the solution for $y(x)$ has another really simple form: $y(x)=-\tanh(x/2)$. $\endgroup$ – Juggler Feb 13 '20 at 17:09
  • $\begingroup$ @Juggler. The function $-\tanh(x/2)$ does not satisfies the initial conditions. $\endgroup$ – Nikos Bagis Feb 13 '20 at 20:11
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Here is also my note (unfortunately, probably also not super helpful).

$$y'''+(y')^2-yy''=0\quad\quad(1)$$ Can be rewritten as: $$y'''=y^2\left(\frac{y'}{y}\right)'$$.

Integrating both sides (using integration by parts), this leads to:

$$y''=yy'-2\int(y')^2dx + C.$$ Thus, $$2\int(y')^2dx=yy'-y''+C=\left(\frac{y^2}{2}\right)'-y''+C=\left(\frac{y^2}{2}-y'\right)'+C.\quad\quad(2)$$ Thus the differential equation is also equal to: $$\int \left(y'\right)^2dx = \left(\left(\frac{y}{2}\right)^2-\frac{y'}{2}\right)'+C$$

From this point on, the derivation gets somehow fishy, but if we assume that the integral $\int (y')^2dx$ is zero (or a constant) (but of course, I don't know why this should be the case), we arrive to a differential equation $$\left(\frac{y^2}{2}-y'\right)'=-C,$$ which has a solution $y(x)=ae^x+be^{-x}-C$. Plugging in the initial conditions, we get $a=0, b=-1, C = 1$ as we guessed at the first place.

But I hope someone will come up with a better solution:)

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  • $\begingroup$ When replacing $y(x)=ae^x+be^{-x}+C$ in $(\frac{y^2}{2}-y')'=yy'-y''$ it doesn't gives $C$ and $\int \left(y'\right)^2dx =C=-1$ is not possible but I like this approach because if the integral stays positive it sounds like conservation of energy ($\int m\dot{r}^2 dt=E$) which could make sense in a physical system +1 $\endgroup$ – Daniel D. Feb 10 '20 at 2:07
  • $\begingroup$ @dandide Yes. That would be also my thinking. Maybe approximate the integral with some other "conservation" would lead to some solvable equation. $\endgroup$ – pisoir Feb 10 '20 at 7:15
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THIS NOTE MAY (NOT) HELP

$y'''-y'=0\rightarrow y=Ae^x+Be^{-x}+Ce^{0x}\rightarrow_{B.C} A=0,B=1,C=-1$ so this equation determines uniquely that $y=e^x-1$ but it also solves $(y^{(6)}-y^{(4)})+(y'''+(y')^2-yy'')=0$ for example so it lives in the space of solution of this last equations but it is not clear it belong to the basis of that space.

Instead of solving $y'''-y'=0$ directly one could use it to reduce the other equation, $$0=(y^{(6)}-y^{(4)})+(y'''+(y')^2-yy'')$$

$$\rightarrow_{[y'''=y'\Rightarrow y^{(6)}=y^{(4)}]} 0=y'''+(y')^2-yy'' \quad\text{(our equation)}$$

$$\rightarrow_{\frac{d}{dx}}0=y''''+y'y''-yy'''$$

$$\rightarrow_{[y'''=y'\Rightarrow y''''=y'' \text{ & } y''=y+D]}0=(y+D)+y'(y+D)-y(y')=Dy'+y+D=De^{-\frac{x}{D}}(e^{\frac{x}{D}}y)'+D$$

$$\rightarrow y=Fe^{-\frac{x}{D}}-D$$

$$\rightarrow_{BC\Rightarrow F=1,D=1} y=e^{-x}-1$$

And we get the same solution again anyway, so $y'''-y'=0$ helps obtain a solution but it doesn't really helps us solve the original equation $0=y'''+(y')^2-yy''$ (or $0=(y^{(6)}-y^{(4)})+(y'''+(y')^2-yy'')$)

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Thanks all for the suggestions. Some interesting ideas.

I noticed that making the substitution $z=(\textrm{d}y/\textrm{d}x)^{2}$, gives the following

$$\frac{\textrm{d}^{2}z}{\textrm{d}y^{2}}+4z\frac{\textrm{d}}{\textrm{d}y}\biggl(\frac{y}{2\sqrt{z}}\biggr)=0.$$

Not really sure that helps though...

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