I'd like to solve a differential equation $$ f^2(x) f''(x)=-x $$ where $f(x)$ is defined on $[0,1]$ and has a boundary condition $f(0)=f(1)=0$.

I somehow found out that the solution is fairly close to $f(x) = x^{1/3} \phi^{2/3}(\Phi^{-1}(1-x))$ where $\phi$ and $\Phi$ are pdf and cdf of a standard normal distribution, but it fails to solve the differential equation exactly.

Thank for all comments!

Based on the solution structure of Emden–Fowler Equation, I was able to identify the values of constants that satisfy the boundary conditions. The followings are the details:

Define \begin{equation} Z_R(\tau) \triangleq \sqrt{3} J_{1/3}(\tau) - Y_{1/3}(\tau) , \quad Z_L(\tau) \triangleq - \frac{2}{\pi} K_{1/3}(\tau) \end{equation} where $J, Y, K$ are Bessel functions. Further define \begin{equation} \bar{\tau} \triangleq \inf\{ \tau > 0; Z_R(\tau) = 0 \} \approx 2.3834 , \quad a \triangleq \frac{1}{ \bar{\tau}^{4/3} Z_R'(\bar{\tau})^2 } \approx 0.2910 , \quad b \triangleq a \left( \frac{9}{2} \right)^{1/3} \approx 0.1763. \end{equation} Then, the solution curve $\{ (x, f(x)) \}_{x \in [0,1]}$ is characterized by \begin{equation} \left\{ \left( x_R(\tau), y_R(\tau) \right) \right\}_{\tau \in [0, \bar{\tau}]} \bigcup \left\{ \left( x_L(\tau), y_L(\tau) \right) \right\}_{\tau \in [0, \infty]} \end{equation} where \begin{equation} x_R(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_R'(\tau) + \frac{1}{3} Z_R(\tau) \right)^2 + \tau^2 Z_R(\tau)^2 \right] , \quad y_R(\tau) \triangleq b \tau^{2/3} Z_R(\tau)^2. \end{equation}

\begin{equation} x_L(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_L'(\tau) + \frac{1}{3} Z_L(\tau) \right)^2 - \tau^2 Z_L(\tau)^2 \right] , \quad y_L(\tau) \triangleq b \tau^{2/3} Z_L(\tau)^2. \end{equation}

In addition to this analytic solution, I also obtained a numerical solution by repeatedly computing $$ f_{k+1}(x) \gets \left[ \left( f_k(x-2h) + f_k(x+2h) \right) + 4 \left(f_k(x-h)+f_k(x+h)\right) + \frac{8 x h^2}{f_k^2(x)} \right] \big/ 10 $$ on the grid $x \in \{2h,3h,\ldots,1-3h,1-2h\}$ for small $h$ with an initialization $f_0(x) \triangleq 0.5(1-(1-2x)^2)$.

The following figure shows these solutions:

enter image description here

  • $\begingroup$ $f(1)=0\ $ would imply $\ 0 = -1.\ $ $\endgroup$ – Wlod AA Mar 4 at 6:05
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    $\begingroup$ @WlodAA We will have $f''(1) = -\infty$. $\endgroup$ – Seungki Min Mar 4 at 6:22
  • $\begingroup$ I see. Thank you. $\endgroup$ – Wlod AA Mar 4 at 6:24
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    $\begingroup$ This is a case of the Emden-Fowler equation, see eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf, which arises in astrophysics and was subject of intensive study. E. Hille wrote that "it has fantastically complicated singularities near x=0". $\endgroup$ – Alexandre Eremenko Mar 4 at 11:51

Surprisingly, this case of the Emden-Fowler equation is explicitly solvable: see formula (2.3.27) in A. Polyanin and V. Zaitsev, Handbook of exact solutions of ordinary differential equations, Chapman & Hill, 2003.

I copy the formula, without verifying it. Let $$Z=C_1J_{1/3}(\tau)+C_2Y_{1/3}(\tau),$$ or $$Z=C_1I_{1/3}(\tau)+C_2K_{1/3}(\tau),$$ where $J,Y$ are Bessel and $I$, $K$ are modified Bessel functions. Then
$$x=a\tau^{-2/3}[(\tau Z^\prime+(1/3)Z)^2\pm\tau^2Z^2],\quad y=b\tau^{2/3}Z^2$$ satisfy $d^2y/dx^2=Axy^{-2}$ with $A=-(9/2)(b/a)^3.$

For the $+$ sign in $\pm$ take the first formula for $Z$, and for the $-$ the second one.

Remark. Emden-Fowler equation appears for the first time in the famous book by R. Emden, Gaskugeln (1907) and since then frequently arises in the study of stars and black holes.

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    $\begingroup$ The first version with $Y$ seems to be correct, but I think it does not lead to a solution of the original problem. However, the version with $K$ instead of $Y$ does not seem to solve the differential equation (with either version of the $\pm$ sign). Perhaps there is some transcription error? $\endgroup$ – Neil Strickland Mar 4 at 13:46
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    $\begingroup$ Maple solves this using the Airy functions Ai and Bi, which are related to Bessel function of order $1/3, 2/3$. $\endgroup$ – Gerald Edgar Mar 4 at 16:45
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    $\begingroup$ @Neil Strickland: Yes, I made a misprint while copying from the book. Now corrected (second definition of Z has I instead of J). $\endgroup$ – Alexandre Eremenko Mar 4 at 18:39

I tried the following approach. Put $y=f(x)$ and $t=2x-1$ so the differential equation becomes $y^2\ddot{y}=-(t+1)/8$ with boundary conditions $y=0$ at $t=\pm 1$. We can then write $y=\sum_ia_it^i$. The differential equation gives a recurrence relation expressing all the coefficients $a_i$ in terms of $a_0$ and $a_1$. We can then truncate the power series to a given order $d$ and solve numerically for the boundary conditions. This seems to work in a well-behaved way, with good convergence at the endpoints and a result that is stable when we increase $d$. It looks like $a_0=0.450$ and $a_1=0.120$ to $3$ decimal places. Maple code is as follows:

Digits := 50:
d := 50:
y := add(a[i] * t^i,i=0..d):
sol0 := solve([coeffs(rem(expand(y^2 * diff(y,t,t) + (t+1)/8),t^(d-1),t),t)] 
y0 := expand(subs(sol0,y)):
sol1 := fsolve({subs(t= 1,y0),subs(t=-1,y0)},{a[0]=0.45,a[1]=0.1}):
aa[d] := subs(sol1,[a[0],a[1]]);
y1 := subs(sol1,y0);
y1x := subs(t = 2*x-1,y1):
Phi := unapply((1 + erf(x))/2,x):
phi := unapply(diff(Phi(x),x),x):
 plot(x^(1/3) * phi(RootOf(1-x-Phi(_Z)))^(2/3),x=0..1,colour=blue)

This generates the following picture. The power series solution is in red and the function $x^{1/3}\phi(\Phi^{-1}(1-x))^{2/3}$ is in blue.

enter image description here

The coefficients $a_{2i}$ lie on a nice smooth curve, and the coefficients $a_{2i+1}$ lie on a similar curve shifted down slightly. Logs of the absolute values can be displayed as follows:


One could probably get further by finding an exact or approximate formula for these curves.

enter image description here

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