second derivative of a function equals a power of said function

Question

Is it possible to find an expression for $f(x)$ so that:

$$f''(x) = (f(x))^{-2}$$

and

$$f(0) = 1$$

I suspect that no such expression exists but i might be wrong.

Thanks in advance.

Answer 1

Multiply by $f^\prime$ you get $$f^\prime f^{\prime\prime}=f^\prime f^{-2},$$ that is $$\frac{d}{dx}\frac12(f^\prime)^2=\frac{d}{dx}(-f^{-1}),$$ Integrate both sides to get $\frac12(f^\prime)^2=-f^{-1}+c.$ Hence $f'=\pm \sqrt{2c-2f^{-1}}$.

Answer 2

One can take the hard path and let $$f(x) = 1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots,$$ since $f(0) = 1 = a_{0}$, such that \begin{align} 1 &= f^2 \, f'' = (1 + a_{1} x + a_{2} x^2 + a_{3} x^3 + a_{4} x^4 + \cdots)^2 \cdot (2 a_{2} + 6 a_{3} x + 12 a_{4} x^2 + \cdots) \\ &= (1 + 2 a_{1} x + (2a_{2} + a_{1}^2) x^2 + \cdots) \cdot (2 a_{2} + 6 a_{3} x + 12 a_{4} x^2 + \cdots) \\ &= 2 a_{2} + 2(2 a_{1} a_{2} + 3 a_{3} ) x + 2 (6 a_{4} + 6 a_{1} a_{2} + a_{1}^2 a_{2} + a_{2}^3) x^2 + \cdots. \end{align} Equating the equations for the coefficients yields \begin{align} f(x) &= 1 + a_{1} x + \frac{x^2}{2!} - 2 a_{1} \, \frac{x^3}{3!} + \left( 6 a_{1}^2 - \frac{1}{2} \right) \, \frac{x^4}{4!} + a_{1} \, (13 - 24 a_{1}^2) \, \frac{x^5}{5!} \\ & \hspace{10mm} + ( 30 \, a_{1}^{2} \, (4 a_{1}^{2} - 5) + 1) \, \frac{x^6}{6!} + \cdots. \end{align} Further terms may be obtained by considering more terms. Here $a_{1}$ is arbitrary unless another condition is applied. It may be of interest that if $a_{1} = 0$, or $f'(0) = 0$, the the function $f(x)$ is an even function given by $$f(x) = 1 + \frac{x^{2}}{2!} - \frac{1}{2} \, \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + \mathcal{O}\left(\frac{x^{8}}{8!}\right).$$

Answer 3

This is too long for a comment.

Using $$\dfrac{d^2x}{dy^2}=-\frac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^3}$$ the differential equation becomes $$\frac{x''}{(x')^3}+\frac{1}{y^2}=0\tag 1$$ and this one can be integrated leading to a very ugly result.

Reducing the order $(z=x')$, this leads to $$\frac{z'}{(z)^3}+\frac{1}{y^2}=0$$ leading to $$z=\pm\frac{\sqrt{y}}{\sqrt{2} \sqrt{c_1 y-1}}$$ Integrating once more $$x=\pm \frac{\sqrt{c_1} \sqrt{y} \sqrt{c_1 y-1}+\log \left(c_1 \sqrt{y}+\sqrt{c_1} \sqrt{c_1 y-1}\right)}{\sqrt{2} c_1^{3/2}}+c_2$$

Try with Wolfram Alpha to see how frustrating is the result using $(1)$.