During my physics research in cosmology I encountered a differential equation of the following type ($f,g$ are functions that only depend on the spherical coordinate $r$, $\partial_i$ means $\partial/\partial x^i$ where $x^i \in \{x,y,z\}$ etc.):

\begin{equation} \nabla^2 f - [a_1(\nabla^2 f)^2 - a_2 \sum^3_{i,j=1}(\partial_i \partial_j f)(\partial_i \partial_j f)] = -g(r) \end{equation}

where $a_1,a_2$ are arbitrary (non-zero) real numbers (constants).

Useful identities are (since $f = f(r)$):

\begin{equation} (\nabla^2 f)^2 = \Big(\frac{d^2 f}{dr^2}\Big)^2 + \frac{4}{r}\Big(\frac{d^2 f}{dr}\Big)\Big(\frac{df}{dr}\Big) + \frac{4}{r^2} \Big(\frac{df}{dr}\Big)^2 \end{equation}


\begin{equation} \sum^3_{i=1} \sum_{j = 1}^3 (\partial_i \partial_j f)(\partial_i \partial_j f) = \frac{2}{r^2}\Big(\frac{df}{dr}\Big)^2 + \Big(\frac{d^2 f}{dr^2}\Big)^2 \end{equation}

We would like to get an analytical solution for for $df/dr$ if it exists and otherwise a numerical one would be useful.

  • $\begingroup$ Are you sure about the second useful identity? With $f(r)=r=\sqrt{x^2+y^2+z^2}$ I get $$ \partial_xf=\frac{x}r\,,\quad \partial_y\partial_xf=-\frac{xy}{r^3}\,. $$ $\endgroup$
    – Kurt G.
    Commented Jan 25 at 14:47
  • $\begingroup$ Excuse me, there is an implicit summation over $i,j$ in my expressions, let me edit it. For me it was obvious since we do in physics this all the time (Einstein summation convention) but indeed this should be mentioned explicitly! $\endgroup$ Commented Jan 25 at 15:31

1 Answer 1


Too long for a comment.

We're looking to solve the nonlinear first-order equation $$(a_2 - a_1) (u')^2 + u' - \frac{4 a_1 u' u}{r} + \frac{2 u}{r} + \frac{2 (a_2 - 2 a_1)}{r^2} u^2 = g(r)$$ for $u(r) := f'(r)$. For general $a_1, a_2$ the equation looks pretty difficult, at least for general $g$, but at in least in a few special cases the equation can be solved:

  • $a_2 = 0$: $$u(r) = \frac{1}{r^2} \left[C + \frac{1}{2 a_1} \int^r s^2 \left(1 \pm \sqrt{1 - 4 a_1 g(s)}\right) \,ds \right]$$
  • $a_2 = a_1 = a$: $$u(r) = \frac{1}{4 a} r \left[1 \pm \sqrt{r^2 - \frac{8}{a r} \left(\int^r s^2 g(s) \,ds + C\right)}\right].$$

In the case that $g$ is constant, say, $g(r) = \lambda$, which o.p. mentioned in the comments, substituting $u(r) = r v(r)$ and regarding $r$ as a function of $v$ gives $$\frac{r'}{r} = \frac{\pm 1 - 4 a_1 v - \sqrt{8 a_2 (3 a_1 - a_2) v^2 - 8 a_2 v + 4 (a_2 - a_1) \lambda + 1}}{4 (2 a_1 - a_2) v^2 - 4 v + \lambda},$$ so $$r(v) = C \exp \int^v h(s) \,ds ,$$ where $h$ is the expression on the right-hand side of the above differential equation in $v$. Moreover, $\int^v h(s) \,ds$ can be integrated in terms of elementary functions for any $a_1, a_2, \lambda$, but the form of the formula will depend on the discriminants of the two quadratics in $v$ that appear. The numerator of the quadratic under the radical is $32 a_2 (a_1 - a_2) (12 a_1 \lambda - 4 a_2 \lambda - 3)$, which gives some additional insight into why the cases $a_2 = 0$ and $a_1 = a_2$ are comparatively manageable.

  • $\begingroup$ Thanks. What about if $g(r)$ is some non-zero constant while having $a_1,a_2$ general? $\endgroup$ Commented Jan 25 at 23:43
  • 1
    $\begingroup$ One can say a little more in that case---I've updated my answer to include that case---but even then an explicit formula for $u(r)$ is probably hopeless. $\endgroup$ Commented Jan 26 at 1:02

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