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I have the following nonlinear ODE relating the functions $A(r)$ and $B(r)$:

$-\frac{A''}{A} + \frac{A'^2}{2A^2} - \frac{A'B'}{AB} + \frac{B'^2}{2B^2} - \frac{2B''}{B} + \frac{2}{B} = k$

with $k$ a known constant. I do know one solution:

$A_0(r) = -\cos^2{\frac{r}{\alpha}},\ \ \ B_0(r) = \alpha^2 \sin^2{\frac{r}{\alpha}}$

for a known constant $\alpha$. In this case $r \in [0,\frac{\pi}{2}\alpha)$ and I believe $k$ works out to be $\frac{12}{\alpha^2}$.

Furthermore, the solution I'd like to find (if there is one) should only differ significantly from $A_0, B_0$ for very small $r$, ie. $r \ll \alpha$. In other words,

$A(r) = A_0(r) + A_1(r),\ \ \ |A_1(r)| \ll |A_0(r)|\ \text{ unless }\ r \ll \alpha \\ B(r) = B_0(r) + B_1(r),\ \ \ |B_1(r)| \ll |B_0(r)|\ \text{ unless }\ r \ll \alpha$

Also, $A(r) \le 0$ and $B(r) \ge 0$ over the whole domain. Finally, $B(0) = 0$ (hence $B_1(0) = 0$) -- and I suspect $B(r)$ may have one more zero, if that helps narrow it down.

With two unknown functions, I guess there's no hope of solving it outright. But is there a way to turn this into a non-differential equation relating $A$ and $B$? Or at least a simpler differential equation? Or to use the $A_0, B_0$ baseline as a guide for a numerical technique or series expansion or something, just to get some better understanding of the solutions that are possible?

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This isn't quite an answer, but I thought I'd share with you what I've got since no one else has.

Starting from your ODE, \begin{align} -\frac{A''}{A}-\frac{1}{2}\left(\frac{A'}{A}-\frac{B'}{B}\right)^2-2\frac{B''}{B}+\frac{2}{B}=k, \end{align} I'll assume $B$ is a function of $A$ ($B(A)$), so via the chain rule : \begin{align} B'&=B_AA'\\ B''&=B_AA''+B_{AA}A'^2 \end{align} where $B_A$ is short for d$B/$d$A$. This turns the ODE into \begin{align} -\left(\frac{1}{A}+\frac{2B_A}{B}\right)A''+\left(\frac{1}{2}\left(\frac{1}{A}-\frac{B_A}{B}\right)^2-\frac{2B_{AA}}{B}\right)A'^2=k-\frac{2}{B}. \end{align} Note that we now have an ODE of only one function, $A$, since the $B$'s are unkown functions of $A$. If we divide by the the $A''$ coefficient and rename the functions of $A$ the ODE becomes \begin{align} A''+m(A)A'^2=n(A). \end{align} This equation has a nice trick to linearize and solve it, let \begin{align} A'^2&=C\\ A''&=\frac{1}{2}C_A, \end{align} you can verify this using the chain rule. So then the equation becomes \begin{align} \frac{1}{2}C_A+m(A)C=n(A), \end{align} using an integrating factor, where I've written $\smallint m(A)\mathrm dA=M$ for short, \begin{align} \frac{\mathrm d}{\mathrm dA}(e^{2M}C)&=2e^{2M}n(A)\\ C=A'^2&=e^{-2M}\left(c_1+2\int e^{2M}n(A)\mathrm dA\right). \end{align} This is separable as well. The resulting integral would give a separable equation, \begin{align} \int\frac{e^M \mathrm dA}{\sqrt{c_1+2\smallint e^{2M}n(A)\mathrm dA}}=r+c_2. \end{align} The problem lies withing solving these two integrals of unknown functions of $A$, ($B(A)$), though $M$ and $n$ contain derivatives of $B$ which is why I think that maybe you could do it. If the two integrals can be solved, then you'd end up with an equation of $A$, $B$ and $r$, which I believe is what you want.

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    $\begingroup$ Wow, this is incredible. I can't imagine how you figured all this out. While I'm trying to understand it, I will leave the question open in case your answer stimulates anyone else to extend it, but otherwise I'll certainly mark it as accepted. Thank you so much!! $\endgroup$ Commented Apr 12, 2021 at 17:08
  • $\begingroup$ Where'd you come up with this equation? $\endgroup$ Commented Apr 12, 2021 at 21:05
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    $\begingroup$ Haha, it's a long story but suffice it to say that, if I didn't screw it up, this is the equation of any spherically symmetric, static spacetime metric whose scalar curvature (LHS) is constant (RHS). I think we can choose $g_{rr} = 1$, $g_{tt} = A(r)$ and $g_{\theta\theta} = B(r)$. In fact, it comes from this idea which is probably loony but I couldn't help trying to pursue it a little further. If the Higgs boson is spherically symmetric, I think the wave equation implies that its $R$ is constant. $\endgroup$ Commented Apr 13, 2021 at 0:13

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