# Making sense of a 2nd-order nonlinear ODE with *2 unknown functions* and known asymptotic behavior

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?

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.
• 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. Commented Apr 13, 2021 at 0:13