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I am working on a project studying the p-Laplace equation, and I think it will be a good idea to look at the solution to the corresponding ODE on Schwarzchild space (given by the metric $\tilde{g}=(1+\frac{m}{2r})^{4}[dr^2+r^2(d\theta^2+ \sin^2(\theta)d\varphi^2)])$. I am looking for boundary conditions: initially a constant c and eventually 0 at infinity. After I set up the ODE using local coordinates for the gradient and divergent, I am left with the following incredibly messy ODE: $$f''(r)+f''(r)^{p-1} f'(r)^{2-p} \frac{1}{p-2}+f'(r)\left(\frac{2}{(p-2)(r+\frac{m}{2})}\right)+f'(r)\left(\frac{m}{r^2}(1+\frac{m}{2r})^{-1}\right)=0$$ Should I bother with continuing to look for a closed form solution to this? It is nonlinear in the worst possible way and is not even separable. One thing I have tried is taking as an ansatz $f'(r)\neq 0$, so that we get $$\frac{y''}{y'}+\left(\frac{y''}{y'}\right)^{p-1} \frac{1}{p-2}+\left(\frac{\frac{2}{p-2}+\frac{m}{r}}{r+\frac{m}{2}} \right)=0$$ Can this only be solved numerically? I am more interested in the properties of the solution than the closed form (what regularity does it have, etc).

Jacob
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    Some thoughts: Is the last term in your first equation meant to be $f'$ or $f$? Your second equation is algebraic in $y''/y'$, so in principle you can solve for $y''/y'$ (for a given number $p$). Perhaps there is a nice transformation that simplifies your second equation? It is probably best to start looking at perturbative rather than exact solutions – Sal Jul 25 '21 at 20:39
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    @Sal thanks for your comment! They are all f' or f''.. this comes from $$\Delta^p=\frac{1}{\sqrt{g}} \sum_{m}{ \sqrt{g}(\sum_}{i,j,k,h} g_{ij}g^{hi}g^{kj} \partial_k f \partial_h f)^{\frac{p-2}{2}} \ \sum_{l}g^{lm}\partial_l f}=0$$. when p=2(we have the standard Laplacian), the form is quite nice actually: $f''(r)+f'(r) \left[ \frac{2}{r+\frac{m}{2}}\right]=0$. – Jacob Jul 25 '21 at 20:50
  • No problem! Given where the equation comes from, you might try a perturbation series $f(r)=\sum\limits_{n \geq 0}f_n(r)\epsilon^n$ starting with $p=2+\epsilon$. The zeroth and first order equations are easily solveable. I have not investigated higher order terms – Sal Jul 25 '21 at 21:29

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Let $m=1$, $f'(r)=e^{G(r)}$, $G'=g$, $p=2+\eta$, and $\eta>0$. Then the first equation is a polynomial in $g$ of degree $\eta+1$.

$$ \frac{2}{r}\frac{2r+\eta}{2r+1}+g(\eta+g^\eta)=0 $$

When $\eta$ is even, the polynomial is of odd degree thus has at least one real root. Consider the large $r$ behaviour

$$ g(\eta+g^\eta)\sim-\frac{2}{r} \qquad ;\qquad r\to\infty $$

The left hand side is positive if $g \in\mathbb{R}$ and $\eta$ is odd, the right hand side is negative; thus there are only complex solutions when $\eta$ is odd. The consistent dominant balances are

$$ g \sim -\frac{2}{\eta r} \qquad ; \qquad r \to \infty $$

For the real root when $\eta$ is even, and

$$ g \sim (-\eta)^{1/\eta} \qquad ; \qquad r \to \infty \tag{*} $$

For the complex roots. Since you are interested in the solutions that remain finite for large $r$, we require $\Re(g)<0$ as $r \to \infty$. In this way you can determine from eq. (*) the imaginary part of the allowed complex roots in terms of $\eta$, which will determine the large $r$ oscillatory behavior of $f'$.

You may perform a similar analysis for $r\to 0$.

Finally, you could consider a perturbation series $g=\sum\limits_{n=0}^\infty g_n\eta^n$. Note $(g_0+\eta g_1+\eta^2g_2+\cdots)^\eta=1+\eta \ln(g_0)+\eta^2(g_1/g_0+\ln^2(g_0)/2)+\cdots$. Matching powers of $\eta$ we find for $\eta^0$

$$ g_0=-\frac{2}{r}\frac{2r}{2r+1} $$

Matching powers of $\eta^1$

$$ g_1=-\frac{2}{r}\frac{1}{2r+1}-\ln(g_0) $$

And so on. You'd need to carefully keep track of the different roots.

Sal
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