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Let $n$ - dimension $\geq 3$.

Consider a compact manifold (M,g). Let $\epsilon_0$ denote the injectivity radius of $(M,g)$. Let $B_\epsilon(0)$ denote a geodesic ball of radius $\epsilon < \epsilon_0$.

Consider the Green's function on $B_\epsilon(0)$ ( i.g. verifies that $\Delta G = \delta_y$ and $G=0$ on the boundary. G is also positive, smooth and well defined of the diagonal).

Is it possible to get the following upper bound

$$ G(x,y) \leq C(n) \rho(x,y)^{2-n}. $$

It is known that this estimate holds near the singularity (even for a general compact subdomain of $(M,g)$); see Schoen and Yau for instance.

Is it true for all $x\neq y$, not just near the singularity?

N.B. In $\mathbb{R}^n$, we know that $G(x,y) \leq \mbox{ Fundamental solution } \leq C r^{2-n}$; it is a consequence of the max principle. In short, I am trying to get such estimates for a geodesic ball on manifolds.

Do we have an explicit formula for the Green's function of a geodesic ball ? Can we derive such bound from it?

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On a compact manifold, how is a bound other than near the singularity an issue? Am I missing something here? –  Michael Renardy May 27 '13 at 15:39
    
I need a global estimate, not a local one. If I have an estimate near the singularity, does that mean that it holds on the whole manifold? –  Henry May 27 '13 at 15:47
    
I must add that if I use compactness to extend, my constant C will depend on $\Omega$ and I do not want that. –  Henry May 27 '13 at 15:50
    
For instance, if I say that $G(x,y) \leq C(n) \rho(x,y)^{2-n} + M Vol (\B_\epsilon\setminus \B_r (y) ),$ it will not be enough. I really need a bound $C(n) \rho(x,y)^{2-n}$ for $B_\epsilon$. –  Henry May 27 '13 at 15:57
    
Maybe I can ask a related question. It is known in $\mathbb{R}^n$ that $G(x,y)$ is given by a regular part (harmonic function) and by the Fondamental solution. Is there something equivalent for a Green's function on a compact subdomain (or geodesic ball) of a given manifold $(M,g)$? –  user34432 May 27 '13 at 16:52

2 Answers 2

up vote 0 down vote accepted

I suggest you look at: http://www.jstor.org/stable/2374588

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Yes. I know about that paper. It gives the bound for the singularity for instance. However, I did not see anything else that could give me an answer. Did I miss something? –  user34432 May 27 '13 at 17:56
    
Define $M(r) = max_{\partial B_r(y)} G(r), r = \rho(x,y)$. It is known that : i) near its singularity, $M(r)\leq C(n) r^{2-n}$, e.g. for small $r$ ii) M(r) decreases as $r$ increases My question is : iii) is the following hold : $M(r) \leq C(n) r^{2-n}$ for all $r\leq R$ of my geodesic ball. Note that in $\mathbb{R}^n, G(x,y) \leq C(n) r^{2-n}$ for all $r\leq R$ in a ball of radius R for instance. This follows from the fact that the Green's function is "bounded" by the fondamental solution of the Laplace operator. –  user34432 May 27 '13 at 18:08

I don't know if this helps you...

Consider the warped product metric on $M^n = R \times S^{n-1}$

$ds^2 = dr^2 + f(r)^2 ds_{S^{n-1}}.$

Assume $u=u(r)$, where $r$ is the distance function from some fixed point.

Then

$\nabla u = u'(r) \nabla r$

$\Delta u = u''(r)\Delta r + u'(r)^2$

Sometimes you can find Green function solving the EDO

$\Delta u = 0.$

Indeed,

$\Delta r = (n-1) \frac{f'(r)}{f(r)}$,

so

$(n-1)u''(r) \frac{f'(r)}{f(r)} + u'(r)^2 =0$.

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Indeed, I knew about that approach. However, I do not want to consider a specific metric. I was trying to find in the literature a formula for the green's function of a geodesic ball, which seems unknown. I was wondering if it were indeed the case and if some bounds were known instead. –  Henry May 27 '13 at 16:26

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