Green’s Functions of the Laplacian

David Glickenstein February 11, 2000

1. Preliminary Calculations. Suppose we are on ℝ^{n}, with n ≥ 3. Let

Γ(x, y) = ω_{n−1}|x − y|^{2−n}

where ω_{n−1} is the volume of the (n-1)-sphere S^{n−1}. We want to compute derivatives of this function.

|\frac{∂}{∂x^{i}} Γ(x, y)| = |ω_{n−1} \frac{∂}{∂x^{i}} (∑_{j}(x^{j} − y^{j})^{2})^{(2−n)/2}|

= ω_{n−1}(n − 2) |\frac{x^{i} − y^{i}}{(∑_{j}(x^{j} − y^{j})^{2})^{n/2}} |

≤ ω_{n−1}(n − 2)\frac{r}{r^{n}}
= ω_{n−1}(n − 2)\frac{1}{r^{n−1}}

and Γ(x, y) = ω_{n−1}r^{2−n} ≤ ω_{n−1}r^{1−n} for r ≤ 1 so for a bounded domain Ω we have that Γ(x, y) and \frac{∂}{∂x^{i}} Γ(x, y) are bounded by (n − 2)r^{1−n}, which is integrable over Ω, so we can interchange the differentiation and integration, so \frac{∂}{∂x^{i}} ∫_{Ω} Γ(x, y)f(y)dy = ∫_{Ω} \frac{∂}{∂x^{i}} Γ(x, y)f(y)dy.

Now,

\frac{∂^{2}}{∂x^{k}∂x^{i}} Γ(x, y) = ω_{n−1} \frac{∂}{∂x^{k}} [(n − 2)\frac{x^{i} − y^{i}}{(∑_{j}(x^{j} − y^{j})^{2})^{n/2}} ]

= \{ \begin{matrix}ω_{n−1}(2 − n)n\frac{(x^{i}−y^{i})(x^{k}−y^{k})}{(∑_{j}(x^{j}−y^{j})^{2})^{(n+2)/2}} & if i ≠ k \\ ω_{n−1}(2 − n)n\frac{(x^{i}−y^{i})^{2}}{(∑_{j}(x^{j}−y^{j})^{2})^{(n+2)/2}} + (n − 2)_{\frac{1}{(∑_{j}(x^{j}−y^{j})^{2})^{n/2}}} & if i = k \end{matrix} \}

so |\frac{∂^{2}}{∂x^{k}∂x^{i}} Γ(x, y)| ≤ 2ω_{n−1}n(n − 2)\frac{1}{r^{n}}, but this is not integrable, so we cannot simply interchange the order of integration to get Δ ∫ Γ(x, y)f(y)dy. We need to cut off the singularity.

2. Two proofs of Δ_{distr(y)}Γ(x, y) = δ_{x}(y) on ℝ^{n}. Let’s first do it directly. Consider ∫ Γ(x, y)Δf(y)dy. We want to use the divergence theorem (but can’t for x = y), so let’s look at ∫_{ℝ^{n}\B_{ϵ}} div_{y} [Γ(x, y)∇f(y)] dy = ∫_{∂B_{ϵ}} Γ(x, y)∇f(y) · ν(y)ds(y) where ν is the outward pointing normal, and B_{ϵ} = B_{ϵ}(x). Notice that the left hand side is

∫_{ℝ^{n}\B_{ϵ}} [∇_{y}Γ(x, y) · ∇f(y) + Γ(x, y)Δf(y)] dy

If we also look at

∫_{ℝ^{n}\B_{ϵ}} div_{y} [∇_{y}Γ(x, y)f(y)] dy = ∫_{∂B_{ϵ}} f(y)∇_{y}Γ(x, y) · ν(y)ds(y)

we can subtract the two and get

∫_{ℝ^{n}\B_{ϵ}} [Γ(x, y)Δf(y) − Δ_{y}Γ(x, y)f(y)] dy = ∫_{∂B_{ϵ}} [Γ(x, y)∇f(y) · ν(y) − f(y)∇_{y}Γ(x, y) · ν(y)] ds(y)

We easily see that Δ_{y}Γ(x, y) = 0. Furthermore,

|∫_{∂B_{ϵ}} Γ(x, y)∇f(y) · ν(y)ds(y)| ≤ sup |∇f(y)| V ol(∂B_{ϵ})\frac{1}{ω_{n−1}ϵ^{n−2}}
≤ sup |∇f(y)| ϵ^{n−1} \frac{1}{ϵ^{n−2}}
≤ sup |∇f(y)| ϵ

which goes to zero as ϵ → 0.

We now have to calculate ∇_{y}Γ(x, y) · ν(y). This is easily done since ν(y) = − |x − y| (the minus is because it is the outward pointing normal for ℝ^{n} \ B_{ϵ}). Thus we really have \frac{d}{dr}(ω_{n−1}r^{2−n}) = (2−n)ω_{n−1}r^{1−n} so^{−} ∫_{∂} B_{ϵ}f(y)∇_{y}Γ(x, y)·ν(y)ds(y) = ∫_{∂} B_{ϵ}f(y)ds(y) → f(x) as ϵ → 0. Thus if we let ϵ → 0 we get ∫_{ℝ^{n}} Γ(x, y)Δf(y)dy = f(x), or Δ_{distr(y)}Γ(x, y) = δ_{x}(y).

Our second method uses the Fourier transform.

3. The Fundamental Solution. We now want to show that u = Qf(x) = ∫_{ℝ^{n}} Γ(x, y)f(y)dy is a solution to the equation Δu = f. First assume that f is C^{∞}, which implies that Qf is C^{∞} (This is because it is a convolution). We first show that the equation is true in the sense of distributions. That is let ϕ ∈ C_{c}^{∞}(ℝ^{n}) and show ∫_{ℝ^{n}} ΔQf(x)ϕ(x)dx = ∫_{ℝ^{n}} f(x)ϕ(x)dx. We do this as follows:

∫_{ℝ^{n}} ΔQf(x)ϕ(x)dx = ∫_{ℝ^{n}} Qf(x)Δϕ(x)dx
= ∫_{ℝ^{n}} ∫_{ℝ^{n}} Γ(x, y)f(y)dyΔϕ(x)dx

= ∫_{ℝ^{n}} ∫_{ℝ^{n}} Γ(x, y)Δϕ(x)dxf(y)dy
= ∫_{ℝ^{n}} ϕ(y)f(y)dy
= ∫_{ℝ^{n}} f(x)ϕ(x)dx

Now we have that Qf is a weak solution and is C^{∞}, so elliptic regularity tells us that it is an actual solution. Thus our Green’s function is G(x, y) = Γ(x, y).

Can we weaken the smoothness conditions?

4. The Green’s Function for Δ_{g} on a Compact Manifold (M, g). We can now study the Green’s function on a Riemannian manifold (M, g). Recall that this means that we would like to find a function G(x, y) such that the function Qf(x) = ∫_{M} G(x, y)f(y)dV(y) is the inverse of the Laplacian, i.e. ΔQf(x) = f(x) for any f ∈ L^{2}(M) and also that QΔu(x) = u(x) for all u ∈ H_{2}(M). Unfortunately, this is not entirely plausible, since it would mean that Δ : H_{2}(M) → L^{2}(M) is injective, which is not true since the constants are in the kernel (note that this is not true in the case ℝ^{n} since the constants are not integrable). Thus we want

ΔQf(x) = f(x) − \frac{(f, 1)}{(1, 1)}1 = f(x) − \frac{1}{V} ∫_{M} f(y)dV(y)

where V is the volume of M, which means that ΔQf(x) is the projection of f(x) onto the orthogonal complement of the constant functions.

We first take a function η(x, y) = \overline{η}(d_{g}(x, y)) where d is the distance and \overline{η} is a function with compact support within the injectivity radius and which is identically 1 in a neighborhood of zero. Let our first approximation be H(x, y) = ω_{n−1}d(x, y)^{2−n}η(x, y). Notice that H is symmetric in the two variables. We want to mimic our approach to ℝ^{n}, so let’s compute Δ_{distr(y)}H(x, y). First we will need some estimates.

5. Estimates on Δ_{y}H(x, y). We first derive the formula for the Laplacian of a radial function. Recall that in polar coordinates, we can write the metric as

g = dr^{2} + r^{n−1}g_{ij}dθ^{i}dθ^{j}

We then compute in polar coordinates, letting \sqrt{g} = \sqrt{det [g_{ij}]}:

Δf(r) = \frac{1}{r^{n−1}\sqrt{g}} ∂_{i}r^{n−1}\sqrt{g}g^{ij}∂_{j}f(r)
= \frac{1}{r^{n−1}\sqrt{g}} ∂_{r}r^{n−1}\sqrt{g}∂_{r}f(r)
= f′′(r) + \frac{n − 1}{r}f′(r) + f(r)∂_{r} log \sqrt{g}

We now apply this to H(x, y) = \frac{1}{(2−n)ω_{n−1}} d(x, y)^{2−n}η(d(x, y)). So H(r) = \frac{1}{(n−2)ω_{n−1}} r^{2−n}η(r),

∂_{r}H(r) = ∂_{r} \frac{1}{(n − 2)ω_{n−1}} r^{2−n}η(r)
= \frac{1}{(2 − n)ω_{n−1}} [(2 − n)r^{1−n}η(r) + r^{2−n}η′(r)]

and

∂_{r}^{2}H(r) = \frac{1}{(2 − n)ω_{n−1}} [(2 − n)(1 − n)r^{^{−}n}η(r) + (2 − n)r^{1−n}η′(r) + (2 − n)r^{1−n}η′(r) + r^{2−n}η′′(r)]

so if you fix x, we find that find that, letting r = d(x, y):

Δ_{y}H(x, y) = ΔH(r)
= \frac{1}{(2 − n)ω_{n−1}} [(2 − n)(1 − n)r^{^{−}n}η(r) + (2 − n)r^{1−n}η′(r) + (2 − n)r^{1−n}η′(r) + r^{2−n}η′′(r)]

+\frac{n − 1}{r} \frac{1}{(2 − n)ω_{n−1}} [(2 − n)r^{1−n}η(r) + r^{2−n}η′(r)]

+\frac{1}{(2 − n)ω_{n−1}} ∂_{r} log \sqrt{g}r^{2−n}η(r)

= \frac{η′(r)}{ω_{n−1}} (2 + \frac{n − 1}{2 − n}) r^{1−n}

\frac{1}{(2 − n)ω_{n−1}} (η′′(r) + ∂_{r} log \sqrt{g}η(r)) r^{2−n}

6. Finding the Green’s Function. We need

∫_{M\B_{ϵ}} div(∇_{y}H(x, y)ϕ(y))dV(y) − ∫_{M\B_{ϵ}} div(H(x, y)∇ϕ(y))dV(y)
= ∫_{∂B_{ϵ}} ∇_{y}H(x, y) · ν(y)ϕ(y)dV(y) − ∫_{∂B_{ϵ}} H(x, y)∇ϕ(y) · ν(y)dV(y)

The left side is ∫_{M\B_{ϵ}} Δ_{y}H(x, y)ϕ(y)dV(y) − ∫_{M\B_{ϵ}} H(x, y)Δϕ(y))dV(y). As for the right side, similar arguments (Do it!!!!!) to the above case show that it goes to ϕ(x) as ϵ→ 0. Thus we find that

∫_{M} H(x, y)Δϕ(y))dV(y) = ϕ(x) + ∫_{M} Δ_{y}H(x, y)ϕ(y)dV(y)

or Δ_{distr(y)}H(x, y) = δ_{x}(y) + Δ_{y}H(x, y).

Now, if H(x, y) were the Green’s function, then we would just form the fundamental solution Q_{1}f(x) = ∫_{M} H(x, y)f(y)dV(y). It is not, however, which we see by computing ΔQ_{1}f(x). Again, let ϕ ∈ C_{c}^{∞}(ℝ^{n}) and we compute:

∫_{M} Q_{1}f(x)Δϕ(x)dV(x) = ∫_{M} ∫_{M} H(x, y)f(y)dV(y)Δϕ(x)dV(x)
= ∫_{M} ∫_{M} H(x, y)Δϕ(x)dV(x)f(y)dV(y)
= ∫_{M} [ϕ(y) + ∫_{M} Δ_{x}H(x, y)ϕ(x)dV(x)] f(y)dV(y)
= ∫_{M} [f(x) + ∫_{M} Δ_{x}H(x, y)f(y)dV(y)] ϕ(x)dV(x)

So we get that ΔdistrQ_{1}f(x) = f(x) + ∫_{M} Δ_{x}H(x, y)f(y)dV(y).

Thus we need to understand the regularity of ∫_{M} Δ_{x}H(x, y)f(y)dV(y) and change our operator Q_{1} to get the fundamental solution. We will try to find a new operator Q_{2} so that Δ (Q_{1} + Q_{2}) = f(x). To do this, we simply need to solve Δu = ^{−} ∫_{M} Δ_{x}H(x, y)f(y)dV(y) weakly.

Now, if we had that f_{2}(x) = ^{−} ∫_{M} Δ_{x}H(x, y)f(y)dV(y) is in L^{2}(M) and that it integrates to zero, then we

Let’s follow the same program we did before. We want to solve Δu = f_{2}. Let Q_{2}f(x) = ∫_{M} H(x, y)f_{2}(y)dV(y). We now check to see how close this is to the solution we want. By the last calculation we see that we get

ΔdistrQ_{2}f(x) = f_{2}(x) + ∫_{M} Δ_{x}H(x, y)f_{2}(y)dV(y)
= f_{2}(x) − ∫_{M} Δ_{x}H(x, y) ∫_{M} Δ_{y}H(y, z)f(z)dV(z)dV(y)
= f_{2}(x) − ∫_{M} ∫_{M} Δ_{x}H(x, y)Δ_{y}H(y, z)dV(y)f(z)dV(z)

So we find that

Δdistr(Q_{1} + Q_{2})f(x) = f(x) − ∫_{M} ∫_{M} Δ_{x}H(x, y)Δ_{y}H(y, z)dV(y)f(z)dV(z)

We can, of course, continue this course of action indefinitely.

Now, we look at Q_{2}:

Q_{2}f(x) = ∫_{M} H(x, y)f_{2}(y)dV(y)
= ^{−} ∫_{M} H(x, y) ∫_{M} Δ_{y}H(y, z)f(z)dV(z)dV(y)
= ^{−} ∫_{M} ∫_{M} H(x, y)Δ_{y}H(y, z)dV(y)f(z)dV(z)

Thus our second approximation to the Green’s function is

G_{2}(x, y) = H(x, y) − ∫_{M} H(x, z)Δ_{z}H(z, y)dV(z)

Now, if we could solve Δ_{distr(x)}F (x, y) = R_{2} = ∫_{M} Δ_{x}H(x, z)Δ_{z}H(z, y)dV(z) where R_{2} is continuous, then we would take G(x, y) = G_{2}(x, y) + F (x, y) and we would be done. Unfortunately, R_{2} is not necessarily continuous, so we continue until we become continuous using the following lemma.

Lemma 6.1. Let F (x, y) = ∫_{M} G(x, z)H(z, y)dV(z) and suppose that |G(x, z)| ≤ Const· d(x, z)^{a−n} and |H(z, y)| ≤ Const · d(z, y)^{b−n}, where 0 < a, b < n, then

|F (x, y)| ≤ \{ \begin{matrix}Const · d(x, y)^{a+b−n} & if a + b < n \\ Const · (1 + |log d(x, y)|) & if a + b = n \\ Const & if a + b > n \end{matrix} \}

Proof: Let d = d(x, y)/2. We now compute the integral in 3 parts:

∫_{m} = ∫_{B_{d}(x)} + ∫_{B_{3d}(y)\B_{d}(x)} + ∫_{M\B_{3d}(y)}

Now we compute separately:

|∫_{B_{d}(x)} G(x, z)H(z, y)dV(z)| ≤ ∫_{B_{d}(x)} |G(x, z)H(z, y)| dV(z)
≤ Const ∫_{B_{d}(x)} d(x, z)^{a−n}d(z, y)^{b−n}dV(z)
≤ Const · d^{b−n} ∫_{B_{d}(x)} d(x, z)^{a−n}dV(z)
= Const · d^{b−n} ∫_{S^{n−1}} ∫_{0}^{d} r^{a−n}r^{n−1}drdθ
= Const · d^{b−n}Vol(S^{n−1})\frac{1}{a}d^{a}
= Const · d^{a+b−n}

where the constant depends on a and n.

|∫_{B_{3d}(y)\B_{d}(x)} G(x, z)H(z, y)dV(z)| ≤ ∫_{B_{3d}(y)\B_{d}(x)} |G(x, z)H(z, y)| dV(z)
≤ Const ∫_{B_{3d}(y)\B_{d}(x)} d(x, z)^{a−n}d(z, y)^{b−n}dV(z)
≤ Const · d^{a−n} ∫_{B_{3d}(y)\B_{d}(x)} d(z, y)^{b−n}dV(z)
≤ Const · d^{a−n} ∫_{S^{n−1}} ∫_{0}^{3d} r^{b−n}r^{n−1}drdθ
= Const · d^{a−n}Vol(S^{n−1})\frac{1}{b}3^{b}d^{b}
= Const · d^{a+b−n}

where the constant depends on b and n. And finally,

|∫_{M\B_{3d}(y)} G(x, z)H(z, y)dV(z)| ≤ ∫_{M\B_{3d}(y)} |G(x, z)H(z, y)| dV(z)
≤ Const ∫_{M\B_{3d}(y)} d(x, z)^{a−n}d(z, y)^{b−n}dV(z)
≤ Const ∫_{M\B_{3d}(y)} (d(z, y) − 2d)^{a−n} d(z, y)^{b−n}dV(z)
= Const ∫_{S^{n−1}} ∫_{3d}^{K} (r − 2d)^{a−n} r^{b−n}r^{n−1}drdθ
≤ Const ∫_{3d}^{K} (r − 2d)^{a+b−2n} r^{n−1}dr

Now, if we change variables to s = r − 2d we get

∫_{d}^{K−2d} s^{a+b−2n}(s + 2d)^{n−1}ds ≤ ∫_{d}^{K−2d} s^{a+b−2n} [(2s)^{n−1} + (4d)^{n−1}] ds
= ∫_{d}^{K−2d} [s^{a+b−n−1} + (4d)^{n−1}s^{a+b−2n}] ds

The second term is

(4d)^{n−1} \frac{1}{a + b − 2n + 1} [(K − 2d)^{a+b−2n+1} − d^{a+b−2n+1}]

where the constant depends on . The third inequality follows from the fact that d(z, y) − 2d ≤ d(x, z).

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7. Axiomatic approach to the Green’s function. It may be easier to understand the derivation by showing the properties that we require of our Green’s Function. We shall need the following:

Theorem 7.1. If we can find a function G(x, y) such that

1. Δ_{distr(y)}G(x, y) = δ_{x}(y) − \frac{1}{V}

2. Δ_{y}G(x, y) = 0

3. G(x, y) ∼ d(x, y)^{2−n}

Then G(x, y) is the Green’s Function for the Laplacian, i.e. if we define Qf(x) = ∫_{M} G(x, y)f(y)dV(y) then

• ΔQf(x) = f(x) − \frac{1}{V} ∫_{M} f(y)dV(y) and

• QΔf(x) = f(x) − \frac{1}{V} ∫_{M} f(y)dV(y) for appropriate f.

Proof: Let us just check to see if QΔf(x) = f(x)− \frac{1}{V} ∫_{M} f(y)dV(y). We first look weakly.

(QΔf(x), ϕ(x)) = ∫ QΔf(x)ϕ(x)dV(y)
= ∫ ∫ G(x, y)Δf(y)dV(y)ϕ(x)dV(x)
= ∫ Δ_{distr(y)}G(x, y)f(y)dV(y)ϕ(x)dV(x)
= ∫ (f(x) − \frac{1}{V} ∫ f(y)dV(y)) ϕ(x)dV(x)

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Proof: Although we have already done the proof, let’s do it again for old time’s sake. Let’s first compute ΔdistrQf(x). We consider the divergence theorem on M \ B_{ϵ}(x):

∫_{M\B_{ϵ}(x)} div_{y}(∇_{y}G(x, y)ϕ(y))dV(y) − ∫_{M\B_{ϵ}(x)} div_{y}(G(x, y)∇ϕ(y))dV(y)
= ∫_{∂B_{ϵ}(x)} ∇_{y}G(x, y) · ν(y)ϕ(y)dV(y) − ∫_{∂B_{ϵ}(x)} G(x, y)∇ϕ(y) · ν(y)dV(y)

and the left hand term is

∫_{M\B_{ϵ}(x)} Δ_{y}G(x, y)ϕ(y)dV(y) − ∫_{M\B_{ϵ}(x)} G(x, y)Δϕ(y)dV(y)
= − ∫_{M\B_{ϵ}(x)} G(x, y)Δϕ(y)dV(y)

because of property 2.

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