Newtonian Potential

— 1. Definition of Newtonian Potential —

Definition 1 Let {\Omega} be a bounded open set, we define the Newtonian potential of {f} is the function {N_f} on {{\mathbb R}^n} by

\displaystyle N_f(x) = \int_\Omega \Gamma(x-y) f(y)\, dy,

where {\Gamma} is the fundamental solution of Laplace’s equation.

We need to check if {N_f } is well-defined. If {f \in L^\infty(\Omega)}, then one can easily see {N_f \in L^\infty ({\mathbb R}^n)}. We can further relax {f} to be in {L^p(\Omega)}, where {p > \frac{n}{2}}. Since the Holder conjugate {p' < \frac{n}{n-2}}, simply by Holder’s inequality, for any {x \in R^n}

\displaystyle |N_f(x)| = \left| \int_\Omega \Gamma(x-y) f(y) \, dy \right| \le C\| f \|_p \left( \int_\Omega \frac{1}{|x-y|^{(n-2)p'}} \, dy \right)^{\frac{1}{p'}} < \infty.

Therefore {N_f \in L^\infty ({\mathbb R}^n)} is well-defined in the whole space. If we only require {N_f} to be defined in {\Omega}, we only need {f} to be integrable plus a little bit local boundedness near the singularity {x}, for example, local Holder continuity or so. In this case, we can see {N_f} is also well defined outside {\Omega}, but not defined on {\partial \Omega}. So this case will become less interesting when we try to solve Dirichlet problem through Newtonian potential, since the function on boundary needs to be defined.

— 2. Regularity of Newtonian Potential —

Theorem 2 (Representation for first derivative) Let {f} be bounded and integrable in {\Omega}. Then {N_f \in C^1({\mathbb R}^n)} and for any {x \in {\mathbb R}^n},

\displaystyle D_i N_f(x) = \int_{\Omega} D_i \Gamma(x-y) f(y)\, dy, \quad i = 1,2, \cdots, n.

Proof: Define a function

\displaystyle v(x):= \int_\Omega D_i \Gamma(x-y) f(y)\, dy.

Since {|D_i \Gamma(x)| \le C|x|^{1-n}}, we know {v} is well-defined. We want to show {v = D_i N_f}.We fix a function {\eta} in {C^1({\mathbb R})} satisfying {0 \le \eta \le 1, 0 \le \eta' \le 2, \eta(t) = 0} for {t \le 1}, {\eta(t) =1} for {t \ge 2}. And define for {\varepsilon >0} small,

\displaystyle N_f^\varepsilon(x) = \int_{\Omega}\Gamma(x-y) \eta_\varepsilon f(y) \, dy,

where {\eta_\varepsilon = \eta (|x-y| / \varepsilon)}. Clearly {N_f^\varepsilon \in C^1({\mathbb R}^n)} since we make the singularity vanished. And we have

\displaystyle |v(x)- D_i N_f^\varepsilon(x)| \le \int_{|x-y| \le 2\varepsilon} D_i [(1-\eta_\varepsilon) \Gamma(x-y)] |f(y)| \, dy\le C\|f\|_{L^\infty} \cdot \varepsilon.

Therefore {D_i N_f^\varepsilon \rightarrow v} uniformly in compact subsets of {{\mathbb R}^n} as {\varepsilon \rightarrow 0}. One can also easily see that {N_f^\varepsilon \rightarrow N_f} uniformly as {\varepsilon \rightarrow 0}. Therefore, {N_f \in C^1} and

\displaystyle D_i N_f(x) =v(x)= \int_{\Omega} D_i \Gamma(x-y) f(y)\, dy.

\Box

Theorem 3 (Representation for second derivative) Let {f \in C_{loc}^\alpha(\Omega)}, with {\alpha \le 1}, and also bounded. Then {N_f \in C^2(\Omega)}, {\Delta N_f = f} in {\Omega}, and for any {x \in \Omega},

\displaystyle D_{x_i x_j} N_f(x) = \int_{\Omega_0} D_{x_i x_j} \Gamma(x-y) (f(y)-f(x)) \, dy - f(x) \int_{\partial \Omega_0} D_{x_i} \Gamma \nu_j(y) \, dS_y,

where {\Omega_0} is any domain containing {\Omega} for which the divergence theorem holds and {f} is extended to be 0 outside {\Omega}.

Proof: The proof follows from the same strategy as the previous theorem. We define {u(x)} to be the RHS of the equation stated in theorem. Since {f} is locally Holder continuous, the integrand is integrable near the singularity, hence {u} is well-defined. Let {\eta_\varepsilon} be the same as defined in last theorem, then define

\displaystyle v_\varepsilon = \int_{\Omega} D_{x_i} \Gamma(x-y) \eta_\varepsilon f(y) \, dy.

We will have

\displaystyle D_{x_j} v_\varepsilon (x) = \int_\Omega D_{x_j} (D_{x_i} \Gamma \eta_\varepsilon) f(y) \, dy

\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy + f(x)\int_{\Omega_0} D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) \, dy

\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy + f(x)\int_{\Omega_0} D_{x_j x_i} \Gamma \eta_\varepsilon + D_{x_i} \Gamma D_{x_j} \eta_\varepsilon \, dy

\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy - f(x)\int_{\Omega_0} D_{y_j x_i} \Gamma \eta_\varepsilon + D_{x_i} \Gamma D_{y_j} \eta_\varepsilon \, dy

\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy - f(x)\int_{\partial \Omega_0} D_{x_i} \Gamma \nu_j(y) \, dS_y,

provided {2\varepsilon < dist(x, \partial \Omega)}. Hence, by subtraction

\displaystyle |u(x) - D_{x_j} v_\varepsilon (x)| = \left| \int_{|x-y| \le 2 \varepsilon} D_{x_j}[(1-\eta_\varepsilon)D_{x_i}\Gamma] (f(y) -f(x)) \, dy \right|

\displaystyle \le \left(\frac{n}{\alpha} + 4 \right) [f]_{\alpha;x} (2\varepsilon)^\alpha.  

Therefore {D_{x_j} v_\varepsilon} converges to {u} uniformly on any compact subset of {\Omega} as {\varepsilon \rightarrow 0}. One can also easily see that {v_\varepsilon} converges uniformly to {D_i N_f} in {\Omega}. Therefore {N_f \in C^2(\Omega)} and {u = D_{x_i x_j} N_f}.Finally, setting {\Omega_0 = B_R(x)}, we have for sufficiently large {R},

\displaystyle \Delta w = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy - f(x) \int_{\partial B_R(x)} D_{x_i} \Gamma \nu_i(y) \, dS_y

\displaystyle = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy + f(x) \int_{\partial B_R(x)} D_{y_i} \Gamma \nu_i(y) \, dS_y

\displaystyle = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy + \frac{f(x)}{n \omega_n R^{n-1}}\int_{\partial B_R(x)} \nu_i(y) \nu_i(y) \, d S_y

\displaystyle = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy + f(x).

But we can estimate

\displaystyle \left| \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy \right| = \left| \int_{|x-y| < \delta} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy \right|

\displaystyle \le \int_{|x-y| < \delta} \frac{1}{|x-y|^{n-\alpha}} \, dy \le \delta^\alpha,

for any {\delta > 0}. Therefore {\int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy = 0} and hence {\Delta w =f}. \Box

Remark 1 If the local Holder continuity of {f} is replaced by Dini continuity, we still have the same result. And this theorem suggests that we can use Newtonian potential to get a solution of Poisson equation.

Theorem 4 (interior Schauder estimate) If {f \in C^\alpha(\bar{\Omega})}, then {N_f \in C^{2,\alpha}(\bar{\Omega'})} for any {\Omega' \subset \subset \Omega}. Furthermore,

\displaystyle \|N_f\|_{C^{2,\alpha} (\bar{\Omega'})} \le C(n, \Omega ,\Omega') \|f\|_{C^\alpha (\bar{\Omega})}.

Proof: Without loss of generality we can assume {\Omega = B_2} and {\Omega' = B_1}. Take {x,z \in B_1}, from the representation we have, we know

\displaystyle D_{x_i x_j} N_f(x) - D_{x_i x_j} N_f(z)

\displaystyle = \int_{B_2} D_{x_i x_j} \Gamma(x-y) (f(y)-f(x)) - D_{x_i x_j} \Gamma(z-y) (f(y)-f(z)) \, dy

\displaystyle - f(x) \int_{\partial B_2} D_{x_i} \Gamma(x-y) \nu_j(y) \, dS_y + f(z) \int_{\partial B_2} D_{x_i} \Gamma(z-y) \nu_j(y) \, dS_y.

If we set {\delta = |x-z|}, {\xi = \frac{x+z}{2}.} We can split the RHS into 6 parts, namely,

\displaystyle D_{x_i x_j} N_f(x) - D_{x_i x_j} N_f(z) = I_1 + I_2 + I_3 + I_4 + I_5 + I_6,

where

\displaystyle I_1 = \int_{B_\delta (\xi)} D_{x_i x_j} \Gamma(x-y) (f(y)-f(x)) \, dy;

\displaystyle I_2 = \int_{B_2 - B_\delta(\xi)} D_{x_i x_j} \Gamma(x-y) (f(z)-f(x)) \, dy;

\displaystyle I_3 = -\int_{B_\delta (\xi)} D_{x_i x_j} \Gamma(z-y) (f(y)-f(z)) \, dy;

\displaystyle I_4 = \int_{B_2 - B_\delta(\xi)} [D_{x_i x_j} \Gamma(x-y) - D_{x_i x_j} \Gamma(z-y)] (f(y)-f(z)) \, dy;

\displaystyle I_5 = [f(z) - f(x)] \int_{\partial B_2} D_{x_i} \Gamma(z-y) \nu_j(y) \, dS_y;

\displaystyle I_6 = f(x) \int_{\partial B_2} [D_{x_i} \Gamma(z-y) - D_{x_i} \Gamma(x-y)] \nu_j(y) \, dS_y.

We will estimate all these 6 terms. Since {B_\delta(\xi) \subset B_{2 \delta}(x)}, we have

\displaystyle |I_1| \le C \int_{B_{2 \delta}(x)} \frac{1}{|x-y|^n} |f(y) - f(x)| \, dy \le C \delta^\alpha.

And {I_3} should have a same estimate by this argument. To estimate {I_2}, notice that if {y \in \partial B_2}, {|x-y|>1}, and if {y \in \partial B_\delta(\xi)}, {|x-y|>\frac{\delta}{2}}. And we will have by divergent theorem

\displaystyle |I_2| \le |f(x) - f(z)| \left( \int_{\partial B_2} |D_{x_i} \Gamma(x-y) \nu_j(y)| \, dS(y) +\int_{\partial B_\delta(\xi)} |D_{x_i} \Gamma(x-y) \nu_j(y)| \, dS(y) \right)

\displaystyle \le \left( \int_{\partial B_2} 1 \, dS(y) + \int_{\partial B_\delta(\xi)} \left(\frac{\delta}{2} \right)^{1-n} \, dS(y) \right) \delta^\alpha \le C \delta^\alpha.

To estimate {I_4}, note that {D_{x_i x_j} \Gamma(x-y)} is smooth when {y \in B_2 - B_\delta(\xi)}, by mean value theorem, we know {|D_{x_i x_j} \Gamma(x-y) - D_{x_i x_j} \Gamma(z-y)| \le |D D_{x_i x_j} \Gamma(\hat{x}-y)| |x-z|} for some {\hat{x}} between {x} and {z}. And for {y \in B_2 - B_\delta(\xi)}, we always have {|\hat{x} - y| \ge \frac{|y-z|}{3}}. Then we will have

\displaystyle |I_4| \le \int_{B_2 - B_\delta(\xi)} |D D_{x_i x_j} \Gamma(\hat{x}-y)| |x-z| |f(y)-f(z)| \, dy

\displaystyle \le C \delta \int_{B_2 - B_0.5\delta(z)} \frac{1}{|z-y|^{n+1-\alpha}} \, dy \le C \delta \int_{\frac{1}{2}\delta}^\infty \frac{1}{r^{2-\alpha}} \, dr \le C \delta^\alpha.

Since {|z - y| < 1} on {y \in \partial B_2}, we have trivial estimate for {I_5}:

\displaystyle |I_5| \le C|f(x) - f(z)| \le C \delta^\alpha.

To estimate {I_6}, we again have mean value theorem as in estimate of {I_4}, and since {DD_{x_i} \Gamma(\hat{x}-y)} and {f} are bounded, we have

\displaystyle |I_6| \le C|x-z| \le C \delta^\alpha.

Combining these 6 estimates we can conclude {N_f in C^\alpha(\bar{B_1})}. From the previous theorems we can bounded the {L^\infty} norm of {N_f} and its gradient by {\|f\|_L^{\infty}}. Therefore we have the inequality

\displaystyle \|N_f\|_{C^{2, \alpha}(\bar{B_1})} \le C(n) \|f\|_{C^\alpha(\bar{B_2})}.

\Box

This tells us if {\Delta u = f} and {f \in C^\alpha}, then {u \in C^{2,\alpha}}. And the following example tells us if {f} only continuous, then {u} is not necessarily {C^2} ({\alpha} cannot be 0).

Remark 2 If {f} is only continuous, {u} is {C^1}, but not necessarily {C^2}.

An example: For {\alpha} with {|\alpha| =2}, let {P} be a homogeneous harmonic polynomial of degree 2 with {D^\alpha P \neq 0}. Choose {\eta \in C_0^\infty (\{ x: |x| < 2\})} with {\eta = 1} when {t_k = 2^k}. Define {f(x) = \sum_0^\infty \frac{1}{k} \Delta(\eta P)(t_k x).} Notice that {\Delta(\eta P) (t_k x) =0}, when {|t_k x|<1} or {|t_k x| > 2}. Therefore

\displaystyle f(x) = \begin{cases} \frac{1}{k} \Delta (\eta P)(2^k x), & \frac{1}{2^k} \le |x| < \frac{1}{2^{k-1}};\\ 0, & x =0. \end{cases}

And since {f} vanishes on the boundary of each annulus {\{ \frac{1}{2^k} \le |x| < \frac{1}{2^{k-1}} \}} and {\frac{1}{k} \rightarrow 0} as {k \rightarrow \infty}, we can easily see that {f} is continuous.

\noindent Now for {x} such that {\frac{1}{2^l} \le |x| < \frac{1}{2^{l-1}}} define {v(x): = \sum_{k=0}^l \frac{1}{2^{2k}} \frac{1}{k} (\eta P)(2^k x)}. Then by a similar argument we can see {v} is well-defined and smooth away from {0}. For any {x \neq 0}, we can find an {l} such that {\frac{1}{2^l} \le |x| < \frac{1}{2^{l-1}}}, and Notice that

\displaystyle D^\alpha v(x) = \sum_{k=0}^l \frac{1}{k} D^\alpha (\eta P)(2^k x) = \sum_{k=0}^{l-1} \frac{1}{k} D^\alpha P + \frac{1}{l} D^\alpha (\eta P).

Since {D^\alpha P} is not {0}, {l \rightarrow \infty} as {|x| \rightarrow 0}, we can see {v(x)} is not {C^2} around {x = 0}. Noticing that

\displaystyle v(x) = \sum_{k=0}^{l-1} \frac{1}{k} P(x) + \frac{1}{l} \eta(2^l x) P(x),

since {P} is homogeneous. We can see as {|x| \rightarrow 0}, {\frac{1}{l} \eta(2^l x) P(x) \rightarrow 0.} Also {|\sum_{k=0}^{l-1} \frac{1}{k} P(x)| \le l |P(x)| \le C x^2 \log |x| \rightarrow 0} as {|x| \rightarrow 0}, therefore {v(x)} stays bounded when {|x| \rightarrow 0}. Now suppose {u} is a {C^2} solution of {\Delta u = f}. Then {\Delta (u-v) = 0} for {x \neq 0}. Since {u-v} is bounded around 0, {0} is a removable singularity for {u - v}, which makes it a harmonic function. But this implies {v \in C^2}, which leads to a contradiction. So there is no {C^2} solutions in any neighborhood of the origin to {\Delta u = f}.

— 3. Solution to Dirichlet problem —

Theorem 5 Let {\Omega} be a bounded domain and suppose each point of {\partial \Omega} is regular (in the sense that Perron’s method will give a harmonic function prescribed continuous boundary value, e.g. {\partial \Omega} is {C^1}.) Then if {f} is bounded and locally Holder continuous, the classical Dirichlet problem : {\Delta u = f} in {\Omega}, {u = g} on {\partial \Omega} is uniquely solvable for any continuous boundary values {g}.

Proof: From Theorem 2 and 3, we know that {N_f \in C^1({\mathbb R}^n) \cap C^2(\Omega)}, {\Delta N_f = f} in {\Omega}. Consider {v = u - N_f}, then the Dirichlet problem is equivalent to {\Delta v = 0} in {\Omega}, {v = g - N_f} on {\partial \Omega}, which has a unique solution by Perron’s method. \Box

The regularity of {f} can be relaxed as following theorem.

Theorem 6 Let {\Omega} be as described above, {f \in L^p(\Omega)} for some {p > \frac{n}{2}} and is local Holder continuous then the classical Dirichlet problem can by uniquely solved.

Proof: From section 1 we know {N_f} is well-defined on {{\mathbb R}^n}. And we need to check {N_f \in C(\bar{\Omega})}. We fix a function {\eta} as in Theorem 2. And define for {\varepsilon >0} small,

\displaystyle N_f^\varepsilon(x) = \int_{\Omega}\Gamma(x-y) \eta_\varepsilon f(y) \, dy,

where {\eta_\varepsilon = \eta (|x-y| / \varepsilon)}. Clearly {N_f^\varepsilon \in C^1({\mathbb R}^n)}. Extending {f} to be 0 outside {\Omega}, then

\displaystyle |N_f(x)- N_f^\varepsilon(x)| \le \int_{|x-y| \le 2\varepsilon} (1-\eta_\varepsilon) \Gamma(x-y) |f(y)| \, dy \le C\|f\|_{p, B_{2\varepsilon} (x)}.

Hence {N_f^\varepsilon(x) \rightarrow N_f(x)} uniformly as {\varepsilon \rightarrow 0}, and thus {N_f(x) \in C({\mathbb R}^n)}. Then follow the proof of Theorem 2 one can see {w \in C^1(\Omega)}. Note that we might not have {w \in C^1({\mathbb R}^n)} anymore since we lose boundedness of {f} near the boundary of {\Omega}. Then we can follow the exact same proof of Theorem 3 and Theorem 8 to get the result. \Box

Remark 3 In order to solve Dirichlet problem, we can’t assume {f \in L^p(\Omega)} for some {p \le \frac{n}{2}}, otherwise {N_f} might not be well-defined on boundary.

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