Greens functions and Existence

Representation formula

Definition 1. Let $U \subseteq \R^d$ be a $C^2$ domain. We say $G$ is the Greens’ function of $U$ if:

  1. For every $x \in U$, $H_x(y) \defeq G(x, y) - \Gamma(x - y)$ extends to a harmonic function in $U$.
  2. For all $x \in U$, $y \mapsto G(x, y)$ is continuous on $\bar U - \set{x}$, and vanishes on $\partial U$.

Remark 2. If $U$ is only a $C^1$ domain, the Greens function may not be differentiable in the classical sense on $\partial U$.

Remark 3. Physically this means $G(x, y)$ is the steady state temperature at $y$ if you put a point source of heat at $x$ and maintain the boundary temperature at $0$.

Proposition 4. If $U \subseteq \R^d$ is a bounded $C^2$ domain, $G$ the Greens function of $U$ and $u \in C^2(\bar U)$ solves the Dirichlet problem \begin{align} \label{e:D1}\tag{D1} -\lap u &= f &&\text{in } U\,,\\ \label{e:D2}\tag{D2} u &= g &&\text{on } \partial U\,, \end{align} for some $f\in C(U)$, $g \in C(\partial U)$, then $$\label{e:G}\tag{G} u(x) = \int_U f(y) G(y, x) \, dy - \int_{\partial U} g(y) \grad_y G(y, x) \cdot \hat n(y) \, dS(y) \,. $$

Lemma 5. If $f$ is continuous at $x$, then $$ \lim_{\epsilon \to 0} \int_{\partial B(x, \epsilon)} f(y) \Gamma(x - y) \, dS(y) = 0 \quad\text{and}\quad \lim_{\epsilon \to 0} \int_{\partial B(x, \epsilon)} f(y) \grad \Gamma(x - y) \cdot \hat n \, dS(y) = -f(x) \,, $$ where $\hat n$ is the outward pointing unit normal.

Proof sketch of Proposition 4. Compute \begin{align} \int_{U - B(x, \epsilon)} &(u(y) \lap_y G(x, y) - G(x, y) \lap_y u(y) ) \\ &= \int_{\partial (U - B(x, \epsilon))} (u(y) \grad_y G(x, y) - G(x, y) \grad_y u(y) ) \cdot \hat n(y) \, dS(y) \end{align}

Proposition 6 (Symmetry). If $x, y \in U$ then $G(x, y) = G(y, x)$.

Proof sketch. Let $x_1, x_2 \in U$, $x_1 \neq x_2$ set $u_i(y) = G(x_i, y)$, $B_\epsilon = B(x_1, \epsilon) \cup B(x_2, \epsilon)$, and compute $$ \lim_{\epsilon \to 0} \int_{U - B_\epsilon} (u_1 \lap u_2 - u_2 \lap u_1) \, dx \,. $$

Poisson formula in upper half space

Proposition 7. If $H = \set{(x’, t) \in \R^d \st x’ \in \R^{d-1}, t > 0 }$ is the upper half space. Then the Greens function on $H$ is given by $$\label{e:H}\tag{H} G(x, y) = \Gamma(y - x) - \Gamma(y - x^*)\,, $$ where $\Gamma$ is the Newton kernel, and $x^*$ is the reflection of $x$ on $x_d = 0$. That is, if $x = (x’, t) \in H$, then $x^* = (x’, -t) \in -H$.

Proposition 8. If $G$ is defined by \eqref{e:H}, $x = (x’, t) \in H$, $y = (y’, 0) \in \partial H$ then $$ \partial_d G(x, y) = P_t(x’ - y’)\,, $$ where $P_t$ is the Poisson kernel defined by $$ P_t(y’) \defeq \frac{2 t}{d V_d (\abs{y’}^2 + t^2)^{d/2} } \,. $$

Remark 9. Notice $$ P_t(y’) = \frac{1}{t^{d-1}} P_1\paren[\Big]{ \frac{y’}{t} } \,. $$

Proposition 10. If $g \in C(\R^{d-1}) \cap L^\infty(\R^{d-1})$, then $$\label{e:PH}\tag{PH} u(x’, t) = P_t * g $$ is harmonic on $H$, continuous on $\bar H$ and $u = g$ on $\partial H$.

Remark 11. Note that Proposition 10 needs proof! Proposition 10 only gives you the converse if you already have a solution). Proposition 10 also uses Greens theorem which requires the domain to be bounded.

If $g \in L^p(\R^{d-1})$ for some $p \in [1, \infty)$, then $u$ is still harmonic on $H$, but need not be continuous on $\bar H$. In this case it’s reasonable to ask how $u$ takes on the boundary conditions $g$.

Proposition 12. If $g \in L^p(R^{d-1})$ for some $p \in [1, \infty)$, and $u \colon H \to \R$ is defined by \eqref{e:PH}, then $u( \cdot, t) \to g(\cdot)$ in $L^p$ as $t \to 0$.

Proof sketch. Remember continuous functions are dense in $L^p$.

Poisson formula in a ball

Let $B_R = B(0,R) \subseteq \R^d$. Given $x \in B(0, R) - \set{0}$, define $x^* \in B(0, R)^c$ by $$ x^* = \frac{R x}{\abs{x}} \cdot \frac{1}{\abs{x / R}} = \frac{R^2x}{\abs{x}^2} \,. $$ This is called Kelvin inversion.

Proposition 13. The Green’s function on $B_R$ is given by $$ G(x,y) = \begin{cases} \displaystyle \Gamma(y-x) - \Gamma\paren[\Big]{\frac{\abs{x}}{R} (y - x^*) } & x \neq 0 \,, \\\displaystyle \Gamma(y) - \Gamma(R) & x = 0 \,. \end{cases} $$

Proposition 14. If $x \in B_R$ and $y \in \partial B_R$, then $$ \frac{\partial G}{\partial \hat n}(x,y) = -P(x,y), $$ where $\hat n$ is the outward pointing unit normal and $P$ is the Poisson kernel $$ P(x,y) = \frac{R^2-\abs{x}^2} {d V_d R \abs{x-y}^d}. $$

Lemma 15. The Poisson kernel in the ball is an approximate identity in the following sense:

  1. For any $x \in B_R$, $$ \int_{\partial B_R} P(x, y) \, dS(y) = 1 \,. $$
  2. For any $\delta > 0$, $y_0 \in \partial B_R$, $$ \lim_{x \to y_0} \int_{ \partial B_R \cap B(y_0, \delta)^c } P(x, y) \, dS(y) = 0 $$

Proposition 16 (Poisson formula). If $g \in C(\partial B_R)$, then the function $u$ defined by $$\label{e:PB}\tag{PB} u(x) = \int_{\partial B_R} P(x,y)g(y)\,dS(y) $$ is harmonic on $B_R$, continuous on $\bar B_R$, and satisfies $u = g$ on $\partial B_R$.

Remark 17. At the center of the ball, $$ P(0,y)=\frac{1}{dV_d R^{d-1}} =\frac{1}{\abs{\partial B_R}}. $$ Thus \eqref{e:PB} at $x = 0$ recovers the mean value property.

Proposition 18. If $p \in [1, \infty)$ and $g \in L^p(\partial B_R)$ and $u$ is defined by \eqref{e:PB} then $u$ is harmonic in $B_R$ and $$ \lim_{\rho \to 1} \int_{\partial B_R} \abs{u(\rho y) - g(y)}^p \, dS(y) = 0 \,. $$

Problem 19. Is there a harmonic function on $B(0, 1)$ such that for a dense set of $y \in \partial B(0, 1)$ we have $u(x) \to \infty$ as $x \to y$? Prove or disprove.