On page 121 of Gilbarg-Trudinger's book (Elliptic PDE of second order) they have the following Green's function in $\mathbb{R}^n (n\geq 3)$: \begin{equation} G(x, y)=\Phi(y-x)-\Phi(y-\tilde{x})-2b_n\int_{0}^{\infty}e^{\ as}D_n\Phi(y-\tilde{x}+bs)\text{ d} s,\quad (x, y\in\mathbb{R}_{+}^{n}, x\neq y), \end{equation} where $\Phi$ is the fundamental solution of Laplace's equation. Here, $x=(x_1,\dots,x_{n-1}, x_n)\in\mathbb{R}_{+}^{n}$ and $\mathbb{R}_{+}^{n}=\{x\in\mathbb{R}^n\vert\ x_n>0\}$. The rest of notation is understood as $D_n=\frac{\partial}{\partial y_n}, \tilde{x}=(x_1,\dots, x_{n-1}, -x_n)$ and $a\leq 0$. Lastly, $b$ is a constant unit vector in $\mathbb{R}^n$ with $b_n>0$.

Next, they say that $G$ is clearly harmonic in $y$ for $y\neq x$.

I am a bit puzzled as to why this is the case.

I know that if $y\in\mathbb{R}_{+}^{n}$ then $\Delta\Phi (y-\tilde{x})=0$ and since $y\neq x$ then $\Delta\Phi (y-x)=0$ so if $G$ is harmonic then it must be because: \begin{equation} H(y)\equiv \int_{0}^{\infty}e^{\ as}D_n\Phi(y-\tilde{x}+bs)\text{ d} s \end{equation} is harmonic (in $y$).

I was thinking along the lines that since $\Phi$ is harmonic then $D_n\Phi$ is harmonic, so if I could show that: \begin{equation} \sum_{i=1}^{n} H_{y_iy_i}(y)=\int_{0}^{\infty}e^{\ as}\sum_{i=1}^{n}\Phi{y_ny_iy_i}(y-\tilde{x}+bs)\text{ d} s=0 \end{equation} then I'd be done.

However, I am not sure if I can or how to justify this step.

I am able to show that $H(y)$ is uniformly bounded. Would this help in anyway?

  • $\begingroup$ It's basically just a matter of justifying differentiation under the integral sign, isn't it? Then $\Delta H = \int_0^\infty e^as D_n (\Delta\Phi)(y-\tilde x + bs) \, ds$, but $\Delta \Phi(y-\tilde x + bs) = 0$ as long as $y-\tilde x + bs\ne 0$. $\endgroup$ – Sam Mar 22 '13 at 11:54
  • $\begingroup$ Yes, though in the above argument I mean $\Delta D_n\Phi (y-\tilde{x}+bs)$. I know that $y_n+x_n+b_ns>0$ so $y-\tilde{x}+bs\neq 0$, but I am not sure how to justify this differentiation. $\endgroup$ – Nirav Mar 23 '13 at 1:43

Maybe something like this is ok.

Define: \begin{equation} F_m(s)\equiv e^{\ as}\frac{D_n\Phi (y+\frac{e_i}{m}-\tilde{x}+bs)-D_n\Phi (y-\tilde{x}+bs)}{\frac{1}{m}}, \end{equation} where $e_i$ is the vector of zeros everywhere except in the $i^{\text{ th}}$ slot. Then we see that $F_m$ is measurable and converges pointwise to: \begin{equation} F(s)\equiv e^{\ as}D_iD_n\Phi(y-\tilde{x}+bs). \end{equation} Now note that we have the following bound on the second derivatives of the fundamental solution: \begin{equation} \vert D^2\Phi(y-\tilde{x}+bs)\vert\leq C\vert y-\tilde{x}+bs\vert^{-n}\quad\forall \ y-\tilde{x}+bs\neq 0. \end{equation}

Since $y\in\mathbb{R}^{n}_{+}$ we have that: \begin{equation} \vert y-\tilde{x}+bs\vert^{-n}\leq(x_n+b_ns)^{-n} \end{equation} and the RHS is integrable wrt s.

For each $s\in [0, \infty)$ we have, \begin{equation} \vert F_m(s)\vert\leq \left|\frac{D_n\Phi (y+\frac{e_i}{m}-\tilde{x}+bs)-D_n\Phi (y-\tilde{x}+bs)}{\frac{1}{m}}\right|\leq \vert D^2\Phi (z)\vert, \end{equation}

where the RHS estimate follows by the mean value theorem for some $z$ in the line segment between $(y+\frac{e_i}{m}-\tilde{x}+bs)$ and $(y-\tilde{x}+bs)$. Thus, $F_m$ is bounded by a positive summable function, namely, $C(x_n+b_ns)^{-n}$ for each $s\in [0, \infty)$. As all the hypotheses for Lebesgue's Dominated Convergence Theorem are satisfied, we conclude that: \begin{equation} D_i H(y)=\lim_{m\rightarrow\infty}\int_{0}^{\infty} F_m(s) \text{ d}s=\int_{0}^{\infty}\lim_{m\rightarrow\infty} F_m(s) \text{ d}s=\int_{0}^{\infty}e^{\ as}D_iD_n\Phi(y-\tilde{x}+bs) \text{ d}s \end{equation}

We can follow a similar argument for $H_{y_iy_i}$ with the bound $\vert D^3\Phi (y-\tilde{x}+bs)\vert\leq C\vert y-\tilde{x}+bs\vert^{-n-1}$, from which the desired result follows.


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