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Consider the function $\psi(\bar x)$ is a solution to Poisson's equation $$\nabla^2 \psi(\bar x) = \rho (\bar x)$$

int the "quarter-space" defined by $V=\{(x,y,z): x\ge 0,y \ge 0, z \in R\}$ with the boundary conditon $\frac{\partial\psi}{\partial n}$= $f(\bar x)$ on $S_1=\{(0,y,z): y \ge 0, z \in R\}$ and $\frac{\partial\psi}{\partial n}$= $g(\bar x)$ on $S_2=\{(x,0,z): x \ge 0, z \in R\}$

Knowing that

$$G(\bar x', \bar x) = -\frac{1}{4\pi\mid\bar x'-x\mid}$$

is a Green's function for Poisson's equation in $R^3$, how would I use the method of Images to construct a Green's Function $G(\bar x', \bar x)$ for the quarter space.

I understand that the method images can be implemented to express $G(\bar x', \bar x)$ as

$$= \frac{1}{4\pi \sqrt{(x-x')^2 +(y-y')^2 +(z-z')^2 }} - \frac{1}{4\pi\sqrt{(x-x')^2 +(y-y')^2 +(z-z')}} $$

However I am not sure how take into account the boundary condition. With the dimensions given, am I right expressing Green's function as follows

$$= \frac{1}{4\pi \sqrt{(x)^2+ (y-y')^2 +(z-z')^2 }} - \frac{1}{4\pi\sqrt{(x)^2+(y-y')^2 +(z-z')}} + \frac{1}{4\pi \sqrt{(x-x')^2 +(y)^2+(z-z')^2 }} - \frac{1}{4\pi\sqrt{(x-x')^2+(y)^2 +(z-z')}} $$

Am I on the right track, or is there something which I am not understanding? Any information or advice would be greatly appreciated.

P.S. $R$ is a Rational number in this context (don't know the symbol for it)

KD10001
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1 Answers1

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The volume of interest is a quarter of the space as you defined it. You need a Green's function that vanishes on the 2 boundaries that you specificed. To interpret it, I'll use electrostatic notation (familiar to physicists - hope it will be also clear to you). If $$ G_1=+\frac1{4\pi|{\bf r}-{\bf r'}|} $$ is the potential due to a "unit charge" locates at ${\bf r'}=(x',y',z')$, then you need 3 more charges to make the total potential (the Green's function) vanish on $V$ - they are located symmetrically to the charge with respect to the 2 boundaries and have opposite charge: $$ G=G_1+G_2+G_3+G_4 $$ where: $$ G_2=-\frac1{4\pi|{\bf r}-{\bf r'}+2x'{\bf{\hat x}}|} $$ $$ G_3=-\frac1{4\pi|{\bf r}-{\bf r'}+2y'{\bf{\hat y}}|} $$ $$ G_4=+\frac1{4\pi|{\bf r}-{\bf r'}+2x'{\bf{\hat x}}+2y'{\bf{\hat y}}|} $$ where $\bf{\hat x},\bf{\hat y}$ are unit vectors along the respective axes. You should check $G$ vanishes on the boundary, that is: $$ G(0,y,z)=G(x,0,z)=0 $$

am301
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