- #1
Peeter
- 303
- 3
Am just playing around, and
following examples of Fourier transform solutions of the heat equation, tried the same thing for
the electrostatics Poisson equation
<br /> \nabla^2 \phi &= -\rho/\epsilon_0 \\<br />
With Fourier transform pairs
<br /> \begin{align*}<br /> \hat{f}(\mathbf{k}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 x \\<br /> {f}(\mathbf{x}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 k \\<br /> \end{align*}<br />
one gets
<br /> \begin{align*}<br /> \phi(\mathbf{x}) &= \frac{1}{\epsilon_0} \int \rho(\mathbf{x}') G(\mathbf{x-x'}) d^3 x' \\<br /> G(\mathbf{x}) &= \frac{1}{(2 \pi)^3} \iiint \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k<br /> \end{align*}<br />
Now it seems to me that this integral G only has to be evaluated around a small neighbourhood of the origin. For example if one evaluates one of
the
integrals
<br /> \int_{-\infty}^\infty \frac{1}{{k_1}^2 + {k_2}^2 + {k_3}^3 } e^{ i k_1 x_1 } dk_1 <br />
using a an upper half plane contour the result is zero unless k_2 = k_3 = 0. So one is left with something loosely like
<br /> G(\mathbf{x}) &= \lim_{\epsilon \rightarrow 0} \frac{1}{(2 \pi)^3} <br /> \int_{k_1 = -\epsilon}^{\epsilon} dk_1<br /> \int_{k_2 = -\epsilon}^{\epsilon} dk_2<br /> \int_{k_3 = -\epsilon}^{\epsilon} dk_3<br /> \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} } <br />
However, from electrostatics we also know that the solution to the Poission equation means that G(\mathbf{x}) = \frac{1}{4\pi\lvert{\mathbf{x}}\rvert}.
Does anybody know of a technique that would reduce the integral limit expression above for G to the 1/x form? Am thinking something residue related, but I'm a bit rusty with my complex variables and how exactly to procede isn't obvious.
following examples of Fourier transform solutions of the heat equation, tried the same thing for
the electrostatics Poisson equation
<br /> \nabla^2 \phi &= -\rho/\epsilon_0 \\<br />
With Fourier transform pairs
<br /> \begin{align*}<br /> \hat{f}(\mathbf{k}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 x \\<br /> {f}(\mathbf{x}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 k \\<br /> \end{align*}<br />
one gets
<br /> \begin{align*}<br /> \phi(\mathbf{x}) &= \frac{1}{\epsilon_0} \int \rho(\mathbf{x}') G(\mathbf{x-x'}) d^3 x' \\<br /> G(\mathbf{x}) &= \frac{1}{(2 \pi)^3} \iiint \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k<br /> \end{align*}<br />
Now it seems to me that this integral G only has to be evaluated around a small neighbourhood of the origin. For example if one evaluates one of
the
integrals
<br /> \int_{-\infty}^\infty \frac{1}{{k_1}^2 + {k_2}^2 + {k_3}^3 } e^{ i k_1 x_1 } dk_1 <br />
using a an upper half plane contour the result is zero unless k_2 = k_3 = 0. So one is left with something loosely like
<br /> G(\mathbf{x}) &= \lim_{\epsilon \rightarrow 0} \frac{1}{(2 \pi)^3} <br /> \int_{k_1 = -\epsilon}^{\epsilon} dk_1<br /> \int_{k_2 = -\epsilon}^{\epsilon} dk_2<br /> \int_{k_3 = -\epsilon}^{\epsilon} dk_3<br /> \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} } <br />
However, from electrostatics we also know that the solution to the Poission equation means that G(\mathbf{x}) = \frac{1}{4\pi\lvert{\mathbf{x}}\rvert}.
Does anybody know of a technique that would reduce the integral limit expression above for G to the 1/x form? Am thinking something residue related, but I'm a bit rusty with my complex variables and how exactly to procede isn't obvious.
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