0
$\begingroup$

Suppose you are presented with the equation ($D=3$) \begin{equation} \nabla^2 A(x) = \nabla^2 B(x). \end{equation} Decompose $A$ and $B$ into their Fourier components, \begin{equation} A(x) = \int d^3k \ e^{ikx} \tilde A(k),\qquad B(x) = \int d^3k\ e^{ikx} \tilde B(k), \end{equation} to yield \begin{equation} \int d^3k \ e^{ikx}\left[-k^2 (\tilde A - \tilde B)\right]=0. \end{equation} Assuming the inverse Fourier transform of zero is uniquely zero (or based on the answers Fourier Transforming the Klein Gordon Equation), we conclude \begin{equation} \tilde A = \tilde B \quad\Rightarrow\quad A(x) = B(x). \end{equation} Yet we know $A(x)$ may vary up to arbitrary harmonic functions.

What gives?

Also consider this:
\begin{equation} 0 = \int d^3k\ e^{ikx} (k^2)^n \delta (k),\qquad\text{for}\qquad n>0. \end{equation}

A closely related problem may be posed simply as \begin{equation} \nabla^2 A(x) = 0 \quad\Rightarrow\quad -k^2 \tilde A(k) = 0\quad\Rightarrow\quad \tilde A(k) = 0\quad\Rightarrow\quad A(x) = 0. \end{equation}

$\endgroup$
4
  • 1
    $\begingroup$ $xf(x)=0$ does not imply $f(x)=0$. It implies that $f(x)=\sum_i c_i\delta^{(i)}(x)$ for some coefficients $c_i$. $\endgroup$ Jun 12, 2018 at 19:48
  • $\begingroup$ @AccidentalFourierTransform what does the superscipt $i$ on the delta function represent? Yes I was thinking that $f(x) = g(x)\delta(x)$ provided $xg(x)|_0 = 0$. $\endgroup$
    – Lone Wolf
    Jun 12, 2018 at 20:11
  • $\begingroup$ $f^{(i)}(x)=\frac{\mathrm d^i}{\mathrm dx^i}f(x)$. $\endgroup$ Jun 12, 2018 at 20:12
  • $\begingroup$ @AccidentalFourierTransform I see. Do you know where else I can read about decompositions such as $f(x) = \sum_i c_i \delta^{(i)}(x)$ ? Also it seems the expression for $f(x)$ is not exactly unique. $\endgroup$
    – Lone Wolf
    Jun 12, 2018 at 21:22

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.