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 '18 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 '18 at 20:11
  • $\begingroup$ $f^{(i)}(x)=\frac{\mathrm d^i}{\mathrm dx^i}f(x)$. $\endgroup$ Jun 12 '18 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 '18 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.