Editing Dirac delta function (section)

===Plane wave decomposition===
One approach to the study of a linear partial differential equation
<math display="block">L[u]=f,</math>

where {{mvar|L}} is a [[differential operator]] on {{math|'''R'''<sup>''n''</sup>}}, is to seek first a fundamental solution, which is a solution of the equation
<math display="block">L[u]=\delta.</math>

When {{mvar|L}} is particularly simple, this problem can often be resolved using the Fourier transform directly (as in the case of the Poisson kernel and heat kernel already mentioned).  For more complicated operators, it is sometimes easier first to consider an equation of the form
<math display="block">L[u]=h</math>

where {{mvar|h}} is a [[plane wave]] function, meaning that it has the form
<math display="block">h = h(x\cdot\xi)</math>

for some vector {{mvar|ξ}}.  Such an equation can be resolved (if the coefficients of {{mvar|L}} are [[analytic function]]s) by the [[Cauchy–Kovalevskaya theorem]] or (if the coefficients of {{mvar|L}} are constant) by quadrature.  So, if the delta function can be decomposed into plane waves, then one can in principle solve linear partial differential equations.

Such a decomposition of the delta function into plane waves was part of a general technique first introduced essentially by [[Johann Radon]], and then developed in this form by [[Fritz John]] ([[#CITEREFJohn1955|1955]]).{{sfn|Courant|Hilbert|1962|loc=§14}} Choose {{mvar|k}} so that {{math|''n'' + ''k''}} is an even integer, and for a real number {{mvar|s}}, put
<math display="block">g(s) = \operatorname{Re}\left[\frac{-s^k\log(-is)}{k!(2\pi i)^n}\right]
=\begin{cases}
\frac{|s|^k}{4k!(2\pi i)^{n-1}}   &n \text{ odd}\\[5pt]
-\frac{|s|^k\log|s|}{k!(2\pi i)^n}&n \text{ even.}
\end{cases}</math>

Then {{mvar|δ}} is obtained by applying a power of the [[Laplacian]] to the integral with respect to the unit [[sphere measure]] {{mvar|dω}} of {{math|''g''(''x'' · ''ξ'')}} for {{mvar|ξ}} in the [[unit sphere]] {{math|''S''<sup>''n''−1</sup>}}:
<math display="block">\delta(x) = \Delta_x^{(n+k)/2} \int_{S^{n-1}} g(x\cdot\xi)\,d\omega_\xi.</math>

The Laplacian here is interpreted as a weak derivative, so that this equation is taken to mean that, for any test function {{mvar|φ}},
<math display="block">\varphi(x) = \int_{\mathbf{R}^n}\varphi(y)\,dy\,\Delta_x^{\frac{n+k}{2}} \int_{S^{n-1}} g((x-y)\cdot\xi)\,d\omega_\xi.</math>

The result follows from the formula for the [[Newtonian potential]] (the fundamental solution of Poisson's equation). This is essentially a form of the inversion formula for the [[Radon transform]] because it recovers the value of {{math|''φ''(''x'')}} from its integrals over hyperplanes.  For instance, if {{mvar|n}} is odd and {{math|1=''k'' = 1}}, then the integral on the right hand side is
<math display="block">
\begin{align}
& c_n \Delta^{\frac{n+1}{2}}_x\iint_{S^{n-1}} \varphi(y)|(y-x) \cdot \xi| \, d\omega_\xi \, dy \\[5pt]
& \qquad = c_n \Delta^{(n+1)/2}_x \int_{S^{n-1}} \, d\omega_\xi \int_{-\infty}^\infty |p| R\varphi(\xi,p+x\cdot\xi)\,dp
\end{align}
</math>

where {{math|''Rφ''(''ξ'', ''p'')}} is the Radon transform of {{mvar|φ}}:
<math display="block">R\varphi(\xi,p) = \int_{x\cdot\xi=p} f(x)\,d^{n-1}x.</math>

An alternative equivalent expression of the plane wave decomposition is:{{sfn|Gelfand|Shilov|1966–1968|loc=I, §3.10}}
<math display="block">\delta(x) = \begin{cases}
  \frac{(n-1)!}{(2\pi i)^n}\displaystyle\int_{S^{n-1}}(x\cdot\xi)^{-n} \, d\omega_\xi & n\text{ even} \\
  \frac{1}{2(2\pi i)^{n-1}}\displaystyle\int_{S^{n-1}}\delta^{(n-1)}(x\cdot\xi)\,d\omega_\xi & n\text{ odd}.
  \end{cases}</math>