An old electrodynamics exam question asks:

"Find the Green function (for the one-dimensional Poisson equation) that solves the equation $$ \frac{d^2}{dx^2}G(x,x') = -\delta(x-x'). $$ Choose the boundary conditions so that $G(x,x')$ is translation invariant (i.e. dependent only on $x-x'$). "

My solution would be to introduce r = x - x' and write for the homogeneous case (r=/=0)

$$ \frac{d^2}{dr^2} G(r) = 0 $$

Then, on either side of zero, G has to be linear in r:

$$ G(r) = (ar+b) \theta(r) + (cr +d) \theta(-r) $$

with the Heavise theta function $\theta$. Integration of the non-homogenous differential equation over the singularity then yields

$$ \left [\frac{d}{dr} G \right]_{-\epsilon}^{\epsilon} = -1 $$

With the expression for the Greens function found above, this is just

$$ a-c=1 $$

So in my opinion, we have found the Green function (just substituting r = x - x' back)

$$ G(x,x') = \left[ a(x-x')+b \right] \theta(x-x') + \left[ (a+1)(x-x')+d\right] \theta(x'-x) $$

However, the solution says that the Green function is actually

$$ G(x,x') = -\frac{1}{2} |x-x'| +const $$

In my opinion, these are not equivalent at all or at least I don't see how (I have three constants and they have just one). How can I get from one expression to the other or what is wrong in my derivation?

  • 1
    $\begingroup$ Should it depend on $x-x^\prime$ or on the absolute value of it?! $\endgroup$ Apr 1 at 11:58
  • 1
    $\begingroup$ The question said it should only depend on x-x' (so not necessarily just the absolute value). $\endgroup$
    – F L
    Apr 1 at 12:00
  • $\begingroup$ I might be wrong here, but isn't the general solution of this differential equation always translationally invariant? What I mean is that the general form of the fundamental solution to $F^{\prime\prime}(x)=\delta(x)$ is given by $F(x)=x\theta(x)+cx+d$ for some constants $c,d\in\mathbb R$. Now my guess is that your case is a trivial modification, i.e. changing the differential equation appropriately, I guess that the solution only changes by the substitution $x\to x-x^\prime$, but I haven't tried. $\endgroup$ Apr 1 at 12:39
  • $\begingroup$ There are boundary conditions in the infinity. Note that this is the solution for the field if an infinite charge sheet. $\endgroup$
    – Roger V.
    Apr 1 at 12:49
  • 1
    $\begingroup$ Re "Heavise": Not "Heaviside"? $\endgroup$ Apr 2 at 19:30

2 Answers 2


Integration over the singularity not only yields $$ \Big[\frac{d}{dr} G \Big]_{-\epsilon}^{\epsilon} = -1 $$ but also: $$ \big[G \big]_{-\epsilon}^{\epsilon} = 0. $$ That will fix $b$ and $d$ to be equal to the "const" of the second solution.

This still leaves your $a$ as a free parameter. So you are right that without further restrictions the second solution is not the only one. A further restriction could be to impose reflection symmetry (which would be natural), but if the text does not say that, $a$ will be undetermined.

  • $\begingroup$ Thanks for the answer, I think I got it. Just so I understand this right: Your second condition follows from your first condition: You are saying the derivative is finite, so the function has to be continuous? Or what do you mean by "integration also yields"? $\endgroup$
    – F L
    Apr 1 at 14:02
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    $\begingroup$ Sorry I forgot to explain that, but as mentioneed in in the other answer by @Hyperion we have to avoid the $\delta'(r)$ term in $G''(r)$, because the purpose of a Green function is to give us the solution for a source term of just $\delta(r)$, nothing else (it's in your first equation). $\endgroup$ Apr 1 at 19:29

The general solution of $$G^{\prime \prime}(r)=-\delta(r) \tag{1} \label{1}$$ for the Green function $G(r)$ can indeed be obtained from your ansatz $$G(r)=(ar+b) \theta(r)+(cr+d) \theta(-r). \label{2} \tag{2}$$ From $$ G^\prime (r)=a \theta(r)+c\theta(-r)+(b-d)\delta(r) \tag{3} \label{3} $$ we infer $b=d$ because of the absence of a term $\sim \delta^\prime(r)$ on the right-hand-side of \eqref{1}. Thus we obtain $$G^{\prime \prime}(r)=(a-c)\delta(r), \tag{4}$$ where comparison with \eqref{1} yields $a-c=-1$. Expressing $c$ and $d$ in terms of $a$ and $b$, the general solution of \eqref{1} is given by $$G(r)= ar +r\theta(-r)+b, \tag{5} \label{5}$$ where the relation $\theta(r)+\theta(-r)=1$ was used. The presence of the two free parameters $a$, $b$ in \eqref{5} reflects the fact that the Green function is not uniquely determined by \eqref{1} but only up to an arbitrary solution of the homogeneous equation. Choosing $a=-1/2$ one finds indeed $$G(r)=-r/2 +r \theta(-r)+b =-|r| /2+b \tag{6}$$ as a possible solution for the Green function.


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