# Wrong solution for Green function of one-dimensional Poisson equation

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?

• Should it depend on $x-x^\prime$ or on the absolute value of it?! Apr 1 at 11:58
• The question said it should only depend on x-x' (so not necessarily just the absolute value).
– F L
Apr 1 at 12:00
• 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. Apr 1 at 12:39
• There are boundary conditions in the infinity. Note that this is the solution for the field if an infinite charge sheet. Apr 1 at 12:49
• Re "Heavise": Not "Heaviside"? Apr 2 at 19:30

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.
• 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). 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.