What does a 4+1D wave look like at the light cone?

For ease of reference in this post equations are numbered as in ref. 1.

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The equation $(36)$ cited in the OP reads

$$G_4(r,t)=frac1{4pi^2c^3}left(frac{delta(t-r/c)}{r(t^2-r^2/c^2)^{1/2}}-frac{eta(t-r/c)}{c(t^2-r^2/c^2)^{3/2}}right).tag{36}$$

Here $delta$ is the Dirac delta, and $eta$ is the Heaviside step function.

By itself, this expression doesn't really help understand what's going on: the Dirac delta in the first term is multiplied by a function that is singular at the delta's distributional singularity, and the second term isn't integrable. So, even if there is some way to evaluate this Green's function as a distribution like $langle G_4,hrangle$ for a smooth test function $h$, I suppose the expression would have to be transformed somehow.

Let's go another way. Namely, start from how equation $(36)$ was derived. The authors in ref. 1 derived it by integrating the Green's function for (5+1)-dimensional wave equation,

$$G_5=frac1{8pi^2c^2}left(frac{delta(tau)}{r^3}+frac{delta'(tau)}{cr^2}right),tag{32}$$

where $tau=t-r/c$, along the line of uniformly distributed sources in 5-dimensional space, using the integral

$$G_{n-1}(r,t)=2int_r^infty s(s^2-r^2)^{-1/2}G_n(s,t)ds,tag{25}$$

where $r=r_{n-1}$ is the radial coordinate in $(n-1)$-dimensional space.

Remember that a Green's function for a wave equation is the impulse response of the equation, i.e. the wave that appears after the action of the unit impulse of infinitesimal size and duration, $f(r,t)=delta(r)delta(t)$. For visualization purposes it's useful to replace this impulse with one that is finite at least in one variable, e.g. time. So let's choose another force function $f(r,t)=delta(r)F(t)$, where $F$ is defined as the Gaussian function

$$F(t)=frac1{sqrt{2pi}sigma}expleft(-frac{x^2}{2sigma^2}right).$$

Then, following equation $(34)$, we'll have the displacement response of the (5+1)-dimensional equation given by

$$phi_5(r,t)=frac1{8pi^2c^2}left(frac{F(tau)}{r^3}+frac{F'(tau)}{cr^2}right).tag{34}$$

Now, to find the displacement response $phi_4(r,t)$ of the (4+1)-dimensional equation, we can use $phi_5$ instead of $G_5$ in $(25)$. The resulting integral doesn't seem to have a closed form, so the following animations of $phi_4(r,t)$ were done using numeric quadrature.

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In the animations below we use $c=1$.

Let's first see how the $phi_5(r,t)$ looks. This will enable us to compare it with $phi_4(r,t)$. The following animation uses $sigma=0.005.$

We can see that it basically has two spikes: one positive and one negative. Let's increase $sigma$ to $sigma=0.04$ to resolve them better:

Now we can finally look at $phi_4(r,t)$. The following animation uses $sigma=0.005$, as the first one above.

Notice the difference of this wave from the $phi_5(r,t)$: the former has a "tail" inside the light cone, i.e. on the LHS of the positive spike here. This tail corresponds (in the limit $sigmato0$) to the second, negative, term in the equation $(36)$, while the positive spike corresponds to the first term with the Dirac delta.

So, the answer to

What does it mean that the Green’s function for the 4+1D wave equation has an opposite sign delta-function exactly on the light cone?

is exactly that: there is a negative wake inside the light cone and the opposite sign, i.e. positive, Dirac delta, i.e. infinitely high and narrow spike, on the light cone.

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References:

1: H. Soodak, M. S. Tiersten, Wakes and waves in N dimensions, Am. J. Phys. 61, 395 (1993)