Why does Green's function for wave equation in one spatial dimension have no singularity?

Question

Green's function is the response to a Dirac-delta source. But the green's function for wave equation in one spatial dimension doesn't seem to have a singularity, whereas the heat equation does. What's the reason for this?

Answer 1

For what it's worth, short of actually calculating the Greens functions, dimensional analysis can give a crude hint/inside of what singularities to expect although it fails to detect e.g. logarithmic singularities, and cannot distinguish between e.g. $1/x$ and $\delta(x)$.

Examples:

1. Dimensional analysis of the $n$-dimensional Laplace operator $-\vec{\nabla}^2 G(\vec{r})~=~\delta^n(\vec{r})$ suggests that $G(\vec{r})~\sim~r^{2-n} $. In OP's 3rd picture with $n=2$ this is constant, i.e. it failed to detect a logarithmic singularity.

2. Dimensional analysis of the $(n+1)$-dimensional d'Alembertian/wave operator $(\partial_t^2-\vec{\nabla}^2)G(\vec{r},t)~=~\delta^{n}(\vec{r})\delta(t)$ suggests that $G(\vec{r},t)~\sim~(r^2-t^2)^{(1-n)/2} $. In OP's 1st picture with $n=1$ this correctly predicts no singularity.

3. Dimensional analysis of the $n$-dimensional heat operator $(\partial_t-\vec{\nabla}^2)G(\vec{r},t)~=~\delta^{n}(\vec{r})\delta(t)$ suggests that $G(\vec{r},t)~\sim~t^{-n/2} $. In OP's 2nd picture with $n=1$ this correctly predicts a singularity.

Of course this answer is an oversimplication of the actual singularity structure, and is only meant as a tool to develop some intuition.

Answer 2

In addition to @Qmechanic's good points, we can also think in terms of (e.g., $L^2$) Sobolev spaces, as characterized by the weighted $L^2$ spaces the Fourier transforms lie in. Thus, considering $\Delta u=\delta$, we have (up to normalization) $-r^2\widehat{u}=\widehat{\delta}=1$. It is elementary that $\int_{\mathbb R^n} 1\cdot (1+|x|^2)^{-{n\over 2}-\epsilon}\,dx<+\infty$, so $\delta\in H^{-{n\over 2}-\epsilon}(\mathbb R^n)$, for every $\epsilon>0$. Since $-\Delta$ is a "positive" second order operator, it shifts the Sobolev space index by $2$. Thus, a fundamental solution is in $H^{-{n\over 2}+2-\epsilon}(\mathbb R^n)$.

At the same time, Sobolev's imbedding/inequality shows that on $\mathbb R^n$ it is essentially sharp that $H^{{n\over 2}+\epsilon}(\mathbb R^n)\subset C^o(\mathbb R^n)$.

So, the fundamental solution is continuous if (let's not quite say "only if") $-{n\over 2}+2-\epsilon > {n\over 2}$ for some small-enough $\epsilon>0$. This is equivalent to $2-\epsilon>n$, so only $n=1$ can succeed.