Norton's dome

Norton's dome is a simple example of a physical system which does not have a well-defined Cauchy problem.

Newtonian mechanics

The usual formulation of Norton's dome is a point mass sliding along a dome of a specific shape. There are many ways to do it, but at its core it reduces to a force of the form

\begin{equation} \vec{F}(r) = \sqrt{r} \end{equation}

Or some other form of root $r^{p/q}$, as long as the derivative is not defined at $0$. This corresponds to

\begin{equation} \frac{d r^{p/q}}{dr}= \frac{p}{q} r^{(p/q) - 1} \end{equation}

which lacks a derivative at $r = 0$ for $p/q < 1$.

In which case, the Newton law gives us

\begin{equation} \ddot{r}(t) = \sqrt{r} \end{equation}

A simple example of a solution is

\begin{equation} \dot{r}(t) = \frac{1}{144} t^4 \end{equation}

As we have that $(r^4)'' = r^2$. But a more general possible solution is

\begin{equation} r(t) = \frac{1}{144} \theta(t - T) (t - T)^4 \end{equation} \begin{equation} \dot{r}(t) = \frac{1}{144} \left[ \delta(t - T) (t - T)^4 + 4 \theta(t - T) (t - T)^3 \right] \end{equation}

As the singular support of $\delta$ is at $t = T$ where $(t - T)^4 = 0$, only the second term remains, so that

\begin{equation} \dot{r}(t) = \frac{1}{144} \left[ 4 \theta(t - T) (t - T)^3 \right] \end{equation}

Using the same method, and the fact that $\theta^n = \theta$, we get

\begin{eqnrray} \ddot{r}(t) = \frac{1}{144} \left[ 4 \delta(t - T) (t - T)^3 + 12 \theta(t - T) (t - T)^2 \right]\\ &=& \frac{1}{12} \theta^2(t - T) (t - T)^2 \\ &=& \sqrt{r(t)} \end{eqnrray} \begin{equation} \dot{r}(t) = \frac{1}{144} \left[ \delta(t - T) (t - T)^4 + 4 \theta(t - T) (t - T)^3 \right] \end{equation}