# Heat equation, heat ball, and level set

I'm currently reading Evans book on PDE, and I've got some trubles in understanding the heat equation.

In the book the heat kernel is derived:

$$\Phi(x, t) = \begin{cases} \frac{1}{(4\pi t)^{\frac{n}{2}}}e^{\frac{\Vert x \Vert ^{2}}{4t}}, &t>0\\ 0, &\text{otherwise}. \end{cases}$$

Now define the heat ball, for fixed $x \in \mathbb{R}^n$, $r > 0$, $t \in \mathbb{R}$ as $$E(x,t,r) = \{ (y, s) \in \mathbb{R^{n+1}}\ :\ s \le t,\ \Phi(x-y,t-s)\ge\frac{1}{r^n}\}.$$

In the book it is said that the boundary of $E$ is a level set for $(y, s) \to \Phi(x-y, t-s)$. Now it is obviously true that if $(y,s)$ is in the boundary and $(y, s) \ne (x,t)$ then $\Phi(x-y,t-s) = \frac{1}{r^n}$. Moreover it is certainly true that $(x, t)$ is in the boundary of $E$, but $(y, s) \to \Phi(x-y, t-s)$ has a singularity in $(x, t)$, so how should I avoid this?

• What you have appears to be a point source with a defined singularity, so I don't see how (or why) you would avoid it. – Sharat V Chandrasekhar May 31 '17 at 17:54
• I don't want to avoid it. I didn't explain well. I mean how can we say that the boundary of E is a level set? In all the points but $(x,t)$ the values of the function is $1/(r^n)$ – jJjjJ Jun 1 '17 at 6:10
• @Sharat V Chandrasekhar – jJjjJ Jun 1 '17 at 6:11