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?

  • $\begingroup$ 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. $\endgroup$ – Sharat V Chandrasekhar May 31 '17 at 17:54
  • $\begingroup$ 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 $1/(r^n)$ $\endgroup$ – jJjjJ Jun 1 '17 at 6:10
  • $\begingroup$ @Sharat V Chandrasekhar $\endgroup$ – jJjjJ Jun 1 '17 at 6:11

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