Maximum principle for the Heat Eqn in R^n

[loesung]

Hi! I am trying to understand the MP for the heat eqn. in $$\mathbb{R}^n$$ (see attached jpegs below). I don't understand why we are working with the function
$$v(x,t):=u(x,t)-\frac{\mu}{(T+\epsilon-t)^{n/2}}e^{\frac{|x-y|^2}{4(T+\epsilon-t)}}$$
Why not just work with u(x,t)? I understand that, eventually we want $$\epsilon\to 0$$ and get $$u$$ back, but I guess I really don't understand the proof (as I can't convince myself that v(x,t) is necessary)!

My second issue is in Part (1) of the proof, we are working in the cylindrical domain $$U_T=B^0(y,r)\times(0,T]$$.
In Part (2), though, we somehow jump from this domain to $$\mathbb{R}^n$$. Why do we need to start with the cylindrical domain in order to show the MP for $$\mathbb{R}^n$$?


Thanks for your time. I look forward to any helpful replies!



los

 

[gtoguy97]

The idea behind the function v(x,t) is to help us prove that u(x,t) satisfies the maximum principle in \mathbb{R}^n. In particular, we want to show that if u(x,t) is a solution of the heat equation in \mathbb{R}^n, then it must be bounded by its initial and boundary values. To do this, we need to approximate u(x,t) with a function that is bounded. The function v(x,t) does this by adding a term that "smoothes out" the behavior of u(x,t). As \epsilon\to 0, the term becomes zero and we recover u(x,t). As for why we start with the cylindrical domain U_T=B^0(y,r)\times(0,T], the reason is that this domain allows us to easily approximate u(x,t) with v(x,t). By starting with this domain, we can use the maximum principle on v(x,t), which is easier to prove than the maximum principle on u(x,t). Then, we can use the maximum principle on v(x,t) to show that u(x,t) must also be bounded by its initial and boundary values. This provides us with the maximum principle on u(x,t) in \mathbb{R}^n.