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On page 57. in Partial Differential Equation by Lawrence C. Evans, he prove the maximum principle for the Cauchy problem of the heat equation, i.e. (I quote)

Suppose $u\in C^2_1(\mathbb{R}^n\times (0,T])\cap C(\mathbb{R}^n\times [0,T])$ solves $u_t-\Delta u= 0$ in $\mathbb{R}^n\times (0,T)$ and $u=g$ on $\mathbb{R}^n\times \{t=0\}$. Moreover, u satisfies the growth estimate

$$u(x,t)\le Ae^{a|x|^2}$$

for $x\in\mathbb{R}^n,0\le t\le T$ for constants $A,a>0$. Then

$$\sup_{\mathbb{R}^n\times [0,T]}u = \sup_{\mathbb{R}^n}g$$

In the proof they define $v(x,t):=u(x,t)-\frac{\mu}{(T+\epsilon -t)^\frac{n}{2}}\exp{\frac{|x-y|^2}{4(T+\epsilon -t)}}$ The proof consists of several steps. First they show for $4aT<1$ that

  1. $\max_{\overline{U_T}} v= \max_{\Gamma_T}v$, where $U_T:=B^0(y,r)\times (0,T]$ for fixed $r>0$.
  2. If $x\in \mathbb{R}^n$ then $v(x,0)\le g(x)$

Now in equation $(29)$ they say: for $r$ selected sufficiently large, we have $v(x,t)\le A\exp{a(|y|+r)^2}-\mu (4(a+\gamma))^{\frac{n}{2}}\exp{(a+\gamma)r^2}\le \sup_{\mathbb{R}^n}g$. Why is this all less or equal the supremum of $g$, for large $r$?

And why can we conclude with all these facts, that $v(y,t)\le \sup_{\mathbb{R}^n} g$ for all $y\in \mathbb{R}^n$ and $0\le t\le T$?

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Hint: compare the coefficients of $r^2$ in the two exponential functions. –  Siminore Jun 18 '12 at 9:00

1 Answer 1

up vote 4 down vote accepted

I guess I was able to solve it by myself. All you have to do is to write the expression in a nice way:

$$v(x,t)\le A\exp{(a(|y|+r)^2)}-\mu (4(a+\gamma))^{\frac{n}{2}}\exp{((a+\gamma)r^2)}=\exp{((a+\gamma)r^2)}[(-\mu (4(a+\gamma))^{\frac{n}{2}}+A\exp{(-\gamma r^2+2ar|y| + a|y|^2)]}$$

This converges to $-\infty$ as $r\to \infty$ and the conclusion follows.

We already know that the max is attained at the "boundary" $\Gamma_T$ and we have found a bound of $v$ by (the supremum) of $g$ on $\Gamma_T$.

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