热方程证明问询：积分守恒性与解的渐近衰减性
Alright, let's work through these two proofs for the heat equation solution step by step. First, let's restate the given setup clearly:
Given $g\in C(\mathbb{R}^n)\cap L1(\mathbb{R}n)$ with $|g|\leq M$, and $u$ is a bounded solution to the heat equation:
$$
\begin{cases}
\Delta u - u_t = 0 & (t>0, x\in\mathbb{R}^n), \
u(x,0) = g(x) & (x\in\mathbb{R}^n)
\end{cases}
$$
1. Prove $\int_{\mathbb{R}^n} u(x,t),dx = \int_{\mathbb{R}^n} g(x),dx$ for all $t>0$
First, recall that for the heat equation on $\mathbb{R}^n$, the unique bounded solution (given our conditions on $g$) can be written as the convolution of $g$ with the heat kernel:
$$
u(x,t) = (g * K_t)(x) = \int_{\mathbb{R}^n} g(y) K_t(x-y),dy
$$
where the heat kernel is defined as:
$$
K_t(z) = \frac{1}{(4\pi t)^{n/2}} e{-\frac{|z|2}{4t}}
$$
A core property of the heat kernel is that its integral over $\mathbb{R}^n$ equals 1 for any $t>0$—this comes from standard Gaussian integral calculations (you can verify it by switching to polar coordinates and evaluating directly).
Now compute the integral of $u(x,t)$:
$$
\int_{\mathbb{R}^n} u(x,t),dx = \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} g(y) K_t(x-y),dy \right) dx
$$
We can swap the order of integration using the Fubini-Tonelli Theorem: since $|g(y) K_t(x-y)| \leq M K_t(x-y)$, and $\int_{\mathbb{R}^n} |g(y)| \left( \int_{\mathbb{R}^n} K_t(x-y),dx \right) dy = \int_{\mathbb{R}^n} |g(y)| dy < \infty$ (because $g\in L^1$), the double integral of the absolute value is finite, justifying the swap.
After swapping, substitute $z = x-y$ (so $dx = dz$):
$$
\int_{\mathbb{R}^n} g(y) \left( \int_{\mathbb{R}^n} K_t(z),dz \right) dy
$$
The inner integral is 1, so we're left with:
$$
\int_{\mathbb{R}^n} g(y),dy = \int_{\mathbb{R}^n} g(x),dx
$$
That's exactly the result we needed to prove.
2. Prove $\lim_{t\to\infty} \sup_{x\in\mathbb{R}^n} |u(x,t)| = 0$
Again, use the convolution representation of $u(x,t)$. First, take the absolute value:
$$
|u(x,t)| = \left| \int_{\mathbb{R}^n} g(y) K_t(x-y),dy \right| \leq \int_{\mathbb{R}^n} |g(y)| K_t(x-y),dy
$$
Split this integral into two parts: a bounded ball around the origin, and the rest of $\mathbb{R}^n$. For any $\epsilon > 0$, since $g\in L1(\mathbb{R}n)$, we can choose a large enough $R > 0$ such that:
$$
\int_{|y| > R} |g(y)| dy < \frac{\epsilon}{2}
$$
Now break the integral into two pieces:
$$
\int_{\mathbb{R}^n} |g(y)| K_t(x-y),dy = \int_{|y| \leq R} |g(y)| K_t(x-y),dy + \int_{|y| > R} |g(y)| K_t(x-y),dy
$$
Analyze the tail integral ($|y| > R$)
Since $K_t(z) > 0$ for all $z$ and $\int_{\mathbb{R}^n} K_t(z),dz = 1$, we have:
$$
\int_{|y| > R} |g(y)| K_t(x-y),dy \leq \int_{|y| > R} |g(y)| dy < \frac{\epsilon}{2}
$$
This bound holds for all $x\in\mathbb{R}^n$ and all $t>0$.
Analyze the bounded integral ($|y| \leq R$)
Since $|g(y)| \leq M$, this integral is bounded by:
$$
M \int_{|y| \leq R} K_t(x-y),dy = M \int_{B(x,R)} K_t(z),dz
$$
where $z = x-y$ and $B(x,R)$ is the ball of radius $R$ centered at $x$.
Make a change of variable: $z = \sqrt{t} w$, so $dz = t^{n/2} dw$. The integral becomes:
$$
\int_{B(x/\sqrt{t}, R/\sqrt{t})} \frac{1}{(4\pi)^{n/2}} e{-\frac{|w|2}{4}} dw
$$
As $t\to\infty$, the radius $R/\sqrt{t}$ tends to 0, so the ball shrinks to a single point. The integral of the Gaussian function over a shrinking ball tends to 0 uniformly in $x$ (since the Gaussian is integrable, the integral over a set of measure approaching 0 goes to 0).
For our fixed $R$, there exists a $T>0$ such that for all $t>T$:
$$
\int_{B(x,R)} K_t(z),dz < \frac{\epsilon}{2M}
$$
for all $x\in\mathbb{R}^n$. This gives:
$$
\int_{|y| \leq R} |g(y)| K_t(x-y),dy \leq M \cdot \frac{\epsilon}{2M} = \frac{\epsilon}{2}
$$
Combine the bounds
For all $t>T$, we have:
$$
|u(x,t)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
$$
for every $x\in\mathbb{R}^n$. Since $\epsilon$ was arbitrary, this means $\sup_{x\in\mathbb{R}^n} |u(x,t)| \to 0$ as $t\to\infty$.
内容的提问来源于stack exchange，提问作者Username Unknown
