能量估计

设 $u \in C^{2, 1} (Q_{T}) \cap C ({\bar{Q}}_{T})$ 是初值问题

{\begin{cases} \partial_{t} u - Δ u = f (x, t), & (x, t) \in Q_{T} = Ω \times (0, T] \\ u (x, 0) = φ (x) \\ u |_{\partial Ω} = 0 \end{cases}

我们建立如下的能量不等式：

$\int_{Ω} u^{2} \d x \leq \nature^{T} (\int_{Ω} φ^{2} \d x + \int_{0}^{T} \int_{Ω} f^{2} \d x \d s) .$
$sup_{[0, T]} \int_{Ω} u^{2} \d x + 2 \int_{0}^{T} \int_{Ω} | \nabla u |^{2} \d x \d t \leq M (\int_{Ω} φ^{2} \d x + \int_{0}^{T} \int_{Ω} f^{2} \d x \d t) .$

在主要方程两侧乘上 $u$ 得： $u u_{t} - u \laplace u = f u$ ， ${(\frac{1}{2} u^{2})}_{t} - \nabla \cdot (u \nabla u) + | \nabla u |^{2} = f u$ ．
再在 $Ω$ 上积分，可得

\partial_{t} \int_{Ω} \frac{1}{2} u^{2} \d x - \int_{Ω} \nabla \cdot (u \nabla u) \d x + \int_{Ω} | \nabla u |^{2} \d x = \int_{Ω} f u d x

其中

\int_{Ω} \nabla \cdot (u \nabla u) \d x = \int_{\partial Ω} u \nabla u \d x = 0.

得

\begin{matrix} (Crucial) & \partial_{t} (\frac{1}{2} \int_{Ω} u^{2} \d x) + \int_{Ω} | \nabla u |^{2} \d x = \int_{Ω} f u \d x \leq \frac{1}{2} \int_{Ω} u^{2} \d x + \frac{1}{2} \int_{Ω} f^{2} \d x . \end{matrix}

第一次估计

忽略第二项，从而

\partial_{t} (\int_{Ω} u^{2} \d x) \leq \int_{Ω} u^{2} \d x + \int_{Ω} f^{2} \d x

令 $y (t) = \int_{Ω} u^{2} d x$ ，乘上积分因子并积分得

y (t) \leq \nature^{t} y (0) + \nature^{t} \int_{0}^{t} \int_{0}^{l} f^{2} (x, t) \d x \d s .

代入边值条件，并将 $t$ 放到最大！

\int_{Ω} u^{2} \d x \leq \nature^{T} (\int_{Ω} φ^{2} \d x + \int_{0}^{T} \int_{0}^{l} f^{2} (x, s) \d x \d s) .

上式代入 $(Crucial)$ ，可得

\begin{aligned} \frac{1}{2} \frac{\partial}{\partial t} \int_{Ω} | u |^{2} \d x + \int_{Ω} | \nabla u |^{2} \d x & \leq \frac{1}{2} \int_{Ω} f^{2} d x + \frac{1}{2} \int_{Ω} u^{2} \d x \\ = \frac{1}{2} \int_{Ω} f^{2} \d x + \frac{1}{2} \nature^{T} (\int_{Ω} φ^{2} \d x + \int_{0}^{T} \int_{Ω} f^{2} \d x \d t) . \end{aligned}

两边从 $0$ 到 $t$ 积分得

\begin{aligned} \frac{1}{2} \int_{Ω} | u |^{2} \d x + \int_{0}^{t} \int_{Ω} | \nabla u |^{2} \d x \d t & \leq \frac{1}{2} \int_{Ω} φ^{2} \d x + \frac{1}{2} \int_{0}^{t} \int_{Ω} f^{2} \d x \d t + \frac{1}{2} (\nature^{T} - 1) (\int_{Ω} φ^{2} \d x + \int_{0}^{T} \int_{Ω} f^{2} \d x \d t) \\ = \frac{1}{2} \nature^{T} (\int_{Ω} φ^{2} \d x + \int_{0}^{T} \int_{Ω} f^{2} \d x \d t) \end{aligned}

两边取 sup 即证．