## 能量不等式证明

news/2024/4/19 13:06:57/文章来源:https://blog.csdn.net/coldasice342/article/details/136476029

### 波动方程初值问题能量不等式的证明

#### Gronwall 不等式

d G ( τ ) d τ ≤ e C τ F ( τ ) \frac{dG(\tau)}{d\tau}\leq e^{C\tau}F(\tau) G ( τ ) ≤ C − 1 ( e C τ − 1 ) F ( τ ) G(\tau)\leq C^{-1}(e^{C\tau}-1)F(\tau)

#### 常用的替换技巧

u t u t t = 1 2 ⋅ 2 u t ⋅ u t t = 1 2 ( u t 2 ) t u_{t}u_{tt}=\frac{1}{2}\cdot 2u_{t}\cdot u_{tt}=\frac{1}{2}(u_{t}^2)_t u u t = 1 2 ⋅ 2 u ⋅ u t = 1 2 ( u 2 ) t uu_t=\frac{1}{2}\cdot 2u\cdot u_t =\frac{1}{2}(u^2)_t u u x x = ( u u x ) x − u x 2 = u x 2 + u u x x − u x 2 uu_{xx}=(uu_x)_x-u_x^2=u_x^2+uu_{xx}-u_x^2 ∵ ( u t u x ) x = u t x u x + u t u x x \because(u_{t}u_{x})_x=u_{tx}u_{x}+u_{t}u_{xx} ∴ u t u x x = ( u t u x ) x − u t x u x \therefore u_{t}u_{xx}=(u_{t}u_{x})_x-u_{tx}u_{x}

u u 二阶导函数连续时， u x t = u t x u_{xt}=u_{tx} ∴ u t x u x = 1 2 ( u x 2 ) t = 1 2 ⋅ 2 u x ⋅ u x t = u x u t x \therefore u_{tx}u_{x} = \frac{1}{2}(u_{x}^2)_t=\frac{1}{2}\cdot 2u_{x}\cdot u_{xt}=u_{x}u_{tx} ∴ u t u x x = ( u t u x ) x − 1 2 ( u x 2 ) t \therefore u_{t}u_{xx}=(u_{t}u_{x})_x-\frac{1}{2}(u_{x}^2)_t

#### 能量不等式证明过程

∂ K τ = Ω 0 ∪ ( − Γ τ 2 ) ∪ ( − Ω τ ) ∪ ( − Γ τ 1 ) \partial K_{\tau} =\Omega_0\cup(-\Gamma_{\tau_{2}})\cup(-\Omega_{\tau})\cup(-\Gamma_{\tau_{1}})

∫ Ω τ [ u t 2 ( x , τ ) + a 2 u x 2 ( x , τ ) ] d x ≤ M [ ∫ Ω 0 ( ψ 2 + a 2 φ x 2 ) d x + ∫ K τ f 2 ( x , t ) d x d t ] \int_{\Omega_\tau} \left[u_t^2(x, \tau) + a^2 u_x^2(x, \tau)\right] dx \leq M \left[ \int_{\Omega_0} (\psi^2 + a^2 \varphi_x^2) dx + \int_{K_\tau} f^2(x,t) dxdt \right] ∫ K τ [ u t 2 ( x , t ) + a 2 u x 2 ( x , t ) ] d x d t ≤ M [ ∫ Ω 0 ( ψ 2 + a 2 φ x 2 ) d x + ∫ K τ f 2 ( x , t ) d x d t ] \int_{K_\tau} \left[u_t^2(x,t) + a^2 u_x^2(x,t)\right] dxdt \leq M \left[ \int_{\Omega_0} (\psi^2 + a^2 \varphi_x^2) dx + \int_{K_\tau} f^2(x,t) dxdt \right]

1. 在波动方程 ∂ 2 u ∂ t 2 − a 2 ∂ 2 u ∂ x 2 = f \frac{\partial^2 u}{\partial t^2}-a^2\frac{\partial^2 u}{\partial x^2}=f 两端同乘以 ∂ u ∂ t \frac{\partial u}{\partial t} 并在区域 K τ K_\tau 上积分并按照前面的常用替换得：

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