题目
问题 12. 对于波动方程 u t t = c 2 u x x u_{tt} = c^2 u_{xx} utt=c2uxx 的解 u ( x , t ) u(x,t) u(x,t)，能量密度定义为 e = 1 2 ( u t 2 + c 2 u x 2 ) e = \frac{1}{2} (u_t^2 + c^2 u_x^2) e=21(ut2+c2ux2)，动量密度定义为 p = c u t u x p = c u_t u_x p=cutux.
(a) 证明：
∂ e ∂ t = c ∂ p ∂ x 和 ∂ p ∂ t = c ∂ e ∂ x . \frac{\partial e}{\partial t} = c \frac{\partial p}{\partial x} \quad \text{和} \quad \frac{\partial p}{\partial t} = c \frac{\partial e}{\partial x}. ∂t∂e=c∂x∂p和∂t∂p=c∂x∂e.
(b) 证明 e ( x , t ) e(x,t) e(x,t) 和 p ( x , t ) p(x,t) p(x,t) 都满足相同的波动方程。
解答
部分 (a)
要证明 ∂ e ∂ t = c ∂ p ∂ x \frac{\partial e}{\partial t} = c \frac{\partial p}{\partial x} ∂t∂e=c∂x∂p 和 ∂ p ∂ t = c ∂ e ∂ x \frac{\partial p}{\partial t} = c \frac{\partial e}{\partial x} ∂t∂p=c∂x∂e，其中 e = 1 2 ( u t 2 + c 2 u x 2 ) e = \frac{1}{2} (u_t^2 + c^2 u_x^2) e=21(ut2+c2ux2)， p = c u t u x p = c u_t u_x p=cutux，且 u u u 满足波动方程 u t t = c 2 u x x u_{tt} = c^2 u_{xx} utt=c2uxx。假设 u u u 的二阶偏导数连续（混合偏导数可交换，即 u t x = u x t u_{tx} = u_{xt} utx=uxt）。
步骤 1: 计算 ∂ e ∂ t \frac{\partial e}{\partial t} ∂t∂e
能量密度 e = 1 2 ( u t 2 + c 2 u x 2 ) e = \frac{1}{2} (u_t^2 + c^2 u_x^2) e=21(ut2+c2ux2)，对 t t t 求偏导：
∂ e ∂ t = ∂ ∂ t ( 1 2 ( u t 2 + c 2 u x 2 ) ) = 1 2 ( ∂ ∂ t ( u t 2 ) + c 2 ∂ ∂ t ( u x 2 ) ) . \frac{\partial e}{\partial t} = \frac{\partial}{\partial t} \left( \frac{1}{2} (u_t^2 + c^2 u_x^2) \right) = \frac{1}{2} \left( \frac{\partial}{\partial t} (u_t^2) + c^2 \frac{\partial}{\partial t} (u_x^2) \right). ∂t∂e=∂t∂(21(ut2+c2ux2))=21(∂t∂(ut2)+c2∂t∂(ux2)).
使用链式法则：
∂ ∂ t ( u t 2 ) = 2 u t ∂ u t ∂ t = 2 u t u t t , ∂ ∂ t ( u x 2 ) = 2 u x ∂ u x ∂ t = 2 u x u x t . \frac{\partial}{\partial t} (u_t^2) = 2 u_t \frac{\partial u_t}{\partial t} = 2 u_t u_{tt}, \quad \frac{\partial}{\partial t} (u_x^2) = 2 u_x \frac{\partial u_x}{\partial t} = 2 u_x u_{xt}. ∂t∂(ut2)=2ut∂t∂ut=2ututt,∂t∂(ux2
