Laplace 变换: 未来不会影响过去

问题. 设连续函数 \(u(t),g(t)\), \(0\leq t <\infty \) 的上界为 \(c e^{\alpha t},\alpha>0\). 假设它们的 Laplace 变换满足如下估计\[ |\hat{u}(s)|\leq c_1|\hat{g}(s)|^2,\quad \Re s>\alpha,\tag{1}\]其中常数 \(c_1>0\). 那么,\[ g(t)=0,\quad 0\leq t \leq T \tag{2}\]表明\[ u(t)=0,\quad 0\leq t \leq T. \tag{3}\]

证明. 令 \(T_1< T\) 且 \(s=\eta+\imath \xi\). 则,\[ \begin{align*} \int_{0}^{T_1}|u(t)|^2dt &\leq \frac{1}{2\pi}e^{2\eta T_1}\int_{-\infty}^{\infty }|\hat{u}(s)|^2d \xi & (\text{Parseval inequlity})\\ &\leq \frac{c_1}{2\pi}e^{2\eta T_1} \int_{-\infty}^{\infty } |\hat{g}(s)|^2d \xi & (\text{by }(1))\\ &\leq c_1 e^{2\eta T_1}\int_{0}^{\infty }e^{-2\eta t}|\hat{g}(t)|^2d t & (\text{Parseval inequlity})\\ &= c_1 e^{2\eta T_1}\int_{T}^{\infty }e^{-2\eta t}|\hat{g}(t)|^2d t & (\text{by (2)})\\ &= c_1 e^{2\eta T_1}\int_{0}^{\infty}e^{-2\eta (t+T)}|\hat{g}(t+T)|^2d t & \\ &\leq c_1c e^{2\eta (T_1-T)}\int_{0}^{\infty}e^{-2\eta t}e^{2 \alpha(t+T)} dt & (\text{boundedness of }g)\\ &= c_1c e^{2\eta (T_1-T)}e^{2 \alpha T} \frac{1}{-2(\eta-\alpha)} e^{-2(\eta-\alpha) t} |_{0}^{\infty} & \\ &= c_1c e^{2\eta (T_1-T)}e^{2 \alpha T} \frac{1}{2(\eta-\alpha)} & (\text{by }\eta>\alpha). \end{align*}\]取极限 \(\eta \to \infty \) 可得\[ \int_{0}^{T_1}|u(t)|^2dt \to 0.\]由 \(T_1 < T\) 的任意性和函数 \(u\) 的连续性可知 (3) 成立, 证明结束.

参考.

[1] [Heinz-Otto Kreiss, Jens Lorenz. Initial-Boundary Value Problems and the Navier-Stokes Equations. SIAM, 2004.](https://doi.org/10.1137/1.9780898719130)
[2] [Fourier 变换和 Laplace 变换.](https://numanal.com/fourier-laplace-transform/)

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