微分方程|MTH2003-JD Differential Equations – DEFERRAL代写

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that any $\lambda \in \mathrm{C}$ such that $\operatorname{Re}(\lambda)<-\mu$ is contained in $\rho(A)$ and in particular
$$R_{A}(\lambda)=\int_{0}^{\infty} e^{k t} T(t) d t$$
Hence it follows for every $\varpi \in \mathbb{R}$ that
$$\omega\left(R_{A}\left(\mu^{\prime}+i \varpi\right) \xi\right)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} e^{i \boxminus t} h(t) d t$$
where
$$h(t):=\sqrt{2 \pi} e^{\mu / t} \omega(T(t) \xi)$$
for all $t \in[0, \infty)$ and $h(t):=0$ for all $t \in(-\infty, 0)$. In particular,
$$|h(t)| \leqslant \sqrt{2 \pi} c\left|k u|\mid \xi \xi| \cdot e^{-|\mu+\mu| t}\right.$$
for every $t \geqslant 0$ and hence $h \in L_{\mathrm{C}}^{2}(\mathrm{R})$. Hence $\left(\mathrm{R} \rightarrow \mathrm{C}, \varpi \mapsto \omega\left(R_{A}\left(\mu^{\prime}+i \varpi\right) \xi\right)\right) \in$ $L_{\mathrm{C}}^{2}(\mathbb{R})$ as well as (4.3.1) follows by the Fourier inversion theorem.

MMTH2003-JD COURSE NOTES ：

we have:
$$\left(A_{r}^{-1} f\right)(x)=\int_{0}^{x} f(y) d y$$
for all $x \in I$ and $f \in X$ and hence
$$\left(\left(A_{r}^{}\right)^{-1} f\right)(x)=\int_{x}^{a} f(y) d y$$ for all $x \in I$ and $f \in X$. Further, by the proof of Corollary 5.1.4 $$\bar{A}{l}^{-1}=U \bar{A}{r}^{-1} U$$
and hence

\begin{aligned}
&\left(\bar{A}{l}^{-1} f\right)(x)=\left(\bar{A}{r}^{-1} U f\right)(a-x)=\int_{0}^{a-x} f(a-y) d y \
&=\int_{x}^{a} f\left(y^{\prime}\right) d y^{\prime}=\left(\left(A_{r}^{}\right)^{-1} f\right)(x)
\end{aligned}