$\widehat{s}(\lambda;\mu)=\frac{1}{\lambda(1+\mu \widehat{a}(\lambda))}, \ Re \lambda>0, Re \mu >0?$

Question

Let $s(t;\mu)$ solution of the one-dimensional Volterra equation $$s(t;\mu)+\mu\int_0^t a(t-\tau)s(\tau;\mu)d\tau=1, \ t\geq 0.$$ where $-\mu \in \sigma (A)$ (spetrum of $A$), $a \in L^1_{loc}(\mathbb{R}^+)$, $A:X \to X$ a closed operator in a Complex Banach space $X$. How show that $$\widehat{s}(\lambda;\mu)=\frac{1}{\lambda(1+\mu \widehat{a}(\lambda))}, \ Re \lambda>0, Re \mu >0?$$ Here $\widehat{f}$ is the Laplace transform of $f$.

Answer 1

You have $$s(t;\mu)+\mu\int_0^ta(t-\tau)s(\tau;\mu) d\tau=1,$$ taking the Laplace transforme above we have $$\widehat{s}(\lambda;\mu)=\int_0^{\infty}e^{-\lambda t} \ dt-\mu \widehat{a*s}(\cdot;\mu)$$ that is $$\widehat{s}(\lambda;\mu)=\frac{1}{\lambda}-\mu \widehat{a*s}(\cdot;\mu)$$ and by the convolution Theorem $$\widehat{s}(\lambda;\mu)(1+\mu \widehat{a}(\lambda))=\frac{1}{\lambda}$$.