I'm an engineer, who works with retarded delay differential equations. So far I've encountered systems with single or multiple point-wise delays, and systems with distributed delays. Someone on a conference told me that I could unify/generalize my models using the two integrals mentioned in the title. The only problem is, that since I'm not a mathematician, I don't know what literature to look up, or where to start.
My systems so far are:
- $\dot{x}(t) = A_0x(t) + \sum_{k=1}^n A_k x(t-\tau_k), \tau_1 < \tau_2 < \dots < \tau_n, \quad x(h) = \phi(h)$ for $h \in [-\tau_n, 0]$
- $\dot{x}(t) = A_0x(t) + \int_{-\tau}^0 A_\tau(\eta) x(t+\eta)d\eta, \quad x(h) = \phi(h)$ for $h \in [-\tau, 0]$
- $\dot{x}(t) = \int_{-\tau}^0 A_\tau(\eta) x(t+\eta)d\eta, \quad x(h) = \phi(h)$ for $h \in [-\tau, 0]$
so my guess would be: $\dot{x}(t) = \int_{-\tau}^0 dB(\eta) x(t+\eta), \quad x(h) = \phi(h)$ for $h \in [-\tau, 0]$, but in the first two cases, this $B$ would be a discontinuous function. What literature would you recommend for me to brush up on my nonexistent functional differential equations knowledge and integrals?
This is intended just as an overview of what these integrals are and how they apply to your situation. It is not intended to tell you everything you need to know. You still need to check texts such as Andrey Yanyuk has suggested. But please feel free to be much more concerned about what the theorems say can be done with the integrals than with why the theorems are true, and all the minutia that comes with it. In particular, wherever the theorems talk about something holding "almost everywhere", you are safe to ignore that part, as for your function $x(t)$, which is continuous, it will hold everywhere.
You do not need to go study all the complexities and minutia about these integrals that we mathematicians love to delve through. The broad strokes are fine.
For your purposes, the advantage to these other forms of integrals is that it converts your multiple cases into a single case, so you only have to develop the theory once, not three times. I do not see any advantage to you in studying Lebesgue-Stieltjes integration. Instead I suggest either Riemann-Stieltjes integration, or straight Lebesgue integration for an abstract measure. Lebesgue-Stieltjes is just the latter, but with the measure defined in a way that mimics the integration of Riemann-Stieltjes. However, measures for your purposes are more easily and intuitively defined in other fashions.
Recall that (Riemann) integration $\int_a^b f(x)\,dx$ is defined as a limit of $\sum_{k=1}^n f(t_k)(x_k - x_{k-1})$ where $a = x_0 < x_1 < \dots < x_n = b$, and each $t_k \in [x_{k-1},x_k]$, and the limit is taken as $\max_k (x_k-x_{k-1})$ goes to $0$.
When $g$ is differentiable, this results in $$\int_a^b f(x)\,dg(x) = \int_a^b f(x)g'(x)\,dx$$ which is nothing new. It becomes useful when $g$ isn't differentiable, or even continuous. Any function $g$ can be used, no matter how nastily behaved. What is affected by the bad behavior of $g$ is which functions $f$ can be integrated with respect to it. But $g$ itself could be anything. For your purposes, simple jump discontinuities would handle all cases. And if $f$ is continuous, it will always be integrable with respect to a $g$ that is continuous except for jump discontinuities at isolated points.
In particular, if $f$ is continuous at $0$, and $$H(x) = \begin{cases}0&x < 0\\1&x\ge 0\end{cases}$$ is the Heaviside function, then $$\int_a^b f(x)\,dH(x) =\begin{cases}f(0)&0 \in [a,b]\\0&0\notin [a,b]\end{cases}$$
So you can express $$\sum_{k=1}^n A_kx(t-\tau_k) + \int_{-\tau}^0A(\eta)x(t - \eta)\,d\eta = \int_{-\tau}^0x(t - \eta)\, dB(\eta)$$ where $$B(\eta) = \int_{-\tau}^\eta A(t)\,dt + \sum_{k=1}^n A_kH(\eta - \tau_k)$$ Which is why Riemann-Stieltjes simplifies and generalizes your situation.
A measure assigns a value $\ge 0$ to subsets of $[a,b]$. In general, it is not reasonable to actually define a measure value for every subset of $[a,b]$, but it will be defined on every subset you will ever have need for, so I will not worry about the difference. The key requirement of a measure is additivity: For all measurable sets $U$ and $V$, then $\mu(U \cup V) = \mu(U) + \mu(V) - \mu(U \cap V)$. In particular, if $U$ and $V$ do not overlap, then $\mu(U \cup V) = \mu(U) + \mu(V)$.
The Lebesgue measure $\lambda$ is completely definable by the requirement that for intervals $[c,d], \lambda([c,d]) = d-c$. For any Riemann-integrable function $f, \int_a^b f(x)\,dx = \int_a^b f(x)\,d\lambda$. In fact, you can think of $dx$ as just another name for $d\lambda$.
You can think of a measure $\mu$ as giving the "weight" of the set, for some mass density distribution. For $\lambda$, the density distribution is a constant $1$, so the weight is the length of the set. But you can also use densities that vary. For some purposes, you need negative weights. So we extend the definition a bit: if $\alpha$ is a set function and $\alpha_+, \alpha_-$ are two measures such that for all measurable sets $S, \alpha(S) = \alpha_+(S) - \alpha_-(S)$, then $$\int f\,d\alpha = \int f\,d\alpha_+ - \int f\,d\alpha_-$$
You can even have point particles: If $T = \{\tau_1, \tau_2, \dots, \tau_n\}$, we can define a measure $\mu_T(S)$ to be the number of elements of $T$ that lie in $S$. More generally, we can define $$\mu_A(S) = \sum_{k=1}^n \begin{cases}A_k&\tau_k \in S\\0&\tau_k \notin S\end{cases}$$ Then $$\int f\,d\mu_A = \sum_{k=1}^n A_kf(\tau_k)$$.
And for your application, $$\sum_{k=1}^n A_kx(t-\tau_k) + \int_{-\tau}^0A(\eta)x(t - \eta)\,d\eta = \int_{-\tau}^0x(t - \eta)\, d\beta(\eta)$$ where the set function $\beta$ is defined by $$\beta(S) = \int_S A(t)\,d\lambda(t) + \mu_A(S)$$