Witam,
czy mógłby ktoś sprawdzić te obliczenia? Wynik ostateczny jest poprawny, nie wiem tylko przekształcenia są dobrze.
\(\displaystyle{ $$u_i (t) = \sum_{j=1}^n \int_{-\infty}^t h(t-\tau)T_{ij} f_j( u_j) d\tau + \int_{-\infty}^t h(t-\tau)I_i (\tau) d\tau}\)
Różniczkujemy stronami:
\(\displaystyle{ $$\frac{d u_i (t)}{dt} = \frac{d}{dt} \left(\sum_{j=1}^n \int_{-\infty}^t h(t-\tau)T_{ij} f_j( u_j(\tau)) d\tau + \int_{-\infty}^t h(t-\tau)I_i (\tau) d\tau \right)}\)
Grupujemy wyrazy
\(\displaystyle{ $$\frac{d u_i (t)}{dt} = \frac{d}{dt} \int_{-\infty}^t h(t-\tau) [\sum_{j=1}^n T_{ij} f_j( u_j(\tau)) + I_i (\tau)] d\tau$$}\)
Zamieniamy całkę niewłaściwą na Riemanna
\(\displaystyle{ $$\frac{d u_i (t)}{dt} = \frac{d}{dt} \lim_{T \to -\infty} \int_{T}^t h(t-\tau) [\sum_{j=1}^n T_{ij} f_j( u_j(\tau)) + I_i (\tau)] d\tau$$}\)
\(\displaystyle{ Niech $g(t, \tau) =h(t-\tau) [\sum_{j=1}^n T_{ij} f_j( u_j(\tau)) + I_i (\tau)] $
$$\frac{d u_i (t)}{dt} = \frac{d}{dt} \lim_{T \to -\infty} \int_{T}^t g(t, \tau) d\tau$$}\)
Stosujemy wzór na różniczkowanie pod znakiem całki, czyli:
\(\displaystyle{ \frac{\mathrm{d}}{\mathrm{d}x} \left (\int_{a(x)}^{b(x)}f(x,t)\,\mathrm{d}t \right) = f(x,b(x))\cdot b'(x) - f(x,a(x))\cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(x,t)\; \mathrm{d}t.}\)
Otrzymujemy
\(\displaystyle{ $$\frac{d u_i (t)}{dt} =\frac{d}{dt}\lim_{T \to -\infty} \left( \int_{T}^t g(t,\tau) \, d \tau\right)
$$=\lim_{T \to -\infty} \left( g(t,t)\cdot \frac{dt}{dt} - g(t,T)\cdot \frac{dT}{dt} + \int_{T}^{t} \frac{\partial}{\partial t} g(t,\tau) \, d\tau\right)$$
$$=\lim_{T \to -\infty} \left( g(t,t) + \int_{T}^{t} \frac{\partial}{\partial t} g(t,\tau) \, d\tau\right) = g(t,t) + \int_{-\infty}^{t} \frac{\partial}{\partial t} g(t,\tau) \, d\tau $$
\\
$$g(t,t) = h(t-t) \left[\sum_{j=1}^n T_{ij} f_j( u_j(t)) + I_i (t)\right] = \frac{1}{\mu _i} e^{-\frac{t-t}{\mu _i}} \left[\sum_{j=1}^n T_{ij} f_j( u_j(t)) + I_i (t)\right] = $$
$$ = \frac{1}{\mu _i} \left[\sum_{j=1}^n T_{ij} f_j( u_j(t)) + I_i (t)\right]$$
Niech $z(\tau) = \left[\sum_{j=1}^n T_{ij} f_j( u_j(\tau)) + I_i (\tau)\right] $
$$\frac{\partial}{\partial t} g(t,\tau) = \frac{\partial}{\partial t} h(t-\tau) z(\tau) = h'(t-\tau) z(\tau) + h(t-\tau)z'(\tau) = \left( \frac{1}{\mu _i} e^{-\frac{t-\tau}{\mu _i}} \right)' z(\tau) = $$
$$ = - \frac{1}{\mu _i}\frac{1}{\mu _i} e^{-\frac{t-\tau}{\mu _i}} z(\tau) = - \frac{1}{\mu _i}\frac{1}{\mu _i} e^{-\frac{t-\tau}{\mu _i}} \left[\sum_{j=1}^n T_{ij} f_j( u_j(\tau)) + I_i (\tau)\right] $$
$$ \int_{-\infty}^{t} - \frac{1}{\mu _i}\frac{1}{\mu _i} e^{-\frac{t-\tau}{\mu _i}} \left[\sum_{j=1}^n T_{ij} f_j( u_j(\tau)) + I_i (\tau)\right] d\tau = -\frac{1}{\mu _i} u_i(t)$$
\\
$$\frac{d u_i (t)}{dt} = - \frac{1}{\mu _i} u_i (t) + \frac{1}{\mu _i} \left[\sum_{j=1}^n T_{ij} f_j( u_j(t)) + I_i (t)\right]}\)
