- #1
perwiradua
- 7
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(The full article is here: https://app.box.com/s/bjb63h4nt12rsvdke6yl. The part I would like to ask is on page 995-996))
I am trying to understand the idea behind the author's method in solving the differential equation
[itex]
I'(t) = -D(t) F \big (e^{\int_0^t \theta (u) du} I(t) \big )- \theta (t) I(t) \qquad (1.5)
[/itex]
where [itex] D(t) [/itex] is positive and [itex] \theta(t) [/itex] is nonnegative for [itex] 0 < t < H [/itex].
{start: How is this part relevant? --------------
Let
[itex]
\mu(x,y) = \int_x^y D(t) e^{\int_0^t \theta (u) du} dt
[/itex]
where [itex] F(v) [/itex] is positive for [itex] 0 < v < \ell [/itex] and [itex] F \in C(0,\ell) [/itex],
for some number [itex] \ell [/itex] for which
[itex]
\mu(0,H) \leq \int_0^\ell \frac{dv}{F(v)} < \infty.
[/itex]
end:-----------}
Equation (1.5) holds for [itex] t_{j-1} \leq t < t_j [/itex], with the boundary condition
[itex]
I(t) \rightarrow 0 \quad \text{as} \quad t \uparrow t_j \qquad (1.8)
[/itex]
for [itex] j = 1, 2, \ldots n [/itex]. To solve the equation subject to this condition, we change the dependent variable to [itex] J(t) = e^{\int_0^t \theta (u) du} I(t) [/itex].
With this as the unknown, (1.5) becomes
[itex]
J'(t) = D(t) e^{\int_0^t \theta (u) du} F \big ( J(t) \big ).
[/itex]
Subsequently. by separation of variables and imposition of (1.8) we find
[itex]
\int_0^{J(t)} \frac{dv}{F(v)} = \mu(t,t_j)
[/itex]
for [itex] t_{j-1} \leq t < t_j [/itex].
{start: This is the bit that I don't understand. Why need to define such
integral?}
Hence, if we define [itex] \psi [/itex] via
[itex]
z = \int_0^{\psi(z)} \frac{dv}{F(v)}
\quad \text{for} \quad
0 \leq z \leq \mu(0,H),
[/itex]
end:-----------}
we have
[itex]
J(t) = \psi \big ( \mu(t,t_j) \big )
\quad \text{for} \quad
t_{j-1} \leq t < t_j .
[/itex]
This gives
[itex]
I(t) = \kappa(t,t_j) \quad \text{for} \quad t_{j-1} \leq t < t_j
[/itex]
where
[itex]
\kappa(x,y) = e^{-\int_0^t \theta (u) du} \psi \big ( \mu(x,y) \big ).
[/itex]
I am trying to understand the idea behind the author's method in solving the differential equation
[itex]
I'(t) = -D(t) F \big (e^{\int_0^t \theta (u) du} I(t) \big )- \theta (t) I(t) \qquad (1.5)
[/itex]
where [itex] D(t) [/itex] is positive and [itex] \theta(t) [/itex] is nonnegative for [itex] 0 < t < H [/itex].
{start: How is this part relevant? --------------
Let
[itex]
\mu(x,y) = \int_x^y D(t) e^{\int_0^t \theta (u) du} dt
[/itex]
where [itex] F(v) [/itex] is positive for [itex] 0 < v < \ell [/itex] and [itex] F \in C(0,\ell) [/itex],
for some number [itex] \ell [/itex] for which
[itex]
\mu(0,H) \leq \int_0^\ell \frac{dv}{F(v)} < \infty.
[/itex]
end:-----------}
Equation (1.5) holds for [itex] t_{j-1} \leq t < t_j [/itex], with the boundary condition
[itex]
I(t) \rightarrow 0 \quad \text{as} \quad t \uparrow t_j \qquad (1.8)
[/itex]
for [itex] j = 1, 2, \ldots n [/itex]. To solve the equation subject to this condition, we change the dependent variable to [itex] J(t) = e^{\int_0^t \theta (u) du} I(t) [/itex].
With this as the unknown, (1.5) becomes
[itex]
J'(t) = D(t) e^{\int_0^t \theta (u) du} F \big ( J(t) \big ).
[/itex]
Subsequently. by separation of variables and imposition of (1.8) we find
[itex]
\int_0^{J(t)} \frac{dv}{F(v)} = \mu(t,t_j)
[/itex]
for [itex] t_{j-1} \leq t < t_j [/itex].
{start: This is the bit that I don't understand. Why need to define such
integral?}
Hence, if we define [itex] \psi [/itex] via
[itex]
z = \int_0^{\psi(z)} \frac{dv}{F(v)}
\quad \text{for} \quad
0 \leq z \leq \mu(0,H),
[/itex]
end:-----------}
we have
[itex]
J(t) = \psi \big ( \mu(t,t_j) \big )
\quad \text{for} \quad
t_{j-1} \leq t < t_j .
[/itex]
This gives
[itex]
I(t) = \kappa(t,t_j) \quad \text{for} \quad t_{j-1} \leq t < t_j
[/itex]
where
[itex]
\kappa(x,y) = e^{-\int_0^t \theta (u) du} \psi \big ( \mu(x,y) \big ).
[/itex]