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Help needed in understanding the author method of solving a DE

  1. Jan 30, 2014 #1
    (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]
     
  2. jcsd
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