# Help needed in understanding the author method of solving a DE

1. Jan 30, 2014

(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

$I'(t) = -D(t) F \big (e^{\int_0^t \theta (u) du} I(t) \big )- \theta (t) I(t) \qquad (1.5)$

where $D(t)$ is positive and $\theta(t)$ is nonnegative for $0 < t < H$.

{start: How is this part relevant? --------------

Let

$\mu(x,y) = \int_x^y D(t) e^{\int_0^t \theta (u) du} dt$

where $F(v)$ is positive for $0 < v < \ell$ and $F \in C(0,\ell)$,
for some number $\ell$ for which

$\mu(0,H) \leq \int_0^\ell \frac{dv}{F(v)} < \infty.$

end:-----------}

Equation (1.5) holds for $t_{j-1} \leq t < t_j$, with the boundary condition

$I(t) \rightarrow 0 \quad \text{as} \quad t \uparrow t_j \qquad (1.8)$

for $j = 1, 2, \ldots n$. To solve the equation subject to this condition, we change the dependent variable to $J(t) = e^{\int_0^t \theta (u) du} I(t)$.

With this as the unknown, (1.5) becomes

$J'(t) = D(t) e^{\int_0^t \theta (u) du} F \big ( J(t) \big ).$

Subsequently. by separation of variables and imposition of (1.8) we find

$\int_0^{J(t)} \frac{dv}{F(v)} = \mu(t,t_j)$

for $t_{j-1} \leq t < t_j$.

{start: This is the bit that I don't understand. Why need to define such
integral?}

Hence, if we define $\psi$ via

$z = \int_0^{\psi(z)} \frac{dv}{F(v)} \quad \text{for} \quad 0 \leq z \leq \mu(0,H),$

end:-----------}

we have

$J(t) = \psi \big ( \mu(t,t_j) \big ) \quad \text{for} \quad t_{j-1} \leq t < t_j .$

This gives

$I(t) = \kappa(t,t_j) \quad \text{for} \quad t_{j-1} \leq t < t_j$

where

$\kappa(x,y) = e^{-\int_0^t \theta (u) du} \psi \big ( \mu(x,y) \big ).$