2.4 Differences Between Linear and Nonlinear DEs

THM-1 (Existence and Unique for First-Order Linear ODE)

Statement

if p(t) is continuous on an open interval$I \quad \exist ! \ y =\phi (t)$satisfying $y' + p(t)y = q(t), \ \forall t \in I \ and \ y = y_0$

Proof

Since P(t) is continuous on I ,we set $\mu(t) = e^{\int_{t_0}^{t}p(t)dt} \neq0 \ ,t \in I$($\mu '(t) = \mu(t) p(t)$)

We have $\mu(t) \in C'$($\mu \ is \ diff.)$

Multiplying (1) by $\mu(t)$, then we obtain $[\mu(t)y]' = \mu(t)q(t)$

Since q(t) is continuous on I ,,thus

$$\mu(t) y(t) = \int _{t_0}^{t} \mu(t)q(t) dt +C$$ $$y(t) - \mu^{-1}(t) \int_{t_0}^{t} \mu(s)q(s)ds + C$$

$y = \phi(t)$ exists , is differential and satisfy (1). Also , from the mitial condition (2), we determine uniquely $c = y_0$. Hence ,there is only one solution of (1)-(2).

THM-2 (First-Order Nonlinear ODE)

Statement

Let $f (t ,y)$and $\frac{\delta f}{\delta y}$be continuous in some rectangle R ,$\alpha < t <\beta$ ,$\gamma < y < \delta$

$R = \{ (t, y)\ | \ \alpha < t < \beta ,\ \gamma < y < \delta \}$

containing $(t_0 ,y_0)$. Then , in some interval $(t_0 -h , t_0 + h) \subset (\alpha ,\beta)$for some $h > 0$ ,

$\exist! \ y = \phi(t) \ s.t. \ y'(t) = f(t ,y) \ and \ y(t_0) = y_0$

Remark

• Thm-2 can imply Thm-1 (linear case)

  In the case of Thm-1 , $f(t ,y) = q(t) = p(t)y \in C$ and $\frac{\delta f}{\delta y}(t ,y) = -p(t) \in C$

  there is only one solution in some interval $(t_0 +h ,t_0 - h) \subset I$

• Uniqueness

  We can make sure that the graph of two solution cannot have any intersection.