MathDB

Let

g(t)

and

h(t)

be real, continuous functions for

t\geq 0.

Show that any function

v(t)

satisfying the differential inequality

\frac{dv}{dt}+g(t)v \geq h(t),\;\; v(t)=c,

satisfies the further inequality

v(t)\geq u(t),

where

\frac{du}{dt}+g(t)u = h(t),\;\; u(t)=c.

From this, conclude that for sufficiently small

t>0,

the solution of

\frac{dv}{dt}+g(t)v = v^2 ,\;\; v(t)=c

may be written

v=\max_{w(t)} \left( c e^{- \int_{0}^{t} |g(s)-2w(s)| \, ds} -\int_{0}^{t} e^{-\int_{0}^{t} |g(s')-2w(s')| \, ds'} w(s)^{2} ds \right),

where the maximum is over all continuous functions

w(t)

defined over some

t

-interval

[0,t_0 ].

B7